Category: Class 12

MP Board Class 12th Physics Important Questions Chapter 3 Current Electricity

Current Electricity Important Questions

Current Electricity Objective Type Questions

Question 1.
Choose the correct answer of the following:

Question 1.
The flow of current through a conduction is due to:
(a) Protons
(b) Positive ions
(c) Free electrons
(d) Positive and negative ions.
Answer:
(c) Free electrons

Question 2.
The specific resistance of a wire depends upon:
(a) Length
(b) Diameter
(c) Mass
(d) Material.
Answer:
(b) Diameter

Question 3.
A wire is stretched redouble its length. Its resistance will be:
(a) Halved
(b) Doubled
(c) One fourth
(d) Four times.
Answer:
(d) Four times.

Question 4.
The unit of specific resistance is :
(a) Ohm
(b) Ohm^-1
(c) Ohm meter
(d) Ohm^-1
Answer:
(c) Ohm meter

Question 5.
Which of the following is an oh-mic resistance:
(a) Junction transistor
(b) Transistor
(c) LED
(d) Copper wire.
Answer:
(d) Copper wire.

Question 6.
On increasing temperature the resistance of the which of the following decreases :
(a) Semiconductor
(b) Metal
(c) Electrolyte
(d) Alloy.
Answer:
(a) Semiconductor

Question 7.
The conductance of super conductor is :
(a) Infinite
(b) Very high
(c) Very low
(d) Zero.
Answer:
(a) Infinite

Question 8.
In the series combination of two resistance which quantity remains same :
(a) Only potential difference
(b) Only current
(c) Current and potential difference both
(d) Neither current nor potential difference.
Answer:
(b) Only current

Question 9.
Electric cell is source of:
(a) Electrons
(b) Electrical energy
(c) Electric charge
(d) Electric current.
Answer:
(b) Electrical energy

Question 10.
The e.m.f. of a cell depends upon:
(a) Quantity of electrolyte filled in it
(b) Distance between the electrodes
(c) Size of electrodes
(d) Nature of electrolyte and electrodes.
Answer:
(d) Nature of electrolyte and electrodes.

Question 11.
Three cells each having e.m.f. E and internal resistance r are joined in series. One cell by mistake is joined in reverse order. The resultant e.m.f. and internal resistance will be:
(a) 3E, 3r
(b) E,3r
(c) E, r
(d) 3E,r
Answer:
(b) E,3r

Question 12.
The internal resistance of a cell can be decreased by :
(a) Decreasing the size of electrodes
(b) Increasing the distance between electrodes
(c) Decreasing the distance between electrodes
(d) None of these.
Answer:
(c) Decreasing the distance between electrodes

Question 13.
Two cells of e.m.f. and internal resistance E₁r₁ and E₂, r₂ respectively are joined in parallel. Their equivalent e.m.f. will be:
(a) E₁ + E₂
(b) E₁ – E₂
(c) \(\frac{\mathrm{E}_{1} r_{2}+\mathrm{E}_{2} r}{r_{1}+r_{2}}\)

(d) \(\frac{\mathrm{E}_{1}+\mathrm{E}_{2}}{r_{1}+r_{2}} \times r_{1} \cdot r_{2}\)
Answer:

(c) \(\frac{\mathrm{E}_{1} r_{2}+\mathrm{E}_{2} r}{r_{1}+r_{2}}\)

Question 14.
Potentiometer measures:
(a) Terminal voltage of cell
(b) Current in circuit
(c) e.m.f. of cell
(d) None of these.
Answer:
(c) e.m.f. of cell

Question 15.
In the null deflection position:
(a) No current flows through the galvanometer
(b) Current flows due to primary circuit through the galvanometer
(c) Current flows through galvanometer due to secondary circuit
(d) Nothing can be said.
Answer:
(a) No current flows through the galvanometer

Question 16.
The SI unit of the potential gradient is:
(a) ohm/cm
(b) volt
(c) volt cm
(d) volt/cm.
Answer:
(d) volt/cm.

Question 17.
To increase the sensitivity of potentiometer:
(a) The potential difference across its wire should be high
(b) The length of its wire should be less
(c) The current through the wire should be high
(d) potential difference across its wire should be low and length of wire should be large.
Answer:
(d) potential difference across its wire should be low and length of wire should be large.

Question 18.
In the balance condition of potentiometer its resistance is :
(a) Zero
(b) Infinite
(c) Very small
(d) Very high.
Answer:
(d) Very high.

Question 2.
Fill in the blanks:

1. According to ……….. law, a if the physical conditions of a conductor remains unchanged then current flowing through it is directly proportional to the applied potential difference.
2. On increasing the length of a conductor its resistance ………..
3. On decreasing the area of cross-section of a conductor its resistance ………..
4. On increasing the temperature of a metallic conductor its resistance ………..
5. KirchhofFs first law is in accordance with the law of conservation of ………..
6. In series grouping of resistors the resistance but in parallel grouping the resistance ………..
7. With increase in length of potentiometer its sensitivity ………..
8. Meter bridge works on the principle of ………..
9. The specific resistance of alloys is ……….. and temperature coefficient of resistance is ………..
10. The Kirchhoff’s first law is in accordance with the law of conservation of ……….. while the second law is in accordance with law of conservation of ………..
11. On increasing the distance between electrodes of a cell, its internal resistance ……….. but on increasing area its internal resistance ………..
12. The phenomenon in which at low temperature the resistivity of a substance becomes zero is celled ………..
13. In Ohm’s law the V-I graph is a ………..

Answers:

1. Ohm’s
2. Increases
3. Increases
4. Increases
5. Charge
6. Increases, decreases,
7. Increases
8. Wheatstone bridge
9. High, low
10. Charge, energy
11. Increases, decreases
12. Super conductivity
13. Straight line.

Question 3.
Match the Column:

Answer:

1. (e) Metre Bridge
2. (c) Law of conservation of charge
3. (b) Ohmic conductor
4. (a) Law of energy conservation
5. (d) Non-ohmic conductor

Answers:

1. (e) coulomb/sec. (or ampere).
2. (d) ampere/metre²
3. (b) joule/coulombx ampere
4. (c) ohm x meter
5. (a) ohm^-1

Answers:

1. (e) I²Rt
2. (a) ohm
3. (b) volt
4. (c) Electrical energy
5. (d) VI

Answer:

1. (c) Potential difference
2. (e) e.m.f.
3. (b) Vector quantity
4. (a) l₁/l₂
5. (d) \(\mathrm{R}\left(\frac{l_{1}}{l_{2}}-1\right)\)

Question 4.
Write the answer in one word / sentence:

1. What is the direction of electric current ?
2. What is the unit of specific resistance ?
3. Kirchhoffs first law is based upon which law ?
4. Kirchhoffs second law is based upon which law ?
5. On which principle does f&etre bridge works ?
6. Does Kirchhoffs law can be applied for both a.c. and d.c. currents ?
7. In a carbon resistance, there is green, violet, red and silver strip’s. What will be the resultant resistance.
8. What is the relation between 1 kWh and joule.
9. When the potentiometer is in equilibrium, what will be its resistance ?
10. The algebraic sum of current at a point is zero. Then what will be that point ?

Answer:

1. Opposite to flow of electrons,
2. Unit is ohm x meter
3. Laws of conservation of charges
4. Laws of conservation of energy
5. Wheatstone bridge
6. Yes
7. 57 x 10² ± 10% ohm
8. 1 kWh = 3.6 x 106 joule
9. Infinity
10. Junction point.

Current Electricity Very Short Answer Type Questions

Question 1.
To flow current in a conducting wire, how much charges is present in it ?
Answer:
On flowing current through the conducting wire, charges becomes zero.

Question 2.
Why Meter bridge is known by these name ?
Answer:
Because in Meter bridge, a wire of one meter length is used.

Question 3.
Two bulb are marked as 25 W and 100 W. Whose resistance will be more ?
Answer:
R ∝ \(\frac {1}{V}\), Resistance of 25 W bulb will be more.

Question 4.
What do you mean by electric power. Write its unit ?
Answer:
In any electrical circuit the rate of energy decay is called as power. Its unit is watt.

Question 5.
On which factor relaxation time depend on ?
Answer:
It depends on nature of the material of the conductor.

Question 6.
Why there is internal resistance in a cell ?
Answer:
Because, inside the cell, the motion of ions get obstacled by the collision of electrolytes molecules.

Question 7.
Why wire of Meter bridge is not made of copper wire ?
Answer:
Since the resistance of copper wire is very less and its temperature coefficient is more, therefore it is not used in Meter bridge.

Current Electricity Short Answer Type Questions

Question 1.
What is current ? Write its unit
Answer:
The electric current is defined as the rate of flow of charge through any section of a
Electric current = \(\frac {Total charge flowing}{ Time taken }\)
l = \(\frac {q}{ t}\) The SI unit of electric current is ampere (A).

Question 2.
Is electric current a scalar or a vector quantity ? Give reason.
Answer:
Electric current is a scalar quantity. This is because laws of ordinary algebra are used to add electric current and laws of vector addition are not applied for the addition of electric current.

Question 3.
Define current density. Is it scalar or vector ? Write its unit.
Answer:
Current density at a point in a conductor is defined as the amount of current flowing per unit area of the conductor held perpendicular to the flow of current.
Current density J = \(\frac {Current (l)}{ Area (S)}\) (or) \(\frac {l}{ S}\)
Current density is a vector quantity. The SI unit of current density is ampere / meter².

Question 4.
What is drift velocity ? What is its value ?
Answer:
Drift velocity is defined as the average velocity with which the velocity free electrons with which they get drifted towards the positive terminal of the conductor under the influence of the external electric field. Its value is 10^-5ms^-1

Question 5.
Is Ohm’s law applicable to all conductors ? Write conditions for its application
Or
Write the condition or under which Ohm’s law is not obeyed.
Answer:
No, Ohm’s law is not applicable for non – ohmic conductors, e.g., semiconductor diode, discharge tube, etc. It is valid only when there is no change in physical conditions of conductor i.e., temperature, length or mechanical strain etc.

Question 6.
Define specific resistance or resistivity. Write its unit and dimensional formula.
Answer:
Specific resistance or resistivity is numerically equal to the resistance offered by a conductor of unit length and unit cross-sectional area. Its unit is ohm – meter. It depends upon temperature. a’ ‘ The dimensional formula of specific resistance is [ML³T³A^-2].

Question 7.
If you are given two wires of same material, having same length but different diameters, then which wire will have higher resistance and which will have high specific resistance ?
Answer:
The thin wire (having less diameter) will have higher resistance than that of thicker one as R ∝ \(\frac { 1 }{ A }\) Both wires will have same specific resistance as they are made of same material and specific resistance is characteristic of the material of wire.

Question 8.
It is easier to start motor car during summer rather than in winter. Why ?
Answer:
The internal resistance of the battery during summer is less as compared to winter, therefore more current can be drawn from the battery.

Question 9.
When a high power heater is connected to mains, bulbs become dim. Why ?
Answer:
All electrical appliances are connected in parallel at home, therefore when power heater is used, very high current passes through it. So, potential difference falls on wires connected through mains (V = E – lr), which results in decreasing intensity of bulbs used in circuit.

Question 10.
What do you measure by meter bridge ? When is its sensitivity maximum ?
Answer:
Resistances are measured by meter bridge. Sensitivity of meter bridge is maximum when the null point is obtained at the center of wire. Moreover, all the resistances must be of same order.

Question 11.
What is the wire of meter bridge made up of ?
Answer:
The wire of meter bridge is made of manganin or constantan because it has low temperature coefficient of resistance and high specific resistance.

Question 12.
How can the sensitivity of potentiometer be increased ?
Answer:
The sensitivity of the potentiometer will be more if the potential gradient is less. For this, V should be less but more than the emf of E. The length of potentiometer should be large.

Question 13.
Write two possible reasons for obtaining deflection on one side in the experiment of potentiometer.
Answer:
The two possible reasons are as follows:

• If the emf of secondary cell in primary circuit is less than the emf of primary cell in the secondary circuit.
• If positive terminals of all the cells are not connected to the same point.

Question 14.
What do you understand by potential gradient of a potentiometer ? Give its unit
Answer:

• Fall in potential per unit length is called potential gradient. If the length of wire be / and potential difference across the wire is v, then
• Potential gradient, K =\(\frac {V}{l}\) Its unit is volt / m.

Question 15.
The potential gradient of potentiometer wire is doubled, what will happen to its null point ?
Answer:
Let emf of cell E is balanced at the length of –
E = Kl = l = \(\frac {E}{ K}\) = l ∝ \(\frac {1}{ K}\)
Hence, the null point will be at the half of the length.

Question 16.
Why the length of wire in potentiometer is more ?
Answer:
For greater sensitivity, the length of null point should be greater, for which the potential gradient should be smaller.
Potential gradient, K =\(\frac {V}{ l}\) To decrease K the value of l should be greater and F should be smaller. But, the value of V can be decreased up to a certain limit, hence the length is increased sufficiently.

Question 17.
Prove that potentiometer is an ideal voltmeter.
Answer:
When the potential difference between two points is measured by the voltmeter, then same current passes through the voltmeter, hence it measures little less than actual potential difference. But, when potentiometer measures the potential difference between two points, no current flows through it. Hence, it measures the accurate potential difference.

Question 18.
A carbon resistance has bands blue, red and green respectively. What is its resistance ?
Answer:

.’. The value of carbon resistance is 62 x 105 ohm with tolerance ± 20% .

Question 19.
A carbon resistance has bands blue, green and red respectively. Write its resistance.
Answer:

The value of carbon resistance is 65 x 102 ohm with tolerance ± 20% .

Question 20.
A carbon resistance has bands red, blue and green. What is its resistance ?
Answer:

The value of carbon resistance is 26 x 105 ohm with tolerance ±20% .

Question 21.
In high tension battery internal resistance should be high. Why ?
Answer:
If high tension battery is used in a circuit having low external resistance, current may go beyond its safety limit. If internal resistance of the battery is high, then current cannot go beyond its safety limit. So, internal resistance of high tension battery must be high.

Question 22.
What is cause of end error in a meter bridge ?
Answer:
The end error in meter bridge is due to following reasons:
The zero mark of the scale provided along the wire may not start from the position where the bridge wire leaves the copper strip and 100 cm mark of the scale may not end at the position where the wire touches the copper strip. The resistance of the copper wire and metal strip of meter bridge has not been taken in account.

Question 23.
What is Ohm’s law ? On what factors does the resistance of a conductor depend upon ?
Answer:
If all the physical conditions of any conductor as length, temperature, etc. remain constant, then the current which flows through it is proportional to the potential difference applied across the ends of the conductor. If I is the current in conductor and V is the potential difference, then
V ∝ l = V=RI
Where, R is a constant, called resistance of conductor.
For factors affecting resistance of conductor:

• Length: The resistance of conductor is directly proportional to the length of the conductor i.e., R ∝ l
• Area of cross – section: The resistance of a conductor is inversely proportional to the area of cross – section of the conductor i.e., R ∝ \(\frac{l}{A}\)
• Temperature: Increase in temperature, increases the resistance.

Combining above two laws, we get,
R ∝ \(\frac{l}{A}\)
R = p \(\frac{l}{A}\)
Where, p= a constant, called specific resistance of the material of the conductor.
Specific resistance:
We have,
R = p \(\frac{l}{A}\)
Let l = 1 And A = l, then
R = P

Thus, the specific resistance of a material is defined by the resistance of unit length and unit area of cross-section of that material.
Unit: Now, p = \(\frac{R.A}{l}\)
= \(\frac{Unit of R x Unit of A}{Unit of l}\)
= \(\frac{\mathrm{ohm} \times \mathrm{m}^{2}}{\mathrm{m}}\) = Ohm x m = Ω x m

Question 24.
What are the possible errors of a meter bridge and how can they be removed ?
Answer:
The possible errors and their removal methods are:
1. It might happen that the wire is not uniform. To remove this error, balance point should be obtained at the-middle.

2. During’the experiment, it is assumed that the resistance of L shaped plates are negligible, but actually it is not so. The error created due to this is called end error. To remove this error, the resistance box and the unknown resistance must be interchanged and then the mean reading should be taken.

3. If the jockey is pressed for a long period of time, then it gets heated and its resistance changes. Hence, jockey must not be pressed for a long interval.

Question 25.
Prove Ohm’s law on the basis of free electron theory.
or
Determine the formula for specific resistance in terms of free electron density and relaxation time.
Answer:
Consider a conductor of length l, area of cross-section A having ‘n’ No. of free electron per unit volume. If potential difference V is applied between its opposite ends then,
\(\frac { V }{ l }\)
Force experienced by each free electron will be
F = eE or F = \(\frac { eV }{ l }\)
The acceleration produced on the electron of mass m is
a = \(\frac { F }{ m}\) = \(\frac { eV }{ ml }\)
If the time interval between two successive collisions (relaxation time) is r then drift velocity is
V_d = \(\frac {eEτ }{ m }\)
= \(\frac {eVτ }{ ml }\)
But cuurent l = neAv_d
l = neA\(\left(\frac{e V \tau}{m l}\right)=\frac{n e^{2} \tau}{m} \frac{A}{l}V\)
\(\frac { V }{ l }\) = \(\frac{m}{n e^{2} \tau} \cdot \frac{l}{A}\)
If temperature remains constant, then all terms on R.H.S. will be constant, so
\(\frac { V }{ l }\) = R
This is Ohm’s law. Where R is constant called resistance.
R = \(\frac{m}{n e^{2} \tau} \cdot \frac{l}{A}\) ……….(1)
But resistance is directly proportional to length / and inversely proportional to area of cross-section A of conductor
R ∝ \(\frac { l }{A }\)
R = p\(\frac { l }{A }\) ………..(2)
From eqns. (1) and (2),
p = \(\frac{m}{n e^{2} \tau}\)

Question 26.
Obtain a relationship between current and drift velocity.
Answer:
Consider a conductor of length /and area of cross-section A. Let n be the number of electrons per unit volume.
Volume of conductor = Al
Total number of electrons = nAl.
Total charge of conductor, Q = nAle.
The conductor is now joined to a cell of potential difference V.
So, the electrons get drifted towards the positive terminal of the cell. Let the drift velocity be v_d.
Time taken to cover length l is –

t = \(\frac{l}{v_{d}}\)
l = \(\frac { Q }{t }\)
= \(\frac { nAel }{t }\)
l = nAev_d,
( v_d = \(\frac { distance}{time }\) = \(\frac { l }{t }\)
This is the required relationship.

Question 27.
Write four differences between electromotive force and potential difference.
Or
Define e.m.f. of a cell and potential difference. Give differences between them.
Answer:
e.m.f.: Work done in flowing 1 coulomb charge through a circuit is called e.m.f. of a cell When the terminals are not connected to an external circuit, the maximum potential difference between the terminals is equal to the e.m.f. of the cell. Unit of e.m.f is volt.

Potential difference:
The difference of potentials of any two points of the circuit is called potential difference.
Or
Work done in bringing unit positive charge from one point to another is called potential difference between these two points.

Differences between e.m.f. and potential difference:
e.m.f.:

• The maximum potential difference bet-ween the terminals of a cell is called its e.m.f., when the cell is in open circuit.
• This term is used for the electric sources as generator, cell, battery, dynamo, etc.
• It is established even the circuit is off.
• e.m.f. does not depend upon the resistance of the circuit.

Potential difference:

• Difference of potential of any two points of the circuit is called potential difference.
• This term is used for any two points of the circuit.
• It is established till the current is flowing through the circuit.
• It depends upon the resistance of the circuit

Question 28.
What do you understand by internal resistance of a cell ? On what factors does It depend and how ?
Or
What do you mean by internal resistance of a cell ? Write the factors affecting it.
Answer:
The resistance offered by the electrolyte of the cell during the flow of current inside the cell is called its internal resistance.
The following factors affect the internal resistance:

• Distance between the electrodes:
  As the distance increases, the internal resistance increases.
• Area of the immersed electrodes:
  As the area increases, the internal resistance decreases.
• Concentration of the electrolyte:
  As the concentration is more, the internal resistance is more.
• Temperature:
  The increase of temperature, decreases the internal resistance.

Question 29.
Establish the relationship between internal resistance of a cell, e.m.f. and extecoafresistance.
Or
Establish the relationship between the electromotive force, potential difference and internal resistance of a cell.
Answer:

Let e.m.f. of a cell be E and its internal resistance is r. If current / is flowing through a resistance R, then by Ohm’s law,
l = \(\frac { V}{R }\) ……………(1)
Where, V is the potential difference across the resistance. Now, total e.m.f. of the circuit = E and total resistance of the circuit = R + r
l = \(\frac {E}{R+r }\)
By eqns. (1) and (2), we get
\(\frac {V}{R}\) = \(\frac {E}{R+r }\)
VR + Vr = ER
Vr = ER – VR
r = \(\frac {R(E – V)}{V}\)
r = R(\(\frac {E}{V}\) – 1)
This is the required relationship.

Question 30.
Explain Kirchhoff’s laws of distribution of current through the different conduct^£
Or
Write down the Kirchhoff’s laws relating to the distribution of electric current.
Or
State and explain Kirchhoff’s laws.
Answer:
Kirchhoff’s laws:
(i) The algebraic sum of current meeting at any junction in a circuit is zero. In this law, the currents flowing towards the junction are considered as positive and those flowing  way from the junction as negative. As shown in the figure, we have

i₁ – i₂ – i₃ – i₄ – i₅ = 0
i₁ + i₄ = i₂ + i₃ + i₅

(ii) in any closed mesh (or loop) of an electrical circuit, the algebraic sum of the product of the currents and resistances is equal to the total e.m.f. of the mesh. If we go along the direction of conventional current, the potential difference will be taken as negative and opposite to it will be positive. Inside the cell, if we move from low to high potential, along the direction of conventional current, the e.m.f. will be positive.

For loop 1,
E₂ – i₂R₂ – (i₁ + i₂ )R₃ = 0
or E₂ = i₂R₂ + (i₁ + i₂ )R<sub>3

For loop 2,
i2R2 – E2 – i1R1 + E1 = 0
or E1 – E2 = i1R1 – i2R2</sub>

Question 31.
Derive the principle of Wheatstone bridge by Kirchhoffs law.
Or
Explain the principle of Wheatstone bridge and obtain the expression for balance condition by Kirchhoff s laws.
Answer:
Principle of Wheatstone bridge:
Four resistances P, Q, R and S are connected to form a quadrilateral ABCD. A cell E is connected across the diagonal AC and a galvanometer across BD. When the current is flown through the circuit and galvanometer does not give any deflection, then the bridge is said to be balanced. In this condition,
\(\frac { p }{ Q }\) = \(\frac { R }{ S }\)
This is the principle of Wheatstone bridge.

Formula derivation:
Let the current i is divided into two parts and i2, flowing through P, Q and R, S respectively. In the position of equilibrium, the galvanometer shows zero deflection, i.e,, the potential of B and D will be equal. In the closed mesh ABDA, by Kirchhoffs second law, we get

i₁P – i₂R = 0
or i₁ P = i₂R ……….(1)
Similarly, in the closed mesh BCDB, we have
i₁Q – i₂S = 0
or i₁Q = i₂S ……….(2)
Dividing eqn. (1) by eqn. (2), we get
\(\frac{i_{1} P}{i_{1} Q}=\frac{i_{2} R}{i_{2} S}\)
\(\frac{P}{Q}\) = \(\frac{R}{S}\)
This is Wheatstone bridge principle or principle of balance.

Question 32.
Explain the principle of potentiometer.
Answer:
Let AB be the wire of potentiometer of length L. A storage cell C, a key K and a rheostat Rh are connected in series with the resistance wire.
When the key is inserted, the current starts flowing through the circuit and a potential difference is established between A and B, let it is V.
∴ Potential gradient, P = \(\frac{V}{L}\)

Now, the +ve terminal of the experimental cell is connected to A and -ve to galvanometer which is connected to jockey J. When jockey is touched near A, we get the deflection in one side and if it is touched near B, the deflection is in other side. Thus, a point on AB is found such that galvanometer gives no deflection. Hence, no current flows through the galvanometer and get the null point J.

In this balanced position,
e.m.f. of the cell = Potential difference between A and J
If the length of AJ = l.
∴ E = pl
Thus, knowing p and E can be calculated.
This is the principle of potentiometer.

Question 33.
On what factors the resistance of a wire or conductor depends ? Define specific resistance and write its unit.
Or
On what factors the resistance of a conductor depends and how ?
Answer:
The factors affecting the resistance are:

• Length: The resistance of conductor is directly proportional to the length of the conductor i.e., R ∝ l
• Area of cross – section: The resistance of a conductor is inversely proportional to the area of cross – section of the conductor i.e., R ∝ \(\frac{l}{A}\)
• Temperature: Increase in temperature, increases the resistance.

Combining above two laws, we get,
R ∝ \(\frac{l}{A}\)
R = p \(\frac{l}{A}\)
Where, p= a constant, called specific resistance of the material of the conductor.
Specific resistance:
We have,
R = p \(\frac{l}{A}\)
Let l = 1 And A = l, then
R = P

Thus, the specific resistance of a material is defined by the resistance of unit length and unit area of cross-section of that material.
Unit: Now, p = \(\frac{R.A}{l}\)
= \(\frac{Unit of R x Unit of A}{Unit of l}\)
= \(\frac{\mathrm{ohm} \times \mathrm{m}^{2}}{\mathrm{m}}\) = Ohm x m = Ω x m

Question 34.
What do you understand by specific resistance or resistivity of a conductor ? Give its unit and dimensional formula.
Answer:
For specific resistance of a conductor: Refer Short Answer: Type Q. No. 33(iii). Dimensional formula of specific resistance is determined as:
As p = \(\frac{RA}{l}\)
[p] = \(\frac{[R][A]}{[l]}\)
[R] = \(\frac{[V]}{[l]}\) = \(\frac{W/P}{[l]}\) = \(\frac{\left[\mathrm{ML}^{2} \mathrm{T}^{-2}\right] /[\mathrm{AT}]}{[\mathrm{A}]}\)
= \(\frac{\left[\mathrm{ML}^{2} \mathrm{T}^{-3} \mathrm{A}^{-1}\right]}{[\mathrm{A}]}\) = [ML²T^-3A^-2
[p] = \(\frac{\left[\mathrm{ML}^{2} \mathrm{T}^{-3} \mathrm{A}^{-2}\right]\left[\mathrm{L}^{2}\right]}{[\mathrm{L}]}\)
= \(\left[\mathrm{ML}^{3} \mathrm{T}^{-3} \mathrm{A}^{-2}\right]\)

Question 35.
Give differences between specific resistance and resistance.
Answer:
Differences between specific resistance (resistivity) and resistance:

Specific resistance:

• It is defined by the resistance offered by the conductor of unit length and unit area of cross – ection.
• Its unit is ohm x m.
• It does not depend upon the length and area of cross – ection.

Resistance:

• The hindrance offered by a conductor to the flow of current is called electrical resistance.
• Its unit is ohm.
• It depends upon the length and area of cross-section of the conductor.

Question 36.
Evaluate It and I₂ as shown in the figure.

Solution:
Equivalent figure of fig. (a) is fig. (b)
In closed current BEFC, by Kirchhoff s second law,
2l₁ – 6I₂ = 0
or I₁ = 3I₂ …….(1)
In closed path ABCD
2I₁ + l(l₁ + l₂) = 10
3I₁ + I₂ = 10 …….(2)

From eqns. (1) and (2),
3 x 3I₂ + I₂ = 10
or 9I₂ + I₂ = 10
or I₂ = 1 amp.
Putting the value of 12 in eqn. (1) we get,
I₁ = 3 x l = 3 amp

Question 37.
Three resistors R₁ R₂ and R₃ are connected in series. Obtain the expression for the equivalent resistance.
Answer:
The series combination of three resistors having resistances R₁ R₂ and R₃ are shown in figure.

When this combination is connected to a cell (C) of e.m.f. E volt, then I current flows through combination.
Let the potential difference across R₁ R₂ and R₃ be V₁, V₂ and V₃ respectively. Then, by Ohm’s law,
V₁= IR₁ V₂ = IR₂ and V₃ = IR₃ .
If potential difference across A and B be V, then
V = V₁ + V₂ + V₃
or V = IR₁ + IR₂ + IR₃
V = I(R₁ + R₂ + R₃) ……(1)
If equivalent resistance of this combination be R, then
V = IR …..(2)
From eqns. (1) and (2), we get
1R = I(R₁ + R₂ + R₃)
or R = R₁ + R₂ + R₃ …(3)
Thus, when a number of resistances are connected in series with each other, the equivalent resistance of the combination is equal to the sum of their individual resistances.

Question 38.
Find out equivalent resistance of the three resistances R₁ ,R₂ and R₃ connected in parallel combination.
Answer:
The resistances R₁ ,R₂ and R₃ are connected in parallel as shown in adjacent figure. As all the resistances are connected between two points A and B, hence the potential difference between two points will be same for all. Let it be V. Let I be the total current. This current is divided into three parts at point A. If through R₁, R₂ and R₃; currents I₁ I₂ and I₃ are respectively flowing. Then,

I = I₁ + I₂ + I₃
By Ohm’s law,
I₁ = \(\frac{V}{R_{1}}\), I₂ = \(\frac{V}{R_{2}}\), I₃ = \(\frac{V}{R_{3}}\)
I = \(\frac{V}{R_{1}}\) + \(\frac{V}{R_{2}}\) + \(\frac{V}{R_{3}}\) ……….(1)
If the equivalent resistance of this combination be R, then
I = \(\frac{V}{R}\) ……….(2)
From eqns. (1) and (2), we have
\(\frac{V}{R}\) = \(\frac{V}{R_{1}}\) + \(\frac{V}{R_{2}}\) + \(\frac{V}{R_{3}}\)
\(\frac{1}{R}\) = \(\frac{1}{R_{1}}\) + \(\frac{1}{R_{2}}\) + \(\frac{1}{R_{3}}\) …………(3)
Thus, if a number of resistances are connected in parallel, the reciprocal of equivalent resistance of the combination is equal to the sum of the reciprocals of their individual resistances.

Current Electricity Long Answer Type Questions

Question 1.
Describe an experiment to compare the e.m.f. of two cells by potentiometer onTne following points :

1. Circuit diagram
2. Derivation of formula
3. Two precautions.

Or
Describe an experiment to compare the e.m.f. of two cells using a potentiometer under the following heads :

1. Labelled diagram of electric circuit
2. Formula used
3. Observation table
4. Two main precautions.

Answer:
1. Circuit diagram:
AB → Potentiometer wire
B₁ → Lead accumulator
K₁ → Plug key
Rh → Rheostat
E₁ E₂ → Experimental cells
K₂ → Two ways key
G → Galvanometer
J → Jockey.

2. Derivation of formula:
Let the first cell is having e.m.f. E, and the balancing point is obtained at distance l₁. Then by the principle of potentiometer,
E₁ = pl₁ ……(1)
Where, p is potential gradient.
Let E₂ is the e.m.f. of second cell whose balancing point is at l₂, then
E₂ = pl₂ ……(2)
Dividing eqn. (1) by eqn. (2), we get,
\(\frac{E_{1}}{E_{2}}=\frac{pl_{1}}{pl_{2}}=\frac{l_{1}}{l_{2}}\)

3. Observation tab:

4. Precautions:
The e.m.f. of lead accumulator should be greater than that of experimental cells. All the positive terminals should be connected to one point

Question 2.
Determine the internal resistance of a cell by potentiometer on the following points:

1. Labelled circuit diagram
2. Derivation of formula
3. Precautions.

Or
Determine an experiment to find out the internal resistance of a cell by potentiometer under the following points :

1. Labelled circuit diagram
2. Formula derivation
3. Observation table
4. Precautions (any two).

Answer:
1. Circuit diagram:
AB → Potentiometer wire
C → Storage cell
K → Plug key
Rh → Rheostat
E → Experimental cell
R.B. → Resistance box
G → Galvanometer
J → Jockey.

2. Formula derivation:
Let the e.m.f. of the cell is E and its internal resistance is r. When it is connected to external resistance R, its potential difference is V, then

r = R (\(\frac{E}{V}\) – 1) ……..(1)
When R is not connected, then let the balance point is found at a distance l₁ from A.
∴ E = pl₁
Similarly, when R is also connected, then let the balanced point is obtained at a distance l₂ from A.
∴ V = pl₂
Now, putting these values in eqn. (1), we get
r = R \(\left(\frac{\rho l_{1}}{\rho l_{2}}-1\right)\) = R \(\left(\frac{l_{1}}{l_{2}}-1\right)\)
This is the required formula

3. Observation table :

4. Precautions:

• The e.m.f. of the storage cell (C) must be greater than that of experimental cell (£).
• All the positive terminals must be connected to a single point

Question 3.
How are the cells connected in mixed combination ? Derive an expression for current flowing through external resistance.
Or
Prove that the current in the external circuit for a mixed combination of cells is maximum, when the internal resistance of the combination equals the external resistance.
Answer:
In this combination, series connections of equal number of cells are connected in parallel and finally providing one +ve and one -ve terminal as shown in Fig. (a). These two terminals are connected to the ends of an external resistance R so that current starts flowing through the resistance.

Suppose n cells are connected in series combination and let m be the number of such series connected in parallel. Let the e.m.f. of each cell be E and internal resistance be r. Total e.m.f. of each series combination i.e, for a row, will be nE and total internal resistance will be nr. Hence, the equivalent becomes as given in Fig. (b).

As all the cells are now in parallel, so total e.m.f. of the combination will also be nE and total internal resistance will be r’. This is shown in Fig. (c). As each resistor is of value nr and nr such resistors are connected in parallel, hence total internal resistance r’ will be calculated as:

\(\frac{1}{r’}\) = \(\frac{1}{nr}\) + \(\frac{1}{nr}\) + ………. + m terms = \(\frac{m}{nr}\) or r’ = \(\frac{nr}{m}\)
As R is the external resistance, then r’ and R are in series. Hence, total resistance of combination becomes,
R_total = R + r’ = R + \(\frac{nr}{m}\)
The current through the circuit will be given by Ohm’s law as

l = \(\frac{Total e.m.f.of battery }{Total resistance of circuit}\)

l = \(\frac{n \mathrm{E}}{\mathrm{R}+\frac{n r}{m}}\) = \(\frac{m n \mathrm{E}}{m \mathrm{R}+n r}\) ……….(1)

This is the expression for the current. For the current to be maximum, mR+nr must be minimum

mR + nr = \((\sqrt{m \mathrm{R}})^{2}+(\sqrt{n r})^{2}-2 \sqrt{m \mathrm{R}} \sqrt{n r}+2 \sqrt{m \mathrm{R}} \sqrt{n r}\) (as subtracting and adding \(2 \sqrt{m \mathrm{R}} \sqrt{n r}\))

= \((\sqrt{m \mathrm{R}}-\sqrt{n r})^{2}+2 \sqrt{m \mathrm{R}} \times \sqrt{n r}\)

But \((\sqrt{m \mathrm{R}}-\sqrt{n r})^{2}\) cannot be negative, as it is a perfect square.

mR + nr will be minimum, only when \((\sqrt{m \mathrm{R}}-\sqrt{n r})^{2}\) = 0

or \(\sqrt{m \mathrm{R}}-\sqrt{n r}\) = 0

\(\sqrt{m \mathrm{R}}\) = \(\sqrt{n r}\) ⇒ mr = nr ………(2)
∴ R = \(\frac{n r}{m}\) ………..(3)

Hence, the external resistance should be equal to the internal resistance of the battery. Under this condition the current supplied by the battery to the resistance R is maximum. The value of maximum current can be obtained by putting eqn. (2) in eqn. (1).

∴ I_max = \(\frac{mnE}{mR + mR}\)

= \(\frac{mnE}{2mR}\) = \(\frac{nE}{2R}\)
On substituting mR = nr, we get

I_max = \(\frac{mnE}{nr + nr}\) = \(\frac{mnE}{2nr}\)

= \(\frac{mE}{2r}\) .

Question 4.
explain the experiment of determining the unknown resistance of a wire with he help of meter bridge on the following points:

1. Electrical circuit
2. Principle.

Or
Describe an experiment to determine the unknown resistance by meter bridge on following points :

1. Circuit diagram
2. Description of apparatus
3. Observation table
4. Precautions.

Or
What are the possible errors in performing the experiment with metre bridge and how they can be removed ?
Answer:
1. Circuit diagram:

Where ,
AC→ wire
E → cell
R → resistance box
S → unknown resistance,
G → galvanometer
K → key.

2. Description of apparatus:
It consists of 1 meter long manganin or constant resistance wire, fixed on a wooden plank. Thick steel or brass strips are fixed as shown in the figure with two gaps. In one gap, resistance box R and in other, unknown resistance S are connected. A, C and D are terminals. A jockey slides on the wire.

3. Formula derivation or principle:
Meter bridge consists of 1 m long constant-an wire AC, fixed on a wooden plank. In two gaps resistance R and unknown resistance S are connected. Galvanometer G is connected between B and D, where B is sliding point i. e., jockey.
A cell E, with a plug key K is connected between A and C.
Let null point is obtained at a distance l from A.
∴ AB = l cm
and BC = (100 – l)cm
If x be the resistance per unit length, then Resistance offered by AB is P = lx
and resistance offered by BC is Q = (100 – l)x
Now, by the principle of Wheatstone bridge,

\(\frac{P}{Q}\) = \(\frac{R}{S}\)

S = \(\frac{R(100 – l)}{l}\)
Wheatstone bridge is sensitive when all the four resistances are of same order. Hence, metre bridge is also suitable for the resistance of same order.

4. Observation table:

5. Precautions:

• The connection should be tight.
• The plugs of R.B. should not be loose.
• The current should be passed only when readings are to be taken.
• Jockey should not be rubbed with the wire.

Possible errors and their removal:
1. It might happen that the wire is not uniform. To remove this error, balance point should be obtained at the-middle.

2. During’the experiment, it is assumed that the resistance of L shaped plates are negligible, but actually it is not so. The error created due to this is called end error. To remove this error, the resistance box and the unknown resistance must be interchanged and then the mean reading should be taken.

3. If the jockey is pressed for a long period of time, then it gets heated and its resistance changes. Hence, jockey must not be pressed for a long interval.

Question 5.
How are cells connected in series ? Derive an expression for current flowing through outer circuit. When is this combination useful ?
Answer:
In this combination, the -ve terminal of one cell is connected to the + ve terminal of second cell, whose -ve terminal is connected to the +ve terminal of third one and so on as shown in Fig. (a). Let n cells each of e.m.f. E and internal resistance r be connected in series through an external resistance R, then

Total e.m.f. of this combination = n E ’ and total internal resistance
r’ = r + r + r +…………….+ n times
= nr
The series combination of n cells is equivalent to a single cell having e.m.f. «E and internal resistance nr. This is shown in Fig. (b).
Now nr and external resistance R are in series. So, the equivalent resistance of circuit becomes R + nr Applying Ohm’s law, current flowing through the circuit is given by –
l = \(\frac{e.m.f. of battery }{Total resistance}\)
= \(\frac{nE }{nr + R}\)
This is the expression for current flowing through the external resistance R. If r « R, then nr + R * R.
Hence, from eqn. (1), we get
l = n. \(\frac{E }{R}\)
= n x Current flowing through each cell.

Utility:
Thus, if the internal resistance of each cell is negligible as compared to external resistance, the current flowing through the battery is n times the current supplied by each cell.
So, the cells should be connected in series only when the internal resistance of each cell is much less than the external resistance.

Question 6.
n cells are connected in parallel combination with internal resistance. Derive an expression for current flowing through external resistance. When is this combination useful?
Answer:
Let n cells be connected in parallel and e.m.f. of each cell be E and internal resistance be r. Since, all the positive terminals of cells connected to point A and -ve terminals connected to the point B, the total e.m.f. of battery will be E (because in parallel combination potential difference remains same).

Since the cells are connected in parallel, therefore their internal resistance will also be in parallel combination. Let the equivalent internal resistance be r’
∴ \(\frac{1 }{r’}\) = \(\frac{E }{R}\) + \(\frac{E }{R}\) + …………. + n times \(\frac{n }{r}\)
or r’ = \(\frac{r }{n}\)
Hence, the equivalent circuit becomes

Where, r’ is the total resistance of the combination of cells. As R and r’ are in series, so the total resistance of the circuit becomes
= R + r’ = R + \(\frac{r }{n}\)
Current through external resistance R will be given by Ohm’s law as
l = \(\frac{Total e.m.f.of battery }{Total resistance of circuit}\)
l = \(\frac{E}{R+\frac{r}{n}}\) = \(\frac{n E}{n R+r}\)
This is the required expression.
If R << r, then nR + r≈r
Now, from eqn. (1), we have
I = n \(\frac{E}{r}\) = n x Current given by one cell .

Utility:
Thus, if internal resistance of each cell is much greater than the external resistance, then the current flowing through the battery is n times the current through each cell i.e., maximum current is obtained. So, the cells should be connected in parallel when internal resistance of each cell is much greater than the external resistance.

Current Electricity Numerical Questions

Question 1.
The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4Ω, what is the maximum current that can be drawn from the battery? (NCERT)
Solution:
Given: E = 12 V; r – 0.4 Ω
Current l = \(\frac{E}{R+r}\)
For current to be maximum R = 0
I_max = \(\frac{E}{r}\) = \(\frac{12}{0.4}\) = 30A

Question 2.
A battery of emf 10V and internal resistance 3) is connected to a resistor. If the current in the circuit ¡s 0.5A. What is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed? (NCERT)
Solution:
Given:
E =10V; r = 3Ω l = 0.5A

l = \(\frac{E}{R + r}\)
R+r = \(\frac{E}{l}\)
R = \(\frac{E}{l}\) – r = \(\frac{10}{0.5}\) – 3 = 20 – 3 = 17Ω
Now, V = E – lr = 10 – 0.5 x 3 = 10 – 1.5 = 8.5V

Question 3.
(a) Three resistance IΩ, 2Ω and 3Ω are combined in series. What is the total resistance of the combination ?
(b) If the combination is connected to a battery of emf 12V and negligible internal resistance, obtain the potential drop across each resistor. (NCERT)
Solution:
(a) Given:
R₁ = IΩ; R₂ = 2Ω; R₃ = 3Ω
R₂ = R₂ + R₂ + R₂ = 1 + 2 + 3 = 6Ω .
(b) E = 12V; r = 0
l = \(\frac{E}{R+r}\) = \(\frac{E}{R_{S}+0} = \frac{12}{6}\)
I = 2A
Potential difference across (R₁)
V₁ = IR₁ = 2 x l = 2V
Potential difference across (R₂)
V₂ = IR₂ = 2 x 2 = 4V
Potential difference across (R₃)
V₃ = IR₃ = 2 X 3 = 6V.

Question 4.
(a)Three resistors 2Ω, 4Ω and 5Ω are combined in parallel. What is the total resistance of the combination ?
(b) If the combination is connected to a battery of emf 20V and negligible internal resistance, determine the current through each resistor and the total current drawn from the battery. (NCERT)
Solution?:
(a) Given:
R₁ = 20; R₂ = 40; R₃ = 50
Equivalent resistance:
\(\frac{1}{R_{p}}\) = \(\frac{1}{R_{1}}\) + \(\frac{1}{R_{2}}\) + \(\frac{1}{R_{3}}\)
= \(\frac{1}{2}\) + \(\frac{1}{4}\) + \(\frac{1}{5}\) = \(\frac{10+5+4}{20}\) = \(\frac{19}{20}\)
∴ \(\frac{1}{R_{p}}\) = \(\frac{20}{19}\) Ω

(b) E = 20V; r = 0
Current through different resistors,
l₁ = \(\frac{E}{R_{1}}\) = \(\frac{20}{2}\) = 10A
l₂ = \(\frac{E}{R_{2}}\) = \(\frac{20}{4}\) = 5A
l₃ = \(\frac{E}{R_{3}}\) = \(\frac{20}{5}\) = 4A
Total current drawn
l = l₁ + l₂ + l₃ = 10 + 5 + 4 = 19A

Question 5.
At room temperature (27°C) the resistance of a heating element is 100Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that temperature coefficient of the resistor material is 1.70 x 10^-4 °C^-1 ? (NCERT)
Solution:
Given:
t₁ = 27 °C
R₁ = 1000
R₂ = 1170
α = 1.70 x 10^-4 °C^-1
We know that, α = \(\frac{R_{2}-R_{1}}{R_{1}\left(t_{2}-t_{1}\right)}\)
t₂ – t₁ = \(\frac{R_{2}-R_{1}}{R_{1} \alpha}\)
= \(\frac{117-100}{170 \times 10^{-4} \times 100}\)
t₂ – t₁ = 1000
t₂ = 1000 + t₁ = 1000 + 27 = 1027 °C

Question 6.
A negligible small current is passing through a wire of length 15 m and uniform cross – section 6.0 x 10^-7 m² and its resistance is measured to be 5.0ΩWhat is the resistivity of the material at the temperature of the experiment ? (NCERT)
Sol. Given:
l = 15m; A = 6.0 x l0^-7m²; R = 5.0Ω
using R = \(\frac { pl }{ A }\)
p = \(\frac {RA}{ l }\) = \(\frac{5 \times 6 \times 10^{-7}}{15}\) = 2.0 x l0^-7Ωm

Question 7.
Determine the current in each branch of the network shown in figure. (NCERT)

solution:
\(\frac {P}{ Q }\) = \(\frac {10}{5}\) = \(\frac {2}{ 1 }\) and \(\frac {R}{S }\) = \(\frac {5}{ 10 }\) = \(\frac {1}{ 2 }\).
i.e \(\frac {P}{ Q}\) = \(\frac {R}{ S}\)

Therefore, bridge is not in equilibrium. Current flowing through different branches is as shown in fig. Now applying Kirchhoffs loop rule.

For loop ABDA
10I₁ + 5I₂ – 5(I – I₁) = 0
3I₁ + l₂ – I = 0 ……(1)

For loop BCDB.
5(I₁ – I₂) – 10 (I – l₁ + I₂) – 5I₂ = 0
or I₁ – I₂ – 2 (I – I₁ + I₂) I₁ = O
or 3I₂ – 4I₂ – 2I = 0 …..(2)
By eqns. (1) and (2), 5I₂ + I = 0
I = – 5I₂
I₂ = – \(\frac {1}{ 5 }\) I …….(3)
And I₁ = \(\frac {2}{ 5 }\) I …….(4)

For loop ADCA,
5(I – I₁) + 10 (1 – l₁ + l₂) + 10I = 10
5I – 3I + 2I₁ = 2

Question 8.
Find out effective resistance between points A and B for circuit given below:

Solution:
\(\frac {1}{ R}\) = \(\frac {1}{ 2 + 2 + 2 }\) + \(\frac {1}{ 3 }\)

\(\frac {1}{ R}\) = \(\frac {1}{6}\) + \(\frac {1}{3}\)

\(\frac {1}{ R }\) = \(\frac {1 + 2}{ 6 }\) = \(\frac {3}{6}\)
R = \(\frac {6}{3}\) = 2Ω

Question 9.
Find out effective resistance between points A and B for circuit given below:

\(\frac{1}{\mathrm{R}_{1}}\) = \(\frac {1}{ 4}\) + \(\frac {1}{ 4}\) = \(\frac{1}{\mathrm{R}_{1}}\) = \(\frac {2}{ 4}\)

R₁ = \(\frac {4}{2}\) = 2Ω

\(\frac{1}{\mathrm{R}_{2}}\) = \(\frac {1}{ 6}\) + \(\frac {1}{ 6}\) =\(\frac{1}{\mathrm{R}_{2}}\) = \(\frac {2}{ 6}\)

R₂ = \(\frac {6}{ 2}\) = 3Ω
Effective resistance R = R₁ + R₂
R = 2 + 3 = 5Ω

Question 10.
10⁹ electrons flows from point A to B in 10^-3 second. Find out magnitude and direction of electric current.
Solution:
Formula: I = \(\frac {ne}{t}\)
Given: n = 10⁹ t = 10^-3 sec.
Putting the given value in the formula, we get
I = \( \frac{10^{9} \times 16 \times 10^{-19}}{10^{-3}}\)
or l = 1.6 x l 0^-7 ampere.
Direction of electric current will be from B to A. Ans.

Question 11.
150 m A current flows through a conductor. How many electrons will flow through it in 20 sec ?
Solution:
Given:
I = 150 mA = 0.15 A, t = 20 sec.
Formula:
n = \(\frac {It}{e}\) = \(\frac{0.15 \times 20}{1 \cdot 6 \times 10^{-19}\)

n = \(\frac{3}{1 \cdot 6} \times 10^{19}\) = 1.875 x 10¹⁹

Question 12.
200 mA current flows through a conductor. How many electrons will flow through it in 5 sec ?
Solution:
Given:
I = 200 mA = 0.2 A, t = 5 sec
Formula:
n = \(\frac {It}{e}\) = \(\frac{0\cdot 2\times 5}{1\cdot 6\times 10^{-19}}\)

n = \(\frac{1}{1 \cdot 6} \times 10^{19}\)
n = 0.625 x 10¹⁹

Question 13.
The length of a wire becomes twice when stretched. How many times resistance will increase ?
Solution:
On stretching the wire, radius of the wire will decrease but volume will remain same. Let l₁ and r₁ be initial length and radius of wire respectively. On stretching, the length becomes l₂ say and radius r₂
Initial volume = Final volume

From the formula, R = p \(\frac {1}{A}\)

Putting the value of \(\left(\frac{r_{2}}{r_{1}}\right)^{2}\) from eqn. (1) in eqn. (2), we get

As per the question, l₂ = 2l₁
\(\frac{R_{1}}{R_{2}}\) = \(\left(\frac{l_{1}}{2 l_{1}}\right)^{2}\) = \(\frac {1}{4}\)
R₂ = 4R₁
Hence, the resistance of wire is increased four times.

Question 14.
The ratio of area of cross-section of two wires made of same metal and equal length is 2 : 1. It applied potential difference between their ends are same then what will be the ratio of current among them ?
Solution:

Question 15.
The ratio of length of two wire, made of same metal and equal area of cross-section is 2 : 1. If the applied potential difference between their ends are same, then what will be the ratio of current following among them ?
Solution:

Question 16.
125 mA current flows through a lamp in 10 sec. How many electrons will flow across it ?
Solution:
Given:
I = 125 mA = 0.125 A. t = 10 sec.

MP Board Class 12th Physics Important Questions

MP Board Class 12th Biology Important Questions Chapter 16 Environmental Issues

Environmental Issues Important Questions

Environmental Issues Objective Type Questions 

Question 1.
Choose the correct answer:

Question 1.
Disease caused due to mercury pollution is:
(a) Methenamine
(b) Minimata
(c) Asbestosis
(d) Liver inflammation.
Answer:
(b) Minimata

Question 2.
Plants are natural air purifiers because they exhibit :
(a) Respiration
(b) Photosynthesis
(c) Transpiration
(d) Drying.
Answer:
(b) Photosynthesis

Question 3.
Metal found in atmosphere is :
(a) Cadmium
(b) Lead
(c) Mercury
(d) Zinc.
Answer:
(b) Lead

Question 4.
Burning of fossil fuel results in:
(a) SO₂ pollution
(b) NO₂ pollution
(c) N₂O pollution
(d) NO pollution.
Answer:
(a) SO₂ pollution

Question 5.
Increase of the concentration of which gas results greenhouse effect :
(a) CO₂
(b) CO
(c) O₂
(d) Nitrogen oxide.
Answer:
(a) CO₂

Question 6.
Gas responsible for depletion of O₃ in atmosphere is :
(a) CFC
(b) NO₂
(c) CO₂
(d) SO₂
Answer:
(a) CFC

Question 7.
Carbon monoxide is the chief pollutant of :
(a) Water
(b) Air
(c) Noise
(d) Soil.
Answer:
(b) Air

Question 8.
Effect of air pollution is seen in :
(a) Leaves
(b) Flower
(c) Forests
(d) None of these.
Answer:
(b) Flower

Question 9.
Shrinking of leaves occurs by :
(a) SO₂
(b) O₂
(C) H₂S
(d) CO.
Answer:
(a) SO₂

Question 10.
Example of greenhouse gas is :
(a) CO₂
(b) CH₄
(c) CFC
(d) All of these.
Answer:
(d) All of these.

Question 11.
Gas produced during combustion of fossil fuels is :
(a) CO₂
(b) CH₄
(c) CFC
(d) N₂O
(a) CO₂

Question 12.
Gas used in refrigeration is :
(a) Freon
(b) N₂O
(c) CH₄
(d) CO₂.
Answer:
(a) Freon

Question 13.
Radiation emitted due to heating of earth is:
(a) UV
(b) Infrared
(c) Gamma
(d) None of these.
Answer:
(b) Infrared

Question 14.
Gas responsible for depletion of atmospheric ozone layer is:
(a) CO₂
(b) CH₄
(c) CFC
(d) O₂
Answer:
(c) CFC

Question 2.
Fill in the blanks:

1. ………………….. is a chief green house gas.
2. ………………….. is the measurement unit of water pollution.
3. ………………….. and ………………….. gases are responsible for acid rain.
4. ………………….. gas is essential for life. (MP 2009 Set B)
5. The substances which produce pollution are called ………………….. (MP 2013)
6. Strong, loud, unpleasant or untolerable sound is called ………………….. (MP 2013)
7. We can ………………….. the quantity of CO₂ from forestation.
8. CO₂, CH₄, N₂O, CFCs are………………….. gases.
9. ………………….. is discharged in Bhopal gas disaster.
10. In human DDT is reached by …………………..
11. ………………….. is also known as chemical weeds.
12. Pollution is the …………………. changes which occurs in atmosphere.
13. Deficiency of soil nutrients is called …………………..
14. The percentage of CO₂ of atmosphere can be ………………….. by forestation. (MP 2009 Set D)

Answer:

1. CO₂
2. BOD
3. SO₂ and NO₂
4. O₂
5. Pollutant
6. Noise
7. Lost
8. Green house
9. Methyl isocyanate
10. Food chain
11. Ozone
12. Unwanted
13. Negative pollution
14. Reduced.

Question 3.
Match the followings:
I.

Answer:

1. (c)
2. (f)
3. (e)
4. (b)
5. (d)
6. (a)
7. (g)
8. (h)

II.

Answer:

1. (b)
2. (c)
3. (d)
4. (a)

Question 4.
Write the answer in one word/sentences:

1. Write full name of BOD.
2. What is the hearing capacity of man?
3. Name the fish used to control mosquito.
4. What is the concentration of CO₂ in environment?
5. When is World Environment Day celebrated?
6. Write the names of two main gases which pollute air.
7. Write the name of highly polluted river of India.
8. Name the term used, for the increase in atmospheric temperature, due to increase in CO₂ concentration.
9. What is the amount of oxygen taken up by microorganisms, that decompose waste matter?
10. Write full name of CFC.
11. Write the name of a chief greenhouse gas.
12. Write the name of gas responsible for ozone layer depletion.
13. Who is the greatest enemy of nature and organism? (MP 2013)
14. When we celebrated ozone day?

Answer:

1. Biological Oxygen Demand
2. 10 – 12 decibel
3. Gambussia
4. 0 03%
5. 5th June
6. SO₂and CO
7. Ganga
8. Green House effect
9. B. O. D
10. Chlorof – luorocarbon
11. CO₂
12. CFC
13. Global warming
14. 16December.

Environmental Issues Very Short Answer Type Questions

Question 1.
What is the definition of pollution?
Answer:
Pollution is defined as an undesirable changes in physical, chemical or biological characteristics of air, land, wafer and soil.

Question 2.
What are the essential components of biodiversity?
Answer:
Essential components of biodiversity are:

1. Genetic diversity.
2. Species diversity.
3. Ecological diversity.

Question 3.
When and how photochemical smog is formed?
Answer:
Photochemical smog is the chemical reaction of sunlight, nitrogen oxides and volatile organic compounds in the atmosphere, which leaves air borne particles and ground level ozone. Peroxyacetyl Nitrate (PAN) is present in the photochemical smog. It causes damage to respiratory system and environment system. (The word smog is derived from smoke and fog.)

Question 4.
Name two plants which uses carbon monoxide (CO).
Answer:
Two plants which uses CO are:

1. Daucus carota,
2. Ficus variegata.

Question 5.
Name the types of water pollution based on source of water.
Answer:
Types of water pollution based on source of water:

1. Underground water pollution
2. Surface water pollution
3. Lake water pollution and
4. River water pollution.

Question 6.
How does fertilizers pollute water?
Answer:
Excess fertilizers used in the field are drained into the pond and river, where they proliferate growth of algae, which spread all over the surface of the water and cut off oxygen supply for lower aquatic organisms. Due which lower aquatic organisms dies. This is known as eutrophication.

Question 7.
Give harmful effects of cutting of forest.
Answer:
Harmful effects of cutting of forest are:

1. It increases landscape.
2. It decreases fertility of the soil.
3. Rainfall reduces and become uncontrolled.

Environmental Issues Short Answer Type Questions

Question 1.
What is acid rain? Write any two effects of acid rain on human beings.
Answer:
Acid rain:
SO₂ gas is released into the environment by burning of fossil fuels containing sulphide from industries, example Coal. In presence of moisture, SO₂ reacts with water and forms the droplets of sulphurous and sulphuric acid in the environment. Likewise nitrogen oxides released from motor vehicles, burning materials and chemical industries also react with water to produce the droplets of nitric acid. The droplets of sulphurous, sulphuric acid and nitric acids fall down on the earth with rainwater and thus, it is known as acid rain.

• SO₂ + \(\frac { 1 }{ 2 }\) O₂ → SO₃
• H₂O + SO₃ → H₂SO₄
• H₂O + NO₂ → HNO₃

Effects on human beings : Skin diseases, irritation, metabolic diseases.

Question 2.
Write the effect of air pollution on plants.
Answer:
Effect of air pollution on plants:

1. Increase in the concentration of SO₂ causes chlorosis of leaves in plants.
2. The cells and chlorophyll of leaves are degraded and fall down.
3. Vegetative and reproductive growth is inhibited.
4. The development of plant is inhibited.

Question 3.
Write short note on pollution caused due to combustion activities.
Answer:
Combustion of fuel, such as wood, coal, natural gas for cooking and for some other puiposes gives out gases like CO₂, CO, SO₂ and consumes oxygen from the air.

Question 4.
Write preventive measures to control air pollution.
Answer:
Preventive measures to control air pollution:

1. Industrial smokes must be filtered before releasing it into the atmosphere.
2. Tree plantation should be increased and deforestation prevented.
3. Use of automobiles should be minimized which reduce the nitrogen contents in the atmosphere.
4. The use of crude fuels should be avoided and use of high quality fuels should be recommended.
5. Nuclear explosions should be avoided.
6. Legal control of air pollution.
7. Plantation of air purifying plants.
8. Development of parks and gardens in cities.

Question 5.
Write any four measures to control greenhouse effect.
Answer:
Following important measures can be sited as the base steps towards controlling greenhouse effect:

1. The aim is achieved to some extent by reducing the consumption of fossil fuels such as coal and petroleum. This can be achieved by depending more on non – conventional renewable sources of energy such as solar, wind, tidal, biogas and nuclear energies.
2. Disposing of the greenhouse gases as they are formed elsewhere than in the atmosphere.
3. By recovering greenhouse gases present already in the atmosphere and disposing them off elsewhere.
4. Prevention of deforestation and planting of more trees.

Question 6.
Write short note:

1. Bio – magnification.
2. UV rays.
3. Biodegradable pollutants.
4. Non – biodegradable.

Answer:
1. Bio – magnification:
Bio – magnification, also known as bio-amplification or biological magnification is the increasing concentration of a substance, such as a toxic chemical, in the tissues of organisms at successively higher levels in a food chain.

2. UV rays:
Sunlight is the chief source of ultraviolet rays. The rays of sunlight having a wavelength range from 200 to 390 nm are called as UV rays.

Effect of UV rays:
They have a direct effect on living cells. DNA and other chemical substances in cells are inactivated due to UV rays. Prevention of replication of the DNA molecule and its distortion cause many ill effects.

3. Biodegradable pollutants:
The pollutants which are degraded by natural factors and decomposed by natural activities are known as biodegradable pollutants, e.g., Domestic sewage, heat. The domestic sewage can be rapidly decomposed by natural processes or in engineered system (sewage treatment plant) that enhance natures great capacity to decompose and recycle.

4. Non – biodegradable:
The pollutants which cannot be purified by natural methods are called non-biodegradable pollutants. Plastic products, many chemicals, DDT, long chain detergents, glass aluminium, mercury salt and other synthetic products manufactured by man come under this category. These non – biodegradable pollutants not only accumulate but often “biologically magnified” as they move in biochemical cycles and along food chains.

Question 7.
What is the effect of SO₂ in the environment?
Answer:

Question 8.
Write only the sources of water and air pollution.
Answer:
1. Sources of water pollution : Following are the chief sources of water pollution:

(a) Human sources:

• Domestic sewage
• Industrial effluents
• Agricultural wastes
• Oil pollution
• Thermal and nuclear power station.

(b) Natural sources:
Water pollution takes place by natural ways like soil erosion, mixing of metallic substances, plant leaves, humus and faecal matter of animals etc.

2. Sources of air pollution : Following are the chief sources of air pollution:

(a) Human sources:

• Combustion activities
• Industrial activities
• Agricultural works
• Use of solvents
• Activities concerned with atomic energy.

(b) Natural sources:
Volcano and its lava, ash, dust, smoke of forest fire, winds, cyclone. Decomposition of matters in the swamp water and liberation of methane gas and different compounds of hydrogen from forests plants, various pollen grains etc.

Question 9.
Write a note on the effect of water pollution on aquatic organisms.
Answer:
Effect of water pollution on aquatic plants:

1. Inorganic nitrates and phosphates in excess amount stimulates excessive plant growth in lakes and reservoirs. These plants deplete the oxygen contents of the water during night. This leads to suffocation of fish and other aquatic life.

2. The rapid algal growth leads to the diminishing of nutrient in the medium causing rapid decay of algal filament. This increases productivity of lake and stream water brought about by nutrient enrichment is called eutrophication.

3. The number of microorganisms increases in polluted water.

4. Siltation occurs in water.

5. Temperature of water increases and O₂ ratio is reduced.

Effect of water pollution on aquatic animals : Animal life depends on aquatic plants. Aquatic animals are affected from water pollution as follows:

1. Decrease in the value of B.O.D. resulting in the number of aquatic animals.
2. Animals found in freshwater and killed.
3. Animal diversity is also decreased and fishes suffer from various types of diseases.
4. Terrestrial animals are also affected by the use of polluted water.

Question 10.
Write the measures for control and prevention of water pollution.
Answer:
Measures for control and prevention of water pollution are:

Preventive measures:

1. Use of harmful pesticides and weedicides must be stopped completely.
2. Discharge of effluents into rivers, lakes and sea should be strictly prohibited without treatment.
3. Oil spill should be prevented.
4. Proper disposal of sewage so, that it does not find its way into water bodies.
5. Preventing bathing, washing cloths, throwing dead bodies and other wastes into water source.

Curative measures:

1. Adequate waste water treatment:
The domestic sewage and the industrial waste should be properly treated before its disposal into water ways.

Waste water treatment involves three steps :

(a) Primary treatment – During this treatment large objects and suspended undissolved solids are removed and converted into the sludge, a valuable fertilizer.

(b) Secondary treatment – Aeration is supplied to promote bacterial decomposition, followed by chlorination to reduce its content of bacteria.

(c) Tertiary treatment – During this phase nitrates and phosphates are removed. The treated water is then released. Sewage treatment is quite expensive and in many developing countries only the first two steps are followed.

2. Treatment of industrial effluents:
Industrial effluents should be properly treated to remove the pollutants. These involve neutralization of acids and alkalies, removal of harmful chemicals, coagulation of colloidal impurities, precipitation of metallic compounds and reducing the temperature of wastes to decrease thermal pollution. Chemical oxidation can be achieved by chlorination through reaction with ozone. However, there are certain chemicals which are difficult to remove.

3. Recycling:
One of the best methods of prevention and control of water pollution is the recycling of the various kinds of pollutants and wastes, example Dung of cow and buffalo can be used for the production of gobar gas a cheap source of fuel and also as manure.

Question 11.
Write short notes on air pollution caused due to industries.
Answer:
Industrial pollution is caused by industrial pollutants which are released by the industries such as SO₂, CO₂, CO, H₂S and hydrocarbons together with dust and smoke. These are produced by the burning of coal and petroleum. The chemical industries releases HC₁, Cl₂, nitrogen oxide, oxides of copper, zinc, lead, arsenic etc. Industrial pollutants causes air pollution and water pollution.

Question 12.
Write the effects of noise pollution.
Answer:
Effects of Noise pollution:

1. The more acute and immediate effect of noise pollution is impairing of hearing leading to auditory fatigue and may finally lead to. deafness.
2. Interference with speech communication.
3. Noise pollution leads to neurosis, anxiety hypertension, cardiovascular disease, hepatic stress, giddiness.
4. Annoyance leading to ill – temper, bickering, mental disorientation and violent behaviour.
5. The high intensity of noise pollution can cause blood vessels to contract, skin becomes pale, muscles to contract and adrenaline to be short into blood stream with consequence rise in blood pressure. This ultimately results in tension and nervousness.
6. Affects different metabolic activities.

Question 13.
What are the various constituents of domestic sewage? Discuss the effects of sewage discharge on a river.
Answer:
Domestic sewage are the waste originating from the kitchen, toilet, laundry and other sources. They contain impurities such as suspended solid (sand, salt, clay), colloidal materials (faecal matters, bacteria, plastic and cloth fiber), dissolved materials (nitrate, phosphate, calcium, sodium, ammonia) and disease – causing microbes.

When organic wastes from the sewage enter the water bodies, they serve as a food source for microorganisms such as algae and bacteria. As a result, the population of these microorganisms in the water body increases. Here, they utilize most of the dissolved oxygen. for their metabolism. This results in an increase in the levels of BOD in river water and results in the death of aquatic organisms. Also, the nutrients in the water lead to the growth of planktonic algal, causing algal bloom. This causes deterioration of water quality and fish mortality.

Question 14.
List all the wastes that you generate at home, school or during your trips to other places, could you very easily reduce. Which would be difficult or rather impossible to reduce?
Answer:
Wastes generated at home include plastic bags, paper napkin, toiletries, kitchen wastes (such as peelings of vegetables and fruits, tea leaves), domestic sewage, glass, etc. Wastes generated at school include waste paper, plastics, vegetable and fruit peels, food wrapping, sewage, etc.

Wastes generated at trips or picnics include plastic, paper, vegetable and fruit peels, disposable cups, plates, spoons etc. Yes, wastes can be easily reduced by the judicious use of the above materials. Wastage of paper can be minimized by writing on both sides of the paper and by using recycled paper.

Plastic and glass waste can also be reduced by recycling and re-using. Also, substituting plastics bags with biodegradable jute bags can reduce wastes generated at home, school or during trips. Domestic sewage can be reduced by optimizing the use of water while bathing, cooking, and other household activities. Non – biodegradable wastes such as plastic, metal, broken glass, etc. are difficult to decompose because microorganisms do not have the ability to decompose them.

Question 15.
Write preventive measures of sound pollution.
Answer:
Preventive measures of noise pollution:

1. Noise pollution can be controlled by designing quieter machines, proper lubrication and better maintenance of machines. Noise producing parts of machines may be covered by suitable insulating materials.
2. Noise producing industries should be installed away from residential areas.
3. Use of public addressing systems should be restricted at a fixed intensity and hours of the day.
4. Stuffing of cotton plugs in the ear or use of earmuffs and minimise the danger of occupational exposure of noise.
5. Trees absorb sound vibrations and thus, reduce the noise pollution. Thus, plantation of trees along highways, establishment of parks help in reducing noise pollution.
6. By use of silencer in all automobiles and engines.

Question 16.
How many types of pollution are there? Explain in brief.
Answer:
Pollution are of following five types:

1. Air pollution
2. Noise pollution
3. Water pollution
4. Radioactive pollution
5. Soil pollution.

1. Air pollution – Pyrotechnics results in the release of harmful gases like CO₂, CO and SO₂ in the environment which produces various types of diseases in living beings.

2. Noise pollution – Pyrotechnics produces the unwanted sound dumped into the atmosphere leading to health hazards.

3. Water pollution – Pyrotechnics produces some waste materials, on reaching water it causes water pollution.

4. Radioactive pollution – Pollution created due to radioactive substance is called radioactive pollution. Atomic energy centres and bombarding also causes radioactive pollution.

5. Soil pollution – Any change in physical and chemical characteristics of soil due to addition of unwanted substances which adversely affects productivity of soil is termed as soil pollution.

Question 17.
Discuss the role of women and communities in protection and conservation of forests.
Answer:
Women and Communities have played a major role in environmental conservation movements. The Bishnoi community in Rajasthan strictly believes in the concept of living peacefully with nature. In 1731, the King of Jodhpur ordered his ministers to arrange wood for the construction of his new palace. For this purpose, the minister and the workers went to bishnoi village. There, a Bishnoi woman called Amrita Devi along with her daughter and hundreds of other Bishnoi’s showed the courage to step forward and stop them from cutting trees.

They embraced the trees and lost their lives at the hands of soldiers of the king. This resistance by the people of the village forced the king to give up the idea of cutting trees. The Chipko movement was started in 1974, in the Garhwal region of the Himalayas. In this movement, the women from the village stopped the contractors from cutting forest trees by embracing them.

Question 18.
What measures, as an individual you would take to reduce environmental pollution?
Answer:
The following initiatives can be taken to prevent environmental pollution:

Measures for preventing air pollution:

• Planting more trees.
• Use of clean and renewable energy sources such as CNG and Bio – fuels.
• Reducing the use of fossil fuels.
• Use of catalytic converters in automobiles.

Measures for preventing water pollution:

• Optimizing the use of water.
• Using kitchen waste water in gardening and other household purposes measures for controlling noise pollution :
  Avoid burning crackers on Diwali , Plantation of more trees.

Measures for decreasing solid waste generation:

• Segregation of waste.
• Recycling and reuse of plastic and paper.
• Composting of biodegrable kitchen waste.
• Reducing the use of plastics.

Question 19.
What are the causes of air pollution?
Answer:
The causes of air pollution:
The main causes of air pollutions are fossil fuels industries, factories and particulate matter are produced due to combustion of petrol, diesel, kerosene etc. in vehicles, houses and factories which pollute air. In thermal power plants, steel and glass industries, paper and sugar mills etc. combustion of coal and furnace oil produce carbon monoxide, carbon dioxide, sulphur dioxide, ash, dust particles and some heavy metals which are released into atmosphere. Similarly various types of chemicals released from cloth mills, cement industries, asbesters industries, pesticides industries etc. also causes air pollution.

Environmental Issues Long Answer Type Questions

Question 1.
Discuss the causes and effects of global warming. What measures need to be taken to control global warming?
Answer:
Global warming:
Global warming is defined as an increase in the average temperature of the Earth’s surface due to greenhouse effect.

Causes of global warming:
Global warming occurs as a result of the increased concentration of greenhouse gases in the atmosphere. Greenhouse gases include carbon dioxide, methane and water vapour. These gases trap solar radiation released back by the earth. Global warming is a result of industrialization, burning of fossil fuels, and deforestation.

Effects of global warming:
Global warming is defined as an increase in the average temperature of the Earth’s surface. It has been observed that in the past three decades, the average temperature of the Earth has increased by 0-6°C. As a result, the natural water cycle has been disturbed resulting in changes in the pattern of rainfall. It also changes the amount of rainwater. Also, it results in the melting of polar ice caps and mountain glaciers, which has caused a rise in the sea level, leading to the inundation of coastal regions.

Control measures for preventing global wanning:

• Reducing the use of fossil fuels.
• Use of bio – fuels.
• Improving energy efficiency.
• Use of renewable source of energy such as CNG, etc.
• Reforestation.
• Recycling of materials.

Question 2.
Write critical notes on the following:

1. Eutrophication,
2. Biological magnification,
3. Groundwater depletion and ways for its replenishment.

Answer:
1. Eutrophication:
It is the natural ageing process of lake caused due to nutrient enrichment. It is brought down by the runoff of nutrients such as animal wastes, fertilizers, and sewage from land which leads to an increased fertility of the lake. As a result, it causes a tremendous increase in the primary productivity of the ecosystem. This leads to an increased growth of algae, resulting into algal blooms. Later, the decomposition of theses algae depletes the supply of oxygen, leading to the death of other aquatic animal life.

2. Biological magnification:
The increase in concentration of harmful non – biodegradable substances into higher tropic level is called biological magnification. DDT used to protect the crops reach the soil and are absorbed by plants with water and minerals from the soil. Due to rain, these chemicals can also enter water sources and into the body of aquatic plants and animals.

As a result, chemicals enter the food chain. Since, these chemicals cannot be decomposed, they keep on accumulating at each trophic level. The maximum concentration is accumulated at the top carnivore’s level. The producers (phytoplankton) were found to have 0 04 ppm concentration of DDT.

Since many, types of phytoplankton were eaten by zooplankton (consumers), the concentration of DDT in the bodies of zooplankton was found to be 0.23ppm. Small fish that feed on zooplankton accumulate more DDT in their body. Thus, large fish (top carnivore) that feed on several small fish have the have the highest concentration of DDT.

3. Groundwater depletion and ways for its replenishment:
The level of ground water has decreased in the recent years. The source of water supply is rapidly diminishing each year because of an increase in the population and water pollution. To meet the demand of water, water is withdrawn from water bodies such as ponds, rivers, etc. As a result, the source of groundwater is depleting.

This is because the amount or groundwater being drawn for human use is more than the amount replaced by rainfall. Lack of vegetation cover also results in very small amounts of water seeping through the ground. An increase in water pollution is another factor that has reduced the availability of groundwater.

Measures for replenishing groundwater:

• Preventing over-exploitation of groundwater.
• Optimizing water use and reducing water demand.
• Rainwater harvesting.
• Preventing deforestation and plantation of more trees.

Question 3.
Why ozone hole forms over Antarctica? How will enhanced ultraviolet radiations affect us?
Answer:
The ozone hole is more prominent over the region of Antarctica. It is formed due to an increased concentration of chlorine in the atmosphere. Chlorine is mainly released from Chlorofluorocarbons (CFC’s) widely used as refrigerants. The CFC’s release chlorine atoms by the action of UV rays on them. The release of chlorine atoms causes the conversion of ozone into molecular oxygen. One atom of chlorine can destroy around 10,000 molecules of ozone and causes ozone depletion.

• CF₂Cl₂ → CF₂Cl + Cl
• Cl + O_{3 →} ClO + O₂
• ClO + O → Cl + O₂

The formation of the ozone hole will result in an increased concentration of UV – B radiation on the Earth’s surface. UV – B damages DNA and activates the process of skin ageing. It also causes skin darkening and skin cancer. High levels of UV – B cause corneal cataract in human beings.

Question 4.
Discuss briefly the following:

1. Radioactive wastes,
2. Defunct ships and e-wastes,
3. Municipal solid wastes.

Answer:
1. Radioactive wastes:
Radioactive wastes are generated during the process of ‘ generating nuclear energy from radioactive materials. Nuclear waste is rich in radioactive materials that generate large quantities of ionizing radiation such as gamma rays. These rays cause mutation in organisms which often results in skin cancer. At high dosage, these rays can be lethal. Safe disposal of radioactive wastes is a big challenge. It is recommended that nuclear wastes should be stored after pre-treatment in suitable shielded containers, which should then be buried in rocks.

2. Defunct ships and e – wastes:
Defunct ships are dead ships that are no longer in use. Such ships are broken down for scrap metal in countries such as India and Pakistan. These ships are a source of various toxicants such as asbestos, lead, mercury etc. Thus, they contribute to solid wastes that are hazardous to health.

E – wastes or electronic wastes generally include electronic goods such as computers, etc. Such wastes are rich in metals such as copper, iron, silicon, gold, etc. These metals are highly toxic and pose serious health hazards. People of developing countries are involved in the recycling process of these metals and therefore, get exposed to toxic substances present in these wastes.

3. Municipal solid wastes:
Municipal solid wastes are generated from schools, offices, homes and stores. It is generally rich in glass, metal, paper waste, food, rubber, leather and textiles. The open dumps of municipal wastes serve as a breeding ground for flies, mosquitoes and other disease causing microbes. Hence, it is necessary to dispose municipal solid waste properly to prevent the spreading of diseases. Sanitary landfills and incineration are the methods for the safe disposal of solid wastes.

Question 5.
What initiatives were taken for reducing vehicular air pollution.in Delhi? Has air quality improved in Delhi?
Answer:
Delhi has been categorized as the fourth most polluted city of the world in a list of 41 cities. Burning of fossil fuels has added to the pollution of air in Delhi. Various steps have been taken to improve the qualify of air in Delhi

Introduction of CNG (Compressed Natural Gas):
By the order of the Supreme Court of India, CNG powered vehicles were introduced at the end of year 2006 to reduce the levels of pollution in Delhi. CNG is a clean fuel that produces very little unbumt particles.

1. Phasing out of old vehicles
2. Use of unleaded petrol.
3. Use of low – sulphur petrol and diesel.
4. Use of catalytic converters.
5. Application of stringent pollution-level norms for vehicles.
6. Implementation of Bharat stage I, which is equivalent to euro II norms in vehicles of major Indian cities.

The introduction of CNG – powered vehicles has improved Delhi’s air quality, which has lead to a substantial fall in the level of CO₂ and SO₂. However, the problem of Suspended Particulate Matter (SPM) and Respiratory Suspended Particulate Matter (RSPM) still persists.

Question 6.
Discuss briefly the following:

1. Greenhouse gases
2. Catalytic converter
3. Ultraviolet – B

Answer:
1. Greenhouse gases:
The greenhouse effect refers to an overall increases in the average temperature of the earth due to the presence of greenhouse gases. Greenhouse gases mainly consist of carbon dioxide, methane and water vapour. When solar radiations reach the Earth, some of these radiations are absorbed.

These absorbed radiations are released back into the atmosphere. These radiations are trapped by the greenhouse gases present in the atmosphere. This helps in keeping our planet warm and thus, helps in human survival. However an increase in the amount of greenhouse gases can lead to an excessive increase in the Earth’s temperature, thereby causing global warming.

2. Catalytic converter:
Catalytic converters are devices fitted in automobiles to reduce vehicular pollution. These devices contain expensive metals such as platinum, palladium and rhodium that act as catalysts. As the vehicular discharge passes through the catalytic converter, the unbumt hydrocarbons present in it get converted into carbon dioxide and water. Carbon monoxide and nitric oxide released by catalytic converters are converted into carbon dioxide and nitrogen gas (respectively).

3. Ultraviolet – B:
Ultraviolet – B is an electromagnetic radiation which has a shorter wavelength than visible light. It is a harmful radiation that comes from sunlight and penetrates through the ozone hole on to the earth’s surface. It induces many health hazards in humAnswer: UV – B damages DNA and activates the process of skin ageing. It also causes skin darkening and skin cancer. High levels of UV – B cause comeal cataract in human beings.

Question 7.
Describe the diseases caused due to radioactive pollution.
Answer:
Following diseases are caused by radioactive pollution:

1. Leukaemia and Bone cancer – In human beings and other animals, radioactive pollution causes blood and bone cancer.
2. Ageing – The reproductive capacity of organisms is decreased due to radioactive pollution and ultimately causes ageing.
3. Epidemic disease – Radioactive pollution causes the decreased rate of antitoxin production which affects the immune system and results in the production of epidemic diseases.
4. Nervous system and sensory cells become irritated.
5. Radioactive pollution results skin cancer.
6. Mutations also take place due to radioactive pollution.
7. Thyroid cancer is also produced due to radioactive iodine.

Question 8.
Burning of crackers spread pollution in the environment. Explain.
Answer:
Explosive substances are used in crackers. Crackers are burned for celebration but it causes many harms to the living organisms. Some of the harm effects of burning of crackers are as follows:

1. Air pollution – It releases harmful gases such as CO₂, CO, SO₂ etc. into the atmosphere which causes respiratory diseases.

2. Noise pollution – Burning of crackers produces noise which causes noise pollution which effects our hearing capacity, causes headache, increases heartbeat and blood pressure.

3. Water pollution – Waste of burning of crackers are drained into the pond, river with rainwater or through drain which caused harm to aquatic organisms.

4. Soil pollution – Burning of crackers not only produces smoke of harmful gases but its harmful chemicals mixes with soil and pollute it, which makes the soil sterile and reduced growth of plants.

Question 9.
Write an essay on air pollution.
Answer:
Air pollution:
Any change in the composition of air is called pollution. The chief reason of air pollution is the releasing of harmful gas in the environment that affect our eyes, lungs, skins, heart and brain and produce various disorders in them.

Sources of air pollution : Following are the chief sources of air pollution:

1. Human sources:

• Combustion activities
• Industrial activities
• Agricultural works
• Use of solvents
• Activities concerned with atomic energy

2. Natural sources:
Volcano and its lava, ash, dust, smoke of forest fire, winds, cyclone. Decomposition of matters in the swamp water and liberation of methane gas and different compounds of hydrogen from forests plants, various pollen grains etc.

Effects of air pollution:

• Air pollutant like SO₂ enter the soft tissues causing drying of the mouth, scratching throat and smarting eyes.
• Hydrocarbons and many other pollutants are responsible for causing cancer.
• Oxides of carbon, sulphur and nitrogen diffuse into the blood stream to combine with haemoglobin causing reduction in its oxygen carrying capacity. CO severely damages cardiovascular system and disturbs psychometry functions.
• RA.N. inhibits Hill reaction and thus, decreases photosynthetic production of an ecosystem.

Control of air pollution:

• Industrial smokes must be filtered before releasing it into the atmosphere.
• Tree plantation should be increased and deforestation prevented.
• Use of automobiles should be minimized which reduce the nitrogen contents in the atmosphere.
• The use of crude fuels should be avoided and use of high quality fuels should be recommended.
• Nuclear explosions should be avoided.
• Legal control of air pollution.
• Plantation of air purifying plants.
• Development of parks and gardens in cities.

Question 10.
Name the major air pollutants and their individual effects.
Answer:
Major air pollutants and their individual effects are as follows:

1. Carbon monoxide:
|It is highly poisonous gas. On entering blood stream, it combines with haemoglobin to block oxygen transport function. It causes laziness, headache, disturbance of psychometry functions, decrease in visual perception, serious effects on cardiovascular system etc.

2. Sulphur dioxide:
The animals and human population when exposed to SO₂, suffer from respiratory diseases. It causes chest constriction, headache, vomiting and death from respiratory ailments.

3. Hydrogen sulphide:
Its rotten egg like smell cause nausea, irritation in eyes and throat.

4. Nitrogen oxide:
It inhibits cilia action so that soot and dust penetrate far into the lungs and finally resulting respiratory diseases in human beings and animals. In plants, it causing chlorosis.

5. Aerosoles:
These are the chemicals affecting ozone layer of the atmosphere due to which UV rays enter in the atmosphere and causing various diseases in plants and animals.

6. Ammonia:
It inflame upper respiratory passage in human beings. In plants, it inhibits seed germination, destruction of chloroplast and inhibition of the growth of roots and shoots.

7. Hydrogen chloride:
It affects the leaves of the plants, eyes, respiratory organs of animals and human beings.

8. Hydrocarbons:
It causes yellowing of plants, drying of buds, whereas in man and other animals, mucous glands of eyes and nose are irritated.

Question 11.
Write down the methods to control soil pollution.
Answer:
Methods to control soil pollution:
By following methods soil pollution can be prevented:

1. Government should make provisions for latrines and people should not be allowed to litter in open fields.
2. Solid wastes such as tin, copper, iron, glass etc., should not be dumped into soil.
3. The solid wastes should be recycled to form new materials.
4. Fertilizers and pesticides should be judiciously used.
5. Instead of pesticides to kill pests, biological control methods should be adopted.
6. The,excretory waste of man and cattle should be used to prepare biogas.
7. Clean and well covered dustbins should be used in villages and cities.
8. Closed hollow pipes should be used to collect and discharge liquid wastes.
9. To prevent soil erosion, grass and small plants should be grown.
10. Soil management should be adopted.

Question 12.
What are the sources of soil pollution?
Answer:
Sources of soil pollution:

1. Acid rainwater and water from mines are main sources of soil pollution.
2. Mixing of debris, waste products with the soil cause soil pollution.
3. By excessive use of pesticides and fertilizers, pollute the soil.
4. When industrial wastes are discharged in the soil, they also pollute the soil.
5. Heavy metals example Cadmium, Zinc, Nickel, Arsenic etc., are mixed with the soil from mines. These metals are harmful for plants as well as primary and secondary consumers, in a food chain.
6. Bones of dead animals, paper putrified flesh and food, iron, copper, mercury etc., pollute the soil.
7. In our villages and rural areas, where there are no latrines, people go in fields for being at ease. From their stool also the soil is polluted.
8. Insecticides like D.D.T. is very dangerous substance. When these enter the body of consumers from producers, their concentration is increased because these are non – degradable substances. Moreover, these can remain in the atmosphere for upto 15 years.

Question 13.
What is visible spectrum? Write short note on spectrum of solar light and also write effect of UV rays on the living organisms.
Answer:
The visible spectrum is the portion of the electromagnetic spectrum that is visible to the human eye. Electromagnetic radiation in this range of wavelength is called visible light or simply light. It ranges from 390 nm to 760 nm wavelength.

Different spectrum of sunlight (solar light):
Spectrum of sunlight can be divided into three parts :

1. Ultraviolet spectrum – The UV spectrum covers the wavelength range 100 – 390 nm and is divided into three bands. It is invisible.
2. Visible spectrum – It include light of ranges from 390 to 760 nm wavelength.
3. Infrared spectrum – Light ranges 760 nm above is called as infrared ray. It is also invisible.

Effect of UV rays in the living organisms:

1. It make DNA abnormal thus, protein sythesis in the body get effected.
2. It may cause skin cancer.
3. It supress immune system of the body.
4. It causes sunburn and premature ageing of skin.
5. It causes formation of harmful photochemical smog which causes respiratory diseases.
6. It damage eyes.

Question 14.
Explain non – ionising and ionising radiation.
Answer:
Types of radiations on the basis of their action on cells, radiations are of two types :

1. Non – ionising radiations:
These include ultraviolet rays (UV rays; 100-300 nm), these have low penetration capacity. These affect only ‘those cells which absorb them. The cause are as following :

• Sunburn – It includes rupturing of sub – cutaneous blood capillaries, blisters, reddening of skin and injury to stratum germinativum.
• Snow blindness – It damages eyesight due to damage of corneal cells.
• Inactivation of organic bio-molecules and formation of thymin dimer in DNA.
• However they cause cancer, tumor, skin disease, etc.

2. Ionizing radiations :
These include X – rays, cosmic rays and atomic radiations.

These radiations have low wavelengths but high penetration power. These damage the living cells shifting the electron from one to other bio – molecule. Their harmful effects are as follows:

1. Short range effects : These may appear within few days or a few weeks after exposure. The effects include loss of hair, nails, subcutaneous bleeding, metabolic changes, change in proportion of blood cells, dead tissues or death of the organisms in high dose.

2. Long range effects : These appear in several months or even several years after their exposure. These cause tumours, cancers, mutations, genetic deformities, shorter life span and developmental defects.

Question 15.
What are sources of noise pollution? Write any four effects of noise pollution.
Answer:
Sources of noise pollution:
Noise is either natural such as thunder or man – made. The common sources of noise pollution are :

1. Industries such as textile mills, printing-press, engineering establishments, etc.
2. Defence equipment and vehicles such as tanks, artillery and rocket launching explosions, practise firing, etc.
3. Transport vehicles as trains, trucks, buses, two – wheelers, jet planes, etc. and accessory noise produced by horns, sirens.
4. Other applications as dynamite blasting, use of jack hammers, pile drivers, bulldozer, lawn mowers, etc.

Effects of noise pollution:

1. The more acute and immediate effect of noise pollution is impairing of hearing leading to auditory fatigue and may finally lead to deafness.
2. Interference with speech communication.
3. Noise pollution leads to neurosis, anxiety, hypertension, cardiovascular disease, hepatic stress, giddiness.
4. Annoyance leading to ill – temper, bickering, mental disorientation and violent behaviour.
5. The high intensity of noise pollution can cause blood vessels to contract, skin becomes pale, muscles to contract and adrenaline to be shot into blood stream with consequence rise in blood pressure. This ultimately results in tension and nervousness.
6. Affects different metabolic activities.

Question 16.
What is Greenhouse effect? Write any four impacts of Greenhouse effect
Answer:
Greenhouse effect:
The Greenhouse effect may therefore, be defined as “The progressive warming up of the earth’s surface due to blanketting effect of man – made CO₂ in the atmosphere. It means the excessive presence of those gases blocked in the infrared radiation from earth’s surface to the atmosphere leading to an increase in temperature, which in turn would make life difficult on earth to forthcoming future generations.

Four impacts of Greenhouse effects are :

1. Effect on global climate:
The climate of earth has never been free of change. Its composition, temperature and self – cleansing ability have all varied since, the planet first formed. Yet the pace in the past two countries has been remarkable. The atmosphere’s composition in particular has changed significantly faster than it has at any time in human history.

Increase in Greenhouse gases does not increase the temperature of the earth i. e., it is not uniform. Temperature is normal at poles and is very less at tropics. In Iceland, Greenland, Sweden, Norway, Finland, Alaska, Siberia, temperature is more, thus at poles, ice starts melting.

2. Effect of Greenhouse Effect on forests:
When the atmospheric temperature increases, only those plants can survive which can bear high temperature. Because of this, new species of plant also come into existence, Herbaceous plant can not survive in this high temperature. Plants with high wood will increase in number. According to a survey if the amount of CO₂ in the atmosphere is doubled, there will be a great decrease in a green biomass.

3. Effect on crops:
The positive aspect of greenhouse effect will be on agriculture. Because, CO₂ is a natural fertilizer, the plants will grow larger and faster with increasing CO₂ in the atmosphere. At first sight, the abnormal growth of plants might be expected to be beneficial because the yields of major crops might increase, however chances of depletion of characters and productivity of soil are also associated with this.

4. Effect on ozone layer:
During depletion, the chlorine, fluorine, or bromine molecules of CFCs and halogens are converted into reactive free radical form by photochemical reactions from their intial non – reactive sites. Oxides of nitrogen generally inactivate Cl but the lowering of stratospheric temperature changes NO₂ into non – reactive nitric acid. Thus, Cl or F are free to react with ozone, distintegrating it into O₂ + O. The tiny ice particles during winter favours the conversion of chlorine into chlorine monoxide, which behaves as a catalytic compound. The abundance of chloromonoxide rich air in stratosphere continues to rise. Chlorine monoxide react with nascent oxygen and thereby gets converted into chlorine. Thus, the cycle continues destroying the ozone level.

MP Board Class 12th Biology Important Questions

MP Board Class 12th Biology Important Questions Chapter 15 Biodiversity and Conservation

Biodiversity and Conservation Important Questions

Biodiversity and Conservation Objective Type Questions 

Question 1.
Choose the correct answer:

Question 1.
Tera biodiversity is used for the first time by :
(a) V. G. Rosseu
(b) Linnaeus
(c) Odum
(d) Theophrastus.
Answer:
(a) V. G. Rosseu

Question 2.
When does wild life conservation Act come in the force in India :
(a) 1883
(b) 1972
(c) 1973
(d) 1972
Answer:
(b) 1972

Question 3.
When does IBWL is started :
(a) 1952
(b) 1981
(c) 1973
(d) 1972
Answer:
(a) 1952

Question 4.
Where is NBPR is situated :
(a) Delhi
(b) Kolkata
(c) Lucknow
(d) Mumbai
Answer:
(a) Delhi

Question 5.
The number of biosphere reserves in India is :
(a) 73
(b) 7
(c) 416
(d) 23
Answer:
(b) 7

Question 6.
Which National Park is associated with white tiger:
(a) Kanger
(b) Satpura
(c) Bandhavgarh
(d) Kanha
Answer:
(c) Bandhavgarh

Question 7.
First National Park of M. P. is :
(a) Shivpuri
(b) Bandhavgarh
(c) Kanha
(d) Kanger
Answer:
(c) Kanha

Question 8.
Where is “Plant Fossil” National Park is situated:
(a) Shivpuri
(b) Mandala
(c) Bastar
(d) Bhopal
Answer:
(b) Mandala

Question 9.
The National Park of M.P. which is marked as biosphere reserve is :
(a) Kanha
(b) Shivpuri
(c) Bandhavgarh
(d) Satpura.
Answer:
(b) Shivpuri

Question 10.
First Tiger Project of M.P. is :
(a) Bandhavgarh
(b) Kanha
(c) Sidhi
(d) Shivpuri.
Answer:
(b) Kanha

Question 2.
Fill in the blanks :

1. FRI is situated is …………………….
2. Kanjiranga sanctuary is associated with the conservation of …………………….
3. Red data book is related with the ……………………. conservation.
4.  ……………………. is the fust biosphere reserve of India.
5. List of the endangered species are available in ……………………. data book.
6.  ……………………. is the important componant of air for life.
7. Mineral is the example of ……………………. sources.
8. ……………………. percent part of the earth has water.
9. Biodiversity is the source of different types of …………………….
10. Species for which there are no living representative is called …………………….
11. Farming which is develop in fresh water is called …………………….
12. Botanical garden is an ……………………. method of plant conservation.
13. ……………………. conservation is the on site conservation.
14. ……………………. percent part of the mangrove plant are found in India.
15. Gir national park in India is famous for …………………….

Answer:

1. Dehradun
2. Rhinoceros
3. Endangered species
4. Nilgiri
5. Red
6. Oxygen
7. Non – renewable
8. 71
9. Germplasm
10. Extinct
11. Aquatic culture
12. Ex – situ
13. In – situ
14. 5
15. Asiatic lion.

Question 3.
Match the followings:
I.

Answer:

1. (b)
2. (c)
3. (d)
4. (a)

II.

Answer:

1. (d)
2. (a)
3. (b)
4. (e)
5. (c)

Question 4.
Write the answer in one word/sentences:

1. Where does Kanha National Park situated?
2. From which person Chipko movement is associated?
3. When does Wild life Conservation Act is emenced in force?
4. How many percent contribution of India is in world’s biodiversity?
5. How many species are found in India?
6. Which are known as the lungs of the earth?
7. How many national park in India?
8. How many wild 1 ife sanctuary in India?
9. What is the basic component of biodiversity?
10. Write the full name of IUCN.
11. Which are most helpful in nature balancing?
12. How many species of bamboo found in India?
13. Name any one book associated with the conservation of wild life.
14. Upper surface of the land is called.

Answer:

1. M.P
2. Sunder Lai Bahuguna
3. 1972
4. 8.1%
5. 1200 species
6. Amazon – rain forest
7. 90
8. 448
9. Gene
10. International Union for conservation of nature and natural resourses
11. Wild animal
12. 100 (about)
13. Red data book
14. Soil.

Biodiversity and Conservation Very Short Answer Type Questions

Question 1.
Name the three important components of biodiversity.
Answer:
Three important components of biodiversity are:

1. Genetic diversity.
2. Species diversity.
3. Ecosystem diversity.

Question 2.
What is forest?
Answer:
A large area of land covered with trees, shrubs and grasses.

Question 3.
As which day 21st March is observed?
Answer:
21st March is observed as “World Forest Day”.

Question 4.
What are extinct species?
Answer:
Species which are not found at present in the earth but survived in the past in this world are called as extinct species.
Example : Dinosaur.

Question 5.
What are mentioned in the Red Data Book of IUCN?
Answer:
The IUCN Red Data Book contains list of threatened species. It was founded in 1964. Correction was done in 2004.

Biodiversity and Conservation Short Answer Type Questions

Question 1.
How do ecologists estimate the total number of species present in the world?
Answer:
The diversity of living organisms present on the earth is very vast.According to an estimate by researchers, it is about seven millions. The total number of species present in the world is calculated by ecologists by statistical comparison between a species richness of a well studied group of insects of temperate and tropical regions. Then, these ratios are extrapolated with other groups of plants and animals to calculate the total species richness present on the earth.

Question 2.
The species diversity of plants (22 %) is much less than that of animals (72 %). What may be the reason that animals achieved greater diversifications?
Answer:
Animals have achieved greater diversification than plants due to following reasons:

1. They are mobile and thus, can move away from their predators or unfavourable, environments. On the other hand plants are fixed and have fewer adaptations to obtain optimum amount of raw materials and sunlight therefore, they show lesser diversity.

2. Animals have well developed nervous system to receive stimuli against external factors and thus, can respond to them. On the other hand, plants do not exhibit any such mechanism, thus they show lesser diversity than animals.

Question 3.
Can you think of a situation where we deliberately want to make a species extinct? How would you justify it?
Answer:
Yes, there are various kinds of parasites and disease causing microbes that we deliberately want to eradicate from the earth. Since, these microorganisms are harmful to human beings, scientists are working hard to fight against them. Scientists have been able to eliminate smallpox virus from the world through the use of vaccinations. This shows that humans deliberately want to make these species extinct. Several other eradication programmes such as polio and hepatitis B vaccinations are aimed to eliminate these disease causing microbes.

Question 4.
Explain conservation of biodiversity.
Answer:
Conservation of biodiversity:
India is one of the richest (among the 12 mega centres of the world) countries in biological diversity. This rich biodiversity is due to a variety of climatic conditions prevailing on different ecological habits ranging from tropical, subtropical, temperate, alpine to desert. These varied conditions harbour a plethora of organisms, which forms an important natures wealth, responsible for socio – economic development of life in our country. But the biodiversity of organisms are under serious threat and there is an urgent need for biodiversity conservation on war foot level.

Question 5.
What is social forestry?
Answer:
The planning of social forestry started in India since 1976, which is related with the conservation of forests. This project ;s useful for local people in various ways, such as it fulfil their requirements, provides work to unemployed, use of wasteland and help to maintain O₂ and CO₂ balance in the atmosphere, etc. Thus, project is started by Indian government, the chief objectives of this project are as follows:

1. Plantation of useful plants in the forest.
2. Development of forests on personal lands by the cooperation of government.
3. To prevent the harmful effects of pollution by development of artificial forest.
4. Preservation of endangered wild animals.

Question 6.
Write importance of forests.
Answer:
Importance of forests:

1. Forests play a vital role in the life and economy of all tribes living in forests, by providing food, medicines and other products of commercial value.

2. Forests are large biotic communities. It provides shelter and sustenance to a larger number of diverse species of plants, animals and microorganisms.

3. Forests prevent soil erosion by wind and water. The trees provide shade which prevents the soil from drying during summer. Trees reduce the velocity of raindrops or wind striking the ground so that dislodging of the slil partiles is reduces. The root system of plants firmly binds the soil.

Question 7.
Write any five features of Indian forests.
Answer:
Indian forests are characterized by:

1. Indian forests are mainly tropical forests.
2. Himalayan forests are characterized by the presence of coniferous trees.
3. In few parts of our country having temperate forests.
4. Our forests contain a large number of useful varieties of plants and animals.
5. A great variations are present in Indian forests.

Question 8.
Enumerate any five reasons for the destruction of wildlife (animals).
Answer:
The main reason of destruction of wildlife (animals) are:

1. Entertainment, personal profit, earning money by wrong methods are the human illattitudes, unkindness toward wild animals has brought the animals at endanger level.
2. Huge reduction in the natural habitat of wild animal so it has reduced their living area due to urbanization, industrialization and deforestation.
3. Exhorbitant extraction and consumption is harmful for wild animals like skin of animals, Teeth of elephant etc.
4. Various types of pollution has force the reduction of wild animals.
5. Very loose and unpunishable wildlife act which has increase poaching, hunting of wild animals.

Question 9.
How is biodiversity important for functioning of ecosystem?
Answer:
An ecosystem with high species diversity is much more stable than an ecosystem with low species diversity. Also, high biodiversity makes the ecosystem more stable in productivity and more resistant towards disturbances such as alien species invasions and floods. If an ecosystem is rich in biodiversity, then the ecological balance would not get affected.

Various tropic levels are connected through food chains. If any one organism or all organisms of any one trophic level is illed, then it will disrupt the entire food chain. For example, in a food chain, if all plants are killed, then all deer’s will die due to the lack of food. If all deer’s are dead, soon the tigers will also die.

Therefore, it can be concluded that if an ecosystem is rich in species, then there will be other food alternatives at each trophic level which would not allow any organism to die due to the absence of their food resource. Hence, biodiversity plays an important role in maintaining the health and ecological balance of an ecosystem.

Question 10.
What are sacred groves? What is their role in conservation?
Answer:
Sacred groves are forest patches for worship in several parts of India. All the trees and wildlife in them are again treated and given total protection. They are found in khasiandjointia hills in Meghalaya. Western Ghat regions of Karnataka and Maharashtra, etc. Tribals do not allow anyone to cut even a single branch of tree in these sacred groves thus, sacred groves have been free form all types of exploitations.

Question 11.
Among the ecosystem services are control of floods and soil erosion, how is this achieved by the biotic components of the ecosystem?
Answer:
Control of soil erosion:
Plant roots hold the soil particles tightly and do not allow the top soil to be drifted away be winds or moving water. Plants increase the porosity and fertility of the soil.

Control of floods:
It is carried out by retaining water and preventing run off rain water. Litter and humus of plants function as sponges thus, retaining the water which percolates down and get stored as underground water. Hence, the flood is controlled.

Question 12.
Give three hypothesis for explaining why tropics show greatest levels of species richness.
Answer:
There are three different hypothesis proposed by scientists for explaining species richness in the tropics:

1. Tropical latitudes receive more solar energy than temperate regions, which leads to high productivity and high species diversity.
2. Tropical regions have less seasonal variations and have a more or less constant environment. This promotes the niche specialization and thus, high species richness.
3. Temperate regions were subjected to glaciations during the ice age, while tropical regions remained undisturbed which led to an increase in the species diversity in this region.

Biodiversity and Conservation Long Answer Type Questions

Question 1.
What is the significance of the slope of regression in a species – area relationship?
Answer:
The slope regression (z) has a great significance in order to find a species – area relationship. It has been found that in smaller areas (where the species-area relationship is analyzed), the value of slopes of regression is similar regardless of the taxonomic group or the region. However, when a similar analysis done in larger areas, then the slope of regression is much steeper.

Question 2.
What are the major causes of species loss in a geographical regions?
Answer:
The following are the major causes for the loss of biodiversity around the world:

1. Habitat loss and fragmentation:
Habitats of various organisms are altered or destroyed by uncontrolled and unsustainable human activities such as deforestation, slash and burn agriculture, mining, and urbanization. This results in the breaking up of the habitat into small pieces, which effects the movement of migratory animals and also, decreases the genetic exchange between populations leading to a declination of species.

2. Over – exploitation:
Due to over – hunting and over – exploitation of various plants and animals by humans, many species have become endangered of extinct (such as; the tiger and the passenger pigeon).

3. Alien species invasions:
Accidental or intentional introduction of non-native species into a habitat has also led to the declination or extinction of indigenous species. For example, the Nile perch introduced in Lake Victoria in Kenya led to the extinction of more than two hundred species of native fish in the lake.

4. Co – extinction:
In a native habitat, one species is connected to the other in an intricate network. The extinction of one species causes the extinction of other species, which is associated with it in an obligatory way. For example, the extinction of the host will cause the extinction of its parasites.

Question 3.
What do you understand by Threatened species? Explain its types.
Answer:
Species which have been greatly reduced in their number or whose natural habitats have been, disturbed due to which these are near extinction and may become extinct if the causative factors continue are called threatened species. It is estimated that about 25,000 plant species and 1,000 vertebrate species and subspecies and many invertebrate species are threatened with extinction. It is believed that at least 10% of the living species are in danger.

The organisms which are near extinction are of following types:

1. Endangered (E) species:
The species which are facing danger of extinction and whose survival is unlikely if the causal factors continue to operate. These are the species whose number have been reduced to a critical level or whose habitats have been so drastically reduced that they are deemed Lto be in immediate danger of extinction. For example, Indian rhinoceros, Asiatic lion and the great Indian bustard, snow leopard etc.

2. Vulnerable (V) species:
These are the species having sufficient number of individuals in their natural habitats. However, in the near future, they might represent the category of endangered species if unfavourable factors in the environment continue to operate, e.g., Musk deer, black buck, golden langur, etc.

3. Rare (R) species:
These are species with small population in the world. At present these are not endangered and vulnerable but are at risk. These species are usually localize within geographical areas or habitats or are thinly scattered over a more extensive range, e.g., Indian elephant, Asiatic wild ass, gharial, wild yak etc.

4. Threatened (T) species:
The species which do not fall under the endangered or vulnerable categories but indications are available that such species may come under any of these two categories if appropriate measures are not taken to protect them.

Question 4.
What are National Parks? Explain any five National Parks found in India.
Answer:
National Park:
Natinal park is an area which is strictly reserved for the betterment of the wildlife and where activities like forestry, grazing or cultivation are not permitted.

Five national parks of India are:

1. Shivpuri national park – It is located at Shivpuri near Gwalior (Madhya Pradesh). Wildlife present in this park tiger, cheetal, sambhar.
2. Guindy deer national park – It is located near Chennai (Madras) in Tamil Nadu. Wildlife found here ore Alvino deer, black buck, cheetal and famous snake park.
3. Betla national park – It is located in Palamu at Bihar. Wildlife found in this national park are elephant, tiger.
4. Dachigham national park – It is located at Srinagar in Jammu and Kashmir. Wildlife leopard, black beer, brown bear, musk deer, hangul, scrow, etc.
5. Bandhavgarh national park – It is located at Shahdol in Madhya Pradesh. Wildlife such as white tiger, panther, cheetal, bison, nilgai, barking deer, wild boar, etc. are found here.

Question 5.
Write in brief, the reasons necessary for conservation of wild species.
Answer:
Necessity for wildlife conservation : The conservation of wildlife is required for the following reasons :

1. To maintain balance in nature:
The wildlife helps us in maintaining the balance of nature. Once this equilibrium is disturbed it leads to many problems. The destruction of carnivores or insectivores often leads to an increase in the herbivores which in turn affects the forest vegetation or crop.

2. Economic value:
The wildlife can be used commercially to earn money example Animal products like hides, ivory, fur etc. are of tremendous economic value. The collection and supply of dead or living specimens of wildlife for museums and zoos fetches good amount of money. Wildlife can increase our earning of foreign exchange if tourism is promoted properly.

3. Scientific value:
The preservation of wildlife helps many naturalists and behaviour biologists to study morphology, anatomy, physiology, ecology and behaviour biology of the wild animals under their natural surroundings.

4. Recreational value:
The wildlife of any country provides best means of sports and recreation. Bird – watching is a hobby of many people all over the world. A visit to the parks and sanctuaries is an enjoyable proposition for children as well as for adults.

5. Cultural value:
The wildlife of India is our cultural asset and has deep rooted impact on Indian art, sculpture, literature and religion. Indus valley civilization shows the use of animals symbols in their seals.

6. Preservation of human race:
The destruction of wildlife in an area may eventually lead to the end of human civilization.

Question 6.
Describe the National and International efforts prescribed for the conservation of forests.
Answer:
The forest conservation is started in India on national level since British government. In 1856, Lord Dalhousi had formulated a policy for the conservation of forest in Burma. In 1894, Indian government also prepared a forest policy on national level. The main points of this policy are:

1. Forest management
2. Proper use of forest land
3. Polity for, protected forests
4. Improved forest production

Indian government established national parks, sanctuaries and zoological parks, The F.A.O. of United Nations is also functioning on forest conservation on international level. This organization also provides financial help for this purpose. In 1952, Indian government also prepared India’s New National Forest Policy under the direction of F.A.O. Forest policy has been planned for :

1. Prevention of deforestation of hill plants.
2. Reforestation of grazing land.
3. Development of grazing land.
4. Plantation of economically useful forest trees.
5. Increase in the profit of government from forests.

Question 7.
Write a short note on wild animals in India.
Answer:
India as a country has a diverse range of wildlife. India is home for many species of wild animals. More than 25% land are dense forest in India and around 400 national parks. Some of the most important and popular wild animals in India are as follows:

(A)
Animals : In Indian forest, below mentioned animals are found:

1. Deer – Its many species are found in India example Musdeer, Sambhar deer, chital, etc.

2. Antelope – These are same as deer example Nilgai, Barasingha (Swamp deer), four homed antelope (Chousingha) etc.

3. Elephant – Elephants are large mammals of the family elephantidae. It is found in Kerela and North India.

4. Rhinoceros – It is found in Himalaya region and in the forest of Bengal and Assam. Humans are the biggest threat to the Indian rhinoceros as the have been hunted to the brink of extinction for their horns.

5. Wild Ass – Wild asses are not found in any part of the world. Now in India, it is found in the little Rann of Kutch in the Gujarat state of India.

6. Carnivorous animals – Some Indian wild carnivores are:

• Indian lion (Asiatic lion) : Now it is confined to forests in the fall.
• Cheetah : It is on the verge of extinction.
• Lion : Lion is a national animal, at present its population in our country are more than 3,000.
• Leopard : It is similar to cheetah but smaller than cheetah.

(B)
Birds:
Peacock, wild fowl, many types of duck, stork, pigeons, partridge, quail, vulture, kite, piquant, owl, indian paradise flycatcher (dudhraj) are found in forests of our country.

(C)
Reptiles:
Crocodiles, alligators, tortoise, lizards, snakes and other reptilians are found in Indian forest. Many vertebrates and invertebrates are also found in Indian forest.

Question 8.
What are the main rules of Indian forest act?
Answer:
The Indian Forest Act, 1927 was largely based on previous Indian forest acts implemented under the British. Things which are included in this act are as follows:

1. Forest arrangement – Due to this Act, give the protection and arrangement to forest and Midlife.
2. Appropriate use of forest land – Uses of extra land for forest animals and grow some plant these are useful in wild animals.
3. Act for protection of forest – This act, stop passion of deforestation and must be conservation of forest and wild animals.
4. Increasing of forest product – Due to this act try to increasing of forest product and discovered new information which are better for wildlife.

Question 9.
Write an essay on measures of forest conservation.
Answer:
Forest conservation:
Forest conservation and management are essential to maintain the forests in their natural state and also to prevent the depletion of wildlife and forest wealth. For the success of conservation it is necessary to know the cause of depletion and destruction of forests. Forests are generally destroyed by fire, improper cutting of trees and by animals.

Essentiality of forest conservation:
Forest is a complex system which is responsible for the ecological balance in nature. Deforestation causing natural imbalance and affects the biotic components of the environment resulting floods, drought, epidemics, environmental pollution. Many ecologically important species of plants and animals are lost due to which economically important substance like wood, medicines, resin, lac and various food materials will not be available for us.

Measures of forest conservation : The following measures or efforts are prescribed for reforestation:

1. Establishment of conserved forests and their conservation in proper way.
2. Reforestation on deforested land. Old and damaged plants would be replaced by new plants.
3. Proper management of forests.
4. By promoting public awareness about forests.
5. Replacement of burnt off areas of the forests.
6. Plantation of trees that increase forest productivity.
7. Forestation of plants on hills and wastelands and prevention of grazing by cattle.
8. Prevent forests from fire, diseases and insects.
9. Providing basic protection for all forests by law.
10. Regulating human activity in the forest such as grazing by cattle and collection of firewood and fodder etc.
11. Provide special attention for the conservation of endangered plant and animal species under the inspection of specialists.
12. Removal of undesirable trees and vegetation for the better growth of desirable species.
13. Forestation of industrially useful plants.
14. The government will arrange the management of useful forests.

MP Board Class 12th Biology Important Questions

MP Board Class 12th Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance

Electrostatic Potential and Capacitance Important Questions Objective Type Questions

Question 1.
Choose the correct answer of the following:

Question 1.
The SI unit of electrical capacitance:
(a) Stat farad
(b) Farad
(c) Coulomb
(d) Stat coulomb.
Answer:
(b) Farad

Question 2.
The potential difference between the plates of a capacitor is constant. A dielectric medium is filled instead of air in between the plates. The intensity of electric field will:
(a) Decrease
(b) Remains unchanged
(c) Become zero
(d) Increase.
Answer:
(b) Remains unchanged

Question 3.
On replacing the air by an insulating material between the plates of a capacitor its capacity:
(a) Remains unchanged
(b) Increases
(c) Decreases
(d) Nothing can be said.
Answer:
(b) Increases

Question 4.
On increasing the separation between the plates of a parallel plate capacitor its capacitance :
(a) Remains unchanged
(b) Increases
(c) Decreases
(d) Nothing can be said.
Answer:
(c) Decreases

Question 5.
When two capacitors are joined in series each capacitor will have the same :
(a) Charge
(b) Potential
(c) Charge and potential
(d) Neither charge nor potential.
Answer:
(a) Charge

Question 6.
When two capacitors are joined in parallel each capacitor will have the same:
(a) Charge and potential
(b) Only charge
(c) Only potential
(d) Neither charge nor potential.
Answer:
(c) Only potential

Question 7.
Two capacitors of equal capacitance first connected in parallel then connected in series. What is the ratio of their capacities in both the cases:
(a) 2 : 1
(b) 1 : 2
(c) 4 : 1
(d) 1 : 4.
Answer:
(c) 4 : 1

Question 8.
The formula of capacitance of a spherical conductor is:
(a) C = \(\frac { 1 }{ 4π{ £ }_{ 0 }R } \)
(b) C = 4πt£₀R
(c) C = 4πr£₀R²
(d) C = 4π£₀R³
Answer:
(b) C = 4πt£₀R

Question 2.
Fill in the blanks:

1. 1 farad = one coulomb/ ……………
2. 1 farad = …………… stat farad.
3. Dimensional formula of capacitance is ……………
4. is a device in which with or out changing in shape or size of a conductor its capacitance can be increased ……………
5. On increasing the distance between the plater of a parallel plate capacitor its capacity ……………
6. Three capacitor each of 3pF are joined in series their equivalent capacitance will be ……………
7. The dimensional formula of electric potential is ……………
8. The potential due to a point charge q at a distance r is given as ……………
9. The potential difference = Intensity of electric field × ……………
10. The increase in kinetic energy of a charge q when it is accelerated by a potential difference V is ……………
11. Due to presence of dielectric medium the potential ……………
12. The work done in moving a charge perpendicular to the electric field is ……………
13. The potential of earth is considered to be ……………

Answers:

1. 1 Volt
2. 9 × 10¹¹
3. [M^-1L^-2T⁴A²]
4. Capacitor
5.  Decreases
6. lµF
7. [ML²T^-3 A^-11]
8. V = \(\frac { 1 }{ 4π{ £ }_{ 0 }R } \) \(\frac { q }{ r}\)
9. Distance between the two point
10. qV
11. Decreases
12. Zero
13. Zero.

Question 3.
Match the Column:

Answers:

1. (c) Q/V
2. (d) \(\frac { 1}{2}\) C/V²
3. (e) 4π£₀R
4. (a) £₀A/d
5. (b) 4π£₀ab / (b – a)

Question 4.
Write the answer in one word / sentence:

1. What is the potential of earth. Write SI units ?
2. What will be the electric field intensity inside a hallow sphere ?
3. In which direction of electric dipole, electric potential is zero ?
4. What is the net charge of a charge condenser ?
5. What quantity remains constant when the condenser are connected in series ?
6. What quantity remain constant when the conductor are connected in parallel ?

Answers:

1. Zero, volt
2. Zero
3. Broad-side-on position
4. Zero
5. Charge
6. Potential difference.

Electrostatic Potential and Capacitance Important Questions Very Short Answer Type Questions

Question 1.
What do you understand by equal potential surface ?
Answer:
The surface of the conductor where potential is in every point is called equal potential surface.

Question 2.
Write the name of the physical quantity whose SI unit in J/C. Is it a scalar or vector ?
Answer:
Electric potential, it is a vector quantity.

Question 3.
Draw a equi potential surface for a unit charge.
Answer:

Question 4.
Define farad.
Answer:
If the potential of a conductor increases by one volt when one coulomb of charges is given to it, then the capacity of the conductor is said to be one farad.

Question 5.
On going in direction of electric lines of force, electric potential decreases or increases.
Answer:
Electric potential decreases.

Question 6.
Give an example in which electric field is non-zero but potential is zero.
Answer:
At broad-side-on position of an electric dipole electric field is non-zero and potential is zero.

Question 7.
Does electron try to go toward high potential area or low potential area ?
Answer:
Since electron is negatively charge so it tries to go toward high potential area.

Question 8.
Potential between two parallel surface are same. The distance between them is R. If a charge q is bought from one surface to another, then what will be the work I done to do this ?
Answer:
Amount of work done will be zero on both the surface are equipotential.

Question 9.
If area of a plate of a parallel plate condenser in made half. Will it behave as condenser.
Answer:
When area of the plate if a parallel plate condenser is made half. Its capacity become half. Therefore it will not act as condenser.

Question 10.
A capacitor of capacity C is charged with potential difference V. What will be the magnitude of electric flux passing through the surface of it ?
Answer:
Zero.

Question 11.
Why condenser are used in computer’s ?
Answer:
Condenser are used as memory chip in computer.

Question 12.
Write one use of capacitor ?
Answer:
To accumulate electric charge.

Electrostatic Potential and Capacitance Important Questions Short Answer Type Questions

Question 1.
What is potential ? Is it a vector or scalar quantity ?
Answer:
Work done in bringing a unit positive charge from infinity to a point in the I electric field is called potential at that point. If charge q is brought from infinity to a point and IT work is done.
∴ V = \(\frac {W}{p}\)
It is a scalar quantity.

Question 2.
Can same amount of charge be given 1 and a solid sphere of same radii, if they have same potential ?
Answer:
No, because capacities of both spheres of same radii are always equal. Therefore i both the spheres can hold same amount of charge at same potential.

Question 3.
What is meant by capacity of a conductor ? Give its unit.
Answer:
The capacity of a conductor is defined by the charge given to the conductor, which increases its potential through unity.
Capacity = \(\frac {Charge}{Potential}\)
or C = \(\frac {q}{v}\)
Its SI unit is farad.

Question 4.
The surface of any conductor is always equipotential. Why ?
Or
The potential at every point on a charged conductor is same. Why ?
Answer:
All the points of the surface of a conductor are in electrical contact with one another. If the potential is not equal then the charges will flow from higher potential to lower potential till the potential of both the points on the surface becomes same. This will give rise to electrodynamics situations. Thus, the surface of a conductor is always equipotential.

Question 5.
What would be the work done if a point charge +q is taken from a point A to point B on the circumference of a circle with another point charge +q at the center:
Answer:
The points A and B are at same distance from the charge + q at the center, so V_A = V_B So, work done, W= q₀ (V_A – V_B) = 0.

Question 6.
Explain the meaning of capacity of a capacitor. What will be the effect on capacity of a parallel plate capacitor if a dielectric medium of dielectric constant k is filled in between the plates ?
Answer:
The capacity of a capacitor, is equal to charge given to one of its plates which produces unit potential difference across the plates. In this case capacity increases, it becomes k times its initial value.

Question 7.
What will be the change in the value of charge and potential difference between the plates of a parallel plate capacitor, if after charging its battery is removed and distance between its plates is reduced ?
Answer:
Charge remains same but potential difference decreases.

Question 8.
Two equipotential surface does not intersect each other, why ?
Answer:
Electric lines of forces are always perpendicular to equipotential surface. If two equipotential surface intersect each other then at the point of intersection there will be two direction of electric fields which is impossible. Therefore they does not intersect each other.

Question 9.
Why must electrostatic field be normal to the surface at every point of a charged conductor ?
Answer:
If electric field is randomly directed, then it can be resolved, into two components. The horizontal component on this surface is E sin θ.
For electrostatic situation
£ sin θ = 0
⇒ sin θ = 0
⇒ θ = 0°

So, the electric field is normal to surface.

Question 10.
The potential at any point inside the hollow conductor remains same. Why ?
Answer:
When charge is given to a hollow conductor then the distribution of charge takes place on its upper surface. Therefore the intensity of electric field inside the conductor is zero. Hence, no work is done in moving unit positive charge inside it. Therefore potential at every point inside the conductor remains same.

Question 11.
Can the potential be zero where electric field is not zero ?
Answer:
Yes, the electric field on the equatorial line of a dipole is not zero but potential is zero.

Question 12.
What will be the effect on electric field, potential, difference, electric capacity and energy if a dielectric of dielectric constant K is filled between the plates of a capacitor ?
Answer:
The electric field will become \(\frac {1 }{ K }\)times, potential difference will become \(\frac {1 }{ K }\) times, electric field will become K time and energy will become \(\frac {1 }{ K }\) times.

Question 13.
Can 1 coulomb charge be given to a sphere of radius 1cm ?
Answer:
As we know that the formula of potential ¡s V = \(\frac { 1 }{ 4π { £ }_{ 0 } } \) \(\frac { q }{ r }\) …(1)
Given,q = lC, r = lcm = 10^-2m
Putting these values in eqn. (1)
V = 9 × 10⁹ × \(\frac{1}{10^{-2}}\) = 9 x 10¹¹volt
Where = \(\frac{1}{4 \pi \varepsilon_{0}}\) =  9 × 10⁹ in SI unit.
This value of potential is greater than barrier potential of air. Therefore IC charge cannot be given to a sphere of radius 1cm.

Question 14.
In the shown figure what will be the work done to bring a z point charge from the point X to Y to Z ?
Answer:
There the point Z and Y are situated on same equaipotential surface. Therefore work done to bring a point charge from A’to Zand from X to Z will be same.
.’. W_y= W_z.

Question 15.
Derive an expression for electric potential due to a point charge. Is it scalar or vector and why ?
Answer:
Consider a point charge q placed at origin O. Potential at P has to be found out. Let the medium between charge ‘q’ and P has dielectric constant E_r.

Electric field at P due to charge q is
E = \(\frac{1}{4 \pi \varepsilon_{0} \varepsilon_{r}} \cdot \frac{q}{r^{2}}\)
The electric field \(\vec{E}\) points away from the charge q. A force \(\vec{F}\) = -q₀ \(\vec{E}\) has to be applied on the charge so that it can be brought near to q. The small work required to move the test charge q₀ from P to Q through a small distance dr is given by dW = Fdr
= -q₀ Edr
The total work done in moving the charge q0 from infinity to point P will be obtained by integrating the above equation as –

But electric potential is defined as work done per unit test charge.

Potential at P is V = \(\frac{1}{4 \pi \varepsilon_{0} \varepsilon_{r}} \cdot \frac{q}{r}\)
If medium between q and q₀ is vacuum then £_r = 1
Then , V =\(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}\)
This is the required expression.

Question 16.
What are the factors affecting the potential of a charged
Answer:
The factors affecting the potential of a charged conductor are:
1. Amount of charge on conductor:
By the formula V = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}\) it is clear that
V ∝ q, hence more is the charge, more will be the potential of charged conductor.

2. Shape of conductor (Area of conductor):
If the charge is kept constant on a conductor and its surface area is decreased then the potential of conductor increases whereas on increasing the surface area its potential decreases. So the potential of a conductor is inversely proportional to the radius.

3. Presence of other conductor near the charged conductor:
If an uncharged conductor is brought near a charged conductor then the potential of the charged conductor decreases.

4. Medium surrounding the conductor:
Due to presence of insulating medium near the charged conductor its potential will decrease.

Question 17.
Define equipotential surface. Write its properties.
Answer:
Equipotential surface:
An equipotential surface is the locus of all those points at which the potential due to distribution of charge remains same.

Properties:

• Potentials on every point are equal
• No work is done in moving a positive charge from one point to another
• The electrical lines of force are normal to the equipotential surface
• Two equipotential surfaces do not intersect each other.
• All the points on the surface of a conductor are in electric contact. If the potentials are not same then the
• charge will flow from higher potential to lower potential till the potential of both the points become same.
• Thus the surface of a conductor is always equipotential.

Question 18.
Obtain a relation between electric Held intensity and potential difference.
Or
Prove that E = \(\frac { dv }{ dr }\)where symbols have their usual meanings.
Answer:
Suppose A and B are two points in the electric field of charge q. The direction of electric field is radially outwards from A to B. Suppose the distance between A and B is very small (i. e., dr) then the electrie field between A and B can be taken as uniform. As the potential is inversely proportional to distance hence potential at A is more than that of B. Let the potential at B is V then that at A is V + dV.

Work done in bringing the test charge q0 from B to A is –
dW = q₀dV …….(1)
Force acting on q₀ will be
F = q₀ E
Work done in bringing the test charge against the repulsion force will be
dW = -q₀Edr
(Work = Force x Displacement) ………(2)
The negative sign shows that the direction of displacement and direction of force are opposite to each other.
From eqns. (1) and (2), we get,
q₀dV = q₀Edr
or dV = -Edr
E = – \(\frac { dv }{ dr }\)
This is required relation between intensity of electric field and potential difference.

Question 19.
Prove that capacity of an isolated spherical conductor is directly proportional to its radius.
Or
Derive an expression for the capacity of a spherical conductor.
Answer:
The capacity of a conductor is its ability to store electrical energy and it is equal
to that charge which increases its potential by unity.
∴ Capacity = \(\frac { Charge }{ Potential }\)

Capacity of an isolated spherical conductor:
Let us consider about a spherical conductor of radius r. The charge + Q is given to it. The charge will be distributed on its surface uni- formly. Therefore the lines of force will be emitted normally to the surface seem to becoming from its center. Hence, we can suppose that all the charges are kept at the centre.
∴ Potential on the surface V = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r}\)
But capacity C = \(\frac { Q }{ V }\)
Putting the value of V, we get
C = \(\frac{Q}{\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r}}\)
C = 4πE₀r
C ∝ r.
Thus, the capacity is proportional to the radius of the spherical conductor.

Question 20.
What do you mean by a capacitor ? Explain its principle.
Answer:
The capacitor is a device by which the capacitance of a conductor is increased without changing its size or volume. Actually it stores electrical energy.
Principle of capacitor:
Let A be a charged conducting plate. Another uncharged conductor plate B is brought near to A, therefore due to induction negative charges will be induced on the front surface and positive charges on the other side of plate B.
Now, the negative charge reduces the potential while the positive charge increases.

As the negative charge is nearer therefore the potential of plate A decreases. Now, the plate B is earthed then the free positive charge will go to earth and hence the potential of A decreases by more value.
C = \(\frac { Q }{V}\)
As V decreases, C will increase. This arrangement is called capacitor or condenser.

Question 21.
Derive an expression for parallel plate capacitor.
Answer;
Let A and B be two plates of a parallel plate capacitor separated by a distance d apart. Area of each plate is A and dielectric constant of the medium between them is E_r Now, plate A is given + Q charge. Therefore, – Q charge will be induced on the nearer surface of the plate B and + Q charge on the other side. As B is connected to earth, + Q charge of B will go to earth. Let the charge density of A is cr, therefore that of B will be -σ.

Now, σ = \(\frac { Q }{A}\)
Intensity between the plates will be given by
E = \(\frac{\sigma}{\varepsilon_{0} \varepsilon_{\mathrm{r}}}\)
E = \(\frac{Q}{A \varepsilon_{0} \varepsilon_{r}}\)
But, potential difference between the plates A and B is
V = Electric field intensity ×
Distance between to plates = Ed
V = \(\frac{Q}{A \varepsilon_{0} \varepsilon_{s}} \cdot d\)
But, C = \(\frac{Q}{V}=\frac{Q}{\frac{Q d}{A \varepsilon_{0} \varepsilon_{r}}}\)
C = \(\frac{\varepsilon_{r} \varepsilon_{0} A}{d}\)
This is the required relation.
For air or vacuum, E_r = 1
C = \(\frac{\varepsilon_{0} A}{d}\)

Question 22.
Three capacitors of capacitance’s C₁ C₂ and C₃ are connected in series. Derive an expression for the equivalent capacitance.
Answer:
The given figure shows three capacitors of capacitances C₁ C₂ and C₃ con – nected in series. A potential difference of V is applied across the combination, charges of + Q and – Q are developed on the plates of the capacitor.
Potential difference across the individual capacitors will be

V₁ = \(\frac{Q}{C_{1}}\) , V₂ = \(\frac{Q}{C_{2}}\), V₃ = \(\frac{Q}{C_{3}}\) …….(1)
The sum of these must be equal to the applied potential difference V.
V = V₁ + V₂ + V₃ ………(2)
Let C be the equivalent capacitance of the series combination
∴ C = \(\frac { Q }{ V }\) or V = \(\frac { Q }{ C }\) ………..(3)
V₁ + V₂ + V₃ = \(\frac { Q }{ C }\) [from equ..(2)]
\(\frac{Q}{C_{1}}\) + \(\frac{Q}{C_{2}}\) + \(\frac{Q}{C_{3}}\) = \(\frac{Q}{C}\) [from equ..(1)]
Q(\(\frac{1}{C_{1}}\) + \(\frac{1}{C_{2}}\) + \(\frac{1}{C_{3}}\)) = \(\frac{Q}{C}\)
\(\frac { 1 }{ c }\) = \(\frac{1}{C_{1}}\) + \(\frac{1}{C_{2}}\) + \(\frac{1}{C_{3}}\)
This is the required expression.

Question 23.
Three capacitors of capacitance’s C₁ C₂ and C₃ are connected in parallel. Derive an expression for the equivalent capacitance C.
Answer:
Consider three capacitor of capacitance’s C₁ C₂ and C₃ connected in parallel. A potential difference V is applied across the combination. Charges set up in the individual capacitor will be.

Q₁ = C₁V, Q₂ = C₂V,Q₃ = C₃V …(1)
Total charge stored in the parallel combination is
Q = Q₁ + Q₂ + Q₃ ……….(2)
If C is the equivalent capacitance of the combination
Then, C = \(\frac { Q }{ V }\) Q = CV …………(3)
Q₁ + Q₂ + Q₃ = CV [from eq. (2)]
C₁V + C₂V + C₃V = CV [from eq. (1)]
V(C₁ + C₂ + C₃) = CV
C = C₁ + C₂ + C₃
This is the required expression.

Question 24.
Derive an expression for the energy of a charged conductor.
Or
Prove that energy of a charge conductor is directly proportional to its square of potential.
Answer:
The work done in charging a conductor is stored as energy in it. This energy is called electrostatic potential of conductor.

Formula derivation:
Let us consider about a conductor of capacity C which is given charge +Q due to which its potential becomes V. As the charge increases work done also increases. Let at any instant the potential of conductor be V due to charge q.
∴ C = \(\frac { q }{ v}\)
or v = \(\frac { q }{ C}\)
Now, at potential Kthe work done in giving the charge dq will be dW
∴ dw = Vdq
or dw = \(\frac { q }{ C}\)dq
Work done in charging the conductor from 0 to Q

This work done is stored as potential energy on the conductor. Energy of a charge conductor
U = \(\frac{1}{2} \frac{Q^{2}}{C}\)
But Q = CV
∴ U = \(\frac{1}{2} \frac{C^{2} V^{2}}{C}\)
or U = \(\frac{1}{2} C V^{2}\)
∴ U ∝ V² because C is constant

Question 25.
Prove that on connecting two charged conductors, charges distribute on them according to their capacities.
Answer:
When two isolated charged conductors A and B are connected by a thin wire, charge flows from the conductors at high potential to the conductor at low potential till the potential of both A and B became equal. The phenomenon involved is called distribution of charges and the total charge of the entire system remains conserved. Let the capacitance of A and B be C₁ and C₂, the charges be Q₁ and Q₂ respectively. Then the potentials are V₁ and V₂ respectively.
∴ Initially, Q₁ = C₁V₁ and Q₂ = C₂V₂

The conductors are joined by a wire of negligible capacitance, the charges flow from- the conductor at higher potential to the conductor at lower potential till the potentials on each conductor become equal.
The net charge on the system,
Q = Q₁ + Q₂
The common potential, V = \(\frac { Total charge }{ Total capacitance }\)

After the potential becomes equal let the charge on A ₁ be Q₁ and charge on A₂ be Q₂.

Dividing eqn. (2) by eqn. (3), we get

When the conductors are joined, then the charges get distributed in the ratio of their capacities.

Question 26.
Obtain an expression for potential due to a group of point charges.
Or
Derive the expression for potential energy.
Answer:
Consider a group of point charges q₁,q₂,q₃……..q_n which are situated at a dis-tance of r₁, r₂, r₃…….. _nn respectively from the point P. The potential due to these point charges is to be obtained at P. Now potential at P due to q₁ is

potential due to q₂,q₃, ……… q_netc

Total potential at P will be V = V₁ + V₂ + V₃ + ……….. + V_n

This is the required expression.

Electrostatic Potential and Capacitance Important Questions Long Answer Type Questions

Question 1.
Derive the expression for the capacity of a parallel plate capacitor, when the medium between the plates is partially filled by a dielectric medium.
Answer:
Let A and B are parallel plates of a capacitor. The distance between the plates is d and plate of thickness t and dielectric constant Er is introduced.

Now, plate A is given charge +Q.
Let the charge density be σ.
∴ σ = \(\frac { Q}{ A}\)
Intensity of field in air ,
E_{0}=\(\frac{\sigma}{\varepsilon_{0}}=\frac{Q}{\varepsilon_{0} A}\)
If the intensity of field inside the dielectric medium be E, then
Dielectric constant = \(\frac {Electric field in vacuum}{ Electric field in medium}\)
or \(\varepsilon_{r}=\frac{E_{0}}{E}\)
or E = \(\frac{E_{0}}{\varepsilon_{r}}=\frac{Q}{\varepsilon_{0} \varepsilon_{r} A}\)
Now, potential difference between A and B,
V=E₀ (d – t) + Et, [(d – t) is vacuum distance]

This is the required expression.
Metal is a conductor. When metal is used in place of the dielectric, it will conduct electricity and the potential difference will become zero. So, capacitor will not work.

Question 2.
Calculate the loss of energy, when two charged conductors are connected.
Or
The capacities of two conductors are C₁ and C₂, Q₁ and Q₂ charges are given to them so that their potentials become V₁ and V₂ respectively. If they are connected by a wire, then calculate the following:

• Common potential
• Loss of energy.

or
prove that when two charged conductors are connected, there will be a loss of energy
Or
In redistribution of charges, is there a loss of energy ? Deduce an expression to confirm the answer.
Answer:
Let A and B be two conductors of capacities C₁ and C₂ respectively. When charges Q₁ and Q₂ are given separately the potentials become V₁ and V₂ respectively.Total charges, Q = Q₁ + Q₂ ………..(1)
But, Q₁ = C₁V₁ and Q₂ = C₂V₂
By eqn. (1), we get
Q = C₁V₁ + C₂V₂
Total capacity, C = C₁ + C₂

(1) Common potential:
Let the conductors are connected by a wire and the common potential becomes-V.
Q₁ + Q₂ = (C₁ + C₂)V
V = \(\frac{Q_{1}+Q_{2}}{C_{1} + C_{2}}\)
V = \(\frac{C_{1} V_{1}+C_{2} V_{2}}{C_{1}+C_{2}}\) ……..(2)
This is the expression for the common potential.

(2) Loss of energy: Total energy of the conductors before connection:

and total energy after connection,

Putting the value of V from eqn. (2) in eqn. (4), we get

Hence, difference of energy’,

(V₁ – V₂)² is positive, hence (V₁ – V₂) is positive. Hence, during redistribution, there will be always loss of energy.
i.e., U₁ – U₂>0 ⇒ U₁>U₂
i.e., energy before joining is greater than energy after joining.
The loss in energy,

Question 3.
Explain the construction and working of Van de Graaff generator. Write its uses.
Answer:
Van de Graaff generator is a machine which produces electricity of about 107 V or more potential difference.
Construction:
It consists of a large metallic sphere S of diameter 5 m, mounted on high insulating support PP about 15 m high. An endless insulating belt made up of rubber passes over the pulleys p₁ and P₂. A motor rotates p₁ C₁ and C₂ are two metallic combs called spray comb and collecting comb respectively. C₁ is connected to S. To prevent the leakage of charge, the generator is put inside a large enclosure filled with gas at 15 atm. pressure. This iron enclosure is connected to earth.

Working:
The comb C₁ is connected to the positive terminal of the battery, therefore the surface density of the points becomes very high which causes the wind present nearby it to get charged. Thus, the spray comb sprays the charge on the belt. Now, the electric wind moves up to the collector comb C₂ When it reaches in front of the collector comb C₂ opposite charge induces on the tip to neutralize the same type of charge. The negative charge wind of C₂, cancels the positive charge of the belt. Thus, by the repeated actions more and more positive charge is induced on sphere, hence its potential increases to about 10⁷volts or more.

Uses:

• To generate high potential.
• To accelerate the positive particles such as protons, Deuteronomy, are particle etc. and used in nuclear disintegration.

Question 4.
When Anil opened the cap of the tap, then he found no water in coming out of it. Then he opened the cap of the water tank and found no water in the tank. To fill up water in the water tank he switch on the switch of the motor and found motor is not starting. Then he called the electric technician. The technician said him on checking that the condenser of the motor is not functioning. On replacing capacitor, the motor start working.

Answer the following questions:

1. What values does Anil exhibits ?
2. What is the function of condenser ?
3. What is total charge on a charged condenser ?
4. The capacity of a capacitor is 3pF. If it is charged up to 100 V potential difference, then what will be charged stored in it ?

Answer:

1. Anil exhibited his presence of mind.
2. It accumulate charge and hence it conserved energy.
3. Net charge on a condenser is zero.
4. C = 3µF = 3 × l0^-6 F, V = 100V

∴ By formula Q = CV = 3 × l0^-6 × l00
or Q = 3 × 10^-4C.

Electrostatic Potential and Capacitance Important Questions Numerical Questions

Question 1.
Can 1 coulomb charge be given to a sphere of radius 1cm ?
Answer:
No.
As we know that the formula of potential is V = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}\) …(1)
Given, q = 1C, r = 1cm = 10^-2m
Putting these values in eqn. (1)
V = 9 × 10⁹ x \(\frac{1}{10^{-2}}\) = 9 × 10¹¹volt
Where = \(\frac{1}{4 \pi \varepsilon_{0}}\) = 9 × 10⁹ in SI unit.
This value of potential is greater than barrier potential of air. Therefore lC charge cannot be given to a sphere of radius 1cm.

Question 2.
You are given three capacitor of 4pF each. How they will be combined to obtain resultant capacity of 6pF ?
Solution:
Given : Q = C₂ = C₃ = 4µF
When two capacitor is joined in series and third capacitor joined parallel with them, then resultant capacity is obtained as 6µF.

C₁ ,C₂ is in series, its resultant (C’) is
\(\frac { 1 }{ c}\) = \(\frac{1}{C_{1}}+\frac{1}{C_{2}}\)
or \(\frac { 1 }{ c}\) = \(\frac { 1 }{ 4}\) + \(\frac { 1 }{ 4}\) = \(\frac { 2 }{ 4}\)
or C = \(\frac { 4 }{ 2}\) = 2µF.
C and C₃ is in parallel combination,
Its resultant C is C = C + C₃
C = 2 + 4 or C = 6µF.

Question 3.
A hollow metallic sphere of radius 0-1 m is given 6pC. Calculate its potential:

1. At the surface of sphere
2. At the center.

Solution
Given, r = 01 m, q = 6µC = 6 x 10^-6 C
Formula: V = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}\)

1. Potential at the surface:
V = 9 × 10⁹ × \(\frac{6 \times 10^{-6}}{0 \cdot 1}\)
V = 54 × 10⁴
V- 5.4 × 10⁵ volt.

2. At the center:
Inside the sphere the potential remains same and equal to that on the surface hence V = 5 . 4 x 10⁵volt.

Question 4.
A test charge is moved from A to B, B to C and A to C in an electric field E as shown in the figure :

Find (1) Potential difference between A and C
(2) At which point electric potential will be high and why ?
Solution:
1. In right angled ∆ABC
AB² = AC² – BC² = 5² – 3²
∴ AB = 4 = dr
BC is perpendicular to electric field, therefore potential will be same at B and C.
V_A – V_C = V_A – V_B = -Edr = -4E
2. Therefore potential at point C will be more than potential of point A.

Question 5.
Identical water droplets, having equal charge on each are combined to form a big drop. Compare the capacity of bigger drop with that of a small drop.
Solution:
Let radius of small droplet = r
Radius of the big drop = R
Volume of big drop = Volume of 27 droplets
\(\frac { 4 }{ 3 }\) πR³ = 27 × \(\frac { 4 }{ 3 }\) πR³
or R³ = 27r³
or R³ = (3r)³
or R = 3r
or \(\frac { R }{ r }\) = \(\frac { 3}{ 1 }\)
Since, C ∝ radius
or \(\frac{C_{1}}{C_{2}}=\frac{r_{1}}{r_{2}}\)
or \(\frac{C_{1}}{C_{2}}\) = \(\frac { 3}{ 1}\) = 3
or C₁ = 3C₂
The capacity of bigger drop is three times that of smaller one.

Question 6.
How three capacitor of 3pF each can be combined such that their resultant capacity is :

1. 9pF,
2. 4.5pF.

Solution:
1. When the three capacitor is joined in parallel, then
C = C₁ + C₂ + C₃
= 3 + 3 + 3 = 9µF.

2. When two capacitor are joint in series, then resultant C’ is
\(\frac { 1 }{ c}\) = \(\frac{1}{C_{1}}+\frac{1}{C_{2}}\)
= \(\frac { 1 }{ 3}\) + \(\frac { 1 }{ 3}\) = \(\frac { 2 }{ 3}\)
C = \(\frac { 3 }{2}\) = 1.5 µF
Now C is joined in parallel with C₃
C = C + C₃ = 1.5 + 3 = 4.5µF.

Question 7.
The potential difference between two points is 10V. How much work is required to move a charge 100 pC from a point to the other ?
Solution:
Given, V= 10 volt, q = l00µC = l00 × l0^-6C
Formula : w = qV
= 100 × 10^-6 × 10
= 10^-3 joule.

Question 8.
Find the area of the plate of a 2F parallel plate capacitor, if the separation between the plates is 0.5 cm ?
Solution:
As C = \(\frac{\varepsilon_{0} A}{d}\)
A = \(\frac{C d}{\varepsilon_{0}}\)
Here, C = 2F, d= 0.5cm = 0.5 x 10^-2m
A = \(\frac{2 \times 0 \cdot 5 \times 10^{-2}}{8.85 \times 10^{-12}}\)
= 1.13 × 10⁹ m² = 1130Km²

Question 9.
Two charges 5 x 10^-8C and -3 x 10^-8C are located 16 cm apart At what point, on the line joining the two charges, is the electric potential zero ? Take the potential at infinity to be zero. (NCERT)
Solution:
Case I.
Let electric potential be zero at point C lying at distance x from the positive charge.

Given, q₁ = 5 × 10^-8 C;
q₂ = -3 × l0^-8 C
AC = x cm : CB = (16 – x) cm
Now, Potential at C is zero i. e.,
V₁ + V₂ = 0

-8x + 80 = 0
8x = 80
x = 19 cm
i.e., electric potential at a distance of 10 cm from positive charge will be zero.

Case II.
The other possibility is that the point C may also lie on produced AB.

Now, V₁ + V₂ = 0

5(x – 16) – 3x = 0
5x – 80 – 3x = 0
2x – 80 = 0
x = 40 cm from the positive charge

Question 10.
Determine the equivalent capacitance between A and B in the following circuits:

Solution:
(i). Mark the junctions as C and D.

(But C will be A and D will be B)
Draw the equivalent network, which is given below

Equivalent capacitance,
C = C₁ + C₂ + C₃
or C = 1 + 1 + 1 = 3µF

(ii). To move from A to B, there are two paths P -1 and P – II.

(As A and B, the path is dislocated temporarily)
The capacitors in P -II are in series. So, the equivalent becomes

The resultant capacity of series combination is
\(\frac { 1 }{ C’ }\) = \(\frac { 1}{ 3 }\) + \(\frac { 1 }{ 3}\) + \(\frac { 1 }{3 }\)
= \(\frac { 3 }{ 3 }\) = 1µF = C’ = 1µF
The equivalent further becomes

Total capacity C = 3 +1 = 4µF.

MP Board Class 12th Physics Important Questions

MP Board Class 12th Biology Important Questions Chapter 14 Ecosystem

Ecosystem Important Questions

Ecosystem Objective Type Questions 

Question 1.
Choose the correct answer:

Question 1.
The flow of energy in ecosystem: (MP2011)
(a) Unidirection
(b) Bidirectional
(c) Tridirectional
(d) Four directional
Answer:
(a) Unidirection

Question 2.
Chief source of energy in ecosystem :
(a) Solar energy
(b) Green plants
(c) Food substances
(d) All of these.
Answer:
(a) Solar energy

Question 3.
The term ‘ecosystem’ was used for the first time by : (MP 2015)
(a) Tansley
(b) Odum
(c) Reiter
(d) Mishra and Puri.
Answer:
(a) Tansley

Question 4.
The pyramid of number of tree ecosystem is :
(a) Inverted
(b) Upright
(c) Both (a) and (b)
(d) None of these.
Answer:
(a) Inverted

Question 5.
Correct food chain is :
(a) Grass → Grasshopper → Frog → Snake → Hawk
(b) Grass → Frog → Snake → Peacock
(c) Grass → Peacock → Grasshopper → Hawk
(d) Grass → Snake → Rabbit.
Answer:
(a) Grass → Grasshopper → Frog → Snake → Hawk

Question 6.
Pyramid of biomass of lake ecosystem is :
(a) Upright
(b) Inverted
(c) Upright or inverted
(d) None of these.
Answer:
(b) Inverted

Question 7.
Vegetation present between two communities :
(a) Ecad
(b) Ecotype
(c) Ecotone
(d) None of these.
Answer:
(c) Ecotone

Question 8.
Xerosere is started from :
(a) Water
(b) Naked rock
(c) Swamp
(d) All of these.
(b) Naked rock

Question 9.
Serial development of plant community is called :
(a) Attack
(b) Succession
(c) Both (a) and (b)
(d) None of these.
Answer:
(b) Succession

Question 10.
Plants starting succession in any area are called :
(a) Pioneer
(b) Sere
(c) Displacer
(d) All of these.
Answer:
(a) Pioneer

Question 11.
In forest ecosystem, the pyramid of energy is : (MP 2012,17)
(a) Always inverted
(b) Always upright
(c) First upright then inverted
(d) None of these.
Answer:
(b) Always upright

Question 12.
The study of ecology of a species is called :
(a) Ecology
(b) Autecology
(c) Synecology
(d) None of these.
Answer:
(b) Autecology

Question 13.
Food chain starts from :
(a) Respiration
(b) Photosynthesis
(c) Decomposers
(d) Nitrogen fixation.
Answer:
(b) Photosynthesis

Question 14.
Man is:
(a) Autotrophic
(b) Carnivorous
(c) Herbivorous
(d) Omnivorous.
Answer:
(d) Omnivorous.

Question 15.
In any ecosystem, solar energy is conserved by :
(a) By producers
(b) By consumers
(c) By decomposers
(d) All of the above.
Answer:
(a) By producers

Question 16.
The two components of ecosystem are: (MP 2013)
(a) Organism and Plant
(b) Weeds and trees
(c) Frog and man
(d) Biotic and Abiotic.
Answer:
(d) Biotic and Abiotic.

Question 2.
Fill in the blanks:

1. The transitional zone present between to adjacent communities is called ………………….
2. All the plants of a particular area constitute …………………. of that place.
3. Only …………………. % energy is transferred from one trophic level to another.
4. Pyramid of …………………. is always upright. (MP 2013)
5.  …………………. and …………………. are the examples of omnivorous animals.
6. The term ‘ecosystem’ was proposed by ………………….
7. The chief source of energy ecosystems is (MP 2016)
8. N₂ fixing bacteria is called …………………. (MP 2009 Set A)
9. All ecosystems are depended for energy on …………………. (MP 2009 Set D)
10. In forest trees basically function as ………………….
11. Establishment of organisms in new habitat is called ………………….
12. A sereal development of plant community is called ………………….
13. Succession which takes place in baren rock is called ………………….
14. Succession in soil is called ………………….

Answer:

1. Ecotone
2. Flora
3. 10
4. Energy
5. Man, Pig
6. Tansley
7. Sunlight
8. Nitrifying bacteria
9. Sun (solar energy)
10. Producers
11. Ecesis
12. Succession
13. Lithosere
14. Psamosere.

Question 3.
Match the followings:
I.

Answer:

1. (e)
2. (d)
3. (a)
4. (c)
5. (b)

II.

Answer:

1. (b)
2. (c)
3. (d)
4. (a)

Question 4.
Write the answer in one word/sentances:

1. What is called the study of ecology of a species?
2. What is called the process of soil formation?
3. What type of energy pyramid found in nature?
4. Give one example of an omnivorous animal.
5. Who proposed the term ecosystem?
6. Who gave the 10% rule of ecosystem?
7. In an ecosystem, producer is known as.
8. What is the direction of the flow of energy in an ecosystem?
9. When many food chains operate simultaneously and interlock such patterns is termed as?
10. Give an example of autotrophic componant.
11. Which gas is used by green plants in photosynthesis?
12. Name the special nutrient which is present in metheonine amino acid.
13. Name the cycle in which nitrogen is converted into multiple chemicals and circulated among atmosphere.
14. Due to an ecosystem fungus and bacteria are known as.
15. What is T₂?

Answer:

1. Autecology
2. Pedogenesis
3. Upright
4. Man
5. A.G. Tansley
6. Lindman
7. Biotic componant
8. Leanear
9. Food web
10. Plants
11. CO₂
12. Sulfer
13. Nitrogen cycle
14. Decomposer
15. Herbivorous animal.

Ecosystem Very Short Answer Type Questions

Question 1.
Write the name and ratio of different components of biosphere.
Answer:
Name and ratio of different components of biosphere is:

Some gases are also found in biosphere example Helium, Neyon and Crypton are found in less quantity.

Question 2.
Differentiate between detritivore and decomposer.
Answer:
Detritivore are organisms which feed on detritus and break them into smaller particles, example earth worm. And decomposer are organisms which by secreting enzymes break down complex organic matter into in organic substance example some bacteria and fungi.

Question 3.
Explain consumers of ecosystem
Answer:

1. Producers – All the green plants.

2. Consumers – Depends on others for food.

• Primary consumer: Depends on plants called herbivores.
• Secondary consumers : Depends on herbivores for food.
• Tertiary consumers : Depends on secondary consumers.

3. Decomposers – They decomposed dead organic matter.

Question 4.
Differentiate between biome and ecosystem.
Answer:
An ecosystem is the interaction of living and non – living things in an environment. A biome is a specific geographic area notable for the species living there.

Question 5.
What are transducers according to some ecologists?
Answer:
Some ecologists call green plants as transducers of the ecosystem as they convert solar energy into chemical energy.

Question 6.
Name four submerged plants.
Answer:

1. Hydrilla
2. Vallisneria
3. Elodea
4. Potamogeton.

Question 7.
Name the stages of xerosere.
Answer:
Stages of xerosere:

1. Crustose lichen stage
2. Foliose lichen stage
3. Moss stage
4. Herb stage
5. Shrub stage
6. Climax forest stage.

Question 8.
When many food chain operate simultaneously and interlock such pattern is formed.
Answer:
Food web.

Question 9.
Name the ecosystem which shows most productivity.
Answer:
Tropical ecology.

Question 10.
What are fungi and bacteria called in an ecosystem?
Answer:
Micro – consumer or Decomposer.

Question 11.
Which part of the energy is transferred from one trophic level to other in ecosystem?
Answer:
10%.

Question 12.
Name the type of chemosynthetic bacteria.
Answer:
Autotrophic.

Question 13.
Name the word which is similar to ecosystem given by Prof. R. Mishra.
Answer:
Ecocosm.

Question 14.
Name the trophic level in which green plants are found.
Answer:
Primary trophic level.

Question 15.
Who gave the word transformer for producer?
Answer:
E.J. Kormondy.

Question 16.
Name any two sedimentary cycle.
Answer:

Phosphorus cycle
Sulphur cycle.

Question 17.
The energy pyramids are always.
Answer:
Upright.

Question 18.
Give examples of decomposers.
Answer:
Bacteria and Fungi.

Question 19.
Who gave 10% rule of energy?
Answer:
Lindeman.

Question 20.
Which form of nitrogen is absorbed by plants?
Answer:
In the form of nitrate ion (NO₃^–)

Ecosystem Short Answer Type Questions

Question 1.
Draw a pyramid of energy of grassland ecosystem.

Answer:

Question 2.
Explain nitrogen cycle in nature.

Answer:

Question 3.
Explain sulphur cycle by diagram.
Answer:

Question 4.
Explain the effect of light on plants.
Answer:
Effect of light on plants:
Light is the source of energy. It is essential for life. It is an important factor of an ecosystem. The existence of life on earth is because of light obtained from sun. Sunlight is essential for photosynthesis, help in preparation of food for the whole living world. Light effects biological activities of plants by its intensity, period and duration. Plants are classified into following two categories on the basis of requirement of light intensity:

1. Heliophytes – Plants, which can grow better in bright light are called heliophytes.
2. Sciophytes – Plants, which require relatively less of light and they can grow better in shades are called sciophytes.

Question 5.
Explain the meaning of food web and draw its diagram.
Answer:
Food Web:
In nature, foodchains are not isolated sequences, but are interrelated and interconnected with one another. When many foodchains operate simultaneously and interlock such pattern is termed as food web. Thus, the food web is a description of feeding connections between the organisms which make up a community. Energy passes through one trophic level to next via these food web links, example a rat feeds on various kinds of grains, fruits, stems, roots, etc.

A rat in its turn is consumed by a snake which is eaten by a falcon. The snakes feed on both frogs and rats. Thus, a network of food – chains exists and this is called food web. The food web gets more complicated because of variability in taste and preference availability and compulsion and several other factors at each level. For example, tigers normally do not feed on fishes or crabs but in Sunderbans they are forced to eat them.

Question 6.
Explain calcium cycle with well labelled diagram.
Answer:

Question 7.
Draw ecological pyramid of number of a tree ecosystem and grassland ecosystem.
Answer:

Question 8.
What do you mean by ecosystem? Describe the important components of a pond ecosystem.
Or
Write about role of decomposers in an ecosystem with example.
Answer:
Ecosystem:
The system resulting from the interaction between organisms and their environment is called as ecosystem.

Components of a pond ecosystem:
A pond ecosystem should contain all components of ecosystems like:

1. Producers:
Organism, which can synthesize their own food are included under producers, example Volvox, Pandorina, Oedogonium, Saggitaria, Utricularia, Azolla, Trapa, Lemna, Typha, Nymphaea etc. form the producer class of the pond ecosystem.

2. Consumers:

• Primary consumer – Animals, which feed on producers are included into this category example Daphnia, Cyclops, Paramoecium, Amoeba and small fishes.
• Secondary consumers – Primary consumers also serve as food for water snakes, a few tortoise, few types of fish etc. hence, these are carnivores.
• Tertiary consumers – Secondary consumers also serve as food for aquatic birds like kingfisher, cranes, big fish and these together form a top class carnivorous group and called as tertiary consumers.

3. Decomposers:
All producers and consumers die and accumulate on the floor of the pond. Even the waste material and faeces of these animals get accumulated on the floor of the pond. Similarly, the floor of pond is also occupied by decomposers, which include bacteria and fungi. These decomposers decompose complex organic compounds of their bodies into simpler forms which are finally mixed with soil of floor of ponds. These are again absorbed by the roots of producer plants and thus matter is recycled.

Question 9.
Explain pyramid of biomass of pond ecosystem.
Answer:
The biomass, i.e., the living weight of the organisms in the foodchain present at different trophic levels in an ecosystem forms the pyramid of biomass. When biomass of consumers is greater than biomass of producer then pyramid is called as inverted pyramid of biomass. example pyramid of biomass of pond ecosysyem is always inverted.

Ecosystem:
The system resulting from the interaction between organisms and their environment is called as ecosystem.

Question 10.
Distinguish between:

1. Grazing food chain and Detritus food chain
2. Production and Decomposition
3. Upright and Inverted pyramid
4. Food chain and Food web
5. Litter and Detritus,
6. Primary and Secondary productivity

Answer:
1. Differences between Grazing food chain and Detritus food chain :

Grazing food chain:

• Energy for the food chain comes from the sun.
• First trophic level organisms are producers.

Detritus food chain:

• Energy comes from detritus (organic matter).
• First trophic level organisms are detritivores and decomposers.

2. Differences between Production and Decomposition :

Production:

• It refers to the process of synthesis of organic compounds from inorganic substances utilising sunlight.
• Example: Plants perform the function of production of food.

Decomposition:

• It is the phenomenon of degradation of waste biomass.
• Example : Bacteria and fungi decompose dead organic matter.

3. Differences between Upright pyramid and Inverted pyramid:
Upright pyramid:
When the number of producers or theirbiomass is maximum in an ecosystem and it decreases progressively at each trophic level in a food chain, an uprightpyramid is formed.

Inverted pyramid:
When the number of individuals or their biomass at the producer level is minimum and it Increases progressively at each trophic level in a food chain, an inverted pyramid is formed.

4. Differences between Food chain and Food web:

Food chain:

• A food chain is a single pathway where energy is transferred from producers to successive orders of consumers.
• All food chains start with green plants which are the original source of all food.
• Energy flow is unidirectional.

Food web:

• A food web is a network of various food chains which are interconnected with each other like an interlocking pattern.
• It has many linkages and intercrosscs among producers and consumers.
• Energy flow in multidirectional.

5. Differences between Litter and Detritus:
Litter:
The dead remains of plants (leaves, flowers etc.) and animals excreta which falls on the surface of the earth in terrestrial ecosystems is called litter.

Detritus:
The dead remains of plants and animals constitute detritus. It is differentiated into litter fall (above ground detritus) and below ground detritus.

6. Differences between Primary and Secondary productivity:

Primary productivity:

• It is the rate at which organic matter is built up by producers.
• It is due to photosynthesis.
• Primary productivity is two types :
  Gross Primary Productivity (GPP) and Net Primary Productivity (NPP) GPP – R = NPP (R = loss in Respiration)

Secondary productivity:

• It is the rate of synthesis of organic matter by consumers.
• It is due to herbivory and predation.
• Secondary productivity is two types :
  Gross Secondary Productivity (GSP) and Net Secondary Productivity (NSP) NSP = GSP – R (Loss in Respiration)

Question 11.
What is primary productivity ? Give brief description of factors that affect primary productivity.
Answer:
The rate of biomass production is called productivity.
It is expressed in terms of g^-2yr^-1 or (kcal – m^-2) yr^-1 to compare the productivity of ecosystems. It can be divided into Gross Primary Productivity (GPP) and Net Primary Productivity (NPP).

Gross primary productivity of an ecosystem is the rate of production of organic matter during photosynthesis. A considerable amount of GPP is utilized by plants in respiration. Gross primary productivity minus respiration losses (R), is the Net Primary Productivity (NPP). GPP – R = NPP

Primary productivity depends on:

• The plant species inhabiting a particular area.
• The environmental factors.
• Availability of nutrients.
• Photosynthetic capacity of plants.

Question 12.
Define decomposition and describe the processes and products of decomposition.
Answer:
Decomposition:
Decomposition is the process that involves the breakdown of complex organic matter or biomass from the body of dead plants and animals with the help of decomposers into inorganic raw materials such as carbon dioxide, water, and other nutrients.

The various processes involved in decomposition are as follows :

1. Fragmentation:
It is the first step in the process of decomposition. It involves the breakdown of detritus into smaller pieces by the action of detritivores such as earthworms.

2.Leaching:
It is a process where the water soluble nutrients go down into the soil layers and get locked as unavailable salts.

3. Catabolism:
It is a process in which bacteria and fungi degrade detritus through various enzymes into smaller pieces.

4. Humification:
The next step is humification which leads to the formation of a dark coloured colloidal substance called humus, which acts as reservoir of nutrients for plants.

5. Mineralization:
The humus is further degraded by the action of microbes, which finally leads to the release of inorganic nutrients into the soil. This process of releasing inorganic nutrients from the humus is known as mineralization. Decomposition produces a dark coloured, nutrient rich substance called humus. Humus finally degrades and releases inorganic raw materials such as CO₂, water, and other nutrient in the soil.

Question 13.
Given account of explain energy flow in ecosystem.
Answer:
Energy flow:
In the ecosystem, energy is transferred in an orderly sequence. The flow of solar energy from producers to consumers and to decomposers subsequently in an ecosystem is known as energy flow. Energy flow is an ecosystem is always unidirectional. Sun is the sole source of solar energy in an ecosystem. Green plants utilize this energy in photosynthesis and convert it in the form of chemical energy and store it.

Plants utilize maximum part of this energy to do its biological functions. Some of it is converted into heat and released in the environment. Remaining part of the energy is stored in various components of the body. When a consumer eats these producer plants, the energy is then transferred into its body.

In any food chain energy flows from primary producers to primary consumers, from primary consumers to secondary consumers and secondary consumers to tertiary consumers and so on. Because every organism of a trophic level continuously converts chemical energy into heat, there is always a loss of energy with each step in a food – chain. According to an estimate only 10% of the total energy obtained is transferred from one trophic level to another.

Question 14.
Write important features of a sedimentary cycle in an ecosystem.
Answer:
Sedimentary cycles have their reservoirs in the earth’s crust or rocks. Nutrient elements are found in the sediments of the earth. Elements such as sulphur, phosphorus, potassium, and calcium have sedimentary cycles. Sedimentary cycles are very slow. They take a long – time to complete their circulation and are considered as less perfect cycles. This is because during recycling, nutrient elements may get locked in the reservoir pool, thereby taking a very long – time to come out and continue circulation. Thus, it usually goes out of circulation for a long – time.

Ecosystem Long Answer Type Questions

Question 1.
Describe various components of ecosystem.
Answer:
Ecosystem has following components:

1. Abiotic components:
The physical conditions of the ecosystem depends upon latitude, overall climatic and edaphic factors. The physical deficiencies are overcome by artificial irrigation and use of fertilizers. Thus, like natural ecosystem all organic, inorganic substances and climatic factors together forms abiotic component of the ecosystem.

• Organic substances : Such as carbohydrates, proteins, lipids, etc.
• Inorganic substances : Such as C, H, N, P, K, Ca, I, etc.
• Climatic factors : Such as temperature, light, water, humidity, wind, pH, minerals, soil structure, etc.

2. Biotic factors:
Three types are there

(a) Producers :
This type of crop (dominant species) depends upon the climate, season and the choice of the farmer.

(b) Consumers:

• Primary consumers : Insects, beetles, fishes, etc.
• Secondary consumers : Frog and fishes.
• Tertiary consumers : Snakes and cranes.

(c) Decomposers:
Bacteria, fungi etc.

Question 2.
Explain consumer components ecosystem of a pond in brief.
Answer:
Consumers:
They feed on producers directly or indirectly. It is of following categories:

1. Primary consumers:
Varied forms of zooplanktons are found in the water surface. Most of them are unicellular protists, such as Amoeba, Paramoecium whereas some are multicellular crustaceans, such as Daphnia, Cyclops, etc. Animals which are found under the surface of water are called as benthos, such as many types of fishes, crustaceans, molluscs, insects, beetles, etc. They too feed on producers.

2. Secondary consumers:
They feed on primary consumers, example big fishes, water snakes, etc.

3. Tertiary consumers:
They feed on secondary consumers, example kingfisher, cranes, omnivorous man, etc.

Question 3.
Describe carbon cycle in an ecosystem.
Answer:
Carbon Cycle:
Importance of carbon:
Carbon is considered as the basis of life. Carbon is the most important constituent of proteins, fats and nucleic acids which form the essential constituents of protoplasm.

Sources of carbon – The three sources of carbon in non – living world are:

• The carbon dioxide of the air and that which is dissolved in water.
• The rocks of the earth crust containing carbonates.
• The fossil fuel like coal and petroleum.

Recycling of carbon:
Carbon dioxide (CO₂) is the main source of carbon for the living beings. The carbon of coal, graphite, petroleum are insoluble and carbonates are not available to the organism until they are burnt or chemically changed. Most of the carbon dioxide enters the living world through photosynthesis. In this process green plants trap CO₂ from atmosphere and convert it into carbohydrates by using water and solar energy.

The amount of the carbon fixed by photosynthesis is nearly 7 x 10¹³ kg/year. The organic compounds synthesized in photosynthesis are passed from plants to the herbivores and carnivores. It is estimated that one hectare of a healthy forest produces about 10 tonnes of oxygen and absorbs 30 tonnes of carbon dioxide annually.

Question 4.
What do you mean by trophic levels? What are ecological pyramids ? Explain various types of ecological pyramids.
Answer:
Trophic level:
In an ecosystem, the producer consumer arrangement is a kind of structure known as trophic structure and each food level in the food chain is called as trophic level or energy level. In other words each level of food in food chain is called its trophic level. The first trophic level (T₁) in an ecosystem is occupied by producers. Herbivores (primary consumers) form second trophic level (T₂), secondary consumers form third trophic level (T₃), tertiary consumers form fourth trophic level (T₄) and decomposers form fifth trophic level (T₅) in an ecosystem.

Food or Ecological pyramids:
If we express the organisms of various trophic levels according to their number, biomass and ratio of energy stores within it, then we obtain a cone or pyramid like structure which is known as food or ecological pyramid. Ecological pyramids represent the trophic structure and function of an ecosystem. In base and successive trophic levels the tiers which make up the apex. Ecological pyramids are of the following three types:

1. Pyramid of biomass
2. Pyramid of number
3. Pyramid of energy.

1. Pyramid of Biomass:
Biomass is the dry weight of living organisms per unit of space. The ecological pyramid, which shows the quantitative relationship of the standing crop at each trophic level/The pyramid of biomass shows gradual reduction in biomass at each trophic level from base to apex.

The pyramid of biomass may be :

• Upright – example all terrestrial ecosystems.
• Inverted – example all aquatic ecosystems.

2. Pyramid of Number:
The ecological pyramid which shows the number of individual organisms at each trophic level. It represents numerical relationship between different trophic level of a food chain. In this pyramid more abundant species from the first trophic level and from the base of pyramid and the less abundant species remain near the top. The pyramid of number may be:

• Upright : example grassland, pond, forest ecosystem.
• Inverted : example ecosystem of tree.

3. Pyramid of Energy:
It indicates the total amount of energy at each trophic level of the food chain. At each producer level, the total energy available is relatively more than at the higher trophic levels because of the loss of the energy at each trophic level. Thus, there is a gradual loss of energy at each trophic level. The pyramid of energy of each types of ecosystem is always upright.

Question 5.
What is meant by terrestrial biomes? What are its types? Explain any one biomes in detail.
Answer:
Terrestrial biome:
Large area occupying ecosystems in nature are called biomes. If biomes are on land than they are called terrestrial biomes.
Terrestrial biomes may be :

1. Forest biomes – They may be as below :

• Topical rain forest
• Cold tropical forest
• Taiga forest.

2. Grassland biomes – They may be as below:

• Tropical rain forest
• Cold tropical forest

3. Desert biomes

4. Tundra biomes.

Grassland biome – Grassland biome or ecosystem has long grasses, Its, land is fertile. It receives approximately 25 to 75 cm average rainfall. Its component are:

1. Abiotic component:
All organic, inorganic substances and climatic factors together form abiotic component.

2. Biotic component:

• Producers – Grasses, herb, shrubs.
• Primary consumers – Herbivore like cow, buffalo, goats, sheep, deer, rabbit, rat insect.
• Secondary consumers – Carnivore animals which eat primary consumers, like snake, birds, foxes, jackal etc.
• Tertiary consumers – These organism which eat secondary consumers because no other one eats them, like Hawk, Peacock etc.
• Decomposers – Micro fungus, Bacteria, Actinomycetes are decomposer of grassland biomes and recycle the material back to soil and used by producers.

Question 6.
What are biogeochemical cycles? Write in short sulphur and calcium cycle.
Answer:
Biogeochemical cycles:
All living organisms get matter from the biosphere components i.e., lithosphere, hydrosphere and atmosphere. Essential elements or inorganic substances are provided by earth and are required by organisms for their body building and metabolism, they are known as biogeochemicals or biogenetic nutrients.

Sulphur cycle:
Producers (green plants) need sulphur in the form of sulphates from soil or from water (aquatic plants). The animals get sulphur through food. Some animals get sulphur from water also. Sulphur is found in three amino acids hence, sulphur is component of most proteins, some vitamins and enzymes. Plants pick up sulphur in the form of sulphates. They are converted to organic form mostly as component of some amino acids. It is found in nature as element and also as sulphates in soil, water and rocks. After the death of plants and animals, they are decomposed by microbes like Asperigillus, Neurospora and Escherichia releasing hydrogen sulphide.

Calcium cycle:
Calcium is slowly released from the rocks by water and wind action. These are either blown into the air or absorbed by plants through their roots. Animals obtain it directly as compounds and also from plants. Calcium is released from plant and animal bodies by decomposition after death. Molluscs and Corals deposit a large quantity of calcium in their shells and skeletons making it unavailable for quick cycling.

MP Board Class 12th Biology Important Questions

MP Board Class 12th Biology Important Questions Chapter 13 Organisms and Populations

Organisms and Populations Important Questions

Organisms and Populations Objective Type Questions 

Question 1.
Choose the correct answer:

Question 1.
Example of photophilous plant is:
(a) Sunflower
(b) Abies
(c)Taxus
(d) All of these
Answer:
(a) Sunflower

Question 2.
Example of saprophytic plant is:
(a) Orobanche
(b) Monotropa
(c) Balanophora
(d) Rafflesia
Answer:
(b) Monotropa

Question 3.
Example of total parasitic plant:
(a) Balanophora
(b) Monotropa
(c) Rafflesia
(d) Drosera
Answer:
(c) Rafflesia

Question 4.
Characteristic features of aquatic plants are:
(a) Undeveloped stem with thorns
(b) Leaves with hairs
(c) Soft and flexible stem
(d) None of the above
Answer:
(c) Soft and flexible stem

Question 5.
Sunken stomata are found in:
(a) Xerophytic plants
(b) Aquatic plants
(c) Terrestrial plants
(d) Floating plants
Answer:
(a) Xerophytic plants

Question 6.
Spongy roots are present in :
(a) Vallisnaria
(b) Hydrilla
(c) Trapa
(d) Pistia
Answer:
(b) Hydrilla

Question 7.
Which soil is best for growth of plants : (MP 2016)
(a) Sandy soil
(b) Silty soil
(c) Loamy soil
(d) Clayey soil.
Answer:
(c) Loamy soil

Question 8.
Hydrophytes are characterised by : (MP 2009 Set A)
(a) Spiny less developed stem
(b) Leathery leaves
(c) Delicate and mucilage bearing stem
(d) Sunken stomata.
Answer:
(c) Delicate and mucilage bearing stem

Question 9.
Sunken stomata is found in :
(a) Xerophytes
(b) Hydrophytes
(c) Terrestrial
(d) Floating hydrophytes.
(a) Xerophytes

Question 10.
Spongy roots are found in :
(a) Jussiaea
(b) Eichhomia
(c) Trapa
(d) Pistia.
Answer:
(a) Jussiaea

Question 11.
Which one is the example of mangrove :
(a) Rhizophora
(b) Eichhomia
(c) Avicennia
(d) Both (a) and (b)
Answer:
(d) Both (a) and (b)

Question 12.
Pneumatophores are the characteristic feature of :
(a) Mangrove plants
(b) Mesophytes
(c) Hydrophytes
(d) Xerophytes
Answer:
(a) Mangrove plants

Question 13.
Vascular bundles are feebly developed in :
(a) Hydrophytes
(b) Xerophytes
(c) Psammophytes
(d) Halophytes.
Answer:
(a) Hydrophytes

Question 14.
Phyllode is a modification of :
(a) Root
(b) Stem
(c) Leaf
(d) None of these
Answer:
(c) Leaf

Question 15.
Phylloclade is a modification of :
(a) Root
(b) Stem
(c) Leaf
(d) Petiole
Answer:
(b) Stem

Question 16.
Rootless plant is :
(a) Wolffia
(b) Eichhomia
(c) Lemna
(d) Jussiaea.
Answer:
(a) Wolffia

Question 17.
Free floating plant is :
(a) Pistia
(b) Eichhomia
(c) Azolla
(d) All of these
Answer:
(d) All of these

Question 18.
Vivipary is found in :
(a) Hydrophytes
(b) Xerophytes
(c) Epiphytes
(d) Mangrove plants.
Answer:
(d) Mangrove plants.

Question 19.
Transpiration Higher among the following plants : (MP 2015)
(a) Mesophytes
(b) Hydrophytes
(c) Xerophytes
(d) Algae cells.
Answer:
(a) Mesophytes

Question 2.
Fill in the blanks:

1. Sandal plant is a …………………… parasite.
2.  …………………… and …………………… are found in lichen’s thallus.
3.  …………………… is the example of rodent.
4. Plants and animals combine togather and make the …………………… factor of the enviroments.
5. Inactivity of animals in summer season is called ……………………
6.  …………………… is prescribed the word ecology.
7. Raffesia is a …………………… plant.
8. Formation of forest ecosystem is the last stage of the ……………………
9. The study of human papulation is called ……………………
10. Respiratony root is found in ……………………  plant
11. Vivipary is the characteristics of …………………… plant
12. Group of animals which are interacted to each other is called ……………………
13. Herbivorous animals which are take the food in the plants is called ……………………
14. Plants they grow low density of light is called ……………………

Answer:

1. Partial parasite
2. Algae and fungi
3. Guinea pig
4. Biotic
5. Summer hybernation
6. Peter
7. Parasitic plant
8. Secession
9. Demography
10. Man – grove
11. Mangrove
12. Community
13. Plant eater
14. Sciophyte.

Question 3.
Match the followings:
I. (MP 2012)

Answer:

1. (d)
2. (a)
3. (b)
4. (e)
5. (c)

II. (MP 2012)

Answer:

1. (c)
2. (b)
3. (a)
4. (e)
5. (d)

Question 4.
Write the answer in one word/sentances:

1. Name any animal which lays eggs in the nest of other animal?
2. Name the plant which has pneumatophores.
3. Name two symbiotic partners of lichen.
4. Write the type of water which is available in plants.
5. Loss of the upper surface of the soil is called.
6. When do we celebrate world environmental day?
7. Name the rootless free floating plant.
8. In which plant, root cap is absent.
9. Give an example of symbiosis.
10. Formation of soil is called.
11. Plants are dependent to other plant for shelter is called.
12. Name an amphibious plant.
13. Name the plant in which stem is modified into leaf like structure and leaves are modified into spines.
14. Give an example of sexual dimorphism.

Answer:

1. Cuckoo
2. Rhizophora
3. Algae and Fungi
4. Capillary water
5. Soil erosion
6. 5 June
7. Wolffia
8. Hydrophytes
9. Pea and Rhizobium bacteria
10. Soil fertility
11. Epiphytes
12. Ranunculus
13. Opuntia
14. Honey bee

Organisms and Populations Very Short Answer Type Questions

Question 1.
Give two examples of each:
Symbiotic, commensal, phytoplankton, zooplankton and rooted floating plant
Answer:

1. Symbiotic – Escherichia, Triconimpha
2. Commensal – Orchid and Trees, Hermit crab
3. Phytoplankton – Nostoc, Anabaena
4. Zooplankton – Paramecium, Euglena
5. Rooted floating plant – Wolffia, Lemna.

Question 2.
Name the speciality of any population.
Answer:
Any population has some speciality, these are :

1. Population density
2. Growth rate
3. Death rate
4. Age distribution
5. Biotic capacity
6. Population growth form
7. Changes of population
8. Population dispersion

Question 3.
Name any two mangrove plants.
Answer:
Rhizophora, Avicennia.

Question 4.
Name the vagetation found in sundarban delta.
Answer:
Mangrove vegetation.

Question 5.
Name the plant in which pneumatophores are found.
Answer:
Rhizophora.

Question 6.
Give the term for scientific study of human population.
Answer:
Demography.

Question 7.
What is sex ratio of India according to 2001 census?
Answer:
933 females per 1000 of njales.

Question 8.
Name any animal which lays egg in the nest of other animal.
Answer:
Koyal.

Question 9.
Give the name of larva of frog, salamander and moths.
Answer:
Tadpole, Axolotal, Catterpillar.

Question 10.
Give the reason for increase in population.
Answer:
Decrease in death rate.

Question 11.
Give example of species polymorphism.
Answer:
Honeybee.

Question 12.
Name the species found in different geographical areas.
Answer: Allopatric

Question 13.
Write the name of any saprophytic angiospermic plant
Answer:
Monotropa.

Question 14.
What is called the relationship between hermit crab and sea – anemone?
Answer:
Protocooperation.

Question 15.
Why insectivores eat insects?
Answer:
To fulfill the need of nitrogen.

Question 16.
Which are main abiotic factor?
Answer:
Temperature, water, light, soil.

Question 17.
Give the name of one insectivores plant
Answer:
Utricularia.

Question 18.
What is ecological competition?
Answer:
Ecological competition is the struggle between two organisms for the same resources within an environment.

Question 19.
A transition area between two biomes.
Answer:
Ecotone.

Question 20.
The number of live births per thousand of population per year.
Answer:
Birthrate.

Organisms and Populations Short Answer Type Question

Question 1.
How is diapauses different from hibernation?
Answer:
Diapauses is a stage of suspended development to cope with unfavorable con-ditions. Many species of zooplankton and insects exhibit diapauses to tide over adverse climatic conditions during their development. On the other hands hibernation or winter sleep is a resting stage where in animals escape winters (cold) by hiding themselves in their shelters. They escape the winter season by entering a state of inactivity by slowing their metabolism. The phenomenon of hibernation is exhibited by bats, squirrels, and other rodents.

Question 2.
If a marine fish is placed in a fresh water aquarium, will the fish be able to survive? Why or why not?
Answer:
If a marine fish is placed in a freshwater aquarium, then its chances of survival will diminish. This is because their bodies are adapted to high salt concentrations of the marine environment. In freshwater conditions, they are unable to regulate the water entering their body (through osmosis). Water enters their body due to the hypotonic environment out¬side. This results in the swelling up of the body, eventually leading to the death of the marine fish.

Question 3.
Give definition of :

1. Population
2. Population density
3. Biotic potential
4. Growth rate
5. Death rate

Answer:

1. Population:
A community of animals, plants or humans among whose members inbreeding occurs.

2. Population density:
Population density is a measurement of population per unit area or unit volume.

3.Biotic potential:
Biotic potential is the ability of a population of living species to increase under ideal environmental conditions sufficient food supply, no predators and a lack of disease. An organisms rate of reproduction and the size of each litter are the primary determining factors for biotic potential.

4. Growth rate:
Human population growth is the increase in the number of individuals in a population. Global human population growth amounts to around 83 million annually.
Growth rate = \(\frac { Annually birth }{ Mid year population}\) x 100

5. Death rate:
The death rate is the number of people per thousand who die in a particular are during a particular period of time.
Death rate = \(\frac { Annually birth }{ Mid year population}\) x 100

Question 4.
What suggestions will you give for awakening people towards population control.
Answer:
Following things can be done for controlling the population:

1. Promotion of education: Misunderstanding can be avoided by promotion of education such as:

• Children are gift of god
• More income from the more children.

2. To introduce for reality of population growth.
3. Uses of family planning techniques.
4. Restriction of more than one marriage.
5. To decrease birth rate.

Question 5.
Write the differences:

1. Species and Population
2. Population growth and Population density,
3. Monospecific and Polyspecific population
4. Competition and Scattering.

Answer:
1. Species and Population:
Population on defined as organisms that belong to the identical species and identical geographical niche or area. The said area should enable these species to interbreed with each other.

2. Population growth and Population density:
In biology, Population growth is the increase in the number of individuals in a population but population density is a measurement of population per unit area, it is a quantity of type number density.

3. Monospecific or Polyspecific population:
Monospecific population is the population of individuals of only one species but poly specific or mixed population is the population of individuals of more than one species and it is generally referred to as a community.

4. Competition and Scattering:
Competition is in general a contest between two or more rivalry between two or more entities, organisms, animals, economic groups or social groups etc. Population scattering is a method which shows equilibrium by interaction of population. In population, scattering seeds away from the parent plant.

Question 6.
Give four reasons of population growth in India.
Answer:
Four reasons of population growth in India are as follows:

1. Increasing growth rate.
2. Decreasing death rate.
3. No importance of education and promotion of education is less.
4. Conservative.

Question 7.
Explain, the formation of new species.
Answer:
Although all life on earth shares various genetic similarities, only certain organisms combine genetic information by sexual reproduction and have offspring that can than successfully reproduce. Scientists call such organisms members of the same biological species.

Question 8.
What do you mean by population equilibrium?
Answer:
Population equilibrium:
When growth pattern of any population is observed then it becomes clear that any population increases at faster rate in the beginning in a new area and reaches to a maximum limit and then remain constant for long time.This phase is called as population equilibrium. At equilibrium birth and death rate of any population become equal.

Question 9.
Write cooperative interactions between the members of a species.
Answer:
The cooperative intraspecific interactions involve help to other members. The cooperative interaction amongst the individuals of a species is necessary for reproduction, perpetuation, parental care of young ones, social organization, territoriality, protection and food, etc.

1. Cooperation for protection and food:
For protection and food, members of a species may form groups. Such organisms which live in group are called as gregarious.

2. Cooperation for reproduction:
It is necessary for reproduction in which adult male and female comes together for mating.

3. Social organization:
Active association for mutual benefits amongst the individuals of same species often bring social organization. Success of these organizations is measured in the terms of the survival or colony. Honeybees, ants, wasps and termites, etc. form well – organized societies showing division of labour and polymorphism. In these social in sects, a large number of individuals of different kinds live together in the colony and work collectively for the benefit of the group.

The insect societies are formed of different castes such as workers, drones (male) and queen which are specialized for the different kind of work. The workers collect and store food, build houses of complicated design and pay special attention to the queen. The drones (male) and queen (female) are mainly concerned with reproduction.

Question 10.
Define phenotypic adaptation. Give one example.
Answer:
Phenotypic adaptation involves changes in the body of an organism in response to genetic mutation or certain environmental changes.These responsive adjustments occur in an organism in order to cope with environmental conditions present in their natural habitats For example, desert plants have thick cuticles and sunken stomata on the surface of their leaves to prevent transpiration. Similarly, elephants have long ears that act as thermoregulators.

Question 11.
Most living organisms cannot surv ive at temperature above 45°C. How are some microbes able to live in habitats with temperatures exceeding 100°C?
Answer:
Archaebacteria (Thermophiles) are ancient forms of bacteria found in hot water springs and deep sea hydrothermal vents. They are able to survive in high temperatures because their bodies have adapted to such environmental conditions. These organisms contain specialized thermo-resistant enzymes, which carry out metabolic functions that do not get destroyed at such high temperatures unlike other enzymes.

Most living – organism can not survive above 45°C because:

• Above 45° C enzymes get denatured.
• Protoplasm precipitates at high temperature.

Question 12.
List the attributes that populations but not individuals possess.
Answer:
A population has the following attributes that an individual does not possess:

1. Birth rates and death rates.
2. Sex ratio.
3. Population density.
4. Age distribution.
5. Population growth.

Question 13.
If a population growing exponentially double in size in 3 years, what is the intrinsic rate of increase (r) of the population?
Answer:
A population grow exponentially if sufficient amounts of food resources are available to the individual. Its exponential growth can be calculated by the following integral form of the exponential growth equation :
N_t = N₀ e^rt
Where,
N_t = Population density after time t.
N_o = Population density at time zero.
r = Intrinsic rate of natural increase.
e = Base of natural logarithms (0.434).
From the above equation, we can calculate the intrinsic rate of increase (r) of a population. Now, as per the question,
Present population density =x
Then, Population density after two years = 2x
t = 3 years
Substituting these values in the formula, we get:
=> 2x = x e^3r
=> 2 = e^3r
Applying log on both sides :
=> log 2 = 3r log e
\(\frac { log 2 }{ 3 loge }\) = r
\(\frac { 0.301 }{ 3 x 0.434 }\) = r
\(\frac { 0.301 }{1.302 }\) = r
0.2311 = r
Hence, the intrinsic rate of increase for the above illustrated population is 0.2311.

Question 14.
Name important defence mechanisms in plants against herbivory.
Answer:
Several plants have evolved various mechanisms both morphological and chemical to protect themselves against herbivory.

1. Morphological defence mechanisms:

• Cactus leaves (Opuntia) are modified into sharp spines (thorns) to deter herbivores from feeding on them.
• Sharp thorns along with leaves are present in Acacia to deter herbivores.
• In some plants, the margins of their leaves are spiny or have sharp edges that prevent herbivores from feeding on them.

2. Chemical defence mechanisms :

• All parts of Calotropis weeds contain toxic cardiac glycosides, which can prove to be fatal in ingested by herbivores.
• Chemical substances such as nicotine, caffeine, quinine and opium are produced in plants as a part of self – defense.

Question 15.
An orchid plant is growing on the branch of mango tree. How do you describe this interaction between the orchid and the mango tree?
Answer:
An orchid growing on the branch of a mango tree is an epiphyte. Epiphytes are plants growing on other plants which however, do not derive nutrition from them. Therefore, the relationship between a mango tree and an orchid is an example of commensalisms, where one species gets benefited while the other remains unaffected. In the above interaction, the orchid is benefited as it gets support while the mango tree remains unaffected.

Question 16.
What is the ecological principle behind the biological control method of managing with pest insects?
Answer:
The ecological principle behind the biological control method of managing with pest insects is to keep their population in check by using their natural predators and parasites.

Question 17.
Distinguish between the following:

1. Hibernation $nd Aestivation
2. Ectotherms and Endotherms.

Answer:
1. Differences between Hibernation and Aestivation :

Hibernation:

• It is called winter sleep.
• Animal rests in warm place.
• It lasts for long winter months.

Aestivation:

• It is called summer sleep.
• Animal rests in cool, moist, shady place.
• It lasts for day time as nights are cooler and animal will come out.

2. Differences between Ectotherms and Endotherms :

Ectotherms:

• These are cold – blooded animals.
• They are unable to regulate their body temperature which changes with temperature of the surrounding environment.
• They hibernate in winters and aestivate in summers.

Endotherms:

• These are warm – blooded animals.
• They are capable of maintaining their body temperature.
• They can regulate their body temperature so, they do not need to show behavioural adaptations like these.

Question 18.
List the various abiotic environmental factors.
Answer:

1. Atmospheric factors : Light, temperature, wind and water.
2. Lithosphere : Rock, soil.
3. Hydrosphere : Pond, river, lake and ocean.
4. Edaphic factors: Soil texture, soil water, soil air, soil microorganism, soil pH, minerals.
5. Topographic factors : Slope, altitude, valley.

Question 19.
Give an example for:

1. An endothermic animal,
2. An ectothermic animal,
3. An organism of benthic zone.

Answer:

1. An endothermic animals : Birds such as crows, sparrows, pigeons, cranes, etc. and mammals such, as bears, cows, rats, rabbits etc. are endothermic animals.
2. An ectothermic animals : Fishes such as sharks, amphibians frogs, and reptiles, tortoise, snakes, and lizards are ectothermic animals.
3. An organism of benthic zone : Decomposing bacteria is an example of an organism found in the benthic zone of a water body.

Question 20.
Define population and community.
Answer:
Population:
Population is a group of individuals of same species, which can reproduce among themselves and occupy a particular area in a given time.

Community:
It is an assemblage of several population in a particular area and time and exhibit interaction and interdependence through trophic relationship.

Question 21.
Define the following terms and give one example for each:

1. Commensalism
2. Parasitism
3. Camouflage
4. Interspecific competition.

Answer:
1. Commensalism:
It is interspecific interaction in which one species is benefited and other the neither harmed nor benefitted. The two individuals may be physically associated. example Sucker fish and shark.

2. Parasitism:
Parasite is an organism which is dependent upon other living organism for their food and other requirements. Parasitism is an interspecific interrelation in which the individual of one species lives at the cost of the individual of another species, its host which it harms and eventually kills. In this case host is always larger than the parasite. The parasite that is seen on the surface of host plant is called ectoparasite or external parasite and if a parasite is found inside the plant body is said to be an endoparasite or internal parasite.

Examples Parasitic bacteria:
Parasitic bacteria cause several diseases in plants and animals. For example, the citrus canker disease of lemon is caused by Xanthomonas citri, black rots of vegetables of Cruciferae group is caused by Xanthomonas campestris. The common diseases caused by bacteria in human beings are diarrhoea, pneumonia, tetanus, leprosy, plague etc.

3. Camouflage:
It is a strategy adapted by prey species to escape their predators. Organisms are cryptically coloured so, that they can easily mingle in their surroundings and escape their predators. Many species of frogs and insects camouflage in their surroundings and escape their predators.

4. Interspecific competition:
It is an interaction between individuals of different species where both species get negatively affected. For example, the competition between flamingoes and resident fishes in South American lakes for common food resources i.e., zooplankton.

Question 22.
Explain population dispersal in brief.
Answer:
Population dispersal:
The movement of individuals of a population from a particular place to another is known as population dispersal. Dispersal of members of population may occur for search of food, escape from enemies or for breeding purpose. There are three types of

population dispersal:

1. Immigration – Inward movement of members of other population and settle with local population. It causes increase in population.
2. Emigration – Outward movement of individuals of a population to shift in some other place. It causes decrease in population.
3. Migration – It can be observed in birds. It is periodic departure of organism and returning back to its original place, example Siberian cranes visit India during winters and go back to north during summer.

Question 23.
With the help of suitable diagram describe the logistic population growth curve.
Answer:
Logistic growth:
1. The resources become limited at certain point of time so, no population can grow exponentially.

2. This growth model is more realistic.

3. Every ecosystem or environment or habitat has limited resources to support a particular maximum number of individuals called is carrying capacity (K).

4. When N is plotted in relation to time t, the logistic growth show sigmoid curve and is also called Verhulst-Pearl logistic growth. It is given by the following equation:
\(\frac { dN }{ dt }\) = rN [ \(\frac{K-N}{K}\) ]
Where, N = Population density at time t
R = Intrinsic rate of natural increase
K = Carrying capacity.

a = When resources are not limiting the growth, plot is exponential
b = When resources are limiting the growth. Plot is logistic, k is carrying capacity.

Question 24.
Write characteristic features of aquatic plants.
Answer:
Characteristic features of aquatic plants (Hydrophytes):

1. The cuticle is either absent or very thinly deposited on the epidermis.
2. Usually stomata are absent, if present they are non – functional.
3. Many air chambers or large intercellular spaces are present inside the plant body of hydrophytes.
4. The whole body of the plant helps in the absorption of water and minerals.
5. Root system is feebly developed and usually non – branched or less branched. Some hydrophytes like Wolffia lack foots.
6. The mechanical tissues are less distributed.
7. The vascular tissues are either absent or feebly developed in the case of hydrophytes.

Question 25.
Write short note on population growth form or pattern.
Answer:
Population growth from or pattern: The way of growth of population is known as population growth pattern. Population growth shows two patterns :

1. S – shaped form : In this pattern, the increase in population is very slow in the beginning, then there is a fast growth rate and then becomes stable.

2. J – shaped form : In some organisms the population increases very fast in the beginning because of some environmental factors, the growth is checked abruptly.

Question 26.
Give only five examples of communication of informations in animals.
Answer:
Communication of information within a species:
Communication may be defined as an act which influences the activity of one individual by some behaviour of another. Most of the animal species have some mechanism of exchanging information among their members through sight, sound, touch or chemicals. The evolution of the techniques of communication in the animal kingdom is progressive but complex.

Following are few examples:

1. The honeybee gives the information to the other worker bees of the hive regarding availability of nectar in the vicinity by dancing in a special manner on the surface of the hive. Through the round dance they communicate that the availability of the food is very near to the hive and the waggle dance (by moving abdomen) shows the direction and the distance of the food source from the hive.

2. Croaking of male frog attract female frog during breeding season.

3. The dogs express their various intensions by facial expressions and movement of the tail and by making typical sound.

4. Rabbits inform their group members about any sort of danger by tapping their
tails.

5. Certain chemical compounds called pheromones are secreted by animals to transmit message to other members of the species. Pheromones are detected by smell or taste.

Question 27.
Write distinguish between:

1. Parasitism and Symbiosis,
2. Mutualism and Commensalism,
3. Hydrosere and Xerosere.

Answer:
1. Differences between Parasitism and Symbiosis:

Parasitism:

• In this case, one organism depends on other for their food.
• Only one organism benefitted.
• It produce only disease.
• example Tapeworm and Man.

Symbiosis:

• In symbiosis two organisms lives together in such way that both are benefitted by each other.
• Both organism benefitted.
• It does not produce any disease.
• example Lichens.

2. Differences between Mutualism and Commensalism:

Mutualism:

• It is an interspecific interaction in which both the species are mutually benefited.
• The two individuals may be physically or physiologically associated.
• example Rhizobium and the legume plants.

Commensalism:

• It is interspecific interaction in which one species is benefited and other the neither harmed nor benefitted.
• The two individuals may be physically associated.
• example Sucker fish and shark.

3. Differences between Hydrosere and Xerosere:

Hydrosere:

• Succession in water is called hydrosere.
• Changes are fast.
• It produce aquatic plants.

Xerosere:

• Succession in deserts is called xerosere.
• Changes are slow.
• It produce xerophytic plants.

Question 28.
Write short notes on:

1. Parasitism
2. Species dominance
3. Biotic potential.

Answer:
1. Parasitism:
Parasite is an organism which is dependent upon other living organism for their food and other requirements. Parasitism is an interspecific interrelation in which the individual of one species lives at the cost of the individual of another species, its host which it harms and eventually kills. In this case host is always larger than the parasite. The parasite that is seen on the surface of host plant is called ectoparasite or external parasite and if a parasite is found inside the plant body is said to be an endoparasite or internal parasite.

Examples : Parasitic bacteria:
Parasitic bacteria cause several diseases in plants and animals. For example, the citrus canker disease of lemon is caused by Xanthomonas citri, black rots of vegetables of Cruciferae group is caused by Xanthomonas campestris. The common diseases caused by bacteria in human beings are diarrhoea, pneumonia, tetanus, leprosy, plague etc.

2. Species dominance:
Presence of any biotic community or presence of more popu¬lation of any species biosphere shown its dominance and it is called species dominance which has capacity to reduce other species in population. The effect of dominant species is more upon environment or other species.

3. Biotic potential:
Biotic potential is the ability of a population of living species to increase under ideal environmental conditions sufficient food supply, no predators and a lack of disease. An organisms rate of reproduction and the size of each litter are the primary determining factors for biotic potential.

Question 29.
Explain population fluctuation.
Answer:
Population is generally a group of individuals of a particular species occupying a particular area at a specific time.

Population fluctuation:
Any increase or decrease in number of individuals in a population from its equilibrium state is known as population fluctuation. It may occur due to various reasons, such as due to change in climate or due to change in physical environment or due to predators.

Question 30.
Give an example for:

1. Heliophytes
2. Sciophytes
3. Viviparous.

Answer:

1. Heliophytes : Sunflower, Amranthus.
2. Sciophytes : Java moss, Lycopodium, Polytrichus.
3. Viviparous : Rhizophora, Salicomia, Sonneratia.

Organisms and Populations Long Answer Type Questions

Question 1.
Write a short note on :

1. Adaptations of deserts plants and animals
2. Adaptations of plants to water scarcity
3. Behavioural adaptations in animals
4. Importance of light to plants
5. Effect of temperature or water scarcity and the adaptations of animals.

Answer:
1. Adaptations of desert plants and animals :

(i) Adaptations of desert plants:
Plants have an extensive root system to tap underground water. They bear thick cuticles and sunken stomata on the surface of their leaves to reduce transpiration. In Opuntia, the leaves are entirely modified into spines and photosynthesis is carried out by green stems. Desert plants have special pathways to synthesize food, called CAM (C₄ pathway). It enables the stomata to remain closed during the day to reduce the loss of water through transpiration.

(ii) Adaptations of desert animals:
Animals found in deserts such as desert Kangaroo rats, Lizards, Snakes, etc. are well adapted to their habitat. The Kangaroo rat found in the deserts of Arizona never drinks water in its life. It has the ability to concentrate its urine to conserve water. Desert lizards and snakes bask in the sun during early morning and bur¬row themselves in the sand during afternoons to escape the heat of the day. These adaptations occur in desert animals to prevent the loss of water.

2. Adaptations of plants to water scarcity :
Plants found in deserts are well adapted to cope with water scarcity and scorching heat of the desert. Plants have an extensive root system to tap underground water. They bear thick cuticles and sunken stomata on the surface of their leaves to reduce transpiration.

In Opuntia, the leaves are modified into spines and the process of photosynthesis is carried out by green stems. Desert plants have special pathways to synthesize food, called CAM (C₄ pathway). It enables their stomata to remain closed during the day to reduce water loss by transpiration.

3. Behavioural adaptations in animals:
Certain organisms are affected by temperature variations. These organisms undergo adoptions such as hibernation, aestivation, migration, etc. to escape environmental stress to suit their natural habitat. These adaptations in the behaviour of an organism are called behavioural adaptations.

For example, ectothermal animals and certain endosperms exhibit behavioural adaptations. Ectotherms are cold blooded animals such as fish, amphibians, reptiles etc. Their temperature varies with their surroundings. For example, the desert lizard basks in the sun during early hours when the temperature is quite low. However, as the temperature begin? to rise, the lizard burrows itself inside the sand to escape the scorching, sun.

Similar burrowing strategies are exhibited by other desert animals. Certain endotherms (warm blooded animals) such as birds and mammals escape cold and hot weather conditions by hibernating during winters and aestivating during summers. They hide themselves in shelters such as caves, burrows, etc. to protect against tem-perature variations.

4. Importance of light to plants:
Sunlight acts as the ultimate source of energy for plants. Plants are autotrophic organisms, which need right for carrying out the process of photosynthesis. Light also plays an important role in generating photoperiodic responses occurring in plants. Plants respond to changes in intensity of light during various seasons to meet their photoperiodic requirements for flowering. Light also plays an important role in aquatic habitats for vertical distribution of plants in the sea.

5. Effects of temperature or water scarcity and the adaptations of animals:
Temperature is the most important ecological factor. Average temperature on the Earth varies from one place to another. These variations in temperature affect the distribution of animals on the Earth. Animals that can tolerate a wide range of temperature are called eurythermals. Those which can tolerate a narrow range of temperature are called stenothermal animals. Animals also undergo adaptations to suit their natural habitats.

Animals found in polar regions have thick layers of fat below their skin and thick coats of fur to prevent the loss of heat. Some organisms exhibit various behavioural changes to suit their natural habitat. These adaptations present in the behaviour of an organism to escape environmental stresses are called behavioral adaptations. For example, desert lizards are ectotherms. This means that they do not have a temperature regulatory mechanism to escape temperature variations.

These lizards bask in the sun during early hours when the temperature is quite low. As the temperature begins to increase, the lizard burrows itself inside the sand to escape the scorching sun. Similar burrowing strategy is seen in other desert animals. Water scarcity is another factor that forces animals to undergo certain adaptations to suit their natural habitat. Animals found in deserts such as desert kangaroo rats, lizards, snakes, etc. are well adapted to stay in their habitat.

The kangaroo rat found in the deserts of Arizona never drinks water in its life. It has the ability to concentrate its urine to conserve water. Desert lizards and snakes bask in the sun during early morning and burrow in the sand as the temperature rises to escape the heat of the day. Such adaptations can be seen to prevent the loss of water.

Question 2.
List any three important characteristics of a population and explain.
Answer:
The three important characteristic of a population are:

1. Population density : The number of individuals of a species per unit area or a volume is called population density.
P_D = \(\frac { N }{ S }\)
Where, PD = Population density
N = Number of individuals in a region S = Number of unit area in a region.

2. Birth rate : It is expressed as the number of births per 1000 individuals of a population per year.

3. Death rate : It is expressed as the number of deaths per 1000 individuals of a population per year.

Question 3.
Describe the anatomical adaptation of xerophytic plants.
Answer:
The anatomical adaptations of xerophytic plants are:

1. The epidermis of xerophytic plant organs is covered by thick cuticle. It protects plants against heavy transpiration and also provides mechanical strength to some extent.

2. Epidermis may be multiple layered which also reduces the rate of transpiration.
Examples : Banyan, Nerium etc.

3. Except some monocotyledonous plants the stomata are restricted to the lower surface of the leaves. The stomata are generally sunken and surrounded by many hairs.

4. Hypodermis in xerophytes is multilayered and sclerenchymatous which checks evaporation of water.
Example: Pinus needle.

5. Mesophyll tissue is well differentiated into palisade and spongy parenchyma but palisade parenchyma is more in number than spongy parenchyma.
Examples : Pinus needle, Nerium leaf.

6. Vascular bundles are well developed and differentiated into xylem and phloem. Vascular bundles are comparatively more in number. Besides vascular bundles, mechanical tissues are well developed.

7. Leaves of some xerophytes have some rolling devices due to presence of motor cells or bulliform cells in the epidermal zone.
Examples : Poapratensis, Agropyron, Ammophila etc.

8. Stomata in certain desert plants such as Capparis spinosa and Aristida ciliata may sometimes get blocked due to deposition of resinous matter or wax. This adaptation is also known to reduce the loss of water during transpiration.

Question 4.
Describe the morphological adaptations of aquatic plants.
Or
Describe morphological adaptations found in the stem and leaves of xerophytic plants.
Or
Explain in brief, xerophytic plants along with examples.
Answer:
The morphological adaptations of aquatic plants are:

1. Adaptations in roots:

(i) Root system is feebly developed and unbranched. Some floating plants like Wolffia, Utricularia etc. and submerged plants like Ceratophyllum lack roots.

(ii) Root hairs are absent except rooted floating hydrophytes. In most of the hydro¬phytes roots are meaningless as the entire surface of the plant body is in direct contact with water and acts as absorptive surface. This may probably be the reason why roots in hydrophytes are reduced or absent.

(iii) True root caps are absent in some free floating hydrophytes and in lieu of that they develop root pockets or root sheaths which protect their tips from injuries.

(iv) If roots are present they are always of adventitious type e.g., Nymphaea, Nelumbo etc. In Jussiaea modified spongy negative geotrophic floating roots are present which pro¬vide buoyancy to the plants and also do the job of gaseous exchange.

2. Adaptations in stems:

(i) In submerged hydrophytes like Hydrilla, Potamogeton etc. stems are long, slender and flexible.

(ii) In free floating hydrophytes, stems are modified as thick, stout, stoloniferous or as offset and occur horizontal on the water surface. It bears vegetative bud at their apex and helps in vegetative propagation.

(iii) In rooted floating hydrophytes or rooted submerged hydrophytes or in marshy plants, stem may be modified as runner or rhizome.

(iv) The stem may be modified as runner, stolon, tuber to perform vegetative propaga¬tion effectively. Usually stems are of perennial nature.

3. Adaptations in leaves:

(i) Leaves are thin, long, ribbon shaped in the submerged forms.
Example: Vallisneria, Ceratophyllum etc. or they are finely dissected as found in Ranunculus aquatilis.

(ii) In rooted floating plants, like Nymphaea, Nelumbo etc. the petioles of leaves show indefinite power of growth and they keep the laminae of leaves always on the surface of water.

(iii) In free floating plants like Eichornia, Trapa etc. the petioles become charac¬teristically swollen and develop sponginess which keep the plant afloat.

(iv) Occurrence of two or more than two different types of leaves in a plant is called heterophylly. Examples are Limnophila, Heterophylla, Sagittaria, Ranunculus aquatilis etc. They bear leaves of many types viz., submerged leaves, floating leaves, aerial leaves, linear, ribbon shaped leaves or dissected leaves. Floating leaves are entire and lobed. The broad leaves found on the surface of water transpire actively and regu¬late the hydrostatic pressure in the plant body. The submerged leaves act as water absorbing organs. Heterophylly is also associated with quantitative reduction in transpiration.

(v) In the amphibious or marshy plants, the leaves that are exposed to air show typical mesophytic features. They are leathery and more tough than the submerged and floating hydrophytes.

Question 5.
What do you mean by biotic community? Describe characteristic features of any biotic community.
Answer:
Biotic community:
A biotic community is a localized association of several populations of different species living in a given geographic area of habitat. It represents heterogeneous assemblage of different groups of organisms both plants and animals. Biotic community is composed of smaller units of intimately associated members belonging to different species.

The different species of a community share common environment and their relationships are based on direct or indirect functional interactions. The nature or relationship is determined by the requirements of the members of a community.

Characteristics of a community:
Each community has its own characteristics which are not shown by its individual component species.

1. Species diversity:
Each community is made up of much different organisms Plants, animals, microbes, which differ taxonomically from each other. The number of species and population abundance in community also vary greatly.

2. Growth form and Structure:
Each community have a definite growth form. This different growth form determines the structural pattern of a community.

3. Dominance:
In each community, all the species are not equally important. There are relatively only a few of these, which determine the nature of community. These few species exert a major controlling influence on the community. Such species are known as dominants.

4. Succession:
Each community has its own development history. It develops as a result of a directional change in it with time.

5. Trophic structure (Self – sufficiency):
Nutritionally, each community, a group of autotrophic plants as well as heterotrophic animals exists as a self – sufficient, perfectly balanced assemblage of organism.

Question 6.
Write cooperative interactions between the members of a species.
Answer:
The cooperative intraspecific interactions involve help to other members. The co – operative interaction amongst the individuals of a species is necessary for reproduction, perpetuation, parental care of young ones, social organization, territoriality, protection and food, etc.

1. Cooperation for protection and food:
For protection and food, members of a species may form groups. Such organisms which live in group are called as gregarious.

2. Cooperation for reproduction:
It is necessary for reproduction in which adult male and female comes together for mating.

3. Social organization:
Active association for mutual benefits amongst the individuals of same species often bring social organization. Success of these organizations is measured in the terms of the survival or colony. Honey – bees, ants, wasps and termites, etc. form well-organized societies showing division of labour and polymorphism. In these social insects, a large number of individuals of different kinds live together in the colony and work collectively for the benefit of the group.

The insect societies are formed of different castes such as workers, drones (male) and queen which are specialized for the different kind of work. The workers collect and store food, build houses of complicated design and pay special attention to the queen. The drones (male) and queen (female) are mainly concerned with reproduction.

Question 7.
Explain interrelationship between various species of a biotic community along with examples.
Answer:
Odum (1971) distinguished interactions into two broad categories :

1. Positive interactions and
2. Negative interactions.

1. Positive Interactions:
In this type, those type of interactions are considered in which both interacting species are mutually involved to help each other. Here, one interacting species helps the other either one way or on reciprocal terms and may be in the form of nutrition or shelter or substratum or transport. Here interaction may be obligatory or facultative. The positive interactions include three types of interactions. They are :

(i) Commensalism:
This type of interaction occurs in between two organisms of two different species in which one species benefits and the other is neither benefited nor harmed, example Lichens.

(ii) Protocooperation:
The type of interaction where both population are ben¬efited but not obligatory i.e., not essential for the survival of either population is called proto – co – operation. This is also known as non-obligatory mutualism. The relationship between hermit crab and sea anemone is an example of proto – cooperation. The crab uses the gastropod (mollusc) shell as a portable shield and the sea anemone eats the leftover food of the crab, which is protected from its predators by the stinging cells of the sea anemone.

(iii) Mutualism or Symbiosis:
When two different species grow together and are mutually benefited, the plants are known as symbiotic plants and the phenomenon is called symbiosis or mutualism. It is a sort of obligatory association. This type the organisms are dependent upon each other for survival. In this type of association, two types of species are physiologically related. This type of relationship may exist in between two plants or in between one plant and one animal or in between two animals.

2. Negative Interactions:
Those interaction of two different species in which both are harmed or one organism is benefited while other is more or less harmed is referred to as negative interac¬tion. In this type of interaction one population eats the other type of population or one organism does not allow other organisms to grow near it by using the food supply of the other or producing toxic substances. The negative interactions be categorised into following types:

• Competition
• Parasitism
• Predation and
• Antibiosis.

Competition:
When interaction occurs between two species for the use of same resources and when resources are in short supply is referred to as competition. It is a relationship which involves struggle amongst the organisms for food, shelter, water, sunlight and climate.

Parasitism:
Parasite is an organism which is dependent upon other living organism for their food and other requirements. Parasitism is an interspecific interrelation in which the individual of one species lives at the cost of the individual of another species, its host which it harms and eventually kills. In this case host is always larger than the parasite. The parasite that is seen on the surface of host plant is called ectoparasite or external parasite and if a parasite is found inside the plant body is said to be an endoparasite or internal parasite.

Examples : Parasitic bacteria :
Parasitic bacteria cause several diseases in plants and animals. For example, the citrus canker disease of lemon is caused by Xanthomonas citri, black rots of vegetables of Cruciferae group is caused by Xanthomonas campestris. The common diseases caused by bacteria in human beings are diarrhoea, pneumonia, tetanus, leprosy, plague etc.

Predation:
To obtain food by hunting is predation. A predator is an animal or plant that kills and feeds on another organism, its prey. It represents a generalised carnivorous habit. Commonly predators are larger than their prey but this is not always true but predator is in any case always occur as a naturally better equipped hunter than its prey. Most of the predatory organisms are animals but there are some plants (carnivores) also especially fungi which feed upon other animals.

MP Board Class 12th Biology Important Questions

MP Board Class 12th Physics Important Questions Chapter 1 Electric Charges and Fields

Electric Charges and Fields Important Questions Objective Type Questions

Question 1.
Choose the correct answer of the following:

Question 1.
The least charge is:
(a) 1 coulomb
(b) 1 stat coulomb
(c) 1 micro coulomb
(d) Electronic charge.
Answer:
(d) Electronic charge.

Question 2.
The number of electrons in one coulomb charge is:
(a) 5.46 × 10²⁹
(b) 6.25 × l0¹⁸
(c) 1.6 × l0¹⁹
(d) 9.0 × l0²⁹
Answer:
(b) 6.25 × l0¹⁸

Question 3.
An object has – 80 µ C.What is the excess number of electrons:
(a) 5 × 10⁸C
(b) 16 × 10¹⁴C
(c) 5 × 10¹⁴ C
(d) 16 × 10⁸C.
Answer:
(c) 5 × 10¹⁴ C

Question 4.
The intensity of electric field E due to charge Q at distance r as compared to its normal state:
(a) E ∝ r
(b) E ∝ \(\frac { 1 }{ { r }^{ 2 } } \)
(c) E ∝ \(\frac { 1 }{ r }\)
(d) E ∝ \(\frac { 1 }{ { r }^{ 3 } } \)
Answer:
(b) E ∝ \(\frac { 1 }{ { r }^{ 2 } } \)

Question 5.
The value of permitivity E₀ of vacuum (or free space) is :
(a) 8.85 × 10^-12 C
(b) 8.85 × 10^-12C²N^-1m^-2
(c) 9 × 10⁸ Nm²C^-2
(d) 9 × 10 C²N^-1m^-2.
Answer:
(b) 8.85 × 10^-12C²N^-1m^-2

Question 6.
The work done in rotating a dipole of dipole moment p through an angle 180° from direction of uniform electric field E is:
(a) 2pE
(b) pE
(c) \(\frac { 1 }{ p}\)pE
(d) Zero.
Answer:
(a) 2pE

Question 7.
The intensity of electric field E due to a dipole of dipole moment p at a point a distance r from center of dipole depends on r as :
(a) E ∝ r
(b) E ∝ \(\frac { 1 }{ { r }^{ 2 } } \)
(c) E ∝ \(\frac { 1 }{ r }\)
(d) E ∝ \(\frac { 1 }{ { r }^{ 3 } } \)
Answer:
(d) E ∝ \(\frac { 1 }{ { r }^{ 3 } } \)

Question 8.
A surface S is placed in an electric field E parallel to the field. The flux linked with the surface be:
(a) ES
(b) \(\frac{ E }{S}\)
(c) Zero
(d) Infinite.
Answer:
(c) Zero

Question 9.
The total electric flux emerges out from unit positive charge in air is:
(a) E₀
(b) \(\frac { 1 }{ { E }_{ 0 } } \)
(d) \(\frac { 1 }{ { 4\pi E }_{ 0 } } \)
(d) 4πE₀
Answer:
(b) \(\frac { 1 }{ { E }_{ 0 } } \)

Question 10.
The dielectric constant of metal is:
(a) Infinite
(b) Zero
(c) One
(d) None of these.
Answer:
(a) Infinite

Question 11.
The dielectric constant of air is:
(a) 8.85 × l0^-12 C² N^-1m^-2
(b) 1
(c) Infinite
(d) None of these
Answer:
(c) Infinite

Question 2.
Fill in the blanks:

1. The dimensional formula of charge is …………….
2. The two like charges …………… each other.
3. Two unlike charges ……………. each other.
4. The SI unit of electric field is …………….
5. The dimensional formula of electric field is …………….
6. The size of an ideal dipole is …………….
7. Two charges – q and + q are situated at a distance / its dipole moment will be …………….
8. The size of an ideal dipole is …………….
9. Two charges – q and + q are situated at a distance lits dipole moment will be …………….
10. The maximum torque acting on a dipole of dipole moments p in a uniform electric field E is …………….

Answers:

1. [AT]
2. Repel
3. Attract
4. NC^-1
5. [ML^-3A^-1’]
6. Point size
7. q.I
8. pE sin G.

Question 3.
Match the Column:

Answers:
(b) newton / coulomb
(e) coulomb
(a) coulomb x metre
(d) N – m²/C
(c) coulomb²/newton x meter ²

answers:

1. (e) \(\frac { 1 }{ { 4\pi E }_{ 0 } }\) \(\frac { { q }_{ 1 }{ q }_{ 2 } }{ { Kr }^{ 2 } }\)
2. (d) \(\frac { 1 }{ { 4\pi E }_{ 0 } }\) \(\frac { q }{ { Kr }^{ 2 } } \)
3. (b) \(\frac { q }{ { E }_{ 0 } } \)
4. (c) pE(1 – cosθ)
5. (a) – pE cos θ

Question 4.
Write the answer in one word / sentence:

1. What type of charge get accumulated in a glass rod rubbed with silk ?
2. What type of charge get accumulated in rubber rod when rubber with fur fabric ?
3. What is the cause of quanti zation of charge ?
4. What is dielectric constant of a metal ?
5. What is the direction of electric dipole moment ?

Answers:

1. Positive charge
2. Negative charge
3. Transfer of electrons takes place in integral multiple. This is the cause of quantization
4. Infinity
5. Direction of dipole moment is from negative charge to positive charge

Electric Charges and Fields Important Questions Very Short Answer Type Questions

Question 1.
What does q₁ + q₂ = 0 signify in electrostatics ?
Answer:
The equation signifies that the electric charges are algebraically additive and here q₁ and q₂ are equal and opposite.

Question 2.
Why electric lines of force do not form closed loop ? (AH India 2014)
Answer:
Because electric lines of force start from the positive charge and end on die negative charge.

Question 3.
Why must electrostatic field at the surface of charged conductor be perpendicular to every point on it ? (Foreign 2014)
Answer:
As, electric field inside a conductor is always zero. So electric lines of force exert lateral pressure on each other and hence repulsion between like charges take place. Thus, in order to stable the spacing, the lines of force are normal to surface.

Question 4.
What quantity is electric flux ? Write its SI unit.
Answer:
It is a scalar quantity. S.I. unit of electric flux is Nm²/C 

Question 5.
Charge of on QNC is placed as centre of a cube. What will be the electric flux passing through a face of it ?
Answer:

Question 6.
Charges +q₁ -5q₁, +2q and +2q is at Gaussian surface. What will be the magnitude of electric flux passing through Gaussian surface.
Answer:
Zero, because total charges is zero as surface of Gaussian surface.

Question 7.
Fig. shows three charges + 2q, – q and+ 3q. Two charges T 2q and – q are enclosed within a surface S. What is the electric flux through the surface S ?
Answer:

Question 8.
Does charge get changes with the velocity of charged conductor ?
Answer:
No? charge does not get change with velocity of charged conductor.

Question 9.
Which scientist first discovered the charge as positive and negative ?
Answer:
Benjamin Franklin.

Question 10.
Is static electric force is Newtonian between two charges ?
Answer:
Yes, static electric force Newtonian between two charges.

Question 11.
What is the relation between dielectric constant in vacuum and medium ?
Answer:
£ = £₀ × £_r

Question 12.
What is the electric force acting upon charge q placed in electric field E ?
Answer:
qE

Electric Charges and Fields Important Questions Short Answer type Questions

question 1.
Write the limits of inverse square law ?
Answer:

• It is true only for point charge.
• It is only for that charges which is at rest.

Question 2.
Write the basic properties of electric charge ?
Answer:

• Quantization of charge i.e. q = ±ne.
• Charge cannot be created nor be destroyed.

Question 3.
What is dielectric constant of a metal ?
Answer:
The dielectric constant of a metal is ∞

Question 4.
Define intensity of electric field and write its unit.
Answer:
The electric field intensity at a point due to a source charge is defined as the force experienced by a unit positive test charge placed at that point. If \(\overrightarrow{\mathrm{F}}\) is the force acting on test charge qo placed at a point in the electric field of intensity\(\overrightarrow{\mathrm{E}}\) then \(\overrightarrow{\mathrm{E}}\) = \(\frac{\overrightarrow{\mathrm{F}}}{q_{0}}\)
Unit: Its S.I. unit is newton/coulomb.

Question 5.
What is an electric dipole ?
Answer:
A system of two equal and opposite charges kept very close to each other is • called an electric dipole.

Question 6.
What is an electric flux ? Give its SI Units.
Answer:
Electric flux associated with electric field is a measure of total lines of force passing normally to the surface, when held in the electric field. Its SI unit is newton meter per coulomb.

Question 7.
What do you mean by quantization of electric charge ?
Answer:
Quantization of electric charge means that the total charge (q) of a body is always ah integral multiple of basic charge (e) which is the charge on the electron.
.’. Charge on any body, q = ± ne
Where, e = 1.6 x 10^-19 coulomb and n = 0,1,2,3 In above relation, +ve sign denotes the loss of electrons and – ve sign denotes gain of electrons. certain particles are known having these charges which are called quarks.
Charge on any body will never be a fraction of e Like \(\frac { 3 }{ 2 }\)e, \(\frac { 5 }{ 2 }\)e, \(\frac { 7 }{ 2 }\)e. But, now days, certain particles are known having these charges which are called quarks.

Question 8.
Ordinary rubber are insulators but the types of an aeroplane are made a little bit conducting. Why ?
Answer:
When the aeroplane lands and takes off, the friction between types and the run-way may cause the electrification of types. Due to conducting nature of the types, the charge so produced is conducted to the earth and electrical sparking is avoided.

Question 9.
State Gauss’ theorem in electrostatics.
Answer:
The net electric flux passing through any closed surface is \(\frac { 1 }{ { E }_{ 0 } } \) times the net charge present inside it. Mathematically,

Question 10.
The mass of an isolated conductor decreases when some positive charge is given to it. Why ?
Answer:
Mass will reduce because when electrons escape out from the conductor then only it becomes positively charged.

Question 11.
Write the nature of the force acting between q₁ and q₂ if –

1. q₁ q₂ > 0
2. q₁q₂ <0
3. q₁q₂ = 0.

Answer:

1. If q₁q₂ > 0, then both charges are similar and repulsive force acts in between them.
2. If q₁ q₂ < 0, then both charges are of opposite nature and attractive force acts in between them.
3. If q₁q₂ = 0 then one of the two charges is neutral and no force acts in between them.

Question 12.
State principle of conservation of charges. Give examples.
Answer:
According to principle of conservation of charge “Net charge in any isolated system always remains constant”. In the other words “Charge can neither be created nor be , destroyed although it may be transferred from one body to another body”.

Example:
1. Pair annihilation:
When an electron and a positron come near to each other, then they get destroyed and form a y ray photon. Thus the total charge becomes zero.

2. Pair production:
A gamma ray photon splitted into a positron and an electron

Question 13.
Define electric dipole moment. Is it scalar or vector ? Write its SI unit and dimensional formula.
Answer:
Electric dipole moment is the product of magnitude of any one charge of dipole and the distance between both the charges. If + q and – q are two charges placed at a distance 2l apart, then electric dipole moment will be
\(\vec{p}\) = q. 2l

It is a vector quantity whose direction is from negative charge to positive charge.
Unit: Its SI unit is coulomb – meter.
Dimension: [p] = [q][2l] = [ATL]
|p] = [M⁰L¹T¹A¹].

Question 14.
Metallic ropes are attached with the vehicles carrying inflammable substances, connected to earth. Why ? (NCERT)
Answer:
Moving vehicles get charged due to friction. The inflammable material may catch fire due to spark produced by charged vehicle. When metallic ropes or chain is used, the charge developed on the vehicle is transferred to the ground and so the fire is prevented.

Question 15.
When the electric dipole becomes in equilibrium state when kept in electric field?
Answer:
We know that U = – pE cosθ
When the angle between electric dipole and electric field is zero then the electric dipole will be in equilibrium.

Question 16.
State Coulomb’s inverse square law of electrostatic charges. On this basis define unit charge. What is the condition for the law to be applicable ?
Answer:
Coulomb’s inverse square law:
According to it, force of attraction or repulsion between two point charges is directly proportional to the product of their magnitude and inversely proportional to the square of distance between them. Force of attraction or repulsion is the central force which acts along the line joining the two charges. Let the charges q₁ and q₂ be placed at a distance r apart from each other. So, the force acting between them according to Coulomb’s law will be
F ∝ q₁q₂ and F ∝ \(\frac { 1 }{ { r }^{ 2 } } \)

Combining the above two relations, we get

To define unit charge : Let q₁ = q₂ = q, r = lm and F = 9 x 10⁹ newton.
Putting these values in eqn. (1), we get
9 x 10⁹N = 9 × 10⁹ × \(\frac { { q }^{ 2 } }{ { 1m }^{ 2 } } \)
q = ±1 coulomb
So, if two charges of same type and equal magnitude are placed at 1 m apart in vacuum and if force of repulsion between them is 9 × 10⁹ newton, then the two charges are said to be unit charges of magnitude 1 coulomb.

Condition for the applicability of the law:

• It is applicable for the point charges which are stationary’, not for moving charges.
• It is not valid for large distances and distances less than 10^-15 meter because at distances less than 10^-15 m, nuclear forces are dominant.

Question 17.
Define intensity of electric field and write its unit Four properties of electric lines of force.
Answer:
The electric field intensity at a point due to a source charge is defined as the force experienced by a unit positive test charge placed at that point. If \(\overrightarrow{\mathrm{F}}\) is the force acting on test charge q placed at a point in the electric field of intensity \(\overrightarrow{\mathrm{E}}\) then \(\overrightarrow{\mathrm{E}}\) =\(\frac{\overrightarrow{\mathrm{F}}}{q_{0}}\)
Unit: Its SI unit is newton / coulomb.

Properties:
(a) Electric lines of force start from positive charge and end on negative charge.
(b) Tangent drawn at any point of the lines of force gives the direction of electric field intensity.
(c) Two lines of force will never intersect each other.
(d) Lines of force try to contract longitudinally. This demonstrates why unlike charges attract each other.
(e) They are emitted normal to the surface of conductor and end normally.

Question 18.
Why the two lines of force do not cross or intersect each other ?
Answer:
If two lines of force intersect, then there would be two tangents and hence two directions of electric field at the point of intersection, which is not possible.

Question 19.
Prove that electrostatic force is much stronger than the gravitational force.
Or
Compare the electrostatic force and gravitational force.(NCERT)
Answer:
Let a proton and an electron be placed in air or vacuum at lm distance apart, mass of proton and electron are mp and me respectively.
m_p = l.67 × 10^-27 kg
m_e = 9.1 × 10^-31 kg
Charge on proton is q_p = 1.6 ×10^-19C. and charge on electron is q_e = 1.6 x 10^-19C The electrostatic force acting between them is –
\(\frac { 1 }{ { 4\pi E }_{ 0 } }\) \(\frac { { q }_{ p }{ q }_{ e } }{ { Kr }^{ 2 } }\)

The gravitational force between proton and electron is –

Dividing eqn. (1) by eqn. (2), we get

F_e = 2.27 × l0³⁹ × F_G
Thus, the electrostatic force of attraction is 1039 times more than the gravitational force for the electron – proton system. Obviously the electrostatic force acting between two charged particles is very much greater than the gravitational force acting between them.

Question 20.
(a) Define electric flux. Write its SI unit and dimensional formula.
Answer:
Electric flux associated with the electric field is a measure of total lines of force passing normally to the surface, when held in the electric field. It is denoted by Φ_E If the normal drawn to the surface dS makes an angle 0 with the electric field E, then the flux through this area is –
dΦ_E = \(\overrightarrow{\mathrm{E}}\).\(\overrightarrow{\mathrm{dS}}\)
Unit: Its SI unit is Nm²/ C.

Dimension:

Question 20. (b)
Under what conditions flux is said to be positive and negative ?
Answer;
Electric flux:
The electric flux associated with a surface is a measure of total lines of force passing normally to the surface when held in the electric field. It is denoted by ∅_E There are two types of electric flux:

• Positive electric flux – When electric lines of force leaves any body through its surface, it is considered as positive flux.
• Negative electric flux – When electric lines of force enters into body through any surface, it is considered as negative flux.

Question 20. (c)
Show that the electric flux passing through a surface parallel to electric field is zero.
Answer:
Let \(\overrightarrow{\mathrm{E}}\) be the intensity of electric field and \(\overrightarrow{\mathrm{dS}}\) be the normal to the arbitary surface. If the normal to the surface (\(\overrightarrow{\mathrm{dS}}\)) marks an angle θ with E, then,
The electric flux can be given by
d∅_E = E. dS cos θ
If surface is parallel to the electric field
i.e., θ = 90°
d∅_E = E dS cos 90°
= E.dS × 0
d∅_E = 0

Thus flux passing through a parallel surface to an electric field is zero.

Question 21.
State the principle of electrostatic shielding.
Answer:
The electric field inside the charged hollow conductor (spherical shell) is zero. Therefore, when an object is placed inside a hollow conductor then there will be no effect of electric field on the object i.e., the hollow conductor shields the object from external electric field.

The concept of electrostatic shielding is used to shield wires carrying audio signal from external fields like atmospheric electricity or due to electric sparks. If such wires are not shielded, the audio signal will give rise to noisy sound or reproduction.

Question 22.
Derive an expression of electric field intensity on a point in axial position (end on position) of an electric dipole.
Answer:
Consider an electric dipole made up of charges + q and – q separated by a distance 2/apart and placed in vacuum. We have to find out the electric field at point P situated at a distance r from the center of dipole system. To find out the electric field intensity, imagine a unit positive test charge situated at P.

Electric field intensity due to charge (+q) situated at A will be

Electric field intensity due to charge (-q) situated at B will be 1

As AP < PB, hence repulsive effect due to charge +q will be more effect then the attractive effect of charge – q. Therefore, E₁> E₂ but their directions are opposite. Hence, the net electric field will be

This is the expression for the electric field which is directed from A to P. For small and strong dipoles,
r>>2l ⇒ r>>l ⇒ r² >>>l²
∴ r² – l² ≈ r²
Hence, eqn. (1) becomes

This is the required expression.

Question 23.
Derive the expression for the intensity of electric field on the equatorial position of the dipole.
Answer:
Equatorial position is a point on the line which lies on the perpendicular bisector of the axial line, then their position is called equatorial position or broadside on position. Let AB be an electric dipole formed by charges +q and -q. The distance between the charges is 21. P is a point on the equatorial position of the dipole, situated at a distance r from the center of the dipole. Sup- pose unit positive test charge is situated at P. Electric field at P due to +q charge will be

Electric field at P due to -q charge will be


Resolving E and E₂ into its components, we get the component of E₁ are E₁cosθ parallel to AB and E₁ sin θ along OP. Also, components of E₂ are E₂ cosθ parallel to AB and E₂ sin O along PO.
From eqns. (1) and (2), we find |E₁| = |E₂|, hence E₁ sinθ = E₂ sinθ , but their directions are opposite. So, they cancel each other. The only resultant electric field will be
E = E₁ cos θ + E₂ cos θ
= E₁ cosθ + E₁ cos θ
E = 2E₁ cos θ  …………(3)

If the charges are kept in any other medium other than vacuum, then the electric field becomes

The direction of electric field is along AB i.e., it is parallel to axial line which acts from positive charge to negative charge. For small and strong dipoles, r >> 2l ⇒ r >> l
r2 + l2 ≈ r2
Hence, eqn. (4) becomes

Question 24.
Obtain an expression for the intensity of electric field due to a charge q at a distance r from it.
Answer:
Consider a point charge q which is placed at the origin O of the generalized coordinate system. We have to find out the electric field at point P at a distance r from it. For this, imagine a test charge q₀ placed at point P. Both the charges q and q₀ have been placed in vacuum. According to Coulomb’s law, the force on charge q₀ will be

Where, r is the unit vector in the direction from q to q₀. But intensity of electric field is given by the formula,

If any other medium is present having dielectric constant then

Where, £0 = k = dielectric constant of the medium. In magnitude,

Question 25.
Derive the expression for the torque acting on the dipole when kept in a uniform electric field and hence define dipole moment.
Answer:
Consider an electric dipole made up of charges +q and -q and length 2l which is placed in a uniform electric field of strength E. The dipole moment P makes an angle Q with respect to the electric field at any instant of time. As a result, a pair of forces +qE (along the direction of electric field) and -qE (opposite to the direction of electric field) acts on the dipole. These two opposite forces are of equal magnitude. So, they apply a couple on electric dipole, which tries to bring it along the direction of electric field. This couple is called restoring couple.

So, moment of couple or torque (τ)
= Magnitude of any force x Perpendicular distance between the forces
= qE × BC …….(1)
In A ABC, sinθ = \(\frac { BC }{ AB }\)
⇒ BC = AB sinθ = 2l sinθ
Putting the value of BC in eqn. (1), we get,
τ= qE × 2l sin θ
⇒ τ = pEsin θ,
(∴q 2l = p, where p is the dipole moment) ……..(2)
In vector form,
\(\overrightarrow{\mathrm{τ}}\) = \(\overrightarrow{\mathrm{p}}\) × \(\overrightarrow{\mathrm{E}}\)
From eqn. (2), if E = 1 and 0= 90°.Then,
τ = p sin 90° = p
Thus, dipole moment of an electric dipole is numerically equal to the torque acting on the dipole when it is placed perpendicular to a unit electric field.

Question 26.
Derive an expression for the potential energy of a dipole in an electric field.
Answer:
Work done in rotating any electric dipole from standard position to any angle with respect to the electric field is called potential energy. The standard position is considered to θ = i.e., perpendicular to the field. Let p be the dipole moment of the dipole, placed in an electric field of intensity E and it makes an angle θ with respect to the electric field, then
Torque, τ = p E sinθ
Also, work done in displacing the dipole through an angle dθ will be
dW = τdθ = pE sinθ dθ
Now, the work done in displacing from θ = \(\frac { π }{ 2 }\) to θ = θ is

This work done is stored in the form of potential energy,
U = – p E cosθ
In vector form, U = \(\overrightarrow{\mathrm{-p}}\).\(\overrightarrow{\mathrm{E}}\)
This is the required expression.

Question 27.
Derive the expression for the amount of work done in rotating a dipole in a uniform electric field.
Answer:
If a dipole placed in a uniform electric field is rotated from the equilibrium position, then work has to be done. Consider an electric dipole of dipole moment p(= 2 ql) which has been rotated in an electric field E by an angle 0 from the equilibrium position, then torque acting on it will be –
\(\overrightarrow{\mathrm{τ}}\) = \(\overrightarrow{\mathrm{-p}}\) × \(\overrightarrow{\mathrm{E}}\) = pE sin θ, (in magnitude form)
Work done in rotating the dipole through an angle dθ is
dw = τ.dθ
Therefore, work done in rotating it by an angle θ can be obtained from the above equation under proper limits.

Above equation is the total work done in rotating any electric dipole through an angle θ

Question 28.
Write and prove Gauss’ theorem.
Or
Prove that net electric flux through any closed surface is 1 / £₀ times of total charge present inside it, where s £₀ is
permittivity of free space.
Answer:
Gauss’ law:
According to Gauss’ law, the net electric flux through any closed surface is 1 / £₀ times of the total charge present inside it.

Where, £₀ is permittivity of vacuum or air.
Proof:
Let at any point O, inside any closed surface, a charge +q is present. A point P lies at a distance r from O on surface dS. Normal to the surface dS can be represented by \(\overrightarrow{\mathrm{dS}}\) (say). Therefore, electric field at point P due to charge +q can be given by

.’. Electric flux passing through area \(\overrightarrow{\mathrm{dS}}\)will be
dΦ_E = \(\overrightarrow{\mathrm{E}}\).\(\overrightarrow{\mathrm{dS}}\) = E dS cos0

Where, 0 is an angle between electric field f and the normal to the surface.
Net flux through the closed surface is.

But \(\frac { dscos\theta }{ { r }^{ 2 } } \) = dw is the solid angle subtended by the small surface dS at O.

As total solid angle subtended by any closed surface at any point inside it is 4π hence –

Above equation is mathematical form of Gauss’ law.

Electric Charges and Fields Important Questions Long Answer Type Questions

Question 1.
Obtain an expression for the intensity of electric field due to linear charge distribution.
Answer:
Consider a thin wire of length ‘l’ Let the charge given to it be q. So, the linear charge density will be λ = \(\frac { q }{ l }\) ⇒ q = λl
If a test charge ¡s brought near to the wire, then it will be repelled. Hence, the electric field (\(\overrightarrow{\mathrm{E}}\)) will be radially out ward. Draw a normal \(\overrightarrow{\mathrm{dS}}\) on the surface at P. The direction of \(\overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{dS}}\) are the same. The electric flux passing through the curved surface of the cylinder is

= E 2πrl (∴\(\overrightarrow{\mathrm{dS}}\) = Total surface areaofcurved surface 2,πrl …(2)
But according to Gauss’ theorem,
Φ = \(\frac { 1 }{ { E }_{ 0 } } \) x (Total charge present inside the Gaussian surface) …(3)
Φ = \(\frac { q }{ { E }_{ 0 } } \)
From eqns. (2) and (3), we have
E 2πrl = \(\frac { q}{ { E }_{ 0 } } \)
E 2πrl = \(\frac { λl}{ { E }_{ 0 } } \)
[∴q = λl from eqn. (1)]

This is the expression for the electric field due to a linear charge distribution having a linear charge density λ.

Question 2.
Derive an expression for the intensity of electric field due to a uniform infinite plane sheet of charge.
Answer:
Consider an infinite plane sheet on which charge given is q. If S is the total surface area of the sheet, then the surface charge density will be σ = \(\frac {q }{ S}\) = \(\frac { charge }{ area }\) the charge will be distributed uniformly on the surface of the sheet. Hence, surface charge density (= σ) remains constant. Any test charge held near to the surface will be repelled on both the sides i.e., either to paper and the other will go inside, hence the Gaussian surface formed will be cylindrical; which is having three surfaces S₁, S₂ and S₃.

The electric field \(\overrightarrow{\mathrm{dS}}\) is directed radially outwards away from the sheet. Moreover, \(\overrightarrow{\mathrm{dS}}\) is the outward drawn normal on the Gaussian surface. Hence, the total electric flux is

The angle between normal to the curved surface S3 in the direction of electric field \(\overrightarrow{\mathrm{E}}\)] is 90°, hence

For the surfaces Sj and S2, \(\overrightarrow{\mathrm{E}}\) and\(\overrightarrow{\mathrm{dS}}\) are along the same direction, hence the angle between \(\overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{dS}}\) is 0°, so

By Guss’ theorem,
Φ_E = \(\frac { 1}{ { E }_{ 0 } } \) q ……(2)
from eqns. (1) and (2) we get
2ES = \(\frac { q}{ { E }_{ 0 } } \)
E = \(\frac{q}{2 \varepsilon_{0} S}\)
Hence, E = \(\frac { σ}{ { 2E }_{ 0 } } \)
(∴ σ = \(\frac { q }{ S }\))
This is the expression for the intensity of electric field due to a uniformly charged Sheet

Question 3.
Prove that intensity of field is zero inside a spherical shell, while in outside it is such that all the charges are concentrated at the center.
Or
Determine the intensity of electric field by Gauss’ law, due to a uniformly charged spherical shell at a point

1. Outside the shell
2. On the surface of shell
3. Inside the shell.

Draw a graph showing the variation of electric field with the distance from the center.
Answer:
Intensity of electric field due to uniformly charged spherical shell:
Suppose that a sphere of radius R is uniformly charged with +q charge. Intensity at any point due to this charged sphere depends on its position relative to the sphere.

(i) When point P lies outside the spherical shell:
Consider a point P which lies outside the spherical shell of radius R. Now, imagine a sphere of radius r which passes through P. This closed surface behaves as Gaussian surface. Since, sphere is uniformly charged, so that electric field at each point on the surface is equal and points radially outward.

Net flux through the Gaussian sphere of radius r will be

But, charge enclosed inside the Gaussian surface is q. Therefore, from Gauss’ law
Φ_E = \(\frac { q }{ { E }_{ 0 } } \) …….(2)
Equating eqns. (1) and (2), we have
E. 4πr² = \(\frac { q }{ { E }_{ 0 } } \)
E = \(\frac { 1 }{ { 4\pi E }_{ 0 } }\) \(\frac { q }{ { r }^{ 2 } } \) ……..(3)

It is clear from above equation that, the electric field outside any uniformly charged spherical shell is exactly to that electric field when we take the same charge at the center of sphere.

(ii) Point P lies on the surface of spherical shell:
In this case, r = R
From eqn. (3),
E = \(\frac { 1 }{ { 4\pi E }_{ 0 } }\) \(\frac { q }{ { R }^{ 2 } } \)

(iii) When point P lies inside the spherical shell:
As we know, the charge given to spherical shell spreads equally all over the surface, there is no charge present inside the sphere. Hence, by Gauss’ law,


From equs. (4) and (5),
E.4πr² =  0
⇒ E = 0
Therefore, electric field inside a spherical shell is always zero. The adjacent figure shows the variation of electric field with distance.

Question 4.
Obtain an expression for the intensity of electric field due to a solid sphere of charge using Gauss’ theorem at the following points:

1. Outside the sphere
2. On the surface of sphere
3. Inside the sphere.

Also prove that the intensity of the electric field at the center will be zero.
Answer:
Suppose that a sphere of radius R is uniformly charged with +q charge. Intensity of electric field at any point depends on its position with respect to sphere. The volume charge density of sphere is
p = \(\frac { Charge }{ volume }\)

(i) When the point lies outside the sphere:
Consider a point P which lies outside the spherical shell of radius R. Now imagine a sphere of radius r which passes through P.This closed surface behaves as a Gaussian surface. The electric field at each point outside the sphere points radially outwards.

.’. Total flux through the Gaussian sphere of radius r will be –

But charge enclosed inside the Gaussian surface is q. Hence, from Gauss’ law
Φ = \(\frac { 1}{ { E }_{ 0 } } \) q …..(3)
From eqns. (2) and (3),
E 4πr² = \(\frac { q}{ { E }_{ 0 } } \)
E = \(\frac { q }{ { 4\pi { E }_{ 0 }r }^{ 2 } }\)
Hence, the electric field outside any uniform charged sphere is same as if the entire charge is present at the centre.
Putting the value of q from eqn. (1) in eqn. (4), we get

(ii) When the point lies on the surface of sphere:
r = R
From eqn. (4), we have E =

and from eqn. (5), we have E =

(iii) When the point lies inside the sphere:
Suppose the point P lies at a distance r from the center of the sphere such that r < R. At radius r draw a Gaussian surface which is a sphere. Then electric field \(\overrightarrow{\mathrm{E}}\) is radially outwards. Moreover, normal to surface \(\overrightarrow{\mathrm{dS}}\) is along the direction of \(\overrightarrow{\mathrm{E}}\). Hence, angle between \(\overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{dS}}\) is 0°
The flux coming out of the Gaussian surface is –

But by Gauss’ law, the flux Φ = \(\frac { 1}{ { E }_{ 0 } } \) x (Total charge inside the Gaussian surface)
Φ = \(\frac { q}{ { E }_{ 0 } } \) q ……(2) (where q’ is the total charge inside the sphere of radius r)
On volume \(\frac { 4 }{ 3 }\) πR³charge given is q
On volume 1, charge given is = \(\frac { q }{ \frac { 4 }{ 3 } \pi R^{ 3 } } \)
On volume\(\frac { 4 }{ 3 }\) πR³ nr charge given is = \(\frac { q }{ \frac { 4 }{ 3 } \pi R^{ 3 } } \) x \(\frac { 4 }{ 3 }\) πR³ = \(\frac { q{ r }^{ 3 } }{ { R }^{ 3 } } \)
q’ = \(\frac { q{ r }^{ 3 } }{ { R }^{ 3 } } \)
‘Putting the value of q’ in.eqn. (2), we get
Φ = \(\frac { q}{ { E }_{ 0 } } \) \(\frac { q{ r }^{ 3 } }{ { R }^{ 3 } } \) …….(3)
from equs. (1) and (3) we get,

At the center of sphere, r = 0, hence E = 0.

Question 5.
Neeta return home from school. When she was taking out the sweater, she heard twinkling sound. She asked her mother but she was unable to answer.Next day she went to school and asked the science teacher, she was satisfied with her answer. Write down the answer of the question given below on the basis of above statement:

1. What values does she show ?
Answer:
She shown the value of answer, questioning.

2. What was the reason for the sound when she was taking out her sweater ?
Answer:
Due to the friction between her school dress and sweater, its acquire negative charge which get discharge. Therefore the sound was been listen by her.

3. Presence of charge can be determined by which apparatus ?
Answer:
Gold leaf electroscope.

4. Does any new charges get created on rubbing ? If yes or not than why ?
Answer:
No, new charges get created, only transfer of electron occur form one body to another. There one become positive charge and other become negative charge.

Electric Charges and Fields Important Questions Numerical Questions

Question 1.
Two positive ions which have same charges having a force of repulsion of
3.7 × 10^-9N. The distance between them is 5 Å. Find the deficiency of electron on each ions.
Solution Given:
F = 3.7 × 10^-9N, r = 5Å = 5 × 10^-10m
q₁ = q₂ = q

= 10.28 × 10³⁸
q = 3.2 × 10^-19
n = \(\frac { q }{ e }\)

There will be deficiency of two electron in each ions.

Question 2.
Calculate force of repulsion between two protons when they are separated with 4.0×10¹⁵m.
Solution:
We know that, \(\frac { 1 }{ 4\pi { E }_{ 0 } } \) = 9 × 10⁹Nm²C^-2
Given: q₁= q₂ = +1.6 × 10^-19 C,r = 4.0 × 10^-15m.
From formula, F = \(\frac { 1 }{ { 4\pi E }_{ 0 } }\) \(\frac { { q }_{ 1 }{ q }_{ 2 } }{ { Kr }^{ 2 } }\)

or F = 14.4 newton.

Question 3.
Can 1 coulomb of charge be gives to a metallic sphere of radius 1cm? Give reason.
Solution:
Here, q = 1 C, r = lcm = 10^-2m
Electric field on the surface of sphere is
E = \(\frac { 1 }{ { 4\pi E }_{ 0 } }\) \(\frac { q }{ { r }^{ 2 } } \)
= 9 × 10⁹ × \(\frac { 1 }{ { \left( { 10 }^{ 2 } \right) }^{ 2 } } \) = 9 × 10³ N /C
But when intensity of electric field exceeds 3 × 10⁶ N / C air gets ionized and hence, electric discharge takes place. So l Cm of charge cannot be given to a sphere of radius 1 cm.

Question 4.
Two charge of one coulomb are separated by a distance of 1 meter from each other, find:

1. Force between the charge when was in air
2. If it is in dielectric medium, calculate its force.

Solution:
Given:
q₁= q₂ = 1 coulomb, r = l m

Question 5.
Two point charges + 9e and + e are placed 8 m apart. Where should the third charge q be placed on the line joining the two charges so that q should be in equilibrium?
Solution:
Let the charge q be placed at distance x from +9 e charge, then its distance from e will be 8 – x.
In equilibrium,


Taking + ve sign \(\frac { 3 }{ x }\) = \(\frac {1 }{ 8 – x}\)
⇒ 24 – 3x = x
24 = 4x or x = 6m
Taking – ve sign, \(\frac { 3 }{ x }\) = – \(\frac {1 }{ 8 – x}\)
⇒ 24 – 3x = -x
24 – 2x or x = 12 m
Since, +9e and+e are similar charges, hence q will be in equilibrium when x = 6 m i.e., q should be placed 6 m apart from +9e between the charges.

Question 6.
What is the force between two small sphere having charge of 2 × 10^-7 C and 3 × 10^-7C ? (NCERT)
Solution:
Given: q₁ = 2 × 10^-7C,
q₂ = 3 × 10^-7C,
and r = 30 cm = 30 × 10^-2
m = 0.3 m

The force will be repulsion force a both the charge are same.

Question 7.
The electrostatic force on a small sphere of charge 0.4µC due to another small sphere of charge 0.8µC in air is 2 N.

1. What is the distance between two sphere ?
2. What is the force on the second sphere due to the first ?

Solution:
Given:
q₁ = 0.4µC = 0.4 × 10^-6 C
q₂ = -0.8µC = -0.8 × 10^-6 C
F = 0.2N

= 9 × 2 × 8 × 10^-4
r = 3 × 4 × 10^-2
= 0.12 m = 12 cm.

(ii) Both the charges apply same force on each other. There force of first sphere due to second sphere will be 0.2 N. The force between them will be force of attraction.

Question 8.
A polythene piece is rubbed with woollen cloth as a result 3 x 10.7 coulomb charge is produced in it, then: .

1. How many electrons are transferred from which to which substance ?
2. Is mass transferred from woollen to polythene ?

Solution:

1. Polythene gets negative charge, so that electron is transferred from woollen to polythene.
We know that, q = ne

or n = 1.875 × 10¹² electrons.

2. As electrons are transferred from woolen to polythene, mass is also transferred in the same way.

Question 9.
A conductor has 14.4 × 10¹⁹ C positive charge. How many electrons are lacking or in excess ?
Solution:
We know that, q = ne
⇒ n =\(\frac { q }{ e}\)

Therefore, body is lacking by 9 electrons (as it has positive charge).

Question 10.
What is the net flux of uniform electric flux there through a cube of field oriented so that its faces are parallel to the coordinate plane ?
Solution:
All the faces of a cube are parallel to the coordinate axis. Therefore the Number of hold lines entering the cube is equal to the number of field lines piercing out of the cube. As a result net flux through the cube in zero.

MP Board Class 12th Physics Important Questions

MP Board Class 12th Biology Important Questions Chapter 12 Biotechnology and its Applications

Biotechnology and its Applications Important Questions

Biotechnology and its Applications Objective Type Questions 

Question 1.
Choose the correct answer:

Question 1.
Restriction endonuclease is employed for cutting :
(a) A single stranded DNA
(b) Double stranded DNA
(c) RNA fregment
(d) m – RNA.
Answer:
(d) m – RNA.

Question 2.
Which enzyme is helpful in genetic engineering:
(a) RNA polymerase
(b) DNA polymerase
(c) Restriction endonuclease
(d) Nuclease.
Answer:
(c) Restriction endonuclease

Question 3.
Extrachromosomal DNA used as vector in gene cloning is :
(a) Transposon
(b) Intron
(c) Exon
(d) Plasmid.
Answer:
(d) Plasmid.

Question 4.
The most extensively used bacteria in genetic engineering is :
(a) Bacillus
(b) Clostridium
(c) Escherichia
(d) Salmonella.
Answer:
(c) Escherichia

Question 5.
Hybridomas are:
(a) Antibodies
(b) Hybrid swarms
(c) Hybrid cultures
(d) Hybrid zone.
Answer:
(c) Hybrid cultures

Question 6.
Vaccine can not protect us from :
(a) Malaria
(b) Polio
(c) Small pox
(d) Tuberculosis.
Answer:
(a) Malaria

Question 7.
Humulin is:
(a) A type of chitin
(b) New digestive enzyme
(c) Powerfull antibiotic
(d) Human insuline
Answer:
(a) A type of chitin

Question 8.
Which is not used in the form of bio weapons :
(a) Bacillus anthrasis
(b) Bacillus thuringiensis
(c) Botulinum toxin
(d) Small pox.
Answer:
(b) Bacillus thuringiensis

Question 9.
Super bug is called :
(a) Pseudomonas pulida
(b) Xanthomonas citrae
(c) Bacillus anthresis
(d) Streptococcus pneumoniae
Answer:

Question 10.
Biopiracy affected plants are :
(a) Neem
(b) Basmati rice
(c) Turmeric
(d) All of these.
Answer:
(d) All of these.

Question 11.
Golden rice will help in :
(a) Producing petrol like fuel
(b) Pest resistance
(c) Herbicide tolerance
(d) Alliviation of Vit – A deficiency
Answer:
(d) Alliviation of Vit – A deficiency

Question 12.
During gene cloning which is called as “gene taxi” :
(a) Vaccine
(b) Plasmid
(c) Bacterium
(d) Protozoa.
Answer:
(b) Plasmid

Question 13.
In transgenics, expression of transgene in target tissue is determied by.
(a) Transgene
(b) Promoter
(c) Reporter
(d) Enhancer.
Answer:
(b) Promoter

Question 2.
Fill in the blanks :

1. Bt is a ………………………..
2. Bt toxin is formed from ……………………….. by the bacillus.
3. The prooduct of Cry – 1 gene is used for ………………………..
4. RNA interpritation is the method of cellular conservation in ……………………….. genes.
5. Both peptide chains are joined by the ……………………….. in insuline.
6.  ……………………….. is the found in proinsuline which is not found in muture insulin.
7.  ……………………….. is found in transgenic cow ‘Rosi’, it is not found in commen cow’s milk.
8. Flavr savr tomato is a transgenic veriety of ………………………..
9.  ……………………….. has two polypeptide chain and 51 amino acids.
10.  ……………………….. gene which found in Rhizobium is converted from nitrogen into nitrates.
11.  ……………………….. is the reason of male infertility in maize.
12. First transgenic organism is petant in ………………………..
13. Monoclsonal antibody is produce with the help of ……………………….. technology.

Answer:

1. Bio Toxin
2. Protoxin
3. Corn bores
4. Eukaryotic
5. Three sulphaid bundles
6. Peptide ‘c’
7. Human α – heredity lectalbumen
8. Tomato
9. Insulin
10. Nif gene
11. Cytoplasmic heredity
12. 1989
13. Hybridoma.

Question 3.
Match the followings:
I.

Answer:

1. (e)
2. (d)
3. (a)
4. (b)
5. (c)

II.

Answer:

1. (d)
2. (a)
3. (b)
4. (c)

Question 4.
Write the answer in one word/sentances:

1. Write the full name of RNAi.
2. Bt Toxic is encrypted by this gene.
3. Is our blood has proteoses and nucleases.
4. Who produce Bt toxin protein?
5. What is humulin?
6. What is SCID (Severe Combined Immunodeficienay)?
7. Which is controled by Trycoderma?
8. Name the organism which bear the other organism obtained gene?
9. Bt. cotton and Golden rice is the examples of.
10. Name the gene which controles the cell cycle?

Answer:

1. RNA interference
2. By Cry genes
3. No
4. Bacillus thuringiensis
5. Human insulin
6. Hereditary disease
7. Biological fungal disease
8. Transgenic organism
9. Transgenic crop
10. p53 gene.

Biotechnology and its Applications Important Questions Very Short Answer Type Questions

Question 1.
Can you suggest a method to remove oil (hydrocarbon) from seeds based on your understanding of r – DNA technology and chemistry of oil?
Answer:
To remove oil from seeds using recombinat DNA technology would involve:

1. Identifying the genes that code for oil production.
2. Deleting these genes from the seed genome.
3. Splicing back together the remaining DNA.
4. Putting it back into the cell.

It will be not be very easy because the oils are made up of fatty acids and glycerol. Since, fatty acids are important components of cell membrane system, deleting or switching off of its genes might affect the cell structure itself.

Question 2.
Find out from internet what is golden rice?
Answer:
Golden rice is a variety of rice produced through genetic engineering to biosynthesize β – carotene, a precursor of vitamin ‘A’ in the edible parts of rice. It is intended to produce a fortified food to be grown and consumed in areas with a shortage of dietary vitamin ‘A’.

Question 3.
Does our blood have proteases and nucleases?
Answer:
No.

Question 4.
Name the organism whose genetic material has been altered using genetic engineering techniques.
Answer:
Genetically Modified Organisms (GMO).

Question 5.
Name the crops which are prepaired with the help of biotechnology.
Answer:
Bt cotton, Bt maize, paddy, tomato, potato and soyabeen.

Question 6.
Through whom Bt toxin protein originates?
Answer:
By Bacillus thuringiensis.

Question 7.
By which Bt toxin is coded?
Answer:
By Cry genes Bt toxin is coded.

Question 8.
Full form ofRNAi.
Answer:
RNA interference (RNAi).

Question 9.
Name the therapy which is help of missing of defective ones in order to correct genetic disorders.
Answer:
Gene therapy.

Question 10.
Name the scientific name of bacteria in which be form organism toxin.
Answer:
Bacillus thuringiensis.

Biotechnology and its Applications Important Questions Short Answer Type Questions

Question 1.
What is genetically modified (GM) food? Give two examples.
Answer:
Genetically Modified (GM) food:
The food substances produced from genetically modified crops or transgenic crops is called GM food. This food differ from conventionally developed varieties in the following aspects :

1. GM food contains antibiotic resistance gene itself.
2. It contains protein produced by transgene, example Cry protein in insect resistance varieties.
3. These GM foods contain enzyme produced by the antibiotic resistance gene that was used during gene transfer by recombinant DNA technology.

Examples of GM Crops, Food and Fruits:

1. Flavr Savr Tomato:
It is the first food containing genetically engineered DNA. These tomatoes contain genes for antibiotic resistance for kanamycin.

2. Maize:
GM maize has a bacterial gene which increases its resistance to pests and diseases. It also has a gene for ampicillin resistance which is harmful for us, therefore introduction of GM maize is opposed by many European countries.

3. Rape oil seed:
It is a new type of plant that contain genes for resistance to the herbicide Basta. It has for more potential, dangers and can become a weed and would be impossible to control with Basta. It could cross fertilize with relatives such as wild mustard, thus, spreading the resistance to wild plants. Such type of environmental risks could occur with genetically modified rapeseed crop. They might also effect food chains in unpredictable ways.

Question 2.
Write down the advantage of GM Crops.
Answer:
Advantages of GM Crops:

• Genetic modification has made crops more tolerant to abiotic stresses (cold, drought, heat, salt.)
• Viral resistance can be introduced.
• Over ripening losses can be reduced. Example : Flavr Savr Tomato.
• Enhanced nutritional value of food. Example : Golden Rice.
• Reduced reliance on chemical pesticides.

Disadvantages of GM crops:

• Transgenes in crop plants can endanger native species. Example : The gene for Bt toxin expressed in pollen may end natural pollinators such as honeybees.
• Weeds also become resistant.
• Products of transgenes may be allergic or toxic.
• They cause damage to the natural environment.

Question 3.
What is perfect agriculture? How is this method better than traditional method? Explain.
Answer:
Perfect agriculture is a method of agriculture which is sustainable, perfect and harmless. Green revolution and there after the production of agricultural crops has definately increased due to use of new and high yielding varieties, development of irrigation facilities, increased irrigated area, use of fertilizers etc. but it results many problems such as loss of soil fertility, pollution of food and water and diseases. The resistance power of plants and human beings falls slowly.

Food and water borne diseases affecting the health of human beings and animals. All of these conditions and events taking place due to the modem commercial agriculture. (MPBoardSolutions.com) Therefore, it would become necessary to develop a method of agriculture which would be free from above mentioned demerits. This kind of agriculture is called to be as perfect or sustainable agriculture. Organic agriculture is the best example of perfect agriculture.

Question 4.
What is organic cropping? What are its basis?
Answer:
Organic agriculture:
Organic agriculture is a method of agriculture which does not allow the use of synthetic fertilizers, pesticides, insecticides, weedicides, plant growth regulators, substances of animal origin and genetically modified bacteria. In this method biofertilizers, crop rotation methods are used to increase crop production and biopesticides are used to control insects and weeds. Thus, organic farming is a holistic way of agriculture which tries to bridge the widening gap between man and nature.

If has the commitment of meeting production needs on one hand and sustaining resources and ecosystem function on the other hand. Thus, organic farming is an alternative agriculture production system which avoid or largely excludes the use of synthetic chemicals, fertilizers, pesticides and growth regulating hormones and live stock additives.

Basics of organic agriculture:

1. Organic agriculture is based on improvement of soil, plants, animals, man and global scinery and make it sustainable.
2. Organic agriculture is based on those ecosystems and biocycles which utilize that organisms which would be promoted.
3. It is based on the principle which are related with making pollution free environment and possibilities of life.
4. It is also based on saving environment and health of present and future generations.

Question 5.
What is eugenics? Write importance of eugenics.
Answer:
Eugenics: The branch of biology which deals with the study of improvements of human race is called eugenics.

Importance:

1. Development of selective reproduction in similar species.
2. Transfer of genetic materials in various organisms.
3. Development ofGM food and GM crops.
4. Gene cloning.
5. Gene therapy, etc.

Question 6.
Crystals of Bt toxin produced by some bacteria do not kill the bacteria themselves because:
(a) Bacteria are resistant to the toxin,
(b) Toxin is immature
(c) Toxin is inactive
(d) Bacteria enclose toxin in a special sac.
Answer:
(c) Toxin is inactive : In bacteria, the toxin is present in an inactive form called prototoxin. This gets converted into the active form when it enters the salivary gland of insects having alkaline medium.

Question 7.
What are transgenic bacteria ? Illustrate using any one example.
Answer:
The becteria whose DNA is manipulated to carry and express a foreign DNA is called transgenic bacteria. These microbes are used for producing important bio – chemicals. They have been synthesizing alcohol, enzymes, steroids and antibiotics. Example  Bacillus thuringiensis for Bt cotton, hirudin from transgenic Brassica napus seed. Hirudin is a protein which prevents blood clotting. Its gene was chemically synthesized and introduced in Brassica napus, in which hirudin accumulates in the seed from where it is extracted, purified and used as a medicine.

Question 8.
Compare and contrast the advantages and disadvantages of production of genetically modified crops.
Answer:
Advantages of GM Crops:

1. Genetic modification has made crops more tolerant to abiotic stresses (cold, drought, heat, salt.)
2. Viral resistance can be introduced.
3. Over ripening losses can be reduced. Example : Flavr Savr Tomato.
4. Enhanced nutritional value of food. Example : Golden Rice.
5. Reduced reliance on chemical pesticides.

Disadvantages of GM crops :

1. Transgenes in crop plants can endanger native species. Example : The gene for Bt toxin expressed in pollen may end natural pollinators such as honeybees.
2. Weeds also become resistant.
3. Products of transgenes may be allergic or toxic.
4. They cause damage to the natural environment.

Question 9.
What are Cry proteins? Name an organism which produces it. How has man exploited this protein for his benefit?
Answer:
Cry proteins are toxic proteins (insecticidal proteins) secreted by Bacillus thuringiensis in crystal form during a particular phase of their growth. The toxin is coded by a gene called Cry. The genes encoding cry proteins called Bt toxin genes were isolated from B. thuringiensis and incorporated into several crop plants such as Bt cotton, Bt com etc. to provide resistance against insect pests.

Question 10.
What is gene therapy? Illustrate using the example of adenosine deaminase deficiency (ADA).
Answer:
It is a collection of methods which allows correction of a gene defect that has been diagnosed in a child or embryo. In gene therapy, normal genes are inserted into a person’s cells or tissues to treat a hereditary defect. Gene therapy is being tried for sickle cell anaemia and Severe Combined Immuno Deficiency (SCID).

In some children, ADA deficiency can be cured by bone marrow transplantation. In others, it can be treated by enzyme replacement therapy, in which functional ADA is given to the patient by injection. (MPBoardSolutions.com) However, both of these approaches are not completely curative. In gene therapy, lymphocytes from the blood of the patient are grown in culture outside the body.

A functional ADA cDNA (using a retroviral vector) is then introduced into these lymphocytes, which are subsequently returned to the patient. Because these cells are not immortal, the patient requires periodic infusion of such genetically engineered lymphocytes. However, if the gene isolated from marrow cells producing ADA is introduced into cells at early embryonic stages, the disease could be cured permanently.

Question 11.
Consult internet and find out how to make orally active protein pharma – eeutical What is the major problem to be encountered?
Answer:
For making any oral drug or nutritional supplement, the action of digestive enzymes has to be taken into account. Most of the antibiotics and vitamin supplements are made in capsule form to prevent the action of HC1 in the stomach. For protein preparation, the major source is groundnut shells. The protein extracted from the source is predigested so, as to make it absorbable by the digestive system.

Biotechnology and its Applications Important Questions Long Short Answer Type Questions

Question 1.
Diagrammatically represent the experimental steps in cloning and expressing a human gene (say the gene for growth hormone) into a bacterium like E. coli.
Answer:

Question 2.
Explain the following in brief:

1. Biopiracy
2. Biopatent

Answer:
1. Biopiracy:
Some organisations and multinational companies exploid biological resources and genetical resources indegenous to a country without proper authorisation. This is called biopiracy. In fact it is illegal removal of biological material. The process of biopyracy involves collection of samples of biological sources, which can be done unnoticed. This biological material is then subjected to product development for use on a commercial scale.

Today a range of biological resources are facing biopiracy. It includes plants and animals, microorganisms genetic materials etc. Western companies are getting great benefits from using the knowledge and biological resources of the third world communities. (MPBoardSolutions.com) While the companies stand to make huge revenue from this process, the local communities are unrewarded and infact, may have to buy the products of these companies at high prices.

To check illegal exploitation of biological resources Government of India has signed the General Agreement on Tariffs and Trade (GATT), which opens country natural resources for foreign exploitation.

2. Biopatent:
The protection given by government to an inventor of biological material to secure him for a specific time the exclusive right of manufacturing, exploiting, using and selling of an invention is called biopotent.

Today manufacturing companies are being granted patents for products and technologies that make of biological resources, such as plants and animals, genetic materials which was identified developed and used by farmers and indegenous peoples.

There is growing worldwide opposition to the granting of patents on biological materials such as genes, plants, animals and human. Farmers and indegenous peoples are outaged that plants that they developed are being ‘hijacked’ by companies. Groups are diverse as religious leaders, parliamentarians and environment NGOs are intensifying campaign against corporate patenting of living things.

Question 3.
What is gene library?
Answer:
Gene library:
Several clones of cells, each clone containing one or a few foreign genes representing almost all the genes of an organism is referred to as genes library. From this gene library it is possible to identify a clone containing gene of interest. In order to obtain gene library of an organism, its genome is first cut into smaller DNA fragments containing one or a few genes such as fragments can be cloned into a cell which may be that of bacteria, yeast, insects, plant or animal cell.

When such a cell multiplies to form a group of cells, all cells will contain the same foreign DNA fragment which was introduced initially. (MPBoardSolutions.com) These cells which have similar foreign DNA fragment are referred to as a clone of cells. Several clones of cells each clone containing one or a few foreign genes are finally obtained and is called gene library.

Question 4.
What is genetic engineering? Describe the application of genetic engineering in the field of Agriculture and Medicine.
Answer:
The branch of biology by which important and permanent changes are created in plants and animals by increasing or decreasing manipulation and transplantation of genetic units is called genetic engineering.

(A) Application of Genetic engineering or Biotechnology in Agriculture: Genetic engineering is found to be very beneficial in agriculture. Its important use in agriculture are:

1. Increase in photosynthetic efficiency:
An increase in photosynthetic efficiency of crop plants can be achieved by introducing suitable Carbon dioxide Fixation Gene (cfx) from any plant into the crop plants.

2. Transfer of nitrogen fixing ability:
Number of symbiotic and non-symbiotic micro¬organism have capacity of fixing atmospheric nitrogen. Nitrogen fixers are found to possess nitrogen fixing gene (nif genes) which are located on chromosomes or plasmids. Introduc¬tion of nif gene in crop plants results in ability in crop plants to fix atmospheric nitrogen and reduction in the use of chemical nitrogen fertilizers.

3. Disease resistance in crop plants:
Plant breeders at present are developing high yield varieties by transferring gene for disease resistance through conventional breeding.

4. Plant tissue in crop improvement:
Some of the areas of plant improvement where tissue culture has been applied with success are as follows :

• Rescuing hybrids through embryo culture.
• Multiplication of germplasm.
• Production of disease free plants.
• Production of haploid through another culture.
• Somaclonal variation.
• Somatic hybridization.
• Cryopreservation of germplasm.

5. VAM (Vesicular – Arbuscular Mycorrhiza) fungi with Rhizobium can boost the yields: Recently there has been a new dimension to this farm practice by the way of increasing Rhizobium inoculation effect by simultaneous inoculating seeds with VAM as well as Rhizobium culture.VAM are structural modification of hyphae helping in absorption and storage of phosphorus.

(B) Application of Genetic engineering in Medical field:

1. The hereditary diseases like colour – blindness, haemophilia which are caused by recessive genes and also many inborn metabolic disorders due to defective genes as alkaptonuria, phenylketonuria can be cured with the gene therapy.

2. Substances like vitamins, hormones, amino acids and antibodies can be synthesized in bacteria by introducing the genes which code these substances. In this way bacteria can be used as biofactories for the synthesis of these substances.

3. Production of insulin: Insulin is medicine used for the treatment of diabetes.
Initially it is derived from animals (pig and cows) but today it is produced by gene splicing.

4. Hepatitis-B vaccine : Hepatitis-B is a viral disease of liver. Today this vaccine is prepared with the help of genetic engineering.

Question 5.
What is forensic science? Explain DNA fingerprinting method of forensic science.
Answer:
Forensic science : Forensic science is the application of scientific knowledge and methodology to criminal investigations and legal problems.

DNA figerprinting : DNA fingerprinting is a hybridization technique used to identify the similarities of DNA molecule of two individuals.

Procedure of DNA fingerprinting : The DNA fingerprinting involves the following steps:

1. First of all, the DNA of the organism which has to be tested in isolated. It is host DNA. The DNA may be obtained from blood, semen, vaginal swabs and bone marrow. The amount of DNA needed for developing fingerprints is very small.

2. The isolated host DNA is then digested with a suitable restriction enzyme (example Restriction endonucleases) and the digest is subjected to gel electrophoresis (Southern hybridization). This enzyme cut the host DNA into small fragments.

3. The fragments of double stranded DNA (ds DNA) are denatured to produce single stranded DNA’by alkali treatment.

4. The DNA fragments are separated by gel electrophoresis.

5. The electrophoresed DNA is then transferred from gel into a nitrocellulose filter paper where it is fixed by baking the filter at 80°C for 2 hours.

6. A known sequence of single stranded DNA is prepared. It is called probe – DNA. The probe is obtained from organisms or prepared by DNA preparation method.

7. The probe DNA is labelled with radioactive isotopes such as P32 and then probe is added to the nitrocellulose filter paper containing host DNA.

8. The single stranded probe DNA base pair with complementary strands of the host DNA. As a result DNA – DNA hybrids are formed on nitrocellulose paper.

9. The nitrocellulose paper is then washed to remove unbound DNA probes from the paper.

10. The nitrocellulose filter paper is photographed on X-ray film through autoradiography. The film is analyzed to determine the presence of hybrid nucleic acid.

A schematic representation of DNA fingerprinting of a person accused of rape, and of the semen stain recovered from the clothes of rape victim. The fingerprints match perfectly; the probabillity of this being due to chance is extremely low. DNA samples from a close relative of the accused and from an unrelated person are also included for comparisons (not shown here).

In case of disputed parentage, the DNA fingerprints of the child, the mother and the suspected father are compared. As a rule, the bands present in a child’s fingerprint must be accounted for by those in the mother’s and the fathers’s fingerprints. (MPBoardSolutions.com) Therefore, the child’s fingerprint is compared with that of the mother and their common bands are marked. The remaining bands in the child’s fingerprint must be present in that of the child’s father.

Applications of DNA Fingerprinting:

1. Identification of criminals in forensic laboratories and legal pursuits.
2. Settlement of disputed paternity i.e., who is the biological father or mother of the child.
3. It is useful in restoring health of patients suffering from leukemia or blood cancer.
4. It can explore the evolutionary pathway of a species.
5. It can verify the kinship of an individuals with another.

On the basis of DNA fingerprints, the killer of Shri Rajiv Gandhi, the former Prime Minister of India was identified as ‘Dhanu’. The sexual relationship of Bill Clinton, the former President of USA with Monika Levinski, a working staff of White House was also concluded to be true on the basis of DNA fingerprinting.

Question 6.
Explain stem cell technique with example and give its uses.
Answer:
Stem Cell Technology:
Stem cell technology is a rapidly developing field that combines the efforts of cell biologists, geneticists, and clinicians and offers hope of effective treatment for a variety of malignant and non-malignant diseases using stem cells.

Stem cell:
A stem cell is a non – specialized,undifferentiated generic cell which can make exact copies of itself indefinitely and can differentiate and produce specialized cells such as a liver cells, muscle cells, blood cells, and other cells with specific functions. The process of changing into a specific cell type is known as differentiation. Stem cells are found in multicellular organisms.
Stem cells divide regularly to renew and repair the existing tissue.

Example:
The bone marrow and gastrointestinal tract in which stem cells function to renew and repair tissue.

Uses of stem cell:

1. Due to capacity of regeneration it is useful for treatment of injured cells by replacement of deflective tissue, teeth, retina, cochlear cells.
2. It is used to treat diseases including spinal cord injury, stroke, bums, heart disease, diabetes, osteoarthritis and rheumatoid arthritis.
3. Treatment of diabetes.

MP Board Class 12th Biology Important Questions

MP Board Class 12th Biology Important Questions Chapter 11 Biotechnology: Principles and Processes

Biotechnology: Principles and Processes Important Questions

Biotechnology: Principles and Processes Objective Type Questions 

Question 1.
Choose the correct answer:

Question 1.
Which is used in DNA cultivation :
(a) Allesa
(b) Bloting technique
(c) PCR
(d) None of these.
Answer:
(c) PCR

Question 2.
Which is stained by Ethidium bromide :
(a) DNA
(b) RNA
(c) Protein
(d) All of these.
Answer:
(a) DNA

Question 3.
(a) Bacteria
(b) Virus
(c) Yeast
(d) Plant cell.
Answer:
(a) Bacteria

Question 4.
Plasmids are:
(a) Circular
(b) Extra – nuclear
(c) Vector
(d) All of these.
Answer:
(d) All of these.

Question 5.
Resistant gene found in pBR 322 :
(a) Ampicillin
(b) Tetracycline
(c) Both (a) and (b)
(d) None of these.
Answer:
(c) Both (a) and (b)

Question 6.
Eco RI is :
(a) A virus
(b) A bacteria
(c) A restriction enzyme
(d) A cyanobacteria.
Answer:
(c) A restriction enzyme

Question 7.
Plasmid and cosmid are :
(a) DNA vector
(b) RNA vector
(c) A bacteria
(d) A virus.
Answer:
(c) A bacteria

Question 8.
When stablised, bio technical department by Indian government :
(a) 1948
(b) 1985
(c) 1986
(d) 1987.
Answer:
(c) 1986

Question 9.
Who gave the name plasmid :
(a) Jacob and Monod
(b) Laderberg
(c) Ruth senger
(d) Khurana.
Answer:
(b) Laderberg

Question 10.
What is vector DNA :
(a) Breaker of DNA
(b) Cutter of DNA
(c) Transfer in DNA in one
(d) None of these.
Answer:
(c) Transfer in DNA in one

Question 11.
RI plasmid is obtained from :
(a) E. coli
(b) Yeast
(c) Agrobacterium
(d) Rhizobium.
Answer:
(c) Agrobacterium

Question 12.
pBR 322 is :
(a) Bacteria
(b) Virus
(c) Plasmid
(d) Recombinant protein.
Answer:
(c) Plasmid

Question 2.
Fill in the blanks:

1. …………….. is used for DNA staining.
2. …………….. joins DNA segments.
3. …………….. are small sized cloning vectors.
4. In …………….. disturbed genes can be replaced by proper genes.
5. Small segments of DNA are moved to …………….. pole through power steering.
6. PCR is first discoverd by ……………..
7. …………….. bacteria is responsible for tumor formation.
8. …………….. enzyme is used in PCR.
9. …………….. is used for gene transfer in dicot.
10. …………….. is done by electrophorasis.
11. …………….. DNA polymerase is also activeted in high temperature.
12. Sticky ends of helpfull in the function of …………….. enzyme.

Answer:

1. Ethidium bromide
2. DNA ligase
3. Cosmid
4. Gene therapy
5. Distant
6. Kary muilis
7. Agrobacterium tumefaciens
8. Taq polymerase
9. Ti plasmid
10. Isolation of DNA segment
11. Taq DNA polymerase
12. DNA Ligase.

Question 3.
Match the followings:

Answer:

1. (c)
2. (a)
3. (b)
4. (e)
5. (d)

Question 4.
Write the answer in one word/sentances:

1. Biomolecule that destroy viruses and produce immunity in human being.
2. Name of the first cloned animal.
3. Where does CCMB situated?
4. What kind of charge found in DNA segment?
5. Name the technique in which seperate DNA segments.
6. Name the molecular scissors.
7. Write the full name of PCR.
8. Which metal is used in gene gun method?
9. Who discover the enzyme restriction?
10. What is meant by GM organism?
11. Write the full name of FEB.

Answer:

1. Interferon
2. Dolly
3. Hyderabad
4. Negative
5. Bio power steering
6. Re – striction, enzyme
7. Polymerase chain reaction
8. Tungstun or Gold
9. Werner Arber
10. Genetically Modified organism
11. European Federation of Biotechnology.

Biotechnology: Principles and Processes Very Short Answer Type Questions

Question 1.
Which enzyme is used for isolation of target gene?
Answer:
Restriction endonuclease is used for isolation of target gene.

Question 2.
Which DNA polymerase is active in high tempereature?
Answer:
Taq DNA polymerase is active in high temperature.

Question 3.
Name three restriction endonuclease enzyme.
Answer:

1. EcoRI
2. Hind II
3. Hind III.

Question 4.
Write full form of PCR. Which enzyme is used in?
Answer:

1. Polymerase Chain Reaction (PCR).
2. Taq DNA is used in PCR.

Question 5.
What is bacteriophage?
Answer:
Viruses, they are infected to bacteria are called Bacteriophage.

Question 6.
First recombinant DNA was formed in.
Answer:
In bacteria Salmonella typhimurium.

Question 7.
What is molecular scissors?
Answer:
Restriction enzyme is called molecular scissors.

Question 8.
Where does Hind II cuts the DNA molecule?
Answer:
It cuts the DNA molecule at a specific 6 base pair sequence.

Question 9.
What is the funtions of sticky ends?
Answer:
It helps enzyme DNA ligase.

Question 10.
Which type of charge found in DNA?
Answer:
Negative charge.

Question 11.
How does ethidium bromide cause DNA to fluorescence in electrophoresis?
Answer:
The most commonly used stain for detecting DNA is ethidium bromide.

Question 12.
What is electrophoresis?
Answer:
Electrophoresis is a technique used in laboratories in order to separate micromolecules based on size.

Question 13.
Give the function of circular DNA which is found in bacterial cell.
Answer:
It works as a vector.

Question 14.
Name the technique in which we should isolate DNA segment.
Answer:
Electrophoresis.

Question 15.
Name two antibiotic restriction gene which is found in plamid pBR 322.
Answer:
Ampicillin and Tetracycline.

Question 16.
What is plasmid?
Answer:
A plasmid is a small DNA molecule within a cell that is physically separated from a chromosomal DNA and can replicate independently.

Question 17.
What is DNA ligase?
Answer:
DNA ligase is a specific type of enzyme, a ligase that facilitates the joining of DNA strands together by catalyzing the formation of a phosphodiesterbond.

Question 18.
What is C – DNA?
Answer:
C – DNA is a single stranded RNA. C – DNA is often used to clone eukaryotic genes in prokaryotes.

Question 19.
What is Ti – plasmid?
Answer:
Ti or tumour inducing plasmid is a plasmid that is a part of the genetic equipment that Agrobacterium tumifaciens use to transduce their genetic material to plants.

Biotechnology: Principles and Processes Short Answer Type Questions

Question 1.
From what you have learnt, can you tell whether enzymes are bigger or DNA is bigger in molecular size? How did you know?
Answer:
DNA is bigger in molecular size than enzymes. Because DNA is a long double stranded molecule which can go up to a few meters in length when stretched end to end but enzymes although variable in size, would still be smaller than the DNA.

Question 2.
What would be the molar concentration of human DNA in a human cell? Consult your teacher.
Answer:
The average molecular weight of a nucleotide in human DNA is 130.86. The molecular weight of human DNA will therefore be 6 x 10⁹ nucleotides (based on the human genome project) x 130.86 = 784.56 x 109 gm/mol. The molar concentration of DNA can be calculated accordingly.

The molarity can be calculated as
Molar concentration = \(\frac { No. of molecules}{ Molecular weight }\)

Question 3.
Do eukaryotic cells have restriction endonucleases? Justify your answer.
Answer:
No, of eukaryotic cells do not have restriction endonucleases. All the restriction endonucleses have been isolated from the various strains oi bacteria arid they arc also named according to the genus and species of prokaryotes. The first letter of the enzyme comes from the genus and the second two letters come from the species of the prokaryotic cell from which they have been isolated.

Question 4.
Besides better aeration and mixing properties, what other advantages do stirred tank bioreactors have over shake flasks ?
Answer:
Shake flask are used for growing microbes and mixing the desired materials on a small scale in the laboratory. However, the large-scale production of a desired biotechnological product requires large stirred tank bioreactors.

Besides better aeration and mixing properties, bioreactors have the following advantages:

1. It has an oxygen delivery system.
2. It has a foam control, temperature and pH control system.
3. Small volumes of culture can be withdrawn periodically.

Question 5.
Name the scientist who discovered artificial DNA synthsizing method.
Answer:
The Nobel prize in physiology 1968 was awarded jointly to Robert W. Holley, Hargobind Khorana and M. Nirenberg for their interpretation of the genetic code.

Question 6.
What is gene manipulation or genetic engineering? Explain it.
Answer:
Genetic engineering, also called genetic modification is the direct manipulation of an organisms genes using biotechnology. It is a set of technologies used to change the genetic makeup of cells, including the transfer of genes and across species boundaries to produce improved organisms. New DNA is obtained by either isolating and copying the genetic material of interest using recombinant DNA methods or by artificially synthesising the DNA.

It may also mean extracting DNA from another organisms genome and combining it with the DNA of that individual. Genetic engineering is used by scientists to enhance or modify the characterstics of an individual organisms genetic engineering can be used to produce plants that have a higher nutritional value or can tolerate exposure to herbicides.

Question 7.
Write application of genetic engineering in crop improvement.
Answer:
Genetic engineering has placed an important role in improvement of plant production. There are following applications of genetics in plant improvement.

1. Production of polyploidy crops.
2. Hybridization : Hybridization is used to produce plants with, desirable traits.
3. Transgenic plants : It helps a lot in improving the yield and quality of crops.
4. Insect and herbicides resistant plants are engineered.

Question 8.
Can you recall meiosis and indicate at what stage a recombinant DNA is made?
Answer:
A recombinant DNA is made in the pachytene stage of prophase I by crossing over during meosis cell division. Recombination nodules are visible in synaptonemal complex in pachytene sub stage. Crossing over occurs in this time between chromatids than recombinant DNA is formed.

Question 9.
Can you think and answer how a reporter enzyme can be used to monitor transformation of host cells by foreign DNA in addition to a selectable marker?
Answer:
Reporter enzyme can differentiate recombinants from non – recombinants on the basis of their ability to produce a specific colour in the presence of a chromogenic substrate. DNA is inserted within the coding sequence of the enzyme β – galactosidase. (MPBoardSolutions.com) This results into inactivation of the enzyme which is referred to as insertional inactivation. The presence of a chromogenic substrate gives blue – coloured colonies if the plasmid in the bacteria does not have an insert. The presence of the insert results in insertional inactivation of β – galactosidase and the colonies do not produce any colour. These are identified as recombinant colonies.

Question 10.
Make a chart (with diagrammatic representation) showing a restriction enzyme, the substrate DNA on which it acts, the site at which it cuts DNA and the products it produces.
Answer:
The substrate DNA on which a restriction enzyme acts :

Question 11.
Write the uses of animal cloning.
Answer:
Applications (Uses) of Animal cloning :

1. By this technique desired genotypes of any organism can be conserved.
2. It produces organisms with better characters.
3. Endangered plant and animal species can be conserved by this technique.
4. Animals with good quality of milk and protein can be produced.

Question 12.
Write applications of genetic engineering in medical field.
Answer:

1. The hereditary diseases like colour – blindness, haemophilia which are caused by recessive genes therapy.
2. Substances like vitamins, hormones, amino acids and antibiotics can be synthe – sized in bacteria by introducing the genes which code these substances.
3. Production of insulin : It is a medicine used for the treatment of diabetes. It is produced by gene splicing.
4. Hepatitis – B vaccine : Hepatitis – B is a viral disease of liver, today this vaccine is prepared with the help of genetic engineering.

Question 13.
What do you understand by gene bank? What are its significances?
Answer:
Gene bank:
The institution which conserves the genes of the organisms is called as gene bank. The genetic material (DNA) found in the cells of organisms are conserved in gene banks. The best measure of conserving genes is to conserve endangered organisms. The tissues or cells of organisms are also conserved in gene banks.

Significance:
Genes stored in gene bank are used for the production of improved varieties of species and for scientific tests.

Question 14.
What do you understand by gene cloning? What are its significance?
Or
What is gene cloning? Write its importance.
Answer:
Gene cloning:
It is a process in which DNA of an organism is cut into smaller DNA fragments by the use of restriction endonuclease enzymes. Each DNA fragment is introduced into a bacterial, yeast, insect, plant or animal cell. The cells are grown on a suitable medium under suitable conditions. Each cell containing a particular DNA fragment multiplies to give rise a group of cells, all containingthe same foreign DNA.

These groups of cells are known as clone of cells. These copies of DNA resulting from the multiplication or recombinant DNA are called as cloned DNA and the process is known as gene cloning,

Significance:

1. Useful hereditary characters are obtained by this process.
2. Many diseases are cured by this process.
3. Many medicines are synthesized with the help of this process.
4. This process should also be used in eugenics.

Biotechnology: Principles and Processes Long Answer Type Questions

Question 1.
Collect 5 examples of palindromic DNA sequences by consulting your teacher. Better try to create a palindromic sequence by following base – pair rules.
Answer:
Some palindromic DNA sequences and the restriction enzymes which act on them are:

1.  5′ – AGCT – 3 Alul (Arthrobacter lutens)
3′ – TCGA – 5′

2. 5′ – GAATTC – 3’EcoRI (Escherichia coli)
3′-CTTAAG-5′

3.  5′- AAGCTT – 3’HindIII (Haemophilus influenzae)
3′ – TTCGAA – 5′.

4. 5′ – GTCGAC – 3’SalI (Streptomyces albus)
3′ – CAGCTG – 5′

5. 5′ – CTGCAG – 3’PstI (Providencia stuartii)
3′ – GACGTC – 5′.

Question 2.
Describe briefly the followings :

1. Origin of replication
2. Bioreactor
3. Downstream processing.

Answer:
1. Origin of replication:
Origin of replication (ori) is a sequence on the chromosome, form where replication starts and any place of DNA when linked to this sequence can be made to replicate within the host cells. This sequence also controls the copy number of the linked DNA. So, if we want to recover many copies of the target DNA it should be linked to the ‘ori’ site and should be cloned in a vector whose origin supports high copy number.

2. Bioreactor:
Bioreactor is large vessels in which raw materials are biologically converted into specific products, individual enzymes, etc., using microbial, plant, animal or human cells. (MPBoardSolutions.com) A bioreactor provides the optimal conditions for achieving the desired production levels by providing optimum growth conditions of temperature, pH, substrate, salts, vitamins, oxygen, etc.

3. Downstream processing:
Downstream processes include separation and purification, formulation with suitable preservatives, etc. which are collectively referred to as downstream processing. Such formulation has to undergo through clinical trials as in case of drugs. Strict quality control testing for each product is also required. The downstream processing and quality control testing vary from product to product.

Question 3.
Explain briefly:

1. PCR
2. Restriction enzymes and DNA
3. Chitinase.

Answer:
1. PCR:
PCR stands for polymerase chain reaction, a method of amplifying fragments of DNA. This method can make multiple copies of even a single DNA fragment or the gene of interest in a test tube. The reaction ntixture requires:

1. Double – stranded DNA fragment (gene of interest).
2. Primers – small chemically synthesized oligonucleotides that are complementary to the regions of this DNA.
3. The special thermostable DNA polymerase (isolated from a bacterium, Thermus aquaticus), that does not denature and remain active even at high temperature.

Unwinding of two strands of DNA by heating the sample at 92 – 94°C helps primers to get positioned on the exposed nucleotides as per base pairing rules. DNA polymerase recognizes primes as ‘start’ tags and begins to extend the primes using the free nucleotides provided in the reaction and the genomic DNA as template. With each round of reactions, the DNA doubles.

2. Restriction enzymes and DNA:
These enzymes are used in genetic engineering to cut the large DNA molecule into smaller fi agments. When DNA from two different sources are cut by the same restriction enzyme, the resultant DNA fragments have the same kind of’sticky – ends’ and these can be joined together (end – to – end) using DNA ligases. This new DNA created by joining fragments, from two different sources/genomes together is recombinant DNA.

3. Chitinase:
Chitinase is an enzyme that breaks down chitin, a component of fungal cell wall. It is useful for isolating the fungal cell DNA.

Question 4.
Discuss with your teacher and find out how to distinguish between :

1. Plasmid DNA and Chromosomal DNA
2. DNA and RNA
3. Exonuclease and Endonuclease.

Answer:
1. Distinguish between Plasmid DNA and Chromosomal DNA :

Plasmid DNA:

• Plasmid DNA is the naked double – stranded DNA which forms a circle
• It is associated with few proteins but contains RNA polymerase enzyme.
• They are smaller than the host chromosomes and can be easily sparated.
• It has no free ends.

Chromosomal DNA:

• Chromosomal DNA is a double – stranded linear DNA molecule.
• It is associated with large proteins.
• This DNA exists in relaxed and super – coiled forms and provides a template for replication and transcription.
• It has free ends represented as 3′ – 5′.

2. Distinguish between DNA and RNA :

DNA:

• It is mainly confined to the nucleus. A small quantity occurs in mitochondria and chloroplasts.
• Its quantity is constant in each cell of a species.
• It contains deoxyribos sugar.
• Its pyrimidines are adenine and thymine.
• The amount of adenine is equal to the amount of thymine. Also, the amount of cytosine is equal to the amount of guanine.
• It can replicate itself.

RNA:

• It mainly occurs in the cytoplasm. A small quantity is found in the nucleus.
• Its quantity varies in different cells.
• It contains ribose sugar
• Its pyrimidines are adenine and uracil. Adenine and uracil are not necessarily in equal amounts, nor are cytosine and
• guanine necessarily in equal amounts.
• It cannot replicate itself. It is formed by DNA. Some RNA viruses (paramyxo virus) can produce RNA from an RNA template.

3. Distinguish between Exonuclease and Endonucleases:
Exonucleases are nucleases which cut off the nucleotides from the 5′ or 3′ ends of a DNA molecule, whereas endonucleases are nucleases which cleave the DNA duplex at any point except at the ends.

Question 5.
Describe the useful and harmful effects of genetic engineering.
Or
Give importance and uses of DNA recombinant technique.
Or
Describe the Genetic engineering. Write the importance of it in human life.
Answer:
The branch of molecular genetics in which we can manipulate or transplant the genes or the genetic material or DNA according to our will is called gene manipulation or genetic engineering. The main objective of genetic engineering is to synthesize recombinant DNA (formed of the DNA segments of two different organisms).

The useful and harmful effects of genetic engineering are as follows :

(A) Useful effects or Utility:
1. Industrial uses:
Various types of substances such as vitamins, hormones and antibiotics can be-synthesized in bacteria by introducing genes that code these substances. In this way, bacteria can function as living factories for the synthesis of these substances. Humulin (human insulin) is synthesized by this method.

2. Treatment of diseases:
A new system of medicines, gene therapy may develop to cure several genetic disorders such as haemophilia, colour-blindness, etc. Also many inborn meta¬bolic disorders due to defective genes such as alkaptonuria, phenylketonuria, etc. can be cured.

3. Use in agriculture:
The genes for N₂ fixation found in symbiotic bacteria Rhizo – bium leguminosarum or blue-green algae may be transferred to the major food crops, increases food production without using expensive fertilizers. Thus, we can save millions of rupees spent otherwise on fertilizers and manures to boost food production.

4. Changes in the structure and expression of genes:
We can obtain new plants, animals having traits tailored according to our will.

(B) Harmful effects:

1. Normal harmless bacteria can be transformed into cancer causing forms thus ushering a new era of biological warfare.
2. During experiments, it is quite possible to obtain super viruses for which we might have no defence.
3. By the use of recombinant DNA, the bacteria may be made resistant to antibiotics.

Question 6.
Explain the mechanism of recombinant DNA technology in genetic engineering by using plasmid as carrier of genes.
Answer:
Mechanism of Recombinant DNA Technology : Mechanism of recombinant DNA technology involves the following steps :

1. Isolation of desired gene or functional DNA segment:
From the eukaryotic cell desired DNA segment is isolated with the help of enzyme restriction endonuclease. Now this segment of DNA is known as foreign DNA.

2. Transfer of DNA segment from one organism to other:
Plasmid is an extra chromosomal circular DNA found mostly in bacteria over and above the main genome. When bacteria multiplies the plasmid DNA also multiplies along with the chromosomal DNA. These plasmids can be easily isolated from the bacterial cell with the help of restriction endonucleases. Plasmid serves as a vector for transferring the foreign DNA into a suitable recipient.

Foreign DNA and plasmid sliced with the help of endonucleases has free sticky ends through which they join each other with complementary base pairing with the help of enzyme DNA ligase. Thus, a recombinant DNA is formed,

Such a recombinant DNA when introduced into a recipient bacterium (transformation), it replicates and expresses itself, within the bacterial cell, the recombinant DNA molecule replicates along with the endogenous DNA of the host cell and produces copies of cloned DNA. This process is known as gene cloning. The cloned recombinant DNA produced in large quantities’can be isolated, purified and analysed.

Question 7.
Describe the role of genentic engineering in artificial synthesis of human insulin.
Answer:

1. Aldric and his supporter prepared two DNA sequences corresponding to the A and B chains of human insulin.
2. Sticky ends were produced in the E.coli plasmid and the insulin gene by treating them both with the same restriction endonucleases.
3. These two are then joined together by the enzyme DNA ligase.
4. The bacteria are then grown in sterilised bioreactors in the appropriate growth medium.
5. The chain A and B are produced separately, extracted and purified.

Question 8.
What is clone? Give its preparation, extraction and purification.
Answer:
An organism or cell or group of oiganisms, produced a sexually from an ancestor, to which they are genetical.

1. Gene cloning : Following steps are used by gene cloning :
(i) Preparation of gene:
DNA extracted from an organism, with the gene of interest is cut into gene size pieces with restriction enzyme.

(ii) Insertion into vector:
Bacterial plasmids are cut with the same restriction enzyme. Plasmids are small circles of DNA in bacterial cells that are naturally present in addition to the bacterial other DNA.

(iii) Transformation of host cells:
The recombinant plasmids are then transferred into bacteria using either electroporation. The plasmid are small enough to pass through the holes into the cells. However rather than using electricity to create holes in the bacterium, it is done by alternating the temperature between hot and cold.

The bacteria is grown on a culture dish and allowed to grow into colonies. All the colonies on all the plates are called a gene library.

2. Plant cloning:
Plant tissue culture is a method of propagation that has been sprouting in popularity throught as an altemative.to cloning. Plant can be cloned artificially using tissue culture. (MPBoardSolutions.com) Vegetative propagation works because the end of the cutting forms a mass of non specialized cells called a callus, the callus will grow divide and form various specialized cells eventually forming a new plant.

Question 9.
Can you list 10 recombinant proteins which are used in medical practice? Find out where they are used as therapeutics (use the internet).
Answer:

MP Board Class 12th Biology Important Questions

MP Board Class 12th Biology Important Questions Chapter 10 Microbes in Human Welfare

Microbes in Human Welfare Important Questions

Microbes in Human Welfare Objective Type Questions 

Question 1.
Choose the correct answer:

Question 1.
Name the place from which viruses entered into the plant:
(a) From lenticels
(b) From wound
(c) From stomata
(d) From root.
Answer:
(b) From wound

Question 2.
Fertilizer elements are :
(a) Nitrogen
(b) Phosphorus
(c) Potassium
(d) All of these.
Answer:
(d) All of these.

Question 3.
Fertilizer that supplies least % of nitrogen is :
(a) Urea
(b) (NH₄)SO₄
(c) (NH₄)NO₃
(d) Organic nitrogen fertilizer.
Answer:
(d) Organic nitrogen fertilizer.

Question 4.
VAM is:
(a) Bacteria
(b) Yeast
(c) Fungi
(d) Virus.
Answer:
(b) Yeast

Question 5.
Nitrifying bacteria converts :
(a) N₂ into nitrates
(b) Ammonia into nitrates
(c) Nitrates into N₂
(d) CO₂ into carbohydrates.
Answer:
(b) Ammonia into nitrates

Question 6.
Who discovered the virus :
(a) Stanley
(b) Ivanowski
(c) Herel
(d) Beijerinck.
Answer:
(b) Ivanowski

Question 7.
Name the virus which has double helical stranded DNA :
(a) Hepatits A
(b) Hepatits B
(c) Hepatits C
(d) Flepatits D.
Answer:
(b) Hepatits B

Question 8.
Spirally coiled bacteria is known as :
(a) Spirilla
(b) Cocci
(c) Bacilli
(d) Vibrio.
Answer:
(a) Spirilla

Question 9.
Name the bacteria in which flagella are found in whole body :
(a) Peritrichous
(b) Atrichous
(c) Monotrichous
(d) Cephalotrichous
Answer:
(a) Peritrichous

Question 10.
Useful activities of microbes is :
(a) Nitrogen fixation
(b) Nitrification
(c) Biogeochemical cycle
(d) All of the above.
Answer:
(d) All of the above.

Question 11.
Which of the following is prokaryotes :
(a) Chlorella
(b) Clamydomonas
(c) Protomyces
(d) Ossiletoria.
Answer:
(d) Ossiletoria.

Question 12.
Yeast is used in the production of :
(a) Ethyl alcohol
(b) Acetic acid
(c) Cheese
(d) Curd.
Answer:
(a) Ethyl alcohol

Question 13.
B.T. cotton provides immunity from :
(a) Pest
(b) Weeds
(c) Salt
(d) Animal.
Answer:
(a) Pest

Question 14.
Which is the used for making bread :
(a) Lactobacillus
(b) Straptobacillus
(c) Aspergillus
(d) S. cerevisiace.
Answer:
(c) Aspergillus

Question 2.
Fill in the blanks:

1. Rhizobium, Azolla and Anabaena are the example of …………………….. fertilizers.
2. …………………….. bacterium is found in the roots of leguminous plants.
3. Nostoc and Anabaena are called ……………………..
4. …………………….. is nitrogen fixing bacteria.
5. …………………….. is the fungal resistant, bacterial antibody component.
6. …………………….. is the other name of mycoplasma.
7. …………………….. is the other name of Archaebacteria.
8. …………………….. type of ribosome is found in Moneran.
9. Bacterial cell wall is made by ……………………..
10. …………………….. is the meaning of word virus.
11. Butter milk is obtained from the activation of ……………………..
12. Ethanol is produced in a large scale through …………………….. yeast.
13. …………………….. bacteria is helpful in vinegar production.

Answer:

1. Bio
2. Rhizobium
3. Cyanobacteria
4. Rhizobium
5. Polymixin
6. PPLO
7. Living fossils
8. 70S type
9. Mucopeptides
10. Poison
11. Streptococcus lactis
12. Saccharomyces cerevisiae
13. Azotobacter aceti.

Question 3.
Match the followings :
I.

Answer:

1. (d)
2. (a)
3. (e)
4. (b)
5. (c).

II.

Answer:

1. (e)
2. (d)
3. (a)
4. (b)
5. (c)

Question 4.
Write answer in one word/ sentences:

1. Name the bacteria which is used in making swiss cheese.
2. Write the name of the microorganism which converted milk into curd.
3. Name the fish which control mosquito.
4. Write the name of two bacteria which are found in soil in free state.
5. Write the full name of IPM.
6. What is yogurt?
7. What is interferon?
8. Write the full name of B.O.D.
9. Write the name of an autotrophic nitrogen fixing microorganism.
10. Which microbes are used in industries for making citric acid?
11. Who discovered the penicillin?
12. Write the name of an alcoholic drink which is made by without distillation.
13. Microbes which is respire without oxgyen is called?
14. Synibiotic relationship which is found between fungi and higher plant root.

Answer:

1. Propionibacterium shermanii
2. Lactobacillus
3. Gambussia
4. Azotoba – cterium and Clostridium
5. Integrated Pest Management
6. Milk product
7. Antiviral protein
8. Biochemical Oxygen Demand
9. Anabaena
10. Aspergillus niger
11. Alexander Fleming
12. Beer
13. Anaerobic
14. Micorrhiza.

Microbes in Human Welfare Very Short Answer Type Questions

Question 1.
Name the fungus which produce these products:

1. Citric acid
2. Vitamin B₂

Answer:

Question 2.
What is yoghurt?
Answer:
Yoghurt is a semi – solid sourish food prepared from milk fermented by added bacteria.

Question 3.
Why does the river Ganga water not spoil even kept for long – time?
Answer:
Bacteriophages are present in Ganga river which feed on bacteria which spoil water. So, this water does not get spoil for long –   time.

Question 4.
Who discover penicillin?
Answer:
Alexander Fleming.

Question 5.
Name the microorganism/fungi which gave penicillin.
Answer:
Penicillium notatum.

Question 6.
Which organism is used for making citric acid in industries?
Answer:
Aspergillus niger.

Question 7.
Write the name of methanogen bacteria.
Answer:
Methano bacterium species.

Question 8.
Which bacteria is found in anaerobic sludge?
Answer:
Methanogen.

Question 9.
What is lady bird?
Answer:
It is an insect which is used as biocontroller for Aphid control.

Question 10.
Give the name of autotrophic nitrogen fixing microbes.
Answer:
Anabaena.

Question 11.
Name the mineral absorbed by mycorrhiza, which is not absorbed by plant roots.
Answer:
Phosphorus.

Question 12.
Name the microorganism which changes to milk into curd.
Answer:
Lactobacillus spp.

Question 13.
Which fungi is used in treatment of plant due to biological control?
Answer:
Trichoderma spp.

Question 14.
Which bacteria is used in swiss paneer?
Answer:
Propionibacterium shermanii.

Question 15.
Which paneer is completed by the growth of fungus?
Answer:
Roquefort cheese.

Question 16.
Name two alcoholic drink which are formed without distillation.
Answer:
Beer and Wine.

Question 17.
Name two alcoholic drink which are formed by distillation.
Answer:
Whiskey and Rum.

Question 18.
What is the full form of BOD?
Answer:
Biochemical or Biological Oxygen Demand.

Microbes in Human Welfare Short Answer Type Questions

Question 1.
Bacteria cannot be seen with the naked eyes, but these can be seen with the help of a microscope. If you have to carry a sample from your home to your biology laboratory to demonstrate the presence of microbes under a microscope, which sample would you carry and why?
Answer:
Curd can be used as a sample for the study of microbes. Curd contains numerous Lactic Acid Bacteria (LAB) or Lactobacillus. These bacteria produce acids that coagulate and digest milk proteins. A small drop of curd contains million of bacteria, which can be easily observed under a microscope.

Question 2.
Give example to prove that microbes release gases during metabolism.
Answer:
The examples of bacteria that release gases during metabolism are :

1. Bacteria and fungi carry out the process of fermentation and during this process, they release carbon dioxide.
2. The dough used for making idli and dosa gives a puffed appearance. This is because of the action of bacteria which releases carbon dioxide. This CO₂ released from the dough gets trapped in the dough, thereby giving it a puffed appearance.

Question 3.
In which food would you find lactic acid bacteria? Mention some of their useful applications.
Answer:
Lactic acid bacteria can be found in curd. It is this bacterium that promotes the formation of milk into curd. The bacterium multiplies and increases its number, which converts the milk into curd. They also increase the content of vitamin B₁₂
in curd. Lactic acid bacteria are also found in our stomach where it keeps a check on the disease causing microorganisms.

Question 4.
Name some traditional Indian foods made of wheat, rice and Bengal gram (or their products) which involve use of microbes.
Answer:

1. Wheat product: Bread, Cake etc.
2. Rice product: Idli, Dosa.
3. Bengal gram product: Dhokla, Khandvi.

Question 5.
In which way have microbes played a major role in controlling diseases caused by harmful bacteria?
Answer:
Several microorganisms are used for preparing medicines. Antibiotics are medicines produced by certain microorganisms to kill other disease causing microorganisms. These medicines are commonly obtained from bacteria and fungi. They either kill or stop the growth of disease causing microorganisms. (MPBoardSolutions.com) Streptomycin, tetracycline and penicillin are common antibiotics. Penicillium notatum produces chemical penicillin, which checks the growth of staphylococci bacteria in the body. Antibiotics are designed to destroy bacteria by weakening their cell walls.

Question 6.
Name any two species of fungus, which are used in the production of the antibiotics.
Answer:
Antibiotics are medicines that are produced by certain microorganisms to kill other disease causing microorgainsms. These medicines are commonly obtained from bacteria and fungi. The species of fungus used in the production of antibiotics are:

Antibiotic:

• Penicillin
• Cephalosporin

Fungus source:

• Penicillium notatum
• Cephalosporium acremonium

Question 7.
What is sewage? In which way can sewage be harmful to us?
Answer:
Sewage is the municipal waste – water collected from city or town homes, that contains toilet, bathroom and kitchen waste.
It contains large amounts of organic matter and many pathogenic microbes which are harmful to humans as they can cause many diseases like cholera, typhoid, polio.

Question 8.
What is the key difference between primary and secondary sewage treatment?
Answer:
Differences between Primary and Secondary sewage treatment:

Primary sewage treatment:

• It is a physical process.
• Primary treatment involves the removal of large sized floating and suspended solids.
• BOD is not affected in this treatment.

Secondary sewage treatment:

• It is a biological process.
• Secondary treatment involves decomposition of organic matter by microbial action which produce methane, hydrogen sulphide and carbon dioxide.
• BOD is less by secondary treatment.

Question 9.
Do you think microbes can also be used as source of energy? If yes, how?
Answer:
Yes, microbes can be used as a source of energy. Bacteria such as; Methane bacterium is used for the generation of gobar gas or biogas. The generation of biogas is an anaerobic process in a biogas plant, which consists of a concrete tank (10 – 15 feet deep) with sufficient outlets and inlets. The dung is mixed with water to form the slurry and thrown into the tank. The digester of the tank is filled with numerous anaerobic methane producing bacteria, which produce biogas from the slurry. Biogas can be removed through the pipe which is then used as a source of energy, while the spent slurry is removed from the outlet and is used as a fertilizer.

Question 10.
Microbes can be used to decrease the use of chemical fertilizers and pesticides. Explain how this can be accomplished.
Answer:
Microbes play an important role in organic farming which is done without the use of chemical fertilizers and pesticides. Biofertilizers are living organisms which help in increasing the fertility of soil. It involves the selection of beneficial microorganisms that help in improving plant growth through the supply of plant nutrients. Biofertilizers are introduced in seeds, roots or soil to mobilize the availability of nutrients. Thus, they are extremely beneficial in enriching the soil with organic nutrients.

Many species of bacteria and cyanobacteria have the ability to fix free atmospheric nitrogen, Rhizobium is a symbiotic bacteria found in the root nodules of leguminous plants. (MPBoardSolutions.com) Azospirilhum and Azotobacter are free living nitrogen – fixing bacteria, whereas Anabena, Nostoc and Oscillitoria are examples of nitrogen – fixing cyanobacteria. Biofertilizers are cost effective and eco – freindly. Microbes can also act as biopesticides to control insect pest in plants. An example of biopesticides is Bacillus thuringiensis, which produces a toxin that kills the insect pests.

Dried bacterial spores are mixed in water and sprayed in agricultural fields. When larvae of insects feed on crops, these bacterial spores enter the gut of the larvae and release toxins, thereby it. Similarly, Trichoderma are free living fungi. They live in the roots of higher plants and protect them from various pathogens. Baculoviruses is another bio – pesticides that is used as a biological control agent against insects and other arthropods.

Question 11.
Three water samples namely river water, untreated sewage water and secondary effluent discharged from a sewage treatment plant were subjected to BOD test. The samples were labelled A, B and C; but the laboratory attendant did not note which was which. The BOD values of the three samples A, B, and C were recorded as 20mg/L, 8 mg/L and 400mg/L, respectively. Which sample of the water is most polluted ? Can you assign the correct label to each assuming the river water is relatively clean?
Answer:
The BOD values of the three samples A, B and C were recorded as 20 mg/L, 8 mg/L and 400 mg/L. Greater the BOD of waste water more is its polluting potential. So, sample C is more polluting as compared to sample A. Hence, sample A is the secondary effluent, sample B is river water and sample C is untreated sewage water.

Question 12.
What is paneer? Give its preparation method.
Answer:
Paneer is a milk product, which has 20 – 30% protein first filtered and the milk in thin cloth than heat on 60°C at 30 minutes and heat on 75°C at 15 second later it cold till 30°C. Add some quantity of lactic acid bacteria Streptococcus lactis, Streptococcus cremoris and enzyme renin. By this fat and casein protein is detached from milk after 45 minutes this mixture became solid. Now, they are cut into small pieces and boild in hot water, when it starts floating, they are separated and squeezed. Then they are put in salt water for treating. Paneer is now ready.

Question 13.
What is the utilization of bacteria in agriculture?
Answer:
Uses of Bacteria : Uses of bacteria for agriculture are as follows :

1. They increase soil fertility due to decomposition of dead organisms.
2. They increase soil fertility through nitrogen fixation in soil.
3. Blue – green algae is used the form of fertilizers.
4. They balance the quantity of mineral salts.

Question 14.
How are the bacteria used in vinegar industry?
Answer:
Fermentation of sugar by yeast in vinegar industry. Due to this manner, wine is prepared. This ethyl alcohol abandoned in direct air for much time now, it is changed in vinegar.

Question 15.
Name live fungi which give antibiotics.
Answer:
The fungi which give antibiotics are as follows:

Question 16.
Name five bacteria which produce antibiotics.
Answer:
The bacteria which produce antibiotics are as follows:

Microbes in Human Welfare Long Answer Type Questions

Question 1.
Write seven useful activities of Bacteria.
Answer:
Useful activities of bacteria :

1. N₂ fixation : Some bacterias play an important role in nitrogen fixation example Azo – tobacter, Clostridium, Rhizobium. These bacterias increase the fertility of soil by the fixation of atmospheric nitrogen.
2. Lactic acid synthesis : Lactobacillus lacti converts the milk sugar into lactic acid.
3. Acetic acid synthesis: Acetobacter aceti takes part in synthesis of acetic acid or vinegar.
4. Rating of fibres: Isolation of wood fibres from the stem of plants is called rating. Clostridium butyricum is used in rating of fibres.
5. Tobacco and Tea industry : Some bacteria like Micococcus candisens is used to increase the flavour of the leaves of tobacco and tea. This process is called as seasoning.
6. Medicine production : Bacteria are the chief source of antibiotics, hence, they are used to extract antibiotics, example Streptomyces gresius (Streptomycine).
7. As symbionts: Bacteria presents in our body, helps in the various metabolic reactions example E.coli.

Question 2.
Write economic importance of fungi.
Answer:
Following are the economic importance of fungi:

1. As food : Mushrooms like Agaricus, Lycoperdon, Romaria, Clavasia are used as food. These mushrooms contain about 50% protein. Morchella is also used as food.
2. As medicine : Fungi provides various types of antibiotics as example Penicillin, Griseofulvine, Citrinine, Clavicine, Gliotoxine, etc. which are used in the treatment of disease.
3. Fertility of soil: Many fungi species increase the soil fertility by decomposition of dead organic matter.
4. Nitrogen fixation : Many fungus like Rhodotorula increases the soil fertility by nitrogen fixation.
5. In bakery industry : Yeast cells are used in bakery to produce spongy breads.
6. In wine industry : Yeast cells contain enzyme zymase which ferment sugar into alcohol.
7. In chemical industry : Many fungi species are used in the production of various types of acid like citric acid.
8. In cheese industry : Fungus Penicillium roqueforti is used in the preparation of camembert type of cheese from milk.

Question 3.
Explain the uses and types of virus.
Answer:
Uses of Viruses:

1. As it forms a link between non – living and living thus, it helps us to understand organic evolution.
2. Cyanophage (Blue – green algae) viruses are used to destroy blue – green algae grown in different areas.
3. Bacteriophages are used as biocides to destroy many harmful bacterias present in polluted area.

For example : Water of the river Ganga always remain pure and clean in bottles year after year due to presence of bacteriophage in it.

Nucleic acids of viruses : Viruses generally contain only one type of nucleic acid DNA or RNA. DNA is found in animal viruses, whereas RNA is found in plant viruses.

Human Immuno Virus (HIV):
It causes AIDS disease in human. Full form of AIDS is Acquired Immune Deficiency Syndrome. This virus destroys immune system of the body, thus, the patient of AIDS suffers from number of diseases at a time and at last he dies. This virus is found surrounded by protein and a ring of RNA is found at the centre. Whole body of virus is surrounded by a covering of glycoprotein.

Type of viruses:
On the basis of host infected, viruses are grouped into following four groups:

1. Animal viruses : Viruses which infect animals are called animal viruses. These viruses contain DNA as genetic material.
2. Plant viruses : Viruses which infect plants are called plant viruses. These viruses contain RNA as genetic material.
3. Cyanophages : Viruses that infect blue – green algae are called cyanophages. These viruses contain RNA as genetic material.
4. Bacterial viruses or Bacteriophages : Viruses that infect bacteria are called as bacteriophages. They contain DNA as genetic materials.

Question 4.
Find out the role of microbes in the following and discuss it with your teacher:

1. Single Cell Protein (SCP)
2. Soil.

Answer:
1. Single Cell Protein (SCP):
It refers to harmless microbial cells that can be used as an alternate source of good protein. Just like mushrooms (a fungus) is eaten by many people and yeast is used by athletes as a protein source; similarly, other forms of microbial cells can also be used as food rich in protein, minerals, fats, carbohydrate and vitamins. (MPBoardSolutions.com) Microbes like Spirullina and Methylophilus methylophus are being grown on an industrial scale on materials containing starch like waste water from potato processing plants, straw, molasses, animal manure and even sewage. These single cell microbes can be used as source of proteins.

2. Role of microbes in soil:
Microbes play an important role in organic farming which is done without the use of chemical fertilizers and pesticides. Biofertilizers are living organisms which help in increasing the fertility of soil. It involves the selection of beneficial microorganisms that help in improving plant growth through the supply of plant nutrients.

Biofertilizers are introduced in seeds, roots or soil to mobilize the availability of nutrients. Thus, they are extremely beneficial in enriching the soil with organic nutrients. Many species of bacteria and cyanobacteria have the ability to fix free atmospheric nitrogen, Rhizobium is a symbiotic bacteria found in the root nodules of leguminous plants. Azospirilhum and Azotobacter are free living nitrogen-fixing bacteria, whereas Anabena, Nostoc and Oscillitoria are examples of nitrogen – fixing cyanobacteria. Biofertilizers are cost effective and eco – freindly.

Microbes can also act as biopesticides to control insect pest in plants. An example of biopesticides is Bacillus thuringiensis, which produces a toxin that kills the insect pests. Dried bacterial spores are mixed in water and sprayed in agricultural fields. When larvae of insects feed on crops, these bacterial spores enter the gut of the larvae and release toxins, thereby it. Similarly, Trichoderma are free living fungi. (MPBoardSolutions.com) They live in the roots of higher plants and protect them from various pathogens. Baculoviruses is another bio – pesticides that is used as a biological control agent against insects and other arthropods.

Question 5.
Arrange the following in the decreasing order (most important first) of their importance, for the welfare of human society. Give reasons for your answer. Biogas, Citric acid, Penicillin and Curd.
Answer:
The order of arrangement of products according to their decreasing importance is:
Penicillin – Biogas – Citric acid – Curd.

Penicillin is the most important product for the welfare of human society. It is an antibiotic, which is used for controlling various bacteral diseases. The second most important product is biogas. It is an eco – friendly source of energy. The next important product is citric acid, which is used as a food preservative. The least important product is curd, a food item obtained by the action of lactobacillus bacteria on milk. Hence, the products in the decreasing order of their importance are as follows :
Penicillin – Biogas – Citric acid – Curd.

Question 6.
How do biofertilizers enrich the fertility of the soil?
Answer:
Biofertilizers are living orgainsms which help in increasing the fertility of soil. It involves the selection of beneficial microorganisms that help in improving plant growth through the supply of plant nutrients. These are introduced to seeds, roots, or soil to mobilize the availability of nutrients by their biological activity. Thus, they are extremely beneficial in enriching the soil with organic nutrients. (MPBoardSolutions.com) Many species of bacteria and cyanobacteria have the ability to fix free atmopheric nitrogen. Rhizobium is a symbiotic bacteria found in the root nodules of leguminous plants. Azospirillium and Azotobacter are free living nitrogen – fixing bacteria, whereas Anabaena, Nostoc and Oscillitoria are examples of nitrogen – fixing cyanobacteria. Biofertilizers are cost effective and eco – friendly.

Question 7.
Describe role of microorganisms in biogas production.
Answer:
The gas produced by anaerobic fermentation of waste biomass or waste bi – product is called as biogas.

Role of microorganisms in biogas production:
In biogas production under anaerobic condition, microorganism including methanogenes convert the bio – degradable organic waste into biogas and humus. The whole process is completed in three stages:

1. Solubilization
2. Acidogenesis
3. Methanogenesis.

1. Solubilization : In this step methanogenic bacteria hydrolyse the complex polymers into organic acids, alcohol by hydrolytic fermentation.

2. Acidogenesis : It is a biological reaction where simple monomers are Converted into volatile fatty acids, then into acetic acid, CO₂ and hydrogen.

3. Methenogenesis: Anaerobic digestion occurs in which acetate and CO₂ are converted into methane.

MP Board Class 12th Biology Important Questions