Category: Class 11

Students get through the MP Board Class 11th Physics Important Questions Chapter 14 Oscillations  which are most likely to be asked in the exam.

MP Board Class 11th Physics Important Questions Chapter 14 Oscillations

Oscillations Class 11 Important Questions Very Short Answer Type

Question 1.
Define simple harmonic motion with example.
Answer:
When a body moves to and fro about a point in a straight line such that its acceleration is always directly proportional to the displacement and directed towards the mean position, then its motion is called simple harmonic motion.
Example: Motion of a body suspended by a spring, motion of simple pendulum.

Question 2.
What is oscillatory motion?
Answer:
When a body moves periodically to and fro or back and forth about a definite point, then its motion is called oscillatory motion.
Motion of simple pendulum, vibration of tuning fork etc. are the examples of oscillatory motion.

Question 3.
For what type of motion of a body acceleration is directly proportional to displacement?
Answer:
It is simple harmonic motion.

Question 4.
When the velocity and acceleration will be maximum in S.H.M.?
Answer:
The velocity is maximum at the mean position and acceleration is maximum at the extreme position.

Question 5.
Write the formula for kinetic energy and potential energy for a body executing S.H. M.?
Answer:
Kinetic Energy = \(\frac{1}{2} m \omega^{2}\left(a^{2}-y^{2}\right)\)
Potential Energy = \(\frac{1}{2} m \omega^{2} y^{2}\)
Where m – particle of mass,
ω – angular frequency,
a – amplitude and
y = displace-ment of the particle form its mean position.

Question 6.
At which position kinetic energy and potential energy is zero for a body executing S.H.M.?
Answer:
At maximum displacement position (y = ±a) kinetic energy is zero and at mean position (y = 0) potential energy is zero.

Question 7.
Define simple pendulum.
Answer:
When a heavy point mass is suspended by a weightless, inextensible and perfectly flexible string by a rigid support, then it is called simple pendulum.

Question 8.
Which of the following relationships between acceleration ‘a’ and the dis-placement x: of a particle involve simple harmonic motion ? (NCERT)
(a) a = 0.7x,
(b) a = 200x²,
(c) a = -10x²
(d) a = 100x³.
Answer:
In a S.H.M. the relation between acceleration and displacement is
a=-ω²x
(a) a = 0.7 x
Comparing with eqn. (1),
-ω² = 0.7
or’
ω = \(\sqrt{-0 \cdot 7}\)
∵ Value of a is not real therefore the motion is not S.H.M.

(b) a = -200x²
This motion is not a S.H.M.

(c) a = -10x
Comparing-with eqn. (1),
ω² =10
or
ω = \(\sqrt{10}\)
∵ ω has a real value therefore, the relation shows a S.H.M.

(d) a = 100 x³
This motion is not a S.H.M.

Question 9.
The length of a spring is increased, how the time-period be affected?
Answer:
The time-period will increase.

Question 10.
On which conservation law S.H.M. is based?
Answer:
S.H.M. is based on law of conservation of energy.

Question 11.
What is second pendulum? What is its length?
Answer:
The pendulum whose periodic time is two seconds is known as second pendulum. The length of second pendulum is about 100 cm.

Question 12.
Write the characteristics of S.H.M.
Answer:
Characteristics of S.H.M.:

• The motion is periodic.
• The particle moves to and fro about a point in a straight line.
• The acceleration is directly proportional to the displacement.
• The direction of acceleration is always towards the mean position.

Question 13.
Define time-period, frequency and amplitude.
Answer:
Time period: The time taken by an oscillating body to complete one oscillation is called time-period.
Frequency: The number of vibrations or oscillations in one second is called frequency.
Amplitude: The maximum displacement of a particle about its mean position is called amplitude. It is denoted by a or A.

Question 14.
What is the relation between S.H.M. and Circular motion?
Answer:
When any particle is performing uniform motion on the circumference of a circle, then the perpendicular dropped on the diameter of the circle from any point of rotation execute S.H.M. along the diameter of circle.

Question 15.
Define spring constant and write its S.I. unit and dimensional formula.
Answer:
Spring constant: The force required to elongate or compress a spring through unit length is called spring constant.
Unit: SI unit of spring constant is Nm1.
Dimensional formula : Dimensional formula is [ML°T^-2].

Question 16.
Write the formula for the time-period of a body suspended by a spring.
Answer:
Where, T= Time-period, m = Mass suspended and k = Force constant.

Question 17.
What is meant by the effective length of a simple pendulum?
Answer:
The length of pendulum from the point of suspension to the centre of bob is called effective length.

Question 18.
Why the length of simple pendulum is measured up to the centre of the bob?
Answer:
The centre of gravity point of spherical bob lies at its centre. The concept of heavy point mass is assumed that the weight of the body acts at the C.G. point. Therefore the effective length is measured up to centre of the bob.

Question 19.
How time period of a simple pendulum change with effective length of a simple pendulum.
Answer:
We know \(T=2 \pi \sqrt{\frac{l}{g}}\)
or
T ∝ \(\sqrt{l}\)
When effective length is increased, time period also get increased.

Question 20.
Can the pendulum clock be used in a satellite?
Answer:
No, because g = 0 inside satellite, therefore T = ∞.
So, in place of pendulum clock, spring clock can be used inside it.

Question 21.
The pendulum clock cannot be used in artificial satellite. Why?
Answer:
In an artificial satellite, the bodies are in the state of weightlessness. Hence, g becomes zero. Thus, the time-period becomes infinite.Therefore, it cannot be used in the satellite.

Question 22.
A boy is swinging, sitting on a swing. If another boy sits beside him, what will happen to its time-period?
Answer:
The time period is independent of mass, hence there will be no effect.

Question 23.
A boy is swinging, standing on a swing. If he sits on it, what happen to the time-period?
Answer:
When the boy will sit on the swing, his C.G. point will shift downwards. Hence, the effective length will increase therefore the time-period will also increase.

Question 24.
What is the reason that the frequency of oscillation clocks depend on the rise and fall of mercury level in the thermometer?
Answer:
If the mercury level increases, it shows that temperature has increased and the decreases in mercury level shows a decrease in temperature. Due to increase in temperature, the length of pendulum increases and \(T \propto \sqrt{l}\) so time period also increases. Hence clock runs slow, i.e., the frequency of oscillation decreases. But with decrease in temperature, length decreases and so time-period decreases. Hence clock runs fast i.e., frequency of oscillation increases.

Question 25.
The pendulum clock becomes fast when it is taken to pole from equator. Why?
Answer:
We have, T = \(2 \pi \sqrt{\frac{l}{g}}\)
If l is constant ,then T ∝ \(\frac{1}{\sqrt{g}}\)
Hence, time-period is inversely proportional to the square root of acceleration due to gravity. At poles the value of g is greatest, hence the time-period will decrease at poles, hence the clock will be fast.

Question 26.
Why pendulum clock does not show time at satellite?
Answer:
At satellite g = 0, therefore, i.e., time period of pendulum become infinity, therefore it does not show time of satellite.

Question 27.
What is damped oscillation?
Answer:
When the amplitude of an oscillating body decrease gradually by some external opposing force, then its oscillation is called damped oscillation.

Question 28.
Define forced vibration.
Answer:
When an oscillating body oscillates under the influence of some external force, then it oscillates with the frequency of external force, then its oscillation is called forced vibration.

Question 29.
What do you mean by forced oscillation?
Answer:
When the body oscillate under the influence of an external periodic force, with a constant amplitude and a frequency equal to the periodic force, the oscillation of the body is called forced oscillation.

Question 30.
Define resonance.
Answer:
When a body vibrates under the influence of a periodic force and its frequency becomes equal to that of periodic force then the amplitude of the body increases. This phenomenon is known as resonance.

Question 31.
Why marching troops are asked to break their steps while crossing the bridge?
Answer:
The footsteps of marching troops produces a periodic force on the bridge. If the frequency of foot-steps becomes equal to the natural frequency of bridge, the amplitude of bridge will increase and hence the bridge may break.

Question 32.
What is the effect of forced oscillation on amplitude and frequency?
Answer:
Due to forced oscillation amplitude of oscillating body go on reducing and frequency become less than the natural frequency.

Question 33.
When forced oscillation become resonance oscillation?
Answer:
When the natural frequency of the body becomes equal to the frequency due to periodic force then forced oscillation is converted into resonance oscillation.

Question 34.
A simple pendulum is suspended in a lift, whose time-period is T. What will aeration ‘a’ (i) upward, (ii) downward.
Answer:
(i) When the lift accelerate with acceleration ‘a’ upward then the time-period will decrease
( T = \(2 \pi \sqrt{\frac{l}{g+a}}\))
(ii) On acceleration with acceleration ‘a’ downward then its timeperiod will increase
( T = \(2 \pi \sqrt{\frac{l}{g-a}}\) )

Question 35.
What will be the time-period of effective length infinity?
Answer:
84.6 minute.

Question 36.
What will be the time period of pendulum at the centre of earth? Why?
Answer:
At the centre of earth the time-period of pendulum will be infinity as ‘g’ is zero at the centre.

Question 37.
A spring of spring constant k is cut into three equal pieces. What will be the time period of each?
Answer:
The spring constant of each piece will be 3k.

Oscillations Class 11 Important Questions Short Answer Type 

Question 1.
The bob of simple pendulum is a ball filled with water, if a small hole is made at the bottom of the ball, how will its time period change as water is drains out of it?
Answer:
When the ball is completely filled with water, its centre of gravity be at its geometric centre. So, it will oscillate with definite time period. But as the water drains out of it through the hole, the centre of gravity is displaced gradually downward. Hence the effective length also increases gradually. Therefore the time period of the pendulum also increases gradually. But when whole of the water is removed then its centre of gravity again shifted to its geometric centre and its time-period become same as was in initial position.

Question 2.
What is simple harmonic motion. Write its four characteristics.
Answer:
When a body moves to and fro about a point in a straight line such that its acceleration is always directly proportional to the displacement and directed towards the mean position, then its motion is called simple harmonic motion.
Example: Motion of a body suspended by a spring, motion of simple pendulum.
Characteristics of S.H.M.:

• The motion is periodic.
• The particle moves to and fro about a point in a straight line.
• The acceleration is directly proportional to the displacement.
• The direction of acceleration is always towards the mean position.

Question 3.
Write the relation between acceleration and displacement of a particle ex-ecuting1 S.H.M. Also deduce expression for time-period from it?
Answer:
Acceleration of the particle with displacement y is α = -ω²y
Where ω is angular velocity of particle here (-) sign indicate the direction of accel-eration only.
Hence, ω² = \(\frac{\alpha}{y}\) = \(\frac{\text { Acceleration }}{\text { Displacement }}\)
or
ω = \(\sqrt{\frac{\alpha}{y}}\)
But T =\(\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{\frac{\alpha}{y}}}=2 \pi \sqrt{\frac{y}{\alpha}}\)
T = \(2 \pi \sqrt{\frac{\text { Displacement }}{\text { Acceleration }}}\)

Question 4.
Write the expression for displacement, velocity and acceleration of a particle executing S.H.M. and say :
(i) When the velocity of the particle will be maximum and zero.
(ii) When the acceleration of the particle will be maximum and zero.
Answer:
Displacement y = a sin ωt
Velocity ν = ω \(\sqrt{a^{2}-y^{2}}\)
Acceleration a = – ω²y
Where a is amplitude and ω is angular acceleration.

(i) Velocity is maximum at mean position (y = 0)
νmax = ω.a
Velocity is zero at maximum displacement position ν = 0 , when y = ±a.

(ii) Acceleration is maximum when displacement is maximum (y = ±a)
α = ± ω² a, when y = ±a
Acceleration is zero at mean position α = 0, when y = 0.

Question 5.
What is meant by simple pendulum? When the bob is displaced from its mean position it starts oscillations. Why?
Answer:
If a heavy point mass be suspended by a weightless, perfectly flexible and inextensible string from a rigid support and an arrangement is made for its frictionless oscillations, then this arrangement is called simple pendulum. As shown in fig. when bob is displaced from its mean position A to a position B, then the bob raised up to height ‘h\ so that its potential energy increases.

Now, if the bob is released, then its centre of gravity falls down to obtain the stable equilibrium and when it reaches to the mean position its potential energy is converted into kinetic energy. At points, it will not stay, but due to inertia moves forward and hence the bob starts to execute simple harmonic motion to and fro about its mean position.

Question 6.
Write down the formula of time-period of simple pendulum and state the factors on which it does depend.
Answer:
The time-period of simple pendulum is given by :
T = 2π \(\sqrt{\frac{l}{g}}\)
From the above formula, it is clear that the time period of simple pendulum depends on its effective length ‘l’ and acceleration due to gravity ‘g’.
(i) Dependence on length f: The time period of simple pendulum is directly proportional to the square root of effective length of pendulum, i.e., T α \(\sqrt{l}\)
Example: When a boy swinging on a swing stand up suddenly, then his centre of gravity rises up and so the effective length of the swing reduces which results an increase in time period.

(ii) Dependence on acceleration due to gravity ‘g’: The time-period T of simple pendulum is inversely proportional to the square root of acceleration due to gravity at that place i.e., T α \(\frac{1}{\sqrt{g}}\)
Example: When a pendulum is taken up to hills or down in a mine get slow because time-period T increases due to decrease in the value of ‘g’.

Question 7.
Write down the laws of simple pendulum. Give practical application of each law.
Answer:
The laws of simple pendulum and their practical applications are given below :
(i) Law of length: If the value of ‘g’ remains constant, then the time period of simple pendulum is directly proportional to the square root of effecti ve length of simple pendulum.
i-e., T α \(\sqrt{l}\)
This law is used to repair the pendulum clocks when they get slow or becomes fast while taken up to hills or down in a mine.

(ii) Law of acceleration due to gravity : If the effective length of simple pendulum remains constant, then the time period T of the simple pendulum is inversely proportional to the acceleration due to gravity, i.e., T α \(\frac{1}{\sqrt{g}}\)
Due to this reason the pendulum clock get slow while taken up to hills or down in mines.

(iii) Law of mass: The time period of simple pendulum does not depend upon the mass of bob or that of thread.
Hence, either bob is heavy or lighter, if the value of 7’ and ‘g’ are constant, then there is no effect in its time-period. .

(iv) Law of isochronism: If the amplitude of oscillations of simple pendulum are very small, then the time-period of pendulum does not depend upon the amplitude of oscillations.
Because of this, the angular displacement of the bob is kept small in the experiment of simple pendulum.

Question 8.
Establish the relation between time-period and frequency.
Answer:
Let the time-period of a particle is T and its frequency is υ.
∴ The particle complete one vibration in T sec.
Hence, in 1 sec the number of vibration, υ = \(\frac{1}{T}\)
or
υT = 1.

Question 9.
Derive an expression for the displacement of a particle executing S.H.M.
Or
Find an expression for the displacement of S.H.M.
Answer:
Let XYX’Y’ is a circle whose centre is O and radius a. A particle is moving with uniform angular speed ω on the circle.
Let at time t = 0, the particle is at X and in time t, it is at P. A perpendicular PN is drawn on YOY’ from P and N is foot of the perpendicular.

The particle takes t second to subtend an ∠POX = θ.
∴ ω = \(\frac{\theta}{t}\)
or
θ = ωt
Let the displacement ON = y
∴ In ∆ NPO, sin NPO = \(\frac{O N}{O P}\)
But, ∠NPO = ∠POX = θ = ωt
∴ sin ωt = \(\frac{y}{a}\)
or
y = a sin ωt
This is the required equation for the displacement of S.H.M.

Question 10.
Derive the expression for the velocity of a particle executing S.H.M.
Or
Derive an expression for velocity of a particle executing S.H.M. When is the velocity maximum and minimum?
Answer:
Expression for velocity: Let XYX’Y’ is a circle of centre O and radius a. A particle is moving on the circle with angular velocity ω.
Let at t = 0, the particle is at X and after t sec, it is at P.
Let at point P, the velocity of the particle is ν along the tangent at P.

Now, velocity v is resolved into two parts :
(i) ν sin θ, parallel to PN and
(ii) ν cos θ, perpendicular to PN.
Since,ν cos θ is parallel to the direction of motion of foot of the perpendicular N.
∴ Velocity of N, υ = ν cosθ
υ = ν cos ωt.
or
υ = ν \(\sqrt{1-\sin ^{2} \omega t}\)
( ∴ sin² ωt + cos² = 1)

This is the required expression.

Case 1. Ify = 0, then from eqn. (1), we get
ν =ω \(\sqrt{a^{2}-0}\) = ωa
∴ Velocity will be maximum at mean position.
Case 2. If y = a, then from eqn. (1), we get
ν =ω \(\sqrt{a^{2}-a^{2}}=\) = 0
∴ Velocity will be minimum at the extreme position.

Question 11.
Derive the expression for the acceleration of a particle executing S.H.M. Find where the acceleration is maximum and minimum.
Answer:
We have by displacement equation,
y = a sin(ωt + Φ) …(1)
If the acceleration is α, then we have


This is the required expression.
Case 1. If y = 0, then by eqn. (3),
α =0
i.e., at the mean position the acceleration will be zero i.e., minimum.
Case 2. If y = a, then by eqn. (3),
α = – ω²a
i.e., at extreme position the acceleration is maximum.

Question 12.
Derive the expression for the time>period and frequency of a particle ex-ecuting S.H.M.
Answer:
Now, we have the magnitude of acceleration is given by

Again, we know that frequency,
υ = \(\frac{1}{T}\)
∴ υ = \(\frac{1}{2 \pi} \sqrt{\frac{\text { Acceleration }}{\text { Displacement }}}\)

Question 13.
Mass m is suspended by an ideal spring. It oscillates up and down. If the force constant of spring Is k, then prove that the time-period,
T =2 π \(\sqrt{\frac{m}{k}}\)
or
A mass is suspended by a spring. It Is pulled downwards and then left. Prove that it executes S.H.M. Find the expression for time-period.
Answer:
Let a spring is suspended by a rigid support and mass ni is suspended at its lower end.
It is displaced downwards at a distance y and left.
The restoring force F = -ky …( 1)
Where, k is force constant.
Also, by Newton’s second law of motion,
F=mα …………. (2)
Where, α is acceleration.
∴ From eqns. (1)and (2), we get
mα =-ky ……….. (3)
or
α =\(-\frac{k}{m}\) .y
Also, we have, for S.H.M.
α = – ω² y ……..(4)
Since, the L.H.S. of eqns. (3) and (4) are equal, hence the R.H.S. must be equal.

Question 14.
Derive the expression for K.E. and P.E. of a particle executing S.H.M. and also prove that total energy of a particle remains constant.
Answer:
K.E. of the particle : We know that velocity of a particle executing S.H.M. is given by,
ν = ω\( \sqrt{a^{2}-y^{2}} \)

K.E.= \( \frac{1}{2}\)mv²
=\( \frac{1}{2}\) m{ ω\( \sqrt{a^{2}-y^{2}} \)²
=\( \frac{1}{2}\)mω²(a²-y²) …(1)
P.E. of the particle: Now, the acceleration of a particle executing S.H.M. is given by α = -ω²y

Also, by Newton’s second law of motion,
Force = Mass x Acceleration
∴ F = mα
or
F = -mω²y
Now initially the displacement was zero and force was zero. The force was gradually increases and hence the displacement also increases to y.

∴Mean force =\(\frac{0+F}{2} \) = mω²y
∵ Work W= Mean force x Displacement
∴ W = \(\frac{1}{2} \) mω²y ×y
or
W = \(\frac{1}{2} \) mω²y²

By equns.(1) and (2), we get
K.E +P.E = \(\frac{1}{2} \) mω² (a²-y²)+ \(\frac{1}{2} \) mω² y²
or
E = \(\frac{1}{2} \) mω²a² –\(\frac{1}{2}\)mω²y² +\(\frac{1}{2}\)mω²y²
or
E = \(\frac{1}{2}\)mω²a²
As ω and a are constant, therefore the total energy remains constant.

Question 15.
What is second’s pendulum? Calculate its effective length.
Answer:
Second’s pendulum: A simple pendulum whose time-period is 2 seconds, is called second’s pendulum.

Oscillations Class 11 Important Questions Long Answer Type

Question 1.
Deduce the expression for time-period of simple pendulum T = 2π \(\sqrt{\frac{l}{g}} \)
Answer:
Motion of a simple pendulum: Suppose that m is the mass of the bob and l is the effective length of a simple pendulum. S is a point of suspension and O is mean position of the bob. Suppose at any instant during oscillation, the bob is in the position P. Then displacement OP = y and ∠ OSP = θ. In this position of the bob, two forces act upon it:
(i) Tension T is the thread, in the direction PS upward along the thread and

(ii) The weight of the bob, mg, vertically downward. The weight mg can be resolved into two components :
(a) mg cos θ, in the direction SP downward along the thread.
(b) mg sin θ, perpendicular to the thread SP.
The component mg cos θ balances the tension Tin the thread. The component mg sin θ tends to bring the bob back to its mean position. This is known as restorting force on the bob. Thus, the restorting force acting on the bob is given as F= -mg sinθ

Where negative sign shows that the direction of force is opposite to the direction which displacement increases, i.e., towards the mean position O.
1f the angular displacement of the bob be small, then sin θ = 0 (for example, if θ = 5⁰, then sin θ= 0.0872 and θ=0.0873 radian). Thus,
F= -mgθ
But,
θ= \( \frac{\mathrm{Arc}}{\mathrm{Radius}} \) = \( \frac{O P}{S P} \) = \( \frac{y}{l} \)
F= -mg.\( \frac{y}{l} \) …….. (1)

If α be the acceleration produced in the bob due to restoring force, then by Newton’s law of motion
F = mα ……. (2)
From eqns. (1) and (2), we get
mα = -mg \( \frac{y}{l} \)
or
α= –\( \frac{g}{l} \).y …(3)

At a given place g is constant and for a given pendulum is constant.
Therefore,
α ∞ -y …….. (4)
Thus, the acceleration of the bob is directly proportional to its displacement. Hence, for small amplitude, the motion of a simple pendulum is simple harmonic motion. Its periodic time is

Oscillations Class 11 Important Numerical Questions

Question 1.
The time-period of a particle executing S.H.M. is 2 sec. Calculate the time after t =0, that its amplitude behalf of the displacement.
Solution:
Given: T= 2 sec and y = \(\frac{a}{2}\)
∵ y = a sin ωt
Where ω =\( \frac{2 \pi}{T} \)
∴ y = a sin \( \frac{2 \pi}{T} \) t ⇒ \( \frac{a}{2}= \) = a sin \( \frac{2 \pi}{T} \) t

Question 2.
The length of a simple pendulum ¡s 39.2 / π² metre. If g = 9.8 ms^-2, then calculate the time-period of simple pendulum.
Solution:
Given:l = \( \frac{39 \cdot 2}{\pi^{2}} \), g = 9.8 ms^-2
Formula:

Question 3.
The mass of a particle executing S.H.M. is 0.4 kg and amplitude and time period are 0.5 m and π/ 2 respectively. Calculate the velocity and ICE. of particle at a displacement 0.3 m.
Solution :
Given:m = 0.4kg , a = 0.5 m, T= π/2 sec, y = 0.3 m.
Npw we have

Again,

Question 4.
A simple pendulum execute 60 oscillation per minute. Find its effective length ? (g = 981 cm/ s²)
Solution:
Given: Time-period T = \( \frac{\text { Time }}{\text { No.of oscillations }} \) = \frac{1 \mathrm{~min}}{60} = 1 sec

Question 5.
The value of acceleration due to gravity on a planet is \( \frac{1}{4} \) th of that of earth.
If the time-period of simple pendulum on earth is 2 sec, then find the time-period on the planet.
Solution:
g₂= \( \frac{1}{4}\) g₁

Question 6.
What will be the time.period of a simple pendulum of length \( \frac{9 \cdot 8}{\pi^{2}} \) m? Name such pendulum.
Solution:
Given: l= \( \frac{9 \cdot 8}{\pi^{2}} \)m g = 9.8 ms^-2
Formula:
T =2π \( \sqrt{\frac{l}{g}} \)

Question 7.
Two pendulum of length 100 cm and 110*25 cm start oscillating at same time. After how much oscillation again they will oscillate at same time.
Solution:
Let the time-period of the 100 cm pendulum be T₁
T₁ = 2 π\( \sqrt{\frac{100}{g}} \)

and if T₂is the time-period for 110.25 cm pendulum, then
T₂ = 2 π\( \sqrt{\frac{110 \cdot 25}{g}} \)
T₁ <T₂
∴ Both the pendulum to oscillate at same time if the big pendulum perform n oscillaion and small pendulum, perform (n + 1) oscillation.
i.e.,(n+1)T₁ = nT ₂

i.e., Big pendulum after 20 oscillation or small pendulum after 21 oscillation will start oscillating together.

Question 8.
Find the length of second pendulum at (i) Surface of earth (g = 9-8 m/s2), (ii) Surface of moon (g = 1.65 m/s2).
Solution:
Time-period for second pendulum is T= 2 sec.
∴From T = 2 π\( \sqrt{\frac{l}{g}} \)

(i) Length at earth surface
2 = 2π² \( \sqrt {\frac{l}{9.8}} \)
or
l = π² \( \ {\frac{l}{9.8}} \)
or
π².l = 9.8
or
l =\( \frac{9 \cdot 8}{(\pi)^{2}} \) = \( \frac{9 \cdot 8}{(3 \cdot 14)^{2}} \) = 0.99m

(ii) Length of moon
l = \( \frac{1 \cdot 65}{\pi^{2}} \)
or
l = \( \frac{1 \cdot 65}{(3 \cdot 14)^{2}} \) = 0.167 m.

Question 9.
A mass of 98 kg suspended in a spring is oscillating whose spring constant is 200 N/m. Find its time-period.
Solution:
We know,
T = 2π\( \sqrt{\frac{m}{k}} \)
or
T = 2 ×3.14 \( \sqrt{\frac{98}{200}} \)
T= 4-396 second.

Question 10.
The total energy of a particle executing S.H.M. is E. What will be the K.E. and P.E. of particle at the displacement half of the amplitude?
Solution:
We know that total energy is
E = \( \frac{1}{2} \) mω² a²
Also, K.E = \( \frac{1}{2} \) mω²( a² – y ²)

Now, at y = \( a / 2 \),
K.E = \( \frac{1}{2} \) mω²\( \left(a^{2}-\frac{a^{2}}{4}\right) \)
= \( \frac{1}{2} \) mω² . \( \frac{3 a^{2}}{4} \)
= \( \frac{3}{8} \) mω²a² = \( \frac{3}{4} \) E.

Again, P.E = \( \frac{1}{2} \) mω² y ²
Now, at y = \( a / 2 \),
P.E = \( \frac{1}{2} \) mω² \( \frac{a^{2}}{4} \)
P.E = \( \frac{1}{8} \) mω² a² = \( \frac{1}{4} \) E.

Question 11.
The amplitude of a particle executing S.H.M. is 0*01 m and frequency is 60 Hz. What will be the maximum acceleration of the particle?
Solution:
Given : a = 0.01m, υ = 60 Hz Now, maximum acceleration is given by
α = ω² a
= (2π υ ) ²a
= 4π²υ ²a
= 4×3.14 ×3.14×60×60×0.01
= 1419.78ms^-2

Question 12.
At a place a body travels 125 m in 5 sec during the free fall under gravity. What will be the time period of a simple pendulum of length 2-5 m at that place?
Solution:
Given : S= 125 m, υ = 0, t = 5 sec.
Now, s = ut + \( \frac{1}{2} \) gt²
or
125 = 0 +\( \frac{1}{2} \)g×25
or
g = 10 ms^-2
Again,
= 2π\( \sqrt{\frac{l}{g}} \)
= 2π \( \sqrt{\frac{2.5}{10}} \)
= 2π ×0.5 = π sec.

Question 13.
A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0-6 s. What is the weight of the body ? (NCERT)
Solution:
Given :m = 50 kg, y = 20 cm = 0.2 m, T- 0.6 s
From F=ky
∴ mg = ky
or
K = \( \frac{\mathrm{mg}}{\mathrm{y}} \) = [ latex] \frac{50 \times 9 \cdot 8}{0 \cdot 2} [/latex] = 2450Nm^-1

Weight of the body
W = m’g
= 22.36×9.8 = 219.17 N
≈ 219N.

Question 14.
A spring having spring constant 1200 Nm’ is mounted on a horizontal table as shown in fig. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled side ways to a distance of 2.0 cm and released. (NCERT)
Determine
(j) The frequency of oscillations.
(ii) Maximum acceleration of the mass and
(iii) The maximum speed of the mass.

Solution:
Given:k=1200Nm^-1,m= 3kg, amplitude a =2 cm =2 × 10²m
(i) Frequency υ =\( \frac{1}{2 \pi} \) \( \sqrt{\frac{k}{m}} \) = \( \frac{1}{2 \times 3 \cdot 14} \sqrt{\frac{1200}{3}} \)
= \( \frac{1}{6 \cdot 28} \) ×\( \sqrt{400}\) = \( \) \frac{20}{6 \cdot 28}
= 3.18 s^-1 ≈3.2 ^-1

(ii) α = -ω ² y
∴α _max = ω ²a = \( \frac{k}{m} \) .a = \( \frac{1200}{3} \) ×2 ×10^-2
= 8.0 ms ^-2

(iii) ∴ν _max=ωa= \( \sqrt{\frac{k}{m}} \) ×a =\( \sqrt{\frac{1200}{3}} \) ×2 ×10^-2
= 20 × 2 × 10^-2 = 0.4 ms ^-1

Oscillations Class 11 Important Questions Objective Type

1. Multiple- choice questions:

Question 1.
Which of the following equation does not represent simple harmonic, motion:
(a) x = asin(ωt+δ)
(b) x = bcos(ωt+Φ)
(c) x = atan(ωt + Φ)
(d) x = asin(ωt cosωt.
Answer:
(c) x=atan(ωt + Φ)

Question 2.
Time-period and amplitude of a particle executing simple harmonic motion is Tand a. The minimum time taken to reach distance will be:
(a) T
(b) \(\frac{T}{4}\)
(c) \(\frac{T}{8}\)
(d) \(\frac{T}{16}\)
Answer:
(c) \(\frac{T}{8}\)

Question 3.
Time-period and amplitude of a particle executing simple harmonic motion is T and a. Its maximum velocity is :
(a) \(\frac{4 a}{T}\)
(b) \(\frac{4 a}{T}\)
(c ) \(2 \pi \sqrt{\frac{a}{T}}\)
(d) \(\frac{2 \pi a}{T}\)
Answer:
(d) \(\frac{2 \pi a}{T}\)

Question 4.
Acceleration in S.H.M. is :
(a) Maximum at amplitude position
(b) Maximum at mean position
(c) Remain constant
(d) None of these.
Answer:
(a) Maximum at amplitude position

Question 5.
The ratio of acceleration to displacement of a particle executing S.H.M. is measurement of:
(a) Spring constant
(b) Angular acceleration
(c) (Angular acceleration)²
(d) Restoring force.
Answer:
(c) (Angular acceleration)²

Question 6.
When displacement is half of amplitude, then ratio of potential energy to the total energy is:
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{4}\)
(c) 1
(d) \(\frac{1}{8}\)
Answer:
(b) \(\frac{1}{4}\)

Question 7.
What remain constant in S.H.M.:
(a) Restoring force
(b) Kinetic energy
(c) Potential energy
(d) Time-period.
Answer:
(d) Time-period.

Question 8.
Graph between ‘f and ‘F obtained as :
(a) Hyperbola
(b) Parabola
(c) Straight line
(d) None of these.
Answer:
(b) Parabola

Question 9.
The time-period of a seconds pendulum is :
(a) 1 second
(b) 2 second
(c) 3 second
(d) 4 second.
Answer:
(b) 2 second

Question 10.
Resonance is example of:
(a) Tuning fork
(b) Forced oscillation
(c) Free oscillation
(d) Damped oscillation.
Answer:
(b) Forced oscillation

2. Fill in the blanks:

1. A periodic motion of constant amplitude and same frequency is called ………… .
Answer:
Simple harmonic motion

2. In simple harmonic motion its total energy is ………. .
Answer:
Constant

3. The restoring force setup per unit extension in the spring is called ………… .
Answer:
Spring constant (Force constant)

4. When a body oscillates on both sides of its mean position in a straight line, then this bind of motion is called ………… .
Answer:
Simple harmonic motion

5. The maximum displacement of a body in oscillatory motion is called ………… .
Answer:
Amplitude

6. In the presence of damping forces, the amplitude of oscillations of a body ………… .
Answer:
Decreases

7. The time-period of second pendulum is ………… second.
Answer:
Two

8. The periodic time of simple pendulum does not depends upon
Answer:
Mass.

3. Match the following:
I.

Column ‘A’	Column ‘B’
1. Maximum kinetic energy in S.H.M.	(a) 2π\( \sqrt{\frac{\text { Displacement }}{\text { Acceleration }}} \)
2.Maximum potential energy in S.H.M	(b) Remain constant
3. Total energy of S.H.M.	(c) At mean position
4. Acceleration of S.H.M.	(d) At maximum displacement
5. Time-period of S.H.M.	(e) Directly proportional to displacement.

Answer:
1. (c) At mean position
2. (d) At maximum displacement
3. (b) Remain constant
4. (e) Directly proportional to displacement.
5.  (a) 2π\( \sqrt{\frac{\text { Displacement }}{\text { Acceleration }}} \).

II.

Column ‘A’	Column ‘B’
1. Motion of moon	(a) Forced oscillation
2. Resonance	(b) Conservation of energy and momentum
3. S.H.M.	(c) Periodic motion
4. Mechanical waves	(d) Main frequency
5. Minimum frequency of vibration	(e) S.H.M.

Answer:
1. (c) Periodic motion
2. (a) Forced oscillation
3. (e) S.H.M.
4. (b) Conservation of energy and momentum
5. ((d) Main frequency.

4. Write true or false:

1. Every S.H.M. is a periodic motion.
Answer:
True

2. Every periodic motion is S.H.M.
Answer:
False

3. In S.H.M. total energy is directly proportional to square of amplitude.
Answer:
True

4. Acceleration is always zero for a particle executing S.H.M.
Answer:
False

5. Time-period of simple pendulum is directly proportion to square root of its length.
Answer:
True

6. Time-period of simple pendulum is inversely proportional to the square root of acceleration due to gravity.
Answer:
True

7. Time-period of simple pendulum depends on mass, material and size of the pendulum.
Answer:
False

8. In S.H.M., velocity is maximum when acceleration is minimum.
Answer:
True

9. The motion of moon around the earth is S.H.M.
Answer:
False

10. Time-period of hard spring is less than soft spring.
Answer:
True

11. Time-period of a simple pendulum cannot be one day.
Answer:
True

Students get through the MP Board Class 11th Physics Important Questions Chapter 3 Motion in a Straight Line which are most likely to be asked in the exam.

MP Board Class 11th Physics Important Questions Chapter 3 Motion in a Straight Line

Motion in a Straight Line Class 11 Important Questions Very Short Answer Type

Question 1.
When do an object of definite size is considered as a point object, in motion?
Answer:
When the distance travelled by the object is very large in comparison to the size of object.

Question 2.
What is one-dimensional motion? Give three examples.
Answer:
When an object moves in a straight line, then its motion is called one-dimensional motion.
Examples :

• A body falling under gravity.
• Train running on straight rails.
• A man moving on a straight road.

Question 3.
Define displacement and distance.
Answer:
Displacement : The change in position in a particular direction is called displacement.
Distance : The length of a path covered by a particle in a given time interval is called its distance.

Question 4.
Direction of motion of an object is determined by velocity or acceleration.
Answer:
Velocity.

Question 5.
Define velocity. Write its SI unit.
Answer:
Displacement per unit time of a moving body is called its velocity.
Or
In a particular direction, the distance travelled in unit time is called velocity.
Formula: Velocity = \(\frac{\text { Displacement }}{\text { Time }}\) Time
The SI unit of velocity is m/s. It is a vector quantity.

Question 6.
Define speed. Is it a scalar or vector quantity?
Answer:
The rate of change of position of a moving body is called its speed.
Or
The distance travelled in unit time by a moving body is called its speed. It is a scalar quantity.
speed = \(\frac{\text { Distance }}{\text { Time }}\)

Question 7.
What do you understand by relative velocity?
Answer:
Relative velocity : The relative velocity of a body w.r.t. another body is the rate of change of position of that body relative to another body.
Let the velocity of bodies A and B are v₁ and v₂ respectively, then
Velocity of B w.r.t. A = v₂ – v₁
and Velocity of A w.r.t. B= v₁ – v₂.

Question 8.
Define acceleration and write its SI unit.
Answer:
Acceleration is defined as rate of change in velocity. Its SI unit is metre/second².

Question 9.
If the velocity of a body is constant, then what will be its acceleration?
Answer:
Zero.

Question 10.
What will be the displacement of a bullet fired vertically upward when it returns to its original point?
Answer:
Displacement of bullet will be zero.

Question 11.
Explain with reason, can a moving body have acceleration when :
(1) It is moving with uniform speed.
Answer:
Yes, because when speed is uniform its direction may be changed.

(2) It is moving with uniform velocity.
Answer:
No, as acceleration is defined as change in velocity.

Question 12.
Can a moving particle has constant speed but variable velocity?
Answer:
Yes, in uniform circular motion.

Question 13.
Can a body moving with constant speed has acceleration?
Answer:
Yes, in uniform circular motion.

Question 14.
Can uniform acceleration change the direction of velocity?
Answer:
Yes, when a body thrown vertically upwards, at its highest point the direction of velocity changes, while the direction of g is always acting downwards.

Question 15.
Can the speed be zero in one-dimensional motion and acceleration be not zero?
Answer:
Yes, at the highest point, when the body is thrown vertically upwards.

Question 16
What does the area enclosed between the velocity-time curve and time axis show?
Answer:
Distance travelled by the body.

Question 17.
(i) Draw velocity-time graph of a body moving uniformly in a straight line,
(ii) Draw position-time graph of a body moving uniformly in a straight line.
Answer:

Question 18.
Draw the graphs of positive and negative accelerated motion of velocity-time graph.
Answer:

Question 19.
What does the slope represent in position-time graph?
Answer:
The slope of a position-time graph represents velocity of an object.

Question 20.
Which quantity is represented by the area under the velocity-time graphs?
Answer:
It represents displacement of an object.

Motion in a Straight Line Class 11 Important Questions Short Answer Type

Question 1.
Differentiate displacement and distance.
Answer:
Difference between displacement and distance :

Displacement	Distance
1. The difference of position coordinates in a given time-interval is called displacement.	The length of the path in a given time interval is called distance.
2. It is a vector quantity.	It is a scalar quantity.
3. Displacement may be positive, negative or zero.	It is always positive.
4. It does not depend upon the nature of the path.	It depends upon the nature of the path.

Question 2.
Differentiate velocity and speed.
Answer:
Difference between velocity and speed :

Velocity	Speed
1. The distance travelled in unit time in a particular direction, by a moving body is called its velocity.	The distance travelled in unit time by a moving body is called its speed.
2. It is a vector quantity.	It is a scalar quantity.
3. Velocity may be positive, negative or zero.	Speed is always positive.

Question 3.
Define uniform motion and uniform velocity.
Answer:
Uniform motion :
When a body travels equal distance in equal interval of time, in the same direction, then its motion is called uniform motion.
Uniform velocity :
When the displacement of a moving body is equal in equal interval of time, then its velocity is called uniform velocity.

Question 4.
In which of the following examples of motion can the body be considered approximately a point object:
(a) A railway carriage moving without jerks between two stations.
(b) A monkey sitting on the top of a man cycling smoothly on a circular track.
(c) A spinning cricket ball that turns sharply on hitting the ground.
(d) A tumbling beaker that has slipped off the edge of a table?
Answer:
In (a) and (b), we can consider the body approximately as a point object because the motion of the object involves changes of position by distance much greater than its size.

Question 5.
A player throws a ball upwards with an initial speed of 29 m/s :
(a) What is the direction of acceleration during the upward motion of the ball?
(b) What are the velocity and acceleration of the ball at the highest point of its motion?
(c)Choose the x = 0 m and t₀ = 0 s to be the location and time of the ball at its highest point, vertically downward direction to the positive direction of X-axis, and give the signs of position, velocity and acceleration of the ball during its upward and downward motion.
(d) To what height does the ball rise and after how long does the ball return to the player’s hand? (Take g = 9.8 ms^-2 and neglect air resistance)
Answer:
(a) Since the ball is moving under the effect of gravity, the direction of acceleration due to gravity is always vertically downwards.

(b) When the ball is at the highest point of its motion, its velocity becomes zero and the acceleration is equal to the acceleration due to gravity = 9.8 ms^-2in vertically downward direction.

(c) When the highest point is chosen as the location for x = 0 and t = 0 and vertically downward direction to be the positive direction of X-axis.
During upward motion : Sign of position is negative, sign of velocity negative and the sign of acceleration positive i.e., v < 0, a > 0.
During downward motion : Sign of position is positive, sign of velocity is positive and the sign of acceleration is also positive i.e:, v > 0, a > 0.

(d) Let, t = Time taken by the ball to reach the highest point.
H = Height of the highest point from the ground.
Taking vertically upward motion of the ball.
∴ Initial velocity, u =-29.4 ms^-1, a =g = 9-8 ms^-2,
Final velocity v = 0, s = H = ?, t =?
Using the relation, v²-u² = 2as, we get
0² – (29.4)² = 2 × 9.8 H
or H = \(\frac{-29 \cdot 4 \times 29 \cdot 4}{2 \times 9 \cdot 8}\) = -44.1 m
Where -ve sign shows that the distance is covered in upward direction. Using equations v=u + at, we get
0 = -29.4 + 9.8 × t
t = \(\frac{29 \cdot 4}{9 \cdot 8}\) = 3s
i.e., Time of ascent = 3s.
Also we know that when the object moves under the effect of gravity alone, the time of ascent is always equal to the time of descent.
∴ Total time after which the ball returns to the player’s hand = 2t = 2 × 3 = 6s.

Question 6.
Derive graphically the first equation of motion.
Answer:
Velocity-time graph for an uniformly accelerated motion is shown in the figure. It is clear from the graph that at t = 0 the initial velocity of a particle is u and after t sec the final velocity is v.
From ΔEBC

∴ Slope of the curve = Acceleration
or a = \(\frac{v-u}{t}\)
or at = v-u or v=u + at.

Question 7.
What is relative velocity? Derive the expression for it.
Answer:
The relative velocity with respect to a body at rest or in motion is defined the rate of change of position of a body with respect to another body.
Let A and B are two bodies moving with velocities v₁and v₂ and at time t their position coordinate are x₁(t) and x₂(t) respectively.
x₁(t) = x₁1(0) + v₁t …(1)
∴ x₂(t) = x₂(0) + v₂t …(2)
Where x₁(0) and x₂ (0) are position coordinates at t = 0.
Subtracting eqn. (1) from eqn. (2),
∴ x₂(t) – x₁(t) = x₂(0) – x₁(0) + (v₂ -v₁)t …(3)
In the eqn. (3), x₂(t)- x₁(t) represents the position of B w.r.t. A at time t.
∴ x₂(0) – x₁(0) represents the relative position or distance of B w.r.t. A,
Putting t = 0 in eqn. (3) then the relative displacement in 1 sec. will be v₂-v₁
∴ Relative velocity of B w.r.t. A = v₂-v₁
∴ By eqn. (3),

Question 8.
Derive graphically the second equation of motion.
Answer:
From the given graph displacement of an object = Area under the curve OACE
or
s = Area of rectangle OABE + Area of ∆EBC
or
s = OA × AB + \(\frac{1}{2}\)EB × BC
or
s = ut+\(\frac{1}{2}\)t(v – u)
v – u=at
s = ut + \(\frac{1}{2}\)t ×at
s = ut + \(\frac{1}{2}\)at ²

Question 9.
Derive graphically the third equation of motion.
Answer:
From the above v-t graph
Displacement of an object = Area of trapezium OACE

Motion in a Straight Line Class 11 Important Questions Long Answer Type

Question 1.
Derive the equation of motion by Integration method.
Answer:
First equation of motion : v = u + at :Let at any instant t, the velocity of a particle is v.
∴ Acceleration, a = \(\frac{d v}{d t}\)
or dv = adt …(1)
Let at t = 0, v = u
∴ Integrating eqn. (1), we get

∴ v-u = a(t- 0)
or v – u- at or v = u + at.
Second equation of motion : Let a particle is moving with velocity u along X-axis and at any instant t, its displacement is x.
∴ v = \(\frac{d x}{d t}\)
or
dx = vdt
Also, we know that,
v = u + at
∴ dx = (u + at)dt ….(1)
Now, at t = 0, x = x₀
∴ Integrating eqn. (1), we get

Third equation of motion :
Accordrng to definition of acceleration

Question 2.
Derive graphically the second and third equation of motion.
Answer:
From the given graph displacement of an object = Area under the curve OACE
or
s = Area of rectangle OABE + Area of ∆EBC
or
s = OA × AB + \(\frac{1}{2}\)EB × BC
or
s = ut+\(\frac{1}{2}\)t(v – u)
v – u=at
s = ut + \(\frac{1}{2}\)t ×at
s = ut + \(\frac{1}{2}\)at ²

From the above v-t graph
Displacement of an object = Area of trapezium OACE

Motion in a Straight Line Class 11 Important Numerical Questions

Question 1.
A car travels from a city A to city B with a speed of 40 km/h and returns back to city A, with the speed of 50 km/h. What is the average speed of car? What will be its average velocity?
Solution:
Let the distance between A and B is x.

Question 2.
A body obtain a velocity of 10m/s in 3 second from rest. Find its acceleration.
Solution:
Given, u = 0, v = 10m/s and t = 3sec.
From Ist equation of motion,
v = u + at
a = \(\frac{v-u}{t}\)
Putting the values in above equation, we get
a = \(\frac{10-0}{3}\) = \(\frac{10}{3}\) m/s²
or
a = 3.33 m/s²

Question 3.
A body describes 45 m in 8^th second with uniform acceleration from rest. Find its acceleration.
Solution:
Given :u-0,t=8s,s = 45m

Question 4.
A car moving with a velocity of 36 km/h. When brakes are applied it stop after 10m. Calculate acceleration and time taken to stop.
Solution:

Question 5.
A body describe 15m in sixth second and 23m in tenth second. Find initial velocity and acceleration of the body.
Solution:

or 2u+19a=46
Solving eqns. (1) and (2),…..(2)
u = 4 m/s and a = 2m/s²

Question 6.
Two trains are running parallely in two tracks in the same direction with a speed of 10 m/s and 15 m/s. Find relative velocity.
Solution:
Given, V₁ = 10 m/s, v₂ = 15 m/s
Relative velocity = v₂ – v₁
= 15-10 = 5 m/s.

Question 7.
A car is running on a straight road with a velocity of 126 km/hr. It stops after travelling a distance of 200 m. Find the retardation of the car. How much time does it take to stop?
Solution:
Given, initial velocity u = 126km/h
= 126 × \(\frac{5}{18}\) = 35m/s.
Distance curved s = 200 m
Final velocity v = 0.
Using third equation of motion,
v² = u² + 2as
or 0 = 35 × 35 + 2 × a × 200
a = \(-\frac{35 \times 35}{2 \times 200}\) = 3.06m/s²
Using first equation of motion,
v = u + at
or 0=35 – 3.06t
or t= \(\frac{35}{3 \cdot 06}\)
=11.43sec.

Question 8.
A car travels first \(\frac{1}{3}\) part of total distance at a speed of 10km/hr, second\(\frac{1}{3}\) part at 20km/hr and last \(\frac{1}{3}\) part at 60 km/hr. Find out average speed of the car.
Solution:
Let the distance travelled by car = x km
Given, v₁ = 10 km/hr, v₂ = 20km/hr, v₃ 60 km/hr

Question 9.
A ball is thrown vertically upward reaches the roof of a house 100m in height. At same instant a second ball is dropped downward from the roof of the house will zero initial velocity. At what height both the ball will cross each other?
Solution:
Given, h = 100m
Let, both the ball cross each other at a distance of x m after t sec.
For vertically upward motion,
u = ? v = 0 a = -g and h = 100 m
v² = u² +2as, we get
0 = u² – 2 × g × 100
or u² = 200g = 200 × 9.8 = 1960
or
for vertically downward motion, .
u = 0, a = g, s = (100 -x)
From s = ut + \(\frac{1}{2}\)at², weget
100 – x = 0 +\(\frac{1}{2}\) × 9.8 × t² =4.9t²
or 100 – x = 4.9 t²
When the ball is thrown upward, distance covered by it
s = ut + \(\frac{1}{2}\) at² ….

or x = 100 – 25 = 75m

Question 10.
A jet plane is moving with a velocity of 500km/hr. Velocity of gas ejecting out with respect to jet plane is 1500 km/hr. find out the velocity of gas ejecting out w.r.t. a man standing on ground surface.
Solution:
Velocity of gas ejecting out from the jet plane w.r.t. man standing on ground surface
= v_g-v_j =1500-500
= 1000 km/hr.

Question 11.
A woman starts from her home at 9-00 am., walks with a speed of 5 kmh^-1 on a straight road up to her Office 2-5 km away, stays at the office up to 5-00 p.m. and returns home by an auto with a speed of 25 kmh^-1. Choose suitable scales and plot the x – t graph of her motion.
Solution:
x -t graph of the motion of woman is shown in fig. V₁ = speed of woman while walking at 5kmh^-1.
x = Distance covered by her = 2-5 km If t¹ = time taken to reach office, then it can be calculated by using
x₁ = v₁t₁
or t₁ = \(\frac{x}{v_{1}}\)
∴ t₁ = \(\frac{2 \cdot 5}{5}\) = \(\frac{1}{2}\) h = 30 minutes

If O is regraded as origin for both time and distance, then at t = 9.00 am., x = 0 at t = 9.30 am., x = 2.5 km and she reaches in her office. So OA represents x – t graph of the motion, when the woman walks from her home to office.
When she stays at her office from 9.30 a.m to 5.00 p.m., then she is stationary and hence her stay is represented by the straight line AB in the graph.
On return, speed of auto, v₂ = 25 km/h
∴ If t₂ = time taken by her i.e., by auto from office to her home, then
t₂ \(\frac{x}{v_{2}}\) = \(\frac{2 \cdot 5}{25}\) = \(\frac{1}{10}\) = 6minutes
Thus, she reaches back to her home at 5.06 p.m.
Her motion on the return journey in shown by BC part of the graph.
Scale chosen :
Time on X-axis, 1 division = 1 hour.
Distance on Y-axis, 1 division = 0.5 km.

Question 12.
A body travels half of its total path in the last second of its fall from rest. Find the time and height of its fall.
Solution:
Let the body falls from a height h and takes time t to fall.
From Second equation of motion, fall from/I to B is given by

Motion in a Straight Line Class 11 Important Questions Objective Type

1. Multiple- choice questions:

Question 1.
If the velocity of an object is doubled then among the following which will be doubled:
(a) Acceleration
(b) Kinetic energy
(c) Momentum
(d) Weight.
Answer:
(c) Momentum

Question 2.
An object dropped from peak of a tower takes 2 second to cross half its height. The height of the tower will be : (If g = 9-8 m/s²)
(a) 9.8m
(b) 19.6m
(c) 20m
(d) 39.2m.
Answer:
(d) 39.2m.

Question 3.
Ratio of instantaneous velocity and instantaneous speed is ……………….
(a) Less than 1
(b) More than 1
(c) Zero (d) Equal to 1.
(d) Equal to 1.
Answer:
(d) Equal to 1.

Question 4.
If acceleration-time graphs, the area represents :
(a) Displacement
(b) Velocity
(c) Change in velocity
(d) Distance covered.
Answer:
(c) Change in velocity

Question 5.
An object is falling freely under the gravitation. The ratio of the distance covered by it in first, second and third seconds will be :
(a) 1 : 3 : 5
(b) 1 : 2 : 3
(c) 1 : 4 : 9
(d) 1 : 5: 6.
Answer:
(a) 1 : 3 : 5

Question 6.
The motion of a particle of a gas are :
(a) One-dimensional
(b) Two-dimensional
(c) Three-dimensional
(d) Vertically upward and downward.
Answer:
(c) Three-dimensional

Question 7.
Which of the following quantity may be zero for any particle in motion :
(a) Displacement
(b) Distance
(c) Speed
(d) None of these.
Answer:
(a) Displacement

Question 8.
If a player throws a ball vertically upward with initial velocity 29 m/s. The velocity of the ball at its peak of motion will be :
(a) 29 m/s
(b) -29 m/s
(c) Zero
(d) None of these.
Answer:
(c) Zero

Question 9.
A car covers half a distance between two points with a speed of 40 km/hr and next half with a speed of 60 km/hr. The average speed will be :
(a) 40 km/hr
(b) 48 km/hr
(c) 50 km/hr
(d) 60 km/hr.
Answer:
(b) 48 km/hr

Question 10.
The position vector of a particle is represented by the equation x = (3t³ + 7t²+14t + 8)m, the acceleration of the particle at t = 1 sec will be :
(a) 10 m/s²
(b) 32 m/s²
(c) 23 m/s²
(d) 16 m/s².
Answer:
(b) 32 m/s²

2. Fill in the blanks:

1. A device used to measure instantaneous velocity is called ………………
Answer:
Speedometer

2. The motion of the Billiard ball is …………….. motion.
Answer:
Two-dimensional

3. The change in position of a body with respect to time is called ………………….
Answer:
Motion

4. When a body moves in a straight line, then its motion is ……………………
Answer:
One-dimensional

5. The distance travelled by a body in a definite direction is called ………………….
Answer:
Displacement

6. The rate of change of position of a body in motion is called its ………………..
Answer:
Velocity

3. Match the following:

Column ‘A’	Column ‘B’
1. First equation of motion	(a) s = ut+\(\frac{1}{2}\)at2
2. Second equation of motion	(b) v = u + at
3. Third equation of motion	(c) \(\frac{\text { Change in velocity }}{\text { Time interval }}\)
4. Average acceleration	(d) u + \(\frac{1}{2}\)a(2t – 1)
5. Distance travelled in tth second	(e) v2= u2 +2as.

 4. Write true or false:

1. From a velocity-time graph, the distance travelled can be measured.
Answer:
True

2. In two dimensional motion, the path of uniform velocity is a straight line.
Answer:
True

3. The displacement can be zero, negative as well as positive.
Answer:
True

4. The unit of acceleration in SI system is not metre/sec².
Answer:
False

5. Displacement is a vector quantity.
Answer:
True

6. The slope of velocity time graph represents acceleration.
Answer:
True

5. Answer in one word:

1. A body is moving with constant velocity. What is the value of its acceleration?
Answer:
Zero

2. What does the slope of position-time graph give?
Answer:
Velocity

3. What does the slope of velocity-time graph give?
Answer:
Acceleration

4. Who had derived the equations of motion?
Answer:
Gallileo,

5. What is the value of acceleration due to gravity on earth’s surface?
Answer:
9.81 m/s²

Students get through the MP Board Class 11th Physics Important Questions Chapter 2 Units and Measurements which are most likely to be asked in the exam.

MP Board Class 11th Physics Important Questions Chapter 2 Units and Measurements

Units and Measurements Class 11 Important Questions Very Short Answer Type

Question 1.
What do you mean by measurement?
Answer:
In physics, the measurement means comparison of the given physical quantity with standard quantity of same nature.

Question 2.
What are the different type of measurement?
Answer:
The different type of measurement are :

1. F.P.S. (Foot Pound Second)
2. C.G.S. (Centimetre Gram Second)
3. M.K.S. (Metre Kilogram Second).

Question 3.
Define unit.
Answer:
The standard value of any physical quantity is known as its unit.

Question 4.
What are fundamental physical quantities?
Answer:
The physical quantities which do not depend upon other quantities are called fundamental.
Example: Mass, Length, Time etc.

Question 5.
What are derived physical quantities?
Answer:
The physical quantities which are derived from fundamental quantities are called derived quantities e.g.,
Volume = length × breadth × height = (length)³
Similarly Force, Momentum, Pressure, Power etc. are derived quantities.

Question 6.
Define metre.
Answer:
The distance travelled by light ray in \(\frac{1}{29,97,92,458}\) seconds in vaccum is called metre.

Question 7.
What are the different units to measure astronomical distance?
Answer:

• Light year: Distance travelled by light rays in one year in vacuum is called one light year.
  1 light year = 3 × 10⁸ × 365 × 24 × 60 × 60 metre = 9.46 × 10¹⁵ m.
• Parsec : It is the unit to measure the distance of stars from the earth.
  1 Parsec = 3.26 light year= 3.08 × 10¹³ km.
• Astronomical unit: Average distance between sun and earth is called astronomical units. It is denoted by A.U.
  1A.U.= 1.496 × 10¹¹metre.

Question 8.
What do you mean by one kilogram?
Answer:
The mass of platinum-iridium cylinder, kept at International Bureau of weights and measure at severes near Paris in France is defined as one kilogram.

Question 9.
Define one second.
Answer:
One second is the time-interval in which Cs – 133 atom execute 9 19 26 31 770 vibrations.

Question 10.
Define the dimensions of a physical quantity with example.
Answer:
The suitable powers, by which a derived unit is expressed in terms of fundamental units are called dimensions of that physical quantity.
Examples :
(i) Area = Length × Breadth
= [L] × [L] = [L]² = [M⁰L²T⁰].

(ii) Velocity = \(\frac{\text { Distance }}{\text { Time }}\)
= \(\begin{array}{l}
{[\mathrm{L}]} \\
\hline[\mathrm{T}]
\end{array}\) = [LT^-1] = [M⁰LT^-1
Thus, dimension of area in length is 2 and that of velocity are, in length 1 and time -1.

Question 11.
What is dimensional equation? Explain with examples.
Answer:
The equation which represents the relation of a derived unit with the fundamental units is called dimensional equation. In mechanics dimensional equation is represented as
[X] = [M^aL^bT^c]
Examples : Velocity = [M⁰LT^-1 ]
and Force = [MLT^-2].

Question 12.
Which of the three physical quantities have the dimensional formula [ML^-1T^-2]?
Answer:
Pressure, Stress and Young’s modulus.
∵ [ML^-1T^-2]=\(\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^{2}\right]}\) = \(\frac{\text { Force }}{\text { Area }}\)
[L2] Area

Question 13.
What are dimensionless quantities? Give three examples.
Answer:
The quantities which are having no dimension are called dimensionless quantities e.g., relative density, strain, relative humidity.

Question 14.
What is significant figure? What is the significant figure in 7.000 × 10³ m?
Answer:
Significant figure :
The number of digits used to express accurately the measured value of the physical quantity is called the significant figures.
The significant figure of 7.000 × 10³ m is 3.

Question 15.
How many significant figures are in the following :
(i) 6.320 joule,
(ii) 0.2370 gm,
(iii) 0.007 kg,
(iv) 2.64 × 10^-5m.
Answer:
(i)  4,
(ii) 4,
(iii) 1,
(iv) 3.

Question 16.
Candela is unit of which physical quantity?
Answer:
Candela is unit of Luminious intensity.

Question 17.
What do you mean by Armstrong unit?
Answer:
It is the unit to measure micro distance 1 Å = 10^-10 metre.

Question 18.
What are the merits of SI system?
Answer:
The main features of SI system are :

• It is a coherent system of units.
• In this system a physical quantity has only one unit, e.g., the unit of energy is joule, whether it is mechanical energy, heat energy or electrical energy.
• It is a decimal system. Smallest and biggest quantities are expressed in the power of 10. e.g.,
  1 kilometre = 10³ m; 1 mm = 10^-3m etc.
• It is closely related to C.G.S. and M.K.S. system.
• All the derived units in this system are accepted on international level.

 Units and Measurements Class 11 Important Questions Short Answer Type

Question 1.
Which of the following is the most precise device for measuring length :
(a) A vernier callipers with 20 divisions on the sliding scale.
(b) A screw gauge of pitch 1mm and 100 divisions on the circular scale.
(c) An optical instrument that can measure length to within a wavelength of light.
Answer:
The most precise device is that whose least count is minimum.
Now, (a) Least count of vernier callipers = 1 MSD – 1 VSD
= 1MSD – \(-\frac{19}{20}\) MSD = \(\frac{1}{20}\) MSD
= \(\frac{1}{20}\) mm =\(\frac{1}{200}\) cm
= 0.005 cm
(b) Least count of screw guage
= 0.001cm
(c) Wavelength of light, λ ≈ 10 ^-5 cm = 0.00001 cm
∴ Least count of optical instrument = 0.00001 cm Thus, clearly the optical instrument is the most precise.

Question 2.
What are the uses of dimensional analysis?
Answer:
Uses :

• To find the unit of a physical quantity.
• To change one system of units into another system.
• To check the correctness of a physical relation.
• To establish relation between different physical quantities.

Question 3.
Distinguish between fundamental units and derived units.
Answer:
Difference between fundamental and derived units :

Fundamental units	Derived units
1. The units of fundamental quantities are called fundamental units.	The units of derived quantities are called derived units.
2. These units are independent.	These units depend upon fundamental units.
3. There are seven fundamental units and two supplementary units in SI system.	These units are not limited in number.

Question 4.
Answer the following:
(a) You are given a thread and a metre scale. How will you estimate the diametre of the thread?
(b) A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?
Answer:
(a) The diameter of a thread is so small that is cannot be measured using a metre scale. We wind the thread on in close turns on a pencil or a cylindrical glass rod so that the turns touch one another and are tight. Count the number of turns (n) and find the average length (l) of the coiled thread using the metre scale. Divide this average length by the number of turns to get the diameter i.e.,
Diameter = \(\frac{\text { Average length }}{\text { Number of turns }}\) = \(\frac{l}{n}\)

(b) Yes, the accuracy of the gauge increases by increasing the number of divisions of the circular scale because the least count is given by :
Least count = \(\frac{\text { Pitch }}{\text { No. of divisions on the circular scale }}\)
So, theoretically speaking, least count decreases on increasing the number of divisions on the circular scale.

Question 5.
Let us choose such a unit accordance to which velocity of light be one unit. On the basis of this new unit, what will be the distance between Sun and Earth if light takes 8 minute and 20 second to reach earth?
Answer:
Velocity of light = 1 (as per new unit)
Time taken by light to reach earth surface = (60 × 8 + 20) = 500 sec.
∴ Distance between Sun and Earth = Velocity of Light × Time
= 1 × 500 = 500 Unit of length.

Question 6.
Check the correctness of formula, v² = u² +2las by dimensional method.
Answer:
v²=u²+las
Now, dimensions ofu and v = [LT^-1], dimensions of a =[LT^-2]
Dimensions of s = [L]
and 2 is dimensionless
∴ [LT^-1]² =[LT^-1]²+[LT^-2L]
or [ L²T^-2 ] = [ L²T^-2] + [ L²T^-2]
Since, the dimensions of each term is same, hence the formula is true.

Question 7.
What do you mean by principle of homogeneity of dimensions?
Answer:
According to this principle the dimensions of all terms on both the sides of a dimensional equation are same.

Units and Measurements Class 11 Important Questions Long Answer Type

Question 1.
The velocity of sound waves in a medium depends upon the coefficient of elasticity of the medium (E) and its density (d). Establish the formula for the velocity of wave by dimensional method.
Answer:
Let the velocity of sound in the air depends upon the coefficient of elasticity and density of medium, then
v ∝ E^ad^b
or v = k ,E^ad^b ……..(1)
Where v = velocity of sound, E = coefficient of elasticity, d – density and a and b are numbers and k is dimensionless constant.
Now, the dimensional formulae of quantities are :
v=[LT^-1]
E = [ML^-1T^-2]
d = [ML^-3]
By eqn. (1) we get
or [LT^-1] =[ML^-1T^-2]^a[ML^-3]^b
or [LT^-1] =[M^aL^-a’T^-2aM^bL^-3b]
or [LT^-1] =[M^a+b + L^-a-3bT^-2a]
Now,by the principle of homogeneity, the dimension of both the sides will be equal.
a + b = 0 …(2)
∴ -a-3b=1 …(3)
-2a = -1 …(4)
Hence, a=\(\frac{1}{2}\)
Putting the value of a in eqn. (1), we get
\(\frac{1}{2}\)+ b = 0
or
b = – \(\frac{1}{2}\)

Putting the values of a and b in eqn. (1), we get
v = k.E^1/2d-d^-1/2
or

This is the required relation.

Question 2.
How much dyne is in one Newton?
Answer:
Dimensional formula for force is [MLT^-2].
Here, a=1,b=1 and c = -2

Putting the value, we get
n₂= n₁[10³] [10² ] [1]
or
n₂ = n₁ × 10n⁵
If n₁ = 1 dyne
Then n₂ = 1 Newton
∴ 1 Newton = 10⁵ dyne.

Question 3.
According to van der Wall relation between pressure P, volume V and temperature T is given by

Where a and b are constant. Find the dimensions of a and b.
Answer:
According to homogineity of dimensional equation, a
Dimension of \(\frac{a}{V^{2}}\)= Dimension of P.
∴ Dimension of a = Dimension of P ×(Dimension of V)²
= [M¹L^-1T^-2][L³]²
∴ Dimension of a =[M¹L⁵T^-2] and Dimension of b = Dimension of V = [L³]
or Dimension of b = [M⁰L³T⁰].

Question 4.
Find the fundamental unit of force and work using dimensional equation.
Answer:
(1) Force = Mass × Acceleration

(2) Work Force Displacement
= Mass x Acceleration x Displacement

∴ Dimensional formula of work = [ML²T^-2]
SI unit = kg-metre²/sec²
C.G.S. unit = g-cm²/sec².

Question 5.
What are the limitations of dimensional equations?
Explain the limitations of dimensional analysis.
Answer:
Limitations :

• The dimensional formulae of only those quantities can be ob- :d which can be expressed in terms of fundamental quantities.
• It is not applicable to trigonometrical, exponential or logarithmic functions.
• The value of constants cannot be determined by this system.
• If an equation has a constant having dimensions e.g., G, Then relation cannot be blished.

Question 6.
The periodic time of a simple pendulum T, depends upon its effective length (l) acceleration due to gravity (g). Derive the formula for time-period by dimensional
Or
Find out an expression for periodic time of a simple pendulum using dimensional sis.
Answer:
Let the formula for periodic time (T) be,
T ∝ l^a g^b
T = K. l^a g^b
Putting the dimensions of periodic time (7), effective length (l) and acceleration due to gravity in eqn. (1).
[M⁰L⁰T] = [L]^a x [LT^-2]^b .
or
[M⁰L⁰T] = [M⁰L^a+bT^2b]
By the principle of homogeneity, comparing the powers of length and time, we get
a + b = 0 …(2)
and – 2b = 1
⇒ b = – \(\frac{1}{2}\)
Putting in eqn. (2), we get
a- \(\frac{1}{2}\) = 0
∴ a = \(\frac{1}{2}\)
Putting the values of a and b in eqn. (1), we get
T = kl^1/2g^-1/2
or

This is the required formula.

Question 7.
What are the different type of errors in measurement? Explain.
Answer:
Random error :
If an observer measures a physical quantity and each time he finds different readings, then such an error is called random error.
Example : Suppose the diameter of a wire is determined by a screw gauge and each time the reading is different because the wire is not uniform.
To remove this error, so many readings are taken and then mean is calculated.
If a physical quantity is measured n times and its readings are a₁,a₂, a₃ …..a_n

Absolute error :
The difference between the observed value and actual value of a physical quantity is called absolute error. It is always positive.
The absolute error of a quantity x is denoted by | Δx |
Let the observed values of a physical quantity are x₁,x₂, x₃ …..x_nand the average value is x.

Relative error :
The ratio of absolute error and actual value is called relative error.
∴ Relative error = \(\frac{\text { Average absolute error }}{\text { Actual value }}\)
= \(\frac{\text { Actual value }-\text { Measured value }}{\text { Actual value }}\)
or

Percentage error: When the relative error is expressed in percentage, then it is called percentage error.
∴ Percentage error = \(\frac{\text { Average absolute error }}{\text { Actual value }}\) ×100

Question 8.
What do you understand by significant figure ? How do you find significant figure of a quantity?
Answer:
The number of digits used to express accurately, the measured value of a physical quantity is called significant figure.

Counting of significant figures :

• By changing the decimal point, the number of significant figures does not change.
  For example; 1.234, 12.34 and 123.4 the number of significant figure is 4.
• All the zeros between two non-zero numbers are significant.
  For example; Significant figures in 3100-05 is 6.
• The zeros before the first non-zero digit are ignored while all the zeros to the right are counted.
  For example; Significant figure in 0 007 is 1 and that of 6-320 is 4,
• The zeros in a whole number at the end are not counted.
  For example; Significant figure in 6320 is 3 and in 13200 is 3.
• The significant figure does not change with the change of units i.e.,
  0.0432m = 4.32 cm, both have the significant figure 3.
• When a number is expressed in the power of 10, then the power is not counted, i.e., 10.52 × 10⁵ has significant figure 4.

Units and Measurements Class 11 Important Numerical Questions

Question 1.
Convert 0.53 Å into metre.
Solution:
We know that, 1Å = 10^-10 m
0.53 Å = 0.53 ×10^-10m
= 5.3 ×10^-11m

Question 2.
A student measures the thickness of a human hair by looking at it through
a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 35mm. What is theestimate on the thickness of hair?
Solution:
Estimate on the thickness of hair is given by:

Question 3.
State the number of significant figures in the following :
(a) 0.007m²,
(b) 2.64 × 10²⁴kg,
(c) 0.2370 g cm^-3,
(d) 6.320 J,
(e) 6.032 Nm^-2,
(f) 0.0006032 m².
Answer:
The number of significant figures is as given below :
(a) 1,
(b) 3,
(c) 4,
(d) 4,
(e) 4,
(f) 4.

Question 4.
The length, breadth and thickness of a rectangular sheet of metal are 4.234m, l-005m and 2.01cm respectively. Give the area and volume of the sheet to correct significant figures.
Solution:
Here, Length, l = 4.234m
Breadth, b = 1.005m
Thickness, h = 2.01cm = 0 0201m
Area of the sheet = 2(lb + bh + hl)
= 2(4.234 × 1.005 + 1.005 × 0.0201 + 0.0201 × 4.234)
= 8.7209468m²
As the least number of significant figures in thickness is 3,
∴ Area = 8.72m²
Volume = l × b × h
= 4.234 × 1.005 × 0.0201m³
= 0-0855m³

Question 5.
A physical quantity P is related to four observables a, b, c and d as follows :

The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2% respectively. What is the percentage error in the quantity P ? If the value of P calcu-lated using the above relation turns out to be 3.763, to what value should you round off the result?
Solution:

The calculation of error clearly shows that the number of significant figures is 2, so the result of P may be rounded off to two significant digits i.e.,
P = 3.763 = 3.8.

Question 6.
The percentage errors in the measurement of length and mass are 2% and 5% respectively. What will be the percentage error in the measurement of density of a cube?
Solution:

Maximum % error in the measurement of density will be 11%.

Question 7.
Given that the period T of oscillation of a gas bubble from an explosion under water depends on P, d and E, where the symbols are pressure, density and total energy of explosion. Find dimensionally a relation for T.
Solution:
Let T∝P^ad^bE^c
or T =kPsup>ad^bE^c, (where His a dimensionless constant) …(1)
Writing dimensional formula of each physical quantity on both sides of eqn. (1), we
[M⁰L⁰T¹] = [ML^-1T^-2 ]^a [ML^-3 ]^b [ML²T^-2 ]^c
= [M^a+b+c][L^-a-3b+2c][T^-2a-2c]
Comparing power of,
M, 0 = a + b + c ……(2)
L, 0 = – a – 3b + 2c ……(3)
T, 1 = -2a – 2c
or a + c = –\(\frac{1}{2}\)
∴ From eqns. (2) and (4), we get
b = – (a + c)
= \(-\left(-\frac{1}{2}\right)\) = \(-\left(\frac{1}{2}\right)\)
Putting the value of b in eqn. (3),
– a + 2c = 3b
=> -a + 2c =3 x \(\frac{1}{2}\)= \(\frac{3}{2}\)
Adding eqns. (4) and (5),

Question 8.
An object of mass ‘m’ is executing simple harmonic motion suspended at one end of a spring. If its spring constant is K and time-period is T, then prove with the help of dimensional method that equation T=2π\(\frac{m}{\mathbf{K}}\) incorrect. Also find its correct
equation.
Solution:
Given, T = 2π\(\frac{m}{\mathbf{K}}\)
Writing dimensional formula of the above equation
[T] = \(\frac{[\mathrm{M}]}{\left[\mathrm{MT}^{-2}\right]}\) = [T²]
Since, the dimension of T on both sides are not same, therefore the given equation is incorrect.
Let us assume that time-period T, a^th power of mass (m) and b^th power of spring constant (K), then
T∝m^aK^b
or T = K.m^aK^b
Let K = 2π
∴ T = 2πm^aK^b
or [T] = 2π[M]^a [MT^-2]^b
= [M] ^(a+b)[T]^-2b
Comparing both sides, we get
a + b = 0
and -2b = 1
or b = –\(\frac{1}{2}\)
∴ a+ \(-\left(-\frac{1}{2}\right)\)=0
or
a = \(\frac{1}{2}\)
Putting the values of a and b in the above equation,
T = 2πm\(\frac{1}{2}\). K-\(\frac{1}{2}\)

This is the correct form of the equation.

Question 9.
Equation of a particle of velocity (v) and time (t) is given by v = A + Bt +\(\frac{C}{D+t}\) find out the dimension of A, B, C and D.
Solution:
Each form of R.H.S. must have dimension of v.
∴ Dimensional formula of A = Dimensional formula of velocity
v = [LT^-1].
Dimensional formula of B.t = Dimensional formula of v = [LT^-1]
∴ Dimensional formula of B =\(\frac{\left[\mathrm{LT}^{-1}\right]}{[\mathrm{T}]}\) [LT^-2]
∴Dimensional formula off D = Dimensional formula off t = [T].
Dimensional formula of\(\frac{C}{D+t}\) = Dimensional formula of v= [LT^-1]
and Dimensional formula of C = [LT^-1] x [Dimensions of D + t] – [L¹].

Question 10.
Assuming energy E, length L and time T as the fundamental unit, what would be dimensions of mass ? If energy E is being replaced by force F then what would be the dimensions of mass ?
Solution:

∴ Dimensions of mass will be = [E¹L^-2T²].
Replacing energy E by force F, we get
Force F = Mass x Acceleration

Question 11.
If m and c represent mass and velocity, then find with the help of dimensional method the quantity mc² is equivalent to which form?
Solution:
Dimensions of mc² – Dimensions of mass x (Dimensions of velocity)²
= M[LT^-1]²
= [M¹L²T^-2 ] = Dimensions of energy.

Question 12.
The centripetal force acting on a particle revolving in a circular path depends upon velocity of the particle v and radius of circle (r). Using dimensional method establish the formula for the centripetal acceleration (a).
Solution:
According to given question
a ∝ v^a r^b
or a = kv^a r^b …(1)
Writing the dimensional formula for the quantities
[M⁰LT^-2] = [M⁰LT⁰]^b
=[M⁰L^a+bT^-a]
Comparing the powers of M, L and T both sides,
a + b= 1 …(2)
and – a = -2 or a = 2
Putting the value of a in eqn. (2), we get
2 + b=1
or b = 1 – 2 = – 1
Hence, from eqn. (1), we get
a = k.v² r^-1 or a = \(\frac{k v^{2}}{r}\)
If k= 1, then a= \(\frac{v^{2}}{r}\)

Question 13.
Diameter of a sphere is 2.348 cm. Find out volume and surface area of it up to required significant number.
Solution:
Given,
2R = 2.348 cm
∴ R= 1.174 cm
Volume of sphere V =\(\frac{4}{3}\)πR³ = \(\frac{4}{3}\) × 3.14 ×(1.174)³
or
V = 6.774 cm³
and Surface area of sphere S = 4πR²
=4 × 3.14 × (1.174)²
= 17.311 = 17.31cm².

Motion in a Straight Line Class 11 Important Questions Objective Type

1. Multiple- choice questions:

Question 1.
Light year is unit of:
(a) Time
(b) Distance
(c) Speed
(d) Intensity of light.
Answer:
(b) Distance

Question 2.
Fermi is unit of:
(a) Energy
(b) Momentum
(c) Impulse
(d) Length.
Answer:
(d) Length.

Question 3.
The dimensional formula of gravitational constant is :
(a) [ML²T^-2]
(b) [M^-1L³T^-2]
(c) [M^-1L²T^-3]
(d) [ML³T²]
Answer:
(b) [M^-1L³T^-2]

Question 4.
Choose a pair of physical quantities having the same dimensional formula:
(a) Pressure and stress
(b) Stress and strain .
(c) Pressure and force
(d) Power and force.
Answer:
(a) Pressure and stress

Question 5.
Units of energy is :
(a) watt
(b) joule
(c) Electron volt
(d) Calorie.
Answer:
(a) watt

Question 6.
Relation between v and t, velocity of sound and time is
v = at+\(\frac{b}{t+c}\) dimensional formula for a, b and c will be:
(a) [L²], [T], [LT^-2]
(b) [LT^-2], [L], [T]
(c) [LT⁺²], [LT], [L]
(d) [L], [LT], [T²].
Answer:
(b) [LT^-2], [L], [T]

Question 7.
Order of 4-12 × 10⁵ is :
(a) 4
(b) 6
(c) 5
(d) None of these.
Answer:
(b) 6

Question 8.
Which among the following have significant number four :
(a) 0.630
(b) 0.0024
(c) 6.023
(d) 0.004.
Answer:
(c) 6.023

Question 9.
Which among the following is not a fundamental unit:
(a) Metre
(b) Ampere
(c) Kelvin
(d) Litre.
Answer:
(a) Metre

Question 10.
Which among the following physical quantities do not have the dimensional
formula [ML^-1T^-2 ] :
(a) Youngs modulus
(b) Stress
(c) Strain
(d) Pressure.
Answer:
(c) Strain

Question 11.
Light year is equal to :
(a) 9.46 × 10¹⁰ km
(b) 9.46 × 10¹² km
(c) 9.46 × 10¹² metre
(d) 9.46 × 10^-12 km.
Answer:
(b) 9.46 × 10¹² km

Question 12.
If the length of a simple pendulum is increased, then its percentage increase in its time-period will be:
(a) 0.5%
(b) 2.0%
(c) 1%
(d) 4%.
Answer:
(b) 2.0%

2. Fill in the blanks:

1. The volume of cube of side 1cm is equal to ……………….. m³.
Answer:
10^-6

2. A vehicle moving with speed of 18 km/hr covers ……………….. m in 1 second.
Answer:
5

3. 3.0 m/s² = ………………….. km/h².
Answer:
3.9 × 10^-4

4. Number of significant figure in 2.64 × 10²⁴ kg is ………………….
Answer:
3

5. Number of fundamental unit in SI system is …………………..
Answer:
7

6. 1 micron = ………………… metre.
Answer:
10^-6

7. 1Å = …………………….. metre.
Answer:
10^-10

8. The error due to observer measurement is called ………………..
Answer:
Individual error

9. 1 newton = ……………………. dyne.
Answer:
10⁵

10. Dimensional formula of charge is ……………………
Answer:
[AT]

3. Match the following:
I.

Column ‘A’	Column ‘B’
1. Dimensional formula of light year	(a) Dimension of momentum
2. Dimensional formula of frequency	(b) Dimensional of planck constant
3. Dimensional formula of angular ‘ momentum will be equal to	(c) [M0LT0]
4. Dimensional formula of impulse	(d) [ML2T-2]
5. Dimensional formula of work	(e) [M0L0T-1].

Answer:
1. (c) [M⁰LT⁰]
2. (e) [M⁰L⁰T^-1].
3. (b) Dimensional of planck constant
4. (a) Dimension of momentum
5. (d) [ML²T^-2]

II.

Column ‘A’	Column ‘B’
1. [ML-1T-2] is dimensional formula of	(a) Light year
2. Newton-metre2 is unit of	(b) Newton × second
3. Unit of luminious intensity is	(c) Pressure
4. Unit of momentum is	(d) Candela
5. Unit of long distance is	(e) Force.

Answer:
1. (c) Pressure
2. (e) Force.
3. (d) Candela
4. (b) Newton × second
5. (a) Light year

4. Write true or false:

1. Dimensional formula for heat and work are same.
Answer:
True

2. The equation v = u + at, can be derived from dimensional method.
Answer:
False

3. Dimensional formula of impulse and momentum are different.
Answer:
False

4. Dimensional formula for pressure, coefficient of Young modulus and stress are [ML^-1T^-2].
Answer:
True

5. Candela is units of Luminious intensity.
Answer:
True

6. Significant figure changes with change in its units of a physical quantity.
Answer:
False

7. It is not necessary that the physical quantity will be the same, if their dimensional formula and units are same.
Answer:
True

8. The value of constant can be determined by dimensional method.
Answer:
False

9. On knowing one dimensions of a physical quantity, its unit can be determined.
Answer:
True

10. The physical quantity is more significant which have more significant figure.
Answer:
True

Students get through the MP Board Class 11th Physics Important Questions Chapter 15 Waves which are most likely to be asked in the exam.

MP Board Class 11th Physics Important Questions Chapter 15 Waves

Waves Class 11 Important Questions Very Short Answer Type

Question 1.
What do you mean by wave motion?
Answer:
It is a method of energy transformation in which the disturbance produced in the medium propagates without the flow of medium.

Question 2.
What is progressive wave. Write the formula for displacement of it?
Answer:
When disturbance is created continuously in a medium, then the particle of medium vibrates to and fro about their mean position without leaving their place. The distur¬bance move forward. The wave thus produced are called progressive wave.
Its displacement equation is :
\(y=a \sin \frac{2 \pi}{\lambda}(v t-x)\)

Question 3.
Write the principle of superposition of waves.
Answer:
When any number of waves meet simultaneously at a point in a medium, the net displacement at a given time is the algebraic sum of the displacement due to each wave at that time.

Question 4.
Write the name of the phenomenon obtained due to superposition of waves.
Answer:
Interference,

• Beats,
• Stationary waves.

Question 5.
What are stationary waves?
Answer:
When two identical waves of equal frequency travel along the same line but in opposite direction superimpose, then the resultant wave thus formed is called stationary waves.

Question 6.
How many types of stationary waves are there? Write one example of them.
Answer:
There are two types of stationary wave :

1. Transverse wave,
2. Longitudinal wave.

1.Transverse wave: If during the propagation of mechanical wave in a medium, the particles of the medium vibrate about their mean position in a direction perpendicular to the direction of wave propagation then the wave is called transverse wave

2. Longitudinal wave: If during the propagation of mechanical wave in a medium, the particles of the medium vibrate about their mean position along the direction of wave propagation then this type of wave is called longitudinal wave.

Question 7.
What are beats?
Answer:
When two waves with slight difference in frequencies travel in a medium simul¬taneously in the same direction then as a result of their superposition, the intensity of sound at any point increase and decreases alternately. This successive rise and fall in the intensity of sound is called beats.

Question 8.
What are the conditions of beats?
Answer:

• Both the waves should travel along the same straight line, in the same medium, with same velocity.
• These should be slight difference in their frequencies.
• The amplitudes of both the waves should be nearly equal.

Question 9.
If two tunning fork of frequency n₁ and n₂ are hammered together. Then how many beats will be obtained per second?
Answer:
Per second number of beat = n₁~ n₂.

Question 10.
Write three applications of beats?
Answer:

• Determination of frequency of a tuning fork.
• Matching the frequency of sound instruments.
• To enquire about poisonous gas in the mines.

Question 11.
On which factors the velocity of transverse wave produced in stretched string depends on?
Answer:

• Tension of string,
• Mass per unit length of string.

Question 12.
Write the formula for frequency of a stretched string
Answer:
\(n=\frac{1}{2 l} \sqrt{\frac{T}{m}}\)
Where, l = Length of stretched string, T = Time-period, m = Mass per unit length of string.

Question 13.
Explain beats period and frequency of beats.
Answer:
One rise and one fall of sound form one beat. The time interval between two consecutive intense sounds is called ‘beat period’.
The number of times the intensity of sound rises and falls in one second is called ‘beat frequency’.
Let the frequencies are υ₁ and υ₂
Beat frequency = \(v_{1}-v_{2}\)
and Beat period = \(\frac{1}{v_{1}-v_{2}}\)

Question 14.
What is Doppler’s effect?
Answer:
When, there is a relative motion between a sound source and a listener, then the change in the frequency of the source appears to the listener. This effect is called Doppler’s effect.

Question 15.
What are the limitations of Doppler’s effect?
Answer:
Limitations :

• The effect expresses the apparent change in the frequency, not the intensity of sound. ,
• The velocity of source or listener should not be greater than the velocity of sound.

Question 16.
What is difference between the Doppler’s effect on sound and in light waves?
Answer:
Doppler’s effect on sound waves depends upon the relative velocity and velocity of source and listener. While in light waves Doppler’s effect depends upon the relative velocity only.

Question 17.
Explain one application of Doppler’s effect.
Answer:
To estimate the speeds of aeroplanes or submarine: When radio waves transmitted from the Radar, come back after reflection from aeroplane, then their wavelength is changed. If an aeroplane is approaching then its wavelength decreases or frequency increases and if the aeroplane receding from the station, the wavelength increases. Thus, by the change of wavelength or frequency, the speed of aeroplane can be calculated.

Question 18.
What do you mean by organ pipe?
Answer:
The pipe in which sound is produced by vibrating the molecules of air or gas-filled in it.

Question 19.
The open organ pipe produces more melody sound than dosed organ pipe. Why?
Answer:
In open organ pipe the harmonics produced are odd and even both while in closed organ pipe only odd harmonics are produced, therefore harmonics produced by open organ pipe are melody.

Question 20.
A sound-producing whistle is whirled fast in a horizontal circle. What change in the frequency will be observed by an observer standing (i) at the centre of the circle, (ii) outside of the circle?
Answer:

• Since, the whistle is always constant distance from the observer, hence no change in frequency will be observed.
• For the observer standing outside the circle, the whistle will approach and move away from him in every half rotation. Hence, the frequency will be observed increasing and decreasing.

Question 21.
Why a flute has several holes in it?
Answer:
A flute is an open organ pipe, melodious sound of different frequencies can be produced by changing the length of air column by putting fingers on the holes.

Question 22.
Why sound travels faster in moist air than in dry air?
Answer:
Presence of water vapour decreases the density of air. Since velocity of sound is inversely proportional to the density. Hence, sound travels faster in moist air than in dry air.

Question 23.
Why the sound heard is more in carbon dioxide than in air?
Answer:
Velocity of sound in a gas is inversely proportional to the square root of the density of gas. Since carbon dioxide is lighter than air. Thus, velocity of sound is greater in carbon dioxide. Hence, sound heard is more in carbon dioxide than in air.

Question 24.
Sound travels faster on a hot day than on a cold day. Why?
Answer:
Velocity of sound is directly proportional to the square root of temperature. Therefore, sound travels faster on a hot day than on a cold day.

Question 25.
Why are transverse waves not produced in gases?
Answer:
Because the gases are not rigid.

Question 26.
Which property is common to all types of mechanical waves?
Answer:
When the waves propagate through a medium, the particles of a medium do not advance along with waves.

Question 27.
What will be the effect on the speed of sound waves in a gas if the absolute temperature of gas is increased to 4 times of its previous value?
Answer:
According to the formula \(v \propto \sqrt{T}\), speed of sound waves will be doubled.

Question 28.
The speed of sound in air is 330m/s. What will be the effect on the speed of sound waves if pressure is doubled, keeping the temperature constant?
Answer:
No effect, because speed of sound does not depend upon pressure.

Question 29.
Solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases? (NCERT)
Answer:
For the propagation of transverse waves through a medium, the medium should posses rigidity. Gases do not have property of rigidity, so the transverse waves cannot propagate through them. Due to rigidity of solids both, transverse as well as longitudinal wave can propagate through them.

Question 30.
The shape of pulse gets distorted during propagation in a dispersive medium. (NCERT)
Answer:
A sound pulse is a combination of waves of different wavelengths. In a dispersive medium, the waves of different wavelengths travel with different speeds in different directions i.e., with different velocities. So the shape of the pulse gets distorted i.e., a plane wavefront in a non-dispersive medium does not remain a plane wavefront in a dispersive medium.

Question 31.
A sitar, a tambour, and a harmonium all the three are made in unison at the same frequency, even then their voice can be recognised separately. Why is it so?
Answer:
Because the sounds of different qualities are produced by them.

Waves Class 11 Important Questions Short Answer Type 

Question 1.
What are progressive waves? Derive its equation.
Answer:
When the disturbance produced from the body executing simple harmonic motion propagates in the medium, then the particles of medium vibrates to and fro about their mean position and the disturbance advances in the medium. The waves thus produced are called progressive waves.

Consider a harmonic wave propagat- g O ing in a medium along the positive direction of x-axis as shown in fig. The speed ofpropagation of wave is v. Its source is at origin ‘O’. As the source is executing
S.H.M., then the displacement of the particle situated at ‘O’ at any instant ‘f be \(y=a \sin \omega t\) ………(1)
Where, ‘a’ is amplitude of vibration and ‘ co ‘ is angular frequency. Now, \(\omega=2 \pi v=\frac{2 \pi}{T}\)

Where, u is frequency and T is time-period. Since the speed of wave is v hence the disturbance will reach the point ‘P’ at a distance V from the origin ‘O’ in \(\frac{x}{v}\) sec. Thus, at any instant ‘f the displacement of the particle situated at P will be \(y=a \sin \omega\left(t-\frac{x}{v}\right)\) …….(2)

The particle of medium situated at ‘P’ will begin to vibrate after \(\frac{x}{v}\) sec. from the particle situated at ‘O’. Thus, in time ‘t’ the displacement of particle at P will be the same as that the particle at O, before \(\frac{x}{v}\) sec.Therefore, instead of ‘t’ in the eqn. (1) we should write \(\left(t-\frac{x}{v}\right)\) Eqn. (2) shows a progressive wave which is propagating along positive direction of X- axis with velocity v. If a progressive wave is travelling along negative direction of A-axis, then its equation will be :
\(y=a \sin \omega\left(t+\frac{x}{v}\right)\) …………. (3)
As ω = 2πυ
∴ From eqn.(2)
\(y=a \sin 2 \pi v\left(t-\frac{x}{v}\right)\) ……………. (4)
But ν = υλ

\(v=\frac{v}{\lambda}\)
∴ \(y=a \sin 2 \pi \frac{v}{\lambda}\left(t-\frac{x}{v}\right)\)
or
\(y=a \sin \frac{2 \pi}{\lambda}(v t-x)\)
This is the equation of a progressive wave.

Question 2.
Explain Newton’s formula for velocity of sound in air.
Answer:
From purely theoretical considerations, Newton gave an empirical relation to calculate the velocity of sound in a gas.
\(v=\sqrt{\frac{B}{\rho}}\) …………. (1)
Where B is bulk modulus and P is density of the gas.

Sound travels through a gas in the form of compressions and rarefactions. Newton assumed that the changes in pressure and volume of a gas, when sound waves are propagated through it, are isothermal. The amount of heat produced during compression is lost to the surroundings and similarly, the amount of heat lost during rarefaction is gained from the surroundings, so as to keep the temperature constant. Using

coefficient of isothermal elasticity, i.e., Bi in Eqn. (1), Newton’s formula becomes: …(2)
\(v=\sqrt{\frac{B_{i}}{\rho}}\)
Calculation of B_i:
Consider a certain mass of the gas.
Let P = initial pressure of the gas, V= initial volume of the gas. Under isothermal conditions PV= constant
Differentiating both sides, we get
PdV + VdP = 0
PdV = -VdP
\(P=-\frac{V d P}{d V}=-\frac{d P}{\frac{d V}{V}}=B_{i}\) (by definition)
Substituting this value in eqn. (2), we obtain
\(v=\sqrt{\frac{P}{\rho}}\)

Question 3.
Write the characteristics of progressive waves.
Answer:
Characteristics of progressive wave are as follows :

• A harmonic (progressive) wave is a disturbance which travels through a medium with a definite velocity, without change in shape.
• Energy is transferred along the direction of propagation of wave, but there is no net transport of the material of the medium.
• The velocity of an oscillating particle is different on different states of oscillation, being maximum at the mean position and momentarily zero at the extreme positions. The velocity of propagation of the wave is however, constant, depending on the nature of the medium.
• At any instant, the phase of the oscillation varies from particle to particle, because each particle starts oscillating a little later than the previous particle.
• For particles separated by an integral multiple of wavelength λ, the displacement, velocity and acceleration of the particle have the same values at any instant.

Question 4.
Write characteristics of stationary waves.
Answer:
Characteristics of stationary waves :

• These waves do not advance in the medium but remain steady at its place. This is why these waves are called stationary waves.
• No energy is transferred by these waves because the flow of energy at any point of the medium in one direction due to incident wave is the same as the flow of energy at that point in opposite direction due to reflected wave.
• To produce these waves, the presence of bounded medium is necessary because the wave travelling in such a medium, after being reflected from the boundary, produce an identical wave moving in opposite direction. As a result of superposition of incident and reflected wave, the stationary wave is produced.
  The stationary waves cannot be produced in vacuum or in unlimited medium.
• In the waves, certain points in the medium are always at rest, i.e., their displacement remain zero. These points are called nodes. These points are situated at equal distances. In case of longitudinal stationary waves, change in pressure and in density is maximum at nodes as compared to other points.
• The displacement of the point in between two consecutive nodes is maximum as compared to other points. These points are called antinodes. In longitudinal stationary waves, there is no change in pressure and in density at antinodes.
• The distance between two consecutive nodes or between two consecutive antinodes is λ / 2. The distance between a node and its neighbouring antinode is λ / 4.
• Except at nodes, all the particles of the medium vibrates but the amplitude of vibration of the particles are different. The amplitude of the vibration is zero at the nodes and maximum at the antinodes.
• All the particles between two consecutive nodes vibrate in the same phase, i.e., they reach simultaneously through their equilibrium positions.
• At any instant, the particles of both the sides of a node are in mutually opposite phases but the particles of both the sides of an antinode are in the same phase.
• All the particles of medium pass through their mean positions simultaneously twice in each period.

Question 5.
Give difference between progressive waves and stationary waves.
Answer:
Comparison of Progressive and Stationary waves

Progressive waves	Stationary waves
1. These waves advance in a medium with a definite velocity.	1. These waves remain stationary between two boundaries in the medium.
2. Crests and troughs in transverse progressive waves and centre of compression and rarefaction in longitudinal progressive waves occur alternately and advance with a definite velocity.	2. Crests and troughs in transverse stationary waves and centres of compression and rarefaction in longitudinal stationary waves occur alternately at definite places and do not advance. These waves transmit energy in the medium.
3. These waves transmit energy in the medium.	3. These waves do not transmit energy in the medium.
4. In these waves, all the particles of the medium vibrate about their mean positions and the amplitude of vibration is the same for all of them.	4. In these waves, all the particles of the medium do not vibrate and amplitude of vibration also is not the same for all particles. The amplitude is zero at the nodes and maximum at the antinodes.
5. In these waves, at no instant all the particles of the medium pass through their mean positions simultaneously.	5. In these waves, all the particles of the medium pass through their mean positions simultaneously twice in each time period.

Question 6.
What are beats? What are the necessary conditions to obtain beats?
Answer:
Beats: When two waves with slight difference in frequencies travel in a medium simultaneously in the same direction then as a result of their superposition, the intensity of sound at any point increase and decreases alternately. This successive rise and fall in the intensity of sound is called beats.

Conditions:

• Both the waves should travel along the same straight line, in the same medium, with same velocity.
• These should be slight difference in their frequencies.
• The amplitudes of both the waves should be nearly equal.

Question 7.
Derive Newton’s formula for velocity of sound. Point our error in Newton’s formula and hence discuss Laplace correction.
Answer:
Newton assumed that sound travels through air under isothermal conditions and for elastic medium velocity of longitudinal waves,
\(v=\sqrt{\frac{E}{\rho}}\)
∴ PV = constant under isothermal condition.
On differentiating, PdV+ VdP = 0
or
PdV = -VdP
\(P=\frac{-d P}{d V / V}=\frac{\text { stress }}{\text { strain }}=E\)
∴ \(v=\sqrt{\frac{P}{\rho}}\) [from eqn. (1)]
This is Newton’s formula for the velocity of sound in air, which given 280 m/s whereas actual value is 332 m/s.
Laplace correction : Laplace suggested that sound waves travel under adiabatic conditions because:
(i) When sound waves travels in air, changes in volume and pressure takes place rapidly.
(ii) ( Air is a bad conductor of heat. Thus, PVγ = constant
On differentiation Pd(Vγ) + VγdP = 0
or
PγVγ-1 dV + VγdP = 0
or
\(\gamma P=\frac{-V d P}{d V}=\frac{-d P}{d V / V}\)

Hence, according to Laplace, the velocity of sound is \(v=\sqrt{\frac{\gamma P}{\rho}}\) (for air, γ =1.4)
∴ \(v=\sqrt{1 \cdot 4} \times 280=331 \cdot 3 \mathrm{~m} / \mathrm{s}\)

Question 8.
Derive the equation of stationary waves when it is reflected by rigid boundary.
Answer:
When a progressive wave is reflected by a rigid surface, then phase difference of n is produced.
Let the wave is travelling in +ve direction along the Y-axis,
Equation of wave will be
\(y_{1}=a \sin \frac{2 \pi}{\lambda}(v t-x)\) ……….. (1)
Where, a = Amplitude, y =Velocity and λ Wavelength.

When this wave is reflected back, then it will travel along -ve direction of X-axis. ……….. (2)
\(y_{2}=-a \sin \frac{2 \pi}{\lambda}(v t+x)\)
Let the two waves superimpose and stationary wave is formed.
∴By the principle of superposition.
y = y₁ + y₂
or

This is the required equation.

Question 9.
Derive the equation of stationary wave, when reflected by free boundry.
Answer:
Let the incident wave is
\(y_{1}=a \sin \frac{2 \pi}{\lambda}(v t-x)\)
∴ Reflected wave, as the phase does not change
\(y_{2}=a \sin \frac{2 \pi}{\lambda}(v t+x)\)
∴ By the principle of superposition,

Question 10.
Explain beats period and frequency of beats.
Answer:
One rise and one fall of sound form one beat. The time interval between two consecutive intense sounds is called ‘beat period’.
The number of times the intensity of sound rises and falls in one second is called ‘beat frequency’.
Let the frequencies are υ1 and υ2
∴ Beat frequency = υ1 – υ2
and Beat period = \(\frac{1}{v_{1}-v_{2}}\).

Question 11.
Differentiate interference and beats.
Answer:
Distinction between Interference and Beats :

Interference	Beats
1. In this phenomenon, the frequencies of two sound waves are exactly the same.	1. In this phenomenon, the frequencies of two sound waves differ slightly.
2. At any point of the medium, the phase difference between two waves remains constant.	2. At any point of the medium, the phase difference between two waves varies with time.
3. The amplitude of the resultant wave remains constant at every point of the medium.	3. The amplitude of resultant wave varies with time at each point of the medium.
4. Though the intensity of sound is different at different points of the medium yet the intensity of sound does not vary with time at any point of the medium.	4. The intensity of sound decreases and increases alternately with times at any point of the medium.

Question 12.
Use the formula \(v=\sqrt{\frac{\gamma P}{\rho}}\)
to explain why the speed of sound in air:
(a) Is independent of pressure,
(b) Increases with temperature,
(c) Increases with humidity.
Answer:
Given : \(v=\sqrt{\frac{\gamma P}{\rho}}\)
By gas equation PV = RT
or
\(P=\frac{R T}{V}\)
∴ \(v=\sqrt{\frac{\gamma R T}{V \cdot \rho}}\)
But V, p = M (molecular weight of gas)
\(v=\sqrt{\frac{\gamma R T}{M}}\)
(a) In the eqn. (1) P does not appear so the speed is independent of pressure at constant temperature.

(b) If y, R and M are constant, then,
\(v \propto \sqrt{T}\)
i.e., the velocity (speed) of sound is directly proportional to the square root of absolute temperature.
∴ With increase in temperature the speed of sound increases.

(c) From the formula \(v=\sqrt{\frac{\gamma P}{\rho}}\) , it is clear that
\(v \propto \frac{1}{\sqrt{\rho}}\)
The density of humid air is less than that of dry air therefore with increase in humidity the speed of sound increases.

Question 13.
What are harmonics? Which harmonics can be produced in a stretched string?
Answer:
The harmonics are those tones, whose frequency is integral multiple of that of fundamental tone. The harmonics whose frequencies are even multiple of that of fundamental tone, are called even harmonics and the harmonics whose frequencies are odd multiple of that of fundamental tone are called odd harmonics.

In a stretched string even and odd both harmonics can be produced. In Fig. the first, second and third harmonics are shown.

Question 14.
How do you determine the frequency of a tuning fork with the help of beats?
Answer:
Determine frequencies of a tuning fork: Suppose that we have to determine the frequency of a tuning fork. For this purpose, we take such a fork of known frequency
υ₂ whose frequency υ₂ differ slightly from unknown frequency υ₁. Now, both the tuning forks are sounded together and the beats are heard. Let the number of beats heard per second be x. Then,
υ₁ = υ₂ +x or υx = υ₂-x

To find which value is correct out of two values, a little wax is attached to the prong of the tuning fork of unknown frequency υ₁.
Again, the two forks are sounded together and beats are heard.
If number of beats per second increase, then the frequency of the first tuning fork will be (υ₂ – ×) because according to
(υ₁ – υ₂) = ×, × increases as υ₁ decreases.
If number of beats per second decreases, then the frequency of the first tuning fork will be (υ₂ + x) because according to
(υ₁ – υ₂) = -x, x decreases as υ₁ decreases.

Question 15.
Prove that the frequency of fundamental frequency of open organ pipe is double of the fundamental frequency of closed organ pipe of same length.
Or
If the frequency of fundamental tone of an open organ pipe is «, and that of closed organ pipe is n₂, then prove that n₁ = 2n₂.
Answer:
Let the length of organ pipe is l. For the fundamental frequency of open organ pipe.

\(l=\frac{\lambda_{1}}{2}\)
Where, λ1 is the wavelength
λ1 = 2l
If the fundmental frequency is n₁,then

Waves Class 11 Important Questions Long Answer Type

Question 1.
Prove that beats obtained by two sound sources per second is equal to the differences of frequencies of the sound source.
Answer:
When two sources of sound of nearly equal frequencies are sounded together, the amplitude of resultant wave due to superposition of waves at a point is in space. Some¬times the amplitude is maximum and sometimes, it is minimum.

As intensity is directly proportional to the square of amplitude, the intensity also varies periodically with time from maximum to minimum and to maximum again. Let two waves be,
\(y_{1}=a \sin 2 \pi v_{1} t\) and \(y_{2}=a \sin 2 \pi v_{2} t \) are superimposed. On superposition, we get a resultant wave as,
y = y₁+y₂

Question 2.
Prove that even and odd harmonics are produced by a stretched string.
Answer:
When a stretched string is plucked, then transverse waves are produced. The wave reflected from both- the ends and produced stationary waves.
∴The wave is reflected from the rigid end, hence the equation of stationazy wave will be
\(y=-2 a \sin \frac{2 \pi x}{\lambda} \cdot \cos \frac{2 \pi v t}{\lambda}\) ……………… (1)

Let one end of string is at x = 0 and another at x =l. where l is the length of string.
Now, at x = 0, from eqn. (1), we get
\(y=-2 a \sin 0^{\circ} \cdot \cos \left(\frac{2 \pi v t}{\lambda}\right)\)
or
y=0
Again, at x = l, from eqn. (1),
\(y=-2 a \sin \frac{2 \pi l}{\lambda} \cos \frac{2 \pi v t}{\lambda}\)
Now, at x =l, y = 0

Now, the mode of vibration will be obtained by putting k = 1, 2, 3,….
First mode of vibration: If k = 1 and the wavelength of stationary wave is λ₁, then from eqn. (2),
λ₁ = 2l
\(l=\frac{\lambda_{1}}{2}\)
Hence, the string vibrates in one segment. Now, if the velocity is and frequency n₁, then
\(n_{1}=\frac{v}{\lambda_{1}}=\frac{v}{2 l}\) ………… (3)

This frequency is called fundamental frequency or first harmonics.

Second mode of vibratioñ : If k =2 and wavelength ‘λ₂, then from eqn. (2),
λ₂ = l
Also, if the frequency is n₂,then
\(n_{2}=\frac{v}{\lambda_{2}}\)
or
\(n_{2}=\frac{v}{l}=\frac{2 v}{2 l}=2 n_{1}\)

Question 3.
Prove that in a dosed organ pipe odd harmonics are produced.
Or
In a closed organ pipe, describe the mode of vibrations and prove that:
(i) The frequency of fundamental note is inversely proportional to the length of the organ pipe.
(ii) Only odd harmonics are produced.
Answer:
Organ pipe is a long cylindrical tube of uniform diameter and made of wood or metal. Sound is produced by this pipe as a result of vibrations in air column inside this. If one end of this tube is closed then it is called closed organ pipe and if both ends are open then it is called open organ pipe.

When we place a source of sound near the open end of a closed organ pipe, a wave travels towards the closed end. This wave, after being reflected by the closed end, returns. A stationary longitudinal wave is produced inside the closed organ pipe as a result of superposition of incident wave and reflected wave. An antinode is always formed at the open end because the air particles are free to vibrate there. A node is always formed at the closed end because at the closed end the air particles are not free to vibrate.

Suppose that the closed end of the closed organ pipe is at x = 0 and open end at x = L, where L is the length of the organ pipe.
\(y=-2 a \sin \frac{2 \pi x}{\lambda} \cdot \cos \frac{2 \pi v t}{\lambda}\)
Case (i): At x = 0, we have y = o
Thus, the wave function vanishes at x = 0.

Case (ii) : At x = L, we have
\(y=-2 a \sin \frac{2 \pi L}{\lambda} \cdot \cos \frac{2 \pi v t}{\lambda}\)
At x =L, the displacement y has its maximum value ,y will be maximum when \(\sin \frac{2 \pi L}{\lambda}\) maximum i.e,

Where, k = 0, 1,2, 3,4,….
\(\lambda=\frac{4 L}{(2 k+1)}\)
Here, k = 0, 1, 2,… correspond to first, second, third,…. mode of vibration.

First mode of vibration: If λ₁ be the wavelength of the wave produced in air column corresponding k = 0, then from eqn. (2), we have
λ₁ = 4L
If n₁ be the frequency of the wave set up, then
\(n_{1}=\frac{v}{\lambda_{1}}=\frac{v}{4 L}\)

It is the tone of minimum frequency which produces in closed end organ pipe. It is called fundamental tone or first harmonic.
Also,
\(n_{1} \propto \frac{1}{L}\)
i.e., Fundamental frequency is inversely proportional to the length of the organ pipe.

Second mode of vibration :
If λ₂ be the wavelength of the wave set up in air column corresponding to k = 1, then from eqn. (2), we have
\(\lambda_{2}=\frac{4 L}{3}\)
If n₂ be the frequency of the wave set up, then

Thus, in this case, the frequency of the tone produced is three times the frequency of fundamental tone. It is called third harmonic or first overtone.

Third mode of vibration : If k = 2 and corresponding wavelength is λ₃, then from eqn. (2), we get
\(\lambda_{3}=\frac{4 L}{5}\)
If n₃ be the frequency of the wave set up, then
\(n_{3}=\frac{v}{\lambda_{3}}=\frac{v}{4 L / 5}=5 \cdot \frac{v}{4 L}\)
or
n₃ = 5n₁
Hence, the fifth harmonics of second overtone is five times of fundamental frequency.
Hence, the harmonics are n₁, 3n₁ 5n₁,….
Thus, odd harmonics are produced.

Question 4.
Prove that in open organ pipe both odd and even harmonics are produced. Ans. When a source of sound is kept at one end of an open organ pipe, the compressions and rarefactions are set up in the pipe. At the open ends antinodes are formed, hence rarefactions will be produced at the open ends and it returns in the form of compression. Thus, incident wave and reflected wave superimpose and stationary wave is formed.
Since, the reflection is, from free boundary,
\(y=2 a \cos \frac{2 \pi x}{\lambda} \cdot \sin \frac{2 \pi v t}{\lambda}\)

Now, the one end of organ pipe is at x = 0 and another end is at x = l.
Putting x = 0, we get
\(y=2 a \cos 0 \cdot \sin \frac{2 \pi v t}{\lambda}=2 a \sin \frac{2 \pi v t}{\lambda}\)
Again, putting x = l, we get
\(y=2 a \cos \frac{2 \pi l}{\lambda} \cdot \sin \frac{2 \pi v t}{\lambda}\)
For y is maximum,
\(\left|\cos \frac{2 \pi l}{\lambda}\right|=1 t\)
or
\(\frac{2 \pi l}{\lambda}=k \pi \) …………. (2)
Where,k= 1,2,3………

First mode of vibration: If k = I and wavelength is λ₁, then from eqn. (2), we get
λ₁ = 2l
Also
\(n_{1}=\frac{v}{\lambda_{1}}=\frac{v}{2 l} \)

n₁ s the least frequency called fundamental frequency or first harmonics.

Second mode of vibration : 1f k = 2 and corresponding wavelength is λ₂.
then
\(\lambda_{2}=\frac{2 l}{2}=l \)
∴ Frequency ,n₂= \(\frac{v}{\lambda_{2}}=\frac{v}{l}=\frac{2 v}{2 l}\)
n₂ = 2n₁
Hence, second harmonic is obtained at double of fundamental frequency.

Third mode of vibration: If k = 3 and corresponding wavelength is λ₃ and frequency n₃, then

Hence, third harmonic is 3 times of fundamental frequency.
Thus, the harmonics are both odd and even multiple of fundamental frequency.

Question 5.
Derive the expression for the apparent frequency, when the source of sound is moving towards a stationary listener.
Answer:
Suppose S is a source moving towards a stationary listener O. Let υ be the real frequency of the source and υ is the velocity of sound wave. Now, the first wave emitted by the source covers a distance υ in 1 sec. If the source is stationary then these υ waves will be
spread in the distance υ. Thus, the length of one wave i.e., the wavelength will be \(\lambda=\frac{u}{v}\)

Now, suppose the sound-source S is moving with velocity us towards the observer. Then the source covers a distance us in one second towards the listener. Therefore, the v waves emitted by the source in 1 second will spread in a distance (υ-υ_s) only. Thus, the wavelength will be shortened. Let the changed wavelength be λ’.
Then, \(\lambda^{\prime}=\frac{u-u_{s}}{v}\) ………….(1)

Thus, the listener will receive waves of wavelength λ’. Hence, the frequency of sound will appear to him changed. If the apparent frequency be υ’, then
\(v^{\prime}=\frac{u}{\lambda^{\prime}}\) (∴ Velocity of sound υ is steady )
Substituting the value of λ’ in eqn. (1), we have
\(\dot{v^{\prime}}=\frac{u}{\left(u-u_{s}\right) / v}\)
or
\(v^{\prime}=\left(\frac{u}{u-u_{s}}\right) v\) …………. (2)
As (υ-υ_s)< υ,therefore υ’ > υ i.e., the apparent frequency is greater than the real frequency.

If the source S is going away from the listener O, then the wavelength will be in-creased and its value will be given by
\(v^{\prime}=\left(\frac{u}{u+u_{s}}\right) v\)

As (υ + υ_s) > υ_s, therefore, υ‘ < υ i.e., the apparent frequency is less than the real frequency.

Question 6.
Derive the expression for the apparent frequency, when the listener is moving toward stationary source.
Answer:
Suppose, S is a sound source of frequency u and u is the velocity of sound. Again suppose that O is a listener who is coming towards the source with the velocity υ₀. If the listener were stationary then he would have received u waves in 1 second. Therefore, the real wavelength is given by
λ = \(\frac{u}{v}\)

But, the listener himself is moving towards the source with a velocity υ₀ that is, he covers a distance υ₀ toward the source in 1 second. Hence the listener, in addition to υ waves, also receives υ₀/ λ waves contained in the distance υ₀. Therefore, the total waves received by the listener in 1 second i.e., the apparent frequency of the sound is

As (υ+ υ₀) > u, hence υ‘ > υ i.e., apparent frequency will be greater than the actual frequency.

Question 7.
Derive the expression for the apparent frequency heared by a listener, when the source and listener both are moving in the same direction.
Answer:
Let the source and listener are moving with the velocities υ_s and υ₀ respectively in the same direction as shown in the figure.
If the source is moving only.
∴ The frequency heard by the listener will be \(n_{1}=\left(\frac{u}{u-u_{s}}\right) n\)
But, listener is moving away from the source

∴ Apparent frequency, \(n^{\prime}=\left(\frac{u-u_{o}}{u}\right) n_{1}\)

Putting the value of n₁, we get
\(n^{\prime}=\left(\frac{u-u_{o}}{u}\right) \times \frac{u}{u-u_{s}} \cdot n\)
or
\(n^{\prime}=\left(\frac{u-u_{o}}{u-u_{s}}\right) n\)
This is the required expression.

Waves Class 11 Important Numerical Questions

Question 1.
The speed of a wave ¡n a medium is 960m/s. If 3600 waves are passing through a point in the medium in 1 minute, then calculate the wavelength.
Solution:
Given: Speed of wave ν = 960 m/s
Frequency of wave n = \(\frac{3600}{60}\) =60/ second
Wavelength λ = \(\frac{v}{n}=\frac{960}{60}\)
or
λ = 16m.

Question 2.
The speed of sound in air at 14°C is 340 m/s. What is the speed of sound in air at 157.5°C when pressure becomes two times of its initial value.
Solution:
The change in pressure has no effect on speed of sound.
Given: t₁= 14°C,ν₁= 340m/s,t₂ = 1575°C,ν₂ =?

Question 3.
The distance between two consecutive nodes of a stationary wave is 25 cm. If the velocity ¡s 300 mc’, then calculate the frequency.
Solution:
Given: λ /2=25cm
or
λ =50cm = 50 ×\(10^{-2}\)
or
ν = 300 ms^-1
ν = nλ
or
300 = n ×50 ×\(10^{-2}\)
or
\(n=\frac{300}{50 \times 10^{-2}}=6 \times 10^{2}\) = Hz

Question 4.
The frequency of a whistle is 256 Hz. It is moving towards a stationary listener with a velocity of \(\frac{v}{3}\) , where v is the velocity of sound waves, then calculate the apparent frequency.
Solution :
Given : n = 256 Hz, \(v_{s}=\frac{v}{3}\)

Question 5.
An observer is moving towards a stationary source of frequency n. The frequency heard by the observer is 2n. If the velocity of sound is 332 ms^-1, then find the velocity of observer.
Solution:
Given : n’ = 2n, ν = 332 ms^-1.

Question 6.
The frequencies of two sources are 512 Hz and 516 Hz. Find the time inter-val between two consecutive beats.
Solution:
Given : n₁ =512 Hz, n₂ =516 Hz.
∴ Beat period = \(\frac{1}{n_{2}-n_{1}}\)
\(=\frac{1}{516-512}=\frac{1}{4}=0 \cdot 25 \mathrm{sec}\).

Question 7.
Two sound waves of wavelength 1 metre and 1.01 metre produce 40 beats in 10 seconds in a gas. Find the speed of sound in gas.
Solution:
Beats per second \(=\frac{40}{10}=4\)
If ν is the velocity of sound in gas, then
Frequency of first wave υ₁ = \(\frac{v}{1}\)
Frequency of second wave υ₂ \(\frac{v}{1.01}\)

Question 8.
Two tuning forks produce 6 beats when they are sounded together. The frequency of one tuning fork is 288 Hz. Now, the second tuning fork is loaded by little wax, then the beats are not heard. Find the frequency of another tuning fork.
Solution:
Let the frequency of second tuning fork is n.
∴ n = 288 + 6 or 288 – 6
or
n = 294 or 282 Hz
Since, second fork is loaded hence its frequency will decrease.
Since, no beat is heard.
Hence, the frequency of second tuning fork = 294 Hz.

Question 9.
Two sources of sound of equal frequencies are kept at a distance 100 m apart. A listener is moving between the sources and hears 4 beats. If the distance between the sources becomes 400 m, what will be the beats heard ?
Answer:
Spl. The frequency of beats depends upon the frequencies of the sources, not the distance. Hence, 4 beats will be heard.

Question 10.
Frequency of open end organ pipe is 256 Hz. What will be the frequency of same length closed end organ pipe ?
Solution:
Frequency of l length open end organ pipe
\(n=\frac{v}{2 l}\)
Frequency of closed ended organ pipe
\(n^{\prime}=\frac{v}{4 l}\)
or
\(\frac{n}{n^{\prime}}=\frac{v / 2 l}{v / 4 l}=\frac{2}{1}\)
Given, n = 256
∴ \(\frac{256}{n^{\prime}}=\frac{2}{1}\)
or
\(n^{\prime}=\frac{256}{2}=128 \mathrm{~Hz}\).

Question 11.
A string of mass 2-50 kg is under a tension of 200 N. The length of the stretched string is 20-0m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end ? (NCERT)
Solution:
Given : M= 2.50 kg, l = 20m, T= 200N
Mass per unit length m = \(\frac{M}{l}=\frac{2 \cdot 5}{20}\) = 0.125kg/m
∴ \(v=\sqrt{\frac{T}{m}}\)
or
\(v=\sqrt{\frac{200}{0 \cdot 125}}=\sqrt{1600}\) = 40 m/sec.
Time taken by distance (jerk) be
\(t=\frac{l}{v}\)
or
(∴ Velocity = \(\frac{\text { displacement }}{\text { time }}\) )
\(t=\frac{20}{40}\) = 0.5 sec.

Question 12.
A stone dropped from the top of a tower of height 300 m high splashes into the water of a pond near the base of the tower. When is the spHsh heard at the top given that the speed of sound in air is 340 ms^-1? (g = 9.8 ms 2) (NCERT)
Solution:
Given : Height of tower h = 300 m Speed of sound v = 340 ms^-1, initial velocity of stone u = 0 Let the time taken by stone to base of tower is t₁ Then by equation of motion
\(h=u t+\frac{1}{2} g t^{2}\)
300 = \(0 \times t_{1}+\frac{1}{2} \times 9 \cdot 8 t_{1}^{2}\)
or
\(4 \cdot 9 t_{1}^{2}=300\)
or
\(t_{1}=\sqrt{\frac{300}{4 \cdot 9}}=\) = 7.82 sec.

If the time taken to reach the sound up to top of tower is t₂, then \(t_{2}=\frac{h}{v}=\frac{300}{340}\) = 0.885
∴ Total time t = t₁+t₂.
= 7.82+0.885=8.705 = 8.7 sec .

Question 13.
A bat emits ultrasonic sound of frequency 1,000 kHz in air. 1f the sound meets a water surface. What is the wavelength of (a) the reflected sound (b) the transmiffed sound? Speed of sound in air is 340 ms^-1 and in water 1486 ms. (NCERT)
Solution:
Given :
n = 1,000 kHz =I000x10³Hz =10⁶Hz
Speed of sound in air ν_a = 340ms^-1 and that in water νw = 1486ms^-1 
(a) Reflected sound propagate in air

(b) Transmitted sound propagate in water

 

Waves Class 11 Important Questions Objective Type

1. Multiple- choice questions:

Question 1.
Sound of which frequency can be heard by human beings :
(a) 5 vib /sec
(b) 500 vib /sec
(c) 27000 vib / sec
(d) 50, 000 vib /sec.
Answer:
(b) 500 vib /sec

Question 2.
Astronaut cannot hear sound of his partner in moon, because :
(a) Frequency produced is more than audible frequency
(b) No medium is there for propagation of sound
(c) Temperature at night is very low
(d) There are lots of reason in moon.
Answer:
(b) No medium is there for propagation of sound

Question 3.
Two sources of frequency f₁ and f₂ will have beat frequency :
(a) \(\text { (a) } \sqrt{f_{2}}-\sqrt{f_{1}}\)
(b) f₁-f₂
(C) \(2\left(f_{1}-f_{2}\right)\)
(d) \(\sqrt{\left(f_{1}^{2}-f_{2}^{2}\right)}\)
Answer:
(b) f₁-f₂

Question 4.
The law relating sound between source and observer is :
(a) Doppler’s law
(b) Huygens law
(c) Newton’s law
(d) Galilean’s law.
Answer:
(a) Doppler’s law

Question 5.
The effect of beats due to two waves is because of:
(a) Refraction
(b) Reflection
(c) Dispersion
(d) Interference.
Answer:
(d) Interference.

Question 6.
As a utensil is filled with water, its frequency :
(a) Increases
(b) Decreases
(c) Remains same
(d) None of the above.
Answer:
(a) Increases

Question 7.
Waves transport from one end to another :
(a) Energy
(b) Amplitude
(c) Wavelength
(d) Matter.
Answer:
(a) Energy

Question 8.
Energy is not carried by :
(a) Transverse progressive waves
(b) Longitudinal progressive waves
(c) Stationary waves
(d) Electromagnetic waves.
Answer:
(c) Stationary waves

Question 9.
Velocity of sound in air is 350 m/s, frequency in open organ pipe of 50 cm length is:
(a) 175 Hz
(b) 350 Hz
(c) 700 Hz
(d) 50 Hz.
Answer:
(b) 350 Hz

Question 10.
y = 0.15 sin 5x cos 3001 represent an equation of stationary wave. Its wavelength will be :
(a) Zero
(b) 1.25m
(c) 2.512 m
(d) None of these.
Answer:
(b) 1.25m

2. Fill in the blanks:

1. The velocity of sound at 0°C is …………….. .
Answer:
332 m/s

2. …………….. waves require a physical medium for their propagation.
Answer:
Mechanical

3. …………….. waves do not require any medium for their propagation.
Answer:
Electromagnetic

4. In transverse waves, the distance between any two consecutive crest or trough is called …………….. .
Answer:
Wavelength

5. After one …………….. the phase of a particle is same as in the initial moment.
Answer:
Time period

6. Sound waves cannot propagate through …………….. .
Answer:
Vacuum.

.

3. Match the following:

Column ‘A’	Column ‘B’
1. Speed of waves	(a) \(\sqrt{\frac{\gamma P}{D}}\)
2.Speed of sound in gas	(b) \(\frac{2 \pi}{\lambda}\) × path difference
3. Phase of difference	(c) Sound navigation and Ranging
4. Principle of super position	(d) Frequency × Wavelength
5. SONAR	(e)\(\vec{y}=\overrightarrow{y_{1}}+\overrightarrow{y_{2}}+\ldots+\overrightarrow{y_{n}}\)

Answer:
1. (d) Frequency × Wavelength
2. (a) \(\sqrt{\frac{\gamma P}{D}}\)
3. (b) \(\frac{2 \pi}{\lambda}\) × path difference,
4.  (e)\(\vec{y}=\overrightarrow{y_{1}}+\overrightarrow{y_{2}}+\ldots+\overrightarrow{y_{n}}\)
5.  (c) Sound navigation and Ranging

4. Write true or false:

1. When two sound sources travelling perpendicularly. Doppler effect will implement.
Answer:
False

2. Propagation of mechanical wave in medium is possible due to elasticity and inertia.
Answer:
True

3. Propagation of wave in air is a isothermal changes.
Answer:
False

4. If the velocity of listener is more than velocity of sound then Doppler’s effect will not implement.
Answer:
True

5. Velocity of sound in moisture is less than velocity of sound in air.
Answer:
False

6. A person is standing at a point and a car is coming toward him blowing horn. The frequency of horn will be less than original frequency.
Answer:
False

7. Stationary wave does not carry energy.
Answer:
True

8. Beats is produced when frequency of two sound source are same.
Answer:
True

9. In a resonance air column, always transverse non- progressive wave is produced.
Answer:
False

10. Expansion of universe is observed by Doppler’s effect. ‘
Answer:
True

Students get through the MP Board Class 11th Physics Important Questions Chapter 1 Physical World which are most likely to be asked in the exam.

MP Board Class 11th Physics Important Questions Chapter 1 Physical World

Physical World Class 11 Important Questions Very Short Answer Type

Question 1.
Define science.
Answer:
The systematic knowledge based on observation, experiment and scientific logic, is called science.

Question 2.
Define physics.
Answer:
Physics is that branch of science in which matter, energy and their interactions are studied.

Question 3.
What are the different branches in which physics is divided?
Answer:

1. Classical physics : All the discoveries before 1900 A.D. is included in it. It deals with mechanics, light, heat, sound, electricity and magnetism.
2. Modern physics: All the discoveries after 1900 A.D. is included in it. It deals with electron, molecule, nucleus, quantum mechanics etc.

Question 4.
Name any four physicists and their discoveries.
Answer:

Physicists	Discovery
1. Sir C.V. Raman	Raman effect
2. Homi J. Bhabha	Cosmic rays
3. Satyendranath Bose	Wireless
4. Meghnath Saha	Theory of thermal-ionization.

Question 5
Name two devices used in daily life and their working principle.
Answer:

Device	Principle
1. Aeroplane	Bernoulli’s theorem
2. Radio and television	Propagation of e

Question 6.
What are the different branches of science?
Answer:
The different branches of science are Physics, Chemistry, Biology, Botony, Maths, Medical science etc.

Question 7.
What do you mean by matter?
Answer:
The substance which occupy space is called matter. It has mass for example : Stone, Iron etc.

Question 8.
What do you mean by energy?
Answer:
Capacity of doing work is called energy. Without energy no work can be perform.

Question 9.
What are different forms of energy?
Answer:
The different forms of energy are: Mechanical energy, Heat energy, Light energy, Electrical energy, Chemical energy, Nuclear energy etc.

Question 10.
Name any three recent inventions in physics.
Answer:
The recent inventions are LASER, Computer and Super conductivity.

Question 11.
Which scientist discovered Mass-energy relation?
Answer:
Mass-energy relation E = mc² was discovered by Einstein.

Question 12.
Write the name of the force which is always attractive?
Answer:
Gravitational force.

Question 13.
Write the following forces in order of their increasing magnitude. Electromagnetic force, Nuclear force, Gravitational force.
Answer:
Gravitational force, Electromagnetic force and Nuclear force.

Question 14.
Write the name of that strong force which is responsible for stability of nucleus.
Answer:
Nuclear force.

Question 15.
What is the range of weakest force?
Answer:
10^-16m.

Question 16.
Write the name of the field-produce due to :
(i) Static charge,
(ii) Moving charge.
Answer:
(i) Static charge : Only electrical field.
(ii) Moving charge : Magnetic field.

Physical World Class 11 Important Questions Short Answer Type

Question 1.
What do you mean by systematic observation?
Answer:
Systematic observations: To work out on a problem, the scientists collected all relevant data concern to the problem, by systematic experiments and observations.

Question 2.
How hypothesis is formulated?
Answer:
Formulation of hypothesis: On the basis of observations working law is formed called hypothesis.

Question 3.
How hypothesis is tested?
Answer:
Testing of hypothesis: On the basis of hypothesis and predictions, conclusions are drawn and it is verified by new experiments. This is called testing of hypothesis.

Question 4.
How theory is established?
Answer:
Final theory: If the conclusions and predictions are verified by the experiments, then the hypothesis is accepted as final theory, otherwise corrections are made and verified again. The phenomena are repeated till the final theory is achieved.

Question 5.
What is theory? When does it need modification? Explain it with one example.
Answer:
The hypothesis, tested by experiments is called theory.
Sometimes a new measurement shows difference between the existing theory and observations then it needs to be modified.
Example: Huygens thought that light waves are longitudinal in nature. But polarization of light could not be explained on the basis of this thought. Therefore, Fresnel assumed that the light waves are transverse in nature.

Question 6.
Why physics is called as basic science?
Answer:
Physics is called basic science because its seek out and understand the basic laws of nature upon which all physical phenomenon is depended. It has brought to us deeper and deeper levels of understanding nature.

Question 7.
What is scientific method? Explain the steps of scientific method.
Answer:
Scientific method: The methods by which a scientist works to establish theories
and laws are called scientific method.
The steps of scientific method are :

• Systematic observations,
• Formulation of hypothesis,
• Testing of hypothesis,
• Final theory.

Physical World Class 11 Important Questions Long Answer Type

Question 1.
What is the relation of physics to other branches of science?
Answer:
The different branches of science in relation to physics are given below :
1. Physics in relation to Mathematics:
The knowledge of mathematics is necessary to understand the laws and theories of physics which is a powerful tool. Without mathematics expansion of physics is impossible. Radio, television, computer, satellite, telecommunication etc. are the result of combined applications of physics and mathematics.

2. Physics in relation to chemistry:
Physics plays an important role in the development of chemistry. The structure of atoms described by physics, the arrangement of atoms in periodic table, the nature of valence electrons and chemical bonds. The X-rays and neutron diffractions are the results of research in physics.

3. Physics in relation to biological science :
The technology of physics are used in Biology. The optical microscopes developed by physics are extensively used in biological science. The electron microscopes which is nearly a million times powerful than optical microscopes used in the study of the structure of the cells with the help of X-rays position of bone fracture is found and also the presence of foreign materials hidden in the body. X-rays are also used in the treatment of cancer.

Question 2.
Explain the influence of Physics on the society.
Answer:
The Physics has influenced the society in two ways:
(i) Living standard :
Physics has developed the technology and hence the inventions of machines, devices and instruments give comfort and protection to human life.
By the television and radio transmission, has made possible to see the happenings live from any part of the world. Computer can calculate very complicated calculations within no time. Thus, the efficiency of men have increased.

(ii) Effect on the thinking of human beings: The contribution of Physics in the field of education, medical sciences etc., developed the scientific aspect and therefore the narrow thinking of caste and creed, religion and superstitions are removed from the society.

Question 3.
Name few scientists and their contribution ¡n physics.
Answer:
Scientists and their contribution in physics:

Scientists	Discovery
1. Sir Isaac Newton	Laws of motion
2. Albert Einstein	Theory of relativity
3. James Chadwick	Discovery of neutron
4. H. A. Becquerel	Radioactivity
5. SirC.V.Raman	Raman effect
6. Homi J. Bhabha	Cosmic rays
7. Satyendranath Bose	Wireless
8. MeghnathSaha	Theory of thermal-ionization.

Physical World Class 11 Important Questions Objective Type

1. Multiple- choice questions:

Question 1.
Which among the following is called basic science :
(a) Physics
(b) Chemistry
(c) Maths
(d) Biology
Answer:
(a) Physics

Question 2.
The oldest branch of physics is :
(a)Thermodynamics
(b) Optics
(c) Mechanics
(d) Sound.
Answer:
(c) Mechanics

Question 3.
The weakest force of the nature is :
(a) Electrical force
(b) Magnetic force
(c) Nuclear force
(d) Gravitational force.
Answer:
(d) Gravitational force.

Question 4.
Rocket propulsion is based on the law :
(a) Faraday’s law
(b) Newton’s law
(c) Galileo’s law
(d) Bernoulli ‘s theorem.
Answer:
(b) Newton’s law

Question 5.
Working principle of aeroplane is based on …………………….:
(a) Laws-of thermodynamics
(b) Faraday’s laws of electromagnetic induction
(c) Bernoulli’s theorem
(d) Newton’s laws of motion.
Answer:
(c) Bernoulli’s theorem

Question 6.
Quantum mechanics was discovered by the scientist:
(a) Michael Faraday
(b) Chandrashekhar Venkat Raman
(c) Satyendranath Bose
(d) Somnath Saha.
Answer:
(c) Satyendranath Bose

Question 7.
Planetory laws of motion was propounded by :
(a) Max-plank
(b) Newton
(c) Kepler
(d) Faraday.
Answer:
(c) Kepler

Question 8.
Who gave the electromagnetic theory of waves :
(a) Niel-Bohr
(b) James dark Maxwell
(c) Michael Faraday
(d) Johans kepler.
Answer:
(c) Michael Faraday

2. Fill in the blanks:

1. Theory of relativity was given by ……………..
Answer:
Albert Einstein

2. Neutron was discovered by…………………..
Answer:
James Chadwick

3. Abdus Salam is from the country ………………….
Answer:
Pakistan

4. Computer is based on the principle of ………………….
Answer:
Digital logic of electronic circuit

5. Sir Issac Newton belonged to the country …………………
Answer:
England

3. Match the following:
I.

Column‘A’	Column‘B’
1. Newton	(a) Principle of electromagnetic waves
2. Einstein	(b) Cosmic ray
3. Maxwell	(c) Dual nature of light
4. de Broglie	(d) Scattering of light
5. Sir C. V. Raman	(e) Laws of gravitation.
6. H. J. Bhabha	(f) Explanation of photoelectric effect.

Answer:
1. (e) Laws of gravitation.
2. (f) Explanation of photoelectric effect
3 (a) Principle of electromagnetic waves
4 (c) Dual nature of light
5 (d) Scattering of light
6. (b) Cosmic ray

II.

Column ‘A’	Column ‘B’
1. Hydroelectric power	(a) Laws of motion
2. Aeroplane	(b) Propogation of electromagnetic waves
3. Electric generator	(c) Conversion of gravitational energy into electrical energy
4. Rocket propulsion	(d) Digital logic of electronic circuits
5. Television	(e) Faraday’s laws of electromagnetic induction
6. Computer	(f) Bernoulli’s

Answer:
1. (c) Conversion of gravitational energy into electrical energy
2. (f) Bernoulli’s principle.
3. (e) Faraday’s laws of electromagnetic induction
4. (a) Laws of motion
5. (b) Propogation of electromagnetic waves
6. (d) Digital logic of electronic circuits

III.

Column ‘A’	Column ‘B’
1. Sir Isaac Newton	(a) Scotland
2. Galileo	(b) Germany
3. Albert Einstein	(c) India
4. Clark Maxwell	(d) Italy
5. Archimedes	(e) England
6. H. J. Bhabha	(f) Greece.

Answer:
1. (e) England
2. (d) Italy
3. (b) Germany
4. (a) Scotland
5. (f) Greece.
6. (c) India

4. Write true or false:

1. Nuclear reactor is based on the concept that a slow moving neutron splits uranium.
Answer:
True

2. Mesons were discovered by Clark David Anderson.
Answer:
False

3. Newton had given the corposcular theory of light.
Answer:
True

4. Tn 1921, Albert Einstein got the noble prize for theory of relativity.
Answer:
False

5. Fast moving particles are studied using the theory of relativity.
Answer:
True

5. Answer in one word:

1. What is Einsteins mass energy equivalence equation?
Answer:
E = mc²

2. Substance which has mass is called?
Answer:
Matter

3. Rocket propulsion is based on which law?
Answer:
Newton’s third law of motion

4. The laws of photoelectric effect is explained by which equation?
Answer:
Einsteins photoelectric of equation

5. Who gave the principle of uncertainty?
Answer:
Heisenberg (Germany)

Students get through the MP Board Class 11th Biology Important Questions Chapter 11 Transport in Plants which are most likely to be asked in the exam.

MP Board Class 11th Biology Important Questions Chapter 11 Transport in Plants

Transport in Plants Class 11 Important Questions Very Short Answer Type

Question 1.
Which factors effect rate of diffusion?
Answer:
Factors affecting the rate of diffusion are :

• Concentration of the medium,
• Solubility of solutes,
• Temperature,
• Size and mass of diffusing particles,
• Permeability of the separating membrane.

Question 2.
What is Porin? Give its role in diffusion.
Answer:
Porin is a type of protein, which forms pore in the outer membrane of plastids, mitochondria and bacteria, so that protein molecules can pass through them.

Question 3.
Which part of the roots absorb water from the soil?
Answer:
Root hair zone of the roots absorb water from the soil.

Question 4.
Which soil water is absorbed by the roots?
Answer:
Roots absorb capillary water.

Question 5.
What is Guttation?
Answer:
The process of loss of water in the form of water droplets through the hydath- odes of herb plants is called as Guttation.

Question 6.
What is semipermeable membrane?
Answer:
Membranes which allow only the solvent to pass through them not the solute particles are called as semipermeable membrane

Question 7.
Name the tissue helps for conduction of water in plants.
Answer:
Xylem helps for conduction of water and minerals in plants.

Question 8.
What is Transpiration pull?
Answer:
Due to continuous loss of water vapour through stomata during transpiration process tension is created on unbroken water column in the Xylem, which remain continuous due to cohesive force of water molecules, thus water ascend in the plants in upward direction through Xylem. This is called as Transpiration pull.

Question 9.
What is Wilting?
Answer:
When rate of transpiration is very high and rate of absorption fails to keep balance with the loss of water, a kind of water deficit takes place in plants. This condition is called as Wilting. Plant cells become flaccid in this condition.

Question 10.
What do you mean by antitranspiration substance?
Answer:
Substance which reduces rate of transpiration without effecting gaseous exchange in plants is called as Antitranspiration substance e.g. Phenyl mercuric acetate.

Question 11.
What is ascent of sap?
Answer:
The upward movement of water from the root towards the top of the plant is known as ascent of sap. Sap is water with dissolved ingredients (Minerals). Ascent of sap takes place through tracheary elements of Xylem.

Question 12.
Define DPD.
Answer:
The amount by which the diffusion pressure of the solution is lower than that of its solvents at the same temperature and atmospheric pressure is called as diffusion pressure deficit (DPD). DPD is the index of absorbing power of a solution, it is also called as suction pressure.

Question 13.
Define osmotic pressure.
Answer:
A pressure which is developed in a solution when it is separated from its solvent by a semipermeable membrane due to presence of dissolved solutes in it. This is called osmotic pressure (OP).

Transport in Plants Class 11 Important Questions Short Answer Type

Question 1.
Explain role of protein pump during active transport in plants.
Answer:
Absorption of substances which occurs with expenditure of energy through plasma membrane is called as active absorption.

Cell is surrounded by a lipo-protein membrane, which is impermeable to free ions. But some carrier compounds (protein in nature) acts as carrier to carry ions by binding with ions. Carrier combines with ions to form carrier-ion complex, which can move across the membrane with the expenditure of metabolic energy (ATP) of respiration. On the inner surface of the membrane this complex breaks releasing ions into the cell while the carrier goes back to the outer surface to pick fresh ion.

Question 2.
Why water potential of pure water is highest? Explain.
Answer:
Kinetic energy of water molecule is more. If concentration of water is more in any system then its kinetic energy will be more, thus its water potential or osmotic potential is high. Addition of solvent to pure water lowers its osmotic potential. It is represented by Greek symbol psi (Ψ) and its value is equal to osmotic pressure. Its unit is Pascal (Pa).

Question 3.
Explain Water Potential. What are the various factors which effect it? Explain relationship between water potential, solute potential and pressure potential.
Answer:
Water travels from high energy area to low energy area. Kinetic energy of pure water is highest. Water move from one place to other place according to concentration gradient or potential gradient. Water potential of pure water is highest, its value is equal to zero. Addition of solute to pure water lowers its free energy therefore reduces value of water potential, thus its value become negative (-ve).
The differences between the free energy of water molecule is pure and any other system is called as water potential. It is represented by (Ψ_w).
The amount of solute present in water is called solute potential (Ψ_s).
Pressure potential is equal to osmotic pressure, when cell is in turgid condition. Pressure potential is represented by sign Ψ_p.
The water potential (Ψ_w) is actually the sum of other two potentials.
Ψ_w = Ψ_s + Ψ_p

Question 4.
What will happen if pressure more than atmospheric pressure is applied to the pure water or solvent?
Answer:
If pressure more than atmospheric pressure is applied in pure water then its water potential increases. Its value is equal to the amount of water pumped from one place to the other place. When water enters into a cell, it apply pressure on the cell wall. Cell become turgid. It increases pressure potential. Generally, water potential is positive. Water potential is represented by sign Ψ_p.

Question 5.
How does mycorrhiza helpful to plant in absorption of water and minerals?
Answer:
There are fungi which live in intimate association with some other unlike living organisms, both being beneficial to each other, the process being called symbiosis. A symbiotic, non-pathogenic or weakly pathogenic association of various fungi and bryophytes, pteridophytes or flowering plants is known as mycorrhiza. Most of these associations are not specific. A fungus can associate with many plants and a plant can associate with many fungi. The fungus may grow within the root surface or may grow around the root surface.

The fungal partners are usually those that can breakdown the plant cell wall only in a limited way. The fungus takes its nutrition from the damaged portion and grow, but the growth of the fungus within the plant does not damage it. The fungus remains in a restricted area. The plants are benefitted as the fungi spread in soil, absorbs water, nitrogen and other minerals from the soil. The fungi also produce various growth promoting substances and antimicrobial substances. The fungi use carbohydrates produced by other plant partner.

Question 6.
Give role of endodermis in absorption of minerals in plants.
Answer:
Many transport proteins are found embedded in the plasma membrane of endodermal cells of the roots of the plants, which allow only few particles to pass through them.
Transport protein of endodermal cells of roots of the plants can bind to selected solute particles only based on the shape and utility of the mineral ions for the plants. They also control quantity and type of ions to be absorbed.

Question 7.
Derive relationship between osmotic pressure, turgor pressure and suction pressure or diffusion pressure deficit.
Answer:
To understand relationship between OP, TP, and DPD, we must see first what are these.
Osmotic pressure (OP) :
A pressure which is developed in a solution, when it is separated from its solvent by a semipermeable membrane.

Turgor pressure (TP):
When a cell is placed in distilled water, it absorbs water by the process of osmosis and become turgid and the contents of cell exert a pressure against cell wall which is known as turgor pressure.

Diffusion pressure deficit (DPD) :
The amount by which the diffusion pressure of the solution is lower than that of its solvent at the same temperature and atmospheric pressure is called as Diffusion pressure deficit or Suction pressure.
Relationship between Diffusion Pressure Deficit, Osmotic Pressure and Turgor Pressure : When a cell absorbs water the turgor pressure of cell is increased but cell wall also applies an equal but opposite pressure called wall pressure (WP) against the turgor pressure (TP). Therefore, actual force responsible for entry of water into the cell will be OP-TP (or WP). The relationship between DPD, OP, TP and WP can be expressed as follows :
DPD(SP) =OP-WP
or
DPD = OP – TP, (∵ TP = WP)
Due to endosmosis of water the OP of the cell sap, decreases while WP is increased, so that in fully turgid cell TP is equal to OP.
OP = TP (in fully turgid cell)
Hence, DPD (SP) = 0 (zero)

Thus, there will be no absorption of water by the cell in fully turgid cell.
On the other hand, the removal of water from cell sap (exosmosis) results in the increase of its OP and decrease of TP so much so that cell becomes flaccid (fully plasmo- lyzed) and the value of turgor pressure becomes zero.
TP = 0 (in fully flaccid cell)
Hence, DPD (SP) = OP
Consider that a plant cell with OP = 10 atm is immersed in pure water. In beginning, TP inside the cell will be zero, i.e., DPD = OP = 10 atm.

Question 8.
What is Osmosis? Give impotance of osmosis process in plants.
Answer:
Osmosis :
The process of movement of solvent from higher concentration to lower concentration through a semipermeable membrane is called as Osmosis.
It is of following types :

1. Endosmosis :
It is the movement of the solvent molecules from outside to inside the cell through semipermeable membrane (SPM). e.g., Some raisins when kept in water for few hours, they swell as they contain sugar solution of higher concentration.

2. Exosmosis :
It is the movement of solvent molecules from inside to outside the cell through semipermeable membrane (SPM). e.g.,

(i) Swollen raisins shrink if kept in a sugar solution of higher concentration.
(ii) If erythrocytes are placed in 2% NaCl solution, they quickly lose water, shrink up and assume the wrinkled appearance.
Significance of Osmosis :

•  It helps in the absorption of water and minerals from the soil by roots.
• Osmosis helps in the movement of water from one cell to another.
• The phenomenon of plasmolysis is dependent on osmosis.
• It develops turgidity and turgor pressure in plant cell and hence the shape or form of organs is maintained.
• Opening- and closing of stomata is regulated by the osmotic entry and exit of water in guard cells.
• The resistance of plants to drought and frost increases with increase in osmotic pressure of their cells.

Question 9.
What is the role of root pressure in transport of water in plants?
Answer:
According to Priestley (1916), upward flow of water is due to development of a hydrostatic pressure, which is developed in root system. The hydrostatic pressure developed due to accumulation of water absorbed by root, is called root pressure. This pressure is developed in the tracheal elements of the Xylem as a result of metabolic activities of roots. This led to the view that root pressure can raise water in tall trees.

Objections to root pressure theory :

• Root pressure has not been found in all plants.
• No or little root pressure has been observed in gymnosperms which have some of the tallest trees of the world.
• Water continues to rise upwards even in the absence of roots.
• It was observed that, root pressure by itself cannot be held responsible for the ascent of sap in tall trees.Because its maximum value does not normally exceed two atmospheric pressure, hence it can raise the sap up to 15-30 m.

Experiment 1.

Demonstration of root pressure in plants.
Requirements :
Potted Tomato, Balsam or Biyophyllum plant, rubber tube, knife, narrow gas tube, coloured water, petridish, non-drinking oil.
Method:
Take a previously well-watered potted plant of Tomato, Balsam or Bryophyllum. Cut their stem 5-8 cm above the soil level particularly in the morning. Fix a narrow glass tube to the cut end of stem with the help of rubber tube and fill it with coloured water. Cover the glass tube with a small petridish in order to prevent evaporation. A drop of non-drying oil can also be used over the surface of water to prevent evaporation.

Question 10.
Describe the factors affecting Ascent of sap through Xylem in plants.
Answer:
Ascent of sap through Xylem in plants depends on following physical factors :

• Cohesion: Water molecules has tendency to link with each other by force of attraction.
• Adhesion: An attraction exists also between water molecules and the elements of the Xylem wall called as adhesion.
• Surface tension : Molecules of water has more attraction in liquid state as compared to gaseous state.

Water molecules are in a continuous movement strongly attracted to each other to form a continuous column of water in Xylem elements, which help them to move in upward direction through Xylem.

Question 11.
Give role of transpiration pull in transport of water. What are the factors affecting transpiration ? Which factors are useful for plants?
Answer:
Transpiration pull or Tension on the unbroken water column :
As a result of transpiration, water is lost from mesophyll cells to the intercellular spaces. The water vapours move out of the plant through stomata. The DPD of mesophyll cells increases.

With the increase in DPD, these cells absorb water from adjoining cells, ultimately the water is absorbed from Xylem elements of vascular bundles of leaf. Since the xylem elements are filled with continuous water column this tension of pull is transmitted down from petiole to stem and finally to roots leading to upward movement of water.

Thus, water ascends in the plants because of transpiration pull and the column of water remains continuous because of cohesive force of water molecules. This theory is the most accepted theory at the present time.
Factors affecting transpiration : Transpiration is influenced by a number of factors both external and internal. A brief account of them are given below :

[A] External factors :
1. Humidity :
If the plant is surrounded by a saturated atmosphere, there can be no escape of water vapour from the intercellular spaces of the leaf to the outside atmosphere. On the other hand, if the atmosphere is less humid, transpiration goes on at a rapid rate.

2. Temperature :
Rise in temperature reduces the relative humidity of the air as a result, the rate of transpiration increases.

3. Wind :
Wind has a direct influence on humidity. Dry air lowers the amount of air moisture and increases the rate of transpiration, whereas humid air decreases the rate of transpiration.

4. Light:
The rate of transpiration increases markedly in light and decreases in the dark. Light affects transpiration into two ways : Firstly, it increases transpiration by raising the temperature of leaves. Secondly, there is a close relationship between the opening of stomata and light as the stomata opens in day and gets closed in night.

5. Available soil water :
If available soil water is such that the rate of absorption of water is slowed down, the rate of transpiration is correspondingly decreased. Higher con-centration of salts in the soil water reduces the absorption rate and thereby the rate of transpiration.

6. Atmospheric pressure:
At lower atmospheric pressure at the altitudes, the rate of transpiration is increased but the decrease in rate occurs due to the prevailing low tempera-ture at these heights.

[B] Internal factors:
The histological and internal physiological conditions influence the rate of transpira-tion to a great extent. The following structures of the leaves influence the rate of transpiration :

• Presence of thick cuticle and deposition of wax on epidermis, decreases the rate of transpiration.
• Presence of hair on the outer surface of stomata also reduces the rate of transpiration.
• The rate of transpiration depends upon size, position and distribution of the leaves.
• The rate of transpiration is directly proportional to the water contents of the mesophyll tissue.

Question 12.
Xylem transport is unidirectional and phloem transport is bidirectional. Explain.
Answer:
In Xylem transport of water and minerals is unidirectional, i.e. from root to top part of the plant (upward) because suberin layer found in the endodermis of roots causes active transport of ions in one direction.

Whereas in phloem transport of food occurs bidirectional. Generally, food is prepared in the green parts of the plants which contain chlorophyll and then it is collected in sink, which is to be distributed to all parts of the plant. Young buds act as sink. In plants, source and sink for transport of food is changeable as it changes with season thus phloem shows bidirectional movement of food.

Question 13.
Give reason for opening and closing of guard cells during transpiration.
Answer:
Opening and closing of stomata :
Opening and closing of stomata depends upon the activity of guard cells. Guard cells are crescent like or kidney shaped. Their concave wall are thicker than convex which face each other. The opening and closing of stomata depend upon the turgidity of the stomata, when guard cells become turgid, stomata open while guard cells become flaccid, stomata closed. Guard cells contain chloroplast hence they synthesize food by photosynthesis, using CO₂ during day time.

Thus, due to lack of CO₂ the pH value of the cell increases (7.5 p^H). At this p^H starch produced during photosynthesis reacts with inorganic phosphate in presence of phosphorylase enzyme to form glucose-1-phosphate, which is soluble in water thus concentration of the cell sap of the guard cell increases. This lead to absorption of water by the guard cells from neighbouring cells by osmosis, thus increasing turgor pressure due to which the outer thinner wall of the guard cells stretches out and stomata open for diffusion of water vapour.

In dark the reverse reactions occur in the guard cells. During this period photosynthesis does not occur and respiration releases CO₂, thus reducing the pH of guard cell towards acidic side (p^H 5.0).
Now soluble carbohydrates are converted into insoluble starch by the enzyme phosphorylase. This brings about rapid fall of the osmotic concentration of cell sap of guard cell which become flaccid, thus closing the stomatal aperture.

Transport in Plants Class 11 Important Questions Long Answer Type

Question 1.
(a) Describe the process of plasmolysis with labelled diagram.
(b) What will happen if a plant cell is kept in high water potential solution?
AnsWER:
(a) Plasmolysis :
When a living cell is placed in a hypertonic solution then exosmosis of water takes place hence the water of cell comes out into outer solution. Due to exosmosis of water the protoplasm shrinks away from the cell wall and an irregular mass at the centre. This shrinkage of protoplasm is known as plasmolysis.

When plasmolysed cell is placed in water, the water enters into the cell sap, the cell becomes turgid and the protoplasm again assumes its normal sap and position. This phenomenon is called as deplasmolysis.

(b) When a plant cell is kept in high water potential solution, i.e. hypotonic solution then endosmosis occurs. Cell become turgid. Protoplasm apply pressure on the cell wall called as Turgor pressure. Pressure of the protoplasm against cell wall is called as pressure potential (Ψ_P). This turgor pressure is responsible for increased size of the cell.

Question 2.
Explain hypothesis of translocation of organic substances in plants.
Answer:

Munch mass flow hypothesis :
This theory was proposed by Munch (1930). According to this theory, translocation occurs on mass along a gradient of turgor pressure from supply end to the consumption end and which can be established by the following arrangement.

Two bulbs ‘A’ and ‘B’ with semipermeable membrane walls are – connected with a tube ‘C’ containing water to form a closed system. Bulb ‘A’ contains more concentrated solution than bulb ‘B’ and are dipped in water. Now due to higher osmotic pressure of the concentrated sugar solution in ‘A’, water enters into it by endomosis, increasing turgor pressure. This will result in mass flow of sugar solution to bulb ‘B’ through the tube ‘C’ till the concentration of sugar solution in both become equal. If in the above, it could be possible to maintain continuous supply of sugar in ‘A’ and its utilization in ‘B’, the process will go on indefinitely.

According to Munch, similar arrangement exists in plants. Due to photosynthesis, the mesophyll cells of the leaves have higher concentration of sugar solution in soluble form corresponding to bulb ‘A’ and is utilized by different parts (bulb ‘B’). The sieve tubes of the phloem act as tube ‘C’.

In plants mesophyll cells draw water from the xylem of leaf, due to high osmotic and suction pressure or DPD of their cell sap, increasing their turgor pressure. The turgor pressure in the cells of stem and roots are comparatively low hence, the soluble organic solute begins to flow from mesophyll through phloem down to the cells of stem and roots under the gradient of turgor pressure. These organic materials are either consumed or stored and water returns back to the xylem.

Question 3.
Write differences between following :
(a) Diffusion and Osmosis,
(b) Transpiration and Evaporation,
(c) Osmotic pre-ssure and Osmotic potential,
(d) Diffusion and Imbibition,
(e) Apoplast and Symplast pathway,
(f) Guttation and Transportation,
(g) Transpiration and Guttation,
(h) Endosmosis and Exosmosis.
Answer:
(a) Differences between Diffusion and Osmosis

Diffusion	Osmosis
1. Molecules move from higher concentration to lower concentration.	Molecules move from higher osmotic pressure to lower osmotic pressure.
2. Semipermeable membrane is not required.	Semipermeable membrane is required.
3. It occurs in solid, liquid and gas.	It occurs only in liquid.
4. It is a fast process.	It is a slow process.

(b) Differences between Transpiration and guttation

Transpiration	Evaporation
1. Transpiration is a vital process.	Evaporation is a physical process.
2. It takes place only in living plants.	It occurs in both, living and non-living things.
3. In this process, loss of water in the form of water vapour takes place specially through the leaves.	Evaporation takes place through any surface of living or non-living thing.
4. The process of transpiration is regulated by the activity of guard cells.	Guard cells are not involved in this process.
5. Transpiration is largely dependent upon absorption of water from the soil.	Evaporation continues as long as water is available on the surface.

(c) Differences between Osmotic Pressure and Osmotic Potential

Osmotic Pressure	Osmotic Potential
1. Pressure which is developed in a solution, when it is.kept separated from its solvent by a semipermeable membrane is called as Osmotic pressure.	Osmotic potential is that quantity of water which may reduce pressure of solute.
2. Its value is positive.	Its value is negative.

(d) Differences between Diffusion and Imbibition

Diffusion	Imbibition
1. Movement of molecules or particles from a region of higher concentration to a region of lower concentration is called as diffusion.	It is the process of absorption of solvent by a solid substance.
2. It helps for gaseous exchange through stomata, lenticels, passive absorption of water by roots etc.	It helps in initiation of water absorption by roots, absorption of water by germinating seeds etc.

(e) Differences between Apoplast and Symplast Pathway

Apoplast Pathway	Symplast Pathway
1. Pathway of movement of water through cell wall without crossing plasma membrane is called as Apoplast pathway.	Pathway of movement of water through plasmodesmata of the plasma membrane is called as Symplast pathway.

(f) Differences between Guttation and Transportation

Guttation	Transportation
1. Water is lost in the form of water droplets  from the margin of leaves through hydathodes in this process.	Movement of materials in plants from one part to other part is called as transportation.
2. Guttated water contains many minerals,salts and forms a solution.	Transported materials are water, minerals, organic food etc.
3. It occurs through hydathode.	It occurs through xylem and phloem.
4. It occurs at night.	It may occur at any time.

(g) Differences between Transpiration and Guttation

Transpiration	Guttation
1. Water is lost in the form of vapour through stomata, cuticle or lenticel.	Water is lost in the form of water droplets from the margins of leaves through hydathodes.
2. Transpired water vapour is pure.	Guttated water contains many mineral salts and forms a solution.
3. This process is regulated by the act kity of guard cells and occurred during day time only.	Guard cells do not involve in this process and it takes place during night time only.
4. Root pressure does not affect this process. Transpiration takes place in all terrestrial plants.	Root pressure affects this process and it takes place only in herbaceous plants.

(h) Differences between Endosmosis and Exosmosis

Endosmosis	Exosmosis
1. The diffusion of water or solvent molecules takes place from external cell to the cell.	The diffusion of water or solvent molecules takes place from cell to external solution.
2. It takes place when the concentration of cell solution is higher than the concentration of outer solution.	It takes place when the concentration of external solution is higher than the concentration of outer solution.
3. Cells become turgid after endosmosis.	Cells become flaccid after exosmosis.

Question 4.
What is ascent of sap? Explain the view of Dixon and Jolly regarding the ascent of sap.
Or,
Explain the cohesion theory of ascent of sap.
Answer:
Ascent of sap :
The mechanism of upward translocation of water from the roots to the top of the plant is still a mystery. The upward movement of water from the root towards the top of the plant is known as ascent of sap. Sap is water with dissolved ingredients (minerals). The ascent of sap takes place through the tracheary elements of xylem. The mechanism involved in the transport of water from the roots to the leaves of plants, some of them being more than 200 metre high (e.g., Sequoia, Gigantia, Eucalyptus), is still unresolved.

Transpiration pull and cohesion of water theory :
This theory was proposed by Dixon and Jolly in 1894. According to this theory, water molecules are in a continuous movement strongly attracted to each other to form a continuous column of water in xylem elements. The process of transpiration pulls up the water. This theory has the following three essential features:

(A) Strong cohesion force or tensile strength of water :
The principle is mairily based on the facts that:

• Water molecules have strong mutual attraction (cohesion) with each other and cannot be easily separated from one another.
• An attraction exists also between water molecules and the elements of the xylem wall (cohesion). These cohesive and adhesive attractions together maintain water column in xylem tracheids.

(B) Continuity of water column in the plant:
A strong objection against the theory has been the evidence for the occurrence of air bubbles in the water column of xylem elements. They disturb the continuity of the water column.

(C) Transpiration pull or tension on the unbroken water column :
As a result of transpiration, water is lost from mesophyll cells to the intercellular spaces. The water va¬pours move out of the plant through stomata. The DPD of mesophyll cell increases. With the increase in DPD These cells absorb water from adjoining cells, ultimately the water is absorbed from xylem elements of vascular bundles of leaf. Since, the xylem elements are filled with continuous water column. This tension or pull is transmitted down from petiole to stem and finally to roots leading to upward movement of water.

Question 5.
By giving significance of transpiration explain that “Transpiration is a necessary harm”. Give factors affecting transpiration.
Answer:
Significance of Transpiration :
It is one of the most controversial issues for the physiologists. Some consider it a very’ significant and beneficial process but many other consider it as entirely harmful and useless as plants. Spend energy to absorb water continuously but 90% of absorbed water is lost by transpiration process. The significant role of transpiration are as follow:

• Excess amount of water is removed from the plant tissues.
• Bring coolness for the foliage and thus help them in maintaining their vigour in sunlight.
• Causes transpiration pull, through which water is conducted to all parts of the plant.
• Helpful in the absorption of water and minerals.
• Helpful to maintain the concentration of cell sap.
• Saturate the atmosphere with water vapour and regulate the water cycle of the atmosphere.
• Certain plants liberate certain hygroscopic salts with transpired water, which accumulates on the leaf surface. These salts absorb water from the atmosphere and this saves the plant from wilting.

Thus, transpiration seems to be of much significance in the life of plants. However, excessive transpiration may cause permanent wilting and death of the plant. Therefore, “transpiration is a necessary harm” for the plants.

Factors affecting Transpiration :
Transpiration pull or Tension on the unbroken water column :
As a result of transpiration, water is lost from mesophyll cells to the intercellular spaces. The water vapours move out of the plant through stomata. The DPD of mesophyll cells increases.

With the increase in DPD, these cells absorb water from adjoining cells, ultimately the water is absorbed from Xylem elements of vascular bundles of leaf. Since the xylem elements are filled with continuous water column this tension of pull is transmitted down from petiole to stem and finally to roots leading to upward movement of water.

Thus, water ascends in the plants because of transpiration pull and the column of water remains continuous because of cohesive force of water molecules. This theory is the most accepted theory at the present time.
Factors affecting transpiration : Transpiration is influenced by a number of factors both external and internal. A brief account of them are given below :

[A] External factors :
1. Humidity :
If the plant is surrounded by a saturated atmosphere, there can be no escape of water vapour from the intercellular spaces of the leaf to the outside atmosphere. On the other hand, if the atmosphere is less humid, transpiration goes on at a rapid rate.

2. Temperature :
Rise in temperature reduces the relative humidity of the air as a result, the rate of transpiration increases.

3. Wind :
Wind has a direct influence on humidity. Dry air lowers the amount of air moisture and increases the rate of transpiration, whereas humid air decreases the rate of transpiration.

4. Light:
The rate of transpiration increases markedly in light and decreases in the dark. Light affects transpiration into two ways : Firstly, it increases transpiration by raising the temperature of leaves. Secondly, there is a close relationship between the opening of stomata and light as the stomata opens in day and gets closed in night.

5. Available soil water :
If available soil water is such that the rate of absorption of water is slowed down, the rate of transpiration is correspondingly decreased. Higher con-centration of salts in the soil water reduces the absorption rate and thereby the rate of transpiration.

6. Atmospheric pressure:
At lower atmospheric pressure at the altitudes, the rate of transpiration is increased but the decrease in rate occurs due to the prevailing low tempera-ture at these heights.

[B] Internal factors:
The histological and internal physiological conditions influence the rate of transpira-tion to a great extent. The following structures of the leaves influence the rate of transpiration :

• Presence of thick cuticle and deposition of wax on epidermis, decreases the rate of transpiration.
• Presence of hair on the outer surface of stomata also reduces the rate of transpiration.
• The rate of transpiration depends upon size, position and distribution of the leaves.
• The rate of transpiration is directly proportional to the water contents of the mesophyll tissue.

Question 6.
Explain the pathway of water absorption in plants with the help of diagram.
Answer:
In plants, water is absorbed from root hairs. These root hairs absorb the capillary water of the soil by the process of osmosis. Each root hair possessing a vacuole which is filled with cell sap. The cytoplasm of the root hairs functions like that of a semipermeable membrane. These root hairs are associated with the cortical cells. Cortical cells are elaborated up to endodermis. The absorbed water is passed into xylem through root hairs to cortical cells. Cortical cells to pericycle and pericycle to vascular tissues of the stem.

Question 7.
What is Transpiration? What are its types?
Or,
Write three types of transpiration process.
Answer:
Transpiration :
The process of loss of water in the form of water vapour from aerial parts of the plant is known as transpiration. There are three types of transpiration based on plant surface :

1. Stomatal or foliar transpiration :
The loss of water which takes place through specialized apertures which are present in leaves is called foliar transpiration. Most of the (80-90%) foliar transpiration takes place through stomata is called stomatal transpi-ration.

2. Cuticuiar transpiration :
Transpiration through cuticle is called the cuticular transpiration. Cuticular transpiration is only 3-10% of the total transpiration. It is continuous throughout day and night.

3. Lenticular transpiration :
Loss of water through the lenticels of fruits and woody stem is called the lenticular transpiration. The lenticular transpiration is only 0-1 % of the total transpiration. It however, continues day and night because lenticels have no mechanism of closure.

Question 8.
Describe Starch ⇌ Sugar interconversion hypothesis of opening and closing of stomata.
Answer:
Starch ⇌ Sugar interconversion hypothesis :
(i) This theory was proposed by Lloyd in 1908. According to him, the turgidity of guard cells is controlled by changes in OP, caused by interconversion of starch and sugar. He found that the amount of starch in guard cells increased by night and decreased by day.

During day time the guard cells of the stomata contain sugar synthesized by their chloroplasts by the use of light. The sugar is soluble and increases the concentration of the sap of guard cells. Due to higher concentra¬tion of the cytoplasm of guard cells, the water comes to them from the neighbouring cells by osmosis and they become turgid with the result the stomata remain open.

(ii) According to Sayre (1926), stomata open at pH 4-2 to 4-4 when starch contents is very low. The closure of stomata is always associated with an increase in starch contents. Sayre thought that utilization of C02 during photosynthesis during day time will cause the increase of pH resulting in the conversion of starch into sugar.

(iii) According to Hanes, enzyme phosphorylase catalyses the interconversion of starch and sugar.

(iv) According to this hypothesis, CO₂ liberated from respiration is used in photosynthesis by mesophyll cells in daytime. This leads to the lowering of acidity (H⁺ ion concen-tration) of the guard cells. As the H⁺ ion concentration (acidity) decreases the pH increases and the enzymatic interconversion of starch into sugar is favoured.

During night (darkness) CO₂ released from respiration, accumulates in the intercellular spaces increasing the H⁺ ion concentration (lowering the p^H) this favours the conversion of sugar into starch.

(v) According to Steward (1964), OP of guard cells is not affected unless glucose-1- phosphate is further converted into glucose and inorganic phosphate.
According to Steward :
(A) At higher p^H stomata opens in the following manner :

(B) At low p^H closing of the stomata will be take place in the following manner:

Transport in Plants Class 11 Important Questions Objective Type

1. Choose the correct answers:

Question 1.
Transpiration occurs through :
(a) Stomata
(b) Lenticel
(c) Cuticle
(d) All of these.
Answer:
(d) All of these.

Question 2.
Stomata opens during day time because the guard cells have :
(a) Outer membrane is thin
(b) Kidney shaped
(c) Chloroplast is present
(d) Large nucleus is found.
Answer:
(a) Outer membrane is thin

Question 3.
Rate of transpiration increases due to :
(a) Drought
(b) High temperature
(c) Moist soil
(d) Slow speed of wind.
Answer:
(c) Moist soil

Question 4.
Rate of transpiration is measured by :
(a) Potometer
(b) Porimeter
(c) Photometer
(d) None of these.
Answer:
(a) Potometer

Question 5.
When temperature rises, rate of transpiration :
(a) Increases
(b) Decreases
(c) Transpiration stops
(d) None of these.
Answer:
(a) Increases

Question 6.
Stomata closes due to lack of water. This condition is created due to :
(a) Formation of cytokinin
(b) Formation of Auxin
(c) Formation of ethylene
(d) Formation of Abscisic acid.
Answer:
(b) Formation of Auxin

Question 7.
Which of the following reason decreases rate of transpiration :
(a) Wind
(b) Increase in temperature
(c) Decrease in light intensity
(d) Absorption of water by plants.
Answer:
(c) Decrease in light intensity

Question 8.
Water reaches the upper part of the plants by :
(a) Root pressure
(b) Cell pressure
(c) Evaporation
(d) Diffusion.
Answer:
(c) Evaporation

Question 9.
Maximum water is absorbed by plants through :
(a) Root cap
(b) Root hair
(c) Root hair area
(d) Division area.
Answer:
(c) Root hair area

Question 10.
Plants only utilize this water present in the soil:
(a) Runaway water
(b) Gravitational water
(c) Vapourised water
(d) Capillary water.
Answer:
(d) Capillary water.

Question 11.
When the cell is completely turgid which of the following pressure becomes zero :
(a) Turgor pressure
(b) Cellular wall pressure
(c) Vapour pressure
(d) Osmotic pressure.
Answer:
(b) Cellular wall pressure

Question 12.
Stomata are open in day time in most of the plants, such stomata are called :
(a) Photoactive type
(b) Potato type
(c) Barley type
(d) Dark active type.
Answer:
(a) Photoactive type

Question 13.
Which type of stomata is found in the lower surface of leaves :
(a) Appie type
(b) Potato type
(c) Oat type
(d) None of these.
Answer:
(b) Potato type

Question 14.
In which type of plants Sunken stomata is found :
(a) Xerophytes
(b) Hydrophytes
(c) Mesophytes
(d) Sporophytes.
Answre:
(a) Xerophytes

Question 15.
What will happen if a plant cell is kept in concentrated salt solution :
(a) Plasmolysis
(b) Turgid
(c) Contraction
(d) Deplasmolysis.
Answer:
(a) Plasmolysis

Question 16.
Reason for Root pressure is :
(a) Passive absorption
(b) Active absorption
(c) Increase in transpiration
(d) Decrease in rate of photosynthesis.
Answer:
(b) Active absorption

Question 17.
Potometer is used for measurement of:
(a) Rate of absorption
(b) Rate of transpiration
(c) Rate of photosynthesis
(d) Phototropism.
Answer:
(b) Rate of transpiration

Question 18.
In guttation process loss of water occurs through :
(a) Stomata
(b) Hydathodes
(c) Wound
(d) Lenticels
Answer:
(b) Hydathodes

Question 19.
Which wall of the guard cells is thick :
(a) Outer
(b) Inner
(c) Lateral
(d) All of these.
Answer:
(b) Inner

Question 20.
When a cell is at equilibrium then :
(a) DPD = 0
(b) DPD = TP
(c) OP = TP
(d) DPD=OP.
Answer:
(c) OP = TP

Question 21.
What will happen if a plant cell is kept in pure water :
(a) Turgid
(b) Flaccid
(c) Plasmolysis
(d) Impermeable.
Answer:
(a) Turgid

Question 22.
Water potential of cell become positive in plant cell due to :
(a) Transpiration
(b) Low transpiration
(c) High absorption
(d) Guttation.
Answer:
(c) High absorption

Question 23.
Meaning of Osmosis is :
(a) Diffusion of dissolved particles from higher to lower concentration
(b) Diffusion of dissolved particles from lower to higher concentration
(c) Diffusion of water from higher to lower concentration
(d) Diffusion of water from lower to higher concentration.
Answer:
(c) Diffusion of water from higher to lower concentration

Question 24.
Guttation occurs due to :
(a) Transpiration
(b) Osmosis DPD
(c) Root pressure
(d) Osmotic pressure.
Answer:
(c) Root pressure

Question 25.
Transfer of water from one cell to other cell and direction of water flow depends on:
(a) WP
(b) TP
(c) DPD
(d) Primary plasmolysis.
Answer:
(c) DPD

2. Fill in the blanks:

1. The cavity and loose tissues present below the hydathodes is called ………………….
Answer:
Epithem

2. …………………. gave the name protoplasm.
Answer:
Purkinje

3. …………………. gave the K+ exchange process for opening and closing of stomata.
Answer:
Levitt

4. The guard cells for dicot stomata are ………………….
Answer:
Kidney shaped

5. Osmoscope is used to measure ………………..
Answer:
Osmosis

6. Parchment paper, is an example of …………………… membrane.
Answer:
Semipermeable

7. DPD = OP – …………………..
Answer:
TP

8. The cavity present below hydathodes is called ………………..
Answer:
Epithem

9. Loss of water from the margin of leaves in the form of water droplets is called………………….
Answer:
Guttation

10. In osmosis, diffusion of solvent takes place through ………………….
Answer:
Semipermeable membrane

11. The DPD of a fully turgid cell is always …………………..
Answer:
0

12. The DPD of a solution is ……………….. to its concentration.
Answer:
Proportional

13. ………………… is an instrument which is used to measure the rate of transpiration.
Answer:
(b) Potometer

14. The transport of water and mineral salts takes place through …………………
Answer:
Xylem

15. Absorption of water with use of energy is known as …………………..
Answer:
Active absorption

16. In plants, rate of transpiration is measured with the help of ……………….
Answer:
Potometer

17. Loss of water in the form droplets from the margin of leaves is called ……………….
Answer:
Guttation

3. Match the following:

(A)

Column ‘A’	Column ‘B’
1. Transpiration	(a) Semipermeable membrane
2. Guttation	(b) Stomata
3. Bleeding	(c) Hydathode
4. Hydathode	(d) Injured part
5. Osmosis	(e) Epithem.

Answer:
1. (b) Stomata
2. (c) Hydathode
3. (d) Injured part
4. (e) Epithem
5. (a) Semipermeable membrane

(B)

Column ‘A’	Column ‘B’
1. Pulsation theory	(a) Pnestley
2. Active water absorption	(b) Dixon and Jolly
3. Root pressure theory	(c) Godlewski
4. Transpiration pull theory	(d) Atkins and Priestley
5. Relay pump theory	(e) J. C. Bose.

Answer:
1. (e) J. C. Bose,
2. (d) Atkins and Priestley
3. (a) Pnestley
4. (b) Dixon and Jolly
5. (c) Godlewski

4. Answer in one word:

1. Which part of the root take part in water absorption?
Answer:
Root hairs

2. What type of water is absorbed by roots?
Answer:
Capillary water

3. From which tissue of the plant body, translocation of water takes place?
Answer:
Xylem

4. Name the pressure responsible for stomatal opening.
Answer:
Turgor Pressure

5. “Transpiration is a necessary evil”, who stated it?
Answer:
Kartis

6. The pressure exerted in plant cells due to the absorption of water.
Answer:
Osmotic pressure

7. What will be the DPD of a fully turgid cells?
Answer:
Zero

8. Name the water absorbing part of the roots.
Answer:
Root hair

9. Name the process of loss of water from the aerial part of the plant body in the form of water vapour.
Answer:
Transpiration

10. Name the enzyme which converts starch into glucose, 1-phosphate.
Answer:
Phosphorylase

Students get through the MP Board Class 11th Biology Important Questions Chapter 12 Mineral Nutrition which are most likely to be asked in the exam.

MP Board Class 11th Biology Important Chapter 12 Mineral Nutrition

Mineral Nutrition Class 11 Important Questions Very Short Answer Type

Question 1.
Write the name of symbiotic bacteria present in the roots of leguminous plant.
Answer:
Rhizobium bacteria are present in the roots of leguminous plant which absorb atmospheric nitrogen.

Question 2.
Write the name of two micronutrients.
Answer:
The two micronutrients present in plants are Manganese (Mn) and Molybdenum (Mo).

Question 3.
Write the deficiency symptoms of potassium.
Answer:
Deficiency of potassium inhibits protein synthesis. Scorching of older leaves is the most important symptoms of trees. The leaf tips and margins show general chlorosis. Shortening of internodes also occurs.

Question 4.
Why nitrogen is essential for growth of plant?
Answer:
Protein is essential for growth of plant and for development of protein nitrogen is essential, so for the proper development of plant nitrogen is very much essential.

Question 5.
What are the drawbacks of soilless culture?
Answer:
Drawbacks :

• It will be very expensive.
• Limited production will be obtained.
• Crops will never be grown by this method.

Question 6.
What are micronutrients? Give two examples.
Answer:
The mineral elements which are required in very low concentration for the plant growth are known as micronutrient or microelements.
e.g., Zn, Mo, Cl, Mn, B, Cu, etc.

Question 7.
Describe the deficiency symptoms of nitrogen in plants.
Answer:
Deficiency symptoms of nitrogen:

• Yellowing i.e., Chlorosis of leaves, older leaves become yellow first.
• Acceleration in the rate of formation of purple pigment anthocyanin in the leaves which become purple in colour.
• Flowering is delayed or is completely suppressed.

Question 8.
Which pigment is present in the roots of leguminous plants?
Answer:
Leghaemoglobin.

Question 9.
Write the importance of molybdenum in plants.
Answer:
Molybdenum act as an activator for the enzyme nitrate reductase involved in nitrogen metabolism.

Question 10.
In plants, the organic solutes are transported through which tissue?
Answer:
In plants, the organic solutes are transported through phloem tissue.

Question 11.
What is Mineral nutrition?
Answer:
The utilization of various kinds of absorbed ions by a plant for its growth and development is called as Mineral nutrition.

Question 12.
What are nutrient elements?
Answer:
Elements which are essential for plant growth are called as nutrient elements.

Question 13.
In study of water culture purity of water and mineral nutrient is essential. Why? (NCERT)
Answer:
In study of water culture purity of water and pure mineral nutrients are essential because plants can absorb only pure water and pure minerals and their essentiality for the plants can be determined correctly. Impure mineral nutrient cannot help to determine essentiality of the minerals for plants.

Question 14.
Explain following With examples: Macronutrients, micronutrients, useful elements, essential elements, toxic elements.
Answer:
1. Macronutrients: Essential elements which are required in large quantity for normal growth and development of plants are called as macronutrients.
e.g. C, H, O, N, P, K, Ca, S, Mg, Fe, etc.

2. Micronutrients: The mineral elements which are required in very low concentration for the plant growth are known as micronutrient or microelements.
e.g., Zn, Mo, Cl, Mn, B, Cu, etc.

3. Useful elements: Some essential elements which are essential for higher plants called as useful elements.
e.g., Na, Si, Ca and Se.

4. Essential elements: Out of approximately 30 elements universally present in the plants, 16 elements are essential for plants, i.e., they require them.
e.g. C, H, O, N, P, K, Ca, S, Mg, Fe, Mn, Zn, B, Cu, Cl, Al.

5. Toxic element: Concentration of any mineral ion which decreases dry weight of plant by 10% is called as Toxic element. It is different for different plants.
e.g. As, Cu, Hg etc.

Question 15.
Give four general importance of mineral salt.
Answer:
Importance of Mineral salt:

1. Nutrition of plants: Some minerals like N, P, S, etc. help to produce protein.
2. As catalyst: Some minerals act as catalyst in biochemical reactions. e.g. Mn.
3. Bodybuilding: Some elements help to produce structural part, such as cell wall of the cell.
4. Balancing work: Some mineral elements stop toxic effect of some other elements. e.g., Calcium, Magnesium.

Question 16.
Write difference between Active and Passive absorption.
Answer:
Differences between Active and Passive absorptions

Active absorption	Passive absorption
1. It requires metabolic energy.	It does not require metabolic energy.
2. It cannot take place in the absence of root.	It can take place even in the absence of root.
3. It is not influenced by transpiration.	It depends upon transpiration.
4. Root cells play an active role, in the absorption of water.	Root cells have no role in the absorption of water.

Mineral Nutrition Class 11 Important Questions Short Answer Type 

Question 1.
Mention the name of any five macroelements and their deficiency symptoms. (NCERT)
Answer:
Macronutrients :

1. Nitrogen
2. Phosphorus,
3. Sulphur,
4. Magnesium and
5. Calcium.

1.Deficiency symptoms of nitrogen:

• Stunted growth with yellowish-green leaves in young plants.
• Yellowing and drying of older leaves.
• Flowering reduced.

2. Deficiency symptoms of phosphorus :

• Stunted growth.
• The leaves become purple coloured due to excessive anthocyanin.
• Premature leaf falls and necrotic spots on the fruits.
• Vascular tissues less developed.
• Low rate of protein synthesis.

3. Deficiency symptoms of sulphur :

• Plants stunted in growth and flowering delayed.
• Chloroplast becomes pale green in colour.
• Chlorosis of leaves.
• Hard woody stem.

4. Deficiency symptoms of magnesium :

• Interveinal chlorosis or molted chlorosis with green veins.
• Brittleness of the leaves and necrosis.
• Wilting, drying off and dropping of the tips of plant.
• Necrotic patches on the leaves.

5. Deficiency symptoms of calcium :

• Calcium deficiency causes malformation of the younger leaves.
• Chlorosis also occurs along the margins of younger leaves.

Question 2.
Describe essentiality of potassium in plants and give its deficiency symptoms.
Answer:
Potassium: It is essential for activating enzymes concerned in the synthesis of polypeptides from amino acids and is also essential for the process of photosynthesis, and respiration.

Deficiency symptoms: Deficiency of potassium inhibits protein synthesis. Scorch-ing of older leaves is the most important symptom on trees. The leaf tips and margins show general chlorosis. Shortening of internodes also occurs.

Question 3.
All elements are not essential for plants. Explain it. (NCERT)
Answer:
All elements absorbed by the plants are not essential.
Criteria for Essentiality: According to Arnon and Stout, 1939, the criteria for essentiality of mineral elements are as follows :

• The plant is unable to grow normally and complete its life cycle, in the absence of the elements.
• The element is specific and cannot be replaced by another element, however close it may be in the periodic table.
• The element plays a direct role in the metabolism of the plant.

Question 4.
Give atleast essential elements. Describe them and give their deficiency symptoms. (NCERT)
Answer:
1. Sulphur: It is absorbed in the form of sulphate ions from the soil. It is important constituent of some amino acids. It is also essential for the synthesis of sulphur con¬taining vitamins like biotin, thiamine and coenzyme A. Disulphide bridge (S-S) plays an important role in determining protein structure and sulphydryl groups (SH) are necessary for the activity of many enzymes.

Deficiency symptoms :

• Reduced leaf, stunted growth, a general chlorosis, followed by the production of anthocyanin pigments in some species. The younger leaves are affected first.
• Inward rolling of leaf margins and tips.
• Due to development of sclerenchymatous tissue, stem becomes hard.

2. Phosphorus: It is an important constituent of phospholipids, nucleic acids and various coenzymes like NAD, NADP and ATP. It is absorbed from soil in the form of phosphate. Phospholipids are the constituents of membranes. NAD, NADP and ATP are required for oxidation-reduction reactions of photosynthesis, respiration and fat metabo-lism. ATP is the chief source of energy.

Deficiency symptoms :

• Premature leaf fall.
• Formation of necrotic areas on leaves and fruits.
• Leaves become dark blue in colour.
• The growth of roots and shoots is completely restricted.

3. Calcium: It is a constituent of the middle lamella, in which it is present in the form of calcium pectate. It helps to stabilize the structure of the chromosomes.
Deficiency symptoms: Calcium deficiency causes malformation of the younger leaves. Chlorosis also occurs along the margins of the younger leaves.

4. Magnesium: It is one and only constituent of the chlorophyll molecule. It acts as activator for many enzymes in phosphate transfer reactions particularly in carbohydrate metabolism and nucleic acid synthesis. It is also believed to be an important binding agent in ribosomes, where protein synthesis takes place.
Deficiency symptoms: Magnesium deficiency results in extensive interveinal chlorosis of leaves. The older leaves are affected first. Ultimately, leaves develop necrotic spots.

5. Iron: It is essential for chlorophyll synthesis. It is important constituent of ferredoxin, flavoprotein, iron-porphyrin, catalase, peroxidase and cytochromes. Iron is primarily concerned in the formation of chloroplast protein in the leaves.
Deficiency symptoms: Pronounced interveinal chlorosis occurs due to deficiency of iron.

Question 5.
Which statement is right out of following if it is wrong, give reason : (NCERT)
(a) Deficiency of boron causes formation of sclerenchymatous axis.
(b) AH elements present in the cell are essential.
(c) Nitrogen is not an essential element for plants.
(d) It is very easy to determine essentiality of micronutrients in plants because it is taken in very less quantity.
Answer:
(a) Wrong, because deficiency of Boron effect on activity of membrane, pollen germination, cell elongation, cell differentiation and transport of carbohydrates, (b), (c) and (d) are right statements.

Question 6.
If plant shows deficiency symptoms of two elements how will you determine deficient element practically? (NCERT)
Answer:
Deficiency symptoms of different elements are different. By providing particu¬lar deficient element if deficiency symptoms eradicates then deficient element can be confirmed.

Question 7.
Why deficiency symptoms are observed in newly formed parts of some plants whereas in some other plants it is observed in mature part? (NCERT)
Answer:
Deficiency symptoms in plant parts depends on mobility of elements. Deficiency symptoms are observed in old tissue first as compared to new tissues of young part. For example, symptoms of deficiency of nitrogen, potassium and magnesium are observed in old leaves first of all.

Biomolecules of old leaves decomposes and element move towards the new leaves.
When elements are non-mobile then they cannot reach to young parts thus deficiency symptoms are observed in new leaves.

Question 8.
Describe passive absorption of minerals in plants. (NCERT)
Answer:
Passive Absorption: In this method, the movement of minerals into the cells or tissue takes place without the expenditure of energy on the part of absorbing cells. This is a purely physical phenomenon and takes place in many ways.

1. Diffusion: It can be observed by first placing a plant cell or tissue in a medium of high concentration of salt and then transferring it into lower salt concentration. During this transfer, some of the ions taken up will diffuse into the lower salt concentration from the higher salt concentration, until the concentration of both the solutions get balanced. The whole process is unaffected by temperature or metabolic inhibitors.

2. Mass flow: Plants which have a higher rate of transpiration contain comparatively high concentrations of minerals. It is well-known that increase in transpiration is accompanied by increased absorption of ions from the soil. In this way the ions are absorbed by the plants in a mass flow due to the higher rates of transpiration.

3. Donnan equilibrium: According to Donnan, cells contain some fixed ions which do not exert any diffusion pressure. Therefore, ions similar in charge to that of fixed ions, move inwardly through diffusion. Oppositely charged ions also pass inwardly to balance them. Such a diffusion continues till the multiple of cations and anions inside cells becomes equal to that of ions present in outside medium.

Question 9.
Describe process of formation of Root nodules. (NCERT)
Answer:
Formation of Root nodule: It is a combined activity of roots of nutrient plants (Leguminous plants) and Rhizobium bacterias.

Steps of formation of Root nodules are as follows :
First of all Rhizobium bacteria multiply and are collected surrounding the roots of leguminous plants and get connected to root hair cells.

Root hair turns and bacteria attack them. An infected thread appears which carry Rhizobium bacteria to the cortex, where they begin to produce nodule. Now bacterias are released from the thread and enters into the cells which help for differentiation of nitrogen fixation cells. In this way, root nodules are formed. Leguminous plants provide nutrients to bacteria and Rhizobium bacteria provides nitrogenous compounds to plant by nitrogen fixation.

Mineral Nutrition Class 11 Important Questions Long Answer Type 

Question 1.
What are the materials required for atmospheric nitrogen fixation? Give their role in Nitrogen fixation. (NCERT)
Answer:
Rhizobium bacteria shows symbiotic relationship with the roots of leguminous plants like pea, moong, gram, beans etc. Symbiosis occurs in the root nodules. Rhizobium bacteria obtains food from the leguminous plants and Rhizobium bacteria help for nitrogen fixation and provides nitrogenous compounds to the plants.
Symbiotic Nitrogen fixation: The process of nitrogen fixation is done by nitrogen-fixing bacterias like Rhizobium, Cyanobacteria, nitrogen-fixing fungi etc.
Mechanism of Nitrogen fixation: Following enzymes and materials are required for nitrogen fixation:

• Nitrogenase and Hydrogenase enzymes.
• Leghaemoglobin: It protects the enzyme from oxygen.
• One non-haem iron protein: e.g. Ferredoxin: It is an electron carrier.
• Hydrogen donor substances like: NADPH₂, FMNH₂, Pyruvate, Sucrose, Glucose etc.
• Energy compound ATP.
• Cofactor like TPP (Thiamine pyrophosphate), CoA, inorganic phosphate, Mg++ etc.
• Co and Mo.
• A compound to fix ammonia produced by reduction of nitrogen.
  Following are the steps of natural nitrogen fixation:
  1. Reduction of molecular nitrogen:
  
  

Question 2.
What is Hydroponics? Give its uses.
Answer:
Hydroponics (Soilless culture): Large- scale cultivation of plants in water culture or soilless culture is known as hydroponics. It was discovered by Prof. W. F. Gariak in 1929. The plants are grown in large, shallow pots which are full of nutrient solutions. The tubs or tanks are covered with wire netting to provide support to the seedlings. The solution is aerated at regular intervals by means of an inlet tube.
There are certain advantages and disadvantages of this technique.

Uses:

• Possibility to provide desirable nutrient environment.
• Regulation of pH of the nutrient solution conductive to specific crop.
• It is possible to replace the nutrient, so more healthy plant as compared to soil culture can be obtained.
• On the roof of multistoried buildings of metropolitan cities, vegetables and fruits may be produced in large scale.
• Ripening time of fruit is less in soilless culture.

Question 3.
Give an experiment to demonstrate the essentiality of minerals required by plants.
Answer:
Essentiality of minerals required by plants can be demonstrated by using a normal culture medium such as Sach’s culture solution. Components of this solution are as follows:
Sach’s culture solution :

This experiment is performed to ascertain the role of individual elements in the normal growth and development of plants and the effect produced by their deficiencies. In water culture (solution culture) experimental plants are grown in a medium, the chemical composition of which is known and kept under control.

Method: Take seven wide-mouthed bottles of neutral glass of the same size. Clean and thoroughly sterilize them with nitric acid and distilled water. Fit each bottle with a cork containing two holes; one in the centre for introducing a seedling and the other at the side for a bent glass tube for aerating the culture solution. Make a slit running from the central hole to the rim of the cork for readily taking out the seedling.

Now, prepare a normal culture solution and pour it into the first bottle. Then, prepare six other solutions from which omit by turn Mg, Ca, Fe, K, P and N, and designate them as Mg, Ca, Fe, K, P and N, and fill the remaining seven jars with these solutions. Take seed¬lings of same kind and more or less same size. Introduce a seedling in each bottle through split cork. Wrap the bottles with black paper and expose them to light. Make arrangement

for proper aeration of roots. Renew the culture solution fortnightly. Observe carefully the symptoms of plants appeared due to the deficiency of various elements.

Question 4.
Name any five micronutrients and give their specific role and deficiency symptoms.
Answer:
Nutrient elements which are required in traces for normal growth and development of the plants are called as Micronutrients.
e.g. Manganese, Boron, Copper, Zinc, Molybdenum and Chlorine. Their specific role and deficiency symptoms are as follows :

Mineral Nutrition Class 11 Important Questions Objective Type

1. Choose the correct answers:

Question 1.
In Nitrogen cycle nitrifying bacteria convert:
(a) NH₃ into N₂
(b) Nitrogen fixation
(c) NH₃ into NO₂
(d) Amino acid into NHr
Answer:
(c) NH₃ into NO₂

Question 2.
Function of Zinc is :
(a) Production of Chlorophyll
(b) Production of IAA
(c) Closing of stomata
(d) Oxidation of Carbohydrates.
Answer:
(b) Production of IAA

Question 3.
Symptoms of destruction of Chlorophyll is called as :
(a) Necrosis
(b) Chlorosis
(c) Stunted shoot
(d) Rotting.
Answer:
(c) Stunted shoot

Question 4.
Plants which can prepare their own food are called as :
(a) Heterotrophs
(b) Autotrophs
(c) Mixotrophs
(d) None of these.
Answer:
(b) Autotrophs

Question 5.
Falling of premature leaves, flowers and fruits is called as :
(a) Abscission
(b) Dieback
(c) Stunted
(d) Hypertrophy.
Answer:
(a) Abscission

Question 6.
In which form ‘C’ is available to plants :
(a) ‘C’ element
(b) CO₂
(c) CO₃
(d) Amino acid.
Answer:
(b) CO₂

Question 7.
Which element is required by the plants in large quantity :
(a) N
(b) P
(c) Ca
(d) S.
Answer:
(a) N

Question 8.
Which element out of following is not macroelement:
(a) P
(b) K
(c) Mg
(d) Mo.
Answer:
(d) Mo.

Question 9.
Hydroponics name was given by :
(a) Knop and Sach
(b) Hoagland
(c) Gariak
(d) Sach.
Answer:
(c) Gariak

Question 10.
Soil element available to plant is called as :
(a) Mineral salt
(b) Micro salt
(c) Nutrient substance
(d) None of these.
Answer:
(a) Mineral salt

Question 11.
How much percentage of dry weight of plants is formed by Carbon, Hydrogen and Oxygen
(a) 10-15%
(b) 15-25%
(c) 25-35%
(d) 85-95%.
Answer:
(d) 85-95%.

Question 12.
Element required for synthesis of IAA :
(a) B
(b) Cu
(c) Zn
(d) Mo.
Answer:
(c) Zn

Question 13.
Active absorption of minerals in plants depends on :
(a) Availability of O₂
(b) Light
(c) Temperature
(d) Availability of CO₂.
Answer:
(a) Availability of O₂

Question 14.
Element which is not essential for the plants out of the following is :
(a) Iron
(b) Magnesium
(c) Lead
(d) Phosphorus.
Answer:
(c) Lead

Question 15.
Trace element out of following is :
(a) Ca
(b) Mg
(c) C
(d) Cu.
Answer:
(d) Cu.

2. Fill in the blanks:

1. Monotropa is an example of ………………………………… plant.
Answer:
Saprophyte

2. Orobanche is an example of ………………………………… plant.
Answer:
Total root parasite

3. Zn, Cu, Mn, B, Mo and Cl are ………………………………… elements.
Answer:
Micro

4. Extensive interveinal Chlorosis is caused due to deficiency of ………………………………… .
Answer:
Mg

5. Element act as activator for the enzyme nitrate reductase is ………………………………… .
Answer:
Mo

6. Rhizobium is a ………………………………… nitrogen fixing bacteria.
Answer:
Symbiotic

7.
Answer:
2HNO₂

8. ……………………………….. element participates in various metabolic activities of plant.
Answer:
Essential

9. Tetra-pyrrol porphyrin ring of chlorophyll has ………………………………… element at the centre.
Answer:
Mg

10. Lack of ………………………………… causes the absence of glands in the roots of leguminous plants.
Answer:
Boron

11. Rhizobium is found in the glands of the leguminous plants situated at ………………………………… .
Answer:
Roots

12. The element which are required for the growth of plants are called ………………………………… element.
Answer:
Essential

13.‘Whiptail disease’ in cabbage is due to deficiency of ………………………………… .
Answer:
Molybdenum.

3. Match the following:

(A)

Column ‘A’	Column ‘B’
1. Drosera	(a) Minor trace element
2. Dry weight	(b) Hydroponics
3. Zn, copper	(c) Critical element
4. Tank cultivation	(d) Plant ash
5. NPK	(e) Insectivorous.

Answer:
1. (e) Insectivorous.
2. (d) Plant ash
3. (a) Minor trace element
4. (b) Hydroponics
5. (c) Critical element

(B)

Column ‘A’	Column ‘B’
1. K	(a) Nitrogen metabolism
2. Mg	(b) Photolysis of water
3. Mo	(c) Amino acid
4. Mn	(d) Stomatal movement
5. S	(e) Chlorophyll.

Answer:
1. (d) Stomatal movement
2.  (e) Chlorophyll.
3.  (a) Nitrogen metabolism
4. (b) Photolysis of water
5.  (c) Amino acid.

(C)

Column ‘A’	Column ‘B’
1. Cuscuta	(a) Symbiotic
2. Loranthus	(b) Total root parasite
3. Orobanche	(c) Insectivorous
4. Dionaea	(d) Total stem parasite
5. Lichen	(e) Partial stem parasite.

Answer:
1.  (d) Total stem parasite
2. (e) Partial stem parasite.
3. (b) Total root parasite
4. (c) Insectivorous
5. (a) Symbiotic

4. Answer in one word:

1. Write two mineral element essential for production of chloroplast.
Answer:
Mg and Fe

2. Name the pigment found in the root nodule of leguminous plant.
Answer:
Leghaemoglobin

3. Full form of NPK is.
Answer:
Nitrogen, Phosphorus, Potassium

4. Name the enzyme induce Nitrogen fixation.
Answer:
Nitrogenase

5. Name one free-floating aquatic insectivorous plant.
Answer:
Utricularia

6. Name the process in which roots of the plants are kept in nutrient solution for plant growth.
Answer:
Hydroponics

7. Deficiency of which element causes abscission of premature leaves.
Answer:
Phosphorus

8. Name the root of the Cuscuta plant which penetrate into the body of host plant.
Answer:
Haustorial root

9. Name the macroelement which forms structural component of plant body.
Answer:
Carbon

10. From where plants obtain hydrogen.
Answer:
Water

Students get through the MP Board Class 11th Biology Important Questions Chapter 10 Cell Cycle and Cell Division which are most likely to be asked in the exam.

MP Board Class 11th Biology Important Questions Chapter 10 Cell Cycle and Cell Division

Cell Cycle and Cell Division Class 11 Important Questions Very Short Answer Type

Question 1.
What is the average cell cycle duration of cells of mammalia?
Answer:
Average cell cycle duration of cells of mammals is 24-25 hours.

Question 2.
What is the difference between cytokinesis and karyokinesis?
Answer:
Division of the cytoplasm is called as cytokinesis, whereas division of the nucleus is called as karyokinesis.

Question 3.
What is G₀ phase of cell cycle?
Answer:
Cells which do not divide further and reaches to inactive phase, is called as G₀ phase of cell cycle. At this stage cell shows high rate of (active) metabolism but does not divide. They divide depending upon requirement of the organism.

Question 4.
Why Mitosis cell division is called as similar division?
Answer:
At the end of this cell division daughter cells have same number of chromosomes like parent cell thus Mitosis division is called as similar cell division.

Question 5.
Where daughter cells formed by meiosis cell division are same and different?
Answer:
Four daughter cells formed by meiosis division in the testis are the spermatids which are same in size whereas in the ovary of female four daughter cells formed by meiosis are different in size, i.e. one ovum which is larger in size and three polar bodies, which are smaller in size.

Question 6.
Is Mitosis division possible without DNA replication in ‘S’ phase of cell cycle?
Answer:
No, Mitosis division is not possible without DNA replication in ‘S’ phase of cell cycle, because duplication of DNA is essential for Mitosis division.

Question 7.
Name the stage of cell division which involves crossing-over.
Answer:
Crossing-over occurs during pachytene stage of Prophase-I of meiosis cell division.

Question 8.
Name the stain used for staining chromosomes of the cells for study.
Answer:
Acetocarmine stain is used for staining chromosomes of the cell for study.

Question 9.
Name the cells exhibiting mitosis cell division.
Answer:
Mitosis is the characteristic feature of somatic cells.

Question 10.
Name the cells exhibiting meiosis cell division.
Answer:
Meiosis takes place in generative cells to produce gametes.

Question11.
What is chiasmata?
Answer:
The place of chromosomes where they come into contact with each other during crossing-over is called as chiasmata.

Question 12.
List the various phases of the first meiotic division.
Answer:
The following stages are found in the first meiotic division :

1. 1. Prophase-I: It has following substages :
   • Leptotene,
   • Zygotene,
   • Pachytene,
   • Diplotene and
   • Diakinesis.
2. Metaphase-I,
3. Anaphase-I,
4. Telophase-I.

Question 13.
What is synapsis?
Answer:
Pairing of homologous chromosomes in zygotene stage of Prophase-I of meiosis cell division is known as synapsis. The paired structure formed by synapsis is called as bivalent.

Question 14.
What do you mean by Mitotic poison?
Answer:
Chemical substances which inhibit the mitotic division are called as Mitotic poison.
e.g.,

1. Cholchicine : It stops formation of spindles during Metaphase.
2. Ribonuclease : It stops Prophase stage.
3. Mustard gas : It breaks chromosomes.

Question 15.
Name the parts of a flowering plant where meiosis cell division occurs.
Answer:
In the pollen sac of androecium and ovary of gynoecium meiosis division occurs to produce male and female gametes respectively.

Cell Cycle and Cell Division Class 11 Important Questions Short Answer Type

Question 1.
Describe the phenomenon occurs during interphase of cell cycle.
Answer:
Interphase is the period between two successive cell divisions. It has following three stages:

• G_1-phase : RNA and protein synthesis occurs during this stage.
• S-phase : DNA synthesis occurs during this stage.
• G₂-phase : Synthesis of protein and RNA continue in this stage.

Question 2.
What is the significance of Meiosis division?
Answer:
Significance of meiosis : The significance of meiosis are as follows :

• The number of chromosomes is stable generation to generation due to the meiotic division.
• The number of chromosomes is reduced to half in the gametes.
• The characters of parents are transmitted into offspring by this process.
• Crossing-over takes place during this stage which can produce new characters in the organisms.
• Variations are also occurred which help in the study of evolution.
• There are four haploid cells formed from one diploid cell.

Question 3.
Name the stage of cell cycle when following phenomenon occurs :
(i) Movement of chromosomes towards the equator of the cell.
(ii) Breaking of centromere and separation of chromosomal half.
(iii) Pairing of homologous chromosomes.
(iv) Exchange between homologous chromosomes.
Answer:
(i) Metaphase,
(ii) Anaphase,
(iii) Zygotene stage of Prophase-I of Meiosis-I,
(iv) Pachytene stage of Prophase-I of Meiosis-I.

Question 4.
Explain the following :
(i) Synapsis,
(ii) Bivalent,
(iii) Chiasmata.
Answer:
(i) Synapsis :
Pairing of homologous chromosomes in zygotene substage of prophase-I of meiosis cell division is known as synapsis. The paired structure formed as result of synapsis is called as bivalent.

(ii) Bivalent:
The pairs of homologous chromosome formed during zygotene stage of meiosis-I. The chromosomes contract and become short and thick.

(iii) Chiasmata :
In Pachytene stage of Prophase-I of Meiosis-I division, non-sister chromatids of two chromosomes, which come in contact of each other and fixes themselves tightly for crossing-over. These fixed parts are called as Chiasmata.

Question 5.
Write differences between cytokinesis in plant and animal cell.
Answer:
Cytokinesis : Division of cytoplasm after Karyokinesis to produce daughter cells is called as Cytokinesis.

(a) Cytokinesis in plant cell:
It immediately follows karyokinesis. In plant cells, a cell plate of dense cistemae is laid down at the equatorial plate. It grows centrifugally (external laterally) towards plasma membrane until it divides the cell into two parts. The space in between the two membranes of cistemae is filled by the deposition of pectates of calcium and magnesium to become middle lamella.

(b) Cytokinesis in animal cell:
In animal cell, cytokinesis occurs by an invagination of cell membrane almost in the middle of the cell. The furrow gradually depens and ultimately divides the cell into two parts.

Question 6.
Write differences between Anaphase of Mitosis and Meiosis cell division.
Answer:
Differences between Anaphase of Mitosis and Meiosis cell division

Anaphase of Mitosis	Anaphase of Meiosis
One chromatid of one chromosome moves towards one pole and another chromatid of it moves towards the other pole.	In Anaphase-I complete chromosomes move towards their poles and in Anaphase-II movement occurs like Anaphase of Mitosis.

Question 7.
Analyse each phase of cell division and explain how does following changes occurs in the cell:
(i) Number of chromosomes (N) in each cell.
(ii) Quantity of DNA in each cell.
Answer:
(i) In all cells during prophase, metaphase and anaphase stage of cell division, number of chromosomes become double and in telophase stage number of chromosomes become half again, i.e., (N).
(ii) During cell division quantity of DNA changes in all stages. During prophase, metaphase and anaphase number of DNA increases, it becomes double, but in telophase stage quantity of DNA becomes half.

Question 8.
Discuss following with your teacher :
(i) When cell division occurs in haploid seed and lower category of plants ?
(ii) Where cell division does not occur in the haploid cells of higher category of plant?
Answer:
(i) Haploid spores of lower category of plant divides by mitosis division to form gametophyte, whereas zygote develops to form sporophyte and meiosis division occurs for formation of haploid spores, e.g., moss, fern etc.
(ii) Synergids and antipodal cells found in the ovules of angiosperms are haploid. They do not show cell division. After completion of life span they dies.

Question 9.
What is crossing-over? Throw light on its significance.
Answer:
Crossing-over : Crossing-over is the exchange of chromosomal segments or genes between homologous chromosomes during diplotene stage of meiosis-I.

Significance of crossing-over :

• New characters are originated by this process,
• Variations occur in organisms,
• It plays an important role in organic evolution,
• It results in the production of adaptation in organisms.

Question 10.
Write the significance of mitosis.
Answer:
Significance of mitosis:

• Growth and development takes place due to mitosis,
• Some micro-organisms use this method for asexual reproduction,
• Dead cells are substituted by new cells by mitotic division,
• General repairing process is result of mitotic division,
• Genetic information are transferred from parents to offspring by this process.

Question 11.
What is the difference between anaphase-1 of meiosis and anaphase of mitosis ? What is its effect on the whole process?
Answer:
In meiotic anaphase-I, the two chromosomes of each bivalent separate from each other and move towards opposite poles of the spindle. But the two chromatids of each chromosome still continue to remain joined with one another with centromere. Thus, only one chromosome of each homologous chromosome pair reaches each pole and the number of chromosomes are reduced to half in daughter nuclei.

In mitotic anaphase-I, centromere of each chromosomes divides and permits the separation of two sister chromatids, each chromatid after separation is called daughter chromosomes. The daughter chromosomes move apart and migrate towards opposite poles. The centromere is pulled towards poles of the spindle and the arms of chromosomes are dragged behind.

Effect of whole process:
Anaphase-I of meiosis resulting in the formation of daughter nuclei having half of the chromosomes to their parents. Thus, the number of chromosomes is constant from generation to generation.

Question 12.
Write a note on necessity of two types of cell divisions in multicellular organisms.
Answer:
Necessity of two types of cell divisions :
There is complexity in structure and physiology of multicellular organisms. Thus, twc types of cell divisions are required in them. The process of meiosis cell division help in the maintenance of chromosome number in the species. Diploid (2x) organism produce haploid (x) gametes, which on fusion form diploid zygote, whereas zygote divides by mitosis division to form diploid organism. Healing of wound in multicellular organisms occurs by mitosis cell division.

Question 13.
Describe the prophase and anaphase of mitosis along with diagram.
Answer:
1. Prophase : (Pro = before, phasis – stage).

• In this stage chromosomes are found in the form of long threads which are called as chromatin net.

• Chromosomes become short and thick and each of them splits lengthwise forming two chromatids.
• At the ending of prophase nuclear membrane and nucleolus start to disappear.

2. Anaphase :

• During this stage the centromere of each chromosome gets divided into two which make the chromatids free from each other.
• Every chromatid with its centromere moves towards its poles.
• Termination of anaphase movement occurs when the chromosomes form a densely packed group at the two poles.

Question 14.
Draw labelled diagram of different stages of Mitosis cell division.
Answer:

Question 15.
Describe the metaphase of mitosis along with diagram.
Answer:

Metaphase : The following events occur during mitotic metaphase: ‘

• Nuclear membrane and nucleolus disappear during this stage.
• Chromosomes move towards equatorial plane of the cell.
• Chromatids of chromosomes become dense and separated from each other except centromere.
• Spindle fibres are formed.
• The centromeres of the chromosomes are arranged on equator and chromatid towards the opposite poles. Fig Metaphase

Question 16.
Write differences between following :
(i) Mitosis and Meiosis cell division,
(ii) Chromatin and Chromatid,
(iii) Centromere and centriole,
(iv) Centromere and Chromomere,
(v) Metaphase-I and Metaphase-II,
(vi) Zygotene and Pachytene,
(vii) Cell furrow and Cell plate.
Answer:
(i) Differences between Mitosis and Meiosis cell division :

(ii) Differences between Chromatin and Chromatid

Chromatin	Chromatid
1. It is the diffused deep staining hereditary material.	It is longitudinally split half of a chromosome. It is light staining hereditary material.
2. It is metabolically inert and its crossing-over frequency is less.	It is very active metabolically and its crossing over frequency is more.

(iii) Differences between Centromere and Centriole

Centromere	Centriole
1. It is the region of attachment of the sister chromatids and also the site of attachment to the spindle fibres.	It is the minute self replicating body lying near the nucleus in the animal cell. It radiates astral rays and spindle fibres during cell division.

(iv) Differences between Centromere and Chromomere

Centromere	Chromomere
1. It is the region of attachment of the sister chromatids and also the site of attachment to spindle fibres.	The beaded and knot like structure present on chromonemata are called as chromomeres.

(v) Differences between Metaphase-I and Metaphase-II

Metaphase-I	Metaphase-II
1. Pair of homologous chromosomes are arranged at the equator of the cell.	Single chromosomes are arranged at the middle part of the cell.
2.Centromere do not divide.	Centromere divides to form chromatids.

(vi) Differences between Zygotene and Pachytene

Zygotenw	Pachytene
1. Pairing of homologous chromosomes occurs (synapsis) to form Bivalent	Chromosomes split lengthwise but still remain attached at the point of centromere to form tetrad followed by crossing-over.

(vii) Differences between Cell furrow and Cell plate

Cell furrow	Cell plate
1. It is a constriction formed at the middle part of a cell, which grows to divide parent cell into two daughter cells.	It is a cell wall formed at the equator formed by accumulation of vesicles of dictyosomes and elements of endoplasmic reticulum to form two daughter cells.
2. Cell furrow formation occurs in animal cells.	Cell plate formation occurs in plant cell.

Question 17.
Define the following :
1. Homologous chromosomes,
2. Synapsis,
3. G₂alent,
5. Tetrad.
Answer:
1. Homologous chromosomes :
Homologous chromosomes are paired during zygotene and then form chiasmata at some places where crossing over takes place between them and finally resulting in the production of new characters and possibilities of organic evolution.

2. Synapsis:
Pairing of homologous chromosomes in zygotene substage of prophase-I of meiosis cell division is known as synapsis. The paired structure formed as result of synapsis is called as bivalent.

3. G₂-Phase :
In this phase, RNA and protein continues to be synthesised. The cell is now ready to enter the mitotic phase. The cells of this stage contain double amount of DNA.

4. Bivalent:
The pairs of homologous chromosome formed during meiosis-I. The chromosomes contract and become short and thick.

5. Tetrad:
In Pachytene stage of Prophase-I of Meiosis-I, Chromosomes split lengthwise but still remain attached at the point of centromere to form four chromatids called as Tetrad.

Question 18.
Which of the following statement is associated with :
(a) Prophase,
(b) Metaphase,
(c) Anaphase,
(d) Telophase,
(e) Interphase of mito-sis.
1. The nuclear membrane reappears.
2. Chromosomes are thickest and shortest.
3. Chromosomes begin to coil.
4. Centromere divides into two.
5. Nucleus is active, but chromosomes are not distinct.
6. Followed by cytokinesis.
7. Each chromosome consists of two chromatids.
Answer:
1. (d) Telophase
2. (b) Metaphase
3. (c) Anaphase
4. (c) Anaphase
5. (e) Interphase of mito-sis
6. (d) Telophase
7. (c) Anaphase

Cell Cycle and Cell Division Class 11 Important Questions Long Answer Type

Question 1.
Write important differences between Mitosis and Meiosis cell division.
Answer:
Differences between Mitosis and Meiosis

Question 2.
Describe different stages of Meiosis and its significance also.
Answer:
Meiosis-I division :
1. Prophase-I (Gk., pro = before, phasis = stage) : The prophase of first meiotic division is a complex phase of longer duration which is further divided into following five substages:
(i) Leptotene or Leptonema (Gk., leptos = thin, nema = thread) :
At this stage condensation of chromatin starts and the fine chromatin fibres appear with granule like chromomeres on them. The chromomeres are the regions where chromatin fibres are highly coiled.

(ii) Zygotene or Zygonema (Gk., zygo = paired, tiema = thread):
At zygotene stage two homologous chromosomes (one paternal and other maternal) lie side by side which is known as pairing of homologous chromosomes or synapsis. Each pair is called as a bivalent, they are similar in length and in position of their centromere.

The process of synapsis can take place in many ways. It may start from the ends of homologous chromosomes and proceed towards the centromere or it may start from centromere and proceed towards the ends. It may also begin at one point or many points simultaneously. It is believed that synapsis is initiated by some sort of attraction force.

(iii) Pachytene or Pachynema (Gk., pachus = yolked or thick, nema = thread): At pachytene chromosomes contract longitudinally, resulting in shorter and thicker threads. Each unit is a bivalent or tetrad which is composed of two homologous chromosomes in close longitudinal union and which contains four chromatids. Each chromatid has its own centromere. The chromatids of each homologue is called sister chromatids.

During pachytene two of the homologous chromosome exchange their segments. This process is called as crossing-over. During this process, two non-sister chromatids, one from each bivalent partially coil around each other. Then both chromatids break at contact place and reunite in this way that exchange of chromosomal segments take place. This event is called crossing-over. The points of crossing-over are called chiasmata.

(iv) Diplotene or Diplonema (Gk., diplos = double, nema = thread):
In diplotene, the two chromosomes of each bivalent move away from each other and the chiasmata finally disappear. This event is terminalization of chiasmata. The two homologous chromosomes, thus separate from each other, however not completely because both remain united at the point of interchange or chiasmata.

(v) Diakinesis (Gk., dia = through, kinesis = division or movement):
During this stage the bivalents condense further, i.e., continue shortening and also become more thicker. Homologous chromosomes move apart. The nucleolus and nuclear membrane begin to dis¬appear. In animal cell both the centrioles travel to opposite poles. Astral rays emerge out from the centriole and the structure appear like aster or star and spindle formation starts.

2. Metaphase-I (Gk., meta = between) :
This stage begins with disappearance of nuclear membrane and nucleolus. The bivalents arrange themselves at the equatorial plane in such a way that one chromosome faces one pole of the spindle apparatus and the other one another pole. The spindle fibres get attached to bivalents at the centromere.

3. Anaphase-I (Gk,, ana = up,phasis = stage):
Due to contraction of spindles the two chromosomes of each bivalent separate from each other and move towards opposite poles of the spindle. But the two chromatids of each chromosome still continue to remain jointed with one another with centromere. Thus, only one chromosome of each homologous pair reaches at each pole and the number of chromosomes are reduced to half in daughter nuclei.

4. Telophase-I (Gk, telo = end,phasis = stage):
As soon as the chromosomes reach the poles decondensation starts and they lose their identity. Nuclear envelope is formed around the daughter nuclei and nucleolus also reappears.

Cytokinesis :
Telophase-I is generally followed by cytokinesis (similar to mitosis) to form two haploid (n) daughter cells.

Interphase:
A short interphase called interkinesis may intervene between meiosis-I and meiosis-II. However, replication of DNA does not take place in this phase. Sometimes the daughter nuclei are not fully constituted in telophase-I. Instead the chromosomes reachng the opposite poles may progress directly to prophase-II.

Meiosis-II division:
It is similar to mitosis and is also known as equational division. It consists of following phases:
1. Prophase-II:
Sister chromatids of each chromosome begin to condense and chromosomes reappear. Nuclear envelope and nucleolus disappear. Spindle fibres begin to disappear.

2. Metaphase-II:
The haploid number of chromosomes become arranged at the equatorial plane of spindle apparatus, each chromosome gets attached to spindle fibre by its centromere.

3. Anaphase-II:
The centromere divides and sister chromatids by the shortening of spindle fibres are pulled apart to their respective poles. Each half chromosome (chromatid) move towards the opposite pole.

4. Telophase-II:
The chromosomes on the respective poles uncoil and form chromatin network again. Nuclear membrane and nucleolus reappear. Finally, at the end of meio- sis-II four haploid nuclei are produced.

Cytokinesis : It is similar to mitosis, in animal cell by furrow formation and in plant cell by cell plate formation.
This way each daughter cell of meiosis-I produces two haploid daughter cells at the end of meiosis-II. In other words we can say that four haploid cells are formed from a diploid cell by the process of meiosis.

Significance of meiosis :

• It plays an important role in sexual reproduction of both the lower as well as higher organisms. Gametes are produced by this process.
• The process of meiosis helps in the maintenance of chromosome number in the species. Diploid (2n) organisms produce haploid (n) gametes which on fusion form diploid organism again.

Question 3.
Give necessity and significance of cell cycle and describe the phenomenon occurs in its various stages.
Answer:
Cell cycle :
Complete life cycle of any cell is called as cell cycle. Cell cycle includes various phases of the growth, development and reproduction of the cell. It is a continuous process.

Significance of cell cycle :
The cell, after its formation from pre-existing cell takes sometime to undergo further division. In this ‘resting period’ the cell grows in size, its nuclear materials increases and thus cell get ready for the next division. Hence, cell cycle is a cyclic process from the existence of a new cell to the division of cell. The total duration of a cell cycle constitutes one generation time.

Duration of cell cycle :
The duration of cell cycle varies from 25-30 hours in different cells. In case of E. coli the cell cycle is completed in 20 minutes, in rat 22-25 hours, in root tip of onion 20 hours and in human it is completed in 26 hours.

Stages of cell cycle :
Howard and Pele in 1953 discovered various stages of cell cycle. According to them cell cycle is completed in two phases :

1. Interpbase: Interphase is the interval period between the two successive divisions, when cell does not show any division. But it prepares itself for it by synthesizing new proteins and nucleic acids. In interphase chromosomes are not distinguished, they are in the fonn of chromatin network. This phase is also called as preparatory phase.

Interphase is divided into following three subphases, i.e., G₁, S and G₂phase :
(i) Presynthetic or G₁– phase : This phase is also known as the period of initial growth as the young daughter cell grow in size during this phase. Synthesis of new proteins and RNA needed for various activities of growing cell also occurs in this phase. G₁ -phase in most of the cells lasts for about 10-12 hours. Non-dividing cells remain permanently in this stage. There is no change in the DNA contents of the cell at this stage.
Mitochondria, chloroplast, lysosome, Golgi body, vacuoles, ribosomes, etc. are formed in this phase.

(ii) Synthetic phase or S-phase :
This phase is characterized by the replication of DNA molecules. Thus, each chromosome now carries a duplicate set of genes. Each chro¬mosome is now composed of two sister chromatids held together by a common centromere. The cell thus retains the original diploid (2n) number of chromosomes but it has duplicate set of genes. It lasts for about 6-8 hours.

(iii) Post-synthetic or G₂-phase :
In this phase nucleus increases in its size. The synthesis of protein and RNA continues in this phase, r RNA and m RNA required for cell division are synthesized in this phase and the cell prepares itself for the cell division. Mito¬chondria and chloroplasts are duplicated in this phase. Centriole begins to form proteins required for the formation of spindle fibres.

2. Phase of cell division or M-Phase :
In this phase, a cell tends to divide and generating two daughter cells. It is completed in following two stages :

• (i) Karyokinesis : It is the stage of division of nucleus, which is completed in four stages :
  • Prophase,
  • Metaphase,
  • Anaphase and
  • Telophase.
• (ii) Cytokinesis : It is the stage of division of cytoplasm to form daughter cells.

Question 4.
Write short note on Karyotype and Idiogram.
Answer:
Karyotype:
We have well studied earlier that each and every eukaryotic organism including plants and animals is characterised by a set of chromosomes which have certain
constant features. The chromosomes of a particular organism or species are identified on the basis of following important features :

• Number of chromosomes,
• Shape and size of chromosomes,
• Position of centromere,
• Length and ratio of two arms of a chromosome, (v) Presence of secondary constriction,
• Size of satellite bodies on chromosomes.
  Example : Karyotype of Ginkgo biloba.

The term karyotype has been proposed for “the group of characteristics that identifies a chromosome set (haploid) of a particular species.” This karyotype is represented by a diagram called idiogram.

Idiogram :
The idiogram of an organism is prepared by arranging the chromosomes of somatic cells in a descending order of size keeping their centromeres in a straight line (Fig.).

Thus, the longest chromosome is placed on the extreme left and the smallest one on the extreme right. In organisms having sex chromosomes, the sex chromosomes are sometimes placed at the extreme right but more commonly they are placed at their appropriate position according to their size and marked by X and Y. Each chromosome in a karyotype is designated by a serial number according to its position. (44 + XY) (44 + XX).

Cell Cycle and Cell Division Class 11 Important Questions Objective Type

1. Choose the correct answers:

Question 1.
Chiasmata formation takes place during:
(a) Diplotene
(b) Leptotene
(c) Pachytene
(d) Diakinesis.
Answer:
(c) Pachytene

Question 2.
Duplication of chromosomes takes place during mitosis in:
(a) Early prophase
(b) Late prophase
(c) Interphase
(d) Late telophase.
Answer:
(b) Late prophase

Question 3.
Centromere in the first metaphase of meiosis:
(a) Divide
(b) Do not divide
(c) Divide but do not separated
(d) Are not similar.
Answer:
(c) Divide but do not separated

Question 4.
Spindle fibres are attached with chromosomes at:
(a) Telomere
(b) Chromomere
(c) Centromere
(d) Kinetochore.
Answer:
(c) Centromere

Question 5.
Part of the chromosome where crossing-over takes place is:
(a) Chromomeres
(b) Bivalent
(c) Chiasmata
(d) Centromere.
Answer:
(c) Chiasmata

Question 6.
Number of chromosomes becomes half during:
(a) Prophase-1
(b) Anaphase-I
(c) Metaphase-1
(d) Metaphase-lI.
Answer:
(b) Anaphase-I

Question 7.
Reduction division takes place ¡n:
(a) Pollen grains
(b) Pollen tube
(c) Pollen mother cell
(d) Generative cell.
Answer:
(c) Pollen mother cell

Question 8.
Spindle fibres are formed from :
(a) Protein
(b) Lipid
(c) Cellulose
(d) Pectin.
Answer:
(a) Protein

Question 9.
Synapsis takes place between :
(a) Similar chromosomes
(b) Two homologous chromosomes
(c) Non-homologous chromosomes
(d) None of these.
Answer:
(b) Two homologous chromosomes

Question 10.
In animals, spindle fibres are formed from :
(a) Centriole
(b) Centromere
(c) Nucleus
(d) Mitochondria.
Answer:
(a) Centriole

Question 11.
Formation of synaptical complex takes place during :
(a) Leptotene
(b) Zygotene
(c) Pachytene
(d) Diplotene.
Answer:
(b) Zygotene

Question 12.
Diploid number of chromosomes in an organism is 8. Then find out the number of chromatids in each daughter cell will be :
(a) 2
(b) 4
(c) 8
(d) 16.
Answer:
(c) 8

Question 13.
Stage of meiosis in which synapsis takes place is :
(a) Leptotene
(b) Pachytene
(c) Zygotene
(d) Metaphase-I.
Answer:
(c) Zygotene

Question 14.
Stage of meiosis in which centromere divide :
(a) Diplotene
(b) Metaphase-I
(c) Pachytene
(d) Anaphase-II
Answer:
(d) Anaphase-II

Question 15.
During interphase RNA and protein are synthesized :
(a) In S-phase
(b) In G₁phase
(c) In G₂-phase
(d) Both (b) and (c).
Answer:
(d) Both (b) and (c).

Question 16.
In which stage of meiosis centromere divides :
(a) Early prophase
(b) Early metaphase
(c) Early anaphase
(d) Post anaphase.
Answer:
(d) Post anaphase.

Question 17.
In which stage chromonemata begins to form paired chromosome :
(a) Zygotene
(b) Leptotene
(c) Pachytene
(d) Diplotene.
Answer:
(a) Zygotene

Question 18.
In which type of cell division number of chromosomes reduces :
(a) Meiosis
(b) Mitosis
(c) Amitosis
(d) Cleavage.
Answer:
(a) Meiosis

Question 19.
Which organ of the cell disappear during mitosis cell division :
(a) Plastid
(b) Plasma membrane
(c) Nucleus
(d) None.
Answer:
(c) Nucleus

Question 20.
How many chromosomes are present in a cell of human being after meiosis : (a) 46
(b) 23
(c) 20
(d) 48.
Answer:
(b) 23

Question 21.
Life of multicellular organisms begin from :
(a) Sperm
(b) Egg
(c) Zygote
(d) None of these.
Answer:
(c) Zygote

Question 22.
Diploid chromosome number in human is :
(a) 23
(b) 46
(c) 92
(D) 48
Answer:
(b) 46

Question 23.
Synapsis between chromosomes takes place during :
(a) Leptotene
(b) Pachytene
(c) Zygotene
(d) Diakinesis.
Answer:
(c) Zygotene

Question 24.
Stage of mitosis in which centromere divides :
(a) Anaphase
(b) Prophase
(c) Metaphase
(d) Telophase.
Answer:
(a) Anaphase

Question 25.
Which of the following divide during karyokinesis :
(a) Pollen grain
(b) Nucleus
(c) Cell wall
(d) Cytoplasm.
Answer:
(b) Nucleus

2. Fill in the blanks:

1. Somatic cells multiply by ……………….
Answer:
Mitosis

2. Mitosis results in the formation of nuclei having ………………. number of chromosomes.
Answer:
Same (identical)

3. The second division of meiosis can be described as ………………. division.
Answer:
Mitotic

4. The region of attachment of chromosomes to spindle fibres is called ………………….
Answer:
Centromere

5. The haploid condition is reached by ……………. stage.
Answer:
Anaphase

6. In parent and daughter cell, number of chromosomes are same, thus this type of cell division is called as ………………….
Answer:
Mitosis division

7. Chromosomes arrange themselves during metaphase at the ………………… of the cell.
Answer:
Equator

3. Match the following:

Column ‘A’	Column ‘B’
1. Pachytene	(a) Mitosis
2. Zygotene	(b) Metaphase
3. Equational division	(c) S-phase
4. Spindle formation	(d) Crossing-over
5. DNA replication	(e) Homologous pairing.

1. (d) Crossing-over
2. (e) Homologous pairing,
3. (a) Mitosis,
4. (b) Metaphase,
5. (c) S-phase.

4. Write true or false:

1. Interphase is the stage in between Prophase-I and Metaphase-I.
Answer:
False

2. Mitosis is also called as Reductional Division.
Answer:
False

3. Meiosis results in the formation of two haploid cells.
Answer:
False

4. Meiosis takes place in reproductive cells.
Answer:
True

5. During Prophase-I of Meiosis-I, the nuclear membrane and nucleolus disappear in Diakinesis stage.
Answer:
True

5. Answer in one word:

1. Short Interphase between Meiosis-I and Meiosis-II.
Answer:
Interkinesis

2. Division which takes place in somatic cells.
Answer:
Mitosis

3. The process of interchange of genetic material between non-sister chromatid.
Answer:
Crossing-over

4. The term given to each bivalent containing four chromatids at pachytene stage.
Answer:
Tetrad

5. The name of cell division taking place in reproductive cells.
Answer:
Meiosis

Students get through the MP Board Class 11th Biology Important Questions Chapter 13 Photosynthesis in Higher Plants which are most likely to be asked in the exam.

MP Board Class 11th Biology Important Chapter 13 Photosynthesis in Higher Plants

Photosynthesis in Higher Plants  Class 11 Important Questions Very Short Answer Type

Question 1.
What is Photosynthesis? Explain it with chemical reaction.
Answer:
Photosynthesis: Photosynthesis is a chemical process by which green plants synthesize carbohydrate like simple food materials from carbon dioxide and water in the presence of sunlight. Water molecules are oxidized and CO₂ molecules are reduced by this process. Photosynthesis is expressed with the help of following chemical reaction :

Question 2.
Name the products of light reaction of photosynthesis process.
Answer:
The products of light reaction of photosynthesis process are ATP and NADPH₂.

Question 3.
What are Hill reagents or Hill’s oxidizing agents?
Answer:
Hill observed that evolution of O₂ was possible when chloroplast of Stellaria media suspended in water in absence of CO₂ were exposed to light and supplied with salts such as ferricyanides, benzoquinones, dichlorophenol, which served as hydrogen acceptors. These substances are called as Hill reagents.

Question 4.
Name the photosynthetic pigments found in bacteria.
Answer:
Photosynthetic pigments found in the bacteria are :

• Bacteriochlorophyll,
• Bacterioviridin
• Carotenoids etc.

Question 5.
Name any two insectivorous plants.
Answer:

1. Nepenthes,
2. Drosera.

Question 6.
Write any two differences between CAM and C₄ plants.
Answer:
Differences between CAM and C₄ plants

CAM plants	C4 plants
1. They are succulent plants of family Crassulaceae.	They are monocot plants with Kranz anatomy in their leaves.
2. Two carbon cycles are separate due today and night.	Two carbon cycles are separate due to Kranz anatomy of leaves.

Photosynthesis in Higher Plants  Class 11 Important Questions Short Answer Type 

Question 1.
Give one example to prove that oxygen released during photosynthesis pro-cess is oxygen of water, not of CO₂.
Answer:
The oxygen liberated during photosynthesis is a part of water molecule. This is proved by Rubem and Camanis experiment. For this process such type of water is taken whose oxygen contains O¹⁸ isotopes and after equation the oxygen liberated is also O¹⁸.

By this experiments it is proved that after photosynthesis the liberated O₂ is the part of water molecule.

Question 2.
What is compensation point of Light intensity?
Answer:
Afternoon when intensity of light is high, rate of photosynthesis is high so plants releases O₂ whereas at night when there is no sunlight, photosynthesis process does not occur but during morning and evening when there is less intensity of light rate of photosynthesis become very low, equal to rate of photosynthesis.

Therefore, during this time O₂ released during photosynthesis process is totally used for respiration process and CO₂ produced during respiration is toally used up for photosynthesis process by the plant, thus there is no net release of any gas during this time. This condition is called as compensation point.

Question 3.
Describe Hill Reaction in 50 words and give required chemical reactions.
Answer:
Hill reaction : It is the first step of photosynthesis process, which occurs in the grana part of the chloroplast.
It occurs in the presence of light and it was discovered by scientist Hill. It is completed in following steps:
1. Photophosphorylation :

2. Photolysis of water :
12H₂O → 12H⁺ +12OH^–
12OH^– -12e^– →12OH
12OH → 6H₂O + 3CO₂

3. Hydrogenation :

Question 4.
Give significance of Photosynthesis process.
Answer:
Significance of Photosynthesis process :

• By this process green plants not only produces food for themselves but also for all other organisms. Thus, green plants are also called as producers.
• Plants uses harmful CO₂ from the atmosphere to prepare organic food and releases 02 into the atmosphere, thus help to maintain O₂-CO₂ balance in the nature.
• It regulates temperature by absorbing sunlight.
• By absorbing CO₂ from the atmosphere it helps to reduce global warming.

Question 5.
Give affect of Light and CO₂ on photosynthesis process.
Answer:

• Affect of Light: Rate of photosynthesis increases with increase in the intensity of light and decreases in less intensity of light. Colour of light also affect rate of photosynthesis. Rate of photosynthesis is highest in red light, in blue light rate of photosynthesis is second highest. Greenlight is reflected back by chlorophyll molecule, thus rate of photosynthesis is nil in green light.
• Affect of CO₂: Rate of photosynthesis increases with increase in the quantity of CO₂ but up to a certain limit (according to law of limiting factor) beyond which there is no effect.

Question 6.
Explain any two internal factors which affect photosynthesis process.
Answer:
Internal factors affecting photosynthesis process :

1. Chlorophyll: The amount of chlorophyll present in the plant part has a direct relationship with the rate of photosynthesis because it is the pigment which is photoreceptive and is directly involved in trapping the light energy.
2. Accumulation of end products: The rate of translocation of food manufactured in the leaves decreases in the afternoon and therefore it starts accumulating in the mesophyll cells. The accumulation of the end product decreases the rate of photosynthesis in the afternoon.

Question 7.
If a leaf is kept in a dark room then its colour gradually become yellow and greenish-yellow. Which pigment is more stable according to your view ? (NCERT)
Answer:
Chlorophyll-b is more stable which is yellow-green coloured pigment and Xanthophyll (Yellow pigment) in the leaves are more stable.

Question 8.
Two sides of dorsiventral leaves are different in colour, Le., upper part is dark as compared to lower part or compare leaves of the plants kept in the shade with plants kept in light. Which one is dark coloured and why? (NCERT)
Answer:
In dorsiventral leaves upper surface of the leaves contain more chlorophyll-a. Place where chlorophyll-a is found increases rate of absorption of light as compared to lower part. Part of the leaf where more chlorophyll-a is found appears bright green coloured whereas part of the leaves which remain in the shade appears less green due to less quantity of chlorophyll-a.

Question 9.
What is Emmerson effect?
Answer:
Emmerson effect: During working on photosynthesis process Robert Emmerson determined quantum production (number of O₂ molecules produced by absorption of one quantum light) of light of various wavelengths and found that quantum production is maximum in red light of680 nm. wavelength but when wavelength of this red light is increased more, then quantum production falls all of a sudden. This is called as Red drop.
Emmerson also observed that when less wavelength light is provided with red light of 680 nm. wavelength then quantum production increases again. This is called as Emmerson enhancement effect.

Question 10.
Can you identify observing external features whether the plant is C_{3 Or} C₄ plant?
Answer:
Generally C₄ plants grow at high intensity of light at day temperature 30-35°C, i.e, in dry atmosphere whereas C₃ plants grows in moderate temperature. C₄ plants has special anatomy called as Kranz anatomy. Thus, unless we observe internal anatomy it is not possible to identify a plant whether it is C₃  or C₄ plant.

Question 11.
Can you identify observing internal structure whether the plant is C₃ or C₄ plant?
Answer:
Anatomy of C₄ shows that it has Kranz anatomy, i.e., (a) Presence of a double spiral bundle sheath cells tightly packed around the vascular bundle.
(b) Bundle sheath cells are connected with mesophyll cells by plasmodesmata.
(c) Leaves of C4 plants contain two types of chloroplasts :

1. Mesophyll chloroplast: It is smaller, grana is present there and starch grains are absent.
2. Bundle sheath chloroplast: It is larger in size, lacking grana and possessing starch grains.
   In C₄ plants, CO₂ fixation occurs by C₃ and C₄ cycle. C₃ cycle occurs in the bundle sheath whereas C₄ cycle occurs in the mesophyll cells.

Question 12.
RuBisCo is an enzyme which acts as carboxylase and oxygenase enzyme. Why you assume that RuBisCo is found in C4 plants does carboxylation in more quantity?
Answer:
RuBisCo enzyme shows more affinity to CO₂ as compared to O₂. Both CO₂ and O₂ may bind to the active site of this enzyme. This binding is competitive type. Binding of O₂ or CO₂ to the enzyme depends on concentration of these substances.
C₄ plants has such a system that increases CO₂ concentration in them. Thus, it acts as RuBisCo carboxylase in them and it cannot act like oxygenase.

Question 13.
Assume that here plants having higher concentration of chlorophyll-b were found, whereas chlorophyll-a deficient plants were less in number, were capable to do photosynthesis? Then why chlorophyll-b is found in the plants and what are the role of other minor pigments ? (NCERT)
Answer:
Chromatography of chlorophyll shows that green colour of leaves is due to presence of four pigments:

1. Chlorophyll-a,
2. Chlorophyll-b,
3. Xanthophyll and
4. Carotene.

Following diagram proof that maximum photosynthesis occurs in the area of blue and red light of the spectrum. Some photosynthesis occurs in other waves of the spectrum. Though chlorophyll-a is the main pigment which absorbs light, then also other pigments found in the thylakoid, such as Chlorophyll-b, Xanthophyll and carotene absorb light and transfer it to the chlorophyll-a. Thus, they are called as accessory pigments.
These pigments increases wave area which also prevent chlorophyll-a from photooxidation.

Question 14.
Rate of photosynthesis is affected by light.

Answer following questions on the basis of graph:
(a) At which points (A, B, C or D) of the graph light is the limiting factor.
(b) Which limiting factor lies at point A.
(c) What is represented by C and D in the graph.
Answer:
(a) At point B and C light is the limiting factor.
(b) At point A of the graph sunlight, temperature, CO₂ concentration, water etc. fac-tors of photosynthesis are present and they altogether affecting photosynthesis at a time. Out of them anyone become main factor affecting photosynthesis process.

(c) C and D of the graph represent that there is linear relationship between low intensity of light and rate of CO₂ assimilation.
At high intensity of light there is no increase in this rate but other factors become limiting.
After a limit incident, ray causes decomposition of chlorophyll molecule, due to which rate of photosynthesis decreases.

Question 15.
Write differences between following : (NCERT)
(a) C₃  and C₄ Pathways.
(b) Cyclic and Non-Cyclic Photophosphorylation.
(c) Morphology of leaves of C₃  and C₄ plants.
Answer:
(a) Differences between C₃  and C₄ Pathways

C3 Pathway	C4 Pathway
1. CO2 acceptor is RuDP [a 5 carbon compound]	CO2 acceptor is PEP (phosphoenol pyruvate) [a 3 carbon compound].
2. First stable product is PGA (Phospho glyceric acid)	First stable product is OAA (oxaloacetic acid.)
3. Excess of atmospheric O2 inhibits photosynthesis process.	Excess of atmospheric O2 has no effect on photosynthesis.
4. There is only Calvin-Benson pathway.	There is both Calvin-Benson and Hatch- Slack pathway.
5. Assimilation of one molecule of CO2 requires 2NADPFH2 and 3ATP molecules.	Assimilation of one molecule of CO2 requires 2NADPH and 5ATP molecules.
6. Optimum temperature for photosynthesis is 10-25°C.	The optimum temperature for photosynthesis is 30-45°C.
7. C3 path can be observed in temperate plants like potato, mango etc.	C4 path can be observed in tropical grasses like sugar-cane, maize etc.

(b) Differences between Cyclic and Non-Cyclic Photophosphorylation

Cyclic Photophosphorylation	Non-Cyclic Photophosphorylation
1. It involves only pigment system-I.	It involves both pigment system-I and pigment system-II.
2. Electrons released by the pigment Electrons released by the pigment system- I return back to it.	II do not return back to it.
3. It produces only ATP.	It produces ATP and NADPH2.
4. Photolysis of water does not occur.	Photolysis of water occur.
5. Oxygen is not given off.	Oxygen is given off.

(c) Differences between Morphology of leaves of C₃ and C4 plants

Morphology of C3 Plants	Morphology of C4 Plants
1. They do not have Kranz anatomy.	They have Kranz anatomy.
2. Thin layered parenchymatous bundle of sheath is found surrounding vascular bundle.	Double spiral bundle sheath cells are found tightly packed around the vascular bundle.
3. Bundle sheath cells are in contact with mesophyll cells.	Bundle sheath cells are connected with mesophyll cells by plasmodesmata.
4. Leaves of C3 plants contain similar chloroplasts.	Leaves of C4 plants contain two types of chloroplasts : (a) Mesophyll chloroplasts which are smaller, grana is present there and starch grains are absent. (b) Bundle sheath chloroplasts, which are larger, lacking grana and possessing starch grains.

Question 16.
Describe Photorespiration in brief.
Answer:
Photorespiration or C2 Cycle: It was normally believed that rate of respiration is equal in day and night. Recently it has been observed that light affects respiration and the rate of respiration in light maybe three to five times higher than the respiration in darkness. Such type of respiration is called photorespiration and is marked as one of the new discoveries of plants physiology. In photorespiration, temperature plays a very vital role, its rate being very high in between 25-35°C. It also depends upon the concentration of oxygen and increases with increasing oxygen concentration even up to 100%. However, the normal respiration is independent of oxygen concentration. In normal respiration the respiratory substrate is glucose while in photorespiration glycolic acid (2-carbon compound) serves as a substrate.

The diagnostic features of photorespiration are as follows :

• The respiratory substrate is glycolate a 2-carbon compound.
• The substrate is always recently formed.
• The entire process of photorespiration occurs in between chloroplast, cytoplasm, peroxisome and mitochondria.
• It shows positive correlationship with O₂ concentration. The ribulose diphosphate reacts with O₂ to produce one molecule of 3-carbon, phosphoglyceric acid (PGA) and one molecule of 2-carbon phosphoglycolic acid.

RuDP + O₂ → phosphoglyceric acid + phosphoglycolic acid

Phosphoglycolic acid is immediately dephosphorylated into glycolate. The latter en-ters the peroxisome, where glycolate is catalyzed to produce glyoxylate and H₂O₂.
The glyoxylate is transmitted to glycine which is also obtained from serine.
The glycine molecule enters the mitochondria where they lose CO₂ to form serine. Serine diffuses into the peroxisome where it is converted into glycerate which then enters the chloroplast and the Calvin cycle to produce sugars.

• In this process ATP molecules are not formed.
• The rate of photorespiration is highly accelerated in between 25° to 35°C.
• This is found only in presence of light and only in green cells.

Question 17.
What do you mean by Total parasitic? Explain with example.
Answer:
Total Parasitic Nutrition: It is a type of nutrition in which parasitic plant totally depend on the host plant for food as well as shelter. Such plants are non-green e.g., Cuscuta orobanche. The total parasites develop haustoria or sucking roots which go into the vascular bundles of the host plant and absorb prepared food and water from there.

Question 18.
“CO₂ is very essential for the production of carbohydrates by the plants.” Prove this statement by describing the experiment with labelled diagram.
Or,
Give one experiment to prove that CO₂ is necessary for photosynthesis pro¬cess.
Answer:
Experiment to demonstrate that CO₂ is necessary for photosynthesis process: It can be demonstrated by Moll’s experiment:

Method: Take a potted plant having long leaves. Keep it in darkness for 24 to 48 hours to make it starch free. In the morning before sunrise, insert the apex of a leaf (through a split cork) in a wide mouthed bottle which has been laid on its side on the table, and in which caustic potash has been introduced.

Now expose the apparatus to sunlight for some-time. Then pluck the leaf and decolourize it with the help of hot alcohol and test with weak iodine solution. It is found that the apical portion of the leaf lying inside the bottle does not give the starch reaction, whereas the portion outside the bottle turns blue. This proves that starch is formed in those parts only which get CO_2.

Question 19.
Write short note on Facultative or Semi-parasitic nutrition.
Answer:
Facultative or Semi-parasitic nutrition: The type of nutrition in which plant may become parasite under certain conditions is called as facultative nutrition. These parasitic plants have chlorophyll and therefore, synthesize their organic food but they depend on the host plants for their water and mineral requirements.

Question 20.
Write differences between Light reaction and Dark reaction.
Answer:
Differences between Light Reaction and Dark Reaction

Light Reaction	Dark Reaction
1. It occurs in presence of light.	It occurs in absence of light.
2. It occurs in the grana part of the chloroplast.	It occurs in the stroma part of the chloroplast.
3. Fixation of CO2 do not occur.	Fixation of CO2 occurs.
4. Photolysis of water occurs and O2 is produced.	Photolysis of water do not occurs.
5. Products of light reaction are ATP and NADPH2.	Product of dark reaction is glucose. e.g. Loranthus, Mistletoe etc.

Question 21.
What are CAM plants? Give characteristic features of CAM plants and explain mechanism of CAM cycle.
Answer:
Crassulacean Acid Metabolism or CAM Cycle: Crassulacean acid metabolism occurs in certain succulent plants of the family Crassulaceae. As they exhibit a special type of CO₂ assimilation, they are called as CAM plants.

1. Dicotyledonous family:

• Crassulaceae (e.g.,Sedum, Opuntia),
• Cactaceae,
• Chenopodiaceae,
• Compositae,
• Convolvulaceae,
• Euphorbiaceae,
• Caryo- phyllaceae.

2. Monocotyledonous family :

• Liliaceae,
• Orchidaceae.

3. Pteridophytic family: Polypodiaceae.
Characteristic features of CAM plants: CAM plants exhibit the following characteristic features:

• Normally stomata are open during night (dark) and closed during day (light)
• CO₂ fixation takes place during day time and malic acid is formed during night.
• Malic acid is stored in the large vacuoles, which is the characteristic of the cells of CAM plants.
• In day light, decarboxylation of malic acid takes place and CO₂ is formed which is used in C₃ cycle to produce storage glycans.

Malic acid + NADP⁺ → Pyruvic acid + CO₂ + NADPH₂

• In next dark period storage glycans are catabolized through glycolysis and produces PEP as CO₂ acceptor molecule.
  
• Stomata remain closed during day and open at night.
• CO₂ is fixed via. PEP-carboxylase initially into OAA and then to other 4 carbon acids i.e., malic acid. This is called as dark
• CO₂ fixation and is more efficient at 10-15°C.
• The larger vacuoles are used for storing malic and other acids in large amount.

Photosynthesis in Higher Plants  Class 11 Important Questions Long Answer Type 

Question 1.
Describe light reaction of photosynthesis process.
Answer:
Light Reaction: In this reaction photolysis of water and production of the reducing power take place.
In this step, which takes place in the presence of sunlight and chlorophyll there is a break down of the water molecules. Oxygen escapes to the air in the molecular form while hydrogen is accepted by Nicotinamide adenine dinucleotide phosphate (NADP). The whole process can be represented by following summary equation.

This part of the photosynthetic mechanism which takes place in light is known as the Light Reaction or Hill Reaction, after its discoverer. Not only is the reducing agent produced but some solar energy get stored up as the chemical energy of NADPH₂ (about 18 molecules of ATP).

According to Robert Emmersion absorption of water in light reaction occurs through two pigment systems :

(a) Pigment System I: This pigment system absorbs only the light of 680 nm wavelength of sunlight. P700 on absorbing photon of light expels two-electron, P700 becomes oxidized. The expelled electron is picked up by ferredoxin. They transfer the electron to cytochrome b6 which supply it to cytochrome f. They transfer the electron to plastocyanin. From here it is handed over back to P700.

The synthesis of ATP in this cyclic electron transport scheme is possible at two locations. Synthesis of ATP may occur between ferredoxin and cytochromes bg and the other when the electron travels from cytochromes fg to plastocyanin.
In this case the electron expelled out by an excited chlorophyll trap is returned to the oxidized chlorophyll after passing through a series of electrons carriers, therefore this process is also called as Cyclic photophosphorylation.

(b) Pigment System II: This pigment system absorbs only the light of wavelength less than 680 nm of sunlight.
In this process pigment chlorophyll-a673 become excited by absorbing 2 photons of light and emit 2 electrons. It then moves through a series of electron carriers, some of which are plastoquinone, cytochrome bg, cytochrome f and plastocyanin. Most probably energy for ATP synthesis is given out between cytochrome bg and cytochrome f. Plastocyanin hand over the electron to P700.

When pigment system II absorbs light then 24 molecules of water dissociates into H⁺ and OH^– ions.
24H₂O → 24OH^– + 24H⁺

OH^– ion from this goes to pigment system II and oxidized to reduce to produce water and oxygen.

24OH^– – 24e^– → 240H
24OH → 12H₂O + 6O₂

24 H⁺ produced by photolysis of water obtain electrons from ferredoxin and reacts with NADP to form NADPH₂.

24H⁺ + 24e^– +12NADP →12NADPH₂
In this electron transfer, one ATP is formed when it transfer from Cyt. bg to Cyt.f. It is the normal process of photophosphorylation in higher plants. The electron expelled by the excited chlorophyll trap is never returned to the same chlorophyll. Therefore this process also called as Acyclic phosphorylation.

In this way in six complete cycle of cyclic phosphorylation 12ATP are formed and in six complete acyclic phosphorylation 6ATP and 12NADPH₂ are formed. Whereas in one acyclic phosphorylation 2NADPH₂ are formed and 24H₂O dissociates. Therefore both the systems of light reaction together produces 18ATPand 12NADPH₂ molecules.
Both the reactions occurs in light reaction simultaneously. To complete photosynthesis process both the pigment systems are essential.

Question 2.
Explain dark reaction of photosynthesis process in brief.
Or,
What is Calvin cycle? Describe it in brief.
Answer:
Dark reaction or Blackman’s reaction or Biosynthesis phase or Calvin’s cycle : The non-photochemical reaction or dark reaction of photosynthesis was first of all established by Blackman (1905). The reactions of dark phase of photosynthesis occur in the stroma of the chloroplast. In this phase the carbon dioxide is converted to carbohydrate through a series of enzyme catalyzed reactions.

The path of carbon in photosynthesis is popularly called Calvin cycle or C₃ cycle. For this great piece of work Calvin was awarded Noble prize in 1961. This cycle is also called as Bassham and Calvin cycle, Blackman reaction, Carbon assimilation, Path of carbon in photosynthesis etc.

The process can be represented by following reaction :

6CO₂ + 12NADPH₂ +18 ATP→ C₆H₁₂O₆ + 12NADP + 18ADP + 6H₂O + 18Pi

Most of the intermediate compounds of Calvin cycle are 3 carbon compounds. Therefore this cycle is also called as C₃ cycle.
Calvin cycle may be explained by following two steps :

1. Synthesis of carbohydrates
2. Regeneration of ribulose diphosphate.

1. Synthesis of carbohydrates: 6 molecules of CO₂ are first accepted by 6 molecules of Ribulose-1, 5-diphosphate and forms an unstable compound which soon breaks into 12 molecules of 3 carbon compound called as Phosphoglyceric acid. The phosphoglyceric acid is then reduced to 12 molecules of phosphoglyceraldehyde by 12 molecules of NADPH₂. 5 molecules from 12 molecules of phosphoglyceraldehyde converts into Dihydroxyacetone phosphate.

Now dihydroxyacetone phosphate reacts with 3 molecules of phosphoglyceraldehyde to form Fructose-1, 6-diphosphate. One molecule from it converts into Fructose-6-phosphate which converts into glucose or starch.

2. Regeneration of ribulose diphosphate: The regeneration of ribulose-1, 5- diphosphate is essential to carry on the process of photosynthesis. Fructose 6 phosphate and phosphoglyceraldehyde combine and break into 4-carbon compound erythrose-4-phosphate and 5-carbon compound xylulose-5-phosphate.

The former combines with a molecule of dihydroxyacetone phosphate to form sedoheptulose-1,7-diphosphate from which one phosphate is removed to form sedoheptulose-7- phosphate. The latter and phospho¬glyceraldehyde combine to produce one molecule each of xylulose-5-phosphate and ribulose-5-phosphate. Both of these compounds get converted into ribulose-5-phosphate which ultimately forms ribulose-1,5-diphosphate using ATP which is obtained from the photophosphorylation process.

Question 3.
Write a short essay on Bacterial photosynthesis.
Answer:
Bacterial photosynthesis: The anaerobic photosynthetic bacteria may be rods, cocci, or spirilla depending on their colouration. They are known as green or purple bacteria. They use sunlight as source of energy for photosynthesis but like other eukaryotic cells they do not ‘split water’ to transfer the energy or to obtain reducing power. Thus no oxygen is evolved by them. This process is therefore called anoxygenic (without producing oxygen) photosynthesis.

In place of water these bacteria obtain reducing power from hydrogen sulphide, thiosulphate, hydrogen or even some organic compound. They possess a pigment called bacterio-chlorophyll which is different from the chlorophyll pigment found in higher plants. This can be summarized as below :

There are three types of photosynthetic bacteria :

1. Green sulphur bacteria: They contain bacterial chlorophyll. These bacterias are found in H2S medium. In presence of light they reduces CO₂ Reaction is exergonic.
   
   Example: Chlorobacteria, chlorobium.
2. Purple sulphur bacteria : Bacteriochlorophyll is found. They can survive as heterotrophs in organic compounds in absence of inorganic sulphur compound.
   
   Example: Chromacium.
3. Purple non sulphur bacteria : These bacteria uses simple organic compounds such as alcohol, organic acid etc.
   
   As these bacterias uses visible light therefore they are also called as photo-autotrophic.

Question 4.
Write differences between photorespiration, true respiration and photosynthesis.
Answer:
Differences between Photorespiration, True Respiration and Photosynthesis

Question 5.
Explain Blackman’s principle of limiting factor.
Answer:
According to Blackman’s principle of limiting factor: When a physiological process is conditioned by number of factors then the rate of physiological process is limited by the factor found in least quantity.
For example, photosynthesis process is conditioned by many factors such as CO₂, water, light etc. If the concentration of CO₂ is least, then the rate of photosynthesis increases with increase in the quantity of CO₂. As the CO₂ supply increases gradually some other factor become limiting factor i.e., in such condition increased quantity of CO₂ does not help to increase rate of photosynthesis.

The principle was explained by Blackman as follows :

Suppose a leaf is exposed to such a light intensity which can allow the leaf to utilize 5mg of CO₂ per hour in photosynthesis. If 1 mg of CO₂ enters the leaf in one hour, the rate of photosynthesis is limited due to CO₂ factor. If the concentration of the CO₂ is increased from 1 to 5mg per hour, the rate of photosynthesis also increases along the line AC. Thus, the increase in photosynthetic rate will be proportionate with the increase in CO₂ concentration up to 5mg. Any further increase in the CO₂ concentration will have no effect on the rate of photosynthesis, which has become constant along the line CD. It is because the light factor (low intensity) has now become the limiting factor.

Now the rate of photosynthesis will increase further along the line CE only if the light intensity is also increased from low to a medium. At point E, the medium light intensity again becomes limiting factor and the rate of photosynthesis will again be- Fig Graphical representation of Blackman„’S come constant along the line EF. In the same way, law of limiting factor, when light intensity is increased from medium to higher, an increase in the rate of photosynthesis takes place along the EG by adding CO₂. When the rate attains maximum at G, further increase in CO₂ will not increase the rate of photosynthesis which become constant along the line GH. Here, higher light intensity again becomes a limiting factor.

Thus, it is quite evident from the above illustration that the rate of photosynthesis cannot be increased by increasing only one factor. The other factor should also be increased in proper proportion for favourable effect. Besides CO₂ and light, other factors such as temperature, water, etc. may also become limiting under certain conditions.

 

Photosynthesis in Higher Plants Class 11 Important Questions Objective Type

1. Choose the correct answers:

Question 1.
Which method was used by Calvin to explain path of photosynthesis :
(a) Chromatography
(b) Electrophoresis
(c) Spectrophotometry
(d) Histochemistry.
Answer:
(a) Chromatography

Question 2.
Highest effective wavelength of light for photosynthesis is of which colour :
(a) Green
(b) Yellow
(c) Red
(d) Violet.
Answer:
(c) Red

Question 3.
Light reaction occurs in :
(a) Stroma
(b) Grana
(c) Membrane of Chloroplast
(d) Endoplasmic reticulum.
Answer:
(b) Grana

Question 4.
After absorption of radiation by pigment system-I electrons are emitted by:
(a) Chlorophyll-683
(b) Chlorophyll-673
(c) Chlorophyll-695
(d) P-700.
Answer:
(d) P-700.

Question 5.
First step of photosynthesis is :
(a) ATP synthesis
(b) Excitement of chlorophyll and emission of electrons
(c) Photolysis of water
(d) Release of oxygen.
Answer:
(b) Excitement of chlorophyll and emission of electrons

Question 6.
Photophosphorylation is the method in which :
(a) Aspartic acid is formed
(b) CO₂ and H₂O combine
(c) Light energy converts into chemical energy
(d) PGA is formed.
Answer:
(c) Light energy converts into chemical energy

Question 7.
Rate of photosynthesis depends on :
(a) Photoperiod
(b) Intensity of light
(c) Quality of light
(d) Temperature.
Answer:
(a) Photoperiod

Question 8.
CO₂ acceptor of C₃ cycle is :
(a) Phosphoglyceraldehyde
(b) Ribulose diphosphate
(c) Phosphoglyceric acid
(d) Pyruvic acid.
Answer:
(c) Phosphoglyceric acid

Question 9.
C₄ cycle of photosynthesis is absent in :
(a) Zea mays
(b) Triticum vulgare
(c) Oryza sativa
(d) Euphorbia.
Answer:
(d) Euphorbia.

Question 10.
During photosynthesis :
(a) Reduction of water and oxidation of CO₂
(b) Reduction of CO₂ and oxidation of water
(c) Oxidation of both CO₂ and H₂O
(d) Reduction of both CO₂ and H₂O.
Answer:
(b) Reduction of CO₂ and oxidation of water

Question 11.
Electron acceptor required for photophosphorylation and oxidative phosphorylation is:
(a) O₂
(b) CO₂
(c) Cytochrome
(d) Water.
Answer:
(c) Cytochrome

Question 12.
First stable compound of C₃ plant during photosynthesis is :
(a) PGA
(b) Starch
(c) Pyruvic acid
(d) Ribulose diphosphate.
Answer:
(a) PGA

Question 13.
Product of photophosphorylation is :
(a) NADP
(b) ADP from ATP
(c) ATP from ADP
(d) PGA.
Answer:
(c) ATP from ADP

Question 14.
COz acceptor of C₃ cycle is :
(a) RuDP
(b) PGA
(c) OAA
(d) PEPA.
Answer:
(a) RuDP

Question 15.
Kranze anatomy is found in the leaves of:
(a) C₂ plants
(b) C₃ plants
(c) C₄ plants
(d) Fleshy plants.
Answer:
(c) C₄ plants

Question 16.
CO2 acceptor of C₄ cycle is :
(a) PEP
(b) RuDP
(c) PGA
(d) OAA.
Answer:
(a) PEP

Question 17.
Solar energy converts into chemical energy in which process :
(a) Digestion
(b) Respiration
(c) Transpiration
(d) Photosynthesis.
Answer:
(d) Photosynthesis.

Question 18.
NADP is reduced to NADPH₂ during :
(a) Cyclic photophosphorylation
(b) Acyclic photophosphorÿlation
(c) Calvin’s cycle
(d) PS – I.
Answer:
(b) Acyclic photophosphorÿlation

Question 19.
In pigment system-il:
(a) Fixation of CO₂ occurs
(b) Reduction of CO₂ occurs
(c) Decomposition of water occurs
(d) All of these.
Answer:
(c) Decomposition of water occurs

Question 20.
Every green plant which do photosynthesis have:
(a) Chlorophyll-a
(b) Chlorophyl1b
(c) Chlorophyll-c
(d) Chlorophyll-d.
Answer:
(a) Chlorophyll-a

Question 21.
Photorespiration is the special quality of these plants:
(a) C₃ plants
(b) C₄ plants
(c) CAM plants
(d) None of these.
Answer:
(a) C₃ plants

Question 22.
in photorespiration the first process is:
(a) Carboxylation
(b) Decarboxylation
(c) Oxygenation
(d) Phosphorylation.
Answer:
(c) Oxygenation

Question23.
In photorespiration glycolate is changed into CO₂ and serin amino acid in:
(a) Chioroplast
(b) Peroxisorne
(c) both (a) and (b)
(d) Mitochondria.
Answer:
(d) Mitochondria.

Question 24.
The end product of the photosynthesis is:
(a) CO₂ and °2
(b) Carbohydrates
(c) CO₂ and carbohydrates
(d) CO₂ and starch.
Answer:
(b) Carbohydrates

Question 25.
The principle of limiting factor was proposed by:
(a) Blackman
(b) Arnon
(c) Living
(d) Hatch and Slack.
Answer:
(a) Blackman

2. Fill in the blanks:

1. In photorespiration the formation of glycine from glycolic acid is done in ……………. .
Answer:
Peroxisome,

2. Photorespiration is also called ……………. metabolism.
Answer:
Glycolate

3. The process of intake or synthesis of food material is called ……………. .
Answer:
Nutrition

4. Ribose diphosphate works in high concentration of ……………. as oxygenate.
Answer:
Oxygenase

5. Monatropa is a main ……………. plant.
Answer:
Saprophytic

6. In pigment system I ……………. photophosphorylation occurs.
Answer:
Cyclic

7. Products of light reaction are ……………. and ……………. .
Answer:
ATP, NADPH2

8. Chemical formula of chlorophyll-a is ……………. .
Answer:
C₅₅H₇₂O₃₅N₄Mg

9. Main pigment required for photosynthesis is ……………. .
Answer:
Chlorophyll

10. ……………. is reduced during photosynthesis.
Answer:
CO₂

3. Match the following:

(A)

Column ‘A’	Column ‘B’
1. Glycolate metabolism	(a) Green chlorophyll
2. Halophyte	(b) C4
3. Chlorophyll	(c) Photorespiration
4. Hatch and Slack	(d) Plants which need light
5. Photosynthesis	(e) Symbiont
6. Lichen	(f) Autotroph.

Answer:
1. (c) Photorespiration
2. (d) Plants which need light
3. (a) Green chlorophyll
4. (b) C₄
5. (f) Autotroph.
6. (e) Symbiont

(B)

Column ‘A’	Column ‘B’
1. PEP	(a) C3 cycle
2. Amon	(b) Red drop
3. Blackman	(c) C4 cycle
4. Calvin and Benson	(d) Photophosphorylation
5. Robert Emerson	(e) Principle of limiting factor.

Answer:
1. (c) C₄ cycle
2. (d) Photophosphorylation
3. (e) Principle of limiting factor.
4. (a) C₃ cycle
5. (b) Red drop

4. Answer in one word:

1. Where does light reaction occurs in the chloroplast?
Answer:
Grana

2. What is the main source of oxygen evolved during photosynthesis?
Answer:
Water

3. Write the names of end-products of light reaction.
Answer:
ATP and NADPH₂

4. Give chemical formula of chlorophyll.
Answer:
C₅₅H₇₂O₃₅N₄Mg

5. Name the element found in the middle of chlorophyll molecule.
Answer:
Magnesium

6. Name the raw materials of photosynthesis process.
Answer:
CO₂, water, chlorophyll, light

7. Which colour of light shows highest rate of photosynthesis?
Answer:
Red light

8. Where does photosynthesis process occur.
Answer:
Chloroplast

9. Name first product of the C₃ cycle.
Answer:
Phosphoglyceric acid (PGA)

10. Name photosynthetic pigments found in the bacteria.
Answer:
Bacteriochlorophyll, Bacterioviridin, carotenoids

11. Name one enzyme found in the cells of C₄ plants.
Answer:
Phosphoenol-pyruvate carboxylase

12. Write full name of CAM.
Answer:
Crassulacean acid metabolism.

Students get through the MP Board Class 11th Biology Important Questions Chapter 9 Biomolecules which are most likely to be asked in the exam.

MP Board Class 11th Biology Important Questions Chapter 9 Biomolecules

Biomolecules Class 11 Important Questions Very Short Answer Type

Question 1.
What is keratin?
Answer:
Keratin is a simple albuminoid protein and is the main constituent of hair, skin, nails, horns, feathers and wool. It is also known as scleroprotein. It is insoluble in neutral solvents but soluble in strong acids and bases.

Question 2.
What is chitin?
Answer:
Chitin is an important polysaccharide of invertebrates. It is found in the exoskeleton of insects and crustaceans and in the cell wall of fungi.

Question 3.
What do you understand by denaturation of proteins?
Answer:
The bonds of proteins are broken at higher temperature and thus structure and specific characters of proteins are changed. This phenomenon is called denaturation of protein.

Question 4.
What is nucleic acid?
Answer:
Nucleic acids are the polymer of nucleotides made up of C, H, O, N and P which controls the basic functions of the cell.

Question 5.
What are catalysts?
Answer:
Substance which can accelerate the rate of a chemical reaction are called as
catalysts.

Question 6.
What is enzyme? Explain.
Answer:
Enzymes are proteinic substances also called as biocatalysts, synthesized in living cells and initiate or accelerate metabolic reactions.

Question 7.
Define isoenzymes.
Answer:
Enzymes having different molecular configuration but having similar functions are known as isoenzymes, e.g., Lactic acid dehydrogenase (L.D.H.) which may occur in 5 possible forms in the blood sera and tissues of most vertebrates.

Question 8.
What is activation energy?
Answer:
The amount of energy required for the activation of a substance is called as activation energy. Enzymes lowering the energy of activation of the substrate so that biochemical reaction can take place at normal body temperature (at 37°C).

Question 9.
What is enzyme inhibition?
Answer:
The substance which reduces or inhibits the activity of enzyme are called enzyme inhibitor and the whole phenomenon is called as enzyme inhibition.

Question 10.
Give any two biological significance of enzymes.
Answer:

1. Enzymes play an active role in all rnetabolic and catabolic reactions and it is required for all the types of activities of the body.
2. Enzymes play an important role in biosynthesis, growth and repair of the body.

Question 11.
What are cofactors?
Answer:
A non-protein moiety attached to the protein part of enzyme is called as cofactor. If the cofactor is of inorganic nature like potassium, calcium, magnesium, manganese it is called prosthetic group. If cofactor attached to an enzyme protein is organic moiety like NAD, NADP or FAD it is known as coenzyme.

Question 12.
Define prosthetic group.
Answer:
The non-protein part of enzyme is called as cofactor. When the cofactor is of inorganic nature like K, Ca, Mg, Mn it is known as prosthetic group. Prosthetic group is generally tightly bounded to the protein part of enzyme.

Question 13.
Write differences between Apoenzyme and Hoioenzyme.
Answer:
All enzymes are proteins. However, there are some enzymes conjugated with a non-protein moiety (cofactor). Protein part of these conjugated enzymes are called as apoenzymes.
Enzymes with prosthetic group and apoenzyme are called as holoenzymes.

Question 14.
Write the names of any two polysaccharides and give their functions.
Answer:

1. Cellulose : It forms cell wall of plant cell. ,
2. Chitin : It forms exoskeleton of insects.

Question 15.
Write one difference in the following pair of words :

• Purineland pyrimidine,
• Starch and glycogen,
• Nucleoside and nucleotide,
• Fibrous and globular proteins.

Answer:

• Difference between purine and pyrimidine : Purine is two ringed heterocyclic nitrogenous base whereas pyrimidine is single ringed nitrogenous base.
• Difference between starch and glycogen : Starch is the storage food product of plants and glycogen of animals.
• Difference between nucleoside and nucleotide : Nucleotides are made up of nitrogenous base, pentose sugar and phosphate group whereas nucleosides are made up of only nitrogenous base and sugar.
• Difference between fibrous and globular proteins : Fibrous proteins are elongated molecules that may be constituted of several coiled peptide chains which are lightly linked whereas globular proteins are spherical or ovoid in shape due to folded peptide chains.

Question 16.
What do you understand by derived proteins?
Answer:
Derived proteins : Proteins which are derived from simple and conjugated proteins are called derived protein. They are not found in nature as such. They are obtained by the partial hydrolysis of natural proteins. These are formed by either change in the structure or breaking of their bonds, e.g., Peptones, protease, dipeptides, tri and tetrapeptides etc.

On complete hydrolysis of any protein the intermediate products are derived proteins whereas end-products are amino acids.

Question 17.
What is Dextrin? Give its importance.
Answer:
Dextrin is obtained by partial hydrolysis of starch, i.e., it is an intermediary substance of starch synthesis. It is used as sticking material in the cell.

Question 18.
What is agar-agar? Explain.
Answer:
Agar-agar is a polysaccharide, obtained from algae Gelidium, Gracilaria, which is mainly used in production of medium for culturing bacteria. It is neutral in nature. Iris- agar obtained from algae Chondrus crispes is used in preparation of chocolate, paint, toothpaste, ice-cream etc.

Question 19.
Write names of macromolecules of cell.
Answer:
Macromolecules of cell are:

• Polysaccharide,
• Protein,
• Nucleic acid etc.

Question 20.
Why are enzymes called biocatalysts?
Answer:
Enzymes are called as biocatalysts because they are the proteinous chemical substances which catalyse or accelerate rate of biochemical reactions occurs in the living body.

Question 21.
Give two important special characteristics of enzymes.
Answer:

1. Catalytic properties : Enzymes are biological catalysts. The small quantity of enzyme may catalyse metabolism of larger quantity of substances.
2. Specificity of enzymes : Enzymes are highly specific in nature, i.e., a particular enzyme can catalyse a particular reaction, e.g., Enzyme sucrase can catalyse only hydrolysis of sucrose.

Question 22.
What is Active site? Explain it.
Answer:
Part of the enzyme to which substrate molecule fit for biochemical reaction in the body of a living organism is called as active site. It is a three-dimensional structure to which substance fit. If geometry of active site changes then substrate fails to fit into the active site of enzyme. Actually active site is the binding site of the enzyme.

Biomolecules Class 11 Important Questions Short Answer Type

Question 1.
Explain any four factors affecting enzyme activity.
Answer:
The following four factors affect the activity of enzyme:
1. Temperature :
Metabolic reaction catalysed by enzymes increases the rate of activity at higher temperature because as temperature increases the kinetic energy of molecules also increases. But as the kinetic energy of molecules increases weak hydrogen bonds are more rapidly ruptured. The loss of hydrogen bonds modify the tertiary structure of enzymes and its catalytic activity, such alteration is called denaturation of enzyme.

2. Enzyme concentration:
As concentration of enzyme increases, the rate of enzyme activity also increases.

3. Substrate concentration:
In general, an increase in substrate concentration increases the rate of reaction but high concentration of the substrate, inhibits the activity of enzyme.

4. Hydrogen ion concentration (p^H) :
An enzyme have an optimum pH, but the changes in hydrogen ion concentration causes change in enzyme activity. Every enzyme requires a specific pH for the maximum activity, e.g., Trypsin in duodenum acts in alkaline medium (p^H = 8.5).

Question 2.
Explain exoenzymes and endoenzymes with suitable examples.
Answer:
Exoenzymes :
Enzymes which act outside the cells of their origin are called as exoenzymes. Enzymes secreted by the alimentary canal and used for digestion such as pepsin, trypsin, etc. are exoenzymes.

Endoenzymes :
Those enzymes which are produced inside the cell and participate in various reactions inside the cell are called intercellular or endoenzymes. They are active only inside the cells. Enzymes form mitochondria which takes part in cellular respiration.

Question 3.
What is the importance of enzymes?
Answer:
Importance of enzymes :

• Enzymes play an active role in all metabolic and catabolic reactions and it is required for every type of activities of the body,
• Enzymes catalyse and regulate the process of digestion, assimilation and absorption like chemical reactions,
• Enzymes play an important role in biosynthesis, growth and repair of the body,
• Enzymes regulate the energy release in respiration which is used in respiration, muscular, physical and mental activities.

Question 4.
Write down the similarities between enzymes and inorganic catalysts.
Answer:
Similarities between enzymes and inorganic catalysts :

• Enzymes and inorganic catalysts both of them remain unchanged at the end of the reaction chemically and quantitatively and can be used again to catalyse another reaction.
• Both are required in small quantity.
• They do not alter the equilibrium of the reversible reaction.
• Both of them increase the rate of reaction by lowering the activation energy but they do not initiate the reaction.
• The complexes formed with the reactants by both of them are short lived.

Question 5.
What is peptide bond?
Answer:
The bond formed between the carboxylic group (- COOH) of one amino acid and amino group (- NH₂) of another amino acid is called as peptide bond. A molecule of water is released during the formation of peptide bond.

Each peptide chain of considerable length may possess 50 to millions of amino acid units. A peptide may be dipeptide (with 2 amino acid units), a tripeptide (with 3 units) and so on. Beyond 10 amino acid units, a peptide is called polypeptide.

Question 6.
What is deoxyribonucleoside? How it is converted into nucleotides?
Answer:
The substances which are formed from deoxyribose sugar and nitrogenous bases are called deoxyribonucleosides on the basis of nitrogenous bases they are of four types.
Nucleosides can be converted into nucleotides after their reaction with phosphoric acid. After condensation they form nucleic acids.

Question 7.
Name the constituents of the nucleotides found in DNA.
Answer:
The following four types of nucleotides are found in DNA :

Nucleotide	Constituents
1. Deoxyadenylic acid.	Deoxyribose sugar, adenine and phosphoric acid.
2. Deoxyguanylic acid.	Deoxyribose sugar, guanine and phosphoric acid.
3. Deoxycytidylic acid.	Deoxyribose sugar, cytosine and phosphoric acid.
4. Deoxythymidylic acid.	Deoxyribose sugar, thymine and phosphoric acid.

Question 8.
Write down the function of proteins.
Answer:
Function of proteins :

• Proteins form various structures of our body such as biomembranes, organelles (ribosome), hair and nails.
  Skin, horns, wool (all contain keratin), cartilage (contains collagen) and bones (contain ossein protein).
• Proteins function as biocatalyst (enzyme) in our body, e.g., Amylase and pepsin.
• The important cellular structures like cell membranes, connective tissue have proteins in their structure in which proteins functioning as a carrier molecule of different elements and compounds.
• Contractile proteins (e.g., Actin and myosin) participate in cellular movements and locomotion.

Question 9.
Write any four differences between DNA and RNA.
Or,
Write five differences between DNA and RNA on the basis of molecular composition and function.
Answer:
Differences between DNA and RNA

DNA	RNA
1. It contains deoxyribose sugar.	It contains ribose sugar.
2. It has adenine, thymine, cytosine and guanine as nitrogenous bases.	It has adenine, uracil, guanine and cytosine as nitrogenous bases.
It has adenine, uracil, guanine and cytosine as nitrogenous bases.	It consists of single polynucleotide chains, which may get folded on itself to form double helix.
3. It consists of two polynucleotide chains, coiled into a double helix.	It is main constituent of ribosome and generally found in cytoplasm.
4 It is main constituent of chromosome which is found in nucleus.	Generally RNA participates in the synthesis of protein.

Question 10.
Define proteins and write their uses.
Answer:
Proteins :
Proteins are the polymer of amino acids. It is complex nitrogenous compound of the protoplasm. About 20 amino acids are joined together by means of peptide bonds form a protein molecule. About 20% part, the body of organisms is made up of proteins.
Uses of proteins :
Tertiary structure: Linear arrangement of amino acids through peptide linkage to form a polypeptide chain is called as Primary structure of protein. Folding of a linear polypeptide chain into specific coiled structure is referred as Secondary structure of protein.
1. Insulin : It is used for treatment of Diabetes mellitus.

2. Immunoglobulin : Immunoglobulin of the blood plasma in mammals and other animals act as antibodies which neutralize the harmful effect of foreign agents like viruses, bacterias etc.

3. Albumin : Albumin protein of egg (egg – white) is used for hair care.

4. Structural components of cells : Some proteins are essential structural compo-nents of cell membranes, organelles, cytoplasm, extracellular material and fibres. Keratin is the main constituent of hair, skin, nails, horns, feathers and wool. Cartilage is made of collagen. In plants protein is found in the walls of pollen grains.

5. Enzymes (Biocatalysts): The most important feature of proteins is their ability to function within the living cell as reaction catalysing enzymes. Enzymes usually increase the reaction rate by manifold. Enzymes play a key role in the metabolism.

6. Hormones : Proteins serve as hormones also. Hormones from pancreatic islets of Langerhans, pituitary, parathyroid and gastrointestinal mucosa are of peptide nature. Hormones of thyroid and adrenal medulla are derivatives of amino acid tyrosine.

7. As carriers : Haemoglobin, the respiratory pigment of many animals is a conju-gated protein composed of colourless basic proteins, the globin and haem. It has unique ability to bind oxygen in a loose and easily reversible combination. Myoglobin transports oxygen in muscles. In plants p-protein is involved in the transport of organic compounds through phloem.

8. Growth and repair : Proteins play a key role in general body growth and in the repair of wear and tear of the cells and body as a whole.

9. Formation of rhodopsin : The visual purple rhodopsin is made up of retinene an aldehyde derivative of vitamin-A and a protein opsin.

10. Melanin synthesis : Melanin, the skin pigment is derived from the tyrosine.

11. Formation of urea : In ureotelic animals, ornithine, citrulline and arginine (all amino acids) take part in the formation of urea by a cyclic process.

12. In some organisms protein is stored food, such as Albumin of egg, yolk of egg, in wheat as Glutein etc.

13. Fibroin protein secreted by silkworm produces silk on drying.

14. It may release energy by decomposing when required.

Question 11.
Draw a well labelled diagram of Watson and Crick model of DNA.
Answer:

Question 12.
Write short note on structural polysaccharides.
Answer:
Structural Polysaccharides : These polysaccharides take part in forming the structural framework of the cell wall in plants and skeleton of animals. They are of two main types:
1. Cellulose :
Cellulose is an important constituent of cell walls of plant, which provides mechanical support to plant cell. It is polymer of glucose and made up of unbranched chain of approximately 6,000 β-D-glucose units linked by 1,4-glycosidic linkage. It consti¬tutes 20-40% of cell wall.

2. Chitin :
Chitin is an important heteropolysaccharide of invertebrates. It is found in the hard exoskeleton of insects and crustaceans and in the cell wall of fungi. It is polyglycol consisting of N-acetyl-D-glucosamine units connected throughβ-1, 4-glycosidic linkage. Although chitin is soft and leathery it becomes hard when calcium carbonate or certain proteins are deposited in it. The insolubility of structural polysaccharides in water helps to retain the form and to strengthen the structure of the organism. In fungal cell wall chitin is often called as fungus cellulose.

Question 13.
Explain DNA replication.
Or,
Explain DNA duplication in short.
Answer:
Watson and Crick after giving the double helix model of DNA, also postulated the mechanism of DNA duplication, also known as replication. According to them, during duplication, the weak hydrogen bonds between the nitrogenous base of the nucleotides get separated, so that two polynucleotide chains of DNA also separate and uncoil. The chains thus, separated are complementary to one another.

These strands act as template and because of the specificity of base pairing each nucleotide of separated chain attracts its complementary nucleotide from the cell cytoplasm. Once the nucleotides are attached by their hydrogen bonds their sugar radicals write through their phosphate components completing the formation of a new polynucleotide chain. This results in the formation of two double helixes of DNA where in each molecule has one old strand contributed by parent DNA and one synthesized new. This method of DNA duplication is known as semi-conservative method.

Question 14.
Describe the functions of nucleic acids.
Or,
Explain the utility of nucleic acids.
Answer:
Utility of nucleic acids:

• Nucleic acids are the hereditary materials of organisms which involve in the transfer of hereditary characters from one generation to the next.
• DNA controls the synthesis of enzymes which control the various activities of the body.
• Nucleic acids also control protein synthesis,
• Nucleic acids form maximum portion of chromatin network,
• It causes mutation in living beings,
• They form enzymes.

Question 15.
Elaborate the term RNA. Also describe the types and functions of RNA.
Or,
Write location and kinds of RNA in the cell.
Answer:
RNA : Ribonucleic acid (RNA) located in the nucleus, cytoplasm, ribosomes and in some other cell organelles.
Types of RNA and their functions : RNA are of three types :

1. Messenger RNA (mRNA):
It makes a small fraction 5-10%. This RNA directs the sequence of amino acids in protein synthesis after joining with ribosomes. It carries the genetic information contained in DNA. It is short lived and has rapid turn over. It is formed of 700-1500 nucleotides and has a molecular weight from 5,00,000 to 20,00,000. The sequence of three nitrogenous bases of tnRNA forms a codon which is responsible for coding of one amino acid.

2. Ribosomal RNA (rRNA) :
It makes 80% of total cell RNA. It is the most stable type of RNA and is associated with ribosomes.

3. Transfer RNA (tRNA) :
It makes a small fraction (10-15%) of RNA. These are smallest molecules formed of 73-93 nucleotides with molecular weight ranging between 25,000 to 30,000. tRNA works as adaptor molecules for carrying amino acids to the TMRNA template during protein synthesis.

Question 16.
Write names of the nucleosides which help in energy transfer.
Answer:

1. ATP: Adenosine triphosphate,
2. ADP: Adenosine diphosphate,
3. GTP : Guanosine triphosphate,
4. GDP : Guanosine diphosphate,

Question 17.
Describe structure of RNA.
Answer:
Ribonucleic acid (RNA) :
These are second type of nucleic acids which are found in the nucleus, cytoplasm, ribosomes and in some other cell organelles.
RNA occurs normally as long unbranched polymeric molecule in the form of a single chain. Generally non-genetic RNA of eukaryotic or prokaryotic cells have single strand except fRNA which has double strand but without helical structure. But the wound tumour virus (a plant virus) and Rheo virus (animal virus) have double strand in their RNA molecule.

Chemical composition of RNA :
RNA molecules are polymers of ribonucleotides. Each ribonucleotide is made up of three different molecules namely a ribose pentose sugar instead of deoxyribose sugar as in DNA, a phosphoric acid and a nitrogenous base.
Two types of nitrogenous bases are found in RNA :

1. Purines : Adenine (A) and guanine (G).
2. Pyrimidines : Uracil (U) and cytosine (C). Thus, uracil substitutes thymine of DNA. In RNA above four nitrogenous bases join with the ribose sugar to form ribonucleoside.

Nucleosides of RNA : There are four types of ribonucleosides are present in RNA :

1. A + Ribose sugar = Riboadenosine
2. G + Ribose sugar = Riboguanosine
3. C + Ribose sugar = Ribocytidine ‘
4. U + Ribose sugar = Ribouridine.

Nucleotides of RNA : Phosphoric acid joins with each ribonucleoside to form ribonucleotide.

• A + Ribose sugar + Phosphoric acid = Riboadenylic acid
• G + Ribose sugar + Phosphoric acid = Riboguanylic acid
• C + Ribose sugar + Phosphoric acid = Ribocytidylic acid
• U + Ribose sugar + Phosphoric acid = Ribouridylic acid.
  These four ribonucleotides are linked together to form a long strand of ribopolynucleotide.

Question 18.
What is enzyme? Explain the characteristic features of enzymes.
Answer:
Enzymes are proteinic substances also called as biocatalysts synthesized in living cells and initiate or accelerate metabolic reactions by lowering activation energy.
Characteristic features of enzymes :
1. Physical properties :
Physically enzymes behave as colloids or as substance of high molecular mass. They are destroyed or inactivated at temperature below the boiling point of water. At 50°C most enzymes in a liquid medium are inactivated. Dried enzyme extract can endure temperature of 100°C to 120°C or even higher. Thus, enzymes are thermolabile.

2. Chemical properties:
(i) Catalytic properties :
Enzymes are biological catalyst. Catalysts are substances which accelerate the speed of chemical reaction without undergoing any permanent changes. The small quantity of enzyme catalyses the larger quantities of substrates.

(ii) Specificity of enzymes :
Enzymes are highly specific in nature, i.e., a particular enzyme can catalyse only a particular reaction at a particular temperature, e.g., Enzyme sucrase, can catalyse only hydrolysis of sucrose.

3. General properties:

• They initiate and accelerate the rate of biochemical reaction.
• The activity of enzymes depends upon the acidity of the medium (p^H specific). Each enzyme is most active at a particular acidity (p^H).
• Enzyme can accelerate the reaction in either directions :

• All enzymes possess active sites which participate in the biochemical reactions.
• They are very unstable compounds mostly soluble in water, dilute glycerol, NaCl and dilute alcohol.
• They act actively at optimum temperature.
• All enzymes are proteins but all proteins are not enzymes.

Question 19.
Give the composition of Triglyceride.
Answer:
Triglycerides are a group of stored lipid which include oil and fats. Chemically, lipids are the esters or glycerides of fatty acid and glycerol. It is made up of three molecules of fatty acid and one molecule of glycerol. Their general formula is CH₃(CH₂)_n„COOH. The carboxylic group (- COOH) of each fatty acid react with alcoholic group (-OH) of glycerol and form ester linkage.

Question 20.
Write short note on Globular protein.
Answer:
Globular protein:
These are spherical or oval in shape because of the presence of peptide folded or coiled chain which forms compact structure, it is called as globular protein. It forms a three-dimensional structure, which is formed by relatively weak non-covalent bond. It is a secondary structure of protein, soluble in water. These include enzymes, protein, hormones like ACTH, oxytocin, glucagon and insulin etc.

Question 21.
Of which substance gum is made up? Is fevicol different than this?
Answer:
Gum is a carbohydrate, such as D-Galactose, D-Galactonic acid, which are natural substances, whereas fevicol is an artificial product which is used as gum.

Question 22.
Can you explain process of conversion of milk into curd or yogurt on the basis of concept of protein?
Answer:
Protein present in the milk is casein, which is a white coloured substance and has nutritional value. It consists of essential amino-acids required for the growth of human body. Conversion of milk into curd is a chemical process. Lactobacillus bacteria ferment lactose sugar of milk into lactic acid and help to settle casein. Settled casein is called as curd or yogurt.

Question 23.
Can you prepare biomolecule model using commercial model of atom (Ball and stick model)?
Answer:
Yes, we can prepare biomolecule model using commercial model of atom.

Question 24.
Which functional groups are released on titration of amino acid with weak base?
Answer:
On titration of amino acid with weak base two functional groups are released :

• – COOH group (Carboxylic group)
• – NH₂ group (Amino group)

Question 25.
Give structural formula of amino acid Alanine.
Answer:

Question 26.
Write short note on conjugated protein and give two examples of it.
Answer:
Conjugated Proteins : These are complex protein molecules, which is formed by combination of simple protein molecule with characteristics non-protein substance called as prosthetic group, i. e., on hydrolysis conjugated protein yield a protein and a non-protein substance.
Protein + Prosthetic group → Conjugated protein
Conjugated proteins are of following types :

1. Nucleoprotein,
2. Mucoprotein,
3. Glycoprotein,
4. Chromoprotein,
5. Lipoprotein,
6. Metalloprotein and,
7. Phosphoprotein.

e.g., Haemoglobin, Lipoprotein
Globin + Haem → Haemoglobin
Protein + lipid → Lipoprotein.

Biomolecules Class 11 Important Questions Long Answer Type

Question 1.
If a given protein has primary structure, in which amino acids are present at both the ends, then can you connect this information with purity or homogeneity of protein?
Answer:
The primary structure of protein refers to the linear sequence of number of amino acids belonging to the polypeptide chain. The amino acids link together by peptide bond only. It is formed by binding of amino group (-NH₂) of one amino acid with carboxylic group (-COOH) of another amino acid. Therefore, a single polypeptide chain has first amino acid at one end and the last amino acid at the other end. Amino group of first amino acid is called as ‘N’ terminal and carboxylic group of last amino acid is called as ‘C’ terminal.
By above information purity or homogeneity of a protein cannot be determined because many other amino acids (50 to millions) are found in between them.

Question 2.
Find out the proteins used as therapeutic agents and make a list. Give other uses of protein (as cosmetics) etc.
Answer:
1. Insulin : It is used for treatment of Diabetes mellitus.

2. Immunoglobulin : Immunoglobulin of the blood plasma in mammals and other animals act as antibodies which neutralize the harmful effect of foreign agents like viruses, bacterias etc.

3. Albumin : Albumin protein of egg (egg – white) is used for hair care.

4. Structural components of cells : Some proteins are essential structural compo-nents of cell membranes, organelles, cytoplasm, extracellular material and fibres. Keratin is the main constituent of hair, skin, nails, horns, feathers and wool. Cartilage is made of collagen. In plants protein is found in the walls of pollen grains.

5. Enzymes (Biocatalysts): The most important feature of proteins is their ability to function within the living cell as reaction catalysing enzymes. Enzymes usually increase the reaction rate by manifold. Enzymes play a key role in the metabolism.

6. Hormones : Proteins serve as hormones also. Hormones from pancreatic islets of Langerhans, pituitary, parathyroid and gastrointestinal mucosa are of peptide nature. Hormones of thyroid and adrenal medulla are derivatives of amino acid tyrosine.

7. As carriers : Haemoglobin, the respiratory pigment of many animals is a conju-gated protein composed of colourless basic proteins, the globin and haem. It has unique ability to bind oxygen in a loose and easily reversible combination. Myoglobin transports oxygen in muscles. In plants p-protein is involved in the transport of organic compounds through phloem.

8. Growth and repair : Proteins play a key role in general body growth and in the repair of wear and tear of the cells and body as a whole.

9. Formation of rhodopsin : The visual purple rhodopsin is made up of retinene an aldehyde derivative of vitamin-A and a protein opsin.

10. Melanin synthesis : Melanin, the skin pigment is derived from the tyrosine.

11. Formation of urea : In ureotelic animals, ornithine, citrulline and arginine (all amino acids) take part in the formation of urea by a cyclic process.

12. In some organisms protein is stored food, such as Albumin of egg, yolk of egg, in wheat as Glutein etc.

13. Fibroin protein secreted by silkworm produces silk on drying.

14. It may release energy by decomposing when required.

Question 3.
Describe Glycosidic, Peptide and Phosphodiester bonds.
Answer:
1. Glycosidic bond :
During the union of monosaccharide units water molecule is eliminated and the units are linked through an oxygen bridge known as glycoside linkage. Depending upon the steric configuration at carbon-1 of monosaccharide unit the bond is called α and β-bond.

By this bond monosaccharides are converted into oligo and polysaccharides. On hydrolysis they yield monosaccharides.

2. Peptide bond :
The bond formed between the carboxylic group (- COOH) of one amino acid and amino group (- NH₂) of another amino acid is called as peptide bond. A molecule of water is released during the formation of peptide bond.

Each peptide chain of considerable length may possess 50 to millions of amino acid units. A peptide may be dipeptide (with 2 amino acid units), a tripeptide (with 3 units) and so on. Beyond 10 amino acid units, a peptide is called polypeptide.

3. Phosphodiester bond :
Nucleic acid, like DNA molecules are polymers of deoxyribonucleotides. Each deoxyribonucleotides is made up of three different molecules :
(i) a decxyribose sugar (Pentose sugar),
(ii) a phosphoric acid and
(iii) a nitrogenous base. Nitrogenous base may be of two types :
1. Purines : Adenine (A) and Guanine (G).
2. Pyrimidines : Cytosine (C) and Thymine (T).
3′ carbon of one nucleotide combine with 5′ carbon of next (adjacent) nucleotide by phosphate group. Ester bond is formed between phosphate of sugar and hydroxyl group. Ester bond is formed on both the sides of phosphate group, thus it is called as Phosphodiester bond.

Nucleotides unite with each other by phosphodiester bond to form long chain of polynucleotide of DNA or RNA, but in RNA instead of thymine, uracil is found.

Question 4.
What do you mean by Tertiary structure of protein?
Answer:

Tertiary structure: Linear arrangement of amino acids through peptide linkage to form a polypeptide chain is called as Primary structure of protein. Folding of a linear polypeptide chain into specific coiled structure is referred as Secondary structure of protein.
The arrangement and interrelationship of twisted chains of proteins into specific loops and bends are called the tertiary structure of protein.
Tertiary structure of protein is maintained by four kinds of bonds :

• Hydrogen bonds,
• Hydrophobic bonds,
• Ionic or electrostatic bonds and
• Disulphide bonds.

Question 5.
Write structural formula of 10 micro biomolecules of low molecular weight. Find out the industries involved in their production. Who are the customer of them? Find out.
Answer:
Structural formula of low molecular weight micro-biomolecules are as follows: (1) Monosaccharides :

(2) Amino acids:

(3) Oil and fat:

(4) Nitrogenous bases:

(5) Nucleoside and Nucleotide

Question 6.
Give analytical test of protein, fat, oil and sugar and test presence of them
in any fruit juice, saliva, sweat and urine sample. (NCERT)
Answer:
(1) Test for Protein in sample of urine

Experiment	Observation	Inference
1. Take 5ml of Conc. HNO3 in a test tube Now slowly drop the urine sample to it with the help of a dropper by bending the test tube, so that slowly urine comes in contact of acid.	At the region of contact of urine white precipitate is formed.	Protein present.
2. The solution containing white ppt. isheated.	Colour of the ppt. becomes yellow.	Protein confirmed.

(2) Test for presence of Oil and Fat in food sample

Experiment	Observation	Inference
1. Put a drop of fat or oil at the centre of a paper. Spread it slowly with the help of finger.	Paper appears translucent.	Presence of oil or fat in food item confirmed.

(3) Test for presence of Starch in food sample

Experiment	Observation	Inference
1. Take sample of food, grind it and dissolve in water to prepare a solution. Take 5ml of sample solution in a test tube, add freshly prepared iodine solution to it.	Solution turn into blue black colour.	Presence of starch in food sample is confirmed.

(4) Presence of Sugar in sample of fruit juice .

Experiment	Observation	Inference
1.Take 5ml of Benedict’s solution in a test tube. Add 8 drops of fruit juice sample in it. Heat the mixture for 2 minutes and cool it.	Green or yellow or orange or red coloured precipi tate is obtained.	Red colour indicates presence of more quantity of sugar and orange colour shows comparatively less sugar whereas yellow or green colour shows presence of less quantity of sugar.

Question 7.
What do you mean by cellular macromolecules? Write the name of different macromolecules and describe any one in detail.
Answer:
Cellular macromolecules : The molecules formed by the process of polymerization are called as macromolecules. Macromolecules of the cell possess the following characteristic features:

• Chemically they are larger in size,
• Their molecular weight is very high,
• Their solubility is very low.
• Their molecular structure may be simple branched or unbranched and coiled,
• They are the polymers of micromolecules, e.g., (I) Polysaccharides, (II) Proteins and (III) Nucleic acids.

Polysaccharides :
They are polymers of monosaccharides and is made up of 10 to several thousand units of monosaccharides joined by glycosidic linkage. After hydrolysis they produce monosaccharides. Their empirical formula is (C₆H₁₀O₅)_n where n ranges from 10 to 10,000. Their molecular weight is very high.
Polysaccharides are of following two types :
1. Homopolysaccharides : These polysaccharides are made up of one kind of monosaccharides as in starch, cellulose, glycogen and insulin.
2. Heteropolysaccharides : These polysaccharides are made up of two or more kinds of monosaccharides as in agar, pectin and chitin.
On the basis of their functions polysaccharides are classified into following two groups:
(a) Storage polysaccharides and
(b) Structural polysaccharides.

(a) Storage polysaccharides : Polysaccharides which are synthesized by the process of photosynthesis and are stored in the form of food materials in plants and animals are called as storage polysaccharides, e.g., Starch, glycogen, gum, insulin, dextrin.
(i) Starch :
It is the storage carbohydrate of plants. It is formed during the process of photosynthesis and serves as an energy storing material. Starch is a polymer of glucose unit linked together in an α-1,4-glycosidic linkage. It yields only glucose on hydrolysis. In fact it is formed by condensation of amylase and amylopectin. Starch grains may be oval, spherical, lens shaped or irregular.

(ii) Glycogen :
Glycogen is the storage polysaccharide of animals and is called as animal starch. It is a branched polymer of glucose. It is stored mostly in muscles and liver of animals. The glycogen of liver supplies glucose to all tissue through blood. It is also found as storage product in blue-green algae, slime, moulds, fungi and bacteria. Their general formula is (C₆H₁₀O₅)_n)„. It consists of α-D glucose units mostly linked by 1, 4 but highly branched via 1,6-linkage each of which starts other α-1,4 chain of some 8-T 2 glucose units.

(b) Structural polysaccharides :
Polysaccharides which take part in the structural organization and protection mechanism of the body are called structural polysaccharides, e.g., Cellulose, chitin, etc.

Utility of polysaccharides :

• Polysaccharides serves as an important structural component in some animals and in all plants where they constitute cellulose framework.
• They are stored as storage food material and whenever required these can be hydrolyzed to yield energy.
• Cellulose forms protective covering as cell wall in plant cells. In some animals chitin forms protective exoskeleton.
• Mucilage acts as binding substance of cell walls and connective tissue.
  Polysaccharides are also used in textile, furniture, printing, varnish, alcohol and film
  industries.

Question 8.
How many kinds of nucleic acids are found in the cell? Give a detailed account of DNA.
Or,
Explain the structure of DNA.
Answer:
Every type of plant and animal cell (except virus) generally possess following two kinds of nucleic acids :
(i) DNA (Deoxyribonucleic acid),
(ii) RNA (Ribonucleic acid).
Where viruses possess only one type of nucleic acid from the above.
Structure of DNA : Structurally DNA is the chain of polydeoxyribonucleotides. It means that DNA is formed by the polymerization of deoxyribonucleotides. Every DNA molecule possesses the following four kinds of nucleotides :

• Deoxyadenylic acid,
• Deoxyguanylic acid,
• Deoxycytidylic acid,
• Deoxythymidyhc acid.

Watson and Crick (1953) suggested a ‘double helix’ model of DNA. According to them, the DNA molecule consists of two long polynucleotide helically twisted strands which are complementary running in opposite direction (3′ – 5′, 5′ – 3 ‘) and connected together by steps.

The vertical bars are formed by alternate phosphate and sugar groups joined by phosphodiester bonds. These strands are formed together by hydrogen bonds which are established between specific nitrogen base in the sequence A = T, T = A, C ≡ G and G ≡ C. The length of one complete turn is 34Å. There are 10 nitrogenous base pairs in each turn. The distance between two nucleotides is 34 Å. The width of DNA is 20Å. Each successive nucleotide turns 36 degree in the horizontal plane. The twisting of strands results in the formation of deep and shallow spiral groove.

Significance of DNA:

• DNA is the genetic material which transmits parental characters from generation to generation,
• DNA together with protein form chromosomes,
• It is the constituent of ribosomes, mitochondria and plastids.
• It is responsible for mutation and organic evolution,
• It synthesizes enzymes required for various biological activities.

Question 9.
Describe important properties of enzyme.
Answer:
Important properties of enzymes are as follows :
(i) Catalytic properties :
Enzymes are biological catalyst. Catalysts are substances which accelerate the speed of chemical reaction without undergoing any permanent changes. The small quantity of enzyme catalyses the larger quantities of substrates,

(ii) Specificity of enzymes :
Enzymes are highly specific in nature, i.e., a particular enzyme can catalyse only a particular reaction at a particular temperature, e.g., Enzyme sucrase, can catalyse only hydrolysis of sucrose.

(iii) Colloidal state :
Physically enzymes behave as colloids or as substances of high molecular mass. They are hydrophilic in nature. They are amphoteric in nature. Thus, behave as alkali with acid and as acid with alkali.

(iv) Reversible in action : Enzymes can accelerate the reaction in either directions.
e.g., Sucrase enzyme hydrolyse sucrose into glucose and fructose. It also catalyse backward reaction.

(v) Sensitivity :
Each enzyme is catalytically active within a limited range of pH. Most enzymes operate effectively in a range of about 5 to 9 pH. For example, the optimal range for the activity of malt amylase is 5-2 pH, for salivary amylase between 6-7 to 6-8 pH, for trypsin 8 to 11 pH. In all these cases the activity gradually declines on either side of the optimum pH. Most intracellular enzymes function best at neutral pH 7.

(vi) High molecular weight:
Enzymes are the proteins of high molecular weight, e.g., molecular weight of feredoxin of bacteria is 6000 and pyruvate dehydrogenase is 46,00,000.

(vii) High reactivity :
Small quantity of enzyme may catalyse reaction at fast rate. Total number of substrate molecule which may convert into product in one minute by one enzyme is called as Turnover number. Turnover number of different enzymes are as follows:

• Carbonic anhydrase : 36 million,
• Catalase : 5 million,
• Sucrase or Invertase : 10,000,
• Flavo protein : 90.

Question 10.
Explain biological significance of enzymes.
Answer:
Following are the biological importance of enzymes :
1. Healing of wound :
A protein digesting enzyme obtained from pancreas of the pig is very useful for skin diseases and healing of wound. These enzymes destroys protein digesting enzymes of human body thus help for healing of wound.

2. Dehairing of hides :
Enzymes obtained from pancreas destroys hair follicles of hair, thus help for removal of hair from hide.

3. Dissolving the blood clot:
An enzyme called as urokinase obtained from urea helps to dissolve blood clot which are formed in the brain or arteries, thus saves our life.

4. Changing blood group :
Prof. Ken Furunkawa in 1981 explained the blood group is determined by the presence of sugar found in R.B.Cs. for which specific enzymes are found. If these enzymes are destroyed, then blood group ‘A’ and ‘B’ can change into blood group‘O’.

5. Analysis of biochemicals :
Some enzymes are used for quantitative analysis of some specific substances found in blood. For example, uricase, urease and sucrase enzymes are used to determine quantity of uric acid, urea and sucrose present in the blood respectively.

6. Cheese making :
Animal renin (milk protein digesting enzyme) is used in cheese making.
7. Manufacture of fruit juice : Some enzymes are used for processing of some fruit juices. Pectic enzymes are added to these fruit juices to make it clean and clear.

Biomolecules Class 11 Important Questions Objective Type

1. Choose the correct answers:
Question 1.
Lipids contain carbohydrates and N₂ are called :
(a) Phospholipids
(b) Chromolipids
(c) Aminolipids
(d) Glycolipids.
Answer:
(d) Glycolipids.

Question 2.
Lecithin is :
(a) Phospholipids
(b) Chromolipids
(c) Aminolipids
(d) Glycolipids.
Answer:
(a) Phospholipids

Question 3.
Fats and oils are decomposed by bases and produce :
(a) Soap
(b) Vegetable ghee
(c) Saturated fat
(d) Unsaturated fat.
Answer:
(a) Soap

Question 4.
Compounds with free amino and carboxylic group are called :
(a) Glucose
(b) Nucleotide
(c) Amino acid
(d) None of these.
Answer:
(c) Amino acid

Question 5.
Nucleotide responsible for energy transfer is :
(a) NAD
(b) FAD
(c) FMN
(d) ATP.
Answer:
(d) ATP.

Question 6.
Unit of protein is :
(a) Fatty acids
(b) Monosaccharides
(c) Amino acids
(d) Glycerol.
Answer:
(c) Amino acids

Question 7.
Nucleic acids are the polymers of:
(a) Amino acids
(b) Nucleosides
(c) Nucleotides
(d) Globulin.
Answer:
(c) Nucleotides

Question 8.
Peptide bonds are found in:
(a) Proteins
(b) Fats
(c) Nucleic acids
(d) Carbohydrates.
Answer:
(a) Proteins

Question 9.
Glycosidic bonds are found in:
(a) Nucleic acids
(b) Proteins
(c) Polysaccharides
(d) Monosaccharides.
Answer:
(c) Polysaccharides

Question 10.
A nucleoside differs from nucleotide ¡s not having:
(a) Sugar
(b) Nitrogen base
(c) Phosphate
(d) Both (a) and (e).
Answer:
(c) Phosphate

Question 11.
A unit composed of sugar and nitrogen base linked by glycosidic bond is:
(a) Purine
(b) Glycoside
(c) Nucleoside
(d) Nucleotide.
Answer:
(c) Nucleoside

Question 12.
Building block of nucleic acid is:
(a) Amino acid
(b) Nucleoprotein
(c) Nucleotide
(d) Nucleoside.
Answer:
(c) Nucleotide

Question 13.
Lactose molecule ¡s composed of:
(a) Glucose + Fructose
(b) Glucose + Glucose
(c) Glucose + Galactose
(d) Fructose + Fructose.
Answer:
(c) Glucose + Galactose

Question 14.
Heredity is controlled by:
(a) DNA
(b) RNA
(c) Generally DNA and sometimes RNA in few organisms
(d) None of these.
Answer:
(c) Generally DNA and sometimes RNA in few organisms

Question 15.
DNA is made up of:
(a) Deoxyribonucleotide
(b) Ribonucleotide
(c) Both (a) and (b)
(d) Polysaccharides.
Answer:
(a) Deoxyribonucleotide

Question 16.
Which base is absent in RNA:
(a) Adenine
(b) Guanine
(c) Thymine
(d) Uracil.
Answer:
(c) Thymine

Question 17.
Protein controlling the chemical reactions of the body is:
(a) Enzyme
(b) Vitamin
(e) Hormone
(d) Globulin.
Answer:
(a) Enzyme

Question 18.
Snake poison is a type of protein:
(a) Fibrous protein
(b) Globular protein
(c) Lipoprotein
(d) None of these.
Answer:
(b) Globular protein

Question 19.
Nucleic acids were discovered by:
(a) Watson
(b) Crick
(c) Miescher
(d) Beadle.
Answer:
(c) Miescher

Question 20.
Wood and cotton having higher amount of:
(a) Cellulose
(b) Cutin
(c) Chitin
(d) All of these.
Answer:
(a) Cellulose

Question 21.
Insulin is a type of:
(a) Pigment
(b) Food
(c) Protein
(d) Gas.
Answer:
(c) Protein

Question 22.
Unit of polysaccharide is:
(a) Monosaccharides
(b) Oligosaccharides
(o) Peptides
(d) Fatty acids.
Answer:
(a) Monosaccharides

Question 23.
Doctor generally prescribing to take oil in place of fats because:
(a) Oil contains unsaturated fatty acids
(b) It reduces the amount of cholesterol in blood
(c) They are quickly digested
(d) All of these.
Answer:
(d) All of these.

Question 24.
Double stranded model of DNA was proposed by:
(a) Watson and Crick
(b) Miescher
(c) Muller
(d) None of these.
Answer:
(a) Watson and Crick

Question 25.
How RNA is synthesized:
(a) By DNA
(b) By RNA
(c) By ribosomes
(d) By lysosomes.
Answer:
(a) By DNA

Question 26.
Energy providing substance in our body is:
(a) Carbohydrate
(b) Protein
(c) Fat
(d) Vitamin.
Answer:
(a) Carbohydrate

Question 27.
Heparin of the blood is a:
(a) Glycoprotein
(b) Nucleoprotein
(c) Chromoprotein
(d) Glycolipid.
Answer:
(a) Glycoprotein

Question 28.
Term protein was coined by:
(a) Camilo Golgi
(b) Bergellius
(c) Bloor
(d) Funk.
Answer:
(b) Bergellius

Question 29
Mucin of saliva is:
(a) Nucleoprotein
(b) Glycoprotein
(c) Lipoprotein
(d) Chromoprotein.
Answer:
(b) Glycoprotein

Question 30.
The unit formed by β-glycosidic linkage between sugar and base is called as:
(a) Nucleotide
(b) Nucleoside
(c) Glycoside
(d) Punne.
Answer:
(b) Nucleoside

Question 31.
mRNAispolymerof:
(a) Deoxyribonucleoside
(b) Ribonucleoside
(c) Deoxyribonuc1eotk
(d) Ribonucleotide.
Answer:
(d) Ribonucleotide.

Question 32.
Purine of RNA are:
(a) Guanine and adenine
(b) Uracil and thymine
(c) Adenine and cytosine
(d) Uracil and guanine.
Answer:
(a) Guanine and adenine

Question 33.
Enzyme which convert the starch into sugar is:
(a) Hydrolases
(b) Lipases
(c) Amylases
(d) Nucleases.
Answer:
(c) Amylases

Question 34.
Protein part 01 enzyme is called:
(a) Apoenzyme
(b) Coenzyme
(c) Holoenzyme
(d) Cofactor.
Answer:
(a) Apoenzyme

Question 35.
Peptide bonds are formed between:
(a) Amino acids
(b) Glucose molecule
(c) Sucrose molecule
(d) Glycerol molecule.
Answer:
(a) Amino acids

Question 36.
How enzymes differ from organic catalysts :
(a) Their higher rate of diffusion
(b) Active under high temperature
(c) Active under low temperature
(d) Protein nature.
Answer:
(d) Protein nature.

Question 37.
Which ¡s not a character of enzyme:
(a) They are protein
(b) They increase the rate of biochemical processes
(c) They are specific for a reaction
(d) They are used and consumed in reaction
Answer:
(d) They are used and consumed in reaction

Question 38.
Protein synthesis occurs in:
(a) Ribosomes
(b) Mitochondria
(c) Chromosomes
(d) Centrosomes.
Answer:
(a) Ribosomes

Question 39.
Digestive enzymes in cells are found in:
(a) Ribosomes
(b) Lysosomes
(c) Golgi body
(d) Cell wall.
Answer:
(b) Lysosomes

Question 40.
Who said that enzymes are protein:
(a) Pasteur
(b) Leeuwenhoek
(c) Miller
(d) Sumner.
Answer:
(d) Sumner.

Question 41.
Non-protein part of enzyme ¡s called:
(a) Apoenzyme
(b) Holoenzyme
(c) Prosthetic group
(d) All of these.
Answer:
(c) Prosthetic group

Question 42.
Inorganic prosthetic group of enzyme is called:
(a) Coenzyme
(b) Activator
(c) Hormone
(d) All of these.
Answer:
(b) Activator

Question 43.
Substrate of the amylase enzyme is:
(a) Fat
(b) Protein
(c) Starch
(d) Sucrose.
Answer:
(c) Starch

Question 44.
Diastase enzyme digest:
(a) Starch
(b) Protein
(c) Fat
(d) Amino acid.
Answer:
(a) Starch

Question 45.
The percentage of enzyme in mitochondria is:
(a) 19%
(b) 70%
(c) 14%
(d) 10%.
Answer:
(b) 70%

Question 46.
Esterase enzyme belong to group known as:
(a) Oxidation-reduction
(b) Carhoxylating
(c) Hydrolyzing
(d) Transferase.
Answer:
(c) Hydrolyzing

Question 47.
NADPisa:
(a) Enzyme
(b) Part of sRNA
(c) Coenzyme
(d) Part of tRNA.
Answer:
(c) Coenzyme

Question 48.
Coenzyme is often a :
(a) Carbohydrate
(b) Protein
(c) Vitamin
(d) Fatty acid.
Answer:
(c) Vitamin

Question 49.
Temperature is increased from 3°C to 40°C. The rate of enzyme controlled
biochemical reaction will:
(a) Not change
(b) Increase
(c) Increase initially and then decrease
(d) Decrease.
Answer:
(c) Increase initially and then decrease

Question 50.
Enzymes, vitamins and hormones are common in :
(a) Being proteinaceous
(b) Being synthesized in the body of organisms
(c) Enhancing oxidative metabolism
(d) Regulating metabolism.
Answer:
(d) Regulating metabolism.

2. Fill in the blanks:

1. The molecule on which an enzyme acts is known as ……………….
Answer:
Substrate

2. The inactive form of an enzyme is known as ……………….
Answer:
Proenzyme

3. ……………… is the respiratory pigment of higher animals.
Answer:
Haemoglobin

4. Uracil nitrogenous bases is only found in ……………….
Answer:
RNA

5. A.T.P. is a ……………… nucleotide.
Answer:
Higher

6. Proteins are molecules of C, H, O and ………………….
Answer:
N(Nitrogen)

7. A vitamin is often associated as a ……………….. with an enzyme.
Answer:
Coenzyme

8. Biochemical reactions are regulated by catalysts called ……………….
Answer:
Enzymes

9. Enzymes which breakdown compounds without the involvement of water are
called ………………..
Answer:
Lyases

10. A compound with almost similar structure to the substrate can act as a ………………
Answer:
Competitive inhibitor

3. Match the following:

(A)

Column ‘A’	Column ‘B’
1. DNA	(a) Mucin
2. Cellulose	(b) Lipid
3. Glycoprotein	(c) Genes
4. Phosphoprotein	(d) Roughage
5. Oily substance	(e) Milk.

Answer:
1. (c) Genes
2. (d) Roughage,
3. (a) Mucin
4. (e) Milk.
5. (b) Lipid

(B)

Column ‘A’	Column ‘B’
1. Chromosome	(a) Protein synthesis
2. Vitamin-A	(b) Nucleus
3. Spirogyra	(c) Schleiden-Schwann
4. mRNA	(d) Retinol
5. Cell theory	(e) Algae.

Answer:
1. (b) Nucleus,
2. (d) Retinol
3. (e) Algae
4. (a) Protein synthesis
5. (c) Schleiden-Schwann

(C)

Column ‘A’	Column ‘B’
1. Coenzyme	(a) Cofactor
2. Amylase	(b) Carbonic anhydrase
3. Cu, Fe	(c) Non-competitive inhibitor
4. Fastest enzyme	(d) NADP
5. Cyanide	(e) Starch.

Answer:
1. (d) NADP
2. (e) Starch
3. (a) Cofactor
4. (b) Carbonic anhydrase
5. (c) Non-competitive inhibitor

4. Write true or false:

1. The wood contains sufficient amount of cellulose.
Answer:
True

2. An element, not present in a nitrogen base is carbon.
Answer:
False

3. Glucose is stored as glycogen in liver.
Answer:
True

4. The steroid, which has antifertility properties is protein.
Answer:
False

5. Most water present in mature plant cell is found in nucleus.
Answer:
False

6. The catalytic activity of inorganic catalyst cannot be changed by any regulating molecule.
Answer:
True
7. Salivary amylase act best at 6.6 p^H.
Answer:
False

8. Phosphohexose isomerase enzyme is classified under transferases.
Answer:
False

9. The enzyme pyruvic decarboxylase catalyses, the removal of C0² from the substrate pyruvic acid.
Answer:
True

10. Enzymes retain their activity even when extracted from cells.
Answer:
True

5. Answer in one word:

1. Process by which edible oils are converted into hard fats is ……………..
Answer:
Hydrogenation

2. Glycogen is stored in animal body in ………………
Answer:
Liver

3. Molecule in the cell, which undergo self replication is ………………..
Answer:
DNA

4. Glycosidic bond is found in ……………….
Answer:
Disaccharide

5. The genetic material in Tobacco Mosaic Virus is …………….
Answer:
RNA

6. Protein part of enzymes is ……………………….
Answer:
Apoenzyme

7. Chemically enzymes are ………………..
Answer:
Proteins

8. Inorganic prosthetic group of enzyme is ………………..
Answer:
Activator

9. Protein synthesis occurs in ………………
Answer:
Ribosome

10. Substrate of amylase enzyme is ………………..
Answer:
Starch