Category: Class 10

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 2 Polynomials Ex 2.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.2

Question 1.
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
Solution:
(i) we have p(x) = x² – 2x – 8
= x² + 2x + – 4x – 8
= x(x + 2) – 4(x + 2)
= (x – 4)(x + 2)
For p(x) = 0, we must have (x – 4)(x + 2) = 0
Either x – 4 = 0 ⇒ x = 4
or x + 2 = 0 ⇒ x = -2
∴ The zeroes of x² – 2x – 8 are 4 and -2
Now, sum of zeroes


Thus, the relationship between the zeroes and the coefficients in the polynomial x² – 2x – 8 is verified.

(ii) We have p(s) = 4s² – 4s + 1
= 4s² – 2s – 2s + 1
= 2s(2s – 1) – l(2s – 1)
= (2s – 1)(2s – 1)

Thus, the relationship between the zeroes and coefficients in the polynomial 4s² – 4s + 1 is verified.

(iii) We have p(x) = 6x² – 3 – 7x
= 6x² – 7x – 3
= 6x² – 9x + 2x – 3
= 3x(2x – 3) + 1(2x – 3)
= (3x + 1)(2x – 3)
For p(x) = 0, we have,

Thus, the relationship between the zeroes and coefficients in the polynomial 6x² – 3 – 7x is verified.

(iv) We have, p(u) = 4u² + 8u = 4u(u + 2)
For p(u) = 0,
we have
Either 4u = 0 ⇒ u = 0
or u + 2 = 0 ⇒ u = -2
∴ The zeroes of 4u² + 8u are 0 and – 2.
Now, 4u² + 8u can be written as 4u² + 8u + 0.

Thus, the relationship between the zeroes and coefficients in the polynomial 4u² – 4s + 1 is verified.

(v) We have, p(t) = t² – 15 = (t)² – \((\sqrt{15})^{2}\)

Thus, the relationship between zeroes and the coefficients in the polynomial t² – 15 is verified.

(vi) We have, p(x) = 3x² – x – 4
= 3x² + 3x – 4x – 4
= 3x(x + 1) – 4(x + 1)
= (x + 1)(3x – 4)
For p(x) = 0 we have,
Either (x + 1) = 0 ⇒ x = -1

Thus, the relationship between the zeroes and coefficients in the polynomial 3x² – x – 4 is verified

Question 2.
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) \(\frac{1}{4}\), -1
(ii) \(\sqrt{2}, \frac{1}{3}\)
(iii) 0, \(\sqrt{5}\)
(iv) 1, 1
(v) \(-\frac{1}{4}, \frac{1}{4}\)
(vi) 4,1
Solution:
(i) Since, sum of zeroes, \((\alpha+\beta)=\frac{1}{4}\)
Product of zeroes, αβ = -1
∴ The required quadratic polynomial is

have samezeroes, therefore (4x² – x – 4) is the required quadratic polynomial.

(ii) Since, sum of zeroes, \((\alpha+\beta)=\sqrt{2}\)
product of zeroes, αβ = \(\frac{1}{3}\)
∴ The required quadratic polynomial is

(iii) Since, sum of zeroes, (α + β) = 0
Product of zeroes, αβ = \(\sqrt{5}\)
∴ The required quadratic polynomial is

(iv) Since, sum of zeroes, (α + β) = 1
Product of zeroes, αβ = 1
∴ The required quadratic polynomial is

(v) Since, sum of zeroes, (α + β) = \(-\frac{1}{4}\)
Product of zeroes, αβ = \(\)[\frac{1}{4}/latex]
∴ The required quadratic polynomial is

same zeroes, therefore, the required quadratic polynomial is (4x² + x + 1)

(vi) Since, sum of zeroes, (α + β) = 4 and
product of zeroes, αβ = 1
∴ The required quadratic polynomial is
x² – (α + β)x + αβ = x² – 4x + 1

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 6 Triangles Ex 6.6 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.6

Question 1.
In the figure, PS is the bisector of ∠QPR of ∆PQR. Prove that \(\frac{Q S}{S R}=\frac{P Q}{P R}\).

Solution:
We have, ∆PQR in which PS is the bisector of ∠QPR.
∴ ∠QPS = ∠RPS
Let us draw RT || PS to meet QP produced at T, such that
∠1 = ∠RPS [Alternate angles]
Also, ∠3 = ∠QPS [Corresponding angles]

But ∠RPS = ∠QPS [Given]
∴ ∠1 = ∠3
∴ PT = PR
Now, in ∆QRT, PS || RT [By construction]
∴ Using the Basic Proportionality Theorem, we have

Question 2.
In the figure, D is a point on hypotenuse AC of ∆ABC, such that BD ⊥ AC, DM ⊥ BC and DN ⊥ AB. Prove that
(i) DM² = DN.MC
(ii) DN² = DM.AN

Solution:

Question 3.
In the figure, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that AC² – AB² + BC² + 2BC.BD

Solution:
∆ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced.
∵ In ∆ADB, ∠D = 90°
∴ Using Pythgoras Theorem, we have
AB² = AD² + DB² ….. (1)
In right ∆ADC, ∠D = 90°
∴ Using Pythagoras Theorem, we have
AC² = AD² + DC²
= AD² + [BD + BC]²
= AD² + [BD² + BC² + 2BD.BC]
⇒ AC² = [AD² + DB²] + BC² + 2BC – BD
⇒ AC² = AB² + BC² + 2BC – BD [From (1)] Thus, AC² = AB² + BC² + 2 BC.BD

Question 4.
In the figure, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that AC² = AB² + BC² – 2BC.BD.

Solution:
We have ∆ABC in which ∠ABC < 90°
and AD ⊥ BC
In right ∆ADB, ∠D = 90°
Using Pythagoras Theorem, we have
AB² = AD² + BD² …… (1)
Also in right ∆ADC, ∠D = 90°
Using Pythagoras Theorem, we have AC² = AD² + DC²
= AD² + [BC – BD]² = AD² + [BC² + BD² – 2BC.BD]
= [AD² + BD²] + BC² – 2BC.BD = AB² + BC² – 2BC.BD [From (1)]
Thus, AC² = AB² + BC² – 2BC.BD, which is the required relation.

Question 5.
In the figure, AD is a median of triangle ABC and AM ⊥ BC. Prove that


Solution:

Question 6.
Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.
Solution:
We have a parallelogram ABCD.

AC and BD are the diagonals of parallelogram ABCD.
∵ Diagonals of a parallelogram bisect each other.
∴ O is the mid-point of AC and BD.
Now, in ∆ABC, BO is a median.

Question 7.
In the figure, two chords AB and CD intersect each other at the point P. Prove that:
(i) ∆APC – ∆DPB
(ii) AP.PB = CP.DP

Solution:
We have two chords AB and CD of a circle. AB and CD intersect at P.
(i) In ∆APC and ∆DPB,
∴ ∠APC = ∠DPB ….. (1)
[Vertically opp. angles]
∠CAP = ∠BDP …… (2)
[Angles in the same segment]
From (1) and (2) and using AA similarity, we have
∆APC ~ ∆DPB

(ii) Since, ∆APC ~ ∆DPB [As proved above]
∴ Their corresponding sides are proportional,

⇒ AP.BP = CP.DP, which is the required relation.

Question 8.
In the figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that
(i) ∆PAC – ∆PDB
(ii) PA.PB = PC.PD

Solution:
We have two chords AB and CD when produced meet outside the circle at P.
(i) Since in a cyclic quadrilateral, the exterior angle is equal to the interior opposite angle,

Question 9.
In the figure, D is a point on side BC of ∆ABC such that \(\frac{B D}{C D}=\frac{A B}{A C}\). Prove that AD is the bisector of ∠BAC.

Solution:
Let us produce BA to E such that AE = AC
Join EC.

Question 10.
Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see figure)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds ?

Solution:
Let us find the length of the string that Nazima has out.
In right ∆OAB, OB² = OA² + AB²
∴ OB² = (2.4)² + (1.8)²
⇒ OB² = 5.76 + 3.24 = 9.00
\(\Rightarrow \quad O B=\sqrt{9.00}=3 \mathrm{m}\)
i.e., Length of string she has out = 3 m

Since, the string is pulled in at the rate of 5 cm/sec,
∴ Length of the string pulled in 12 seconds

In the ∆PBC, let PB be the required horizontal distance of fly.
Since, PB² = PC² – BC² [By Pythagoras theorem]
∴ PB² = (2.4)² – (1.8)² = 5.76 – 3.24 = 2.52

Thus, the horizontal distance of the fly from Nazima after 12 seconds
= (1.59 + 1.2) m (approximately)
= 2.79 m (approximately)

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 14 Statistics Ex 14.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.2

Question 1.
The following table shows the ages of the patients admitted in a hospital during a year:

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Solution:
Mode:
Here, the highest frequency is 23.
The frequency 23 corresponds to the class interval 35 – 45.
∴ The modal class is 35 – 45
Now, class size, h = 10
Lower limit, l = 35
Frequency of the modal class (f₁) = 23
Frequency of the class preceding the modal class (f₀) = 21
Frequency of the class succeeding the modal class (f₂) = 14
∴ Mode = l + \(\left[\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right]\) × h

Mean: Let assumed mean, a = 40
Class size, h = 10
∴ u_i = \(\frac{x_{i}-a}{h}=\frac{x_{i}-40}{10}\)

∴ Required mean = 35.38 years.
Interpretation:
The maximum number of patients admitted in the hospital are of age 36.8 years while the average age of patients is 35.37 years.

Question 2.
The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:

Determine the modal lifetimes of the components.
Solution:
Here, the highest frequency = 61.
∵ The frequency 61 corresponds to class 60 – 80
∴ The modal class is 60 – 80
∴ We have l = 60, h = 20, f₁ = 61, f₀ = 52, f₂ = 38

Thus, the modal lifetimes of the components is 65.625 hours.

Question 3.
The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Solution:
Mode:
∵ The maximum number of families i.e., 40 have their total monthly expenditure is in interval 1500 – 2000.
∴ The modal class is 1500 – 2000 and l = 1500, h = 500, f₁ = 40, f₀ = 24, f₂ = 33

Thus, the required modal monthly expenditure of the families is ₹ 1847.83.
Mean: Let assumed mean, a = 3250
and class size, h = 500
∴ u_i = \(\frac{x_{i}-a}{h}=\frac{x_{i}-3250}{500}\)
∴ We have the following table;

∴ \(\overline{x}\) = a + h × [\(\frac{1}{N}\) Σf_iu_i] = 3250 + 500 × \(\left[\frac{-235}{200}\right]\)
= 3250 – \(\frac{1175}{2}\) = 3250 – 587.50 = 2662.5
Thus, the mean monthly expenditure = ₹ 2662.50

Question 4.
The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Solution:
Mode : Since greatest frequency i.e., 10 corresponds to class 30 – 35.
∴ Modal class = 30 – 35 and h = 5, l = 30, f₁ = 10, f₀ = 9, f₂ = 3
∴ Mode = l + \(\left[\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right]\) × h
= 30 + \(\left[\frac{10-9}{20-9-3}\right]\) × 5
= 30 + \(\left[\frac{10-9}{20-9-3}\right]\) × 5 = 30 + 0.625 = 30.6 (approx)
Mean: Let the assumed mean, a = 37.5 and class size, h = 5
∴ u_i = \(\frac{x_{i}-a}{h}=\frac{x_{i}-37.5}{5}\)
∴ We have the following table:


Interpretation:
The maximum teacher-student ratio is 30.6 while average teacher-student ratio is 29.2

Question 5.
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Find the mode of the data
Solution:
∵ The class 4000 – 5000 has the highest frequency i.e., 18
∴ Modal class = 4000 – 5000.
Also, h = 1000, l = 4000, f₁ = 18, f₀ = 4, f₂ = 9

Thus, the required mode is 4608.7 runs.

Question 6.
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data:

Solution:
∵ The class 40 – 50 has the maximum frequency i.e., 20
∴ Modal class = 40 – 50
∴ f₁ = 20, f₀ = 12, f₁ = 11 and l = 40. Also, h = 10

Thus, the required mode is 44.7 cars.

MP Board Class 10th Science Solutions Chapter 12 Electricity

MP Board Class 10th Science Chapter 12 Intext Questions

Class 10th Science Chapter 12 Intext Questions Page No. 200

Question 1.
What does an electric circuit mean?
Answer:
A continuous and closed path of an electric current is called an electric circuit.

Question 2.
Define the unit of current.
Answer:
The SI unit of electric current is ampere (A).
When I coulomb of electric charge flows through any cross. Section of a conductor in I second, the electric current flowing through it is said to be 1 ampere.
∵ 1 ampere = 1C/1s

Question 3.
Calculate the number of electrons constituting one coulomb of charge.
Answer:
The SI unit of electric charge is column, which is equivalent to the charge contained in nearly 6 × 10¹⁸ electrons.

Class 10th Science Chapter 12 Intext Questions Page No. 202

Question 1.
Name a device that helps to maintain a potential difference across a conductor.
Answer:
A source of electricity such as battery, cell, power supply etc. can maintain a potential difference.

Question 2.
What is meant by saying that the potential difference between two points is 1 V?
Answer:
One Volt is the potential difference between two points in a current-carrying conductor when 1 joule of work is done to move a charge of 1 coulomb from one point to the other.

1 V = 1 Jc^-1

Question 3.
How much energy is given to each coulomb of charge passing through a 6 V battery?
Answer:
W = VQ
= 6 × 1 = 6 joules
Hence 6 joules of energy is given to each coulomb of charge passing through a 6 V battery.

Class 10th Science Chapter 12 Intext Questions page No. 209

Question 1.
On what factors does the resistance of a conductor depend?
Answer:
The resistance of a conductor depends upon main four factors which are given below:

1. Length of the conductor.
2. Cross-sectional area of the conductor.
3. Natural of material of the conductor.
4. Temperature of the conductor.

Question 2.
Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Answer:
Thicker the wire, lower is the resistance of the wire. Hence, current flows more easily through a thick wire. It is because the resistance of a conductor is inversely proportional to its area of cross-section.

Question 3.
Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?
Answer:
As per ohm’s law, V = IR
\(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}\) —— (i)
Potential difference is half

∴ Current flowing is also half of ts former value.

Question 4.
Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Answer:
a) Resistivity of iron = 10.0 × 10^-8
Resistivity of Mercury = 94.0 × 10^-8
Resistivity of an alloy is greater than iron. By this we conclude that, Iron is good conductor of heat comparing to Mercury.
b) Resistivity of silver is less, hence it is a good conductor of heat.

Question 5.
Use the data in Table 12.1 to answer the following:
Electrical resistivity of some substances at 20°C

(a) Which among iron and mercury is a better conductor?
(b) Which material is the best conductor?
Answer:
(a) Resistivity of iron = 10.0 × 10^-8 Ωm
Resistivity of mercury = 94.0 × 10^-8 Ωm
Hence, resistivity of mercury is more than that of iron. This implies that iron is a better conductor than mercury.
(b) It can be observed from table that the resistivity of silver is the lowest among the listed materials. Hence, it is the best conductor.

Class 10th Science Chapter 12 Intext Questions page No. 213

Question 1.
Draw a schematic diagram of a circuit consisting of a battery of ‘three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.
Answer:
Schematic diagram of a circuit:

Question 2.
Redraw the circuit of question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?
Answer:

Hence, reading in the ammeter would be 0.24 A and reading in voltmeter would be 2.88 V.

Class 10th Science Chapter 12 Intext Questions Page No. 216

Question 1.
Judge the equivalent resistance when the following are connected in parallel:
(a) 1Ω, and 106 Ω
(b) 1 Ω, and 103 Ω and 106 Ω
Answer:
(a) 1 ohm and 10⁶ times ohm
R₁ = 1 ohm
R₂ = 10⁶ times = 1000000 ohm
Total resistance (parallel)
1/R = 1/R₁ + 1/R₂
= 1/1 + 1/1000000
= 1000000 + 1 /1000000
= 1000000/ 1000000
1/R = 1/1 ohm (approx.)
∴ Resistance = 1 ohm (approx.)

(b) Given R₁ = 1 ohm
R₂ = 10³ ohm
R₃ = 10⁶ ohm

Question 2.
An electric lamp of 100Ω, a toaster of resistance 50Ω and a water filter of resistance 500Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
Answer:
Resistance of Electric 1 amp R₁ = 100Ω
Resistance of Tosser, R₂ = 50Ω
Resistance of water filter, R₃ = 50Ω
Potential difference, V = 220V
When these are connected in parallel,


7.04A of electricity is obtained by three appliances
Resistance of an electric iron connected to the same source that takes as much current as all three appliances

∴ Resistance of iron box = 31.25Ω
Electricity flowing through this = 7.04A.

Question 3.
What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Answer:
(i) In parallel circuit, if one electrical appliance stops working due to some defect, then all other appliances keep working normally. In series circuit, if one electrical appliance stops working due to some defect, then all other appliances also stop working.

(ii) In parallel circuits, each electrical appliance gets the same voltage as that of the power supply line. In series circuit, appliances do not get the same voltage, as that of the power supply line.

(iii) In the parallel connection of electrical appliances, the overall resistance of the household circuit is reduced due to which the current from the power supply is high. In the series connection, the overall resistance of the circuit increases too much due to which the current from the power supply is low.

Question 4.
How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4Ω, (b) 1Ω?
Answer:
Given,
R₁ = 2 Ω
R₂ = 3 Ω
R₃ = 6 Ω
(a) When R₂ and R₃ are connected in parallel with R₁ in series we receive

(b) When R₁, R₂, R₃ are connected in parallel we receive

Question 5.
What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?
Answer:
(a) For highest resistance according to question resistances must be connected in series:

4 Ω, 8 Ω, 12 Ω, 24 Ω
R₁ = 4 Ω
R₂ = 8 Ω
R₃ = 12 Ω
R₄ = 24 Ω
Total resistance in series = R₁ + R₂ + R₃ + R₄
= 4 + 8 + 12 + 24
= 48 Ω
Now, resistance is maximum when connected in series.

(b) For lowest resistance, resistances must be connected in parallel.

Resistance is lowest when connected in parallel.

Class 10th Science Chapter 12 Intext Questions Page No. 218

Question 1.
Why does the cord of an electric heater not glow while the heating element does?
Answer:
In cord of an electric heater, as current flows these become hot arid glows but in case of electric heater this will not happen.

Question 2.
Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Answer:
Given, Charge = 96000 coulomb
Time = 1 hour = 3600 seconds
Potential difference = 50 V Ω
As we know, I = Q/T
I = 96000/36000 = 80/3 amp.
V = I × R
50 = 80/3 × R
So, R = 15/8 Ω
Now, Heat = I²RT
= (80/3)² × 15/8 × 3600
800 × 6000
= 4800000 joules or 4.8 × 10⁶ J

Question 3.
An electric iron of resistance 20Ω takes a current of 5 A. Calculate the heat developed in 30 s.
Answer:
As per Joule’s law
H = VIt
H = IR = 5A × 20Ω = 100V
I = 5A, t = 30 sec.
∴ H = 100 × 5 × 30 J
= 15000 J = 1.5 × 10⁴ J.

Class 10th Science Chapter 12 Intext Questions Page No. 220

Question 1.
What determines the rate at which energy is delivered by a current?
Answer:
The rate at which electric energy is dissipated or consumed in an electric circuit is termed as electric power.
P = VI.

Question 2.
An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Answer:
P = VI
V = 220V, and I = 5A.
Power P = 220 × 5 = 1100 W
Power of the motor = P × t
P = 1100 W.
t = 2 Hr 2 × 60 × 60 W
= 7200 S
∴ Energy consumed, E = 1100 × 7200 J
= 7920000
= 7.92 × 10⁶J.

MP Board Class 10th Science Chapter 12 NCERT Textbook Exercises

Question 1.
A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio \(\frac { R }{ R` } \) is –
(a) \(\frac { 1 }{ 25 }\)
(b) \(\frac { 1 }{ 5 }\)
(c) 5
(d) 25
Answer:
(d) 25

Question 2.
Which of the following terms does not represent electrical power in a circuit?
(a) I²R
(b) IR²
(c) VI
(d) \(\frac { { V }^{ 2 } }{ R } \)
Answer:
(b) IR²

Question 3.
An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
Answer:
(d) 25 W

Question 4.
Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be-
(a) 1 : 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1
Answer:
(c) 1 : 4

Question 5.
How is a voltmeter connected in the circuit to measure the potential difference between two points?
Answer:
Voltmeter is connected in parallel in the circuit to measure the potential difference between two points.

Question 6.
A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10-8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
Answer:
Given, Diameter = 0.5 mm = 0.0005 m

So, the resistance is one-fourth if the diameter is doubled.

Question 7.
The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistors are given below:

Plot a graph between V and I and calculate the resistance of that resistor are given below:
Answer:

Question 8.
When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Answer:
Given, V = 12v
I = 2.5 mA = 2.5 × 10^-3 A
As we know, R = \(\frac { V }{ 1 } \)
R = \(\frac{12 \mathrm{V}}{2.5 \times 10^{-3} \mathrm{A}}\) = 48000 Ω
So, R = 4.8 kΩ

Question 9.
A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω, and 12 Ω respectively. How much current would flow through the 12 Ω resistor.
Answer:
R₁ = 0.2 Ω R₂ = 0.3 Ω R₃ = 0.4 Ω R₄ = 0.5 Ω R₅ = 12 Ω

Given, V = 9 V
R5 = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω
= 13.4 Ω
According to formula I = \(\frac { V }{ R } \)
= \(\frac{9 \mathrm{V}}{13.4 \Omega}\)
= 0.67 A
0.67 A current would flow through the 12 Ω resistor.

Question 10.
How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Answer:
Given
I = 5 A V = 220 V
Now, Let the number of 176 Ω resistors be n.

So, four resistors of 176 Ω are required to carry 5 A on a 220 V line.

Question 11.
Show how you would connect three resistors, each of resistance 6Ω, so that the combination has a resistance of (i) 9Ω, (ii) 4Ω.
Answer:
If resistors are connected in series 6Ω + 6Ω + 6Ω =18Ω This ia not correct
When they are connected in parallel
\(\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=3\) This is also wrong,
i) When they are connected in parallel

Two 6Ω resistors are connected in parallel

If 3rd resistor of 6Ω and 3Ω are connected in series, it becomes 6Ω + 3Ω = 9Ω.
ii) When they are connected in series

Resistance = 6Ω + 6Ω
= 12Ω
If 3rd resistor 6Ω is connected to 12Ω in parallel

Total resistance = 4Ω.

Question 12.
Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
Answer:
Here V = 220V
I – 5A

∴ 110 lamps can be connected in parallel with each other across the two wires.

Question 13.
A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Answer:
(i) If coils are connected separately V = 220 V
Resistance R₁ = 240
As per ohm’s law, V = IR

∴ If coils are connected separately 9.16A electricity flows in the coil.
(ii) If coils are connected in series
Resistance R₂ = 24Ω + 24Ω = 48Ω
As per ohm’s law V = IR

∴ If coils are connected in series 4.58A electricity flows.

Question 14.
Compare the power used in the 20 resistors in each of the following circuits:
(i) a 6 V battery in series with 1Ω and 2Ω resistors, and
(ii) a 4 V battery in parallel with 12Ω and 2Ω resistors.
Answer:
(i) Potential Difference V = 6V
If 1Ω and 2Ω resistors are connected in series, then Resistance
R = 1 + 2 = 3W.
As per ohm’s law
\(\mathrm{I}=\frac{6}{3}=2 \mathrm{A}\)
P(I²)R = (2)² × 2 = 8W.
(ii) Potential difference V = 4V
If 12Ω and 2Ω resistors are connected in parallel, voltage is equal
Voltage of resistance 2 W is 4 volts

∴ Power of 2Ω is 8 W.

Question 15.
Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Answer:
Both lamps are connected in parallel potential difference = 220 V
Power = V × I.

Question 16.
Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
Answer:
250W TV set in used in 1 Hr, then its energy
= 250 × 3600 = 9 × 10⁵
energy of Toaster = 1200 × 600
If it is used in 10 minutes, then its power = 1200 × 600 × 7.2 × 10⁵ J
∴ Energy of 250 W TV set is used in 1 Hr is greater than 1200 W toaster used in 10 minutes.

Question 17.
An electric heater of resistance 8Ω draws 15 A from the service mains 2 hours.
Calculate the rate at which heat is developed in the heater.
Answer:
P = I²R
R = 8Ω, I = 15A
P= (15)² × 8 = 1800 J/s.
∴ Rate at which heat is developed in the heater
= 1800 J/s.

Question 18.
Explain the following:
(a) Why is the tungsten used almost exclusively for filament of electric lamps?
(b) Why arc the conductors of electric heating devices, such as bread-toasters and electric irons, majle of an alloy rather than a pure metal?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why are copper and aluminium wires usually employed for electricity transmission?
Answer:
(a) Tungsten used almost exclusively for filament of electric lamps because its melting point is very high and resistivity of tungsten is low, it has special property of glowing on heating, it does not oxidized.

(b) The conductors of electric heating devices such as bread toasters and electric irons are made of alloy because resistivity of an alloy is higher than metal so alloy produces large amount of heat. They do not get oxidized. Melting point is also high.

(c) As in series, circuit voltage is divided. Components of a series circuit receives only small voltage so the amount of current decreases and the device turn hot and do not work properly. Hence, series arrangement is avoided being used in domestic circuits.

(d) Resistance (R) of a wire is inversely proportional to its area of cross-section (A) i.e., when area of cross section decreases the resistance increases or vice versa.

(e) Copper and aluminium are usually used for electricity transmission because they are good conductors of electricity and also have low resistivity and do not get heated.

MP Board Class 10th Science Chapter 12 Additional Important Questions

MP Board Class 10th Science Chapter 12 Multiple Choice Questions

Question 1.
SI unit of an electric current is equal to
(a) 6 × 10⁸ C
(b) 9.8 × 10¹⁰ C
(c) 1.6 × 10^-19 C
(d) None of these
Answer:
(d) None of these

Question 2.
Which materials have changing resistivity?
(a) Conductors
(b) Semiconductors
(c) Non-metals
(d) Super metals
Answer:
(b) Semiconductors

Question 3.
If wire in a circuit is doubled its resistivity will be?
(a) Two times
(b) Ten times
(c) Zero
(d) half
Answer:
(d) half

Question 4.
A wire used in circuit must be
(a) Highly resistant
(b) Non resistance
(c) Have high melting point
(d) None of these
Answer:
(b) Non resistance

Question 5.
A heating w ire must have
(a) High melting point
(b) High conductivity to heat
(c) Both
(d) None
Answer:
(b) High conductivity to heat

Question 6.
SI unit of resistance is:
(a) Coulomb
(b) Ampere
(c) Ohm
(d) Ohm – meter
Answer:
(c) Ohm

Question 7.
Symbol for use of voltmeter in circuit is:

Answer:

Question 8.
Symbol for use of battery in circuit is:

Answer:

Question 9.
Relation of potential difference and current is represented by:
(a) Joule’s law
(b) Ohm’s law
(c) Dalton’s law
(d) None
Answer:
(b) Ohm’s law

Question 10.
Formula which represents the relation among potential difference and current following in a circuit is:
(a) V = IR
(b) V = I/R
(c) l = Q/t
(d) R = p \(\frac { l }{ A } \)
Answer:
(a) V = IR

Question 11.
Kilo watt is a unit for:
(a) Power
(b) Electrical energy commercial
(c) Resistance
(d) Current flow in circuit
Answer:
(a) Power

Question 12.
Which one has less resistance?
(a) 10 watt bulb
(b) 15 watt LED
(c) 60 watt bulb
(d) 60 watt LED.
Answer:
(a) 10 watt bulb

Question 13.
In which type of combinations resistance is more?
(a) Series Combination
(b) Parallel combination
(c) both
(d) Single resistance unit.
Answer:
(a) Series Combination

Question 14.
Joule’s law formulation is
(a) H = IRT
(b) H = I2RT
(c) P = VI
(d) RT = PV
Answer:
(d) RT = PV

Question 15.
Hardest metal which can resist maximum heat is:
(a) Lead
(b) Tungsten
(c) Silver
(d) Carbon
Answer:
(b) Tungsten

Question 16.
If current passing a circuit is 10 A and wire is connected to 100 volt line resistant in the wire will be:
(a) 10 A
(b) 10 Ω
(c) 100 Ω
(d) O Ω
Answer:
(b) 10 Ω

Question 17.
An insulator has a resistivity equals to:
(a) 10^-8 Ω m
(b) 10 Ω m
(c) 5 Ω m
(d) 10^-18 Ω
Answer:
(a) 10^-8 Ω m

Question 18.
Calculate current-(I) in circuit given below:
(a) 0.6 A
(b) 20 A
(c) 50 A
(d) 10 A

Answer:
(a) 0.6 A

Question 19.
If resistors are arranged in parallel with value 1 Ω, 2 Ω and 3 Ω Total resistant will be:
(a) more than 3 Ω
(b) less than 3 Ω
(c) less than 1 Ω
(d) equal to 1 Ω
Answer:
(b) less than 3 Ω

Question 20.
The changing 1 kWh to joule will be equal to
(a) 3.6 × 10⁶ J
(b) 3.6 × 10^-6 J
(c) 1.6 × 10⁶ J
(d) 1 × 10⁶ J
Answer:
(a) 3.6 × 106 J

MP Board Class 10th Science Chapter 12 Very Short Answer Type Questions

Question 1.
What kind of circuit is required for flow of current ?
Answer:
Closed.

Question 2.
Draw a schematic diagram of a simple electric with battery, bulb, ammeter and key.
Answer:

Question 3.
Express in formula form:

1. net charge of a circuit
2. time taken for production of a unit amount of current.

Answer:

1. Q = It.
2. t = Q/I.

Question 4.
What is the value of an electric charge for an electron?
Answer:
-1.6 × 10^-19 C.

Question 5.
Formulate relationship between potential difference and current.
Answer:
V = IR.

Question 6.
Draw the graph verifying ohm’s law.
Answer:

Question 7
Draw circuit with three resistor in series combination.
Answer:

Question 8.
Represent with diagram:
(i) A cell
(ii) A bulb.
Answer:

Question 9.
Which material has highest conductivity and least resistance?
Answer:
Super conductors.

Question 10.
Which quality of electrical energy is inversely proportional to electrical power?
Answer:
Resistance.

Question 11.
Calculate resistance between terminal P & Q of following diagram.
Answer:

First R₁ and R₂ are in series.
So, R’ = R₁ + R₂ = 1 Ω + 1 Ω = 2 Ω.
then, R₃ and R₄ are in series
R” = R₃ + R₄ = 1 Ω + 1 Ω = 2 Ω
Now both R’ and R” are in parallel.
\(\frac { 1 }{ R } \) = \(\frac{1}{R^{\prime}}+\frac{1}{R^{\prime \prime}}+\frac{1}{2}+\frac{1}{2}=1\)
So, R = 1 Ω

Question 12.
If a circuit has open key will current flow from it?
Answer:
No.

Question 13.
If ammeter is not placed in circuit will current flow?
Answer:
Yes.

Question 14.
Which gas is filled in electric bulb?
Answer:
Argon.

Question 15.
Write the formula representing relationship between heat produced and current flowing in a circuit.
Answer:
H = I²RT.

Question 16.
Write the formula to calculate power generated by a circuit a with current M.
Answer:
Since, P = VI
Putting value of I, P = VM (I = m, Given)

Question 17.
Convert 5 kWh in joule.
Answer:
1 kWh = 3.6 × 10⁶ Joule
5 kWh = 5 × 3.6 × 10⁶ J.
= 16 × 10⁶ J
or 1.6 × 10⁷ J

Question 18
One volt ampere is equal to how many watt power?
Answer:
1 Watt Power.

MP Board Class 10th Science Chapter 12 Short Answer Type Questions

Question 1.
What is electric current?
Answer:
Total charge which pass through a particular area in unit time is called electric current of that particular conductor.

Question 2.
What is potential difference?
Answer:
Movement of electron need a pressure difference among conductors. Hence, battery or current flow suppliers are added to circuit. This difference in potential is called potential difference.

Question 3.
Define Ohm’s law.
Answer:
Ohm’s law states that current flowing in a circuit is directly proportional to its potential difference. It is represented by formula,
V α I
or V = RI
or V = IR
Here, R is a constant known as resistance.

Question 4.
Why resistance is applied to a circuit?
Answer:
To control the current flow in desired way resistance is applied. Example- Regulator of fan give desired speed of wind flow by controlling flow of current which is being converted to physical energy by fan.

Question 5.
Write down the factors at which the resistance of the conductor depends.
Answer:

1. Length of wire: Resistance is directly proportional to length R ∝ L.
2. Area cross section: Resistance is inversely proportional to area cross section of a wire R ∝ 1/A.
3. Nature of material.
4. Temperature of the conductor.

Question 6.
Arrange the resistivity in decreasing order of following material,
Iron, Silver, Tungsten, Manganin and Glass
Answer:
Glass > Manganin > Iron > Tungsten > Silver.

Question 7.
Give reason for the following:

1. Electric bulbs are usually filled with chemically inactive gases like nitrogen and argon.
2. Copper and aluminium wires are usually employed for electricity transmission.
3. Fuse wire is placed in series with the device.

Answer:

1. Electric bulbs are usually filled with chemically inactive gases like nitrogen and argon because these gases do not react with the hot tungsten filament and hence, prolong the life of filament of the electric bulb.
2. Copper and aluminium wires are usually employed for electricity transmission because copper and aluminium have low resistivity and thus, they are very good conductors of electricity.
3. Fuse wire is placed in series with the device because when large current passes through the circuit the fuse wire gets heated up and melts and whole circuit breaks and the device is protected from the damage.

Question 8.
If R₁ = 10 Ω, R₂ = 20 Ω and R₃ = 30 Ω, calculate the effective resistance when they are connected in series to a battery of 6 V. Also find the current flowing in the circuit.
(a) Current flows through different resistances, when these are joined in series, as shown in the below figure:

(b) R₁ = 10 Ω, R₂ = 20 Ω, R₃ = 30 Ω
Effective resistance, R = R₁ + R₂ + R₃
R 10 + 20 + 30 = 60 Ω
Potential difference, V = 6V,
Current, I = ?
According to Ohm’s law
V = IR

Question 9.
Find the resistance between points A and B in the circuit diagram given below

Answer:
R₂, R₃, R₄ are in series and have resultant resistance R
R’ = R₂ + R₃ + R₄
= 2 + 2 + 2 = 6 Ω
R’ is in parallel combination with R₁
∴ Resultant resistance of the circuit (R’)
⇒ \(\frac { 1 }{ R } \) = \(\frac{1}{R_{1}}+\frac{1}{R_{1}}\) = \(\frac { 1 }{ 6 } \) + \(\frac { 1 }{ 6 } \) = \(\frac { 2 }{ 6 } \) = \(\frac { 1 }{ 3 } \)
∴ Resistance, R = 3 Ω.

MP Board Class 10th Science Chapter 12 Long Answer Type Questions

Question 1.
An electric iron consumes energy at a rate of 840 W when heating is at the maximum and 360 W when the heating is at the minimum. The voltage at which it is running is 220 V. What are the current and resistance in each case?
Solution:
At maximum heating:
The consumption of energy (electric) is given at the rate of 840 W at voltage 220 V.
P = 840 Ω, V = 220 V
Then current, I₁ = ?
∴ P = V × I₁

Question 2.
Three resistors of 5 Ω,10 Ω and 15 Ω are connected in series and the combination is connected to battery of 30 V. Ammeter and Voltmeter are connected in the circuit. Draw a circuit diagram to connect devices all the in correct order. What is the current flowing and potential difference across 10 Ω resistance?
Answer:
Current flowing I = ?
V₂ = ?
Total resistance, R = R₁ + R₂ + R₃
= 5 + 10 + 15 = 30 Ω
Total potential difference, V = 30 volts
According to Ohm’s law V = IR

Potential difference across 10 Ω is 10 volts.

Question 3.
Nichrome wire of length ‘l’ and radius ‘r’ has resistance of 10 Ω. How would the resistance of the wire change when:
(a) (i) Only length of the wire is doubled?
(ii) Only diameter of the wire is doubled? Justify your answer.
(b) Why element of electrical heating devices are made up of alloys?
Solution:
(a) Resistance, R ∝ 1

(i) Resistance is directly proportional to the length of the conductor. If length of nichrome wire (I) is doubled its resistance also gets doubled.
∴ R’ new resistance = 20 Ω

(ii) The resistance of the wire is inversely proportional to the square of its diameter. If the diameter of the wire is doubled, its resistance becomes one-fourth.
∴ R’, new resistance = \(\frac { 10 }{ 4 } \) Ω = 2.5 Ω.

(b) The heating elements of electrical heating appliances are made up of nichrome alloy because,

• Nichrome has very high resistivity due to which it produces a lot of heat on passing current.
• Nichrome does not undergo oxidation easily even at high temperature, it can be kept red hot without burning.

Question 4.
The resistance of a wire of 0.01 cm radius is 10 Ω. If the resistivity of the material of the wire is 50 × 10-8 ohm meter, find the length of the wire.
Answer:
Resistance of a wire, R = 10 Ω Radius, r = 0.01 × 10^-2 m
Resistivity, p = 50 × 10^-8 ohm meter Length of the wire, l = ?
Area of cross section, A = πr² = 3.14 × (0.01 × 1o-2)² m²
= 3.14 × 0.01 × 0.01 × 10^-4 m²
= 3.14 × 10^-8 m²

Question 5.
(a) Calculate the resistance of the wire using the graph.
(b) How many 176 Ω resistors in parallel are required to carry 5 A on a 220 V line?
(c) Define electric power. Derive relation between power, potential difference and resistance.

Answer:
(a) Resistance of wire = Slope of the graph
According to Ohm’s law,
V = IR or R = \(\frac { V }{ I } \)

(b) Resistance, R¹ = 176 Ω, No. of resistors = n,
Current, I = 5A
Potential difference, V = 220 volts, Resultant resistance = R According to Ohm s law,
V = IR, R = \(\frac { V }{ 1 } \) = \(\frac { 220 }{ 5 } \) = 44 Ω

Thus, 4 resistors of 176 Ω in parallel combination are required to carry 5 A on a 220 V line.

(c) Electric power is defined as the electrical work done per unit-time.

The work done, W by current, I when it flows for time t under potential difference V is given by

MP Board Class 10th Science Chapter 12 NCERT Textbook Activities

Class 10 Science Activity 12.1 Pages No. 203.204

• Set up circuit as shown in figure, consisting of a mchrome wire XY of length, say 0.5 m, an ammeter, a voltmeter and four cells of 1.5 V each (Nichrome is an alloy of nickel, chromium, manganese, and iron metals.)
• First use only cell as the source in the circuit. Note the reading of the voltmeter V for the potential difference across the nichrome wire XY in the circuit. Tabulate them in table given.
• Next connect two cells in the circuit and note the respective readings of the ammeter and voltmeter for the values of current through the nichrome wire and potential difference across the nichrome wire.
• Repeat the above steps using three cells and then four cells in the circuit separately.
• Calculate the ratio of V to 1 for each pair of potential difference V and current I.
  
• Plot a graph between V and 1, and observe the nature of the graph.

Observations:

Approximately same value for \(\frac { V }{ I } \) will be obtained in each case. The straight line is obtained in the V – I graph. Thus, \(\frac { V }{ I } \) is a constant ratio.

Class 10 Science Activity 12.2 Page No. 205

• Take a nichrome wire, a torch bulb, a 10 W bulb and an ammeter (0-5 A range), a plug key and some connecting wires.
• Set up the circuit by connecting four dry cells of 1.5 V each in series with the ammeter leaving a gap XY in the circuit as shown in Fig. 12.1.
  

• Complete the circuit by connecting the nichrome wire in the gap XY. Plug the key. Note down the ammeter reading. Take out the key from the plug. [Note: Always take out the key from the plug after measuring the current the circuit.
• Replace the nichrome wire with the torch bulb in the circuit and find the current through it by measuring the reading of the ammeter.
• Now repeat the above step with the 10 W bulb in the gap XY.
• Are the ammeter readings differ for different components in the gap XY? What do the above observations indicate?
• You may repeat this Activity by keeping any material component in the gap. Observe the ammeter readings in each case. Analyse the observations.

Observations:

• The current is different for different components as certain components offer an easy path for the flow of electric current while the others resist the flow.

Class 10 Science Activity 12.3 Page No. 206

• Complete an electric circuit consisting of cell, an ammeter, a nichrome wire of length l [say, marked (1)] ana a plug key, as shown in Fig. 12.2.
  
  Electric circuit to study the factors on which the resistance of conducting wires depends.

• Figure 12.2. Electric circuit ti study the factors on which the resistance of conducting wires depends
• Now, plug the key. Note the current in the ammeter.
• Replace the nichrome wire by another nichrome wire of same thickness but twice the length, that is 2l [marked (2) in the Fig. 12.2],
• Note the ammeter reading.
• Now replace the wire by a thicker nichrome wire of the same length / [marked (3)]. A thicker wire has a larger cross-sectional area. Again note down the current through the circuit.
• Instead of taking a nichrome wire, connect a copper wire [marked (4) in Fig. 12.2] in the circuit. Let the w ire be of the same length and same area of cross-section as that of the first nichrome wire [marked (1)]. Note the value of the current.
• Notice the difference in the current in all cases.
• Does the current depend in the length of the conductor?
• Does the current depend on the area of cross-section of the wire used?

Observations:

• The length of the conductor affects the current flow. Ammeter reading decreases to one-half when the length of the wire is doubled.
• The cross-section of the wire also affects the current flow. The length and cross-section changes affect the resistance also.

Class 10 Science Activity 12.4 Page No. 210

• Join three resistors of different values in series. Connect them with a battery, an ammeter and a plug key. as shown in Fig. 12.3. You may use the resistors of values like 1Ω, 2Ω, 3Ω etc, and a battery of 6 V for performing this Activity.
• Plug the key. Note the ammeter reading.
• Change the position if ammeter to anywhere in between the resistors. Note the ammeter reading each time.
• Do you find any change in the value of current through the ammeter?

Observations:

• The value of the current in the ammeter remains unchanged. This shows that it is independent of its position in the electric current. This shows that the current is the same in every part of a series combination.

Class 10 Science Activity 12.5 Page No. 211

• In Activity 12.4, insert a voltmeter across the ands X and Y of the series combination of three resistors, as shown in Fig. 12.4.
• Plug the key in the circuit and note series combination of resistors. Let it be V. Now measure the potential difference across the two terminals of the battery. Compare the two values.
• Take out the plug key and disconnect the voltmeter. Now insert the voltmeter across the ands X and P the first resistor, as shown in Fig. 12.4.
  
• Plug the key and measure the potential difference across the first resistor. Lei it be V₁.
• Similarly, measure the potential difference across the first resistor, separately. Let these values be V₃, and V₂, respectively.
• Deduce a relationship between V, V₁, V₂ and V₃.

Observations:

• The potential difference (V) is equal to the sum of potential differences V₁, V₂, and V₃.
• Total potential difference across a combination of resistors in series is equal to the sum of potential difference across the individual resistors,
V = V₁ + V₂ + V₃

Class 10 Science Activity 12.6 Page No. 211

• Make a parallel combination. X Y of three resistors having resistances R₁, R₂, and R₃, respectively Connect it with a battery, a plug key and an ammeter, as shown in Fig. 12.5. Also connect a voltmeter in parallel with the combination of resistors
  

• Plug the key and note the ammeter reading. Let the current be I. Also take the voltmeter reading. It gives the potential difference V, across the combination. The potential difference across each resistor is also V. This can be checked by connecting the voltmeter across each individual resistor (see Fig. 12.6).
• Take out the plug from the key. Remove the ammeter and voltmeter from the circuit. Insert the ammeter in series with the resistor Rl, as shown in Fig. 12.6. Note the ammeter reading, 11.
  

Similarly, measure the currents through R₂ and R₃ Let these be I₂, and I₃, respectively. What is the relationship between I, I₁, I₂ and I₃?

Observations:

• The total current I, is equal to the sum of the separate currents through each branch of the combination,
  I = I₁ + I₂ + I₃

MP Board Class 10th Science Solutions

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.2

Question 1.
Evaluate the following:
(i) sin60° cos30° + sin 30° cos 60°
(ii) 2tan² 45° + cos²30° – sin²60°

Solution:
(i) We have




= \(\frac{\frac{1}{12}\times67}{\frac{4}{4}}=\frac{67}{12}\)

Question 2.
Choose the correct option and justify your choice:
(i) \(\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}=\)
(A) sin 60°
(B) cos 60°
(C) tan 60°
(D) sin 30°

(ii) \(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=\)
(A) tan 90°
(B) 1
(C) sin 45°
(D) 0

(iii) sin 2A = 2sin A is true when A =
(A) 0°
(B) 30°
(C) 45°
(D) 60°

(iv) \(\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}=\)
(A) cos 60°
(B) sin 60°
(C) tan 60°
(D) sin 30°
Solution:

Question 3.
If tan (A + B) = \(\sqrt{3}\) and tan (A – B) = \(\frac{1}{\sqrt{3}}\) 0°< A + B ≤ 90°; A > B, find A and B.
Solution:
We have,
tan 60° = \(\sqrt{3}\), tan 30° = \(\frac{1}{\sqrt{3}}\) ……………. (1)
Also, tan(A + B) = \(\sqrt{3}\) and tan (A – B) = \(\frac{1}{\sqrt{3}}\) …………… (2)
From (1) and (2), we get
A + B = 60° …………… (3)
and A – B = 30° ………….. (4)
On adding (3) and (4), we get
2A = 90° ⇒ A =45°
On subtracting (4) from (3), we get
2B = 30° ⇒ B = 15°

Question 4.
State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin6 increases as θ increases.
(iii) The value of cosθ increases as θ increases.
(iv) sin θ = cos θ for all values of θ.
(v) cot A is not defined for A = 0°.
Solution:
(i) False:
Let us take A = 30° and B = 60°, then
L.H.S = sin (30° + 60°) = sin 90° = 1

∴ L.H.S. ≠ R.H.S.
(ii) True:
Since, the values of sin θ increases from 0 to 1 as θ increases from 0° to 90°.
(iii) False:
Since, the value of cos θ decreases from 1 to 0 as θ increases from 0° to 90°.
(iv) False:
Let us take θ = 30°
sin 30° = \(\frac{1}{2}\) and cos 30° = \(\frac{\sqrt{3}}{2}\)
⇒ sin 30° ≠ cos 30°
(v) True:
We have, cot 0° = not defined

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 6 Triangles Ex 6.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.3

Question 1.
State which pairs of triangles in the given figures are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form.


Solution:
(i) In ∆ABC and ∆PQR,
We have : ∠A = ∠P = 60°
∠B = ∠Q = 80°
∠C = ∠R = 40°
∴ The corresponding angles are equal.
∴ Using the AAA similarity rule,
∆ABC ~ ∆PQR

(ii) In ∆ABC and ∆QRP,

i.e., Using the SSS similarity, ∆ABC ~ ∆QRP

(iii) In ∆LMP and ∆DEF,

∴ Triangles are not similar,

(iv) In ∆MNL and ∆QPR,
\(\frac{M L}{Q R}=\frac{M N}{Q P}\) [\(\frac{1}{2}\) each]
and ∠NML = ∠PQR
∴ Using SAS similarity, we have
∆MNL ~ ∆QPR.

(v) In ∆ABC and ∆FDE,
\(\frac{A B}{D F}=\frac{B C}{E F}\) [\(\frac{1}{2}\) each]
Now, angle between DF and EF is 80°. But angle between AB and BC is unknown.
∴ Triangles are not similar.

(vi) In ∆DEF and ∆PQR,
∠D = ∠P = 70°
[∵ ∠P = 180° – (80° + 30°) = 180° – 110° = 70°]
∠E = ∠Q = 80°
∠F = ∠R = 30° [∵∠F = 180°]
∴ Using the AAA similarity rule,
∆DFE ~ ∆PRQ.

Question 2.
In the figure, ∆ODC ~ ∆OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.

Solution:
We have, ∠BOC = 125° and ∠CDO = 70°
since, ∠DOC + ∠BOC = 180° [Linear pair]
⇒ ∠DOC = 180° – 125° = 55° ……………… (1)
In ∆DOC,
Using the angle sum property for ∆ODC, we get
∠DOC + ∠ODC + ∠DCO = 180°
⇒ 55° + 70° + ∠DCO = 180°
⇒ ∠DCO = 180° – 55° – 70° = 55°
Again,
∠DOC = ∠BOA ……………. (2) [vertically opposite angles]
and ∠OCD = ∠OAB = 55° ………….. (3)
[corresponding angles of similar triangles]
Thus, from (1), (2) and (3)
∠DOC = 55°, ∠DCO = 55° and ∠OAB = 55°.

Question 3.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that \(\frac{O A}{O C}=\frac{O B}{O D}\).
Solution:
We have a trapezium ABCD in which AB || DC. The diagonals AC and BD intersect at O.

In ∆OAB and ∆OCD,
∠OBA = ∠ODC (Alternate angles)
and ∠OAB = ∠OCD(Alternate angles)
Using AA similarity rule, ∆OAB ~ ∆OCD
So, \(\frac{O B}{O D}=\frac{O A}{O C}\) (Ratios ot corresponding sides of the similar triangles)
⇒ \(\frac{O A}{O C}=\frac{O B}{O D}\)

Question 4.
In the figure, \(\frac{Q R}{Q S}=\frac{Q T}{P R}\) and ∠1 = ∠2. show that ∆PQS ~ ∆TQR.

Solution:
In ∆PQR
∵ ∠1 = ∠2 [Given]
∴ PR = QP ……………… (1)
[∵ In a ∆ sides opposite to equal angles are equal]

and ∠SQP = ∠RQT = ∠1
Now, using SAS similarity rule,
∆PQS ~ ∆TQR.

Question 5.
S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ∆RPQ ~ ∆RTS.
Solution:
In ∆PQR,
T is a point on QR and S is a point on PR such that ∠RTS = ∠P.

Now in ∆RPQ and ∆RTS,
∠RPQ = ∠RTS [Given]
∠PRQ = ∠TRS [Common]
Using AA similarity, we have ∆RPQ ~ ∆RTS.

Question 6.
In the figure, if ∆ABE ≅ ∆ACD, show that ∆ADE ~ ∆ABC.
Solution:
We have,
∆ABE ≅ ∆ACD
Their corresponding parts are equal, i.e.,
AB = AC, AE = AD


and ∠DAE = ∠BAC (common)
∴ Using the SAS similarity, we have ∆ADE ~ ∆ABC.

Question 7.
In the figure, altitudes AD and CE of ∆ABC intersect each other at the point P. Show that:
(i) ∆AEP ~ ∆CDP
(ii) ∆ABD ~ ∆CBE
(iii) ∆AEP ~ ∆ADB
(iv) ∆PDC ~ ∆BEC

Solution:
We have a ∆ABC in which altitude AD and CE intersect each other at P.
⇒ ∠D = ∠E = 90° …………. (1)
(i) In ∆AEP and ∆CDP,
∠AEP = ∠CDP [From (1)]
∠EPA = ∠DPC [Vertically opp. angles]
∴ Using AA similarity, we get
∆AEP ~ ∆CDP

(ii) In ∆ABD and ∆CBE,
∠ADB = ∠CEB [From (1)]
Also, ∠ABD = ∠CBE [Common]
Using A A similarity, we have
∆ABD ~ ∆CBE

(iii) In ∆AEP and ∆ADB,
∵ ∠AEP = ∠ADB [From (1)]
Also, ∠EAP = ∠DAB [Common]
∴ Using AA similarity, we have
∆AEP ~ ∆ADB

(iv) In ∆PDC and ∆BEC,
∵ ∠PDC = ∠BEC [From (1)]
Also, ∠DCP = ∠ECB [Common]
∴ Using AA similarity, we have
∆PDC ~ ∆BEC

Question 8.
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ABE ~ ∆CFB.
Solution:
We have a parallelogram ABCD in which AD is produced to E and BE is joined such that BE intersect CD at F.

Now, in ∆ABE and ∆CFB
∠BAE = ∠FCB [Opp. angles of a || gm are always equal]
∠AEB = ∠CBF [∵ Parallel sides are intersected by the transversal BE]
Now, using AA similarity, we have ∆ABE ~ ∆CFB.

Question 9.

Prove that:
(i) ∆ABC ~ ∆AMP
(ii) \(\frac{C A}{P A}=\frac{B C}{M P}\)
Solution:
We have ∆ABC, right angled at B and ∆AMP, right angled at M.
∴ ∠B = ∠M = 90°
(i) In ∆ABC and ∆AMP,
∠ABC = ∠AMP [From (1)]
and ∠BAC = ∠MAP [Common]
∴ Using AA similarity, we get
∆ABC ~ ∆AMP

(ii) ∵ ∆ABC ~ ∆AMP [As proved above]
∴ Their corresponding sides are proportional.
⇒ \(\frac{C A}{P A}=\frac{B C}{M P}\)

Question 10.
CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that 0 and H lie on sides AB and FE of ∆ABC and ∆EFG respectively. If ∆ABC ~ ∆FEG, show that:
CD AC
(i) \(\frac{C D}{G H}=\frac{A C}{F G}\)
(ii) ∆DCB ~ ∆HGE
(iii) ∆DCA ~ ∆HGF
Solution:
We have, two similar ∆ABC and ∆FEG such that CD and GH are the bisectors of ∠ACB and ∠FGE respectively.

(i) In ∆ACD and ∆FGH,
Since ∆ABC ~ ∆FEG
∠A = ∠F …………….. (1)
and ∠ACB = ∠FGE ⇒ \(\frac{1}{2}\) ∠ACB = \(\frac{1}{2}\) ∠FGE
⇒ ∠ACD = ∠FGH ……………. (2)
From (1) and (2),
∆ACD ~ ∆FGH [AA similarity]
∴ Their corresponding sides are proportional,
∴ \(\frac{C D}{G H}=\frac{A C}{F G}\)

(ii) In ∆DCB and ∆HGE,
Since ∆ABC ~ ∆FEG
⇒ ∠B = ∠E …………….. (1)
and ∠ACB = ∠FGE
∴ \(\frac{1}{2}\)∠ACB = \(\frac{1}{2}\)∠FGE
⇒ ∠DCB = ∠HGE ……………. (2)
From (1) and (2),
∆DCB ~ ∆HGE [AA similarity]

(iii) From (i) part, we get
∆ACD ~ ∆FGH
⇒ ∆DCA ~ ∆HGF

Question 11.
In the figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD⊥BCand EF⊥AC, prove that ∆ABD ~ ∆ECF.

Solution:
We have an isosceles ∆ABC in which
AB = AC.
⇒ Angles opposite to them are equal
∠ACB = ∠ABC ……………. (1)
In ∆ABD and ∆ECF,
∠ECF = ∠ABD [from (1)]
and AD⊥BC and EF⊥AC
⇒ ∠ADB = ∠EFC = 90°
∴ ∆ABD ~ ∆ECF [AA similarity]

Question 12.
Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆PQR (see figure). Show that ∆ABC ~ ∆PQR.

Solution:
We have ∆ABC and ∆PQR in which AD and PM are medians corresponding to sides BC and QR respectively such that
\(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A D}{P M}\)

∴ Using SSS similarity, we have ∆ABD ~ ∆PQM
∴ Their corresponding angles are equal.
⇒ ∠ABD = ∠PQM ⇒ ∠ABC = ∠PQR
Now, in ∆ABC and ∆PQR, \(\frac{A B}{P Q}=\frac{B C}{Q R}\) …………. (1)
Also, ∠ABC = ∠PQR …………… (2)
From (1) and (2),
∆ABC ~ ∆PQR. (SAS similarity)

Question 13.
O is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA² = CB CD.
Solution:
We have a ∆ABC and a point D on its side BC such that ∠ADC = ∠BAC.

In ∆BAC and ∆ADC,
∵ ∠BAC = ∠ADC [Given]
and ∠BCA = ∠DCA [Common]
Using AA similarity, we have
∆BAC ~ ∆ADC.
∴ Their corresponding sides are proportional.
⇒ \(\frac{C A}{C D}=\frac{C B}{C A}\)
⇒ CA × CA = CB × CD
⇒ CA² = CB × CD

Question 14.
Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ∆ABC ~ ∆PQR.
Solution:
Given : ∆ABC and ∆PQR in which AD and PM are medians.
Also, \(\frac{A B}{P Q}=\frac{A C}{P R}=\frac{A D}{P M}\) …………….. (1)
To Prove: ∆ABC ~ ∆PQR
Construction: Produce AD to E and PM to N such that AD = DE and PM = MN. Join BE, CE, QN and RN.

Proof: Quadrilaterals ABEC and PQNR are parallelograms, since their diagonals bisect each other at point D and M respectively.
⇒ BE = AC and QN = PR

⇒ ∆ABE ~ ∆PQN ⇒ ∠1 = ∠3 …………. (4)
Similarly, we can prove
⇒ ∆ACE ~ ∆PRN ⇒ ∠2 = ∠4 …………….. (5)
From (4) and (5)
⇒ ∠1 + ∠2 = ∠3 + ∠4
⇒ ∠A = ∠P ………….. (6)
Now, in ∆ABC and ∆PQR, we have
\(\frac{A B}{P Q}=\frac{A C}{P R}\) [From (1)]
and ∠A = ∠P [From (6)]
∴ ∆ABC ~ ∆PQR

Question 15.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Solution:
Let AB = 6 m be the pole and BC = 4 m be its shadow (in right ∆ABC), whereas DE and EE denote the tower and its shadow respectively.
∵ EF = Length of the shadow of the tower = 28 m
and DE = h = Height of the tower

In ∆ABC and ∆DEF, we have ∠B = ∠E = 90°
∠C = ∠F [∵ Angular elevation of the sun at the same time is equal]
∴ Using AA similarity, we have
∆ABC ~ ∆DEF
∴ Their sides are proportional i.e., \(\frac{A B}{D E}=\frac{B C}{E F}\)
⇒ \(\frac{6}{h}=\frac{4}{28}\) ⇒ h = \(\frac{6 \times 28}{4}\) = 42
Thus, the required height of the tower is 42 m.

Question 16.
If MD and PM are medians of triangles ABC and PQR, respectively where, ∆ABC ~ ∆PQR, prove that \(\frac{A B}{P Q}=\frac{A D}{P M}\).
Solution:
We have ∆ABC ~ ∆PQR such that AD and PM are the medians corresponding to the sides BC and QR respectively.

∵ ∆ABC ~ ∆PQR
And the corresponding sides of similar triangles are proportional.
∴ \(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{C A}{R P}\) ……………. (1)
∵ Corresponding angles are also equal in two similar triangles.
∴ ∠A = ∠P, ∠B = ∠Q and ∠C = ∠R …………… (2)
Since, AD and PM are medians.
∴ BC = 2BD and QR = 2QM
∴ From (1) \(\frac{A B}{P Q}=\frac{2 B D}{2 Q M}=\frac{B D}{Q M}\) ………… (3)
And ∠B = ∠Q ⇒ ∠ABD = ∠PQM
∴ From (3) and (4), we have
∆ABD ~ ∆PQM (SAS similarity)
∴ Their corresponding sides are proportional.
⇒ \(\frac{A B}{P Q}=\frac{A D}{P M}\)

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.1

Question 1.
In ∆ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:
(i) sin A, cos A
(ii) sin C, cos C
Solution:
In right angle ∆ABC, we have
AB = 24 cm, BC = 7 cm

∴ Using Pythagoras theorem,
AC² = AB² + BC²
⇒ AC² = 24² + 7² = 576 + 49 = 625 = 25²
⇒ AC = 25 cm

Question 2.
In the figure, find tan P – cotR?

Solution:
In right angle ∆PQR
Using the Pythagoras theorem, we get
QR² = PR² – PQ²
⇒ QR² = 13² – 12² = (13 – 12)(13 + 12) = 1 × 25 = 25
∴ QR = \(\sqrt{25}\) = 5 cm
Now, tanP = \(\frac{Q R}{P Q}=\frac{5}{12}\) , cotR = \(\frac{Q R}{P Q}=\frac{5}{12}\)
∴ tanP – cotR = \(\frac{5}{12}-\frac{5}{12}\) = 0.

Question 3.
If sinA = \(\frac{3}{4}\), calculate cosA and tanA.
Solution:
Let us consider, the right angle ∆ABC, we have
Perpendicular = BC and Hypotenuse = AC

Question 4.
Given 15 cot A = 8, find sin A and sec A.
Solution:
In the right angle triangle ABC, we have 15 cot A = 8

Question 5.
Given sec θ = \(\frac { 13 }{ 12 } \), calculate all other trigonometric ratios.
Solution:

Question 6.
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Solution:

From equations (i) and (ii) we get:
\(\frac{\mathrm{CD}}{\mathrm{EF}}=\frac{\mathrm{AC}}{\mathrm{BE}}=\frac{\mathrm{AD}}{\mathrm{BF}}\)
⇒ ∆CDA ~ ∆EFB [By SSS similarity]
⇒ ∠A = ∠B Hence Proved

Question 7.
If cot θ = \(\frac { 7 }{ 8 }\), evaluate:
(i) \(\frac { \left( 1+sin\theta \right) \left( 1-sin\theta \right) }{ \left( 1+cos\theta \right) \left( 1-cos\theta \right)}\)
(ii) cot²θ
Solution:

Question 8.
If 3 cot A = 4, check whether \(\frac { 1-tan^{ 2 }A }{ 1+tan^{ 2 }A }\) = cos² A – sin² A or not.
Solution:
In right angled ∆ABC, ∠B = 90°
For ∠A, we have:
Base = AB and perpendicular = BC. Also, Hypotenuse = AC

Question 9.
In triangle ABC, right angled at B, if tan A = \(\frac{1}{\sqrt{3}}\) find the value of:
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C
Solution:
In right ∆ABC, ∠B = 90°
For ∠A, we have
Base = AB, Perpendicular = BC,
Hypotenuse = AC

Question 10.
In ∆PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sinP, cosP and tanP.
Solution:
In right ∆PQR, Q = 90°

PR + QR = 25 cm and PQ = 5 cm
Let QR = x cm ⇒ PR = (25 – x) cm
∴ By Pythagoras theorem, we have
PR² = QR² + PQ²
⇒ (25 – x)² = x² + 5²
⇒ 625 – 50x + x² = x² + 25
⇒ – 50x = – 600
⇒ x = \(\frac{-600}{-50}\) = 12 i.e., QR = 12 cm
⇒ RP = 25 – 12 = 13 cm

Question 11.
State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) secA = \(\frac{12}{5}\) for some value of angle A.
(iii) cosA is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sinθ = \(\frac{4}{3}\) for some angle θ.
Solution:
(i) False
∵ A tangent of an angle is the ratio of sides other than hypotenuse, which may be equal or unequal to each other.
(ii) True
∵ cos A is always less than 1.
∴ \(\frac{1}{\cos A}\) i.e., sec A will always be greater than 1.
(iii) False
∵ ‘cosine A’ is abbreviated as ‘cosA’
(iv) False
∵ ‘cot A’ is a single and meaningful term whereas ‘cot’ alone has no meaning.
(v) False
∵ \(\frac{4}{3}\) is greater than 1 and sinθ cannot be greater than 1

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 12 Areas Related to Circles Ex 12.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.1

Unless stated otherwise, use π = \(\frac{22}{7}\).

Question 1.
The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.
Solution:
We have, r₁ = 19 cm and r₂ = 9 cm
∴ Circumference of circle – I = 2πr₁ = 2π(19) cm
and circumference of circle – II = 2πr₂ = 2π(9) cm
Sum of the circumferences of circle-I and circle-II
= 2π(19) cm + 2π(9) cm
= 2π(19 + 9) cm
= 2π(28) cm
Let R be the radius of the circle – III.
∴ Circumference of circle – III = 2πR
According to the condition, 2πR = 2π(28)
⇒ R = \(\frac{2 \pi(28)}{2 \pi}\) = 28 cm
Thus, the radius of the new circle = 28 cm.

Question 2.
The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.
Solution:
We have,
Radius of circle – I, r₁ = 8 cm
Radius of circle – II, r₂ = 6 cm
∴ Area of circle – I = πr₁² = π(8)² cm²
Area of circle-II = πr₂² = π(6)² cm²
Let the radius of the circle – III be R cm.
∴ Area of circle-III = πR²
Now, according to the condition,
πr₁² + πr₂² = πR²
⇒ π(8)² + π(6)² = πR²
⇒ π(8² + 6²) = πR²
⇒ 8² + 6² = R²
⇒ 64 + 36 = R²
⇒ 100 = R²
⇒ 10² = R² ⇒R = 10
Thus, the radius of the new circle = 10 cm.

Question 3.
The given figure depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.

Solution:
Diameter of the innermost (Gold scoring) region = 21 cm
Radius of the innermost (Gold scoring) region = \(\frac{21}{2}\) = 10.5 cm
∴ Area of Gold scoring region = π(10.5)² cm²
= \(\frac{22}{7} \times\left(\frac{105}{10}\right)^{2} \mathrm{cm}^{2}=\frac{22}{7} \times \frac{105}{10} \times \frac{105}{10} \mathrm{cm}^{2}\)
= \(\frac{22 \times 15 \times 105}{100}\) cm² = 346.50 cm²
Since, each band is 10.5 cm wide.
∴ Radius of Red scoring region
= 10.5 cm + 10.5 cm
= 21 cm
Area of Red scoring region
= π(10.5 + 10.5)² cm² – π(10.5)² cm²
= [π(21)² – π(10.5)²] cm²
= π[(21)² – (10.5)²] cm²

Area of White scoring region
= π[(42 + 10.5)² – (42)²] cm²
= π[(52.5)² – (42)²] cm²
= π[(52.5 + 42)(52.5 – 42)] cm²
= \(\frac{22}{7}\) × 94.5 × 10.5 = 22 × \(\frac{945}{10} \times \frac{15}{10}\)
= 3118.5 cm²

Question 4.
The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?
Solution:
Diameter of a wheel = 80 cm
∴ Radius of the wheel = \(\frac{80}{2}\) cm = 40 cm
So, circumference of the wheel
= 2πr = 2 × \(\frac{22}{7}\) × 40 cm
⇒ Distance covered by a wheel in one revolution = \(\frac{2 \times 22 \times 40}{7}\) cm
Distance travelled by the car in 1 hour (i.e., in 60 mins)
= 66 km = 66 × 1000 × 100 cm
∴ Distance travelled in 10 minutes

Thus, the required number of revolutions = 4375

Question 5.
Tick the correct answer in the following and justify your choice: If the perimeter and the area of a circle are numerically equal, then the radius of the circle is
(A) 2 units
(B) π units
(C) 4 units
(D) 7 units
Solution:
(A): We have
Numerical area of the circle
= Numerical circumference of the circle
⇒ πr² = 2πr
⇒ πr² – 2πr = 0
⇒ r² – 2r = 0
⇒ r(r – 2) = 0
⇒ r = 0 or   r = 2
But r cannot be zero
∴ r = 2
Thus, the radius of circle is 2 units.

In this article, we will share MP Board Class 10th Hindi Book Solutions Chapter 16 तुम वही दीपक बनोगे (दिवाकर वर्मा) Pdf, These solutions are solved subject experts from latest edition books.

MP Board Class 10th Hindi Vasanti Solutions Chapter 16 तुम वही दीपक बनोगे (दिवाकर वर्मा)

तुम वही दीपक बनोगे पाठ्य-पुस्तक के प्रश्नोत्तर

तुम वही दीपक बनोगे लघु-उत्तरीय प्रश्नोत्तर

प्रश्न 1.
कवि प्रतिपल सजग रहने की सलाह क्यों देता है?
उत्तर
कवि प्रतिपल सजग रहने की सलाह देता है। यह इसलिए कि वायुमण्डल विषैला हो गया है।

प्रश्न 2.
विषधरों को कीलने के लिए कवि कौन-सी युक्ति सुझाता है?
उत्तर
विषधरों को कीलने के लिए कवि मधुर-मादक-मत्त ध्वनि-सी युक्ति सुझाता है।

प्रश्न 3.
कवि को ऐसा क्यों लगता है कि प्राण आहादित नहीं है?
उत्तर
आज रागिनी बेसुरी है। संवेदनाएँ क्षत-विक्षत हैं और मन की बाँसुरी चुप है। इसलिए कवि को ऐसा लगता है कि प्राण आहादित नहीं है।

प्रश्न 4.
दामन बचाना कवि को कठिन क्यों लगता है?
उत्तर
दामन बचाना कवि को कठिन लगता है। यह इसलिए कि चारों ओर अग्नि की ज्वाला जल रही है।

प्रश्न 5.
कवि चारों दिशाओं में जलन क्यों अनुभव करता है?
उत्तर
कवि चारों दिशाओं में जलन अनुभव करता है। यह इसलिए कि मन मरुस्थल बन रहे हैं और तन की प्यास नहीं बुझ रही है।

तुम वही दीपक बनोगे लघु-उत्तरीय प्रश्नोत्तर

प्रश्न 1.
अमावस की कालिमा से कवि का क्या तात्पर्य है?
उत्तर
अमावस की कालिमा से कवि का तात्पर्य है-द्वैष और अविश्वास का अंधकार।

प्रश्न 2.
‘पोटली विष की भरी है’ का आशय स्पष्ट कीजिए।
उत्तर
पोटली विष की भरी है’ का आशय है। ईया, द्वेष, नफ़रत, स्वार्थ आदि का विस्तृत वातावरण।

प्रश्न 3.
वर्तमान स्थिति में मानव-संबंध के बारे में कवि के विचारों का उल्लेख कीजिए।
उत्तर
वर्तमान स्थिति में मानव-संबंध के बारे में कवि के विचार सुस्पष्ट हैं। उसका यह मानना है कि आज चारों ओर द्वैष और अविश्वास का इतना विषेला वातावरण फैल चुका है कि उससे निजात पाना न केवल कठिन है, अपितु अपने-आप में एक बहुत बड़ी चुनौती भी है।

प्रश्न 4.
जमाने के चलन को सुधारने के लिए कवि की युवाओं से क्या अपेक्षाएँ हैं?
उत्तर
जमाने के चलन को सुधारने के लिए कवि की युवाओं से अपेक्षाएँ हैं कि वे अमृतमयी मनुहार से प्राण संपादित करके बासंती बनेंगे।

तुम वही दीपक बनोगे भाषा-अनुशीलन

प्रश्न 1.
निम्नलिखित शब्दों के विलोम शब्द लिखिए
अमावस्या, मधुर, मूक, अमृत।
उत्तर
शब्द – विलोम
अमावस्या – पूर्णिमा
मधुर – कठोर
मूक – वाचाल
अमृत – विष।

प्रश्न 2.
निम्नलिखित सामासिक पदों का विग्रह कर समासों के नाम लिखिए
विषधर, वायुमण्डल, क्षत-विक्षत, अग्नि-ज्वाला, चतुर्दिश।
उत्तर

प्रश्न 3.
निम्नलिखित वाक्यांश के लिए एक शब्द लिखिए
उत्तर
वाक्यांश – एक शब्द
जो विष से भरा – विषैला
बसंत से सम्बंधित – वासंती
जहाँ कुछ उगता नहीं – मरुस्थल
अँधेरे से भरी रात्रि। – अमावस्या।

तुम वही दीपक बनोगे योग्यता-विस्तार

प्रश्न 1.
युवाओं को संबोधित कवियों की रचनाओं का संग्रह कीजिए एवं कक्षा में सुनाइए।
प्रश्न 2.
‘युवा देश की तस्वीर बदलते हैं’ इस विषय पर अपने विचार लिखिए।
प्रश्न 3.
आकाशवाणी और दूरदर्शन के ‘युवा कार्यक्रम’ को देखिए और उस में भाग लीजिए।
उत्तर
उपर्युक्त प्रश्नों को छात्र/छात्रा अपने अध्यापक/अध्यापिका की सहायता से हल करें।

तुम वही दीपक बनोगे परीक्षोपयोगी अतिरिक्त प्रश्नोत्तर

अर्थग्रहण संबंधी प्रश्नोत्तर

प्रश्न 1.
‘तुम वही दीपक बनोगे’ कविता का प्रतिपाय स्पष्ट कीजिए।
उत्तर-
‘तुम वही दीपक बनोगे’ कविता कविवर दिवाकर वर्मा की एक मार्मिक और हृदयस्पर्शी कविता है।
प्रस्तुत कविता देश की वर्तमान युवा पीढ़ी को समर्पित और संबोधित है। कवि का यह मानना है कि वर्तमान में चारों ओर द्वैष और अविश्वास का अंधकार छाया हुआ है। उसको भेदकर युवा वर्ग ही दीप-सा प्रकाश दे सकता है। कवि को यह पूरा-पूरा विश्वास है कि युवा वर्ग आज के विषैले समाज को अपने मधुर राग से, त्रसित मानवता को मलय । पवन के समान शीतलता से, खण्डित रिश्तों को प्रेम के सेतु से, तीक्ष्ण ताप से प्रताड़ित मानव को प्रेमपूर्वक तथा प्यासे हुए प्राणों को बासंती स्पंदन से अमृतदान दे सकता है।

प्रश्न 2.
कवि युवा वर्ग को कौन-सा दीपक बनने के लिए कह रहा है?
उत्तर
कवि युवा वर्ग को अमावस्या की कालिमा को धूप के समान उजियार कर देने वाला दीपक बनने के लिए कह रहा है।

प्रश्न 3.
आज मनुष्य के संबंध परस्पर कैसे हो रहे हैं?
उत्तर
आज मनुष्य के संबंध परस्पर खटाई पड़ने से फटे हए ध के समान हो रहे हैं।

प्रश्न 4.
रिक्त स्थानों की पूर्ति दिए गए विकल्पों में से उचित शब्दों के चयन से कीजिए।
1. है मुझे विश्वास दृढ़, तुम बन वही ………….. जलोगे। (आग, दीपक)
2. पोटली ………….. की भरी है। (अमृत, विष)
3. …………… भी बेसुरी है। (बाँसुरी, रागिनी)
4. …………… मन की बाँसुरी है। (प्राण, मूक)
5. …………… ही बस फट रहे हैं। (बम, संबंध)
उत्तर
1. दीपक
2. विष
3. रागिनी
4. मूक
5. संबंध।

प्रश्न 5.
दिए गए विकल्पों में से सही विकल्प का चयन कीजिए।
1. दिवाकर वर्मा का जन्म हुआ था-
1.1 जनवरी को,
2. 25 दिसम्बर को,
3. 20 जनवरी को,
4. 20 दिसम्बर को।
उत्तर
2. 25 दिसम्बर को

2. दिवाकर वर्मा की मुख्य विधा है
1. गीत
2. नवगीत
3. दोनों
4. कोई नहीं।
उत्तर
3. दोनों

3. दिवाकर वर्मा का नाटक है
1. रत्नावली
2. चंदनवन में आग
3. सुंदर बन
4. अब तो खामोशी तोड़ो।
उत्तर

4. दिवाकर वर्मा का जन्म हुआ था
1. 1920 में
2. 1930 में
3. 1940 में
4. 1941 में
उत्तर
4. 1941 में

5. दिवाकर वर्मा को पुरस्कार मिला है
1.कलश-सम्मान
2. कला-मंदिर
3. भोपाल का पवैया
4. उपर्युक्त सभी।
उत्तर
4. उपर्युक्त सभी।

प्रश्न 6.
सही जोड़ी का मिलान कीजिए।
कन्यादान – तुलसीदास
एक कंठ विषपापी – डॉ. वासुदेवशरण अग्रवाल
जानकी मंगल – महावीर प्रसाद द्विवेदी
कला और संस्कृति – दुष्यंत कुमार
अद्भुत आलाप – सरदार पूर्ण सिंह।
उत्तर
कन्यादान – सरदार पूर्ण सिंह
एक कंठ विषपापी – दुष्यंत कुमार
जानकी मंगल – तुलसीदास
कला और संस्कृति – डॉ. वासुदेवशरण अग्रवाल
अद्भुत आलाप – महावीर प्रसाद द्विवेदी

प्रश्न 7.
निम्नलिखित वाक्य सत्य हैं या असत्य? वाक्य के आगे लिखिए।
1. वायुमण्डल विषैला है।
2. प्राण आह्लादित हैं।
3. प्रतिपल सजगता चाहिए।
4. आज दूरियाँ घट रही हैं।
5. आज आदमी अंगार बनता जा रहा है।
उत्तर

1. सत्य
2. असत्य
3. सत्व
4. असत्य
5. सत्य।

प्रश्न 8. एक शब्द में उत्तर दीजिए
1. विष की क्या भरी है?
2. रागिनी भी क्या है?
3. आज क्षत-विक्षत क्या हैं?
4. आज क्या बढ़ रही हैं।
5. कौन अंगार बनता जा रहा है।
उत्तर

1. पोटली
2. बेसुरी
3. संवेदनाएँ
4. दूरियाँ
5. आदमी।

तुम वही दीपक बनोगे लघु-उत्तरीय प्रश्नोत्तर

प्रश्न 1.
किसका किससे विश्वास है?
उत्तर
कवि का आज के युवावर्ग से विश्वास है।

प्रश्न 2.
संजीवन जगाने के लिए कवि ने युवा वर्ग से क्या कहा है?
उत्तर
संजीवन जगाने के लिए कवि ने युवा वर्ग से तन में प्राण फूंकने के लिए कहा है।

प्रश्न 3.
आज क्या फट रहे हैं?
उत्तर
आज संबंध ही बस फट रहे हैं।

प्रश्न 4.
जमाने का चलन क्या हो गया है?
उत्तर
प्राण में कोकर उग रहे हैं। यही जमाने का चलन हो गया है।

तुम वही दीपक बनोगे कवि-परिचय

जीवन-परिचय-हिन्दी साहित्य के विशिष्ट सर्जक के रूप में दिवाकर वर्मा का सुनाम है। आपका जन्म 25 दिसंबर, 1941 को उत्तर-प्रदेश के सोरो, जिला एटा में हुआ था। शिक्षा-प्राप्ति के समय से ही आप साहित्य-रचना के क्षेत्र में सक्रिय हो गए। आपका साहित्य क्षेत्र मुख्य रूप से भारतीय संस्कृति और साहित्य है। इसके अतिरिक्त समाज और दर्शन भी आपके साहित्य की रचना की परिधि में आते हैं।

रचनाएँ-दिवाकर वर्मा की प्रमुख विधा गीत और नवगीत हैं। गीत रचनाओं में आस्था के स्वर, सूर्य के वंशज सुनो, और उलझते गए जाल में आदि उल्लेखनीय हैं। इसके अतिरिक्त सुंदरवन (बालगीत), अब तो खामोशी तोड़ो (गजल-संग्रह), चंदनवन में आग (दोहा-संग्रह) और रत्नावली (नाटक) भी उनकी सृजनात्मकता की उपलब्धियाँ हैं।

भावपक्ष-चूँकि दिवाकर वर्मा कवि हैं अतएव उनकी भावधारा सरल, सरस और सपाट है। उसमें तेज है, गति है, निरंतरता है और ताजगी है। इससे प्रस्तुत हुआ कथ्य अपने तथ्य को आसानी से स्पष्ट कर पाया है। इस प्रकार दिवाकर वर्मा का भावपक्ष रोचक और आकर्षक है।

कलापक्ष-दिवाकर वर्मा का कलापक्ष अलंकृत और चमत्कृत है। रसों में वीर रस और श्रृंगार रस के अधिक प्रवाह हैं। अलंकारों में अनुप्रास, रूपक, प्रतीक, उठोक्षा, मानवीकरण आदि अधिक प्रयुक्त हुए हैं। बिंबों और प्रतीकों को यथास्थान दिया गया है। मक्तक छंद की योजना सटीक और यथोचित रूप में है।

साहित्य में स्थान-दिवाकर वर्मा के साहित्य में ‘मानस’ की गंभीरता के साथ ही ‘मानव’ की उदारता का विशिष्ट गुण है। वे जीवन और काव्य में छद्म के स्थान पर सच्चाई के पक्षधर हैं। उन्होंने साहित्य, समाज और दर्शन पर गंभीर आलेख प्रस्तुत किए हैं, उनकी काव्य-रचनाएँ और समीक्षाएँ हिन्दी में विशेष ख्यात हुई हैं।
दिवाकर के महत्त्वपूर्ण साहित्यिक योगदान के लिए उन्हें अखिल भारतीय भाषा साहित्य सम्मेलन का रंजन कलश सम्मान, कला-मंदिर, भोपाल का पवैया, पुरस्कार एवं अन्य संस्थाओं से ‘रत्न भारती’ तथा ‘कला गुरु साहित्य सम्मान’ प्रदान किए गए हैं। दिवाकर वर्मा अपनी सतत साहित्य, रचनाधर्मिता के कारण अनेक संस्थाओं से संबद्ध रहकर साहित्य और संस्कृति की सेवा कर रहे हैं।

तुम वही दीपक बनोगे कविता का सारांश

कविवर दिवाकर वर्मा विरचित कविता ‘तुम वही दीपक बनोगे’ वर्तमान युवा-पीढ़ी के सोए हुए भावों को जगाने वाली कविता है। कवि को यह आशा ही नहीं पूरा विश्वास है कि आज का युवा वर्ग ही चारों ओर फैले हुए अंधकार को दूर करने के लिए वही दीपक बनकर प्रकाश फैलायेगा। वही आज के विषधरों को कील देने वाले बीन से ध्वनि करेगा। वही आज क्षत-विक्षत हो रही संवेदना को प्राण फूंक देने वाले संजीवन जगाने हेतु मलय समीर के समान चलेगा। वही बढ़ रही विजन की बस्ती बनाने के लिए आगे पैर रखोगे। वही शमन पर होला-हवाला और अंगार बनते जा रहे आदमी के लिए ताप का मर्दन करोगे। आज चारों ओर हो रहे जलन में अमृतमयी मनुहार से प्राण स्पदित करने वाले बसंती हवा बनोगे।

तुम वही दीपक बनोगे संदर्भ, प्रसंग सहित व्याख्या

1. जो अमा की कालिमा भी
धूप सी उजियार कर दे
है मुझे विश्वास दृढ़, तुम बन वही दीपक जलोगे!

वायुमण्डल है विषेला विषधरों की भी बहुलता,
पोटली विष की भरी है,
चाहिए प्रतिपल सजगता,
मधुर-मादक-मत्त ध्वनि से विषधरों को कील दे जो
है मुझे विश्वास तुम उस बीन से निश्चित बजोगे!

शब्दार्व-अमा-अमावस्या। विषधर-साँप।

संदर्भ-प्रस्तुत पद्यांश हमारी पाठ्य-पुस्तक ‘हिंदी सामान्य’ 10वीं में संकलित कवि दिवाकर वर्मा विरचित कविता ‘तम वही दीपक बनोगे’ से है।

प्रसंग-प्रस्तुत पद्यांश में कवि ने आज के युवावर्ग से वर्तमान समय में फैले हुए अंधकार के लिए दीपक बनने का विश्वास रखते हुए कहा है कि

व्याख्या-अमावस्या की काली रात को तुम धूप की तरह उजाला से भर दो। मुझे दृढ़ विश्वास है कि तुम इस प्रकार का अवश्य दीपक बनोगे। कवि का पुनः कहना है कि आज सारा वातावरण विषैला हो चुका है। इससे विषधरों की भरमार हो रही है। विष की पोटली भर चुकी है। इसके प्रति हर क्षण सजग रहने की आवश्यकता है। आज मधुर मादक मत्त ध्वनि से इन फैले हुए विषधरों को कील देने की आवश्यकता है। मुझे विश्वास है कि तुम उस बीन से निश्चित ध्वनि निकालोगे।

विशेष-

1. सामयिक दशा पर ज्वलंत विचार प्रस्तुत है।
2. भाषा लाक्षणिक है।

सौंदर्य-बोध पर आधारित प्रश्नोत्तर

(क) भाव-सौंदर्य
प्रश्न 1.
उपर्युक्त पद्यांश के भाव-सौंदर्य पर प्रकाश डालिए।
उत्तर
उपर्युक्त पद्यांश का भाव-स्वरूप ओजस्वी है। समय की बदलती तीखी दशा का तीव्रोल्लेख है। आज विषैले वातावरण पर सीधा प्रकाश डालकर कवि ने समय की नब्ज को न केवल पहचानने की कोशिश की है, अपितु उसको दूर करने की भी प्रेरणा दी है।

(ख) शिल्प-सौंदर्य
प्रश्न 1.
उपर्यक्त पयांश के शिल्प-सौंदर्य पर प्रकाश डालिए।
उत्तर
उपर्युक्त पद्यांश का शिल्प-सौंदर्य मिश्रित शब्दों का है। संपूर्ण कथ्य सरल, सपाट और सटीक भाषा में प्रस्तुत है। व्यंजना शब्दावली से प्रस्तुत हुई व्यंजनात्मक शैली प्रभावशाली रूप में है।

विषय-वस्तु पर आधारित प्रश्नोत्तर

प्रश्न 1.
उपर्युक्त पयांश का मुख्य भाव लिखिए।
उत्तर
उपर्युक्त पद्यांश का मुख्य भाव आज के विषैले वातावरण को समाप्त करके शांत और सुखद वातावरण की स्थापना का है। इसके लिए कवि ने आज के युवा वर्ग के प्रति दृढ़ विश्वास व्यक्त कर उन्हें प्रेरित करने का प्रयास किया है।

2. प्राण आहादित नहीं औ’
रागिनी भी बेसुरी है,
क्षत-विक्षत संवेदनाएँ हैं,
मूक मन की बाँसुरी है,
आज संजीवन जगाने
फूंक दे जो प्राण तन में
है मुझे विश्वास दृढ़ तुम मलय-मारुत सम चलोगे!

बढ़ रही हैं दूरियाँ
औ’ वर्ग नित नव बन रहे हैं,

व्याख्या-अमावस्या की काली रात को तुम धूप की तरह उजाला से भर दो। मुझे दृढ़ विश्वास है कि तुम इस प्रकार का अवश्य दीपक बनोगे।
कवि का पुनः कहना है कि आज सारा वातावरण विषैला हो चुका है। इससे विषधरों की भरमार हो रही है। विष की पोटली भर चुकी है। इसके प्रति हर क्षण सजग रहने की आवश्यकता है। आज मधुर मादक मत्त ध्वनि से इन फैले हुए विषधरों को कील देने की आवश्यकता है। मुझे विश्वास है कि तुम उस बीन से निश्चित ध्वनि निकालोगे।

विशेष-

1. सामयिक दशा पर ज्वलंत विचार प्रस्तुत है।
2. भाषा लाक्षणिक है।

सौंदर्य-बोध पर आधारित प्रश्नोत्तर

(क) भाव-सौंदर्य
प्रश्न 1.
उपर्युक्त पद्यांश के भाव-सौंदर्य पर प्रकाश डालिए।
उत्तर
उपर्युक्त पद्यांश का भाव-स्वरूप ओजस्वी है। समय की बदलती तीखी दशा का तीव्रोल्लेख है। आज विषैले वातावरण पर सीधा प्रकाश डालकर कवि ने समय की नब्ज को न केवल पहचानने की कोशिश की है, अपितु उसको दूर करने की भी प्रेरणा दी है।

(ख) शिल्प-सौंदर्य
प्रश्न 1.
उपर्यक्त पयांश के शिल्प-सौंदर्य पर प्रकाश डालिए।
उत्तर
उपर्युक्त पद्यांश का शिल्प-सौंदर्य मिश्रित शब्दों का है। संपूर्ण कथ्य सरल, सपाट और सटीक भाषा में प्रस्तुत है। व्यंजना शब्दावली से प्रस्तुत हुई व्यंजनात्मक शैली प्रभावशाली रूप में है।

विषय-वस्तु पर आधारित प्रश्नोत्तर

प्रश्न 1.
उपर्युक्त पयांश का मुख्य भाव लिखिए।
उत्तर
उपर्युक्त पद्यांश का मुख्य भाव आज के विषैले वातावरण को समाप्त करके शांत और सुखद वातावरण की स्थापना का है। इसके लिए कवि ने आज के युवा वर्ग के प्रति दृढ़ विश्वास व्यक्त कर उन्हें प्रेरित करने का प्रयास किया है।

3. किस तरह दामन बचायें
प्रज्वलित है अग्निज्वाला,
तीलियाँ तो संवरित हैं
शमन पर हीला-हवाला,
आदमी अंगार बनता जा रहा ।
ऐसे समय में
है मुझे विश्वास तुम ही ताप का मर्दन करोगे!

उग रहे मन-प्राण में कीकर
जमाने का चलन है,
मन बने मरुस्थल, तृषित तन
औ’ चतुर्दिश ही जलन है,
प्राण स्पंदित करे
अमृतमयी मनुहार से जो
है मुझे विश्वास दृढ़ तुम पवन बासंती बनोगे!

शब्दार्च-दमन-वस्त्र। शमन-शांति। ताप-गर्मी। मर्दन-नाश। कीकर-चुभन । चतुर्दिश-चारों दिशाओं। तृषित-प्यासा। स्पंदित-गतिशील । मनुहार-मनाना।

संदर्भ-पूर्ववत्।

प्रसंग-पूर्ववत्।

व्याख्या-आज की कठिन स्थिति यह है कि आज चारों ओर दखों और विषमताओं की अग्निज्वाला प्रज्वलित हो रही है। शांति के नाम पर होला-हवाला हो रहा है। आज आदमी एक-दूसरे के प्रति अंगार बनते जा रहा है। ऐसे समय में मुझे पूरा भरोसा है कि तुम ही अपेक्षित ताप का नाश कर डालोगे। आज यह भी हो रहा है कि चारोंओर मन-प्राण में कीकर उग रहे हैं। शायद यही जमाने का प्रचलन हो गया है। आज प्रायः मन मरुस्थल बन गया है, जिससे तन की प्यास बुझ नहीं पा रही है। इस प्रकार चारों दिशाओं में प्यास की जलन बढ़ रही है। आज प्राणों की अमृतमयी मनुहार से जो गतिशील कर सकता है, तो केवल तुम्हीं कर सकते हो। मुझे पूरा-पूरा भरोसा है कि तुम्हें बसंत हवा बनकर इस तीखे वातावरण को रसमग्न कर सकोगे।

विशेष-

1. वर्तमान समाज की विडंबनाओं का सपाट चित्र है।
2. व्यंग्यात्मक शैली है।

सौंदर्य-बोध पर आधारित प्रश्नोत्तर
(क) भाव-सौंदर्य

प्रश्न 1.
उपर्युक्त पद्यांश के भाव-सौंदर्य पर प्रकाश डालिए।
उत्तर
उपर्युक्त पद्यांश का भाव-योजना तत्सम प्रधान तद्भव शब्दों से पुष्ट है। भावों की क्रमबद्धता, सहजता, प्रवाहमयता और उपयुक्त देखते ही बनती है। ये भाव बड़े ही सुपरिचित और विश्वसनीय ढंग से प्रस्तुत किए गए हैं। इसलिए रोचक बन गए हैं।

(ख) शिल्प-सौंदर्य
प्रश्न 1.
उपर्युक्त पयांश के शिल्प-सौंदर्य पर प्रकाश डालिए।
उत्तर
उपर्युक्त पद्यांश का भाषा-शैली लाक्षणिक और अलंकृत है। व्यंजना शब्द-शक्ति की प्रधानता है तो रूपक और अनुप्रास अलंकार का मण्डन देखने योग्य है। करुण और वीर रस का मिला-जुला प्रवाह भाव और भाषा की सजीवता में वृद्धि कर रहा है।

विषय-वस्तु पर आधारित प्रश्नोत्तर

प्रश्न 1.
उपर्युक्त पयांश के भाव को सस्पष्ट कीजिए।
उत्तर
उपर्युक्त पद्यांश में कवि ने आज के मनहूस, विषम और दुखद वातावरण का चित्र खींचते हुए कठिन जीवन के विविध पक्षों को सामने लाने का प्रयास किया है। इस प्रकार की विडंबनापूर्ण जिंदगी को सखद बनाने के लिए वर्तमान युवा पीढ़ी को प्रेरित करते हुए आत्म-विश्वासपूर्वक आह्वान किया है।

In this article, we will share MP Board Class 10th Social Science Book Solutions Chapter 21 Globalisation Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Social Science Solutions Chapter 21 Globalisation

MP Board Class 10th Social Science Chapter 21 Text book Exercises

Objective Type Questions

Question 1.
Multiple Choice Questions
(Choose the correct answer from the following)

Question (i)
Globalisation has improved the standard of living of –
(a) Poor class
(b) Higher class
(c) Rural areas
(d) All classes of the society
Answer:
(c) Rural areas

Question (ii)
Which industries are closed due to globalisation –
(a) Large – scale industries
(b) Multinational companies
(c) Small – scale industries
(d) Industries of all type
Answer:
(d) Industries of all type

Question (iii)
The process of globalisation began in India from the year –
(a) 1947
(b) 1951
(c) 1991
(d) 2001.
Answer:
(c) 1991

Question (iv)
The World Trade Organisation was established in –
(a) 1985
(b) 1995
(c) 2001
(d) 2005.
Answer:
(b) 1995

Question (v)
The main basis of globalisation is –
(a) foreign trade
(b) internal trade
(c) agricultural trade
(d) small scale industry.
Answer:
(a) foreign trade

Question 2.
Fill in the blanks: (MP Board 2009)

1. At present the number of reserved industries is …………….
2. Under globalisation the transportation of goods and services is ……………. between various countries.
3. The companies who produce goods in different countries are called ……………. (MP Board 2009)

Answer:

1. Eight
2. Liberalised
3. Mulitnational Companies.

MP Board Class 10th Social Science Chapter 21 Very Short  Answer Type Questions

Question 1.
What was the foreign policy of India prior to the year 1991?
Answer:
Liberalisation and Globalisation.

Question 2.
What are those companies called who produce in more than one country?
Answer:
Multinational Company.

Question 3.
Which consumer class is benefited more by globalisation?
Answer:
Higher Class.

Question 4.
What do you understand by multinational companies?
Answer:
A company whose work is related to production and sale spread all over the world is known as multinational companies.

MP Board Class 10th Social Science Chapter 21  Short  Answer Type Questions

Question 1.
State the meaning of globalisation. Who have been benefited by globalisation?
Answer:
Globalisation is meant the working of whole world together with co – operation and co-ordination in the form of a market. Under the process of globalisation the restrictions on the inflow and outflow of goods and services from one country to another are withdrawn. Thereby the market prices start working freely in whole of the world. As a result the prices in all the countries becomes near about equal. In this way as a result of globalisation all the markets of the world are unified. The upper strata of our society is the most beneficiary of this system.

Question 2.
What is foreign trade?
Answer:
The foreign trade has linked all the countries of the world together. There are several big companies of the world, which are called multinational companies. These companies sell their products in several countries of the world. This is noticeable here that a multinational company is that who produces goods in more than one country. These companies produce goods on large scale and sell these produced goods in several countries.

Question 3.
What is the unification of market?
Answer:
India has adopted the New Economics Policy since 1991. The main objective of this policy is to take advantage of the progress of the world and technical knowledge and thus accelerate the economic development of the country. This policy has opened the way of liberalisation by avoiding the existing administrative restriction. Along with this the efforts has been made to motivate private investment and to attract the foreign capital. It can be said in brief that through the new policy a new chapter has began by linking the India market economy with the economy of the world.

Question 4.
How are the small producers affected by globalisation?
Or
Write the effect of Globalization on the small producers? (MP Board 2009)
Answer:
All people have not been benefited by the globalisation, thus now it is clear that globalisation has affected the Indian economy in both the ways inversely and positively. The reality is this, that globalisation has benefited the industries and business and we have approached in the world market. Several Indian producers have got the form of multinational company.

The development rate of Indian economy has also become more than 9 per cent. Consumers are now getting world – level standardised goods. But in reality the people of all classes have not got the benefit of globalisation equally. Several small and tiny industries have been closed due to competition. The problems like poverty and unemployment prevailing in the developing countries like India have become more complicated.

Along with this due to the influence of developed countries upon. The World Trade Organisation, developing and backward countries are not getting the benefits of globalisation. The new labour – laws are not affecting the labour class favourably. Therefore the efforts are needed.

MP Board Class 10th Social Science Chapter 21 Long Answer Type Questions

Question 1.
What do you understand by globalisation? Explain the causes that motivate the process of globalisation.
Or
What is globalisation? Explain the causes that motivate this process. (MP Board 2009)
Answer:
Globalisation means opening ftp the economy to facilitate its integration with the world economy. In this situation economics are integrated with each other. In case of globalisation it becomes easy to sell goods and services abroad.

Factors Inspiring Globalisation:

1. Expansion of Technical Knowledge:
During last 50 years the technical knowledge has developed rapidly. The transport technology has made it possible to send goods upto distant places in a lower cost. Telecommunication facilities such as internet, mobile phone, fax etc. has made the task of connecting people with °ach other throughout the world very easy. Communication satellite has brought a revolutionary change by expanding these facilities. As a result globalisation has expanded rapidly.

2. Process of Liberalisation:
Till the middle of twentieth century the production was limited mainly within the boundaries of the countries. So many countries imposed strict restrictions to protect the goods produced by them from competition. India also during the decades of 1950 and 1960 had permitted only to import the necessary goods as machineries, fertilizers and petroleum etc. Several industries developed due to this policy and India became self – reliant in several sectors.

3. Expansion of Competition and Market:
Competition has a special importance in capitalist economic system. In this system different producing compaines in order to seek the hold on market take support of competition. For this purpose these companies along with cutting down the prices use advertisements and various medium of convincing and canvassing the buyers.

4. Expansion of Multinational Companies:
Multinational companies play a significant role in linking the distant countries with each other. These companies set up their factories for production purpose in those countries where they get cheaper labour and other means of production and due to this the capacity of these companies increasesto compete.

Question 2.
How do foreign trade helps in unifying the markets of various countries?
Answer:
During the decades of 1970 and 1990 some such changes took place due to which began the process of liberalising foreign trade. For example, dissolution of Union of Socialist Soviet of Russia, economic unification of Europe, emergence of Japan as a major power of the world and economic development of Korea, Singapore and Hongkong. As a result several countries agreed to liberalise the world trade.

This strengthen, the process of liberalization. After the establishment of The World Trade Organisation1 in 1955, almost all the countries of the world have reduced their taxes on imports and have opened the markets of their countries for other countries. So, the process of globalisation has gained speed.

Question 3.
Explain the economic condition of India after the globalisation, and discuss the problems created by gloabaissation?
Or
Explain the main problems which are created by globalisation? (MP Board 2009, 2010)
Answer:
The following aspects are there undertaking The system of globalisation in India:
1. Industrialisation:
The industrialisation in the country has been accelerated due to the industrial policy declared in the year 1991 and the reforms done afterwards. Now number of industries reserved exclusively for public sector is reduced only to three. It means that the private sector has got sufficient opportunities for expansion.

2. Increase in Foreign Investment:
After the globalisation the multinational companies have increased their investment in India. These companies have shown their interest in such investment as cell phones, motorcars, electronic equipments, cold drinks, junk food materials and banking services. New opportunities of employment have been created by these industries and services.

3. Advantage to Indian Companies:
Globalisation has set up several Indian companies as multinational companies. These companies have expanded their activities at world level. For example, Tata Motars, Infosys, Ranbaxy, Asian Paints, Sunderm Fastners can be taken as examples which have now become multinational companies.

The Indian companies, in order to compete with foreign companies, have adopted the latest technology and have raised the standards of their production. There are some companies who have improved their condition by collaborating with multinational companies.

Problems Created by Globalisation:
This is not true to say that India has only benefited by globalisation. The reality is this, the globalisation has also created several problems. Following are these problems.

1. Impact on Small Producers:
Globalisation has adversely affected the several small industries of India. Small industries are not capable to compete with goods produced in foreign. As a result several small industries have closed.

2. Uncertainty of Employment:
Lives of labourers have been greatly affected by globalisation. These days due to growing competition, maximum employers like flexibility in providing employment to labourers. It means that the jobs of labourers are not secure. The factory owners, to minimize the cost, provide temporary employment to the labourers; so that they may not have to pay them salary round the year.

1. Benefit not to all
2. Regional disparities.

MP Board Class 10th Social Science Chapter 21 Additional Important Questions

Objective Type Questions

Question 1.
Multiple Choice Questions:
(Choose the correct answer from the following)

Question (i)
The New Industrial Policy was adopted since –
(a) 1989
(b) 1990
(c) 1991
(d) 1992.
Answer:
(c) 1991

Question (ii)
The main base of globalisation is –
(a) Multinational Companies
(b) Foreign Trade
(c) America
(d) European Union.
Answer:
(b) Foreign Trade

Question (iii)
New economic policy liberated the –
(a) Private sector
(b) Public sector
(c) Industrial sector
(d) Tertiary sector.
Answer:
(a) Private sector

Question (iv)
W.T.O. means –
(a) World Terrorist Organisation
(b) World Transport Organisation
(c) World Trade Organisation.
(d) World Temperate Organisation
Answer:
(c) World Trade Organisation.

Question 2.
Fill in the blanks:

1. World bank is an International ……………… institution that extends financial assistance to member nation for development purposes.
2. Financial Crunch is a situation in which the government ……………… falls drastically short of government expenditure.
3. Privatisation is defined as transfer of ownership and control from the public sector to the ……………… sector.
4. The private sector allowed to establish industries and
   business but subject to ……………… and ………………

Answer:

1. Financial
2. Revenue
3. Private
4. Control regulation

Question 3.
True and False type questions:

1. New Economic Policy or N.E.P. was adopted in 1995.
2. Globalization means decreasing integration between different economics of the world.
3. Privatisation means to bring most of the enterprises of the country under the ownership, control and management of the private sector.
4. Liberalisation means to rid industry and trade of unnecessary restrictions and make them more competitive.

Answer:

1. False
2. False
3. True
4. True.

Question 4.
Match the following:

Answer:

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

Answer in One – Two Words or One Sentence:

Question 1.
Define World Trade Organisation (WTO)?
Answer:
WTO is the global international organisation dealing with the rules of trade between nations.

Question 2.
When was WTO established?
Answer:
WTO was established on January 1, 1995.

Question 3.
Where is WTO located?
Answer:
WTO is located at Geneva, Switzerland.

Question 4.
What is GATT?
Answer:
GATT (General Agreement on Trade and Tariffs) was the forum for negotiating trade agreement between nations between 1947 – 1994.

Question 5.
When did India become the member of WTO?
Answer:
India became the member of the WTO on January 1, 1995.

Question 6.
Mention two functions of WTO?
Answer:

1. Administering WTO trade agreements.
2. Forum for trade negotiations between countries.

MP Board Class 10th Social Science Chapter 21 Very Short  Answer Type Questions

Question 1.
Define sustainable development.
Answer:
Sustainable development is an economic development where necessities of generations are not compromised by the pleasure of present generation.

Question 2.
Explain the concept of sustainable economic development. Give two points?
Answer:

1. Development should take place without polluting and damaging the environment.
2. Development should be continuous and it should fulfil the needs and aspiration of the present and future generations.

Question 3.
What is the role of WTO? What two benefits will India get by being a member of WTO?
Answer:
WTO is the global trade organisation dealing with the rules of trade between nations. India, as a member of WTO will be benefited in the following ways:

1. Promotion and expansion of trade among countries.
2. (a) Less restrictions on Indian exports.
   (b) Getting technology from developed countries.

Question 4.
Define Liberalisation?
Answer:
Liberalisation means freedom to the private sector to run those activities which were earlier confined to only public sector. Secondly, Private sectors have given many relaxations from the rules and regulations.

Question 5.
What is NRI?
Answer:
NRI (Non-Resident Indian) is an Indian who does not normally live and work in India, but in some other country. He holds Indian citizenship and Indian passport.

MP Board Class 10th Social Science Chapter 21 Short Answer Type Questions

Question 1.
Mention three freedoms given to the private sector industry by liberalisation?
Answer:

1. Industrial licensing has been abolished except for five industries.
2. Number of public sector industries has been reduced from 17 to 3.
3. Freed from regulation like permission for importing materials.

Question 2.
Enlist main objectives of New Economic Policy?
Answer:
Following were the objectives of new economic policy:

1. Liberalisation of the economy.
2. Expansion of private sector.
3. Encouragement of private foreign investment.
4. Modernisation of agriculture.
5. Controlling fiscal deficit.

Question 3.
What do you mean by disinvestment? How far did we succeed in this programme?
Answer:
Disinvestment means selling shares of its performing enterprise by the government. It is one method of privatization. Our programme of privatization through investments has succeeded to some extent only.

Question 4.
Mention the functions of WTO?
Answer:
Functions of World Trade Organisation (WTO):

1. Administering trade agreements between nations.
2. Forum for trade negotiations.
3. Handling trade disputes between nations.
4. Monitoring national trade policy.
5. Technical assistance and training for developing countries and.
6. Cooperation with other international organisation.

Question 5.
How can you justify the presence and impact of globalisation?
Answer:
The presence and impact of globalisation can be justified by the following facts:

1. Development of social consciousness.
2. Fast and quick technological changes.
3. Global form of modern business.
4. Formation of International Monatory Fund (IMF).
5. Foreign collaboration and joint ventures, financial and assistance.
6. Globalisation of marketing through Cable TV network and satellite links and
7. Introduction of internet facilities like e-mail and e- commerce services.

Question 6.
Explain in short meaning of the term ‘Globalization’. Write its main characteristics (features on related points) also?
Answer:
Meaning:
Globalization means integration, unification or integration of domestic economy with the world economy. Characteristics (Features or Related Points):

1. Due to globalisation producers from outside can sell their goods and services in India. We can do the same with our goods and services.
2. Entrepreneurs from other countries can invest in India and the Indian entrepreneur can also do the same in other countries.

Question 7.
Discuss some positive aspects of Globalisation and Liberalisation?
Answer:
Globalisation and Liberalisation have helped in rapid economic development.

1. A large industrial base created; increase in industrial production.
2. Proportion of people living below the poverty line less.
3. Self – sufficient in food.
4. Mobilised its savings.
5. Generated its own resources for development.
6. Large pool of scientists and technically skilled working person.
7. Export – oriented industries.

Question 8
Differentiate between policy of restriction and policy of liberalization?
Answer:
Liberalization:
It means removing unnecessary trade restrictions and making the economy more competitive. New Economic policy liberated the private sector from strict control and licensing.

Restriction:
If restrictions are imposed on economic activities by government policies, it is called the policy of restriction or restrictive policy. No real economy is completely free of restrictions. When these restrictions are removed, it is called the policy of liberalization.

Question 9.
What measures have been taken for globalisation of the economy of India?
Answer:
Free interaction among economies of the world in the field of trade finance, production techlnologies and investment is termed as globalisation of the economy. It encourages foreign trade and institutional investment. Following measures are adopted for it:

1. Devaluation of rupee by 20% in July 1990 – 91.
2. Full convertibility was offered in 1993-94 to encourage exports earnings.
3. Long period trade policy of remove restrictions.
4. Encouragement to open competition.
5. Modification of custom and tariff.

Question 10.
What is sustainable development? What are its important features?
Answer:
The development, which takes care of the needs of present generation without compromising with the needs of future generation is termed as sustainable development.

Important features of sustainable development:

1. Economic growth
2. No or the least compromise with the necessities of the future generation
3. Pollution free economic development
4. No or the least depletion of non – renewable resources and
5. Preservation of environmental and exhaustible resources.

Question 11.
Explain the meaning of World Trade Organisation?
Answer:
World Trade Organisation (WTO):
The global international organisation, working on multilateral trading system, where trade agreemeftts are negotiated and signed by a large majority of world’s trading nations and ratified in their parliaments are known as W.T.O. Its objective is to assist trade flowing smoothly, freely, fairly and predictable among nations. WTO was established on Jan 1, 1995 in Geneva, Switzerland. India also became its member on January 1,1995.

MP Board Class 10th Social Science Chapter 21 Long Answer Type Questions

Question 1.
Review the status of Indian economy before the new economic policy – 1991?
Answer:
1. Self – reliance:
The five – year plans imed at obtaining self – reliance. The dependency upon foreign aid was reduced in fulfilment of this aim. During this period self – reliance in agriculture sector was achieved and vast industrial sector was developed.

2. Foreign Trade:
In the year 1991 imports were kept under control. During this period only necessary goods like machineries, fertilizer and petroleum were mainly imported. To protect the domestic producers from foreign competition the policy of protection was adopted. Therefore, the trade during this period increase slowly. The contribution of India in total world trade in 1951 was near about. 1 per cent which reduced upto 0.6 percent in 1991.

3. National and Per Capita Income:
During the period of 1951 to 1991 the national income increased at the average rate of 4.0 per cent. But due to the rapid growth in population during this period the per capita income increased at a very slow speed.

4. Increase in the Opportunities of Employment:
During this period though the efforts were made to increase the job opportunities even then the problem increased day – by – day. The problem of unemployment became very complicated till 1991.

5. Crisis of Foreign Currency:
India adopted the policy of reducing imports between the years 1951 to 1991. But India needed foreign currency for the import of petroleum products machinery and other necessary goods. India had to take international loan to import these goods. Therefore, India was trapped in the crisis of foreign exchange.

6. Price rise:
During the period of planning India had to face the problem of continuous price rise. After the First Five-Year Plan during the years 1956 to 1991 the rate of inflation in India was between 5 to 6 per cent.

Question 2.
Explain briefly the key features of Indian economy after adoption of the New Economic Policy?
Answer:
Key Features of Indian Economy after adoption of New Economic Policy:
1. Industrial activities which were restricted for public sector were also opened up to private sector except for industries having national importance tike defence, space research etc.

2. Relaxation in regulations tike quota system, industrial licensing, concentration of economic power etc. This was meant to give freedom to the businesses to undertake activities having growth potential.

3. Permission to import raw material was eased to make/it more competitive for businesses to reduce cost and undertake better technology.

4. Pricing and distribution was made free to give businesses free hand to deal with pricing and distribution strategies.

5. Restriction on investement and increase in production capacity was eased to pave the way for industrial growth.

6. The economy was opened to integrate it with world economy in terms of flow of goods and services investment etc.

Question 3.
Explain the process of globalisation?
Or
Give the meaning of globalisation and describe the steps taken in this direction?
Or
Write the main five factors which promote globalization? (MP Board 2009)
Answer:
Process of Globalisation or Steps taken for Globalisation:
1. Raising Foreign Equity Participation:
Prior to July 1999 f foreign equity participation was subjected to lot of approvals, sanctions and constraints. It was restricted to 40%. Now it has been increased to 51% and the approvals have been made routine work.

2. Devaluation of Rupee:
Rupee was devalued by 20% in July 1990 – 91. The devaluation was made to encourage exports and discourage imports. It also aimed at inflow of more foreign capital.

3. Convertibility of Rupee:
The government offered partial convertibility of rupee through the budget of 1992 – 93. Full convertibility was offered in 1993 – 94. Convertibility of rupee was aimed at encouraging export earnings.

4. Long Period Trade Policy:
The government announced foreign trade policy for a period of five years i.e. 1992 – 97. The sole purpose of this policy was liberalisation. It also removed restrictions on external trade.

5. Encouragement to Open Competition:
Exports and imports were left to market forces, government control was minimised.

6. Modification of Customs and Tariff:
In order to build up our competitive strength, customs and tariff policies were modified to promote international trade.

7. Modernisation of the Economy:
The new economic policy accords top priority to modern techniques and technology. It also promotes computers and electronics industries. It has made the Indian industries dynamic. All foreign collaborations concerning higher technology have been (created by the government.

8. Privatisation of the Company:
It means removing strict control over private sector and making them free to take necessary decisions. Now, the new policy tries to expand private sectors.