MCQ Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 10

If the kinetic Energy possessed by a man of

$50$ kg is

$625J$ , the speed of man is

0%
$5m{s^{ - 1}}$
0%
$50m{s^{ - 1}}$
0%
$0.5m{s^{ - 1}}$
0%
$500m{s^{ - 1}}$

A ball of mass 3 kg moving with a velocity of 4 m/s undergoes a perfectly- elastic collision with a stationary ball of mass m. After the impact is over, the kinetic energy of the 3 kg ball is 6 J. The possible value of m is/are :

0%
1 kg only
0%
1 kg , 9kg
0%
1 kg, 6kg
0%
6kg only

A rubber band of length

$\lambda$ has a stone of mass

$m$ tied to its one end. It is whirled with speed

$v second$ that the stone described a horizontal circular path. The tension

$T$ in the rubber band is -

0%
zero
0%
$mv^2/\lambda$
0%
$>(mv^2)/\ell$
0%
none

Explanation

As there is only Tension T is theonly force. So it is the necessary centripetal force

$T= \dfrac{mv^2}{\lambda}$

Water is dragged from a well of hepth 30 m using bucket of weight 50kg with water If the weight of the rope is 0.4 kg per metre, the amount of work done is

0%
18228J
0%
16464J
0%
14700J
0%
11760J

Two bodies traveling with velocity

$20ms^{-1}$ and

$40ms^{-1}$ have same K.E. If the mass of the body having velocity

$40ms^{-1}$ is 8 kg the mass of other body is

0%
8 Kg
0%
16 kg
0%
24 kg
0%
32 kg

Explanation

Given

$V_1 = 20 m/s$ and

$V_2 = 40 m/s$

$M_2 = 8\ kg$

$K.E_1 = K.E_2$

$\dfrac{1}{2}M_1V_1^2 = \dfrac{1}{2}M_2V_2^2$

$\implies M_1 = M_2 \times \dfrac{V_2^2}{V_1^2} = 8\times \dfrac{40^2}{20^2} = 32\ kg$

A body of mass 2

$\mathrm { Kg }$ moving (initially) with 10

$\mathrm { m } / \mathrm { s }$ is acting upon by a resultant constar which is opposite to its initial velocity. Its speed decreases to 4

$\mathrm { m } / \mathrm { s }$ in 1 s. Then the force

0%
12 $N$
0%
28 $N$
0%
8 $N$
0%
none

A body of mass 1 kg falls from rest through the air, a distance of 200 m and acquires a speed 50 m/s. Work done against air friction is

0%
$750 J$
0%
$1250 J$
0%
$1960 J$
0%
$3210 J$

A block of mass m is oscillating on smooth between two light springs of spring constant K separated by a distance I colliding elastically with the spring.If the velocity of the blocks is increased by an external impulse when it is not touching either of the spring then time period.

0%
Increases
0%
Decreases
0%
Remain same
0%
Time period is independent of I

A perfectly elastic ball falls on a horizontal floor from a height in a time

$t$ . It will hit the floor again after a time

$t'$ . The ratio of

$t'$ and t is

0%
$1:1$
0%
$1:2$
0%
$2:1$
0%
$1:4$

As shown in figure, ball of of mass

$m \,kg$ is suspended by a string

$\ell \,cm$ long. Keeping the string always taut, the ball describes a horizontal circle of radius

$\dfrac{\ell}{2}$ . Calculate the angular speed.
1576173_1d09b453431b4f70a8fe50b4dab79948.1576173-Q

0%
$\sqrt{\dfrac{2g}{3\ell}}$
0%
$\sqrt{\dfrac{2g}{\sqrt{3}\ell}}$
0%
$\sqrt{\dfrac{3g}{2\ell}}$
0%
None of these

Explanation

Let

$\theta$ be the angle made by the string and with the vertical

$\implies tan \theta= \dfrac1 {\sqrt{3}}$

Let

$T$ be the tension in the string and

$\omega$ is the angular speed.

From Free Body Diagram

$T cos \theta =mg$

$T sin \theta =m\omega^2 \dfrac l2$

Form above two equations we get

$\omega =\sqrt{\dfrac{2tan \theta g}{l}}$

using the value of

$tan \theta$

Angular speed ,

$\omega= \sqrt{\dfrac{2 g}{\sqrt{3}l}}$

A man of mass 20 kg is running in a straight line with velocity 10 m/s. What is value of his kinetic energy?

0%
$10\ J$
0%
$100\ J$
0%
$1000\ J$
0%
$10000\ J$

Explanation

The kinetic energy of the particle is given as-

$K.E. = \dfrac{1}{2}mv^2$

$m$ ; mass of the object

$v$ ; speed of the object

$\Rightarrow K.E. = \dfrac{1}{2} \times 20 \times 10^2J$

$\Rightarrow K.E. = 1000 J$

A ball of mass

$m_{1}$ is moving with velocity

$v$ . It collides head on elastically with a stationary ball of mass

$m_{2}$ . The velocity of ball becomes

$\dfrac{v}{3}$ after collision, then the value of the ratio

$\dfrac{m_{2}}{m_{1}}$ is:

0%
$1$
0%
$2$
0%
$3$
0%
$4$

A body of mass

$8\ kg$ collides elastically with a stationary mass of

$2\ kg$ . If initial

$KE$ of moving mass be

$E$ , the kinetic energy left with it after the collision will be:

0%
$0.80\ E$
0%
$0.64\ E$ `
0%
$0.36\ E$
0%
$0.08\ E$

How much height must an object of mass

$5kg$ be lifted to have potential energy

$5000J$ ?

$[g=10 m/s^2]$

0%
$100m$
0%
$50m$
0%
$120m$
0%
$110m$

Explanation

Gravitation potential energy is given by

$P.E = mgh$

Given:

Mass

$m=5 \ kg$

$g=10 \ m/s^2$

$5000 = 5 \times 10 \times h$

$\implies h = 100 m$

The block will again collide after time:

0%
$6\ sec$
0%
$4\ sec$
0%
$8\ sec$
0%
$\infty$

How many collision are possible between the blocks?

0%
$2$
0%
$4$
0%
$27$
0%
$infinite$

A car is accelerated on a levelled road and attains a speed of

$5$ times its initial speed. In this process , the kinetic energy of the car:

0%
does not change
0%
becomes $4$ times that of initial kinetic energy
0%
becomes $8$ times that of initial kinetic energy
0%
becomes $25$ times that of initial kinetic energy

Explanation

Lets initial velocity of car is $v$ , final velocity of this car will be $5\ v$

$K.E = \dfrac{1}{2}$ $mv ^{2}$

$\therefore \ K.E\propto v^2$

$K.E_{f}=K.E_i \left(\dfrac{v_f}{v_i}\right)^2$ where

$f$ stand for final and

$i$ for initial.

$K.E_{f}=K.E_i \left(\dfrac{5v}{v}\right)^2=25\ K.E_i$

So, when the speed increases to 5 times of its initial velocity, the kinetic energy will become 25 times.

The correct answer is d) becomes

$25$ times that of initial kinetic energy

Write true or false for the following statements:
If we know the speed and mass of an object, we can find out its kinetic energy.

0%
True
0%
False

Explanation

Kinetic energy KE of a body of mass m moving with velocity or speed v is given by the formula:

$KE = \dfrac{1}{2} mv^2$
The formula suggests that kinetic energy depends on mass and speed of the object. So, if we know the speed v and mass m of an object, we can find out its kinetic energy.

Write true or false for each statement
A gum bottle lying on a table has no energy.

0%
True
0%
False

Explanation

A gum bottle lying on a table has potential energy by virtue of its position. It is at some height

$h$ above the reference line.

So energy possessed by it =

$P.E.=mgh$

So the statement is false. Answer is option B.

When a stone of mass 'm' falls through a vertical height 'd', the decrease in the gravitational energy is:

0%
$mg/d$
0%
$mg^2/2$
0%
$mgd$
0%
$md/g$

Explanation

Potential energy stored in a body of mass m and at height 'd' will be given by

$m \times g \times d$
This gravitational potential energy keeps on decreasing with the decrease in height of stone, and when the stone falls completely, this potential energy is lost . So when object loses height

$d$ the corresponding loss in potential Energy is

$= mgd$

Potential energy of a person is minimum when:

0%
Person is standing
0%
Person is sitting on a chair
0%
Person is sitting on the ground
0%
Person is lying on the ground

Explanation

Potential energy of a body is defined as energy of a body due to its position in gravitational field.

In general,
Potential energy

$= M \times g \times h$ where: h is the height above ground.
If the person will be lying on ground then it will have minimum height above the ground therefore potential energy of the person will also be minimum.

A car is accelerated on a leveled road and attains a velocity

$4$ times its initial velocity. In this process, the potential energy of the car:

0%
Does not change
0%
Becomes twice to that of initial
0%
Becomes $4$ times that of initial
0%
Becomes $16$ times that of initial

An object of mass

$5 \,kg$ falls from a height of

$5 \,m$ above the ground. The loss of potential energy of the mass is:

0%
$250 \,J$
0%
$25 \,J$
0%
$2.5 \,kJ$
0%
$50 \,J$

Explanation

Given,

$Mass = 5 \,kg$

$Height = 5 \,m$
Let

$g = 10 \,ms^{-2}$
Potential energy stored in a body of mass m and height h will be given by

$= m \times g \times h$

$= 5 \,kg \times 10 \,ms^{-2} \times 5 \,m$

$= 250 \,J$
When the mass falls completely, same potential energy is lost

$= 250 \,J$

When a boy triples his speed, his kinetic energy becomes

0%
half
0%
double
0%
nine times
0%
no charge

Explanation

Given that, speed of particle is becoming 3 times of its initial value while moving. ie $v'=3v$

As we know $K.E = \dfrac{1}{2}$ $mv ^{2}$ ..........(1)

After velocity tripled $K.E' = \dfrac{1}{2}$ $m(3v) ^{2}=\dfrac92mv^2$ ...........(2)

From equation 1 and 2

$K.E' = 9 \times K.E$

When a boy tripled his speed, his kinetic energy will becomes nine times.

The energy stored in water of a dam is the kinetic energy.

0%
True
0%
False

A position dependent force

$F = 7 - 2x + 3x^2$ newton acts on a small body of mass

$2$ kg displace it from

$x = 0$ to

$x = 5 m$ the work done in joules is

0%
$70$
0%
$270$
0%
$35$
0%
$135$

Explanation

It is known that

$W=\int Fdx$

$W = \int^5_0 Fdx = \int^5_0 ( 7- 2x +3x^2) dx$

$W = [ 7x - x^2 + x^3]^5_0$

$W= 35 -25 + 125 = 135 J$

The energy stored in wound watch spring is

0%
$K.E.$
0%
$P.E.$
0%
Heat energy
0%
Chemical energy

Explanation

Energy stored in spring is potential energy, and it is defined as

$E=\dfrac{1}{2}kx^2$

where k is spring constant , and x is the extension/compression in spring.

Two springs have their force constant as

$k_1$ and

$k_2( k_1 > k_2)$ When they are stretched by the same force

0%
No work is done in case of both the springs
0%
Equal work is done in case of both the springs
0%
More work is done in case of second spring
0%
More work is done in case of first spring

Explanation

Relation between Work done and force is defined as

$W = \dfrac {F^2}{ 2k}$

If both springs are stretched by same force then

$W \propto \dfrac {1}{k}$

$k_1 < k_2$ therefore

$W_1 < W_2$

i.e. more work is done in case of second spring.

The force constant of a wire is k and that of another wire is

$2k$ When both the wires are stretched through same distance, then the work done.

0%
$W_2 = 2W^2_1$
0%
$W_2 =2W_1$
0%
$W_2 =W_1$
0%
$W_2 = 0.5 W_1$

Explanation

Work required = change in energy

$W=\Delta E$

$W =\frac {1}{2} kx^2$

If both wires are stretched through same distance then

$W \propto k$ as

$k_2 = 2k_1$ so

$W_2 = 2W_1$

A particle moves under the effect of a force

$F =Cx$ from

$x = 0$ to

$x =x_1$ The work done process is

0%
$Cx^2_1$
0%
$\frac {1}{2} Cx^2_1$
0%
$Cx_1$
0%
$Zero$

Explanation

Consider small displacement dx,

work done for this small displacement

$dW=F.dx$

$W =\int^{x_1}_0 F.dx$

$W= \int^{x_1}_0 Cx dx = C [ \dfrac {x^2}{2}]^{x_1}_0$

$W = \dfrac {1}{2} Cx^2_1$

If velocity of a body is twice of previous velocity, then kinetic energy will become

0%
$2$ times
0%
$\frac{1}{2}$ times
0%
$4$ times
0%
$1$ times

Explanation

Kinetic energy=

$\dfrac{1}{2} mv^2 \therefore K.E. \propto v^2$

$v\rightarrow 2v$

$K \rightarrow 2^2=4K$

If velocity is doubled then kinetic energy will become four times.

A spring

$40$ mm long is stretched by the application of a force. If

$10$ N force required to stretch the spring through

$1$ mm, then work done in stretching the spring through

$40$ mm is

0%
$84 J$
0%
$68 J$
0%
$23$ J
0%
$8 J$

Explanation

we know

$F=kx$

$k = \dfrac {F}{x}$

putting values:

$k= \dfrac {10}{ 1 \times 10^{-3} } = 10^4 N/m$

$W = \dfrac {1}{2} kx^2 = \dfrac {1}{2} \times 10^4 \times ( 40 \times 10^{-3})^2 = 8J$

An object of

$m\ kg$ with speed of

$v\ m/s$ strikes a wall at an angle

$\theta$ and rebounds at the same speed and same angle. The magnitude of the change in momentum of the object will be

0%
$2mv\cos \theta$
0%
$2mv\sin \theta$
0%
$0$
0%
$2\ mv$

Explanation

$\vec P_1=mv\sin \theta \hat i-mv\cos \theta \hat j$
and

$\vec P_2=mv\sin \theta \hat i+mv\cos \theta \hat j$
So change in momentum

$\overrightarrow{\Delta P}=\vec P_2-\vec P_1=2mv\cos \theta \hat j, | \Delta \vec P| =2mv\cos \theta$

A body of mass

$3$ kg is under a force, which causes a displacement in it is given by

$S = \frac {t^3}{3} ( in m )$ find the work done by the force in first

$2$ seconds

0%
$2 J$
0%
$3.8 J$
0%
$5.2 J$
0%
$24 J$

Explanation

Given:

$S = \dfrac {t^3}{3}$

$\therefore dS = t^2 dt$

$a = \dfrac {d^2 S}{ st^2} = \dfrac {d^2}{dt^2} [ \dfrac {t^3}{3} ] = 2t m/s^2$

Now work done by the force

$W = \int^2_0 F.dS = \int^2_0 ma.dS$

$\int^2_0 3 \times 2 t \times t^2 dt = \int^2_0 6t^3 dt = \dfrac {3}{2} [ t^4]^2_0= 24 J$

A spring of spring constant

$5 \times 10 N/m$ is stretched initially by

$5$ cm form the unstretched position. Then the work required to stretch it further by another

$5 Cm$ is

0%
$6. 25 N-m$
0%
$12.50 N-m$
0%
$18.75 N-m$
0%
$25.00 N-m$

Explanation

Work required = change in energy

$W=\Delta E$

$W = \dfrac {1}{2} k (x^2_2 - x^2_1) = \dfrac {1}{2} \times 5 \times 10^3 ( 10^2 - 5^2 ) \times 10^{-4}$

$= 18.75 J$

The decrease in the potential energy of a ball of mass

$20 kg$ which falls from a height of

$50 cm$ is

0%
$968 J$
0%
$98 J$
0%
$1980 J$
0%
None of these

Explanation

Decrease in potential energy

$\Delta U =mg(h_2-h_1)$

$\Delta U= mgh =20 \times 9.8 \times 0.5=98 J$

The maximum velocity at the lowest point, so that the string just slack at the highest point in a vertical circle of radius l

0%
$\sqrt{gl}$
0%
$\sqrt{3gl}$
0%
$\sqrt{5gl}$
0%
$\sqrt{7gl}$

A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time

$t$ is proportional to

0%
$t^{1/2}$
0%
$t^{3/4}$
0%
$t^{3/2}$
0%
$t^2$

Explanation

$P = Fv = mav = m( \dfrac {dv}{dt} ) v \Rightarrow \dfrac {P}{m}dt = vdv$

$\Rightarrow \dfrac {P}{m} \times t = \dfrac {v^2}{2} \Rightarrow v = ( \dfrac {2P}{m})^{ 1/2} (t)^{1/2}$ now

$s = \int v dt = \int ( \dfrac {2P}{m})^{1 /2} t^{1/2}dt$

$\therefore s = ( \dfrac {2P}{m})^{ 1/2} [ \dfrac { 2t^{ 3/2} }{3} ] \Rightarrow s \propto t^{ 3/2}$

The relation between the displacement

$X$ of an object produced by the application of the variable force

$F$ is represented by a graph shown in the figure. If the object undergoes a displacement from

$X=0.5m$ to

$X=2.5m$ the work done will be approximately equal to

0%
$16 J$
0%
$32 J$
0%
$1.6 J$
0%
$8 J$

Explanation

Work done = Area under curve and displacement axis

= Area of trapezium

$= \dfrac {1}{2} \times$ ( sum f two parallel lines )

$\times$ distance between them

$= \dfrac {1}{2} ( 10 +4) \times ( 2.5 -0.5)$

$= \dfrac {1}{2} 14 \times 2 = 14 J$

As the area actually is not trapezium so work done will be more than

$14$ J i.e. approximately

$16$ J

The displacement x of a particle moving in one dimension under the action of a constant force is related to the time t by the equation

$t = \sqrt { x} +3$ where x is in meters and t is in seconds . the work done by the force in the first

$6$ second is

0%
$9 J$
0%
$6 J$
0%
$0 J$
0%
$3 J$

Explanation

According to the question,

$x =( t-3)^2 \Rightarrow v = \dfrac {dx}{dt} = 2(t -3)$

$t = 0 ; v_1 = -6 m/s$ and at

$t =6 sec , v_2 = 6 m/s$

So change in kinetic energy

$= W = \dfrac {1}{2} mv^2_2 - \dfrac {1}{2} mv^2_1 = 0 J$

A ball is moving to and fro about the lowest point A of a smooth hemispherical bowl. If it is able to rise up to a height of 20 cm on either side of A, its speed at A must be

0%
$0.2 \,m/s$
0%
$2 \,m/s$
0%
$4 \,m/s$
0%
$4.5 \,m/s$

A body of mass

$0.1$ kg moving with a velocity of

$10$ m/s hits a spring (fixed at the other end) of force constant

$1000$ N/m and comes to rest after compressing the spring. The compression of the spring is

0%
$0.01 m$
0%
$0.1 M$
0%
$0.2 m$
0%
$0.5 m$

Explanation

From Energy conservation we can say that

The kinetic energy of mass is converted into potential energy of a spring

$KE=PE$

$\dfrac {1}{2} mv^2 = \dfrac {1}{2} kx^2$

$\dfrac {1}{2} mv^2 =\dfrac {1}{2} kx^2 \Rightarrow x = v \sqrt { \dfrac {m}{k} } = 10 \sqrt { \dfrac {0.1}{1000}}= 0.1 m$

The spring will have maximum potential energy when

0%
it is pulled out
0%
it is compressed
0%
both (a) and (b)
0%
neither (a) nor (b)

When spring is compressed its potential energy........

0%
Remains constant
0%
Reduces
0%
Increases
0%
Nothing can be said about it.

Masses of two bodies are

$1$ kg and

$4$ kg respectively. If their kinetic energies are in

$2:1$ proportion, the ratio of their speeds is .........

0%
$2\sqrt{2}:1$
0%
$1:\sqrt{2}$
0%
$1:2$
0%
$2:1$

Explanation

We know that,

$K.E.=\dfrac{1}{2}mv^2$

$m_1=1kg$

$m_2=4kg$

$\therefore \dfrac{K_1}{K_2}=\Bigg(\dfrac{m_1}{m_2}\Bigg)\Bigg(\dfrac{v_1}{v_2}\Bigg)^2$

$\Rightarrow \dfrac{2}{1}=\Bigg(\dfrac{1}{4}\Bigg)\Bigg(\dfrac{v_1}{v_2}\Bigg)^2$

$\Rightarrow \Bigg(\dfrac{v_1}{v_2}\Bigg)^2=\dfrac{8}{1}$

$\Rightarrow \Bigg(\dfrac{v_1}{v_2}\Bigg)=\dfrac{2\sqrt{2}}{1}$

A pile driver drives postes into the ground by repeatedly dropping a heavy object on them. Assume the object is dropped from the same height each time. By what factor does the energy of the pile driver-Earth system change when the mass of the object being dropped is doubled?

0%
$\dfrac{1}{2}$
0%
$1$ ; the energy is the sane
0%
$2$
0%
$4$

$10.0-g$ bullet is fired into a

$200-g$ block of wood at rest on a horizontal surface. After impact, the block slides

$8.00\ m$ before coming to rest. If the coefficient of friction between the block and the bullet before impact?

0%
$106\ m/s$
0%
$166\ m/s$
0%
$226\ m/s$
0%
$286\ m/s$
0%
none of those answer is correct

Calculate the energy possessed by a stone of mass

$10\ kg$ kept at a height of

$5\ m$ .
Take

$g = 9.8\ m/s^{2}$

0%
$196\ J$
0%
$49\ J$
0%
$490\ J$
0%
$980\ J$

Explanation

The energy that stone may posses = kinetic energy + potential energy

Since, stone at rest

$(v = 0)$

$KE=\frac{mv^2}{2}$ So kinetic energy=0

$\therefore$ Total energy = Potential energy

$(E) = mgh$ [

$m$ : mass ,

$h$ : height ,

$g$ : gravity ]

$m = 10\ kg, g = 9.8\ m/s^{2}$
and

$h = 5\ m$

$E = mgh = 10 \times 9.8 \times 5$ Joules

$= 490\ J$

The figure shows a block having a small basket in it and an small inclined plane (with fixed angle) is rigidly attached to the block. The combined mass (including incline plane) of all these is M = 5 kg. Initially they are at rest. Now identical balls each of mass m=1 kg are thrown horizontally with velocity v = 5m/s (with respect to ground) which strike the inclined plane and then are collected in the small basket. Assume that in the basket balls come to rest with respect to basket. Choose the correct statement(s) from the following :

0%
After collecting 5 balls the speed of the block will be 2.5 m/s
0%
After collecting 10 balls the speed of the block will be (10/3) m/s
0%
The velocity of the block will never be exactly 5 m/s no matter how many balls are collected in the basket.
0%
Increase in speed of the block will be same every time for collection of each ball in the basket

Explanation

Total momentum of

$N$ balls

$=N m v=N (1\times5)=5N$

$\therefore$ Momentum of the block along with

$N$ balls

$=5N$ or

$(M+Nm)$

${v}'=5N$ where

${v}'$ is the velocity of the block
After

$N$ balls are collected in the basket.

$\therefore {v}' =\displaystyle \dfrac{5N}{5+N}=\dfrac{5}{\dfrac{5}{N}+1}$

For

$N=5$ ;

${v}'$

$=2.5\ m/s$
For

$N=10$ ;

${v}'$

$=10/3\ m/s$
Also

${v}'$ is never exactly equal to

$5 m/s.$

From above we can say that increase in speed of the ball is not same every time for collection of each ball in the basket, it varies.

A particle of mass m moving along a straight line experiences force F which varies with the distance x travelled as shown in the figure. If the velocity of the particle at

$x_{0}$ is

$\sqrt{\dfrac{2F_{0}x_{0}}{m}}$ , then velocity at

$4x_{0}$ is:

0%
$2\sqrt{\dfrac{2F_{0}x_{0}}{m}}$
0%
$\sqrt{\dfrac{2F_{0}x_{0}}{m}}$
0%
$2\sqrt{\dfrac{F_{0}x_{0}}{m}}$
0%
none of these

Explanation

Increase in KE = work done by all forces

$\Delta KE=F_{net}\cdot ds$

$\dfrac{1}{2}mv_{2}^{2}-\dfrac{1}{2}m \times \left ( \dfrac{2F_{0}x_{0}}{m} \right )=\dfrac{1}{2}(2F_{0}+F_{0})3x_{0}$

$\Rightarrow v_{2}=\sqrt{\dfrac{11F_{0}x_{0}}{m}}$

CBSE Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 10 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 10 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

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