MCQ Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 1

This question has statement 1 and statementOf the four choices given after the statements, choose the one that best describes the two statements.
If two springs

$\mathrm{S}_{1}$ and

$\mathrm{S}_{2}$ of the force constants

$\mathrm{k}_{1}$ and

$\mathrm{k}_{2}$ , respectively, are stretched by the same force, it is found that more work is done on spring

$\mathrm{S}_{1}$ than on spring

$\mathrm{S}_{2}$ .
Statement-l : When stretched by the same amount, work done on

$\mathrm{S}_{1}$ , will be more than that of

$\mathrm{S}_{2}$
Statement-2:

$\mathrm{k}_{1}<\mathrm{k}_{2}$ .

0%
Statement-1 is false, statement-2 is true
0%
Statement-1 is true, statement-2 is false
0%
Statement-1 is true, statement-2 is true, statement 2 is the correct explanation of statement- 1
0%
Statement-1 is true, statement-2 is true, statement-2 is not the correct explanation of Statement-1

Explanation

$k_1x_1=k_2x_2=F$

$\displaystyle W_1= \frac{1}{2}kx_1^2= \frac{F^2}{2k_1}$

Similarly for

$\displaystyle W_2= \frac{F^2}{2k_2}$

Since

$\displaystyle W \propto \frac{1}{k}$

$W_1.W_2$

$\Rightarrow k_1<k_2$ , Statement-2 is true.

From Statement-1,

$W_2>W_1$ , hence it is false.

When a rubber-band is stretched by a distance

$x$ , it exerts a restoring force of magnitude

$F = ax + bx^2$ where

$a$ and

$b$ are constants. The work done in stretching the unstretched rubber band by

$L$ is

0%
$\displaystyle \dfrac{aL^2}{2} + \dfrac{bL^3}{3}$
0%
$\displaystyle \dfrac{1}{2} \left ( \dfrac{aL^2}{2} + \dfrac{bL^3}{3}\right )$
0%
$aL^2 + bL^3$
0%
$\displaystyle \dfrac{1}{2} (aL^2 + bL^3)$

Explanation

$F = ax + bx^2$

$dw = Fdx$

$W = \displaystyle \int_0^L (ax + bx^2) dx$

$W = \displaystyle \frac{aL^2}{2} + \frac{bL^3}{3}$

If collision between the block and the incline is completely elastic, then the vertical (upward) component of the velocity of the block at point B, immediately after it strikes the second incline is

0%
$\sqrt{30}\mathrm{m}/\mathrm{s}$
0%
$\sqrt{15}\mathrm{m}/\mathrm{s}$
0%
0
0%
$-\sqrt{15}\mathrm{m}/\mathrm{s}$

Explanation

After collision, vertical velocity will be

$vsin({ 30 }^{ 0 })cos({ 30 }^{ 0 })-vcos({ 30 }^{ 0 })cos({ 60 }^{ 0 })=0$ .

The balls, having linear momenta

$\vec{\mathrm{p}}_{1}=\mathrm{p}\hat{i}$ and

$\vec{\mathrm{p}}_{2}=-\mathrm{p}\hat{\mathrm{i}}$ , undergo a collision in free space. There is no extemal force acting on the balls. Let

$\vec{\mathrm{p}}_{1}$ and

$\vec{\mathrm{p}}_{2}$ be their final momenta. The following option(s) is (are) NOT ALLOWED for any non-zero value of

$\mathrm{p},\ \mathrm{a}_{1},\ \mathrm{a}_{2},\ \mathrm{b}_{1},\ \mathrm{b}_{2},\ \mathrm{c}_{1}$ and

$\mathrm{c}_{2}$ .

0%
$\vec{\mathrm{p}}_{1}=\mathrm{a}_{1}\hat{i}+\mathrm{b}_{1}\hat{\mathrm{j}}+\mathrm{c}_{1}\hat{\mathrm{k}}$ ;

$\vec{\mathrm{p}}_{2}=\mathrm{a}_{2}\mathrm{i}+\mathrm{b}_{2}\hat{\mathrm{j}}$
0%
$\vec{\mathrm{p}}_{1}=\mathrm{c}_{1}\hat{\mathrm{k}}$ ;

$\vec{\mathrm{p}}_{2}=\mathrm{c}_{2}\hat{\mathrm{k}}$
0%
$\vec{\mathrm{p}}_{1}=\mathrm{a}_{1}\mathrm{i}+\mathrm{b}_{1}\hat{\mathrm{j}}+\mathrm{c}_{1}\hat{\mathrm{k}}$ ;

$\vec{\mathrm{p}}_{2}=\mathrm{a}_{2}\hat{\mathrm{i}}+\mathrm{b}_{2}\hat{\mathrm{j}}-\mathrm{c}_{1}\hat{\mathrm{k}}$
0%
$\vec{\mathrm{p}}_{1}=\mathrm{a}_{1}\hat{\mathrm{i}}+\mathrm{b}_{1}\hat{\mathrm{j}}$ ;

$\vec{\mathrm{p}}_{2}=\mathrm{a}_{2}\mathrm{i}+\mathrm{b}_{1}\hat{\mathrm{j}}$

Explanation

Since there is no net external force acting on the two body system, the net momentum of the system must remain conserved. The final momentum must not have any components in

$j$ and

$k$ directions as both momenta are in

$i$ direction. Hence options

$A$ and

$D$ are not possible.

What is the minimum velocity with which a body of mass

$m$ must enter a vertical loop of radius

$R$ so that it can complete the loop?

0%
$\sqrt{gR}$
0%
$\sqrt{2gR}$
0%
$\sqrt{3gR}$
0%
$\sqrt{5gR}$

Explanation

To just complete the circle, at the highest point, tension is zero and the gravitational force provides the necessary centripetal force.

Hence,

$\dfrac{mv_{top}^2}{R}=mg.............(i)$

Applying conservation of energy at the top and bottom points,

$\dfrac{1}{2}mv_{bottom}^2=mg \times 2R + \dfrac{1}{2}mv_{top}^2.........(ii)$

Substituting

$mv_{top}^2$ from (i) in (ii) and solving,

$v_{bottom}=\sqrt {5gR}$

Two similar spring P and Q have spring constants

$K_P$ and

$K_Q$ , such that

$K_P > K_Q$ . They are stretched, first by the same amount (case a), then by the same force (case b). The work done by the spring

$W_P$ and

$W_Q$ are related as, in case (a) and case (b), respectively:

0%
$W_P > W_Q; W_Q > W_P$
0%
$W_P < W_Q; W_Q < W_P$
0%
$W_P = W_Q; W_P > W_Q$
0%
$W_P = W_Q; W_P = W_Q$

Explanation

Case a) both the springs are stretched by same amount, energy stored springs in P and Q will be

$W_P=\dfrac{1}{2}K_P\Delta x^2$ and

$W_Q=\dfrac{1}{2}K_Q\Delta x^2$ ...(i)

given

$K_P>K_Q \Rightarrow W_P>W_Q$ ...(ii)

Case b) both the springs are stretched by same force, energy stored springs in P and Q will be

$W_P=\frac{1}{2}K_P\Delta x_P^2$ and

$W_Q=\frac{1}{2}K_Q\Delta x_Q^2$ ...(iii)

given

$K_P>K_Q \Rightarrow \Delta x_P<\Delta x_Q$ ...(iv)
and

$F=K_P\Delta x_P=K_Q\Delta x_Q$ ...(v)

$W_P=\dfrac{1}{2}K_P\Delta x_P \Delta x_P$ and

$W_Q=\dfrac{1}{2}K_Q\Delta x_Q \Delta x_Q$ ...(vi)

from equation (v) substituting

$F=K_P\Delta x_P=K_Q\Delta x_Q$ in equation (vi)

$W_P=\frac{1}{2}Fx_P$ and

$W_Q=\dfrac{1}{2}F x_Q$

given

$\Delta x_P>\Delta x_Q \Rightarrow W_P<W_Q$ ...(vii)
hence, correct answer is option A.

The force on a particle as the function of displacement x(in x-direction) is given by

$F=10+0.5x$
The work done corresponding to displacement of particle from

$x=0$ to

$x=2$ unit is?

0%
$25$ J
0%
$29$ J
0%
$21$ J
0%
$18$ J

Explanation

Given that

$F=10+0.5x$

Work done corresponding to the particle from

$x=0$ to

$x=2$ unit is,

$W= \int_{0}^{2} F.dx= \int_{0}^{2} (10+0.5x)dx= [10x+ \dfrac{0.5x^2}{2}]_0^2= 20+1= 21 J$

A ball is suspended by a thread of length L at the point O on a wall which is inclined to the vertical by

$\alpha$ . The thread with the ball is displaced by a small angle

$\beta$ away from the vertical and also away from the wall. If the ball is released, the period of oscillation of the pendulum when

$\beta > \alpha$ will be

0%
$\displaystyle \sqrt{\frac{L}{g}}\left[\pi+2sin^{-1}\frac{\alpha}{\beta}\right]$
0%
$\displaystyle \sqrt{\frac{L}{g}}\left[\pi-2sin^{-1}\frac{\alpha}{\beta}\right]$
0%
$\displaystyle \sqrt{\frac{L}{g}}\left[2sin^{-1}\frac{\alpha}{\beta}-\pi\right]$
0%
$\displaystyle \sqrt{\frac{L}{g}}\left[2sin^{-1}\frac{\alpha}{\beta}+\pi\right]$

Explanation

The motion of the string can be represented as an angular displacement from the mean position,given as:

$\theta=\beta sin(\omega t)$

Time taken from displacement from mean position to

$\beta$ and back to mean position=

$t_1=\dfrac{T}{2}$

where

$T$ is the time period of the oscillation without the wall barrier.

Thus

$t_1=\dfrac{1}{2}\times 2\pi\sqrt{\dfrac{L}{g}}=\pi\sqrt{\dfrac{L}{g}}$

Time taken to displace from mean position to

$\alpha$ can be found by:

$\alpha=\beta sin(\omega t)$

$\implies t=\dfrac{1}{\omega}sin^{-1}\dfrac{\alpha}{\beta}$

$=\sqrt{\dfrac{L}{g}}sin^{-1}\dfrac{\alpha}{\beta}$

Thus time taken to displace from mean position to

$\alpha$ and back to mean position =

$t_2=2\sqrt{\dfrac{L}{g}}sin^{-1}\dfrac{\alpha}{\beta}$

Thus the time period of the oscillation =

$t_1+t_2=\sqrt{\dfrac{L}{g}}[\pi+2sin^{-1}\dfrac{\alpha}{\beta}]$

How much work must be done by a force on

$50\$

$kg$ body in order to accelerate it from rest to

$20\ m/s$ in

$10\$

$s$ ?

0%
$10^3\$ $J$
0%
$10^4\$ $J$
0%
$2\times 10^3\$ $J$
0%
$4\times 10^4\$ $J$

Explanation

Acceleration that causes the given rise in speed in given time =

$a=\dfrac{v}{t}=\dfrac{20\ m/s}{10\ s}=2\ m/s^2$

Thus, force acting on the body =

$F=ma=100\ N$

Distance travelled in the given time =

$s=\dfrac{1}{2}at^2=100\ m$

Thus, work done in given time =

$F.s=10^4\ J$

The amount of work done in stretching a spring from a stretched length of 10 cm to a stretched length of 20 cm is-

0%
Equal to the work done in stretching it from 20 cm to 30 cm
0%
Less than the work done in stretching it from 20 cm to 30 cm.
0%
More than the work done in stretching it from 20 cm to 30 cm.
0%
Equal to the work done in stretching it from 0 to 10 cm.

Explanation

Work done in stretching a spring to x length is

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

$\therefore$ Work one in stretching from

$x_1$ to

$x_2$ is proportional to

$W \propto x_2^2-x_1^2$
hence

$\dfrac{W_{20,30}}{W_{10,20}}=\dfrac{500}{300}$
so, 10 to 20 takes less work when compared to 20 to 30.

A ball tied to a string is swung in a vertical circle. Which of the following remains constant?

0%
Tension in the string.
0%
Speed of the ball.
0%
Centripetal force.
0%
Earth's pull on the ball.

Explanation

Earth's pull on the ball remains constant

$(W=mg)$ .
Tension in the string changes along the vertical circular path.
Speed of the ball changes as sum of kinetic energy and potential energy remains constant during motion.
Speed changes, hence, centripetal force changes.

The K.E. of a body can be increased maximum by doubling its:

0%
mass
0%
weight
0%
speed
0%
density

Explanation

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

$K.E \propto m$ &

$K.E \propto v^2$

So doubling mass will double the kinetic energy and doubling speed will make kinetic energy

$4$ times.

$\therefore$ The K.E. of a body can be increased maximum by doubling its speed.

A lighter body moving with a velocity

$v$ collides with a heavier body at rest. Then :

0%
the lighter body rebounces with twice the velocity of bigger body
0%
the lighter body retraces its path with the same velocity in magnitude
0%
the heavier body does not move practically
0%
both (2) and (3)

Explanation

In a collision system where

$m_1$ moves with

$u_1$ initially and

$m_2$ is at rest with

$m_2 >>> m_1$ .
Let the final velocity of

$m_1$ be

$v_1$ and

$m_2$ be

$v_2$ .
Assumption: Let the collision be elastic. Using linear momentum conservation and equation for coefficient of restitution,

$m_1u_1 = m_2v_2 + m_1v_1$

$v_2 - v_1 = u_1$

We get

$v_1 = \dfrac {m_1 - m_2 }{m_1 + m_2} u_1$

$v_2 = \dfrac {2m_1}{m_1 + m_2} u_1$

Using

$m_2 >>> m_1,$ we get

$v_1 = -u_1$ and

$v_2 = 0$ .

A certain force acting on a body of mass 2kg increase its velocity from 6m/s to 15 m/s in 2s. The work done by the force during this interval is ?

0%
27J
0%
3J
0%
94.5J
0%
189J

Explanation

we know that

v=u+at

15=6+a(2)

$a= 4.5 m/s^2$

$s=ut+0.5at^2= 6(2)+ 0.5(4.5)(4)= 21 m$

$W= mas= 2(4.5)(21)= 189 J$

Name the type of energy (kinetic energy

$K$ or potential energy

$U$ ) possessed in a compressed spring:

0%
$U$
0%
$K$
0%
Both $U$ and $K$
0%
None

Energy equals of a mass of one microgram in kilo joules is

0%
$9 \times 10^{7} kJ$
0%
$10 \times 10^3 kJ$
0%
$8 \times 10^2 kJ$
0%
$7 \times 10^4 kJ$

Explanation

Given

Energy of 1 microgram=?

Solution

We will use Energy mass equivalence

$E=mc^{2}$

$E=10^{-6}kg\times (3 \times10^{8})^{2}m/s^{2}$

$E=9 \times 10^{10}J$

Option $A$ is correct.

A particle moves under the effect of a force

$F=Cx$ form

$x=0$ to

$x=x_{1}$ . The work done in the process is-

0%
$Cx_{1}^{2}$
0%
$\displaystyle \frac{1}{2}Cx_{1}^{2}$
0%
$Cx_{3}^{2}$
0%
zero

Explanation

Similar to a spring with neutral point at x=0

$W = \int \bar F.\bar ds = \int_0^{x_1} Cx dx = 0.5Cx_1^2$

A stone tied to a string is rotated in a vertical circle. The minimum speed with which the stone has to be rotated in order to complete the circle :

0%
decreases with increasing the mass of the stone
0%
is independent of the mass of the stone
0%
decreases with increasing the length of the string
0%
is independent of the length of the string

Explanation

$\textbf{Step 1: Energy Conservation [Ref. Fig.]}$

$\text{Dependence on m}$

Let

$u$ be the minimum speed required at

$B$ .

Apply energy conservation between point

$A$ and

$B$ we get

$KE_A + PE_A = KE_B + PE_B$

Taking point

$A$ as zero potential level

$\Rightarrow$

$\dfrac{1}{2} mu^2 = mg (R + R) + \dfrac{1}{2} mv^2$

$....(1)$

From equation

$1$ , we observe that mass

$m$ is cancelled on both sides.

$\Rightarrow u$ is independent of

$m$ .

$\textbf{Steo 2: Solving Equation}$

$\text{Dependence on R}$

At highest point, Tension will be zero for minimum velocity

$\Rightarrow$ gravitational pull will provide necessary centripetal force

$\Rightarrow mg = \dfrac{mv^2}{R}$

$\Rightarrow v^2 = Rg$ .... Putting in equation

$(1)$

We get,

$\dfrac{1}{2} mu^2 = mg (R + R) + \dfrac{1}{2} m (Rg)$

$\Rightarrow u = \sqrt{5Rg}$

$\Rightarrow u$ is proportional to the square root of

$R$ .

Hence option

$B$ is correct.

Complete the following sentence:
The kinetic energy of a body is the energy by virtue of its______ .

0%
force
0%
motion
0%
roughness
0%
all the above

Explanation

The kinetic energy of a body is the energy by virtue of its motion. This is the definition of kinetic energy. If it has motion then it will have velocity, hence kinetic energy exists.

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

A simple pendulum, composed of a bob of mass

$m$ connected to the end of a massless rod, executes simple harmonic motion as it swings through small angles of oscillation.The maximum angular displacement with the vertical is denoted by

${ \theta }_{ max }$ . Frictional effects and variation of acceleration due to gravity are negligible.
Find out the correct statement?

0%
At $\theta=0$ , the tangential acceleration is $0$
0%
At $\theta={\theta}_{max}$ , the tangential acceleration is $0$
0%
At $\theta=0$ , the speed is $0$
0%
At $\theta=0$ , the restoring force is maximized
0%
At $\theta={\theta}_{max}$ , the speed is maximized

Explanation

We know the tangential acceleration in SHM is given by

$a=-{\omega}^{2}A\sin{\theta}$

where

$A=amplitude$

therefore at

$\theta=0$

$a=-{\omega}^{2}A\sin{0^0}$

$a=0$ because

$\sin{0^0}=0$

Which of the following will lead to a change in kinetic energy of a body?

0%
change in its mass
0%
change in its velocity
0%
all of the above
0%
none of the above

Explanation

Kinetic energy of the body

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

$\implies$

$K \propto m$ and

$K \propto v^2$

Thus changing either the mass of the body or its velocity will lead to a change in kinetic energy of a body.

What is potential energy?

0%
Energy of an object due to its position or arrangement in a system
0%
Energy of an object due to its nature or arrangement in a system
0%
Energy of an object due to its shape or arrangement in a system
0%
None

A physics student is performing a experiment in which she twirls rubber stopper at the end of a string around in a circle at nearly constant speed and tries to determine its acceleration. Which of the following single changes would enables her to most nearly double the acceleration of the rubber stopper?

0%
Double the speed of the rubber stopper.
0%
Use a string that is twice as long as the original string.
0%
Reduce the speed of the rubber stopper to half its original value.
0%
Use a string that is half as long as the original string
0%
Use a rubber stopper with twice as much mass as the original rubber stopper.

Explanation

Due to constant speed , student is doing uniform circular motion so tangential acceleration is zero .

as the tangential acceleration is zero , therefore total acceleration will be equal to centripetal acceleration , which is

$a=v^{2}/r$

which can be doubled by using a string that is half as long as the original string .

$a'=\dfrac{v^{2}}{\frac{r}{2}}=2a$

A ride in an amusement park called scream machine swings the riders around a complete vertical circle during the course of the ride.
Identify where on ride rider feels the maximum speed?

0%
Point A
0%
Point B
0%
Point C
0%
Point D
0%
Point E

Explanation

At the lowermost position of the circular motion, the potential energy is zero and thus the kinetic energy of the rider is maximum at that position.

Using

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

Hence the rider feels maximum speed at the lowermost position of motion i.e point A.

A ball of mass

$m$ is attached with the string of length

$R$ , rotating in a circular motion, with instantaneous velocity

$v$ and centripetal acceleration

$a$ .
Calculate the centripetal acceleration of the ball if its mass is doubled?

0%
${a}/{4}$
0%
${a}/{2}$
0%
$a$
0%
$2a$
0%
$4a$

Explanation

As centripetal acceleration is given by

$a=v^{2}/R$

therefore centripetal acceleration does not depend upon mass of ball, so it will remain constant ,equal to

$a$ .

A simple pendulum, composed of a bob of mass

$m$ connected to the end of a massless rod, executes simple harmonic motion as it swings through small angles of oscillation.The maximum angular displacement with the vertical is denoted by

${ \theta }_{ max }$ . Frictional effects and variation of acceleration due to gravity are negligible.
Which one of the following would enable you to calculate the length of the pendulum?

0%
The mass of the bob
0%
The period of the oscillations
0%
The tangential acceleration at $\theta = 0$
0%
The maximum speed of the bob
0%
The acceleration at $\theta = {\theta}_{max}$

Explanation

For the small oscillation, the motion of the pendulum can be treat as a simple harmonic motion. So, time period of a SHM is

$T=2\pi\sqrt{\frac{l}{g}}$

Thus, time period would enable you to calculate the length of the pendulum.

In the phenomenon of work done by variable forces, the forces:

0%
remain constant
0%
don't remain constant
0%
increase
0%
decrease

A ball with mass m and speed

$V_0$ hit a wall and rebounds back with same speed.
Calculate the change in the object's kinetic energy.

0%
$-mv_0 ^2$
0%
$- \frac{1}{2}mv_0 ^2$
0%
Zero
0%
$\frac{1}{2}mv_0 ^2$
0%
$mv_0 ^2$

Explanation

The speed of the ball remains the same

$v_0$ before and after the collision with wall.

Thus the kinetic energy remains

$\dfrac{1}{2}mv_0^2$ .

Since kinetic energy is a scalar quantity, the change in it is zero.

What is kinetic energy?

0%
Energy possessed by a body by the virtue of its motion
0%
Energy possessed by a body by the virtue of its shape
0%
Energy possessed by a body by the virtue of its size
0%
None

Explanation

Correct Option: A
Explanation: The energy which a body possesses by virtue of being in motion is called as kinetic energy. If we want to stop a moving body then we must do some work on the body to stop it which means that a moving body possesses energy because of motion.

If an object of mass

$m$ moving with the velocity

$v$ , The kinetic energy is defined as,

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

Hence, Option A is correct.

A particle moves from

$X = 0\$ to

$X = 2\ m$ on X-axis under the effect of

${ F } (x) = ({ 4x }^{ 3 } - { 3x }^{ 2 } + 2x + 5)\widehat { i }$ Newton. The work done on the particle is

0%
$22\ Joule$
0%
$36\ Joule$
0%
$46\ Joule$
0%
None of these

Explanation

Work done by a force =

$\int Fdx=x^4-x^3+x^2+5x|^2_0=22 \ Joule$

A force of

$16N$ is distributed uniformly on one surface of a cube of edge

$8cm$ . The pressure on this surface is

0%
$3500Pa$
0%
$2500Pa$
0%
$4500Pa$
0%
$5500Pa$

Explanation

$F=16N$

$A=8\times 8\times { 10 }^{ -4 }{ m }^{ 2 }$

$P=\cfrac { F }{ A } =\cfrac { 16 }{ 64\times { 10 }^{ -4 } } =\cfrac { 1000 }{ 4 } =2500Pa$

A brick of mass

$m$ , tied to a rope , is being whirled in a vertical circle, with a uniform speed. The tension in the rope is:

0%
The same throughout
0%
Largest when the brick is at the highest point of the circular path and smallest when it is at the lowest point
0%
Largest when the rope is horizontal and smaller when it is vertical.
0%
Largest when the brick is at the lowest point of the circular path and smallest when it is at the highest point

Explanation

$T_H + mg = m \omega^2 r$

$T_L - mg = m\omega^2 r$

$\therefore T_L > T_H$

1175239_1075462_ans_a1d2e5bc4d054f8a98ea5dd7597336ca.PNG

$\vec{F}=(5\hat{i}+3\hat{j}+2\hat{k})\ N$ is applied over a particle with displaces it from its origin to the point

$\vec{r}=(2\hat{i}-\hat{j})m$ . The work done on the particle is joule is

0%
$-7\ J$
0%
$+7\ J$
0%
$+10\ J$
0%
$13\ J$

Explanation

Work done by a force =

$\int \vec{F}.\vec{dr}=(5\hat{i}+3\hat{j}+2\hat{k}).(2\hat{i}-\hat{j})=7 \ Joule$

A force F =

$-\frac{K}{X^2} (X \neq 0)$ acts on a particle in x-direction. The work done by this force in displacing the particle from x = +a to x = +2a is. (Where k is a positive constant)

0%
$\frac{-k}{2a}$
0%
$\frac{+k}{2a}$
0%
$\frac{k}{a}$
0%
$-\frac{k}{a}$

Explanation

Given that ,

$F= - \dfrac K{x^2}$ , Where

$x$ is the position of particle .

Work done by this force ,

$W=\int _a^{2a} F.dx=\int _a^{2a} -\dfrac K{x^2} dx= [\dfrac Kx ]_a ^{2a}= \dfrac K{2a}- \dfrac Ka = -\dfrac K {2a}$

Given

$k_1 = 1500 \ N/m,$

$k_2 =500 \ N/m,$

$m_1 = 2 \ kg$ and

$m_2 = 1 \ kg$ . Find potential energy stored in equilibrium: (Take

$g = 10 \ m/s^2$ )

0%
2.3 J
0%
0.4 J
0%
3.4 J
0%
0.6 J

Explanation

Soppose the elongation in first string be

$x_1$ and in the second string be

$x_2$

Now for second string, in equilibrium

$K_2x_2=m_2g$

$x_2= m_2g/k_2$ (1)

For first string in equilibrium

$K_1x_1=(m_1+m_2)g$

$x_1= (m_2+m_1)g/k_1$ (2)

Now potential energy stored will be

$U=\frac{K_1x_1^2}{2}$ +

$\frac{K_2x_2^2}{2}$

Putting the values of

$k_1$ ,

$k_2$ ,

$x_1$ and

$x_2$

We get U= 0.4J

A ball of mass

$'m'$ moves towards a wall with a velocity

$'u'$ , the direction of motion making an angle

$\theta$ with the surface of the wall and rebounds with the same period. The change in momentum of the ball during the collision is:

0%
$2mu\ sin$ $\theta$ towards the wall
0%
$2mu \ sin$ $\theta$ away from the wall
0%
$2mu\ cos$ $\theta$ towards the wall
0%
$2mu\ cos$ $\theta$ away from the wall

Explanation

As the ball hits the wall at an angle and rebounds, there is a change in the momentum only along the direction perpendicular to the wall. The change along the wall is zero. In diagram

$\vec p_1$ and

$\vec p_2$ is the momentum before and after the collision with wall.
Since , the angle with the wall is

$\theta$ , the change is only in

$sin\theta$ component, which is across

$\vec F_{wall}$ .
Change in momentum

$=m[usin\theta -(-usin\theta )]=2mu\ sin\theta$ away from the wall, as the final velocity is away from the wall.

A car of mass 400 kg travelling at 72 kmph crashes a truck of mass 4000 kg and travelling at 9 kmph in the same direction. The car bounces back with a speed of 18 kmph. The speed of the truck after the impact is

0%
9 kmph
0%
18 kmph
0%
27 kmph
0%
36 kmph

Explanation

Using momentum conservation,

$\displaystyle m_{car} \displaystyle u _{car} + \displaystyle m_{ truck} \displaystyle u_{ truck}= \displaystyle m_{ car} \displaystyle \displaystyle v _{car} + \displaystyle m_{truck} \displaystyle v_{truck}$

$400 \times 72 + 4000 \times 9 = -18 \times 400 + 4000 \times v$

$v = 18 \ kmph$

A perfectly elastic ball

$p_{1}$ of mass 'm' moving with velocity v collides elastically with three exactly similar balls

$p_{2}$ ,

$p_{3}$ ,

$p_{4}$ lying on a smooth table. Velocities of the four balls after collision are

0%
0,0,0,0
0%
v,v,v,v
0%
v,v,v,0
0%
0,0,0,v

Explanation

$\textbf{Hint}$ : Use law of conservation of momentum

$\textbf{Step}$ :

Here when

$P_{1}$ and

$P_{2}$ will collide, their momentum will be exchanged because the collision is elastic and the two bodies are similar.

When

$P_{2}$ and

$P_{3}$ collide same happens.

So, momentum of

$P_{1}$ is transferred to

$P_{4}$

So,

$P_{1}$ ,

$P_{2}$ ,

$P_{3}$ will have zero velocity and

$P_{4}$ has a velocity of

$V$

$\textbf{Step 2}: \textbf{Calculation}$ :

Let us take two balls. Let the first ball has a mass of 'm' and velocity 'v'. Let the second ball has a mass of 'm' and velocity is

$0$ that is it is at rest.

For perfectly elastic collision coefficient of restitution is

$e=1$

Total momentum before collision (Initial momentum)=Total momentum after collision(Final momentum)

$P_i=P_f$

$mV=mV_1+mV_2$

$\implies V_1+V_2=V$

$(1)$

We know coefficient of restitution

$e=1$

So,

$e= \dfrac{V_{separation}}{V_{approach}}$

$1=\dfrac{V_1-V_2}{V}$

$V_1-V_2=V$

$(2)$

Solving equation

$(1)$ and

$(2)$ we get

$2V_1=2V$

$\implies V_1=V$

$V_2=0$

$P_{1}$ ,

$P_{2}$ ,

$P_{3}$ will have zero velocity and

$P_{4}$ has a velocity of

$V$

Thus option D is correct.

A particle moves under the effect of a force

$F = C x$ from

$x = 0$ to

$x = x_{1}$ . The work done in the process is (treat

$C$ as a constant):

0%
$\dfrac {C^{2}}{x^{2}_{1}}$
0%
$Cx^{2}_{1}$
0%
$\dfrac {Cx^{2}_{1}}{2}$
0%
$\dfrac {C}{x^{2}_{1}}$

Explanation

Work Done

$\displaystyle =\int_{0}^{x_1} \vec{F}.d\vec{x}$

$\displaystyle=\int_{0}^{x_1}Cx.dx$

$= \dfrac {Cx^{2}_{1}}{2}$

A ball of mass M moving with a velocity V collides head on elastically with another of same mass but moving with a velocity v in the opposite direction. After collision,

0%
the velocities are exchanged between the two balls
0%
both the balls come to rest
0%
both of them move at right angles to the original line of motion
0%
one ball comes to rest and another ball travels back with velocity 2V

Explanation

Using conservation of linear momentum we have:

$M_1u_1-M_2u_2=-M_1v_1+M_2v_2$

$MV-Mu_2=-Mv+Mv_2$

$V-u_2=-v+v_2$ .............(i)

and as for a perfect elastic collision we have

$e=1$ , thus we get

$v_2+v=u_2+V$ .................(ii)

Using these two equations we get:

$u_2=v$ and

$v_2=V$

Thus, the velocities are exchanged between the two balls.

A heavy steel ball of mass greater than 1 kg moving with a speed of 2m/ s collides head on with a stationary ping pong ball of mass less than 0.1 g. The collision is elastic. After the collision the ping pong ball moves approximately with a speed

0%
$2 m / s$
0%
$4 m/ s$
0%
$2\times10^{4}m / s$
0%
$2\times10^{3}m / s$

Explanation

Since the body is much heavy these won't be much change in velocity
&

$e = 1$

i.e.,

$\dfrac{v-2}{0-2} = -1$

$\Rightarrow v = 4 m/s$

A heavier body moving with certain velocity collides head on elastically with a lighter body at rest. Then

0%
smaller body continues to be in the same state of rest
0%
smaller body starts to move in the same direction with same velocity as that of bigger body
0%
the smaller body starts to move with twice the velocity of the bigger body in the same direction
0%
the bigger body comes to rest

Explanation

$\textbf{Hint}$ : Apply Law of conservation of momentum

$\textbf{Step 1:Assumptions}$

Let us assume that there are two bodies. Let the first body has a mass of 'M' and initial velocity=

$V$ and final velocity=

$V_1$ (after collision).

Let the second body has a mass of 'm' and initial velocity=0 ; final velocity=

$V_2$ (after collision)

$\textbf{Step 2:Apply Law of conservation of Linear Momentum}$

We know according to the law of conservation of momentum

Initial momentum=Final Momentum

$MV+m(0)=MV_1+mV_2$

$(1)$

$\textbf{Step3:Coefficient of Restitution}$

We know for elastic collision e=1

$e=1=\dfrac{V_2-V_1}{V-0}$

$V=V_2-V_1$

$V_1=V_2-V$

$(2)$

$\textbf{Step 4:Solve above equations to obtain answer}$

Substitute

$V_1=V_2-V$ in

$(1)$

we get

$MV=M(V_1)+mV_2$

$MV=M(V_2-V)+mV_2$

$MV=MV_2-MV+mV_2$

$2MV=MV_2+mV_2$

$V_2=\dfrac{2VM}{M+m}$

$V_2=2V$ if

$M>>m$

Thus option C is correct

A body of mass 2 kg is at rest. A force of

$4\hat{i} + 3\hat{j} - 5\hat{k}$ N acts on it and displaces through

$2\hat{i} + \hat{j} + 2\hat{k}$ m. The velocity acquired by the body is:

0%
$0\ ms^{-1}$
0%
$1\ ms^{-1}$
0%
$2\ ms^{-1}$
0%
$10\ ms^{-1}$

Explanation

$W.D=\Delta \left ( K.E \right )=\dfrac{1}{2}mv^{2}=\vec{F}.\vec{s}$

$=( 4\hat{i}+3\hat{j}-5 \hat{k} ).( 2\hat{i}+1\hat{j}+2\hat{k} )$

$=8+3 -10$

$=1$

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

$\Rightarrow \dfrac{1}{2}\times 2v^{2}=1$

$\Rightarrow v=1 \ m/ s$

A rain drop of mass (1/10) gram falls vertically at constant speed under the influence of the forces of gravity and viscous drag. In falling through 100 m, the work done by gravity is

0%
0.98 J
0%
0.098 J
0%
9.8 J
0%
98 J

Explanation

work done = mgh

m = mass of drop = 0.1 g = 0.00001kg

g =9.8m/s

h = 100m

work done =

$0.00001 \times 9.8 \times 100$

= 0.098 J

A sphere of mass m moving with constant velocity u, collides with another stationary sphere of same mass. If e is the coefficient of restitution, the ratio of the final velocities of the first and second sphere is

0%
$\dfrac{1+e}{1-e}$
0%
$\dfrac{1-e}{1+e}$
0%
$\dfrac{e}{1-e}$
0%
$\dfrac{1+e}{e}$

Explanation

using momentum conservation

$v_{1}+v_{2}=u$

$e=\dfrac{v_{2}-v_{1}}{u-0}$

$\Rightarrow v_{2}-v_{1}=eu$

$\Rightarrow v_{2}=\dfrac{(1+e)}{2}u$

$\Rightarrow v_{1}=\dfrac{(1-e)}{2}u$

$v_{1} : v_{2}=\dfrac{1-e}{1+e}$

A ball moving with a speed of 2.2 m/sec strikes an identical stationary ball. After collision the first ball moves at 1.1 m/sec at

$60^{0}$ with the original line of motion. The magnitude and direction of the ball after collision is

0%
$5 m/sec,90^{0}$
0%
$2 m/sec,60^{0}$
0%
$\sqrt{39(1.1)}m/sec,30^{0}$
0%
$10m/sec,60^{0}$

Explanation

Conserving momentum in X - direction

$m \times 2.2 = 1.1 \times m \times cos 60 + v\:cos\:\theta \times m$

$2.2 = \dfrac{1.1}{2} + v\:cos\:\theta$

$\dfrac{3.3}{2} = v\:cos\:\theta$ ........................(1)

Similarly in Y - direction

$v\:sin\:\theta = 1.1 sin \;60$

$v\:sin\:\theta = \dfrac{1.1\sqrt{3}}{2}$ ........................(2)

from (1) & (2)

$v = \sqrt{3.9(1.1)}m/s$

$\theta = 30^{0}$

A 50 gm ball collides with another ball of mass 150gm, moving in its direction of motion, After collision the two balls move at a an angle

$30^{0}$ with their initial direction. Ratio of their velocities after collision is

0%
3 : 1
0%
1 : 3
0%
2 : 3
0%
1 : 1

Explanation

The momentum of the system is conserved in every direction. Since the initial vertical momentum is zero, the final vertical momentum is also zero.
Therefore,

$50v_1 \times sin\,30^{\circ} = 150v_2 \times sin\,30^{\circ}$

$\dfrac{v_1}{v_2} = \dfrac{3}{1}$ .

A body of mass 5 kg moving with a speed of

$3 ms^{-1}$ collides head on with a body of mass 3 kg moving in the opposite direction at a speed of

$2 ms^{-1}$ . The first body stops after the collision. Find the final velocity of the second body.

0%
$3 ms^{-1}$
0%
$5 ms^{-1}$
0%
$-9 ms^{-1}$
0%
$30 ms^{-1}$

Explanation

$\underline {Given:}$

$m_1=5kg$

$m_2=3kg$

$u_1=3ms^{-1}$

$u_2=-2ms^{-1}$

Before collision, the 2nd body is moving backwards. Hence its velocity is negative.

$v_1=0ms^{-1}$

After collision, the 1st body stops moving. So its velocity is zero.

$To$

$find:$

$v_2=$

$?$

$\underline {Solution:}$

$m_1 u_1+m_2 u_2$

$=$

$m_1 v_1+m_2 v_2$

Substituting the values, we get,

$(5\times3)$

$+$

$(3\times(-2))$

$=$

$5\times 0$

$+$

$3\times v_2$

$\therefore15$

$-$

$6$

$=$

$3\times v_2$

$\therefore$

$9$

$=$

$3 v_2$

$\therefore$

$v_2$

$=$

$9$

$\div$

$3$

$=$

$3$

$\therefore$

$v_2$

$=$

$3ms^{-1}$

Since we got a positive value for

$v_2,$ it means that the 2nd body moves forwards after collision.

Velocity being a vector quantity, the direction is also important.

After collision, the 2nd body moves with a velocity

$3ms^{-1}$ in the forward direction.

A body is moving in a vertical circle such that the velocities of body at different points are critical. The ratio of velocities of body at angular displacements

$60^{0}$ and

$120^{0}$ from the lowest point is:

0%
$\sqrt{5}:\sqrt{2}$
0%
$\sqrt{3}:\sqrt{2}$
0%
$\sqrt{3}:1$
0%
$\sqrt{2}:1$

Explanation

Given that velocities at different points are critical

so velocity at top point (

$T$ )

$\Rightarrow$

${ V }_{ T }$

$=\sqrt { gR } ...(1)$

From the diagram

in right triangle

$\triangle OMA$

$\Rightarrow$

$MA=OP=OP\cos { \theta }$

At point

$A$ ,

$\theta =60°$

${ \Rightarrow H }_{ A }=OG-OP\\ { \Rightarrow H }_{ A }=R-R\cos { 60° } =\frac { R }{ 2 } ...(2)$

in right triangle

$\triangle OMB$

$\Rightarrow$

$MB=QP=QP\cos { (180-\theta) }$

At point

$B$ ,

$\theta =120°$

${ \Rightarrow H }_{ B }=OG+OQ\\ { \Rightarrow H }_{ B }=R+R\cos { 60° } =\frac { 3R }{ 2 } ...(4)$

Using energy conservation equation at point

$T$ and

$A$

$\Rightarrow \frac { 1 }{ 2 } m{ { V }_{ T } }^{ 2 }+mg(2R)=\frac { 1 }{ 2 } m{ { V }_{ A } }^{ 2 }+mg{ H }_{ 1 }...(2)$

putting the value of eqn (1), eqn (2) in eqn (4)

$\Rightarrow \frac { 1 }{ 2 } m{ (\sqrt { gR } ) }^{ 2 }+mg(2R)=\frac { 1 }{ 2 } m{ { V }_{ A } }^{ 2 }+mg\frac { R }{ 2 }$

$\Rightarrow V_{ A }=2\sqrt { gR }$

Using energy conservation equation at point

$T$ and

$B$

$\Rightarrow \frac { 1 }{ 2 } m{ { V }_{ T } }^{ 2 }+mg(2R)=\frac { 1 }{ 2 } m{ { V }_{ B } }^{ 2 }+mg{ H }_{2}...(2)$

putting the value of eqn (1), eqn (3) in eqn (4)

$\Rightarrow \frac { 1 }{ 2 } m{ (\sqrt { gR } ) }^{ 2 }+mg(2R)=\frac { 1 }{ 2 } m{ { V }_{ B} }^{ 2 }+mg\frac { 3R }{ 2 }$

$\Rightarrow V_{ B }=\sqrt { 2gR }$

Ratio of

${ V }_{ A }$ and

${ V }_{ B }$

$=\sqrt { 2 } :1$

So the correct option is (

$D$ )

A simple pendulum is oscillating with an angular amplitude

$60^{0}$ . If mass of bob is

$50\ g$ , the tension in the string at mean position is:

Consider:

$g=10ms^{-2}$ , length of the string,

$L = 1\ m$ .

0%
$0.5\ N$
0%
$1\ N$
0%
$1.5\ N$
0%
$2\ N$

Explanation

Apply work energy theorem between A and B

$mg L(1-cos\theta)=\dfrac{1}{2}mv^{2}$

$10\times 1\times (1-\dfrac{1}{2})=\dfrac{1}{2}mv^{2}$

$v^{2}=10$

Then, the tension in,

$T=mg+\dfrac{mv^{2}}{R}$

$=\dfrac{50\times 10}{1000}+\dfrac{50}{1000}\times \dfrac{10}{1}$

$=1 N$

CBSE Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 1 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 1 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.