MCQ Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 3

One end of a light spring constant

$k$ is fixed to a wall and the other end is tied to a block placed on a smooth horizontal surface. In a displacement, the work done by the spring is

$\frac {1}{2} kx^{2}$ . The possible cases are

0%
The spring was initially compressed by a distance $x$ and was finally in its natural length
0%
It was initially stretched by a distance $x$ and finally was in its natural length
0%
It was initially in its natural length and finally in a compressed position
0%
It was initially in its natural length and finally in a stretched position

A locomotive of mass

$m$ has a velocity

$v = a \sqrt{x}$ . Find the work done by all the forces acting on locomotive in first

$t$ sec.

0%
$\displaystyle \frac{ma^2 t^2}{4}$
0%
$\displaystyle \frac{ma^4 t^2}{4}$
0%
$\displaystyle \frac{ma^4 t^2}{8}$
0%
$\displaystyle \frac{ma^4 t^2}{2}$

Explanation

$\displaystyle \dfrac{dv}{dt} = \dfrac{dv}{dx} \cdot \dfrac{dx}{dt} = \dfrac{a}{2 \sqrt{x}} (a \sqrt{x}) h = \dfrac{a^2}{2}$

$\therefore F = \displaystyle \dfrac{ma^2}{2}$

$\displaystyle S = u t + \dfrac{1}{2} Ft^2 = 0 + \dfrac{a^2}{4} t^2$

$W = \displaystyle F.s = \dfrac{ma^4 t^2}{8}$

Let

$\theta$ denote the angular displacement of a simple pendulum oscillating in a vertical plane. If the mass of the bob is m, then tension in the string is mg

$\cos\theta$

0%
always
0%
never
0%
at the extreme position
0%
at the mean position

Explanation

For simple pendulum

$\dfrac { m{ v }^{ 2 } }{ r } =T-mg\cos { \phi }$

But when

$\phi =\theta$ , i.e., the bob is at extreme position. its velocity is zero, hence the equation becomes:

$T-mg\cos { \theta }=0$

$\Rightarrow \quad T = mg\cos { \theta }$

Water in a bucket is whirled in a vertical circle with a string to it. The water does not fallen even when the bucket is inverted at the top of its path . We conclude that in this position:

0%
$\displaystyle mg = \frac{mv^2}{r}$
0%
$mg$ is greater than $\displaystyle \frac{mv^2}{r}$
0%
$mg$ is not greater than $\displaystyle \frac{mv^2}{r}$
0%
$mg$ is not less than $\displaystyle \frac{mv^2}{r}$

Explanation

Its because a centrifugal force

$\dfrac { m{ v }^{ 2 } }{ r }$ acts on the water which is away from the center and opposite to force of gravity

$mg$ . If the

$mg$ is greater then

$\dfrac { m{ v }^{ 2 } }{ r }$ water will spill out. but if

$\dfrac { m{ v }^{ 2 } }{ r }$ is greater then

$mg$ water will not spill out of the bucket.

Note: for keeping the water in bucket while rotating it in a vertical circle you will have to keep the velocity above a certain value to make

$\dfrac { m{ v }^{ 2 } }{ r }$ greater then

$mg$ .

A ball with velocity

$9\ m/s$ collides with another similar stationary ball. After the collision the two balls move in directions making an angle of

$30^o$ with the initial direction. The ratio of the speeds of balls after the collision will be :

0%
$\displaystyle \dfrac{v_1}{v_2} = 1$
0%
$\displaystyle \dfrac{v_1}{v_2} > 1$
0%
$\displaystyle \dfrac{v_1}{v_2} < 1$
0%
$\displaystyle \dfrac{v_1}{v_2} = 0$

Explanation

Let

$v_{1}$ and

$v_{2}$ be the velocities of above and below particles respectively, as there is no external force .momentum is conserved.
conserving the momentum in vertical direction,

$m v_{1}\sin(30) =mv_{2}\sin(30)$

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

$\Rightarrow \dfrac{v_{1}}{v_{2}} =1$

A particle P of mass m attached to a vertical axis by two strings AP and BP of length 1m each. The separation

$AB = l$ . P rotates around the axis with an angular velocity

$\omega$ . The tension in the two strings are

$T_1$ and

$T_2$ .

0%
$T_1 = T_2$
0%
$T_1 + T_2 = m\omega^2l$
0%
$T_1 - T_2 = 2mg$
0%
BP will remain taut only if $\omega \ge \sqrt{2g/l}$

Explanation

As the triangle

$ABP$ is equilateral. all the angles are of

$60^o$ .

Balancing the forces in vertical direction,

$T_1sin30^o-T_2sin30^o=mg\Rightarrow T_1-T_2=2mg\ -(1)$

The radius of rotation of circle,

$r=lcos30^o$

Balancing forces in horizontal direction,

$T_1cos30^o+T_2cos30^o=m\omega^2r\Rightarrow T_1+T_2=m\omega^2l\ -(2)$

For BP to be taut,

$T_2$ should be greater than 0.

From

$(1)-(2)$ ,

$2T_2=m\omega^2l-2mg>0\Rightarrow \omega \geq \sqrt{2g/l}$

A stone of mass

$1000g$ tied to a light string of length

$10/3m$ is whirling in a vertical circle. If the ratio of the maximum tension to minimum tension is

$4$ and

$g=10{ ms }^{ -2 }$ , then the speed of stone at the highest point of circle is :

0%
$20{ ms }^{ -1 }$
0%
$10\sqrt { 3 } { ms }^{ -1 }$
0%
$5\sqrt { 3 } { ms }^{ -1 }$
0%
$10{ ms }^{ -1 }$

Explanation

Minimum tension is at topmost point of the circle
Maximum tension is bottom point of the circle.

$\dfrac{Max. Tension}{Min. Tension} = \dfrac { \dfrac { m{ { v }_{ b }^{ 2 } } }{ r } +mg }{ \dfrac { m{ v }_{ t }^{ 2 } }{ r } -mg } =4$
and
Conservation of energy is :

$\dfrac { { v }_{ b }^{ 2 }-{ v }_{ t }^{ 2 } }{ 2 } =2gr$
Using both equations,

${ v }_{ t }^{ 2 }=3gr$
So

${v}_{t} = \sqrt { 3gr } = 10\ m/s$

A toy car is tied to the end of an unstretched string of a length,

$a$ . When revolved, the toy car moves in a horizontal circle of radius

$2a$ with time period,

$T$ . If it is now revolved in a horizontal circle of radius

$3a$ with a period

$T'$ with the same force, then

0%
$\displaystyle T' = \dfrac{\sqrt{3}}{2} T$
0%
$\displaystyle T' = \sqrt {\dfrac{3}{2}} T$
0%
$T' = T$
0%
$\displaystyle T' = \dfrac{3}{2} T$

Explanation

$\displaystyle m(2a)\left(\dfrac{2\pi}{T}\right)^2 = m(3a) \left(\dfrac{2\pi}{T'}\right)^2$ .

So,

$\dfrac{T'}{T}=\sqrt{\dfrac{3}{2}}$

In the elastic collision of heavy vehicle moving with a velocity 10 ms

$^{-1}$ and a small stone at rest, the stone will fly away with a velocity equal to :

0%
40 ms $^{-1}$
0%
20 ms $^{-1}$
0%
10 ms $^{-1}$
0%
5 ms $^{-1}$

Explanation

In the elastic collision between a heavy object and a very light object at rest, the velocity of particles after collision is
for heavy particle,

$v_1 = u_1$
for light particle,

$v_2 = 2u_1 - u_2$
since,

$u_2 = 0$ hence,

$v_2 = 2u_1$
Therefore, the stone will fly away with a velocity equal to

$v_2 = 2u_1 = 2(10) = 20 ms^{-1}$

A sphere of mass m, moving with a speed v, strikes a wall elastically at an angle of incidence

$\theta$ . If the speed of the sphere before and after collision is the same and the angle of incidence and velocity normally towards the wall the angle of rebound is equal to the angle of incidence and velocity normally towards the wall is taken as negative then, the change in the momentum parallel to wall is :

0%
mv cos $\theta$
0%
2mv cos $\theta$
0%
-2mv cos $\theta$
0%
zero

A body of mass

$1\ kg$ is rotating in a vertical circle of radius

$1\ m$ . What will be the difference in its kinetic energy at the top and bottom of the circle?

Take

$g = 10\ m/s^{2}$

0%
$50\ J$
0%
$30\ J$
0%
$20\ J$
0%
$10\ J$

Explanation

According to work energy theorem,

$\Delta K.E.=W$ and here work is done by the gravitational force.

$\Rightarrow \Delta K.E.=W = mg\ (2r) = 1 \times 10 \times 2(1)=20\ J$

The work done in stretching a spring of force constant

$K$ from length

$l_1$ to

$l_2$ is :

0%
$K (l_2 - l_1)$
0%
$\displaystyle \frac{K}{2} (l_2 + l_1)$
0%
$K (l_2^2 - l_1^2)$
0%
$\displaystyle \frac{K}{2} (l_2^2 - l_1^2)$

Explanation

Here, a spring of force constant K is stretched from length

$l_1$ to

$l_2$ .
Let, the spring is stretched through a small distance

$x$
The restoring force exerted by spring against the displacement x is

$F' = - Kx$
Hence, to stretch the spring the required force is equal and opposite to restoring force i.e.

$F = -F' = Kx$
Now, work done by applied force in stretching the spring is

$\displaystyle W = \int_{{l}_{1}}^{{l}_{2}} F dx$

$\displaystyle W = \int_{{l}_{1}}^{{l}_{2}} Kx dx$

$\displaystyle W = \frac{K}{2}\left | x^2 \right |_{l_1}^{l_2}$

$\displaystyle W = \frac{K}{2}(l_2)^2 - (l_1)^2$

A ball of mass

$m$ moving with velocity

$v$ collides elastically with wall and rebounds. The change in momentum of the ball will be :

0%
$4 mv$
0%
$2 mv$
0%
$mv$
0%
$zero$

Explanation

Here, a ball of mass

$m$ moving with velocity

$v$ collides elastically with wall hence, momentum is

$p_i = mv$
The ball rebounds from wall hence, final momentum is

$p_f = -mv$
Change in momentum is

$\Delta p = p_i - p_f$

$\Delta p = mv - (-mv) = 2mv$

The velocity of a particle at highest point of the vertical circle is

$\sqrt{3rg}$ . The tension at the lowest point, if mass of the particle is

$m$ , is

0%
$2\ mg$
0%
$4\ mg$
0%
$6\ mg$
0%
$8\ mg$

Explanation

Let the velocity at lowest point be

$v$ . then according to law of conservation of energy.
energy at lowest point= energy at highest point

$\Rightarrow \dfrac{1}{2}mv^2=mg(2r)+\dfrac{1}{2}m(\sqrt{3rg})^2 \\ \Rightarrow v=\sqrt{7rg}$

At lowest point,

$\dfrac { m{ v }^{ 2 } }{ r } =T-mg \\ \Rightarrow T=mg+ \dfrac { m{ v }^{ 2 } }{ r }=mg+ \dfrac { m (\sqrt{7rg} )^{ 2 } }{ r }$

$\Rightarrow T=mg+7mg=8mg$

A stone of mass

$1\ kg$ is tied of the end of a string

$1\ m$ long. It is whirled in a vertical circle. If the velocity of stone at the top is

$4\ m/s$ . What is the tension in the string at the lowest point?

Take

$g = 10\ m/s^{2}$

0%
$6\ N$
0%
$66\ N$
0%
$5.2\ N$
0%
$76\ N$

Explanation

Let the velocity at lowest point is

$v$ and at highest point be

$u$ , then according to law of conservation of energy.
energy at lowest point= energy at highest point.

$\Rightarrow\dfrac{1}{2}mv^2=mg(2r)+\dfrac{1}{2}mu^2 \\ \Rightarrow v=\sqrt{u^2+4rg}=\sqrt{4^2+(4\times 1 \times 10)}=\sqrt{56}$
At lowest point,

$\dfrac { m{ v }^{ 2 } }{ r } =T-mg$

$\Rightarrow T=mg+ \dfrac { m{ v }^{ 2 } }{ r }=mg+ \dfrac { m (\sqrt{56} )^{ 2 } }{ r }$

$\quad$

$(r=1)$

$\Rightarrow T=mg+56\ m=66 \times m =66 \times 1 = 66\ N$ .

A bullet weighing 10 g and moving at 300 ms

$^{-1}$ strikes a 5 kg block of ice and drops dead. The ice block is sitting on frictionless level surface. The speed of the block, after the collision is :

0%
60 ms $^{-1}$
0%
3 ms $^{-1}$
0%
6 ms $^{-1}$
0%
0.6 ms $^{-1}$

Explanation

Here, a bullet weighing 10 g and moving at 300

$ms^{-1}$ strikes a 5 kg block of ice and drops dead hence

$m_1u_1 + m_2u_2 = m_2 v$

$10 \times 10^{-3} (300) + (5)(0) = 5 v$

$3000 \times 10^{-3} = 5 v$

$v = \dfrac{3}{5} = 0.6 ms^{-1}$

A massive ball moving with a speed

$v$ collide with a tiny ball having a very small mass , immediately after the impact the second ball will move at speed approximately equal to :

0%
$\infty$
0%
$\dfrac{v}{2}$
0%
$v$
0%
$2v$

Explanation

In an elastic collision where the projectile is much more massive than the target, the velocity of the target particle after the collision will be about twice that of the projectile and the projectile velocity will be essentially unchanged. Hence, the second ball will move at a speed approximately equal to

$2v$ .

A spring of spring constant

$5\times 10^3N/m$ is stretched initially by 5 cm from the unstretched position. Then the work required to stretch it further by another 5 cm is

0%
$18.75 J$
0%
$25.00 J$
0%
$6.25 J$
0%
$12.50 J$

Explanation

$\displaystyle W_1=\frac{1}{2 }\times5\times 10^3(0.05)^2$

$\displaystyle \Rightarrow W_2=\frac{1}{2 }\times5\times 10^3(0.10)^2$

$\displaystyle \therefore \Delta W= W_2 -W_1 =\frac{1}{2 }\times5\times 10^3(0.10)^2 -\frac{1}{2 }\times5\times 10^3(0.05)^2 \\ =\frac{1}{2}\times5\times 10^3\times 0.15\times 0.05=18.75J$

A pump is used to lift

$500\ kg$ of water from a depth of

$80\ m$ in

$10\ s$ .

(Take

$g=10\ ms^{-2}$ ). Calculate the work done by the pump.

0%
$16 \times 10^5J$
0%
$4\times 10^5J$
0%
$4\times 10^8J$
0%
$2\times 10^5J$

Explanation

Given,

Mass of water lifted,

$m=500\ kg$

Displacement,

$d=80\ m$

Time taken,

$t=10\ s$

Force,

$F = m\times g$

$F = 500\times 10$

$F= 5000\ N$

Work done,

$W= F\times d$

$W = 5000\times 80$

$W = 4\times 10^{5}\ J$ .

What will be the potential energy of a body of mass 5 kg kept at a height of 10 m ?

0%
50 J
0%
0.5 J
0%
500 J
0%
25 J

A particle moves under the effect of a force

$F=cx$ from

$x=0$ to

$x=x_1$ , the work done in the process is

0%
$cx_1^2$
0%
$\displaystyle \frac{1}{2} cx_1^2$
0%
$2cx_1^2$
0%
zero

Explanation

$\displaystyle W=\overset{x_1}{\underset{0}{\int}}Fdx=\overset{x_1}{\underset{0}{\int}}cxdx=\left [ \frac{1}{2}cx^2 \right ]_0^{x_1}$

$\displaystyle =\frac{1}{2}c(x_1^2-0)=\frac{1}{2}cx_1^2$

Two solid rubber balls

$A$ and

$B$ having masses

$200\ \&\ 400\ \text{gm}$ respectively are moving in opposite direction with velocity of

$A$ equal to

$0.3\ \text{m/sec}.$ After collision the two balls come to rest when the velocity of

$B$ is :

0%
$0.15\ \text{m/sec}$
0%
$1.5\ \text{m/sec}$
0%
$-0.15\ \text{m/sec}$
0%
$\text{None of these}$

Explanation

$\text{Let the velocity of 2nd ball be}\ v\ \text{m/s}$

${ m }_{ A }=0.2\ \text{kg};\quad { v }_{ A }=0.3\ \text{m/s};\quad { m }_{ 2 }=0.4\ \text{kg};\quad { v }_{ 2 }=v$

$\text{Applying Momentum Conservation:}$

$\text{Initial Momentum}=\text{Final Momentum}$

$\Rightarrow { m }_{ A }{ v }_{ A }+{ m }_{ B }{ v }_{ B }=0\quad (\text{final momentum})\\ \Rightarrow 0.2\times 0.3+0.4\times v=0\\ \Rightarrow -0.06/0.4=v\\ \Rightarrow v=-0.15\ \text{m/s}$

Name the type of energy (kinetic energy

$K$ or potential energy

$U$ ) possessed in the following case.

The bob of a simple pendulum at its extreme position.

0%
$K$
0%
$U$
0%
$K$ and $U$
0%
No energy

Explanation

Here simple pendulum is at a certain height and at extreme position it will not have any velocity. Hence , kinetic energy will be zero. Whereas

$U = m\times g\times h$ .

As it is at certain height it has some value for

$h$ . Hence, it will have certain potential energy.

A man raises a box of mass

$50 \ kg$ to a height of

$2 \ m$ in

$2 \ minutes$ , while another man raises the same box to the same height in

$5 \ minutes$ . What is the ratio of work done by them ?

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

Explanation

$Work = Fs\cos\theta$

where,

$F$ is the force applied,

$s$ is the displacement, and

$\theta$ is the angle between the force applied and displacement

Hence, work done is independent of time taken.

In the given cases,

$\theta = 0^{\circ}$ as the force applied are in the same direction. Also,

$F=mg$

$W=Fs=mgs$

In both the cases, mass and displacement are the same.

$W =50\times 10\times 2 = 1000 J$

Hence work done is in the ratio of

$1000 : 1000 = 1:1$ .

Two objects that are moving along an xy-plane on a frictionless floor collide. Assume that they form a closed, isolated system. The following table gives some of the momentum components (in kilogram meters per second) before and after the collision. What are the mission values (a, b):

	Before collision		After collision
Object	P $_x$	P $_y$	P $_x$	P $_y$
A	-4	5	3	a
B	b	-2	4	2

0%
10, 11
0%
1, 11
0%
5, 7
0%
6, 4

Explanation

Applying conservation of momentum in x-axis :

$-4+ b = 3+4$

$\implies b = 11$

Applying conservation of momentum in y-axis :

$5-2 = a+2$

$\implies a = 1$

The hydroelectric plants do not produce electricity, if the water level in the dam is less than 34 m.

0%
True
0%
False
0%
Ambiguous
0%
Data insufficient

A block is acted upon by a force, which is inversely proportional to the displacement

$x$ . The work done will be proportional to

0%
$x$
0%
$x^{1/2}$
0%
$x^2$
0%
none of these

Explanation

Given,

$F=\dfrac{a}{x}$ where a is a constant

$dW= \vec{F} \cdot \vec{ds}= \dfrac{a}{x}dx$

$\therefore W = \int_{}^{}\dfrac{a}{x}dx=a \ln x$

A small sphere is attached to a cord and rotates in a vertical circle about a point

$O$ . If the average speed of the sphere is increased, the cord is most likely to break at the orientation when the mass is at

0%
Bottom point $B$
0%
Top point $A$
0%
Point $D$
0%
Point $C$

Explanation

Tension in the cord is maximum (for a given average speed of rotation) when the mass,

$m$ , is at the bottom points B, as the gravitational force is in the downward direction and tension of the cord is directly opposing it.

Name the type of energy (kinetic energy

$K$ or potential energy

$U$ ) possessed in the following case.

A piece of stone placed on the roof.

0%
$U$
0%
$K$
0%
$U$ and $K$
0%
No energy

Explanation

When a stone is placed at the roof it is at a certain height that is given by

$U = m\times g\times h$

As we have a certain value for

$h$ there would be some value for potential energy.

As the stone is at rest at the roof it will not have any kinetic energy as its velocity is zero,

$K = 1/2 (m\times v^{2})=0$ .

Hence, only potential energy will be there.

During the displacement, which of the curves shown in the graph best represents the work done on the spring block system by the applied force ?

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

By stretching the rubber strings of a catapult we store .......... energy in it.

0%
$potential$
0%
$electrical$
0%
$heat$
0%
$kinetic$

A body starts from rest and acquires a velocity

$V$ in time

$T$ at constant rate. The work done on the body in time

$t$ will be proportional to

0%
$\displaystyle \frac{V}{T}t$
0%
$\displaystyle \frac{V^2}{T}t^2$
0%
$\displaystyle \frac{V^2}{T^2}t$
0%
$\displaystyle \frac{V^2}{T^2}t^2$

Explanation

Acceleration:

$a = \dfrac{V}{T}$

Velocity at time t:

$v=at=\dfrac{Vt}{T}$

Change in KE:

$\Delta K = \dfrac{1}{2}mv^2 -0 = \dfrac{1}{2}m(\dfrac{Vt}{T})^2= \dfrac{mV^2t^2}{2T^2}$

Work done on the body is equal to change in its kinetic energy.

$\therefore W = \dfrac{mV^2t^2}{2T^2}$

$\Rightarrow W \propto \dfrac{V^2t^2}{T^2}$

A simple pendulum is vibrating with angular amplitude of

$\theta=90^{o}$ as shown in figure.
For what value of

$\theta$ is the acceleration directed

$(i)$ Vertically upwards

$(ii)$ Horizontally

$(iii)$ Vertically downwards

0%
$0^{o},90^{o} \ cos^{-1}\displaystyle\frac{1}{\sqrt {3}},\ 90^{o}$
0%
$\cos^{-1}\displaystyle\frac{1}{\sqrt {3}},\ ,\ 0^{o},90^{o}$
0%
$0^{o},90^{o}, \cos^{-1}\displaystyle\frac{1}{\sqrt{2}},$
0%
$\cos^{-1}\displaystyle\frac{1}{\sqrt {3}},\ 90^{o},\ 0^{o}$

Explanation

Acceleration of a simple pendulum is

$a = -ω^{2}\times A\times cos (ωt + θ)$ , where ω is angular velocity,

$A$ is amplitude,

$t$ is time and θ is angle.

If θ is

$0$ , then

$a = -ω^{2}\times A\times cos (ωt)$ so direction is vertically upwards, when θ is

$cos^{-1} (1/3^{1/2})$ then direction is horizontally, when θ is

$90^o$ , then

$a = ω^{2}\times A\times sin (ωt)$ so direction is vertically downwards.

A stone tied to a string of length

$L$ is swired in a verticle circle with the other end of the string at the centre. At a certain instant of time the stone is at it lowest position and has a speed

$u$ . Find the magnitude of the change in its velocity as it reaches a position, where the string is horizontal.

0%
$\displaystyle \sqrt{(u^{2}-gL)}$
0%
$\displaystyle \sqrt{2(u^{2}-gL)}$
0%
$\displaystyle \sqrt{(u^{2}-2gL)}$
0%
$\displaystyle \sqrt{2(u^{2}-2gL)}$

Explanation

By using conservation of energy,

$\frac { 1 }{ 2 } m{ u }^{ 2 }=mgL+\frac { 1 }{ 2 } m{ v }^{ 2 }$

$\Rightarrow v=\sqrt { { u }^{ 2 }-2gL }$

Change in velocity =

$\sqrt { { u }^{ 2 }-2gL } \hat { i } -u\hat { j }$

Magnitude of change in velocity =

$\sqrt { { u }^{ 2 }-2gL+{ u }^{ 2 } } =\sqrt { 2\left( { u }^{ 2 }-gL \right) }$

A spring of force constant k is cut in two parts at its one-third length. When both the parts are stretched by same amount. The work done in the two parts will be:
(Note- Spring constant of a spring is inversely proportional to length of spring.)

0%
equal in both
0%
greater for the longer part
0%
greater for the shorter part
0%
data insufficient

Explanation

As we know that spring constant k is inversely proportional to length of spring

$k\alpha \dfrac { 1 }{ L }$

work done = change in potential energy

$work\quad done=\Delta PE=\dfrac { 1 }{ 2 } k{ x }^{ 2 }$

so work done is directly proportional to k

it work done is inversely proportional to length

$work\quad done\quad \alpha \quad k\quad \alpha \dfrac { 1 }{ L }$

hence work done is more for small length of spring as compared to larger.

A spring of force constant

$800\ N/m$ has an extension of

$5\ cm$ . The work done in extending it from

$5\ cm$ to

$15\ cm$ is

0%
$16\ J$
0%
$8\ J$
0%
$32\ J$
0%
$24\ J$

Explanation

$\displaystyle Workdone, W=\frac{1}{2}k\left ( x_2^2-x_1^2 \right )$

$\displaystyle =\frac{1}{2}k\left [ (0.15)^2-(0.05)^2 \right ]$

$\displaystyle =\frac{1}{2}\times 800 \times 0.02=8J$

A pendulum bob is raised to a height,

$h$ , and released from rest. At what height will it attain half of its maximum speed?

0%
$\displaystyle\frac{3h}{4}$
0%
$\displaystyle\frac{h}{2}$
0%
$\displaystyle\frac{h}{4}$
0%
$0.707h$

Explanation

By energy conservation,

$\dfrac{1}{2}mv^2_{max} = mgh$
Let at height

$h'$ , it attain velocity

$v'=\dfrac{v_{max}}{2} = \sqrt{gh/2}$
Now,

$mgh = \dfrac{1}{2}mv'^2 + mgh'$

$mgh = \dfrac{mgh}{4} + mgh'$

$mgh'=\dfrac{3}{4}mgh$

$\therefore h' = \dfrac{3h}{4}$

A simple pendulum of length

$l$ has maximum angular displacement

$\displaystyle \theta .$ Then maximum kinetic energy of a bob of mass

$m$ is

0%
$\displaystyle \frac{1}{2}mgl$
0%
$\displaystyle \frac{1}{2}mgl\cos \theta$
0%
$\displaystyle mgl\left ( 1-\cos \theta \right )$
0%
$\displaystyle \frac{1}{2}mgl\sin \theta$

Explanation

At point B,

$v=0$ and

$P.E$ is maximum.

At point A,

$P.E= 0$ and

$K.E$ is maximum.

Height of bob at B,

$AP= l-lcos\theta= l(1-cos\theta)$

Thus

$P.E= mgl(1-cos\theta)$

Applying conservation of energy at A and B,

$K.E_m +0 = 0 + P.E$

$\implies K.E_m=mgl (1-cos\theta)$

A particle of mass

$m$ moves on a straight line with its velocity varying with the distance travelled according to the equation

$\displaystyle v= \alpha \sqrt{x},$ where a is a constant. Find the total work done by all the forces during a displacement from

$x=0$ to

$x=d$ .

0%
$\displaystyle \frac{1}{2}ma ^{2}b$
0%
$\displaystyle \frac{1}{4}ma ^{2}b$
0%
$ma ^{2}b$
0%
$2\ ma ^{2}b$

Explanation

$v=\alpha\sqrt{x}$

$\implies \dfrac{dx}{dt}=\alpha\sqrt{x}$

$\implies 2\sqrt{x}=\alpha t$

$\implies x=\dfrac{\alpha^2 t^2}{4}$

Work done by forces

$=$

$\int_0^b Fdx=_0^b\int madx$

$=_0^b\int m\dfrac{d^2x}{dt^2}dx$

$=_0^b\int m(\dfrac{2\alpha^2}{4})dx$

$=\dfrac{1}{2}m\alpha^2 b$

A block of mass

$m=2$ kg is pulled by a force

$F=40$ N upwards through a height

$h=2 m$ . Find the work done on the block by the applied force F and its weight mg.

$\displaystyle \left ( g= 10 m/s^{2} \right )$

0%
$\displaystyle W_F = 80 J;W_{mg}= -40 J$
0%
$\displaystyle W_F = 80 J;W_{mg}= 40 J$
0%
$\displaystyle W_F = -80 J;W_{mg}= -40 J$
0%
$\displaystyle W_F = -80 J;W_{mg}= 40 J$

Explanation

Weight

$mg=(2)(10)=20$ N

Work done by the applied force

$\displaystyle W_{F}= Fh\cos 0^{\circ}$

As the angle between force and displacement is

$\displaystyle 0^{\circ}$

$\displaystyle W_{F}= \left ( 40 \right )\left ( 2 \right )\left ( 1 \right )= 80 J$

Similarly, work done by its weight

$\displaystyle W_{mg}= \left ( mg \right )\left ( h \right )\cos 180^{\circ}$

$\displaystyle W_{mg}= \left ( 20 \right )\left ( 2 \right )\left ( -1 \right )= -40 J$

A force

$F$ acting on a particle varies with the position

$x$ as shown in figure. Find the work done by this force in displacing the particle from

$\displaystyle x= 0\ m\:to\:x= 2\ m.$

0%
$-10J$
0%
$10J$
0%
$5J$
0%
$-5J$

Explanation

From

$\displaystyle x= 0\ m\:to\:x= 2 \ m.$ , displacement of particle and force acting on the particle both are along positive x-direction. Therefore, work done is positive and given by the area under

$F-x$ graph,
or

$\displaystyle W= \frac{1}{2}\left ( 2 \right )\left ( 10 \right )= 10J$

A block is constrained to move along x-axis under a force F=-2x. Here, F is in newton and x in metre. Find the work done by this force when the block is displaced from x=2 m to x=-4 m.

0%
-4 J
0%
-8 J
0%
-12 J
0%
-16 J

Explanation

Work done by a body in moving an object from

$x_1$ to

$x_2$

$=\int_{x_1}^{x_2}Fdx$

$=\int_{x_1}^{x_2}(-2x)dx$

$=x_1^2-x_2^2$

$=-12J$

A bob is suspended from a crane by a cable of length

$5m$ . The crane and load are moving at a constant speed

$v_{0}$ . The crane is stopped by a bumper and the bob on the cable swings out an angle of

$60^{\circ}$ . Find the initial speed

$v_{0}.(g=9.8 m/s^{2})$

0%
$2 ms^{-1}$
0%
$3 ms^{-1}$
0%
$5 ms^{-1}$
0%
$7 ms^{-1}$

Explanation

By using conservation of energy, from the initial point to the highest point

$\frac { 1 }{ 2 } m{ { v }_{ 0 } }^{ 2 }=mgl(1-cos{ 60 }^{ 0 })$

$\Rightarrow { v }_{ 0 }=\sqrt { 2gl(1-cos{ 60 }^{ 0 }) } =\sqrt { gl } =\sqrt { 50 } =7m/s$

A force

$F=(2+x)$ acts on a particle in x-direction where

$F$ is in newton and

$x$ in meter. Find the work done by this force during a displacement from

$x=1.0$ m to

$x=2.0\ m$ .

0%
$3.5 \ J$
0%
$7 \ J$
0%
$4.5 \ J$
0%
$5.5 \ J$

Explanation

As the force is variable, we shall find the work done in a small displacement from

$x$ to

$x+dx$ and then integrate it to find the total work. The work done in this small displacement is

$\displaystyle dW= F\:dx= \left ( 2+x \right )dx$

Thus,

$\displaystyle W= \int_{1.0}^{2.0}dW= \int_{1.0}^{2.0}\left ( 2+x \right )dx$

$\displaystyle = \left [ 2x+\frac{x^{2}}{2} \right ]_{1.0}^{2.0}= 3.5J$

A force

$\displaystyle \vec{F}= \left ( 3t\hat{i}+5\hat{j} \right )N$ acts on a body due to which its displacement varies as

$\displaystyle \vec{S}= \left ( 2t^{2}\hat{i}-5\hat{j} \right )m.$ Work done by this force in 2 second is:

0%
32 J
0%
24 J
0%
46 J
0%
20 J

Explanation

$W = \int \bar F.\bar {ds}$

$\bar ds = 4t\hat i dt$

$dW = (3t\hat i + 5\hat j).(4t\hat i)dt$

$W(t=2)=\int 12t^2dt = 4t^3 = 4(2^3) = 32J$

A block is constrained to move along x-axis under a force

$\displaystyle F= \frac{4}{x^{2}}\left ( x\neq 0 \right ).$ Here, F is in newton and x in metre. Find the work done by this force when the block is displaced from x=4 m to x=2 m.

0%
-1 J
0%
-2 J
0%
-3 J
0%
-4 J

Explanation

Work done in moving an object from

$x_1$ to

$x_2$

$=\int_{x_1}^{x_2}Fdx$

$=\int_{x_1}^{x_2}\dfrac{4}{x^2}dx$

$=4(\dfrac{1}{x_1}-\dfrac{1}{x_2})$

$=4(\dfrac{1}{4}-\dfrac{1}{2})$

$=-1J$

if AB is a massless string.

0%
$\displaystyle \frac{10L}{27}$
0%
$\displaystyle \frac{20L}{27}$
0%
$\displaystyle \frac{30L}{27}$
0%
$\displaystyle \frac{40L}{27}$

Explanation

When the string has traversed an angle $\theta$ , then

$T-mgcos\theta =\dfrac { m{ v }^{ 2 } }{ l }$ , and conservation of energy gives

$\dfrac { m{ u }^{ 2 } }{ 2 } =\dfrac { m{ v }^{ 2 } }{ 2 } +mgl(1-cos\theta ).$ then the string becomes slacked tension in the string becomes zero. Solving the above two equations to get the value of $\theta$

$\Rightarrow cos\theta =\dfrac { -1 }{ 3 } and { v }^{ 2 }={ u }^{ 2 }-2gl(1-cos\theta )=\dfrac { gl }{ 3 }$ from that point the maximum height reached is $\dfrac { { v }^{ 2 }{ sin }^{ 2 }\left( \pi -\theta \right) }{ 2g }$ maximum height $= l(1-cos\theta )+\dfrac { { v }^{ 2 }{ sin }^{ 2 }\left( \pi -\theta \right) }{ 2g } =\dfrac { 4l }{ 3 } +\dfrac { 4l }{ 27 } =\dfrac { 40l }{ 27 }$

The kinetic energy of particle at the lower most position is

0%
$\displaystyle \frac{4mgL}{3}$
0%
$2mgL$
0%
$\displaystyle \frac{8mgL}{3}$
0%
$\displaystyle \frac{2mgL}{3}$

Explanation

$v= 2\sqrt{\dfrac{gL}{3}}$

Also,

$u= 2v \implies u= 4\sqrt{\dfrac{gL}{3}}$

Thus

$K.E_ A= \dfrac{1}{2}mu^2= \dfrac{8mgL}{3}$

Velocity of particle when it is moving vertically downward is

0%
$\displaystyle \sqrt{\frac{10gL}{3}}$
0%
$\displaystyle 2\sqrt{\frac{gL}{3}}$
0%
$\displaystyle \sqrt{\frac{8gL}{3}}$
0%
$\displaystyle \sqrt{\frac{13gL}{3}}$

Explanation

Work-energy theorem for A to C,

$-mgL= \dfrac{1}{2}m(v')^2- \dfrac{1}{2}mu^2$ where

$u= 4\sqrt{\dfrac{gL}{3}}$

$-mgL= \dfrac{1}{2}m(v')^2- \dfrac{8mgL}{3}$

$\implies v'= \sqrt{\dfrac{10 gL}{3}}$

Minimum velocity of the particle is

0%
$\displaystyle 4\sqrt{\frac{gL}{3}}$
0%
$\displaystyle 2\sqrt{\frac{gL}{3}}$
0%
$\displaystyle \sqrt{\frac{gL}{3}}$
0%
$\displaystyle 3\sqrt{\frac{gL}{3}}$

Explanation

The maximum and minimum velocities will be at A and B respectively.

Work-energy theorem,

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

Given:

$u:v= 2:1$

$-mg(2L)= \dfrac{1}{2}mv^2-\dfrac{1}{2}m(2v)^2$

$\implies v= 2\sqrt{\dfrac{gL}{3}}$

CBSE Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 3 - 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 3 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

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