MCQ Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 4

State whether the given statement is True or False :

The energy of an object that is due to the object's motion is called kinetic energy.

0%
True
0%
False

Explanation

The kinetic energy of an object is the energy it possesses because of its motion. The kinetic energy of a point mass m is given by
K.E

$=\frac1 2m\times v^2$

A boy is whirling a stone tied at one end such that the stone is in uniform circular motion.Which of the following statement is correct?

0%
The velocity of stone at A is equal to the velocity of stone B
0%
The speed of stone at A is equal to the speed of stone at B.
0%
The centripetal force is radially outward in the string.
0%
All of the above statements are correct.

A car is accelerated on a levelled road and attains a velocity 4 times of its initial velocity. In this process, the potential energy of the car

0%
does not change
0%
becomes twice of initial
0%
becomes 4 times of initial
0%
becomes 16 times of initial

A mass is performing vertical circular motion (see figure). If the average velocity of the particle is increased, then at which point is the maximum breaking possibility of the string:

0%
A
0%
B
0%
C
0%
D

Explanation

Tension at any point in vertical motion is given by :

$T=\dfrac{mv^{2}}{l}+mg\cos\theta$

where

$\theta=$ angular displacement from lowest point ,

$l=$ length of string

$m=$ mass of string

It is clear that tension at the lowest point (

$B$ ) is greatest than at other points (

$A,C,D$ ). If we increase average velocity, tension will increase at lowest point, therefore at point B, string has maximum possibility of break.

A stone of mass

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

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

$4\;m/s$ . What is the tension in the string?

0%
$\;6\;N$
0%
$\;16\;N$
0%
$\;5\;N$
0%
$\;10\;N$

Explanation

Given :

$v =4$

$m/s$

$m =1 kg$

$r = 1$ m

Using circular motion equation :

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

$\implies$

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

$\therefore$

$T =\dfrac{1 \times 4^2}{1} -1 \times 10$

$\implies T = 6$

$N$

Two weights of 5 kg and 10 kg are placed on a horizontal table of height 1.5 m. Which will have more potential energy?

0%
5 kg
0%
10 kg
0%
Both will have equal energy
0%
None of the above

Explanation

We know that , P.E

$= mgh$
So, It is directly proportional to height and mass.

Since both the weights are at the same height, so the weight with a larger mass will have more potential energy.

Since

$10$ kg object has a larger mass than a

$5$ kg,

so, the potential energy of a

$10$ kg mass will be greater.

Which one of the following possesses potential energy?

0%
Moving vehicle on the road
0%
A running athlete
0%
Stone on the road
0%
A stretched rubber band

An overhead tank having some water possesses .......... energy.

0%
Kinetic
0%
Potential
0%
Thermal
0%
Electrical

Explanation

We know that Potential energy is the energy possessed by a body by the virtue of its position.

$\therefore$ An overhead tank having some water possesses Potential energy, as it is at a height.

When an arrow is released from a bow, potential energy changes into kinetic energy.

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

Energy possessed by a body by virtue of its motion is :

0%
Potential energy
0%
Kinetic energy
0%
Chemical energy
0%
Electrical energy

Explanation

Kinetic energy is defined as the energy possessed by a body by virtue of its motion. and it is equal to

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

$m$ and

$v$ are mass and Velocity of moving body.

A student sitting at the top of a tree has ............ than the student who is sitting on the ground.

0%
less potential energy
0%
more potential energy
0%
more kinetic energy
0%
more gravitational energy

Explanation

Potential energy =

$mgh$

$m\rightarrow mass$

$g\rightarrow acceleration\ due\ to\ gravity$

$h\rightarrow height\ from\ ground$

P.E. is directly proportional to the height from the ground. Hence a student sitting at the top of a tree has more potential energy than the student who is sitting on the ground.

The form of energy present in a wound spring is:

0%
chemical
0%
heat energy
0%
magnetic
0%
mechanical

The energy directly related to the speed of a moving body and its mass is:

0%
Kinetic
0%
Potential
0%
Solar
0%
Electric

Explanation

The kinetic energy of a body is the energy by virtue of its motion.

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

where

$m$ is mass and

$v$ is velocity.

Hence, Kinetic energy is directly related to the speed of a moving body and its mass.

A boy holds a pendulum in his hand while standing at the edge of a circular platform of radius

$r$ rotating at an angular speed

$\omega$ . The pendulum will hang at an angle

$\theta$ with the verticle such that

0%
$\;tan\,\theta=0$
0%
$\;tan\,\theta=\displaystyle\frac{\omega^2r^2}{g}$
0%
$\;tan\,\theta=\displaystyle\frac{r\omega^2}{g}$
0%
$\;tan\,\theta=\displaystyle\frac{g}{\omega^2r}$

Explanation

A pseudo force

$(ma_r)$ acts on the bob due to the rotation of circular platform.

In bob's frame, the bob is at rest (or equilibrium).

$\therefore$

$T sin \theta = ma_r$ where

$a_r = rw^2$

$\implies$

$T sin\theta = mrw^2$ ........(1)

Also

$T cos\theta =mg$ ..........(2)

Divide (2) by (1) we get,

$tan\theta =\dfrac{rw^2}{g}$

A mass tied to a string moves in a vertical circle with a uniform speed of

$5\;m/s$ as shown. At the point

$P$ the string breaks. The mass will reach a height above

$P$ of nearly:

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

0%
$\;1\;m$
0%
$\;0.5\;m$
0%
$\;1.75\;m$
0%
$\;1.25\;m$

Explanation

Given :

$u =5 m/s$

$a = - g = -10 m/s^2$

When the string breaks at point P, from there onwards the mass has the free falling motion.

At the maximum height attained, the velocity of the mass is zero i.e

$v = 0 m/s$

Using

$v^2 - u^2 = 2aS$

$\therefore$

$0 - 5^2 = 2 (-10) h$

$\implies h = 1.25$ m

A particle of mass

$'m'$ describes a circle of radius

$(r)$ . The centripetal acceleration of the particle is

$\displaystyle\frac{4}{r^2}$ . The momentum of the particle:

0%
$\;\displaystyle\frac{2m}{r}$
0%
$\;\displaystyle\frac{2m}{\sqrt{r}}$
0%
$\;\displaystyle\frac{4m}{r}$
0%
$\;\displaystyle\frac{4m}{\sqrt{r}}$

Explanation

Centripetal acceleration of the particle,

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

$\therefore$

$\dfrac{4}{r^2}= \dfrac{v^2}{r}$

$\implies v = \dfrac{2}{\sqrt{r}}$

Thus the momentum of the particle,

$P = mv = \dfrac{2m}{\sqrt{r}}$

In a vertical circle of radius

$(r)$ , at what point in its path a particle may have tension equal to zero:

0%
highest point
0%
lowest point
0%
at any point
0%
at a point horizontal from the centre of radius

Explanation

Let the tension in the string at the highest point be

$T$

Minimum speed required by the particle at the highest point to complete the vertical circular motion is

$\sqrt{gr}$

$\therefore$

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

$\dfrac{m (gr)}{r} = T+mg$

$\implies T =0$

Thus tension can be zero at the highest point.

A stone attached to one end of a string is whirled in a vertical circle. The tension in the string is maximum when

0%
the string is horizontal
0%
the string is vertical with the stone at highest position
0%
the string is vertical with the stone at the lowest position
0%
the string makes an angle of $45^{\circ}$ with the vertical

Explanation

$Answer:-$ C

When the stone is at the bottom position forces acting on stone are:

Upward force tension (T), Downward forces are weight (mg) and also

$\dfrac{mv^2}{R}$ centrifugal force

So balancing forces we get :

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

And minimum tension is obtained at highest point which will be:

$T=mg-$

$\dfrac{mv^2}{R}$

A position dependent force

$F=7-2x+3x^{2}$ N acts on a small object of mass 2kg to displace it from

$x=0$ to

$x=5m$ . The work done in joules is-

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

Explanation

$W=\int F\mathrm{d} x$

$\displaystyle \int_{0}^{5}(7-2x+3x^{2})\mathrm{d} x=\left ( 7x-\frac{2x^{2}}{2}+\frac{3x^{3}}{3} \right )_{0}^{5}=135J$

A particle is moving in a vertical circle, the tension in the string when passing through two position at angle

$30^{\circ}$ and

$60^{\circ}$ from vertical from lowest position are

$T_1$ and

$T_2$ respectively, then

0%
$T_1 = T_2$
0%
$T_1 > T_2$
0%
$T_1 < T_2$
0%
$T_1\ge T_2$

Explanation

In equilibrium,

$T=mv^2/r+mg\cos\theta$
Since,

$\theta$ increases

$\cos\theta$ and

$v$ both will decrease.
Hence, for

$\theta=60^o$ , the tension will be less.
Thus,

$T_1>T_2$ .

A pendulum bob has a speed

$3\;m/s$ while passing through its lowest position, length of the pendulum is

$0.5\;m$ then its speed when it makes an angle of

$60^{\circ}$ with the vertical is

0%
$\;2\;m/s$
0%
$\;1\;m/s$
0%
$\;4\;m/s$
0%
$\;3\;m/s$

Explanation

$\textbf{Step 1: Apply energy conservation}$

$\Delta ABD$ ,

$AB = AD \cos \theta$

$\Rightarrow AB = l \cos \dfrac{\pi}{3} = \dfrac{l}{2}$

$\Rightarrow h = l - \dfrac{l}{2} = \dfrac{l}{2}$

As there are no dissipative force energy will be conserved.

$\Rightarrow (E)_i = E_f$

$\Rightarrow K_1 + U_1 = K_2 + U_2$

Taking point

$C$ is zero potential level

$\Rightarrow \dfrac{1}{2} mu^2 = \dfrac{1}{2} mv^2 + (mgh)$

$\textbf{Step 2: Calculations}$

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

Putting given values

$v = \sqrt{(3)^2(m/s)^2 - 2 \times 9.8(m/s^2) \times \dfrac{0.5}{2}(m)} = 2 m/s$

Hence

$A$ is the correct option.

2115111_280800_ans_d10d9bce5ded4e0a9afbe5fe1264a62f.png

A particle slides from rest from the topmost point of a vertical circle of radius (

$r$ ) along a smooth chord making an angle

$(\theta)$ with the vertical the time of decent is

0%
Least for $\theta=0^{\circ}$
0%
Maximum for $\theta=0^{\circ}$
0%
Least for $\theta=45^{\circ}$
0%
Independent of $\theta$

Explanation

Force on particle along the cord = $mg\cos { \theta }$

Distance travelled by the particle = $\quad \dfrac { d }{ \cos { \theta } }$ , where $d$ is the diameter or the vertical circle.

$s = \dfrac { a{ t }^{ 2 } }{ 2 }$

$\Rightarrow\ t =\sqrt { \dfrac { 2s }{ a } }$ .

Hence, $\ t$ is independent of $\theta$ .

A particle of mass

$m_{1}$ moving with a velocity of

$5m/s$ collides head on with a stationary particle of mass

$m_{2}$ . After collision both the particle move with a common velocity of

$4m/s$ , then the value

$m_ {1}/m_{2}$ is:

0%
4:1
0%
2:1
0%
1:8
0%
1:2

Explanation

Conservation of Momentum principle,
Initial momentum is

$M_i=5m_1$
Final momentum is

$M_f=4(m_1+m_2)$
By the above stated principle

$M_i=M_f=5m_1=4(m_1+m_2)$

$\implies m_1=4m_2$

$\therefore m_1:m_2=4:1$

A time dependent force F = 10 t is applied on 10 kg block. Find out the work done by F in 2 seconds.

0%
40 J
0%
20 J
0%
120 J
0%
60 J

Explanation

$10\, \times\, \displaystyle \dfrac{dv}{dt}\, =\, 10\, t$

$\int_0^v \ dv =\int_{0}^{t} tdt$

$\Rightarrow\, v\, =\, \displaystyle \dfrac{t^2}{2}$ ..........(1)

$dW\, =\, \vec{F}\, .\, \vec{ds}$

$dW\, =\, 10 t. dx$

$dW\, =\, 10 t. v. dt$ .......... (2)

$\because\, dx\, =\, vdt$

from (1) & (2)

$dW\, =\, 10\, t\, .\, \left ( \displaystyle \dfrac{t^2}{2} \right )dt\, \quad\, dW\, =\, 5t^3dt$

$W\, =\, \displaystyle \dfrac{5}{4}\, [t^4]_0^2\, =\, 20\, J$

Force shown acts for

$2$ seconds. Find out work done by force

$F$ on

$10$ kg in

$3$ seconds.

0%
30 J
0%
20 J
0%
50 J
0%
60 J

Explanation

Work done,

$W=Fd$
Displacement

$d$ is given by,

$d=\dfrac{1}{2}a{t}^{2}$

$F=ma$

$10=10a$ ,

$a=1m/s^2$

$d=\dfrac{1}{2}(1){(2)}^{2}=2m$

$W=10\times2=20J$

A force acts on a

$3$ gm particle in such a way that the position of the particle as a function of time is given by

$x=3t-4t^2+t^3$ , where x is in meters and t is in seconds. The work done during the first

$4$ second is:

0%
$384 mJ$
0%
$168 mJ$
0%
$528 mJ$
0%
$541 mJ$

Explanation

Mass of the particle,

$m=0.003kg$

$x=3t-4t^2+t^3$

$\therefore a=-8+6t$

Thus force acting on the particle is

$ma$

Thus the work done on the particle

$=\int Fdx$

$=\int_0^4 m(-8+6t)(3-8t+3t^2)dt$

$=528mJ$ -m represents mass here.

Under the action of a force, a 2kg mass moves such that its position x as a function of time t is given by

$x=t^3/3$ where x is in meters and t in second. The work done by the force in first two seconds is:

0%
1600 joule
0%
160 joule
0%
16 joule
0%
1.6 joule

Explanation

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

$a=\dfrac{d^2x}{dt^2}=2t$

Thus force acting on the particle=

$ma=2mt$

Work done

$=\int Fdx$

$=\int_0^2 (2mt)t^2dt$

$=2m\int_0^2 t^3dt$

$=16J$

The work done by the frictional force on a surface in drawing a circle of radius r on the surface by a pencil of negligible mass with a normal pressing force N (coefficient of friction

$\mu_k$ ) is :

0%
$4\pi r^{2} \mu_K N$
0%
$-2\pi r^{2} \mu_K N$
0%
$-2\pi r \mu_K N$
0%
Zero

A bead is free to slide down a smooth wire tightly stretched between points

$A$ and

$B$ on a verticle circle of radius

$R$ . If the bead starts from rest at '

$A$ ', the highest point on the circle, its velocity when it arrives at

$B$ is

0%
$2\sqrt{gR}$
0%
$2\sqrt{gR} \cos{\theta}$
0%
$\displaystyle \dfrac{2\sqrt{R}}{g}$
0%
None of the above

Explanation

$a= g\cos{\theta}\\AB = 2R\cos{\theta}\Rightarrow v^2 =u^2 + 2as \\v^2 = 0 + 2 (g\cos{\theta}) 2R\cos{\theta}\Rightarrow v^2 = 4gR \cos ^{2 }{\theta}\\ \Rightarrow v = 2\sqrt{ gR} \cos{\theta}$

A bob hangs from a rigid support by an inextensible string of length

$\ell$ . If it is displaced through a distance

$\ell$ (from the lowest position) keeping the string straight, & then released. The speed of the bob at the lowest position is:

0%
$\sqrt{g\ell}$
0%
$\sqrt{3g\ell}$
0%
$\sqrt{2g\ell}$
0%
$\sqrt{5g\ell}$

Explanation

Distance=

$l$
change in height=

$lcos60 = l/2$

From energy conservation,
PE at max height= KE at lowest point=

$mgl/2= mv^2/2$

$v=\sqrt{gl}$

A ball of mass 3 kg collides with a wall with velocity

$10\:m/sec$ at an angle of

$30^{\circ}$ with the wall and after collision reflects at the same angle with the same speed. The change in momentum of ball in MKS unit is-

0%
20
0%
30
0%
15
0%
45

Explanation

$\Delta p=2mv\:\cos 60^{\circ}=mv= 2 \times 3 \times 10 \times \frac{1}{2}=30\: kg \:m/s$

Two equal masses are attached to the two ends of a spring of spring constant k. The masses are pulled symmetrically to stretch the spring by a length x over its natural length. The work done by the spring on each mass during the pulling is

0%
$\displaystyle \frac{1}{2}kx^{2}$
0%
$\displaystyle -\frac{1}{2}kx^{2}$
0%
$\displaystyle \frac{1}{4}kx^{2}$
0%
$\displaystyle -\frac{1}{4}kx^{2}$

Explanation

Hint:

Whenever we extend or compress a spring of spring constant $k$ , spring exerts a force on us that opposes the change in length of spring.

Step 1: Calculate the initial and Final Energy of the system.

Let the mass of two identical blocks attached to two ends of spring is

$m$ . Let spring constant of spring is

$k$ , and length of spring is

$L$ .

When the spring is in its original length, the Energy stored in the system is,

$E_1 = 0$ .

When spring is stretched by length x, by pulling both the blocks symmetrically, Energy stored in the system is given by,

$E_2 = \dfrac{1}{2}kx^2$

Step 2: Calculate Work done by spring.

By conservation of Potential Energy, we know that work done by a conservative force is equal to change in Potential Energy of the system.

Therefore, work done by spring is,

$W = -(\dfrac{1}{2}kx^2 - 0)$

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

Since, both the blocks are pulled symmetrically, work done by spring on each block will behalf of this work done.

Therefore Work done by spring on each block is,

$w = -\dfrac{1}{4}kx^2$

Thus, Work done by spring on each mass is, $w = -\dfrac{1}{4}kx^2$

Option D is correct.

The velocity of a bus, moving on a smooth road, is increased from 8 m/s to 32 m/s in 120s. During this process, the potential energy of the bus

0%
Does not change
0%
Becomes twice that of initial potential energy
0%
Becomes four times that of initial potential energy
0%
Becomes sixteen times that of initial potential energy

Explanation

As potential energy,

$P=mgh$
Since, there is no vertical displacement so,

$h=0$ .
Hence, potential energy change

$P=mgh=0$

The moving bus has only change in its Kinetic energy.

Two bodies P and Q of equal masses are kept at heights

$x$ and

$4x$ respectively. What will be the ratio of their potential energies?

0%
$1:8$
0%
$4:1$
0%
$1:4$
0%
$8:1$

Explanation

Potential energy

$P=mgh$
Given,

$h_1=x ; h_2=4x$

Since, the masses are same,
then

$\dfrac{P_1}{P_2}=\dfrac{h_1}{h_2} = \dfrac{x}{4x}=1:4$

An object of mass 500 g falls from a height of 2m. If

${g = 9.8 m/s^2}$ , what is its kinetic energy just before touching the ground?

0%
Zero
0%
19.6 J
0%
9.8 J
0%
9800 J

Explanation

We know,

Acceleration,

$a = \dfrac{v^2-u^2}{2s}$

where

$v$ and

$u$ are final and initial velocities respectively &

$s$ is the displacement.

Here,

$g = \dfrac{v^2-u^2}{2h}.$ Also,

$u=0$

$\implies$ Velocity when it touches ground,

$v=\sqrt{2gh}$

$v=\sqrt{2gh}=\sqrt{2\times9.8\times2}$

$K.E=\dfrac{1}{2}mv^2=\dfrac{1}{2}\times\dfrac{500}{1000}\times (\sqrt{2\times9.8\times2})^2$

$=\dfrac{1}{2}\times\dfrac{500}{1000}$

$\times2\times9.8\times2=9.8J$

What is work done in holding a body of mass

$20\ kg$ at a height of

$2\ m$ above the ground?

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

0%
$40\ J$
0%
$400\ J$
0%
$10\ J$
0%
Zero

Explanation

A body of mass

$20\ kg$ is held at a height of

$2\ m$ above the ground means there is no displacement because of which there is no change in its potential energy. Hence work done is zero.

Find out work done by applied force to slowly extend the spring from x to 2x.

0%
$\dfrac{1}{2}kx^2$
0%
$\dfrac{3}{2}kx^2$
0%
$\dfrac{5}{2}kx^2$
0%
$\dfrac{7}{2}kx^2$

Explanation

Initially the spring is extended by x

$W\, =\, \vec{F}\, .\, \vec{ds}$

$W\, =\,\displaystyle \int_k^{2x}Kx.dx$

$W\, =\,\left [ \displaystyle \frac{Kx^2}{2} \right ]_x^{2x} \,=\,\displaystyle \frac{3}{2}Kx^2$
It can also found by difference of PE.
i.e.

$U_f\, =\,\displaystyle \frac{1}{2}K(2x)^2\, =\,2Kx^2\, \quad\, \Rightarrow\, \quad\, U_i\, =\, \displaystyle \frac{1}{2}Kx^2\, \quad\, \Rightarrow\, \quad\, U_f\, -\, U_i\, =\, \displaystyle \frac{3}{2}Kx^2$

Initially spring is relaxed. A person starts pulling the spring by applying a variable force

$F$ . Find out the work done by

$F$ to stretch it slowly to a distance by

$x$ .

0%
$\dfrac{Kx^2}{1}$
0%
$\dfrac{Kx^2}{2}$
0%
$\dfrac{Kx^2}{4}$
0%
$\dfrac{Kx^2}{8}$

Explanation

$\int dW\, =\, \int F\, . \, ds\, =\, \int_0^x Kxdx \quad\, \Rightarrow \quad\, W\, =\,\displaystyle { \left ( \frac{kx^2}{2} \right )^x_{\circ}\, =\, \frac{Kx^2}{2}}$

How fast should a girl of 40 kg run so that her kinetic energy is 320J?

0%
64 m/s
0%
8 m/s
0%
16 m/s
0%
4 m/s

Explanation

Given, mass

$m=40kg , K.E=320J , v=?$

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

$320=\dfrac{1}{2}\times 40v^2$

$v^2=\dfrac{640}{40}=16$

$v=4m/s$
So, the girl of

$40$

$kg$ should run at

$4m/s$ so that her kinetic energy is

$320J.$

Which one of the following possesses both kinetic and potential energies?

0%
A man walking on the road
0%
A man climbing a hill
0%
A man lying on his bed
0%
A man running in the garden.

A wound watch spring has ______ energy.

0%
Mechanical
0%
Kinetic
0%
Potential
0%
Kinetic and potential

If the speed of a motor car becomes six times, then the kinetic energy becomes

0%
6 times
0%
36 times
0%
12 times
0%
24 times

Explanation

Kinetic energy, K.E =

$\dfrac{1}{2}$ mv

$^{2}$

If speed becomes 6 times, then

New K.E =

$\dfrac{1}{2}$ m (6v)

$^{2}$ =

$\dfrac{1}{2}$ m x 36 v

$^{2}$ = 36 x (

$\dfrac{1}{2}$ mv

$^{2}$ ) = 36 x K.E

Hence, kinetic energy becomes 36 times.

Asha lifts a doll from the floor and places it on a table. If the weight of the doll is known, what else does one need to know in order to calculate the work Asha has done on the doll?

0%
the time required
0%
the height of the table
0%
the power she could deliver
0%
cost of the doll or the table

The spring of the winding knob of a watch has

0%
mechanical energy
0%
only kinetic energy
0%
only potential energy
0%
kinetic or potential energy

Explanation

We know that total energy is kinetic energy plus potential energy.

Now, kinetic energy is

$E=\frac { 1 }{ 2 } M{ V }^{ 2 }$

Which depends on velocity. In watch there is no displacement in the knob, hence velocity is zero. So there is no kinetic energy. Only potential energy is there.

You lift a suitcase from the floor and keep it on the table. The work done by you on the suitcase depends on

0%
the path taken by the suitcase
0%
your weight
0%
the weight of the suitcase
0%
the time taken by you in doing so

Explanation

The work done by a person in lifting an object is stored as its potential energy

$mgh$ .

Hence, work done depends on the weight of the object

$mg$ .

Thus option C is correct.

The work done in lifting 1 kg mass to a height of 9.8 m is about

0%
1 J
0%
9.8 J
0%
$(9.8)^2 J$
0%
$\displaystyle \frac{1}{(9.8)^2} J$

Explanation

The work done in lifting an object is stored as its potential energy.

Mass of object

$m = 1$ kg

Height upto which it is lifted

$h = 9.8$ m

Hence work done

$W= m g h = 1 \times 9.8 \times 9.8 = (9.8)^2 J$

Thus option C is correct.

A car is moving at

$100\ km/h$ . If the mass of the car is

$950\ kg$ , Its kinetic energy is:

0%
$367\ MJ$
0%
$0.367\ MJ$
0%
$3.67\ MJ$
0%
$3.67\ J$

Explanation

Mass of the car,

$m=950\ kg$

velocity of the car,

$v= 100\ km/h = { \left( 100\times \dfrac { 5 }{ 18 } \right) }$ m/s.

kinetic energy

$\displaystyle KE=\dfrac { 1 }{ 2 } \times m\times { v }^{ 2 }$

$\displaystyle =\frac { 1 }{ 2 } \times 950\times { \left( 100\times \frac { 5 }{ 18 } \right) }^{ 2 }J$

$\displaystyle =366512.35J=0.367\quad MJ$ .

Hence, the kinetic energy of the car is 0.367 MJ.

A girl weighing 50 kg makes a high jump of 1.2 m.What is her kinetic energy at the highest point?

${(g= 10 ms^{-2}}$ )

0%
6000 J
0%
600 J
0%
60 J
0%
Zero

Explanation

Given,

mass of girl

$M=50\ kg\\h=1.2\ m$

A girl is jumping vertically upward, when it will reach at maximum Hight its velocity will become zero

$v_f=0$

$K.E=\dfrac12 mv^2=\dfrac12m\times 0=0$

Option D

The kinetic energy of a body is increased by increasing its:

0%
Mass
0%
Weight
0%
Velocity
0%
Potential energy

Explanation

Kinetic Energy, $KE=\dfrac{1}{2}mv^{2}$

(where, $m=$ mass of the body and $v=$ velocity)

$KE\propto mv^{2}$

Kinetic energy can be increased by either increasing mass or velocity.

The moon revolves around the earth because the earth exerts a radial force on the moon. Does the earth perform work on the moon?

0%
no
0%
yes, sometimes
0%
yes, always
0%
cannot be decided

Explanation

No, the earth does not perform any work on the moon.

Work done(W) is defined as the scalar product of force(F) and displacement(s).

So, W = F

$\ast$ s = FsCos

$\theta$ where \theta is the angle between force and displacement vector.

The radial force exerted on the moon by earth i.e attractive force due to gravity acts in direction perpendicular to which the moon suffers the displacement during rotation.

So,

$\theta$ = 90 hence Cos

$\theta$ = 0

So, W = |F||s|

$\ast$ 0 = 0

Hence work done by earth is zero.

CBSE Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 4 - 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 4 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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