MCQ Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 9

A bod of mass M is suspended by a massless string of length L.The horizontal velocity V at position A is just sufficient to make real the point B.The angle

$\theta$ at which the speed of the bob is half of that it A satisfies:

0%
$\theta=\dfrac{\pi}{4}$
0%
$\dfrac{\pi}{4} < \theta<\frac{\pi}{2}$
0%
$\dfrac{\pi}{2} < \theta<\dfrac{3\pi}{4}$
0%
$\dfrac{3\pi}{4} < \theta<\pi$

Explanation

Use conservation of energy,

$\cfrac { 1 }{ 2 } m{ v }_{ o }^{ 2 }=\cfrac { 1 }{ 2 } m{ v }^{ 2 }+mgl \left( 1-\cos { \theta } \right) -------(i)$

where

$v_{o}=$ horizontal velocity

$mg(2l)=\cfrac { 1 }{ 2 } m{ v }_{ o }^{ 2 }-\cfrac { 1 }{ 2 } m{ v }_{ top }^{ 2 }-------(ii)$

Since

$v_{o}$ is just sufficient

$\cfrac { m{ v }_{ top }^{ 2 } }{ l } =T+mg\\ T=0,{ v }_{ top }=\sqrt { gl }$

Then equation

$(ii)$ ,

$\Longrightarrow { v }_{ o }=\sqrt { 5gl }$

According to the question,

$v=\cfrac { { v }_{ o } }{ 2 }$

From equation

$(i),$

$\cfrac { 1 }{ 2 } m5gl=\cfrac { 1 }{ 2 } m\left( \cfrac { 5gl }{ 4 } \right) +mgl\left( 1-\cos { \theta } \right) \\ \Longrightarrow \cfrac { 20mgl-5mgl }{ 8 } =mgl\left( 1-\cos { \theta } \right) \\ \Longrightarrow \cos { \theta } =\cfrac { 7 }{ 8 }$

Hence,

$\cfrac{3 \pi}{4} < \theta < \pi$ .

A body of mass 1 kg begins to move under the action of a time dependent force

$\overrightarrow { F } =(2t\hat { i } +3t^{ 2 }\hat { j } )$ N, where

$\hat { i }$ and

$\hat { j }$ are unit vectors along x and y axes. What power will be developed by the force at the time t?

0%
$(2t^2+3t^2)W$
0%
$(2t^4+4t^4)W$
0%
$(2t^3+3t^4)W$
0%
$(2t^3+3t^5)W$

Explanation

As given

$\overrightarrow{F}=(2t\hat{i}+3t^2\hat{j})N$

Where

$\hat{i}$ and

$\hat{j}$ are unit vectors along x and y axis

$F=ma$

$\Rightarrow a=\cfrac{F}{m}=\cfrac{(2t\hat{i}+3t^2\hat{j})}{2}=[m=1kg]\\ \Rightarrow a=(2t\hat{i}+3t^2\hat{j})m/s^2$

Acceleration

$a=\cfrac{dv}{dt}$

$\Rightarrow dv=a.dt\rightarrow(i)\\ \int{dv}=\int{a.dt}=\int{(2t\hat{i}+3t^2\hat{j})}dt\\v=t^2\hat{i}+t^3\hat{j}\\P=F.v=(2t\hat{i}+3t^2\hat{j})(t^2\hat{i}+t^3\hat{j})\\ \quad=2t.t^+3t^2t^3\\P=(2t^3+3t^5)W$

Two wires of the same diameter of the same material having the length

$\ell$ and

$2\ell$ . If the force is applied on each, the ratio of the work done in the two will be

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

Explanation

$\begin{array}{l} W=\dfrac { 1 }{ 2 } Fl\, \, \, \, \, \, \, \, \, \, \, \left( { W\, \alpha \, l } \right) \\ \dfrac { { { W_{ 1 } } } }{ { { W_{ 2 } } } } =\dfrac { { { l_{ 1 } } } }{ { { l_{ 2 } } } } =\dfrac { l }{ { 2l } } =\dfrac { 1 }{ 2 } =1:2 \end{array}$

Hence,

option

$(A)$ is correct answer.

By applying a force

$F=(3xy-5z) j +4zk$ a particle is moved along the path

$y=x^2$ from point

$(0,0,0)$ to the point

$(2,4,0)$ . The work done by the F on the particle is (all values are in SI units)

0%
$\dfrac { 280 }{ 5 } J$
0%
$\dfrac { 140 }{ 5 } J$
0%
$\dfrac { 232 }{ 5 } J$
0%
$\dfrac { 192 }{ 5 } J$

Explanation

$\bar{F} = (3xy-5z)\hat{j}+4z\hat{k}$

$x^{2} = y$

$x = y ^{1/2}$

Substitute in above equation

$\bar{F} = (3y^{3/2}-5z)\hat{j}-4z\hat{k}$

dw = F.dx

$(\hat{j})$ direction

$W = _{0}^{4}\int 3y^{3/2}dy$

$\rightarrow \frac{3y^{5/2}}{5/2}|_{0}^{4}$

$W = \frac{192}{5}$ joule

(in 'z' work done is always zero)

1222129_1270771_ans_4f9feac6281b4ada9bda63b6d15973d2.jpg

Two masses

$10 \ gm$ and

$40 \ gm$ are moving with kinetic energies in the ratio

$9:25$ . The ratio of their linear momentum is:

0%
$5:6$
0%
$3:10$
0%
$6:5$
0%
$10:3$

A force is acting on a particle varies with the displacement x as

$F=ax-bx^2$ . Where a=1 N/m and

$b=1 N/m^2$ . The work done by this force for the first one meter (F is in newtons, x is in meters) is :

0%
$\dfrac{1}{6}J$
0%
$\dfrac{2}{6}J$
0%
$\dfrac{3}{6}J$
0%
None of these

The work done in placing a charge of 8

$\times$

${ 10 }^{ -18 }$ coulomb on a condenser of capacity 100 micro-farad is

0%
16 $\times { 10 }^{ -32 }$ joule
0%
3.2 $\times { 10 }^{ -26 }$ joule
0%
4 $\times { 10 }^{ -10 }$ joule
0%
32 $\times { 10 }^{ -32 }$ joule

A particle of mass 1 kg is projected under gravity at an angle

${ 37 }^{ o }$ with horizontal with 50 m/s. Instantaneous power delivered by gravity at t=1s is

$\left( sin{ 37 }^{ o }=\frac { 3 }{ 5 } ,cos{ 37 }^{ o }=\frac { 4 }{ 5 } \right)$

0%
200 J/s
0%
-200 J/s
0%
Zero
0%
-400 J/s

Mass of

$B$ is four times that of

$A . B$ moves with a velocity half that of

$A$ . Then,

$B$ has

0%
kinetic energy equal to that of $A$
0%
half the kinetic energy of $A$
0%
twice the kinetic energy of $A$
0%
kinetic energy one-fourth of $A$

Explanation

According to question given,

$M_B = 4 \times M_A$ and

$V_B = \dfrac{V_A}{2}$

$K.E_A = \dfrac{1}{2}M_AV_A^2$ .........(1)

$K.E_B = \dfrac{1}{2}M_BV_B^2$ =

$\dfrac{1}{2}\ (4 \times M_A) \left(\dfrac{V_A}{2}\right)^2=\dfrac{1}{2}M_AV_A^2$ ........(2)

From above equation

$K.E_A=K.E_B$

A particle is moving in a vertical circle. The tension in the string when passing through two positions at angles

${30}^{o}$ and

${60}^{o}$ from vertical (lowest position) are

${T}_{1}$ and

${T}_{2}$ respectively, then

0%
${T}_{1}={T}_{2}$
0%
${T}_{2}> {T}_{1}$
0%
${T}_{1}> {T}_{2}$
0%
tension in the string always remains the same

Explanation

From the Free Body Diagram we get,

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

$\theta= 30 ^0 \quad T = T_1= \dfrac{mv^2}{r}+ \dfrac{\sqrt{3}mg}{2}$

$\theta= 60 ^0 \quad T= T_2= \dfrac{mv^2}{r}+ \dfrac{mg}{2}$

From Above we get

$T_1 > T_2$

1263782_1287520_ans_6a9341e5be2d4687a8edae8f90075cc6.png

A body of mass

$2\ kg$ moving under a force has relation between displacement

$x$ and time

$t$ as

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

$x$ is in metre and

$t$ is in sec. The work done by the body in first two second will be

0%
$1.6\ joule$
0%
$16\ joule$
0%
$160\ joule$
0%
$1600\ joule$

Explanation

work done = change in kinetic energy

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

velocity

$v=\dfrac{dx}{dt}=\dfrac{d\left ( \dfrac{t^3}{3} \right )}{dt}=t^2$

given,

$t=2s$

Velocity

$=t^2=2^2=4$

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

$=\dfrac{1}{2} \times 2 \times 4^2$

$=16J$

Work done by a force

$\vec{F}=(\hat{i}+2\hat{j}+3\hat{k})\ N$ action on a particle in displacing it from the point

$\vec{r}_{1}=\hat{i}+\hat{j}+\hat{k}$ to the point

$\vec{r}_{1}=\hat{i}-\hat{j}-2\hat{k}$

0%
$-1\ J$
0%
$2\ J$
0%
$-13\ J$
0%
$zero$

Explanation

work = force * displacement,

$W=F\times s$

$s=(i-j-2k)-(i+j+k)=-2j-3k$

$F=i+2j+3k$

$W=(i+2j+3k)\times (-2j-3k)$

$\therefore W=-13J$

A particle moves along positive X-axis from x=

$0$ to x=

$5$ under influence of force is given by

$F={ x }^{ 2 }-2x-8.$ The work done by the force is

0%
-69 units
0%
-23.33 units
0%
-5 units
0%
zero

The displacement (x) and time (t) for a particle are related as

$t=\sqrt { x } +3$ . What is the work done in the first six second of its motion?

0%
3 J
0%
4 J
0%
Zero
0%
5 J

Explanation

Given,

$t=\sqrt x +3$

Then,

$\sqrt x=t-3 \implies x=t^2+9-6t$

Also,

$v=\dfrac{dx}{dt}= 2t-6$

$t=0, v=-6\,ms^{-1}$

$t=6\,sec, v=6\, ms^{-1}$

Work done = Change in K.E

$=K.E_f-K.E_i$

$=\dfrac 12 m(6)^2-\dfrac 12 m(6)^2=0$

Five identical balls each of mass m and radius r are strung like beads at random and at rest along a smooth, rigid horizontal thin rod of length L, mounted between immovable supports. Assume

$10r<L$ and that the collision between balls or between balls and supports are elastic. If one ball is struck horizontally so as to acquire a speed v, the average force felt by the support is

0%
$\dfrac { 5m }{ L-5r }$
0%
$\dfrac { m{ v }^{ 2 } }{ L-10r }$
0%
$\dfrac { 5m{ v }^{ 2 } }{ L-10r }$
0%
$\dfrac { m{ v }^{ 2 } }{ L-5r }$

When a 4 kg mass is hung vertically on a light spring obey's Hooks law , the spring stretches by 2 cms. The work required to be done by an external agent in stretching this spring by 5 cm will be ( g =9.8

${ ms }^{ -2 }$ )

0%
4.9 J
0%
2.4 J
0%
0.495 J
0%
0.24 J

A position dependent force

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

$2 kg$ and displaces it from

$x = 0$ to

$x = 5$ . The work done in joule is :

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

Two bodies of masses

$4kg$ and

$16kg$ are at rest. The ratio of times for which the same force must act on them to produce the same kinetic energy in both of them is

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

Explanation

kinetic energy is given by

$:$

so taking velocity for mass

$1kg$ as

$v_1,$ we get

and taking velocity for mass

$4kg$ as

$v1,$ we have

given that

$e_1 = e_2,$

equating the above terms and simplifying, we get

$,$

now , momentum is given by

$,$

let us assume

$, p_1$ as momentum for mass

$1kg$ and

$p_2$ for mass

$4kg$

$,$

$p_1 = 1 × v_1$

and

$p_2 = 4×v_2$

then

$,$

$\dfrac{{{p_1}}}{{{p_2}}} = \dfrac{1}{4} = 1:4$

Hence,

option

$(A)$ is correct answer.

$\overrightarrow { F } =x\hat i+y\hat j$ , is this force:

0%
a conservation force
0%
a non conservation
0%
depends upon X
0%
depends upon $y \quad \& \quad x$

A force of

$1200 \mathrm { N }$ acting on a stone by means of a rope slides the stone through a distance{ of

$10 \mathrm { m }$ in a direction inclined at

$60 ^ { \circ }$ force. The work done by the force is

0%
$6000 \sqrt { 3 } J$
0%
6000 J
0%
12000 J
0%
8000 J

Explanation

Hence, the option

$B$ is the correct answer.
1380140_1320760_ans_c629284accb34c459d054f61bd8c4a86.png

$\begin{array} { l } { \text { A man weighing } 80 \mathrm { kg } \text { climbs a staircase } } \\ { \text { carrying a } 20 \mathrm { kg } \text { load. The staircase has } 40 } \\ { \text { steps, each of } 25 \mathrm { cm } \text { height. If he takes } 20 } \\ { \text { seconds to climb, the work done is } } \end{array}$

0%
$9800 \mathrm { J }$
0%
$245 \mathrm { J }$
0%
$98 \times 10 ^ { 4 } J$
0%
7840 J

Explanation

As the man climb up the stair he does the work against the gravity

$,$ the total gravitational force act on it in downward direction

$=(80+20)g$

here

$g$ is the acceleration due to gravity

$,$

height raised by the man

$25cm=0.25m$

work done by the man

$\left( {80 + 20} \right) \times g \times 0.25 = 245J$

Hence,

option

$(B)$ is correct answer.

The tension in the string revolving in a vertical circle with a mass m at the end which is at the lowest position

0%
$\dfrac{mv^2}{r}$
0%
$\dfrac{mv^2}{r}-mg$
0%
$\dfrac{mv^2}{r}+mg$
0%
$mg$

Explanation

R.E.F image

$F_{c}$ Fore of centrifygal

$T=F_{c}+mg$

$=\frac{mv^{2}}{12}+mg$

so, option (c)

1167816_1339259_ans_7999f5f5e9ea4ca8bf04bb632276d1ac.png

The work done by a force $F=(-6x^3\hat i)$ in displacing a particle from $x = 4m$ to $x = -2m$ is:

0%
$- 240$ J
0%
$360$ J
0%
$420$ J
0%
will depend upon the path

A variable force given by

$F= ( 3x^2 \hat i + 4 \hat j)$ acts on a particle. the force is in newton and x is in meter. what is the change in KE of particle as it moves from point (2, 3) to (3, 0)

0%
-7J
0%
zero
0%
7J
0%
19J

Work done by air when it expands from 50 litres to 150 litres at a constant pressure of 2 atmospheres is:

0%
$2\times { 10 }^{ 4 }J$
0%
$2\times { 10 }0J$
0%
$2\times { 10 }^{ 5 }\times 100J$
0%
$2\times { 10 }.5\times 100J$

does not vary from point to point in space Which pair of the following forces will never give resultant force of 2 N?

0%
1 N and 3 N
0%
2 N and 2 N
0%
1 N and 1 N
0%
1 N and 4 N

Explanation

$1N$ and

$4N$ will never give

$2N$

mind it

$,$ if the side lengths of a triangle is

$a,b,c$

$a+b>=c$

$b+c>=a$

$c+a>=b$

Hence,

option

$(D)$ is correct answer.

A force

$F_1$ on a body of 2 kg produces an acceleration of

$2.5 m/m^2$ and other force

$F_2$ acting on another body of mass 5 kg produces an acceleration of

$2 m/s^2$ . The ratio of

$F_2/F_1$ is

0%
$2$
0%
$4$
0%
$6$
0%
$8$

Explanation

Given,

$m_1=2kg$

$a_1=2.5m/s^2$

$m_2=5kg$

$a_2=2m/s^2$

Ratio,

$\dfrac{F_2}{F_1}=\dfrac{m_2a_2}{m_1a_1}$

$\dfrac{F_2}{F_1}=\dfrac{5\times 2}{2\times 2.5}=2$

The correct option is A.

A hollow smooth uniform sphere

$A$ of mass

$m$ rolls without sliding on a smooth horizontal surface. It collides elastically and head-on with another stationary smooth hollow sphere

$B$ of the same mass mm and same radius. The ratio of the kinetic energy of

$B$ to that of

$A$ just after the collision is

0%
$1 : 1$
0%
$2 : 3$
0%
$3 : 2$
0%
$\rm{None}$

The speed of a motor increases from 1200 rpm to 1800 rpm in 20 s. How many revolutions does it make during these second?

0%
400
0%
600
0%
500
0%
700

If body of mass 2 kg is moving along X axis under an external force, whose displacement is given by

$x(t) = 3t^2 - 2t + 5 m$ . Find the work done by force from t = 0 to t = 5 sec.

0%
$78\quad J$
0%
$780\quad J$
0%
$840\quad J$
0%
$200\quad J$

AB is long frictionless horizontal surface. One end of an ideal spring of spring constant K is attached to a block of mass m, which is being moved left moved left with constant velocity v, and another end is free.Another block of mass 2m is given a velocity 3v towards the spring. Work done by an external agent in moving m with constant velocity v in a long time will be:

0%
$- 5m{v^2}$
0%
$- 8m{v^2}$
0%
$- 3m{v^2}$
0%
None of these

A force

$\vec { F } = 2 x \hat { i } + 3 y ^ { 2 } \hat { j } \mathrm { N }$ where x and y are in meter, acts upon a particle. Find work done (in J) by this force in displacing the particle from (1 m, 2 m, 5 m) to (2 m, 3 m, 2 m).

0%
24
0%
16
0%
28
0%
22

When a simple pendulum is rotated in a vertical plane with constant angular velocity, centripetal force is:

0%
maximum at highest point
0%
maximum at lowest point
0%
same at all points
0%
zero

In the figure, at the free end of the light string, a force

$F$ is applied to keep the suspended mass of

$18$ \mathrm { kg }$$ at rest. Assuming

0%
$240 N$
0%
$120 M$
0%
$180 M$
0%
$240 M$

Explanation

$3T=180$

or,

$T=60 N$

Now,

$F=T=60 N$

Therefore ,total force on ceiling is

$4T=240 N$

Hence, Option

$A$ is correct.

A block of mass 10 kg is moving up on an inclined plane with constant acceleration 2 m/sec by applying a constant force F as shown in the fig. The work done on the block by applied force F in 2 s is

0%
480
0%
240
0%
960
0%
none

A force

$\vec { F } =\left( { 3x }^{ 2 }+2x-7 \right)$ N acts on a 2 kg body as a result of which the body gest displaced x=0 to x=5 cm. The work done by the force will be-

0%
35 Joule
0%
70 Joule
0%
115 Joule
0%
270 Joule

A particle moves along the x-axis from x=0 to x=5 m under the influence of a force given by

$F=\left( { 7-2x+3x }^{ 2 } \right) N.$ The work done in the process is

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

The amount of work done in stretching a wire by 10N force can cause an elongation of

$2×{ 10 }^{ -2 }m$ will be

0%
10 J
0%
1 J
0%
0.1 J
0%
0.01

Velocity-time graph of a particle of mass 2 kg moving in a straight line is as shown in figure. Work done by all the forces on the particle during the motion is:

0%
$400 J$
0%
$-400 J$
0%
$-200 J$
0%
$200 J$

A particle is displaced from a position

$(2\hat { i } -\hat { j } +\hat { k } )$ metre to another position

$(3\hat { i } +2\hat { j } -2\hat { k } )$ meter under the action of force

$(2\hat { i } +\hat { j } -\hat { k } )$ N. Work done by the force is:

0%
8 J
0%
10 J
0%
12 J
0%
36 J

A cone filled with water is whirled in a vertical circle of radius 4 m. What will be the time period of rotation so that water does not fall ?

0%
4 s
0%
6 s
0%
2 s
0%
1 s

A balloon with its contents weighing 160 N is moving down with an acceleration of g/2

${ ms }^{ -2 }$ the mass to be removed from it so that the balloon moves up with an acceleration of g/3

${ ms }^{ -2 }$ is

0%
5 kg
0%
10 kg
0%
6 kg
0%
3 kg

$10 s$ , a body of

$6 kg$ mass is dragged

$8 m$ with uniform velocity across a floor by a steady force of

$20 N$ . The K.E of the body is:

0%
$2.92 J$
0%
$0.92 J$
0%
$1.92 J$
0%
$0J$

The breaking strength of a string is 55 kg wt. The maximum permissible speed of a stone of mass 5 kg which revolved in a vertical circle of radius 4 m with the help of is the string is g=10

${s}^{2}$ .

0%
$10 \mathrm{m} / \mathrm{sec}$
0%
$15 \mathrm{m} / \mathrm{sec}$
0%
$20 \mathrm{m} / \mathrm{sec}$
0%
$25 \mathrm{m} / \mathrm{sec}$

A force of

$\left(4 x^{2}+3 x\right)$ N acts on a particle which displaces it from

$x=2 m$ to

$x=3 \mathrm{m}$ .
The work done by the force is

0%
$32.8 J$
0%
$3.28 J$
0%
$0.328 J$
0%
zero

A body of mass m is rotating in a vertical circle of radius 'r' with critical speed. The difference in its K.E. at the top and the bottom is _____.

0%
2mgr
0%
4 mgr
0%
6 mgr
0%
3 mgr

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

The work done is stored as elastic potential energy in the spring.

It is given by

$W = \dfrac12 \times k ( x_2^2 - x_1^2)$

$x_1 = 5\, cm$

$x_2 = 15\, cm$

Force constant,

$k = 800 \,N/m = \dfrac{800}{100} N/cm = 8 N/cm$

$W = \dfrac 12 \times 8 ( 15^2- 5^2) = \dfrac 12 \times 8 ( 225 - 25)$

$= \dfrac 12 \times 8 \times 200 = 800 \,N cm$

$W = 800\, N cm = 8 \,N m = 8\, J$

Therefore,

The work done in extending it from

$5\,cm$ to

$15\,cm$ is

$8\, J$ .

Two springs of force constants

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

$\mathrm{K}_{2}$

$\left(\mathrm{K}_{1}>\mathrm{K}_{2}\right)$ are stretched by same force. If

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

$W_{2}$ be the work done stretching the springs then

0%
$W_{1}=W_{2}$
0%
$W _1<W_2$
0%
$\mathrm{W}_{1}>\mathrm{W}_{2}$
0%
$W_{1}=W_{2}=0$

Explanation

As the force on both the springs is same, we have

$F=k_1x_1=k_2x_2$ .....(1)

Now, work done on first spring,

$W_1=\dfrac 12k_1x_1^2$ .......(2)

Work done on second spring,

$W_2=\dfrac 12k_2x_2^2$ ...(3)

On dividing equation (2) by (3), we get

$\dfrac{W_1}{W_2}=\dfrac{\dfrac 12 k_1x_1^2}{\dfrac 12 k_2x_2^2}$

Hence using equation (1), we get

$\dfrac{W_1}{W_2}=\dfrac{F_{x1}}{F_{x_2}}=\dfrac{x_1}{x_2}=\dfrac{k_2}{k_1}$

$k_1>k_2$ , therefore, we get,

$W_1<W_2$

Hence, more work is done on spring with spring constant

$k_2$ .

A vertical spring of force constant

$100 N/m$ is attached with a hanging mass of

$10 kg$ . Now an external force is applied on the same so that the spring is stretched by additional

$2m$ . The work done by the force is (

$g= 10 m/s^2$ )

0%
$200 J$
0%
$400 J$
0%
$450 J$
0%
$600 J$

Explanation

Given,

Spring Constant,

$k=100\,N/m$

Additional elongation,

$x=2\,m$

Work done,

$W=\dfrac 12kx^2$

$=\dfrac 12\times 100\times 2^2=200\,J$

At the instant speed of block is maximum, the magnitude of force exerted by spring on the block is

0%
$\dfrac{mg}{2}$
0%
$mg$
0%
Zero
0%
None of these

CBSE Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 9 - 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 9 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

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