MCQ Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 8

A particle of mass m strikes a wall at an a wall at angle of incidence

$60^o$ with velocity v elastically. The change in momentum is:

0%
$mv$
0%
$m v/2$
0%
$-2 m v$
0%
$Zero$

A stone of mass

$1\ kg$ is tied to a string

$4\ m$ long and is rotated at constant speed of

$40\ ms^{-1}$ in a vertical circle. The ration of the tension at the top and the bottom is

0%
11:12
0%
39:41
0%
41:39
0%
12:11

A ball is dropped from certain height, after striking the gourn it rebounds till

$\frac{2}{5}th$ of the initial height. The ration of its speed just before and just after striking the ground is [assume no loss of energy due to air friction]

0%
$\sqrt{\dfrac{2}{5}}$
0%
$\dfrac{2}{5}$
0%
$\sqrt{\dfrac{5}{2}}$
0%
$\dfrac{5}{2}$

A force acts on a

$30 gm$ particle in such a way that the position of the particle as a function of time is given by

$x=3t-4{t}^{2}+{t}^{3}$ , where

$x$ is in metres and

$t$ is in seconds. The work done on the particle during the first

$4$ second is

0%
$3.84J$
0%
$1.68J$
0%
$5.28J$
0%
$5.41J$

Explanation

Given,

$m=30g=\dfrac{30}{1000}=0.03kg$

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

$dx=(3-8t+3t^2)dt$

Velocity,

$v=\dfrac{dx}{dt}$

$v=3-8t+3t^2$

Acceleration,

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

$a=-8+6t$

Work done on the particle during the first

$4$ second

$W=\int F.dx=m\int adx$

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

$W=0.03\int _0^4 (18t^3-72t^2+82t-24)dt$

$W=0.03[\dfrac{18}{4}t^4-\dfrac{72}{3}t^3+\dfrac{82}{2}t^2-24t]_0^4$

$W=0.03\times 176$

$W=5.28J$

The correct option is C.

A horizontal

$50\ N$ force acts on a

$2\ kg$ crate which is at rest on a smooth horizontal surface. At the instant the particle has gone

$2\ m$ , the rate at which the force is doing work is

0%
$2.5\ W$
0%
$25\ W$
0%
$100\ W$
0%
$500\ W$

Circular flexible current loop of radius R carrying current I is placed in an inward magnetic field B. If we spin the loop with angular speed

$/omega$ , then tension in string

0%
Is zero
0%
is more than iBR
0%
is less than iBR
0%
Does not depend on rotation

A body covers a distance of

$4m$ under the action of force

$F=\left( 17-4x \right) N$ where x is the metres. The work done by the force is

0%
$32J$
0%
$20J$
0%
$36J$
0%
None

Explanation

Given,

$F=(17-4x)N$

Work done by the force,

$dW=F.dx$

$W=\int_0^4 (17-4x)dx=17x|_0^4-\dfrac{4x^2}{2}|_0^4$

$W=17(4-0)-2(16-0)$

$W=68-32=36J$

The correct option is C.

A particle is projected on friction less inclined plane of inclination

$90^{\circ}$ from the horizontal, with the projection angle

$45^{\circ}$ from the inclined plane as shown in the figure. After one collision from the plane. It reaches to its initial point of projection. Coefficient of restitution between particle and plane is

0%
$\frac{2}{3}$
0%
$\frac{1}{3}$
0%
$\frac{3}{25}$
0%
$\frac{1}{\sqrt{2}}$

When two springs A and B with force constant

$K_A and K_B$ are stretched by same force,the respective work done on them is:

0%
$K_B : K_A$
0%
$K_A : K_B$
0%
$K_A K_B :1$
0%
$\surd K\_ 4:\surd K\_ B$

Explanation

It is given that two springs A and B are there.

Work done on first spring ${{W}_{1}}=\dfrac{1}{2}{{K}_{A}}x_{1}^{2}.........(1)$

Work done on second spring ${{W}_{2}}=\dfrac{1}{2}{{K}_{B}}x_{2}^{2}..........(1)$

Since, force is same $F={{K}_{A}}{{x}_{1}}={{K}_{B}}{{x}_{2}}.........(3)$

Taking ratios of equation (1) and (2)

$\dfrac{{{W}_{1}}}{{{W}_{2}}}=\dfrac{{{K}_{A}}x_{1}^{2}}{{{K}_{B}}x_{2}^{2}}=\dfrac{F{{x}_{1}}}{F{{x}_{2}}}=\dfrac{{{K}_{B}}}{{{K}_{A}}}\,\,\,\,(from\,(1))$

Two particles A and B equal masses are respectively tied to the centre and one end of a string, whose other end 0 is fixed. Both the particles always revolve in concentric circles of centre O. The ratio of tensions in both the parts of the string will be

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

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

$F=7-2x+3x^2$ .Work done in the process is

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

Explanation

$\displaystyle w = \int F dx$

$\displaystyle =\int_{0}^{5}7-2x+3x^2 \,dx$

$\displaystyle = [7x-\frac{2x^2}{2}+\frac{3x^2}{3}]_{0}^{5}$

$\displaystyle = [7x-x^2+x^3]_{0}^{5}$

$\displaystyle = [7\times5-5^2+5^3]-[0-0+0]$

$\displaystyle = [35-25+125]-0$

$= 135J$

In this question, answer is 135 J not 70J

1178849_1135506_ans_c61274bc329040fb86dafa6445d37bac.jpg

The disc is at rest at the top of a rough inclined plane. It rolls without slipping. At the bottom of inclined plane there is a vertical groove if radius

$R$ . In order to loop the groove, the minimum height of incline required is:

0%
$\dfrac{15 R}{4}$
0%
$\dfrac{9 R}{4}$
0%
$\dfrac{5 R}{2}$
0%
$\dfrac{7 R}{5}$

Explanation

Hence, option

$(A)$ is correct answer.
1380481_1132935_ans_19e2235363804e628b47ae162028617f.jpg

The force required to stretch a spring varies with the distance as shown in the figure. If the experiment is performed with the above spring of half the length, the line

$OA$ will:-

0%
shift towards $F-$ axis
0%
shift towards $X-$ axis
0%
remains as it is
0%
become double in length

Explanation

It is known that

$k \propto \dfrac {1}{x} \propto \dfrac {1}{L}$

$\therefore k \propto \dfrac {1}{L}$

When the length of the spring is halved, its spring constant will become double.

Slope of the force displacement graph gives the spring constant (k) of spring.

If k becomes double then slope of the graph increases, i.e. Graph shifts towards force-axis.

A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from set, the work done by the force during the first 1 sec. will be :

0%
22 J
0%
9 J
0%
18 J
0%
4.5 J

Explanation

Given,

$F=6t$

$m=1kg$

Force,

$F=ma=6t$

$1\times \dfrac{dv}{dt}=6t$

$dv=6tdt$

Integrating both side,

$v=\int dv=\int _0^1 6tdt$

$v=6[\dfrac{t^2}{2}]_0^1$

$v=3m/s$

Initial velocity,

$u=0m/s$

From work energy theorem,

$W=K_f-K_i$

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

$W=\dfrac{1}{2}\times 1\times 3\times 3-\dfrac{1}{2}\times 1\times 0\times 0$

$W=4.5-0=4.5J$

The correct option is D.

Identify the wrong statement

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A body can have momentum without energy
0%
A body can have energy without momentum
0%
the momentum is conserved in an elastic collision
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Kinetic energy is not conserved in an inelastic collision

If two balls each of mass 0.06 Kg moving in opposite directions with speed 4 m/sec collides and rebound with the same speed,then coefficient of restitution for the collision will be:-

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

Explanation

Coefficient of restitution, $e$

$e=-\dfrac{relative\,velocity\,of\,seperation}{relative\,velocity\,of\,appproach}$

$e=-\dfrac{4(-\hat{i})-4(+\hat{i})}{4\hat{i}-4(-\hat{i})}=1$

Hence, coefficient of restitution is $1$

1043430_1139635_ans_9f500b9946d5416592383b152c83f2b6.jpg

$10$ kkg object attached to a nylon cord outside a space vehicle is rotating at a speed of

$5 m/s$ . If the force acting on the cord is

$125 N$ its radius of path is

0%
2 m
0%
4 m
0%
6 m
0%
1 m

A gun fires a shell of mass

$1.5$ kg with velocity of

$150$ m/s and recoils with a velocity of

$2.5$ m/s.

Calculate the mass of the gun.

0%
$20$ kg
0%
$30$ kg
0%
$90$ kg
0%
$60$ kg

Explanation

Mass of shell

$=1.5$ kg

Velocity of shell

${V_1} = 159\,m/s$

Recoil velocity

$=2.5$ m/s

mass of gun

$=M$

According to the conservation of momentum,

$m{v_1} = m{v_2}$

$\begin{array}{l} 1.5\times 150=M\times 2.5 \\ M=\dfrac { { 1.5\times 159 } }{ { 2.5 } } \\ M=90\, kg \end{array}$

A spring of natural length

$L$ compressed to length

$7\ L/8$ exerts a force

$F_0$ . The work done by the spring in restoring itself to natural length is

0%
$F_0\ L/25$
0%
$F_0\ L/16$
0%
$3F_0\ L/25$
0%
$F_0\ L/8$

The

$P.E.$ and

$K.E.$ of a helicopter flying horizontally at a height

$400\ m$ are in the ratio

$5:2$ . The velocity of the helicopter is:

0%
$56\ m/s$
0%
$28\ m/s$
0%
$14\ m/s$
0%
$42\ m/s$

Explanation

Let mass and velocity of the helicopter be

$'m'$ and

$'v'$

So,

$KE=\dfrac{1}{2}mv^2$ and

$PE=mgh$ , where

$[h=400\ m]$

Given ratio,

$\dfrac{PE}{KE}=\dfrac{5}{2}$

$\Rightarrow \dfrac{mgh}{\dfrac{1}{2}mv^2}=\dfrac{5}{2}$

$\Rightarrow v^2=\dfrac{4}{5} gh$ .

$\Rightarrow v=\sqrt{\dfrac{4}{5}\times 10\times 400}$

$\Rightarrow v=40\sqrt{2}\simeq 56.7\ m/s$ .

At the instant t= 0 a force F=kt( k is a constant) acts on a small body of mass m resting on a smooth horizontal surface.The time,when body leaves the surface is:

0%
$mg\ k \ sin \alpha$
0%
$\dfrac{k }{mg \ sin \alpha}$
0%
$\dfrac{mg \ sin \alpha}{k }$
0%
$\dfrac{mg}{k \ sin \alpha}$

A body of mass tied at the end of a strong of length

$l$ is projected with velocity

$\sqrt{4lg}$ , at what height will it leave the circular path:

0%
$\dfrac{5}{3}l$
0%
$\dfrac{3}{5}l$
0%
$\dfrac{1}{3}l$
0%
$\dfrac{2}{3}l$

Explanation

REF.Image

In a vertical circle the tension at the string is given by :-

$T= \dfrac{mu^{2}}{l}-mg(2-3 cos \theta )$

Here

$u^{2}= 4gl$ , and particle stop circular motion

when

$T=0$

$\Rightarrow T=0\Rightarrow \dfrac{mu^{2}}{L}= mg(2-3 cos\theta )$

$\Rightarrow \dfrac{m.4gL}{L}= mg(2-3 cos \theta )\Rightarrow cos\theta = -2/3$

$\therefore$ Height,

$h=l\left | cos\theta \right |=\frac{2}{3}l$

Hence it attains maximum height,

$H= h+L= \dfrac{5}{3}L$ (Ans)

1102738_1174252_ans_2513c8f5ddca48d3b02195a944deccc4.JPG

A ball of mass m approaches a wall of mass M (>> m) with the speed 4 m/s along normal to the wall. The speed of wall is 1m/s towards the ball . The speed of the ball after an elastic collision with the wall is-

0%
5 m/s away from the wall
0%
3 m/s away from the wall
0%
9 m/s away from the wall
0%
6 m/s away from the wall

Explanation

$\textbf{Step 1: Initial & final situation [Ref. Fig.]}$

Right direction is positive and left direction is negative.

$\textbf{Step 2: Apply law of restitution}$

Final velocity of wall be same as initial as

$M>>> m$

Therefore

$V_{wall} = -1m/s$

$e = \dfrac{V(seperation)}{V(approach)}$ (Along line of collision)

For elastic collision

$e=1$

$\Rightarrow e = 1 = \left(\dfrac{V_{wall}-V_{ball} }{u_{ball} - u_{wall}} \right)$

$u_{wall} =- 1m/s$

$u_{ball} = +4 \, m/s$

$\Rightarrow 1 = \left(\dfrac{-1-V_{ball} }{4 + 1} \right) \Rightarrow V_{ball} = -6 \,m/s$

Negative sign means left direction (away from wall)

Hence option D is correct.

2114887_1172497_ans_331f36a2e6b34ec5a46ea4097acc1257.png

A ball falls from a height such that it strikes the floor of lift at

$10\ m/s$ , if lift is moving in the upward direction with a velocity

$1\ m/s$ , then velocity with the ball rebounds after elastic collision will be then

0%
$11\ m/s$
0%
$12\ m/s$
0%
$13\ m/s$
0%
$9\ m/s$

Explanation

Assume upwards is positive, the lift (object 1) is moving up with speed

$1 m/s$ .

$e = \dfrac {v_2 - v_1}{u_1 - u_2}$ and collision is elastic.

$u_1 = 1m/s$ ,

$u_2 = -10m/s$

$v_1 = 1m/s$

$1 = \dfrac {v_2 - 1}{1 - (-10)}$

$v_2 = 11\ m/s$

The length of simple pendulum is 1 m and mass of its bob is 50 g. The bob is given sufficient velocity so that the bob describe vertical circle whose radius equal to length of pendulum. The maximum difference in the kinetic energy of bob during one revolution is.

0%
0.98 J
0%
1.96 J
0%
4.9 J
0%
9.8 J

Explanation

A bob of mass

$m = 0.05\ Kg$ is performing vertical circle of radius

$(r) = 1m$ .

In the boundary case.

a minimum velocity of

$\mu = \sqrt{5gr}$ has to provided to it

so it completes a vertical circle.

So, kinetic energy

$(K_1)$ at the lowest point,

$K_1 = \dfrac{1}{2}m\mu^2 = \dfrac{5}{2}mgr$

and

$PE = 0$ , raking it as reference at highest point

$PE : \mu_2 =mgh = 2mgr$

$(as \, h = 2r)$

From conservation of energy.

$K_1 + \mu_1= K_2+\mu_2 \Rightarrow K_2 = K_1 - \mu_2 = \dfrac{1}{2}mgr$

Thus, loss in

$KE$ :

$\Delta K = K_1-K_2 = 2mgr = 2\times 0.05 \times 9.8 \times 1 = 0.98\ J$ (Ans)

1373012_1162993_ans_f8a6ff3e4dc94002a20245803142ea11.png

A particle is acted upon by a force F which varies with position x as shown in the figure. If the particle at

$x=0$ has the kinetic energy of

$25$ J, then the kinetic energy of the particle at

$x=16$ m is?

0%
$45$ J
0%
$30$ J
0%
$70$ J
0%
$135$ J

When a spring is stretched by a distance

$x,$ it exerts a force given by

$F$

$=$ (

$-5x - 16x^3$ )

$N,$ where

$x$ is in

$m.$ The work done, when the spring is stretched from

$0.1\ m$ to

$0.2\ m$ is :

0%
$6.9\times 10^{-2}\ J$
0%
$12.2 \times 10^{-2}\ J$
0%
$8.1 \times 10^{-2}\ J$
0%
$12.2 \times 10^{-1}\ J$

Explanation

Given,

$F=-5x-16x^3$

$-kx=-5x-16x^3$

$k=5+16x^2$

work done,

$=\dfrac{1}{2}(k_1x_1^2+k_2x_2^2)$

upon solving the above equation, we get,

work done

$=8.7 \times 10^{-2}J$

The work done by a force

$\vec{F}=\left ( -6x^3\hat{i} \right )\ N$ is displacing a particle from

$x=4\ m$ to

$x=-2\ m$ is

0%
$-240\ J$
0%
$360\ J$
0%
$120\ J$
0%
$420\ J$

Explanation

$F=-6{ x }^{ 3 }\hat { i }$

$x=4$ to

$x=-2$

$\int { dw } =\int { f.dx }$

As displacement is along

$\hat { i }$ so

$F=-6{ x }^{ 3 }$

$-\int { _{ 4 }^{ -2 } } 6{ x }^{ 3 }dx=\left[ \cfrac { 6{ x }^{ 4 } }{ 4 } \right] ^{ -2 }_{ 4 }$

$\left[ \cfrac { -6{ x }^{ 4 } }{ 4 } +\cfrac { 6{ x }^{ 4 } }{ 4 } \right] _{ 4 }^{ -2 }$

$\cfrac { -6\times { (4) }^{ 4 } }{ 4 } +\cfrac { 60{ (-2) }^{ 4 } }{ 4 }$

$=360J$

Velocity-time graph of a particle of mass

$2\,\,kg$ moving in a straight line as shown in figure. Work done by all the forces on the particle is :

0%
$400\,J$
0%
$- 400\,J$
0%
$- 200\,J$
0%
$200\,J$

Explanation

From the graph we have

at t =0

Velocity of particle

$(V_1) = 20 \, ms^{-1}$

$\Rightarrow KE_1 = \frac {1}{2}mv_1^2 = \frac {1}{2}\times 2 \times 20 \times 20$

$= 400\, J$

Now,

at t = 2 sec.

Velocity of particle

$(V_2)=0$

$\Rightarrow KE_2=0$

From, work - energy theorem

Work done by all forces (W) is equal to the change in kinetic enery

$(\Delta KE)$ of the body.

i.e.

$W = \Delta KE$

$\Rightarrow W = KE_2 - KE_1$

$=0-400$

$\Rightarrow W= - 400 \, J$

1172774_1206603_ans_0db7cb826a9d435fb54a43a8309f72e0.jpg

A body of mass

$6\,kg$ is under a force which causes displacement in it given by

$S = \dfrac{{{t^2}}}{4}$ metres where

$t$ is time. The work done by the force in

$2$ seconds is:-

0%
$12\,J$
0%
$9\,J$
0%
$6\,J$
0%
$3\,J$

Explanation

we have

displacement (

$s$ ) given as

$s=\cfrac{{t}^{2}}{4}$

so, velocity

$v=\cfrac{ds}{dt}$

i.e

$v=\cfrac{2t}{v}$

$v=\cfrac{t}{2}{ms}^{-1}$

Now at

$t=0$ . velocity

${v}_{1}=0$

and hence kinetic energy

${k}_{1}=0$

$t=2sec$

velocity

${v}_{2}=\cfrac{2}{2}=1{ms}^{-1}$

So, kinetic energy,

${k}_{2}=\cfrac{1}{2}m{v}_{2}^{2}=\cfrac{1}{2}\times 6\times {(1)}^{2}$

$=3J$

By work energy theorem

Work done by the force

$=$ change in kinetic energy

i.e.,

$w=\Delta K$

$w={K}_{2}-{K}_{1}=3-0$

Work done

$($ w

$)=3$ joule

A light particle moving horizontally with a speed of

$12\ m/s$ strikes a very heavy block moving in the same direction at

$10\ m/s$ . The collision is one-dimensional and elastic. If the speed of the block does not change after the collision, the particle will

0%
Move at $2\ m/s$ in its original direction
0%
Move at $8\ m/s$ in its original direction
0%
Move at $8\ m/s$ opposite to its original direction
0%
Move at $12\ m/s$ opposite to its original direction

Explanation

Conservation of momentum,

$m_1v_1+m_2v_2=m_1v_1'+m_2v_2'$

$\Rightarrow m_1(v_1-v_1')=m_2(v_2-v_2')$ .......(1)

Conservation of kinetic energy requires,

$\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)=\dfrac{1}{2}(m_1{{v_1'}^2}+m_2{{v_2'}^2})$

$\Rightarrow m_1(v_1^2-{{v_1'}^2})=m_2(v_2^2-{{v_2'}^2})$

$\Rightarrow m_1(v_1-v'_1)(v_1+v'_1)=m_2(v_2-v'_2)(v_2+v'_2)$ .........(2)

dividing (2) and (1)

$v_1+v'_1=v_2+v'_2$

$12+v'_1=10+10$

$v'_1=8m/s$

Find the reading of machine. Given each box of mass

$=15 kg$ ,mass of man

$1=30kg$ , mass of man

$2=40 kg$ and of each weighing machine

$=5 kg$

0%
$360/11 kg, 400/11 kg$
0%
$330/11 kg, 440/11 kg$
0%
$360/11 kg, 440/11 kg$
0%
$350/11 kg, 240/11 kg$

Explanation

$Given:$ Each box of mass

$=15kg$ ; mass of man

$1 = 30kg$ ; mass of man

$2 = 40kg$ ; each Weighing machine

$=5kg$

$Solution:$

$T-50 g=50a ....(i)$

$60g -T=60a ....(ii)$

$\Rightarrow 10g =110a$

$\Rightarrow a=\dfrac{g}{11}$

$For (i)$

$Reading =m g+m a$

$=30 g+\dfrac{30}{11}g$

$=\dfrac{360}{11} g$ .

$kg=\dfrac{360}{11}kg$

$For(ii)$

$\Rightarrow mg-ma$

$=\dfrac{400}{11} k g$ .

$So,the$

$correct$

$option:A$

2010756_1190764_ans_ae72def8bcae466888c20f32de2a5992.jpeg

When a massive body suffers an elastic collision with a stationary light body, then massive body approximately comes to rest and light body-

0%
acquires velocity greater than initial velocity of massive body
0%
sticks to the massive body and remains at rest.
0%
acquires half the initial velocity of the massive body
0%
remains at rest but does not stick to the massive body.

Five equal force of

$10$ N each applied at one point and all are lying in one plane. If the angles between them are equal, the resultant force will be

0%
zero
0%
$10$ N
0%
$20$ N
0%
$10\sqrt2$

Explanation

It is clear that all the vectors are inclined at equal plane w.r.t each other. Also since all the forces lie in the same plane the resultant will be a zero vector would cancel each other. Even in real life if we pull an object with the same force from different positions with equal angles between each force the object would not move.

If mathematically calculated then,

Angle between the forces

$Q=\dfrac { { 360 } }{ 5 } ={ 72 }^{ 0 }$

Resultant can be calculated

${ F }_{ R }=\dfrac { F\sin\dfrac { N\theta }{ 2 } }{ \sin\dfrac { \theta }{ 2 } }$

On Substituting we get

${ F }_{ R }=0$ .

A body is moving along y-axis and force acting on it is given by F = sin Ky, where K is a constant . The work done by the force from y = 0 to y = 1 is

0%
$\dfrac { 1 }{ K(1-sinK) }$
0%
$\dfrac { { 2sin }^{ 2 }\dfrac { K }{ 2 } }{ K }$
0%
$\dfrac { cosK-1 }{ K }$
0%
cos K - 1

A force

$F = -kx^3$ is acting on a block moving along x-axis. Here, k is a positive constant. Work done by this force is:

0%
positive in displacing from x = 3 to x = 1
0%
positive in displacing from x = -1 to x = 3
0%
negative in displacing from x = 3 to x = 1
0%
negative in displacing from x = -1 to x = 3

A body of mass 'M' collides against a wall with a velocity v and retraces its path with the same speed. the change in momentum is ............. (take initial direction of velocity as positive)

0%
Zero
0%
2Mv
0%
Mv
0%
-2 Mv

Explanation

Taking +x direction to be positive , and assuming ball was travelling in +x direction initially.

$P_i=Mv$

After collision ball will move in -x direction

$P_f=-Mv$

Change in momentum:

$\Delta P= P_i-P_f$

$\Delta P= Mv+Mv=2Mv$

A force

$F=Kx^{2}$ acts on a particle at angle of

$60^{o}$ with

$x-$ axis. The work done in displacing the particle from

$x_{1}$ to

$x_{2}$ will be:

0%
$\dfrac{kx^{2}}{2}$
0%
$\dfrac{k}{6}(x_{3}^{2}-x_{1}^{3})$
0%
$\dfrac{k}{2}(x_{2}^{3}-x_{1}^{3})$
0%
$\dfrac{k}{3}(x_{2}^{3}-x_{1}^{3})$

Gravel is dropped on a conveyor belt at the rate of

$2kg/s$ . The extra force required to keep the belt moving at

$3 ms^{-1}$ is

0%
$1 N$
0%
$3 N$
0%
$4N$
0%
$6 N$

A car is travelling at 5 m/s up a gradient of 1 inThe car weight 6 tonnes and the friction is 48 kg wt. The work done per second in maintaining the motion of car up the gradient is

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

0%
12400 W
0%
1362 W
0%
17400 W
0%
1520 W

A spring with spring constant

$k$ is extended from

$x=0$ to

$x={x}_{1}$ . The work done will be

0%
${kx}_{1}^{2}$
0%
$\dfrac{1}{2}{kx}_{1}^{2}$
0%
$2{kx}_{1}^{2}$
0%
$2{kx}_{1}$

Explanation

we know that

work done = change in Enegy

$W=\Delta E$

and energy of spring is defined as

$\dfrac{1}{2}kx^2$ where

$x$ is stretch in spring.

$\therefore W=\dfrac{1}{2}kx_1^2$

A force acts on a

$20g$ 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

$4sec$ is:

0%
$-1.6J$
0%
$-1600J$
0%
$2.6 J$
0%
$1600J$

Explanation

Mass

$\left(m\right) =20gm=0.020kg$

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

$\Rightarrow v=\dfrac{dx}{dt}=3-8t+3t^2m/s$

$\Rightarrow a=\dfrac{dv}{dt}=-8+6tm/s^2$

Force

$=ma=0.120t-0.160N$

Work done

$=dw=Fdx=F\times \left( \dfrac{dx}{dt}\right)\times dt$

$\Rightarrow dw=\left( 0.120+0.160\right) \times \left( 3-8t+3t^2\right)dt$

$=0.360t^3-1.440t^2+1.640t-0.450$

$\Rightarrow w=integral\;of\;dw\;from\;\;\;\;t=0\;to\;4\;sec$

$=\left[ 0.090t^4-0.490t^3+0.82t^2-0.450t\right],t=0\;to\;4\;s$

$=2.88J.$

Hence, the answer is

$2.88J.$

If a resultant force of

$30N$ acts on a body of mass

$6kg$ .What is the acceleration of the body?

0%
$20m/sec^2$
0%
$12m/sec^2$
0%
$5m/sec^2$
0%
$8m/sec^2$

A body of mass

$5 kg$ under a force which causes displacement in it given

$S=\dfrac{t^2}{4}$ meter where

$'t'$ is time. The work done by the force in

$4 seconds$ is:

0%
$12J$
0%
$20 J$
0%
$2.55 J$
0%
$10 J$

Explanation

After

$2s$

$\Rightarrow s=\dfrac{t^2}{4}$

$\Rightarrow s=\dfrac{2^2}{4}=1m$

$\Rightarrow v=\dfrac{ds}{dt}$

$\Rightarrow v=\dfrac{2t}{4}$

$v=\dfrac{t}{2}$

$\Rightarrow a=\dfrac{dv}{dt}$

$=\dfrac{1}{2}m/s^2$

$\Rightarrow F=ma$

$=5\times \dfrac{1}{2}$

$\Rightarrow F=2.5N$

$\Rightarrow w=F\times s$

$=2.5\times 1$

$=2.55$

Work done by the force in 4 seconds is

$2.55.$

A force of

$F = 2x\hat{i} + 2\hat{j} + 3z^{2}\hat{k} N$ is acting on a particle. Find the work done by this force in displacing the body from (1,2,3) into (3,6,1) m.

0%
$-10 J$
0%
$100$
0%
$10 J$
0%
$1 J$

When the mass of body is halved and velocity is doubled, then the kinetic energy of the body

0%
remains same
0%
is doubled
0%
is 4 times
0%
is $\frac{1}{4}$ th

Explanation

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

$New\ K.E = \dfrac{1}{2} (\dfrac{m}{2})\ (2v)^2$ =

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

$2 \times K.E$

Hence, kinetic energy of body is doubled.

A force

$F=Ay^{2}+By+C$ acts on a body in the

$y$ direction, The work done by this force during a displacement from

$y=-a$ to

$y=a$ is

0%
$\dfrac{2Aa^{2}}{3}$
0%
$\dfrac{2Aa^{2}}{3}+2Ca$
0%
$\dfrac{2Aa^{2}}{3}+\dfrac{Ba^{2}}{2}+Ca$
0%
$None\ of\ these$

A body of mass

$15\ kg$ moving with a velocity of

$10\ ms^{-1}$ is bought to rest. The work done by the brake is

0%
$-250\ J$
0%
$-500\ J$
0%
$-750\ J$
0%
$-1000\ J$

Explanation

$W = change\ in\ kinetic\ energy =\dfrac{1}{2}m(v^2-u^2)$

$W = \dfrac{1}{2} \times 15 \times (0^2-10^2)$

$W = - 750J$

Hence, option

$C$ is correct answer.

A ball of mass

$5$ kg experiences a force

$F=2x^{2}+x$ . Work done in displacing the ball by

$2$ m is

0%
$22/3$ J
0%
$44/3$ J
0%
$32/3$ J
0%
$16/3$ J

A body of mass 0.5 kg travels in a straight line with velocity

$v=ax^{3/2}$ where

$a=5m^{-1/2}s^{-1}$ . the work done by the net force during its displacement from

$x=0$ to

$x=2m$ is

0%
$1.5J$
0%
$50J$
0%
$10J$
0%
$100J$

CBSE Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 8 - 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 8 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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