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

$4\ kg$ particle moves along the

$X-$ axis. Its position

$x$ varies with time according to

$x\left( t \right) = t + 2{t^3}$ , where

$X$ is in

$m$ and

$t$ is in seconds. Compute kinetic energy of the particle in time t

0%
$KE=72t^2+24t+2$
0%
0
0%
20t
0%
40t

Explanation

Kinetic energy is given by,

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

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

$=\dfrac{d}{dt}(t+2t^3)$

$v=1+6t^2$

$\Rightarrow v^2=(1+6t^2)^2=1+36t^4+12t^2$

$KE=\dfrac{1}{2}\times 4\times (1+36t^4+12t^2)$

$=2(1+36t^4+12t^2)$

$\therefore KE=72t^2+24t+2$

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 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 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%
$500\ W$

When water falls from a height, the energy changes from,

0%
kinetic to potential
0%
potential to kinetic
0%
chemical to mechanical
0%
Mechanical to chemical

A particle of mass

$4m$ which is at rest explodes into masses

$m,m$ &

$2m$ . Two of the fragments of masses

$m$ and

$2m$ are found to move with equal speeds

$v$ each in opposite directions. The total mechanical energy released in the process of explosion is

0%
$m{v}^{2}$
0%
$2m{v}^{2}$
0%
$\dfrac{1}{2}m{v}^{2}$
0%
$4m{v}^{2}$

Explanation

Assume right direction

$=$ positive

using momentum conservation

$p_i=p_f$

$o=2m(-v)+mv_1+mv$

$v_1=v$

K.E final

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

$=\dfrac{1}{2}mv^2(4)$

$=2mv^2$ .

1459351_1113027_ans_cfbc23fc0b4244ca8fbde3563c7456bf.png

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%
$\frac{1}{\sqrt{2}}$

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

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

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,

$m{v_1} = m{v_2}$

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

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 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

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

Two identical spheres move in opposite direction with speed

$v_1$ and

$v_2$ and pass behind an opaque screen, where they may either cross without touching ( Event 1) or make an elastic head-on collision ( Event 2)

0%
We can never make out which events has occured
0%
We cannot make out which event has occured only if $v_1= v_2$
0%
We can always make out which event has occured
0%
We can make out which event has occured only if $v_1= v_2$

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

$P.E.$ 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,

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

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

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

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

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

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

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)

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

$12+v'_1=10+10$

$v'_1=8m/s$

Potential energy of a particle is given as

$U(x)=2{x}^{3}-9{x}^{2}+12x$ where

$U$ is in joule and

$x$ is in metre. If the motion of a particle is S.H.M then find the approx. potential energy of the particle is:

0%
$-36J$
0%
$4J$
0%
$5J$
0%
None of these

Explanation

$\begin{array}{l} u=2{ x^{ 2 } }-9{ x^{ 2 } }+12x \\ F=6{ x^{ 2 } }+18x+12 \\ =-6\left[ { { x^{ 2 } }-3x+2 } \right] \\ =-6\left[ { \left( { x-2 } \right) \left( { x-1 } \right) } \right] \\ Hence, \\ option\, \, D\, \, is\, correct\, \, answer. \end{array}$

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.

When a conservation force does positive work on a body then

0%
its potential energy must increase
0%
its potential energy must decrease
0%
its kinetic energy must increase
0%
its total energy must decrease

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 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 2.0 kg makes an elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its original speed v. What is the mass of other body and the speed of the center of mass of two bodies ?

0%
$1.0 kg and \frac {2}{3}v$
0%
$1.2 kg and \frac{5}{8}$
0%
$1.4 kg and \frac {10}{17} v$
0%
$1.5 kg and \frac {4}{7} v$

Explanation

$m_{1}=2\ kg$

$u_{1}=4\ m/s$

$v_{1}=\dfrac {u_{1}}{4}=\dfrac {4}{4}=1\ m/s$

$u_{2}=0$

since nothing is given so we would take it as elastic collision.

hence conservation of momentum is given by:

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

$2\times 4+0=2\times 1+m_{2}v_{2}$

$m_{2}v_{2}=6$

according to the conservation kinetic energy:

$\dfrac {1}{2}m_{1}u_{1}^{2}+\dfrac {1}{2}m_{2}u_{2}^{2}=\dfrac {1}{2}m_{1}v_{1}^{2}+\dfrac {1}{2}m_{2}v_{2}^{2}$

$2\times 16+0=2\times 1+m_{2}v_{2}^{2}$

$m_{2}v_{2}^{2}=30$

$\dfrac {m_{2}v_{2}^{2}}{m_{2}v_{2}}=\dfrac {30}{6}$

$v_{2}=5\ m/s$

$m_{2}=1.2\ kg$

Work done in lifting a body is calculated by

0%
Mass of the body $\times$ vertical distance moved
0%
Force acting on a body $\times$ vertical distance moved
0%
Weight acting on the body $\times$ vertical distance moved
0%
None of the above.

When linear momentum of a body becomes

$n$ times, for the same applied stopping force, its stopping distance becomes

0%
$n$ times
0%
$\dfrac{1}{n}$ times
0%
$n^{2}$ times
0%
$\dfrac{1}{n^{2}}$ times

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

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 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

$2.55.$

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 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%
$100J$

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 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%
$1 J$

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%
$10:3$

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

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Practice Class 11 Medical Physics Quiz Questions and Answers

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

0:0:2

Practice Class 11 Medical Physics Quiz Questions and Answers

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