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

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%
$C{x}_{1}^{2}$
0%
$\cfrac{1}{2} C{x}_{1}^{2}$
0%
$C{x}_{1}$
0%
Zero

Explanation

Here F = Cx

Work done = F.d Where d is displacement

$w = \int^{x_2}_{x_1} F.dx$

$w = \int^{x_1}_{0} Fdx$

$w = \int^{x_1}_{0} cx \space dx$

$w = \dfrac{1}{2} cx^2_1$

Here (B) is correct answer

An engine draws water from a depth of

$10\ m$ with constant speed

$2\ m/s$ at the rate of

$10\ Kg\ per\ 10\ second$ The power of the engine is (in

$watt$ ): (Take:

$g=9.8\ m/{s}^{2}$ )

0%
$102$
0%
$98$
0%
$100$
0%
$200$

Explanation

$p=\dfrac{mgh}{t}=\dfrac{10\times 9.8\times 10}{10}=98W$

A force

$F=(3{x}^{2}+2x-7)N$ acts on a

$2kg$ body as a result of which the body gets displaced from

$x=0$ to

$x=5m$ . The work done by the force will be:

0%
$5J$
0%
$70J$
0%
$115J$
0%
$270J$

Explanation

Given,

$=3x^2+2x-7, mass=2kg$

So the work-done

$W=\int_{0}^{5}(3x^2=2x-7) dx=[x^3+x^2-7x]_{0}^{5}=[5^3+5^2-7\times5=115J$

A body is dropped from a height

$h$ . When loss in its potential energy is

$U$ then its velocity is

$v$ . The mass of the body is:

0%
$\dfrac {{U}^{2}}{2v}$
0%
$\dfrac {2v}{U}$
0%
$\dfrac {2v}{{U}^{2}}$
0%
$\dfrac {2U}{{v}^{2}}$

Explanation

loss in PE=gain in KE

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

$m=\dfrac{2U}{v^2}$

mm kg be the mass of piece of ice and it falls through a height of

hh meter, then the loss in potential energy will be?

mg?

0%
$J$ ${mgh}^{2}$
0%
$mgh \ joules$
0%
$L/g$
0%
$1/Lg$

Explanation

$m$ kg be the mass of piece of ice and it falls through a height of

$h$ meter, then the loss in potential energy will be

$mgh$ joule

A particle of mass

$m$ initially moving with speed

$v$ . A force acts on the particle

$f=kx$ where

$x$ is the distance travelled by the particle and

$k$ is constant. Find the speed of the particle when the work done by the force equals

$W$ .

0%
$\sqrt { \frac { k }{ m } +{ v }^{ 2 } }$
0%
$\sqrt { \frac { 2W }{ m } +{ v }^{ 2 } }$
0%
$\sqrt { \frac { 2W }{ k } +{ v }^{ 2 } }$
0%
$\sqrt { \frac { W }{ 2m } +{ v }^{ 2 } }$

Explanation

Given,

Final velocity, ${{v}_{1}}$

Change in kinetic Energy = work done by force

$\dfrac{1}{2}mv_{1}^{2}-\dfrac{1}{2}m{{v}^{2}}=W$

${{v}_{1}}=\sqrt{\dfrac{2W}{m}+{{v}^{2}}}$

Hence, final velocity is $\sqrt{\dfrac{2W}{m}+{{v}^{2}}}$

A wheel rotating at an angular speed of 20 rad

${ s }^{ -1 }$ , is brought to rest by a constant torque in 4s.If the M.I is 0.2 kg

${ m }^{ 2 }$ , the work done in first 2s is:

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

Explanation

Angular acceleration

$\alpha=\dfrac{20}{4}=5$

Torque=

$I\alpha=5\times 0.2=1$

Angular displacement in first 2 secs=

$20\times 2+\dfrac{1}{2}5\times 2^2=50$

Work done=Torque×angular displacement

$=1×50=50J$

Bullets of mass

$40\ g$ each are fired from a machine gun with a velocity of

${10}^{3}\ m/s$ . If the person firing the bullets experience an average force of

$200N$ , then the number of bullets fired per minute will be:

0%
$300$
0%
$600$
0%
$150$
0%
$75$

Explanation

$\textbf{Step 1: Newton's Second Law}$

$\Rightarrow \ \ F = \dfrac{\Delta p}{\Delta t} = \dfrac{m\Delta v}{\Delta t} \times n$ [

$n$ is the number of bullets per minute]

$\Rightarrow\ \ 200 N = n \times 0.04(kg) \times \left(\dfrac{10^3 (m/s) - 0}{60(s)}\right)$

$\Rightarrow\ \ n = 300$

Hence the correct option is

$A$ .

A force of

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

$x=2m$ to

$x=3m$ . The work done by the force is:

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

Explanation

$W=\int Fdx=\int (4x^2+3x)dx=\dfrac{4x^3}{3}+\dfrac{3x^2}{2}$

$W=\dfrac{4\times 3^3}{3}+\dfrac{3\times 3^2}{2}-\dfrac{4\times 2^3}{3}-\dfrac{3\times 2^2}{2}=32.8J$

A ball is dropped from a

$45\ m$ high tower while another is simultaneously thrown upward from the foot at

$20\ m/s$ , along the same vertical line. If the collision is perfectly elastic, first ball reaches ground after time-

0%
$2s$
0%
$3s$
0%
$4s$
0%
$5s$

A position dependent force F =

$3x^{2} - 2x + 7$ acts on a body of mass 7 kg and displace it from

$x$ = 10 m to

$x$ = 5 m. The work done on the body is

$x'$ joule. If both

$F$ and

$x'$ are measured in SI units, the value of

$x'$ is :

0%
-835
0%
235
0%
335
0%
935

Explanation

Given that

$F= 3x^2-2x+7$

Mass of body =

$7\ kg$

Work done by force,

$W=\int_{10}^5 F\cdot dx=\int_{10}^5(3x^2-2x+7)dx= [x^3- x^2+7x]_{10}^5= -835\ J$

The potential energy of a force field is: U =

$\frac{A}{r^{2}}-\frac{B}{r}$ B are positive constants and r is the distance of particle from the field. For stable equilibrium, the distance of the particle is :

0%
A/B
0%
B/A
0%
A - B
0%
2A/B

Explanation

The correct option is D.

Given,

The potential of the particle in the force field is:

$u=\dfrac{A}{r^2}-\dfrac{B}{r}$

For stable in equilibrium,

$f=\dfrac{-du}{dr}=0$

$\dfrac{2A}{r^3}+\dfrac{B}{r^2}$

$B=\dfrac{2A}{B}$

A ball dropped from height H, loses 50% of its energy on each collision with the ground. height to which the ball rises after second impact with ground is-

0%
H
0%
$\cfrac {3H} {4}$
0%
$\cfrac {H} {2}$
0%
$\cfrac {H} {4}$

Explanation

We know that energy of an object at height

$H = mgH$

So, energy of a ball after second collision

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

So, the height to which the ball rises after second impact with ground is

$\dfrac{H}{4}$

Therefore, D is correct option.

A dam is situated at a height of

$550\ m$ above sea level and supplies water to a power house which is at a height of

$50\ m$ above sea level.

$2000\ kg$ of water passes through the turbines per second. What would be the maximum electrical power output of the power house if the whole system were

$80\%$ efficient.

0%
$8\ MW$
0%
$10\ MW$
0%
$12.5\ MW$
0%
$16\ MW$

Explanation

Maximum electrical power output,

$P=\dfrac{mg\Delta h}{t}$

$\Rightarrow P=\dfrac{m}{t}(g\Delta h)$

$=2000\times 9.80 \times (550-50)$

$=9.80\times 10^6$

At 80% efficiency,

$0.80 \times 9.80\times 10^6=7.84\times 10^6M\approx 8MW$

In stretching a spring by

$2\ cm$ energy stored is given by

$U$ , then stretching by

$10\ cm$ energy stored will be :

0%
$U$
0%
$5U$
0%
$\dfrac {U}{25}$
0%
$25U$

Explanation

Given Data

$x_{1} = 2cm$

Energy stored

$U_{1} = U$

$x_{2} = 10\,cm$

$U_{2} = ?$

Let Extension Produced as a spring be x initially

in streched condition spring will have

potentiate energy

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

$k$ is spring constant

$u\,\alpha\, x^{2}$

$\dfrac{u_{1}}{u_{2}} = \dfrac{x_{1}^{2}}{x_{2}^{2}}$

$\boxed {-: u_{1},2u}$

$\dfrac{u}{u_{2}} = \dfrac{2^{2}}{10^{2}} \Rightarrow u_{2} = \dfrac{10^{2}u}{4}$

$u_{2} = 25\,u$

$\boxed{\therefore u_{2} = 25\,u}$

$\boxed{Answer\,is\,D}$

1185946_1031724_ans_5c8aa7a8186940e298a10c68b484fc92.JPG

If the potential energy of two molecules is given by.

$U=\dfrac { A }{ { r }^{ 12 } } -\dfrac { B }{ { r }^{ 6 } }$
Then at equilibrium position, its potential energy is equal to:

0%
$\dfrac { { A }^{ 2 } }{ 4B }$
0%
$-\dfrac { { B }^{ 2 } }{ 4A }$
0%
$\dfrac { 2B }{ A }$
0%
$3A$

A force of

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

$x=2m$ to

$x=3m$ . Te work done by the force is:

0%
$32.8\ J$
0%
$3.281\ J$
0%
$0.328\ J$
0%
$zero$

Explanation

$W=F dx=\int (4x^2+3x)dx=\dfrac{4x^3}{3}+\dfrac{3x^2}{2}=32.8J$

A uniform cylinder of radius

$r$ and length L and mass

$m$ is lying on the ground with the curved surface touching the ground. If it is to be oriented on the ground with the flat circular end in contact with the ground the work to be done is:

0%
$mg\left[ \left( \cfrac { L }{ 2 } \right) -r \right]$
0%
$mg\left[ \left( \cfrac { g }{ 2 } \right) -r \right]$
0%
$m(gL-1)$
0%
$MgLr$

Explanation

Work done=change in potential energy of center of mass

Initially when cylinder is lying with curved surface touching the ground, COM is at height R (R is radius of cylinder)

When it is oriented on the ground with the flat surface touching the ground, COM is at height L/2 (L is length of cylinder)

Change in height of COM is

$\left(\dfrac{L}{2}-R\right)$

Change in potential energy

$=mg\left(\dfrac{L}{2}-R\right)$

Work done

$=mg\left(\dfrac{L}{2}-R\right)$

$'n'$ identical cubes each of mass

$'m'$ and edge

$'L'$ are on a floor. If the cubes are to be arranged one over the other in a vertical stack, the work to be done is

0%
$Lmng(n-1)/2$
0%
$Lg(n-1)/mn$
0%
$(n-1)/Lmng$
0%
$Lmng/2(n-1)$

A machine rated as $150 W$ , changes the velocity of a $10kg$ mass from $4 ms^{-1}$ to $10 ms^{-1}$ in $4s$ . The efficiency of the machine is nearly :

0%
70%
0%
30%
0%
50%
0%
40%

Explanation

Work done by the machine

$=\dfrac{mv^2}{2}-\dfrac{mv\rho^2}{2}$

there,

$m=10\ kg,\quad v_f=10\ m/s$ and

$v_i=4\ m/s$

So,

$\Delta w=1/2[10\times 10^2-10\times (4)^2]$

$=1/2\times 10\times 84=4205$

Avg Power applied by machine

$=\dfrac{\Delta w}{time\ taken}$

$=\dfrac{420}{4}=105\ w$

Efficiency

$=\dfrac{Applied\ Power}{Rated\ Power}$

$=\dfrac{105}{150}=0.7=70\%$

Option

$A$ is correct

A particle moves along the

$x-$ axis from

$x=0$ to

$x=5m$ under the influence of a force given by

$F=$

$7 - 2x + 3{x^2}N$ . The work done in the process is

0%
$107\ J$
0%
$270\ J$
0%
$100\ J$
0%
$135\ J$

Explanation

$W=\int_{0}^{5}Fdx$

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

$=35-25+125$

$=135J$

A body mass of

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

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

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

$2$ seconds is

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

A circular coil of radius 4 cm has 50 turns. In this coil a current of 2 A is flowing. It is placed in a magnetic field of

$0.1 \ weber / m^2$ . the moment of work done in rotating it through

$180^\circ$ from its equilibrium position will be

0%
0.1 J
0%
0.2 J
0%
0.4 J
0%
0.8 J

Explanation

Magnetic moment of current carrying loop

$= M = iAN = i\left( {\pi {r^2}} \right)N$

$W=MB \left( {\cos {\theta _1} - \cos {\theta _2}} \right) = MB\left( {\cos {0^0} - \cos {{180}^0}} \right) = 2MB$

$=0.1 J$

$\therefore$ Option

$A$ is correct.

A particle of mass

$2 kg$ travels along straight line with velocity

$v=a\sqrt{x}$ , where

$x$ is constant. The work done by net force during the displacement of particle from

$x=0$ to

$x=4m$ is

0%
$a^2$
0%
$2a^2$
0%
$4a^2$
0%
$\sqrt{2}a^2$

Explanation

$v=a\sqrt{x}\quad \Rightarrow \cfrac{dx}{dt}=ax^{1/2}\quad \Rightarrow \int x^{1/2} dx=\int a dt$

$2\sqrt{x}=at\quad \Rightarrow x=\cfrac{a^2t^2}{4}$

$\Rightarrow v=\cfrac{a^2t}{2}$

$\Rightarrow a^{\prime}=\cfrac{a^2}{2}$

$F=ma^{\prime}=\cfrac{ma^2}{2}$

$Work = Force\times Displacement$

$\Rightarrow Work =\cfrac{ma^2}{2}\times 4=2ma^2$
As,

$m=2$

$\therefore Work = 4a^2$

A sphere, a cube and a thin circular plate; all are of the same material and same mass and all of them are initially heated to same high temperature. Then:

0%
plate will cool fastest and cube the slowest
0%
sphere will cool fastest and cube the slowest
0%
plate will cool fastest and sphere the slowest
0%
cube will cool fastest and plate the slowest

Explanation

1. The surface area of sphere < surface area of the cube for the given same mass and same density.

2. The rate of cooling

$\propto$ area of contact with surroundings.

$\therefore$ the plate will cool faster than the sphere.

The energy possessed by a body due to its change in position or shape is called:

0%
Potential energy
0%
nuclear energy
0%
kinetic energy
0%
chemical energy

When spring is compressed its potential energy........

0%
Remains constant
0%
Reduces
0%
Increases
0%
Nothing can be said about it.

A force

$\vec{F}=x\hat{i}+2y\hat{j}$ is applied on a particle. Find out work done by

$F$ to move the particle from point

$A$ to

$B$

0%
$-3.5\ J$
0%
$-2.5\ J$
0%
$-4.5\ J$
0%
$-4\ J$

Explanation

It is given that,

$F=x\hat{i}+2y\hat{j}$

$Let,\,\,dx=dx\hat{i}+dy\hat{j}$

$dw=\int{F.dx}$

$=\int{(x\hat{i}+2y\hat{j}})(dx\hat{i}+dy\hat{j})$

$=\int\limits_{1}^{0}{xdx}+2\int\limits_{2}^{1}{ydy}$

$={{\left. \dfrac{{{x}^{2}}}{2} \right|}_{1}}^{0}+{{\left. {{y}^{2}} \right|}_{2}}^{1}$

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

$=-0.5-3$

$=-3.5\,\,J$

The velocity

$(v)$ of a particle of mass

$m$ moving along x-axis is given by

$v = \alpha \sqrt {x}$ , where

$\alpha$ is a constant. Find work done by force acting on particle during its motion from

$x = 0$ to

$x = 2m$ .

0%
$m\alpha^{2}$
0%
$m\alpha$
0%
$\dfrac {m\alpha}{2}$
0%
None of these

Explanation

$v=\alpha\sqrt{x}$

$\cfrac{dx}{dt}=\alpha\sqrt{x}$

$\Rightarrow \int_{0}^{x}{\cfrac{dx}{\sqrt{x}}}=\int_{0}^{t}{\alpha dt}$

$x=\cfrac{\alpha^2t^2}{4}$

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

$a=\cfrac{\alpha^2}{2}$

$F=ma=\cfrac{m\alpha^2}{2}$

$W=F\cdot \Delta x=\cfrac{m\alpha^2}{2}\times 2=m\alpha^2$

The distance

$(x)$ converted by a body of

$2\ kg$ under the action of a force is related to time

$t$ as

$x=t^{2}/4$ . What is the work done by the force in first

$2$ seconds?

0%
$4\ J$
0%
$2\ J$
0%
$1\ J$
0%
$0.5\ J$

Explanation

Given that,

Mass $m=2\,kg$

Distance $x=\dfrac{{{t}^{2}}}{4}$

Now, on differentiate

$\dfrac{dx}{dt}=\dfrac{1}{2}t$

Now, velocity and acceleration is

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

$a=\dfrac{dv}{dt}=\dfrac{1}{2}$

Now, the force is

$F=ma$

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

$F=1\,N$

Now, the work done is

$dW=F\centerdot dx$

$\int{dW}=\int\limits_{0}^{2}{1}\times \dfrac{t}{2}dt$

$W=\left[ \dfrac{{{t}^{2}}}{4} \right]_{0}^{2}$

$W=\left[ \dfrac{4}{4}-0 \right]$

$W=1\,J$

Hence, the work done is $1\ J$

A body of mass 15 kg is raised from certain depth. If the work done in raising it by 10 m is 1620 J, its velocity at this position is

0%
$2 \ ms^{-1}$
0%
$4 \ ms^{-1}$
0%
$1\ ms^{-1}$
0%
$8 \ ms^{-1}$

Explanation

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

$1620 = 15 \times10 \times10 + \dfrac{1}{2} \times 15 \times v^2$

$\implies v =4\ m/s$

Which of the following bodies has the largest kinetic energy?

0%
Mass $3M$ and speed $V$
0%
Mass $3M$ and speed $2V$
0%
Mass $2M$ and speed $3V$
0%
Mass $M$ and speed $4V$

Explanation

We know that, the kinetic energy is

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

Checking all options given -

A) $m=3M$ & $v=v$

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

$K.E=\dfrac{1}{2}\times 3M\times {{v}^{2}}$

$K.E=\dfrac{3M{{v}^{2}}}{2}$

B) $m=3M$ & $v=2v$

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

$K.E=\dfrac{1}{2}\times 3M\times 4{{v}^{2}}$

$K.E=6M{{v}^{2}}$

C) $m=2M$ & $v=3v$

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

$K.E=\dfrac{1}{2}\times 2M\times 9{{v}^{2}}$

$K.E=9M{{v}^{2}}$

D) $m=M$ & $v=4v$

$K.E=\dfrac{1}{2}\times M\times 16{{v}^{2}}$

$K.E=8M{{v}^{2}}$

is $9M{{v}^{2}}$ when $m=2M$ and $v=3v$ .

A body mass $10 kg$ is raised from a certain depth. By the time it is raised by $10 m$ , If its velocity is $2ms^{-1}$ , work done during this time is,

0%
$980 J$
0%
$20 J$
0%
$960 J$
0%
$1000 J$

Explanation

Given that,

Mass of body $m=10\,kg$

Height from depth $h=10\,m$

Velocity $v=2\,m/s$

Now,

Work done = work done to change in P.E + work done in providing K.E to body

$W=mgh+\dfrac{1}{2}m{{v}^{2}}$

$W=10\times 9.8\times 10+\dfrac{1}{2}\times 10\times 2\times 2$

$W=980+20$

$W=1000\,J$

Hence, the work done is $1000\ J$

A body of mass

$m$ starts moving from rest along x-axis so that its velocity varies as

$v = a{s}^{1/2}$ where

$a$ is a constant and

$s$ is the distance covered by the body. The total work done by all the forces acting on the body in the first

$t$ seconds after the start of the motion is

0%
$\dfrac {ma^{4}t^{2}}{4\sqrt {2}}$
0%
$8ma^{4}t^{2}$
0%
$4ma^{4}t^{2}$
0%
$\dfrac {1}{8}ma^{4}t^{2}$

A force

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

$\vec { S }=(2{t^2}\hat { i } -5\hat { j })$ . Find the work done by this force in initial

$2s$ .

0%
$23\ J$
0%
$32\ J$
0%
$Zero$
0%
$48\,J$

Explanation

Given,

Force, $F=\left( 3t\hat{i}+5\hat{j} \right)N$

Displacement, $s=\left( 2{{t}^{2}}\hat{i}-5\hat{j} \right)\,m$

$ds=\left( 4t\hat{i}+0\hat{J} \right)\,dt$

$dw=F.ds$

$\int_{0}^{w}{dw}=\int_{0}^{2}{\left( 3t\hat{i}+5\hat{j} \right)}\left( 4t\hat{i}+0\hat{j} \right)\,dt$

$W=\left. 12{{t}^{2}} \right|_{0}^{2}=48\,J$

Net work done is $48\,J$

A body of mass

$2kg$ makes an elastic collision with another body at rest and comes to rest .The mass of the second body which collides with the first body is

0%
$2 kg$
0%
$1.2 kg$
0%
$3 kg$
0%
$1 kg$

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

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

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

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

$=16J$

Force acting on a particle moving in a straight line varies with the velocities of the particle as

$F=K.V$ . Where

$K$ is constant. The work done by this force in time

$t$ is

0%
$KVt$
0%
$K^{2}V^{2}t^{2}$
0%
$K^{2}Vt$
0%
$KV^{2}t$

Explanation

Given that,

Force $F=KV$

Velocity= $V$

We know that,

$F=Ma$

Now,

$Ma=KV$

$M\left( \dfrac{dV}{dt} \right)=KV$

$MdV=KVdt$

Multiply by v in both side

$MVdV=K{{V}^{2}}dt$

On integrating both side

$\int\limits_{{{v}_{1}}}^{{{v}_{2}}}{MVdV=\int\limits_{0}^{t}{K{{V}^{2}}dt}}$

$M\left[ \dfrac{V_{2}^{2}}{2}-\dfrac{V_{1}^{2}}{2} \right]=K{{V}^{2}}t$

$\dfrac{M}{2}\left[ V_{2}^{2}-V_{1}^{2} \right]=K{{V}^{2}}t$

Now, the change in K.E

Now, from work energy theorem

Change in energy = work done

$K.E=K{{V}^{2}}t$

Hence, the work done is $K{{V}^{2}}t$

There are two mass less springs

$A$ and

$B$ of spring constant

$K_A$ and

$K_B$ respectively and

$K_A>K_B$ if

$W_A$ and

$W_B$ be denoted as work done on

$A$ and work done on

$B$ respectively, then:

0%
If they are compressed by same distance $W_A=W_B$
0%
If they are compressed by same force(upto equilibrium state) $W_A=W_B$
0%
If they are compressed to same distance $W_A>W_B$
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If they are compressed by same force(upto equilibrium state) $W_A>W_B$

Explanation

According to question

$K_A>K-B$

$\Rightarrow W_A=+\cfrac{1}{2}K_Ax^2$

$[\therefore$ Work is done by spring is

$=-\cfrac{1}{\sum}Kx^2]$

$W_B=+\cfrac{1}{2}K_Bx^2$

$[$ Here, work is done on spring. Hence force, and displacement is in same direction

$]$

$\because K_A>K_B$

$W_A>W_B$

$F_A=F_B$

$K_Ax_A=K_Bx_B\Rightarrow x_A=\left(\cfrac{K_B}{K_A}\right)x_B \Rightarrow x_A<x_B$

$W_A=+\cfrac{1}{2}K_Ax^2_A=-\cfrac{1}{2}(K_Ax_A)x_A=-\cfrac{1}{2}F_Ax_A$

$W_B=+\cfrac{1}{2}K_Bx^2_B=-\cfrac{1}{2}(K_Bx_B)x_B=-\cfrac{1}{2}F_Bx_B$

$\Rightarrow W_B>W_A$

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

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

$k$ is a constant. The work done in displacing the body from

$x=0$ to

$x$ is proportional to:

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$x^{1/2}$
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$x^{2}$
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$x^{3}$
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$x^{5/2}$

Explanation

Given,

Velocity, $\dfrac{dx}{dt}=v=k{{x}^{3/2}}$

Acceleration,

$a=\dfrac{dv}{dt}=\dfrac{d\left( k{{x}^{3/2}} \right)}{dt}=\dfrac{1}{2}k{{x}^{1/2}}\dfrac{dx}{dt}=\dfrac{1}{2}k{{x}^{1/2}}v$

$a=\dfrac{1}{2}k{{x}^{1/2}}\left( k{{x}^{3/2}} \right)=\dfrac{1}{2}{{k}^{2}}{{x}^{2}}$

$work\,done\,=\,force\times displacemet\,=Mass\times acceleration\times displacement$

$\Rightarrow dW=m.a\,.dx=m.\dfrac{1}{2}{{k}^{2}}{{x}^{2}}dx$

On integrating

$W=m\dfrac{1}{2}{{k}^{2}}\left( \dfrac{{{x}^{3}}}{3} \right)$

Hence, Work done is proportional to ${{x}^{3}}$

A spring gun has a spring constant of

$80\ N/m$ . The spring is compressed

$12\ cm$ by a ball of mass

$15\ gm$ . How much is potential energy of the spring and what will be the velocity of the ball, if the trigger is pulled:

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$57.6\ J, 87.6\ m/s$
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$67.6\ J, 97.6\ m/s$
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$0.576\ J, 76.8\ m/s$
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$none\ of\ these$

Explanation

Spring energy $E=\dfrac{1}{2}k{{x}^{2}}=\dfrac{1}{2}80\times {{0.12}^{2}}=0.576\,J$

From conservation of energy

Maximum Spring energy = Maximum kinetic energy of ball

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

$V=\dfrac{2E}{M}=\dfrac{2\times 0.576}{0.015}=76.8m{{s}^{-1}}$

Velocity of ball is $76.8\,m{{s}^{-1}}$

A particle moves from origin to position

$\vec {r}_{1}=3\hat {i}+2\hat {j}-6\hat {k}$ under the action of force

$4\hat {i}+\hat {j}+3\hat {k}N$ . the work done will be

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$10\ J$
0%
$5\ J$
0%
$2\ J$
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$-4\ J$

Work done from

$d=0\ m$ to

$d=4\ m$

0%
$12.5\ J$
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$15\ J$
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$17.5\ J$
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$20\ J$

Under influence of a force

$\overrightarrow { F } \left( x \right) =\left( 3{ x }^{ 2 }-2x+5 \right) \hat { i } N$ there is displacement of a particle from

$x=0$ and

$x=5m$ on

$x-$ axis. So work done is

$J$ .

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$100$
0%
$150$
0%
$125$
0%
$120$

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

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$3.84J$
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$1.68J$
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$5.28J$
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$5.41J$

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 particle moves under the effect of a force

$F=c\ x$ from

$x=0$ to

$x=x_{1}$ . The work done in the process is

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$\dfrac{c{ { x }_{ 1 } }^{ 2 }}{2}$
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$c{ { x }_{ 1 } }^{ 2 }$
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$c{ { x }_{ 1 } }^{ 3 }$
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$Zero$

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

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$32J$
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$20J$
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$36J$
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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.

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

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

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

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 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 :

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$70$
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$270$
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$35$
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$135$

A force acts on a

$3$ g 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 :

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$2.88$ J
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$450$ mJ
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$490$ mJ
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$530$ mJ

Explanation

We have,

mass,

$m=3\ g=0.003\ kg$

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

Now,

$v=\dfrac{dx}{dt}=3-8t+3t^2\Rightarrow dx=(3-8t+3t^2)dt$

$\Rightarrow a=\dfrac{dv}{dt}=0-8+6t$

Now,

$dw=Fdx$

$\Rightarrow dw=(ma)dx$

$\Rightarrow dw=(0.003)(-8+6t)(3-8t+3t^2)dt$

$\Rightarrow dw=(0.003)(18t^3-72t^2+82t-24)dt$

$\Rightarrow w=(0.003)\displaystyle \int_{0}^{4}{(18t^3-72t^2+82t-24)dt}$

$\Rightarrow w=0.003\times 176=0.528\ J$

$\Rightarrow w=530\ mJ$

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

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Practice Class 11 Medical Physics Quiz Questions and Answers

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