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

A car is accelerated on a leveled road and attains a velocity

$4$ times its initial velocity. In this process, the potential energy of the car:

0%
Does not change
0%
Becomes twice to that of initial
0%
Becomes $4$ times that of initial
0%
Becomes $16$ times that of initial

N identical balls are placed on a smooth horizontal surface. Another ball of same mass collides elastically with velocity

$u$ with first ball of N balls. A process of collision is thus started in which first ball collides with second ball and the second ball with the third ball and so on. The coefficient of resulting for each collision is

$e$ . Find speed of Nth ball :

0%
$(1+e)^{N}u$
0%
$u(1+e)^{N-1}$
0%
$\cfrac{u(1+e)^{N-1}}{2^{N-1}}$
0%
$u^{N}(1+e)^{N-1}$

When a boy triples his speed, his kinetic energy becomes

0%
half
0%
double
0%
nine times
0%
no charge

Explanation

Given that, speed of particle is becoming 3 times of its initial value while moving. ie $v'=3v$

As we know $K.E = \dfrac{1}{2}$ $mv ^{2}$ ..........(1)

After velocity tripled $K.E' = \dfrac{1}{2}$ $m(3v) ^{2}=\dfrac92mv^2$ ...........(2)

From equation 1 and 2

$K.E' = 9 \times K.E$

When a boy tripled his speed, his kinetic energy will becomes nine times.

The energy stored in water of a dam is the kinetic energy.

0%
True
0%
False

A position dependent force

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

$2$ kg displace it from

$x = 0$ to

$x = 5 m$ the work done in joules is

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

Explanation

It is known that

$W=\int Fdx$

$W = \int^5_0 Fdx = \int^5_0 ( 7- 2x +3x^2) dx$

$W = [ 7x - x^2 + x^3]^5_0$

$W= 35 -25 + 125 = 135 J$

If the stone is thrown up vertically and return to ground, its potential energy is maximum

0%
During the upward journey
0%
At the maximum height
0%
During the return journey
0%
At the bottom

Explanation

Potential energy =

$mgh$

$PE \propto h$

Potential energy is maximum when

$h$ is maximum

A particle moves under the effect of a force

$F =Cx$ from

$x = 0$ to

$x =x_1$ The work done process is

0%
$Cx^2_1$
0%
$\frac {1}{2} Cx^2_1$
0%
$Cx_1$
0%
$Zero$

Explanation

Consider small displacement dx,

work done for this small displacement

$dW=F.dx$

$W =\int^{x_1}_0 F.dx$

$W= \int^{x_1}_0 Cx dx = C [ \dfrac {x^2}{2}]^{x_1}_0$

$W = \dfrac {1}{2} Cx^2_1$

If velocity of a body is twice of previous velocity, then kinetic energy will become

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

Explanation

Kinetic energy=

$\dfrac{1}{2} mv^2 \therefore K.E. \propto v^2$

$v\rightarrow 2v$

$K \rightarrow 2^2=4K$

If velocity is doubled then kinetic energy will become four times.

An object of

$m\ kg$ with speed of

$v\ m/s$ strikes a wall at an angle

$\theta$ and rebounds at the same speed and same angle. The magnitude of the change in momentum of the object will be

0%
$2mv\cos \theta$
0%
$2mv\sin \theta$
0%
$0$
0%
$2\ mv$

Explanation

$\vec P_1=mv\sin \theta \hat i-mv\cos \theta \hat j$
and

$\vec P_2=mv\sin \theta \hat i+mv\cos \theta \hat j$
So change in momentum

$\overrightarrow{\Delta P}=\vec P_2-\vec P_1=2mv\cos \theta \hat j, | \Delta \vec P| =2mv\cos \theta$

The decrease in the potential energy of a ball of mass

$20 kg$ which falls from a height of

$50 cm$ is

0%
$968 J$
0%
$98 J$
0%
$1980 J$
0%
None of these

Explanation

Decrease in potential energy

$\Delta U =mg(h_2-h_1)$

$\Delta U= mgh =20 \times 9.8 \times 0.5=98 J$

A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time

$t$ is proportional to

0%
$t^{1/2}$
0%
$t^{3/4}$
0%
$t^{3/2}$
0%
$t^2$

Explanation

$P = Fv = mav = m( \dfrac {dv}{dt} ) v \Rightarrow \dfrac {P}{m}dt = vdv$

$\Rightarrow \dfrac {P}{m} \times t = \dfrac {v^2}{2} \Rightarrow v = ( \dfrac {2P}{m})^{ 1/2} (t)^{1/2}$ now

$s = \int v dt = \int ( \dfrac {2P}{m})^{1 /2} t^{1/2}dt$

$\therefore s = ( \dfrac {2P}{m})^{ 1/2} [ \dfrac { 2t^{ 3/2} }{3} ] \Rightarrow s \propto t^{ 3/2}$

The relation between the displacement

$X$ of an object produced by the application of the variable force

$F$ is represented by a graph shown in the figure. If the object undergoes a displacement from

$X=0.5m$ to

$X=2.5m$ the work done will be approximately equal to

0%
$16 J$
0%
$32 J$
0%
$1.6 J$
0%
$8 J$

Explanation

Work done = Area under curve and displacement axis

= Area of trapezium

$= \dfrac {1}{2} \times$ ( sum f two parallel lines )

$\times$ distance between them

$= \dfrac {1}{2} ( 10 +4) \times ( 2.5 -0.5)$

$= \dfrac {1}{2} 14 \times 2 = 14 J$

As the area actually is not trapezium so work done will be more than

$14$ J i.e. approximately

$16$ J

The displacement x of a particle moving in one dimension under the action of a constant force is related to the time t by the equation

$t = \sqrt { x} +3$ where x is in meters and t is in seconds . the work done by the force in the first

$6$ second is

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

Explanation

According to the question,

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

$t = 0 ; v_1 = -6 m/s$ and at

$t =6 sec , v_2 = 6 m/s$

So change in kinetic energy

$= W = \dfrac {1}{2} mv^2_2 - \dfrac {1}{2} mv^2_1 = 0 J$

The force constant of a wire is k and that of another wire is

$2k$ When both the wires are stretched through same distance, then the work done.

0%
$W_2 = 2W^2_1$
0%
$W_2 =2W_1$
0%
$W_2 =W_1$
0%
$W_2 = 0.5 W_1$

Explanation

Work required = change in energy

$W=\Delta E$

$W =\frac {1}{2} kx^2$

If both wires are stretched through same distance then

$W \propto k$ as

$k_2 = 2k_1$ so

$W_2 = 2W_1$

A body of mass

$3$ kg is under a force, which causes a displacement in it is given by

$S = \frac {t^3}{3} ( in m )$ find the work done by the force in first

$2$ seconds

0%
$2 J$
0%
$3.8 J$
0%
$5.2 J$
0%
$24 J$

Explanation

Given:

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

$\therefore dS = t^2 dt$

$a = \dfrac {d^2 S}{ st^2} = \dfrac {d^2}{dt^2} [ \dfrac {t^3}{3} ] = 2t m/s^2$

Now work done by the force

$W = \int^2_0 F.dS = \int^2_0 ma.dS$

$\int^2_0 3 \times 2 t \times t^2 dt = \int^2_0 6t^3 dt = \dfrac {3}{2} [ t^4]^2_0= 24 J$

The spring will have maximum potential energy when

0%
it is pulled out
0%
it is compressed
0%
both (a) and (b)
0%
neither (a) nor (b)

Work done on a body is equal to change in .......

0%
only Kinetic energy
0%
only potential energy
0%
only mechanical energy
0%
energy

Masses of two bodies are

$1$ kg and

$4$ kg respectively. If their kinetic energies are in

$2:1$ proportion, the ratio of their speeds is .........

0%
$2\sqrt{2}:1$
0%
$1:\sqrt{2}$
0%
$1:2$
0%
$2:1$

Explanation

We know that,

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

$m_1=1kg$

$m_2=4kg$

$\therefore \dfrac{K_1}{K_2}=\Bigg(\dfrac{m_1}{m_2}\Bigg)\Bigg(\dfrac{v_1}{v_2}\Bigg)^2$

$\Rightarrow \dfrac{2}{1}=\Bigg(\dfrac{1}{4}\Bigg)\Bigg(\dfrac{v_1}{v_2}\Bigg)^2$

$\Rightarrow \Bigg(\dfrac{v_1}{v_2}\Bigg)^2=\dfrac{8}{1}$

$\Rightarrow \Bigg(\dfrac{v_1}{v_2}\Bigg)=\dfrac{2\sqrt{2}}{1}$

$5-kg$ cart moving to the right with a speed of

$6\ m/s$ collides with a concrete wall and rebounds with a speed of

$2\ m/s$ . What is the change in momentum of the cart?

0%
$0$
0%
$40\ kg.m/s$
0%
$-40\ kg.m/s$
0%
$-30\ kg.m/s$
0%
$-10\ kg.m/s$

In one experiment, two balls of clay of the same mass travel with the same speed v toward each other. They collide head-on and come to rest. In a second experiment, two clay balls of the same mass are again used. One ball hangs at rest, suspended from the ceiling by a thread. The second ball is fired toward the first at speed v, to collide, stick to the first ball, and continue to move forward. Is the kinetic energy that is transformed into internal energy in the first experiment:

0%
one-fourth as much as in the second experiment,
0%
one-half as much as in the second experiment,
0%
the same as in the second experiment,
0%
twice as much as in the second experiment, or
0%
four times as much as in the second experiment.

Explanation

Kinetic energy is transformed into internal energy:

$Q =-\Delta K$ . In the first experiment, momentum conservation requires the final speed be zero:

$p_1=m-mv=2mv_f\,\rightarrow\,v_f=0$

The kinetic energy converted into internal energy

$mv^2$ :

$\Delta K_1=K_f-K_i=0-(\frac{1}{2}mv^2+\frac{1}{2}mv^2)=-mv^2\,\\ \\ \rightarrow Q_1=mv^2...........(1)$

In the second experiment, momentum conservation requires the final speed be half the initial speed:

$p_2=mv+m(0)=2mv_f\,\rightarrow v_f=\dfrac{v}{2}$

The kinetic energy converted into internal energy

$\dfrac{mv^2}{4}$ :

$\Delta K_2=K_f-K_i=\dfrac{1}{2}(2m)\Bigg(\dfrac{v}{2}\Bigg)^2-\dfrac{1}{2}mv^2=-\dfrac{mv^2}{4}\,\\ \rightarrow Q_2=\dfrac{mv^2}{4}............(2)$

From equation 1 and 2

four times as much as in the second experiment.

A pile driver drives postes into the ground by repeatedly dropping a heavy object on them. Assume the object is dropped from the same height each time. By what factor does the energy of the pile driver-Earth system change when the mass of the object being dropped is doubled?

0%
$\dfrac{1}{2}$
0%
$1$ ; the energy is the sane
0%
$2$
0%
$4$

$10.0-g$ bullet is fired into a

$200-g$ block of wood at rest on a horizontal surface. After impact, the block slides

$8.00\ m$ before coming to rest. If the coefficient of friction between the block and the bullet before impact?

0%
$106\ m/s$
0%
$166\ m/s$
0%
$226\ m/s$
0%
$286\ m/s$
0%
none of those answer is correct

Calculate the energy possessed by a stone of mass

$10\ kg$ kept at a height of

$5\ m$ .
Take

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

0%
$196\ J$
0%
$49\ J$
0%
$490\ J$
0%
$980\ J$

Explanation

The energy that stone may posses = kinetic energy + potential energy

Since, stone at rest

$(v = 0)$

$KE=\frac{mv^2}{2}$ So kinetic energy=0

$\therefore$ Total energy = Potential energy

$(E) = mgh$ [

$m$ : mass ,

$h$ : height ,

$g$ : gravity ]

$m = 10\ kg, g = 9.8\ m/s^{2}$
and

$h = 5\ m$

$E = mgh = 10 \times 9.8 \times 5$ Joules

$= 490\ J$

A bar of mass

$M$ and length

$L$ is in pure translatory motion and its centre of mass has velocity

$V$ . It collides and sticks to a second identical bar which is initially at rest. (Assume that it becomes one composite bar of length

$2L$ ). The angular velocity of the composite bar after collision will be :

0%
$\dfrac{3}{4}\dfrac{V}{L}$
0%
$\dfrac{4}{3}\dfrac{V}{L}$
0%
Counterclockwise
0%
Clockwise

Explanation

Since there is no external torque,

$\vec{L}$ will be conserved.

$L_i= MVL/2$

$L_f= I\omega$
Since both rods stick together it becomes a rod of mass

$2M$ and length

$2L$
Hence,

$L_f= I\omega= (2M)\cfrac{(2L)^2}{12}\omega$
Equating

$L_i=L_f$

$\Rightarrow MVL/2 =\cfrac{2}{3}ML^2\omega$

$\Rightarrow \omega =\cfrac{3}{4}\cfrac{V}{L}$
The rod which has the velocity is sticking to the stationary rod which is at the top. After sticking the inertia of the stationary rod will make the whole rod to rotate in counterclockwise direction. Hence A and C is correct.

The work done by the force

$\vec{F}=A(y^{2}\hat{i}+2x^{2}\hat{j})$ , where

$A$ is a constant and

$x$ &

$y$ are in meters around the path shown is :

0%
$zero$
0%
$Ad$
0%
$Ad^{2}$
0%
$Ad^{3}$

Explanation

Variable force is apllied across the path

$OA$ to

$AB$ to

$BC$ to

$CO$

$W=\int \bar { F } .d\bar { r }$

$\\ =\int A(y^{ 2 }\bar { i } +2x^{ 2 }\bar { j } ).(dx\bar { i } +dy.\bar { j } )$

$\\ =\int A(y^{ 2 }dx+2x^{ 2 }dy)$

Now following the path of displacement along

$OA$ the

$y=0$

$\\ W_{ OA }=\int _{ x=0 }^{ x=d }{ A(0\quad +\quad 2 } x^{ 2 }.0)\quad =\quad 0+0\quad =0$

Similerly, for

$W_{ AB },\quad x=d$

$\\ W_{ AB }=A[0+2d^{ 2 }d]\quad =\quad 2Ad^{ 3 }$

Similarly, for

$W_{ BC },\ x=d,\quad y=d$

$\\ W_{ BC }=\int _{ x=d }^{ x=0 }{ A(d^{ 2 }dx\quad +\quad } 0)$

$\\ W_{ BC }=A[d^{ 2 }(-d)+0]=-Ad^{ 3 }$

$\\ W_{ CO }=A[0+0]$

$\\ W=0+2Ad^{ 3 }-Ad^{ 3 }+0=Ad^{ 3 }$

A particle of mass

$m$ is moving with speed

$u$ . It is stopped by a force

$F$ in distance

$x$ . If the stopping force is

$4F$ then

0%
work done by stopping force in second case will be same as that in first case
0%
work done by stopping force in second case will be $2$ times of that in first case
0%
work done by stopping force in second case will be $\dfrac{1}{2}$ of that in first case
0%
work done by stopping force in second case will be $\dfrac{1}{4}$ of that in first case

Explanation

$\Delta KE$ is same in both the cases, work done will be same.

A ball of mass m moves towards a moving wall of infinite mass with a speed 'v' along the normal to the wall. The speed of the wall is 'u' toward the ball. The speed of the ball after elastic collision with wall is

0%
$u + v$ away from the wall
0%
$2u + v$ away from the wall
0%
$|u - v|$ away from the wall
0%
$|v- 2u |$ away from the wall

Explanation

In the frame of wall, the ball moves towards it with speed

$u + v$ and due to momentum conservation velocity after collision returns with velocity

$u + v$ in the direction in which wall is moving.

So from the ground frame, the velocity of ball after collision with wall is

$2u + v.$

A force

$\vec{F}=(3t\hat{i}+5\hat{j})$ N acts on a particle whose position vector varies as

$\vec{S}=(2t^{2}\hat{i}+5\hat{j})$ m, where

$t$ is time in seconds. The work done by this force from

$t=0$ to

$t=2s$ is:

0%
23 J
0%
32 J
0%
zero
0%
can't be obtained

Explanation

The work done by the force

$\vec{F}=(3t\hat{i}+5\hat{j})$ is

$W=\int \vec{F}.d\vec{s}=\int (3t \hat{i}+5\hat{j}).(4t dt \hat{i})$

$=\int_{0}^{2}12t^{2}dt=\dfrac{12[t^{3}]_{0}^{2}}{3}=\dfrac{12(8-0)}{3}=32 \ J$

A uniform bar of mass

$M$ and length

$L$ collides with a horizontal surface. Before collision, velocity of centre of mass was

$v_{0}$ and no angular velocity. Just after collision, velocity of centre of mass of bar becomes

$v$ in upward direction as shown. Angular velocity

$\omega$ of the bar just after impact is:

0%
$\dfrac{3\left ( v_{0}+v \right )cos\theta }{2L}$
0%
$\dfrac{6\left ( v_{0}-v \right )cos\theta }{L}$
0%
$\dfrac{\left ( v_{0}+v \right )cos\theta }{L}$
0%
$\dfrac{\left ( v_{0}-v \right )cos\theta }{6L}$

Explanation

Angular momentum will be conserved about point A just before and just after collision.

$\displaystyle mv_{0.}\frac{L}{2}cos\theta =\frac{mL^{2}}{3}.\omega -mv.\frac{L}{2}cos\theta$

$\displaystyle \omega =\frac{3\left ( v_{0}+v \right )cos\theta }{2L}.$

A particle of mass

$m_0,$ travelling at speed

$v_0,$ strikes a stationary particle of mass

$2m_0.$ As a result, the particle of mass

$m_0$ is deflected through

$45^o$ and has a final speed of

$\dfrac{v_0}{\sqrt2}$ . Then the speed of the particle of mass

$2m_0$ after this collision is

0%
$\dfrac{v_0}{2}$
0%
$\dfrac{v_0}{2\sqrt2}$
0%
$\sqrt2v_0$
0%
$\dfrac{v_0}{\sqrt2}$

Explanation

A particle of mass

$m_0,$
Before collision

$mv_{0}=2mv cos\theta+\displaystyle \dfrac{mv_{0}}{2}..........(i)$

$\displaystyle \theta=\dfrac{mv_{0}}{2}-2mv sin \theta............(ii)$

By (i) & (ii),

$\theta=45^{0}$
Now again momentum conservation in y direction

$\displaystyle \dfrac{mv_{0}}{2}=2mv.\displaystyle \dfrac{1}{\sqrt2}\Rightarrow v=\displaystyle \dfrac{v_0}{2\sqrt2}$

A particle of mass

$m$ moves along the quarter section of the circular path whose centre is at the origin. The radius of the circular path is

$a$ . A force

$\vec{F}=y\hat{i}-x\hat{j}$ newton acts on the particle, where

$x, y$ denote the coordinates of position of the particle. The work done by this force in taking the particle from point A (

$a, 0$ ) to point B (

$0, a$ ) along the circular path is

0%
$\dfrac{\pi a^{2}}{4}J$
0%
$\dfrac{\pi a^{2}}{2}J$
0%
$\dfrac{\pi a^{2}}{3}J$
0%
$None\ of\ these$

Explanation

$\displaystyle W=\mathbf{\int F.dr}=\int (y\hat{i}-x\hat{j}).(dx\hat{i}+dy\hat{j})$

$=\displaystyle \int (ydx-xdy)$ .........(1)

$\because x^{2}+y^{2}=a^{2} \therefore xdx+y dy=0$

$\displaystyle \Rightarrow W=\int \left ( y\left ( \frac{-ydy}{x} \right )-xdy \right )=-\int \dfrac{x^{2}+y^{2}}{x}dy$

$\displaystyle =-\int_{0}^{a}\dfrac{a^{2}}{\sqrt{a^{2}-y^{2}}}dy=-\dfrac{\pi a^{2}}{2}$

Alternate Method :
It can be observed that the force is tangent to the curve at
each point and the magnitude is constant. The direction of
force is opposite to the direction of motion of the particle.

$\therefore$ work done = (force) (distance)

$=-\sqrt{x^{2}+y^{2}}=\dfrac{\pi a}{2}=-a\times \dfrac{\pi a}{2}=-\dfrac{\pi a^{2}}{2}J$

$W=-\dfrac{\pi a^{2}}{2}J$

A force

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

$\vec S=(2t^2\hat i-5\hat j)m$ . Work done by this force in

$2$ second is

0%
32 J
0%
24 J
0%
46 J
0%
20 J

Explanation

$\vec v=\dfrac {d\vec S}{dt}=(4t)\hat i$

$P=\vec F\cdot \vec v=12 t^2$

$\therefore W=\int_0^2Pdt=\int_0^2 (12)t^2dt=32J$ .

The world class Moses Mabhida football stadium situated in Durban has a symmetrical arc of length 350 m and a height of 106 m, as shown in the picture below on the left.
The picture on the right shows a funicular (Skycar), which takes tourists to the top of the arc. Suppose that the Skycar with tourists inside, starts from the base of the arc and travels a distance of 175 m along the arc to the viewing platform at the top. Assume that the work done by friction during the Skycars complete ascent is

$5.8\times 10^5J$ . If the combined mass of the Skycar and tourists is 5000 kg, then the work done by the motor that lifts the Skycar is approximately equal to,

0%
$4.6\times 10^6 J$
0%
$5.8\times 10^6 J$
0%
$8.0\times 10^6 J$
0%
$9.2\times 10^6 J$

In using an axe to split firewood, the following energy forms are involved

(i) Chemical (muscle) energy
(ii) Mechanical potential energy of the axe;
(iii) Chemical (binding) energy of wood, heat energy, sound energy and kinetic
energy of wood fragments;
(iv) Mechanical kinetic energy of the axe.
Which is the most likely sequence of the energy exchanges?

0%
(i), (ii), (iv), (iii)
0%
(i), (iv), (iii), (ii)
0%
(iv), (i), (ii), (iii)
0%
(i), (ii), (iii), (iv)

The figure shows a block having a small basket in it and an small inclined plane (with fixed angle) is rigidly attached to the block. The combined mass (including incline plane) of all these is M = 5 kg. Initially they are at rest. Now identical balls each of mass m=1 kg are thrown horizontally with velocity v = 5m/s (with respect to ground) which strike the inclined plane and then are collected in the small basket. Assume that in the basket balls come to rest with respect to basket. Choose the correct statement(s) from the following :

0%
After collecting 5 balls the speed of the block will be 2.5 m/s
0%
After collecting 10 balls the speed of the block will be (10/3) m/s
0%
The velocity of the block will never be exactly 5 m/s no matter how many balls are collected in the basket.
0%
Increase in speed of the block will be same every time for collection of each ball in the basket

Explanation

Total momentum of

$N$ balls

$=N m v=N (1\times5)=5N$

$\therefore$ Momentum of the block along with

$N$ balls

$=5N$ or

$(M+Nm)$

${v}'=5N$ where

${v}'$ is the velocity of the block
After

$N$ balls are collected in the basket.

$\therefore {v}' =\displaystyle \dfrac{5N}{5+N}=\dfrac{5}{\dfrac{5}{N}+1}$

For

$N=5$ ;

${v}'$

$=2.5\ m/s$
For

$N=10$ ;

${v}'$

$=10/3\ m/s$
Also

${v}'$ is never exactly equal to

$5 m/s.$

From above we can say that increase in speed of the ball is not same every time for collection of each ball in the basket, it varies.

A bullet moving with a speed of

$100\; ms^{-1}$ can just penetrate into two planks of equal thickness. Then, the number of such planks the bullet will penetrate, if speed is doubled will be

0%
6
0%
20
0%
4
0%
8

Explanation

$\frac{1}{2}mv^2 = n(F.d)$

$\Rightarrow n \,\, \propto v^2$
Thus,
If v is doubled, n will become four times.
Therefore the bullet will penetrate 8 planks.

A block of mass

$10kg$ is released on a fixed wedge inside a cart which is moved with constant velocity

$10 \ ms^{-1}$ towards right. There is no relative motion between block and cart. Then work done by normal reaction on block in two seconds from ground frame will be

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

0%
$1320 J$
0%
$960 J$
0%
$1200 J$
0%
$240 J$

Explanation

Constant velocity means net force

$=0$
Using Lami's theorem in the figure,
We have,

$\displaystyle \dfrac {N}{sin (180^o- 53^o)}=\frac {100}{sin 90^o}$

$\therefore N=100 sin 53^o=80 N$

Now,

$W_N=NS cos 53^o$

$=(80)(20)(0.6)$

$=960 J$

A particle of mass m, moving with velocity v collides a stationary particles of mass 2m. As a result of collision, the particle of mass m deviates by

$45^o$ and has final speed of

$\dfrac {v}{2}$ . For this situation mark out the correct statement (s).

0%
The angle of divergence between particles after collision is $\dfrac {\pi}{2}$
0%
The angle of divergence between particles after collision is less than $\dfrac {\pi}{2}$
0%
Collision is elastic
0%
Collision is inelastic

Explanation

Let velocity of mass 2m is

$\vec{{v}'}$ after collision.;
From conservation of momentum in the xy plane,

$\vec{p_i}=\vec{p_f}$ ;

$mv\hat{i}=\dfrac{mv}{2}(\cos 45^0\hat{i}+\sin 45^0\hat{j})+2m\vec{{v}'}=\dfrac{mv}{2\sqrt2}(\hat{i}+\hat{j})+2m\vec{{v}'}$ ;

$\therefore 2\vec{{v}'}=(\dfrac{2\sqrt2-1}{2\sqrt2})v\hat{i}-\dfrac{v}{2\sqrt2}\hat{j}$

$\Rightarrow \vec{{v}'} =0.32v\hat{i}-0.18v\hat{j}$ ;

$\therefore \tan \theta =\dfrac{0.18}{0.32}=0.56$ ;

$\therefore\theta =29.35^0$ ;

$\left | \vec{v}' \right |= \sqrt{(.32)^2+(.18)^2}v=0.37v$ ;

Since

$( \theta +45^0) < 90^0$ , the angle of divergence between particles after collision is less than

$90^0$ ;

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

${KE}_f=\dfrac{1}{2}m(\dfrac{v}{2})^2 + \dfrac{1}{2}2m(0.37v)^2$ ;

$\therefore {KE}_f < {KE}_i$ ;

$\therefore$ collision is inelastic.

Sunil rolls a marble ball down a frictionless inclined plane as shown above. What kind of energies are possessed by the rolling marble ball?

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(i) and (ii) only
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(ii) and (iii) only
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(iii) and (i) only
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(i), (ii) and (iii)

The force exerted by a compression device is given by

$F(x) = kx (x-l)$ for

$0 \leq x \leq l$ , where

$l$ is the maximum possible compression,

$x$ is the compression and

$k$ is the constant. Work done to compress the device by a distance

$d$ will be maximum when

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$d = \dfrac {l}{4}$
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$d = \dfrac {l}{\sqrt{2}}$
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$d = \dfrac {l}{2}$
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$d = l$

Explanation

For W to be maximum ;

$\dfrac{dW}{dx} = 0$ ;

i.e.

$F(x) = 0 \Rightarrow x= l, x = 0$

clearly for

$d = l$ , the work done is maximum.

Alternate Solution :
External force and displacement are in the same direction

$\therefore$ Work will be positive continuously so it will be maximum when displacement is maximum.

The relationship between the force

$F$ and position

$x$ of a body is as shown in figure. The work done by force

$F$ , in displacing the body from

$x=1 \ m$ to

$x=5 \ m$ will be

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30 J
0%
15 J
0%
25 J
0%
20 J

Explanation

Hint: Work done is the area under F-x graph.

Correct Option: B

$\textbf{Step1: Calculation of work done}$

As work done = area enclosed by F-x graph

Work done in displacing the body from x=1 m to x=5 m will be:

W = Area of $ABNM$ + Area of $CDEN$ - Area of $EFGH$ + Area of $HIJ$

(Area of $EFGH$ is taken negative as the force is in negative direction)

$\Rightarrow W = \left( {1 \times 10} \right) + \left( {1 \times 5} \right) - \left( {1 \times 5} \right) + \left( {\dfrac{1}{2} \times 10 \times 1} \right) = 15J$

So, the work done in displacing the body from x=1 m to x=5 m will be $15J$ .

A uniform solid sphere of radius

$r$ is rolling on a smooth horizontal surface with velocity

$V$ and angular velocity

$\omega$ (where

$V=\omega r)$ . The sphere collides with a sharp edge on the wall as shown in the figure. The coefficient of friction between the sphere and the edge

$\mu=1/5$ . Just after the collision the angular velocity of the sphere becomes equal to zero. The linear velocity of the sphere just after the collision is equal to :

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$V$
0%
$\dfrac {V}{5}$
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$\dfrac {3V}{5}$
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$\dfrac {V}{6}$

Explanation

Let the velocity of sphere after the collision be V'.

Using conservation of momentum,

$mV + 0=mV'+0$

$\implies V'=-V$ where -ve sign suggests that sphere will reverse its direction.

A mass

${ m }_{ 1 }$ with initial speed

${ v }_{ 0 }$ in the positive

$x$ -direction collides with a mass

${ m }_{ 2 }=2{ m }_{ 1 }$ which is initially at rest at the origin, as shown in figure. After the collision

${ m }_{ 1 }$ moves off with speed

${ v }_{ 1 }={ v }_{ 0 }/2$ in the negative

$y$ - direction, and

${ m }_{ 2 }$ moves off with speed

${ v }_{ 2 }$ at angle

$\theta$ . Find the velocity (magnitude and direction) of the centre of mass after the collision :

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${ v }_{ 0 }/3$
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${ v }_{ 0 }/2$
0%
${ v }_{ 0 }/5$
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${ v }_{ 0 }/6$

Explanation

Since there is no external force acting on the 2 mass system, the momentum of the system and the velocity of the centre of mass remains constant.
Therefore,

$V_{cm}$ is the same before and after the collision, i.e;

$V_{cm,x} = \dfrac {m_1v_1+m_2v_2}{m_1 +m_2}$ , as

$m_2=2m_1$ &

$V_{cm,y}=0.$

$V_{cm,x}= \dfrac { m_1v_o+2m_1 \times 0}{m_1+2m_1} = \dfrac{v_o}{3} .$

A body of mass

$m$ is hauled from the Earth's surface by applying a force

$\vec{F}$ varying with the height of ascent

$y$ as

$\vec{F}=2(ay-1)mg$ , where

$a$ is a positive constant. Find the work performed by this force

$W$ and the increment in the body's potential energy

$\Delta U$ in the gravitational field of the Earth over the first half of the ascent.

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$\displaystyle W=\frac{3mg}{4a}$ , $\displaystyle\Delta U=\frac{mg}{2a}$
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$\displaystyle W=\frac{3mg}{a}$ , $\displaystyle\Delta U=\frac{mg}{2a}$
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$\displaystyle W=\frac{3mg}{4a}$ , $\displaystyle\Delta U=\frac{mg}{a}$
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$\displaystyle W=\frac{3mg}{4a}$ , $\displaystyle\Delta U=\frac{mg}{3a}$

Explanation

Force acting on the body of mass m=

$\vec { F } =2\left( a\times y-1 \right) \times m\times g$

Body will be ascenting until

$\vec { F } =0\ \Rightarrow 2\left( a\times y-1 \right) \times m\times g=0\ \Rightarrow a\times y-1=0\ \Rightarrow a\times y=1\ \Rightarrow y =\dfrac { 1 }{ a }$

So, the body will ascent until the height of y=

$\dfrac { 1 }{ a }$

Now, the work done by the force

$\vec {F}$ for half of the ascent height

$=\int _{ 0 }^{ \dfrac { 1 }{ 2a } }{ \vec { F } } dy\ =\int _{ 0 }^{ \dfrac { 1 }{ 2a } }{ 2\left( a\times y-1 \right) } \times m\times g\times dy\ =\dfrac { 3mg }{ 4a }$

Now, the potential energy stored in the body due to gravitational field force for half of ascent height

$=\dfrac{ mg }{ 2a }$

A force

$F = - K (y \widehat i + x \widehat j)$ (where

$K$ is positive constant) acts on a particle moving in the

$x-y$ plane. Starting from the origin, the particle is taken along the

$x$ -axis to the point

$(a, 0)$ and then parallel to

$y$ -axis to the point

$(0, a)$ . The total work done by the force

$F$ on the particle is:

0%
$- 2 Ka^2$
0%
$2 Ka^2$
0%
$- Ka^2$
0%
$Ka^2$

Explanation

Given

$\vec F = -K(y \vec i+x \vec j)$
Consider the small work done by the force

$F$ in displacing the particle through area

$dS$ as

$dW = \vec F.\vec {dS}$

where,

$\vec {dS} = dx \hat{i} + dy \hat j$

$\therefore dW = ( -K(y \hat i+x \hat j)).( dx \hat i + dy \hat j)$

$\therefore dW = -K(y dx + x dy) = - Kd(xy)$ .....................(1)

Since, first the particle is taken along the position x-axis to the point (a, 0) and then parallel to y-axis to the point (0, a) hence, integrating for whole area we get

$\displaystyle W = \int_{x=0}^{x=a} \int_{y=0}^{y=a} dW$

$\displaystyle W = \int_{x=0}^{x=a} \int_{y=0}^{y=a} (- Kd(xy))$ .............from(1)

$\displaystyle W = -K(\left | x \right |_{0}^{a})(\left | y \right |_{0}^{a})$

$W = -Ka^2$

A force

$F = - K (y \widehat i + x \widehat j)$ , (where

$K$ is a +ve constant) acts on a particle moving in xy plane starting from origin, the particle is taken along the positive x-axis to the point (

$a, 0$ ) and then parallel to y axis to the point (

$a, a$ ). The total work done by force

$F$ on the particle is

0%
$- 2 Ka^2$
0%
$2 Ka^2$
0%
$- Ka^2$
0%
$Ka^2$

Explanation

Work was done

$\int Fdt$

When particle moves from

$0$ to

$a$ along x-axis

Then

$W_1 = \displaystyle \int_0^a - (Ky \widehat i) dx = 0$

$\because\,\, y=0$

Now, when the particle moves parallel to the y-axis

$W_2 = \displaystyle \int_0^a - (Ka \widehat j) dy$

$\because\,\, x=a$

$\therefore W_2= - Ka \displaystyle \int_0^a dy = - Ka^2$

$\therefore W = W_1 + W_2 = - Ka^2$

How soon will the frame come to the orientation shown in figure (b) after collision?

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$\cfrac { \pi l }{ 4v }$
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$\cfrac {7 \pi l }{ 8v }$
0%
$\cfrac { \pi l }{ v }$
0%
$\cfrac { 7\pi l }{ 4v }$

Explanation

Consider the L to be a single system. Its COM lies in midway of the diagonal.
The initial velocity of ball be

$u$ and final velocity (in same direction) be

$v$ .
Linear momentum is always conserved. Momentum imparted to the L-system after collision is =

$m(u-v) = 2m(v_{Lcom})$

$v_{Lcom} = (u - v)/2$
About COM of L-system(just consider an axis about it) angular momentum of the 3 body system is conserved.

$L_i = mu(l/2)$ .

$L_f = mv(l/2) + 2m(v_{com})(0) + I_L\omega_L$

$I_L = 2 \times m(l/\sqrt{2})^{2} = ml^{2}$

$=> (u-v)/2 = l\omega$
velocity of separation of the balls is

$v_{Lcom}+(l/2)\omega-v$

$l\omega/2$ is horizontal component of velocity obtained by angular motion.
In elastic collision, velocity of approach = velocity of separation.
Therefore,

$u = v_{Lcom} +(u-v)/4 - v = 3(u-4)/4 - v$

$4(u+v) = 3(u-v) => v = -u/7$

$\omega = (u-v)/2l = 4u/7l$
time taken to reach orientation = angle/

$\omega$

$t = \pi/\omega = 7\pi l/4u$

A solid sphere rolls without slipping on a rough horizontal floor, moving with a speed

$v$ . It makes an elastic collision with a smooth vertical wall. After impact,

0%
it will move with a speed $v$ initially.
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its motion will be rolling without slipping.
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its motion will be rolling with slipping initially and its rotational motion will stop momentarily at some instant.
0%
its motion will be rolling without slipping only after some time.

Explanation

In rolling without slipping , no work is done against friction, hence loss in gravitational potential energy of a body is equal to its total kinetic energy, i.e. linear plus rotational kinetic energies.
Also

$v = \omega r$
Total kinetic energy =

$\dfrac{1}{2} mv^2 +\dfrac {1}{2} I \omega^2$

$= \dfrac {1}{2} mv^2 + \dfrac{1}{2}( mK^2) ( \dfrac {v^2}{r^2} )$

$\dfrac {1}{2} mv^2 [ 1 + \dfrac {K^2}{r^2} ]$
where K = radius of gyration
As all the bodies have the same final total kinetic energy, their final velocities will depend upon the ration K/r. bodies with the smaller value of K/r will have the greater v and hence will reach the bottom earlier.

Two blocks A and B of masses m and 2m respectively placed on a smooth floor are connected by a spring. A third body C of mass m moves with velocity

$v_0$ along the line joining A and B and collides elastically with A. At a certain instant of time after collision it is found that the instantaneous velocities of A and B are same then :

0%
the common velocity of A and B at time t $_0$ is v/3.
0%
the spring constant is k $= \displaystyle \frac{3mv_0^2}{2x_0^2}$ .
0%
the spring constant is k $= \displaystyle \frac{2mv_0^2}{3x_0^2}$ .
0%
none of these

Explanation

Since the masses of A and C are same and the collision is elastic, their velocities are exchanged after the collision. So, now the system is (A+B) and initially B is at rest wheras, A has velocity

${ v }_{ 0 }$ . After some time they have same velocities, so conserving momentum

${ mv }_{ 0 }=(m+2m)v$

$v =\dfrac { { v }_{ 0 } }{ 3 }$

The speed of the striking mass after collision is

0%
$u/2$ backwards
0%
$0$
0%
$u/3$ in same direction
0%
$u/7$ backwards

Explanation

Consider the L to be a single system. Its COM lies in midway of the diagonal.
The initial velocity of ball be

$u$ and final velocity (in same direction) be

$v$ .
Linear momentum is always conserved.

Momentum imparted to the L-system after collision is

$=m(u-v) = 2m(v_{Lcom})$

$v_{Lcom} = (u - v)/2$

About COM of L-system (just consider an axis about it) angular momentum of the 3 body system is conserved.

$L_i = mu(l/2)$

$L_f = mv(l/2) + 2m(v_{com})(0) + I_L\omega_L$

$I_L = 2\times m(l/\sqrt{2})^{2} = ml^{2}$

$=> (u-v)/2 = l\omega$

Velocity of separation of the balls is

$=v_{Lcom}+(l/2)\omega-v$

$l\omega/2$ is horizontal component of velocity obtained by angular motion.
In elastic collision, velocity of approach = velocity of separation.
Therefore,

$u = v_{Lcom} +(u-v)/4 - v = 3(u-4)/4 - v$

$4(u+v) = 3(u-v)\\ => v = -u/7$

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

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Practice Class 11 Medical Physics Quiz Questions and Answers

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