MCQ Questions for Class 11 Medical Physics Work, Energy And Power Quiz 12

A sphere

$A$ moving with speed

$u$ and rotating with an angular velocity

$\omega$ makes a head-on elastic collision with an identical stationary sphere

$B$ . There is no friction between the surfaces of

$A$ and

$B$ . Choose the correct alternative(s). Discard gravity.

0%
$A$ will stop moving but continue to rotate with an angular velocity $\omega$
0%
$A$ will come to rest and stop rotating
0%
$B$ will move with speed $u$ without rotating
0%
$B$ will move with speed $u$ and rotate with an angular velocity $\omega$ .

Explanation

Let

$m$ be the mass of sphere and

$v_1$ and

$v_2$ be the velocities of the spheres A and B respectively after the collision.

Applying conservation of linear momentum:

$P_i = P_f$

$m u + 0 = mv_1 + mv_2 \implies v_1 + v_2 = u$ ........(1)

Also

$\dfrac{v_2- v_1}{ u-0} = -1 \implies v_1 - v_2 = -u$

On solving we get,

$v_1 = 0$ and

$v_2 = u$

Thus A will stop and B will move with

$u$

Also as there is no friction between the A and B, thus there will be no torque. Hence angular velocities of the respective spheres must remains the same as it was initially.

So, A will continue to rotate with

$w$ whereas B will not rotate.

Find the horizontal component of the resultant force, which the axis will exert on the plate after the impact.

0%
$\displaystyle F=\frac{8Mv^2}{\displaystyle l{\left(1+\frac{4M}{3m}\right)}^2}$
0%
$\displaystyle F=\frac{4Mv^2}{\displaystyle l{\left(1+\frac{4M}{3m}\right)}^2}$
0%
$\displaystyle F=\frac{4Mv^2}{\displaystyle l{\left(1-\frac{4M}{3m}\right)}^2}$
0%
$\displaystyle F=\frac{8Mv^2}{\displaystyle l{\left(1-\frac{4M}{3m}\right)}^2}$

Explanation

Angular momentum conservation:

$\dfrac{l}{2}m(v-v^{'})=\dfrac{Ml^2}{3}w$
From above solution,

$v^{'}=\dfrac{3m-4M}{3m+4M}v$
Substituting in momentum equation, we get,

$w=\dfrac{4v}{l(1+\dfrac{4M}{3m})}$

Horizontal reaction force acting on the plate will be the centripetal force given by,

$dF=\dfrac{Mdx}{l}w^2x$

$F=\dfrac{Ml}{2}w^2$

$F=\dfrac{8Mv^2}{l(1+\dfrac{4M}{3m})^2}$

Answer: A

The work done by the tension T in the above process is

0%
$Zero$
0%
$T(L-L\cos \theta )$
0%
$-TL$
0%
$-TL\,\sin \theta$

Explanation

The instantaneous displacement is directed normal to

$T$ all the time.

Therefore

$W_T=0$ .

A force

$F=-6{x}^{3}$ is acting on a block moving along x-axis. Work done by this force is:

0%
Positive in displacing the block from $x=3$ to $x=1$ .
0%
Positive in displacing the block from $x=-3$ to $x=-1$ .
0%
Negative in displacing the block from $x=0$ to $x=4$ .
0%
Zero in displacing the block from $x=-2$ to $x=+2$ .

Explanation

For $(A):$ Displacement $< 0$ and $F<0$ so $W>0$

For $(B):$ Displacement $>0$ and $F>0$ so $W>0$

For $(C):$ Displacement $>0$ and $F<0$ so $W<0$

For $(D):$ $W=\int _{ -2 }^{ 2 }{ -6{ x }^{ 3 } } dx=\left( -\cfrac { 6 }{ 4 } \right) { \left( { x }^{ 4 } \right) }_{ -2 }^{ 2 }=0$

Find the angular velocity of the rod after the collision.

0%
$\displaystyle\omega=\frac{3v}{(4+\eta)l}$
0%
$\displaystyle\omega=\frac{12v}{(4+\eta)l}$
0%
$\displaystyle\omega=\frac{3v}{(4-\eta)l}$
0%
$\displaystyle\omega=\frac{12v}{(4-\eta)l}$

Explanation

Let mass of the disc be one unit.

By conservation of linear momentum,

$v=v^{'}+\eta v_o$

$v^{'}$ is the velocity of disc after collision

$v_o=\dfrac{v-v^{'}}{\eta}$

Conservation of angular momentum,

$v\dfrac{l}{2}=v^{'}\dfrac{l}{2}+\dfrac{\eta l^2}{12}w$

$w$ is the angular velocity of rod.

$w=\dfrac{6(v-v^{'})}{\eta l}$

Conservation of energy,

$\dfrac{1}{2}v^2=\dfrac{1}{2}v^{'2}+\dfrac{1}{2}\eta v_o^2+\dfrac{1}{2}\dfrac{\eta l^2}{12}w^2$

Substituting

$v_o$ and

$w$ in energy equation

$(v^2-v^{'2})= \eta(\dfrac{(v-v^{'})^2}{\eta^2}+3\dfrac{(v-v^{'})^2}{\eta^2})$

$(v^2-v^{'2})=\dfrac{4(v-v^{'})^2}{\eta}$

Since

$v \neq v^{'}$

$v+v^{'}=\dfrac{4(v-v^{'})}{\eta}$

$v^{'}=\dfrac{4-\eta}{4+\eta}v$

Thus we get

$w=\dfrac{6(v-v^{'})}{\eta l}$

and

$v^{'}=\dfrac{4-\eta}{4+\eta}v$

Substituting

$v^{'}$ in above equation, we get

$w=\dfrac{12v}{(4+\eta)l}$

Answer: B

Calculate the work done on the tool by

$\vec{F}$ if this displacement is along the straight line

$y =x$ that connects these two points.

0%
$2.50 J$
0%
$500 J$
0%
$50.6 J$
0%
$2 J$

Explanation

as ,

$dW=f.ds$

$W=\int { f.ds }$

and force is only along the y-axis, work-done along the x-axis is zero

$\displaystyle W=\int_{0,0}^{3,3} { -\alpha x{ y }^{ 2 }.dy } \cos 180^{\circ}$ and

$x= y$

$\displaystyle W=\int_{0}^{3} { \alpha { y }^{ 3 }.dy }$

$\displaystyle W=\alpha \int _{ 0 }^{ 3 }{ { y }^{ 3 }dy }$

$\displaystyle W=2.5*\{ \frac { { 3 }^{ 4 } }{ 4 } -0\}$

$W=50.6 J$

A particle slides along a track with elevated ends and a flat central part as shown in the Figure below. The flat portion BC has a length

$l=3.0 m$ . The curved portions of the track are frictionless. For the flat part the coefficient of kinetic friction is

$\displaystyle \mu _{k}= 0.20,$ the particle is released at point A which is at height

$h=1.5 m$ above the flat part of the track. Where does the particle finally comes to rest?

0%
The particle comes to rest at the center. of the flat part.
0%
The particle comes to rest at $\frac{1}{4}$ th distance from the point B of the flat part.
0%
The particle comes to rest at $\frac{3}{4}$ th distance from the point B of the flat part.
0%
The particle comes to rest at point B

Explanation

Work done by the particle in the flat part due to friction:

$W_{flat} = \mu mg s= 0.6mg$
KE at the point B:

$mg \times 1.5$
Hence, 0.6mg energy is lost in moving from B to C or C to B.
Hence, the particle will move from A->B->C->D->C->B->A->B->E where E is the mid point of BC since 1.2mg will be lost from A->B->C->D->C->B->A->B and 0.3mg will be lost in path B->E.
Thus, the particle will come to rest at the center.

The value of ratio

$M/m$ is

0%
$2:3$
0%
$3:2$
0%
$4:3$
0%
$3:4$

Explanation

$\Rightarrow \dfrac{M}{m}=\dfrac{4}{3}\Rightarrow \dfrac{4}{3}m$
1078625_218592_ans_db7e9b2ed9a3426da4cebad1748fa836.png

A force

$\displaystyle F= -\frac{k}{x^{2}}\left ( x\neq 0 \right )$ acts on a particle in x-direction. Find the work done by this force in displacing the particle from.

$\displaystyle x= +a\:$ to

$\:x= +2a.$ Here,

$k$ is a positive constant.

0%
$\displaystyle -\frac{k}{2a}$
0%
$\displaystyle \frac{k}{2a}$
0%
$\displaystyle -\frac{k}{a}$
0%
$\displaystyle +\frac{k}{a}$

Explanation

$\displaystyle W= \int F\:dx= \int_{+a}^{+2a}\left ( \frac{-k}{x^{2}} \right )dx= \left [ \frac{k}{x} \right ]_{+a}^{+2a}$

$\displaystyle = -\frac{k}{2a}$
Note: It is important to note that work comes out to be negative which is quite obvious as the force acting on the particle is in negative x-direction

$\displaystyle \left ( F= -\frac{k}{x^{2}} \right )$ while displacement is along positive x-direction. (from x=a to x=2a)

An object is displaced from position vector

$\displaystyle \vec{r}_{1}= \left ( 2\hat{i}+3\hat{j} \right )m\:$ to

$\:\vec{r}_{2}= \left ( 4\hat{i}+6\hat{j} \right )$ m under a force

$\displaystyle \vec{F}= \left ( 3x^{2}\hat{i}+2y\hat{j} \right )N.$ Find the work done by this force.

0%
$83 J$
0%
$41.5 J$
0%
$166 J$
0%
$164 J$

Explanation

$\displaystyle W= \int_{\vec{r}_{1}}^{\vec{r}_{2}}\vec{F}\cdot \vec{dr}$

$\displaystyle = \int_{\vec{r}_{1}}^{\vec{r}_{2}}\left ( 3x^{2}\hat{i}+2y\hat{j} \right )\cdot \left ( dx\hat{i}+dy\hat{j}+dz\hat{k} \right )$

$\displaystyle = \int_{\vec{r}_{1}}^{\vec{r}_{2}}\left ( 3x^{2}dx+2ydy \right )= \left [ x^{3}+y^{2} \right ]_{\left ( 2,3 \right )}^{\left ( 4,6 \right )}$

$=83 J$

Which of the following statements is correct regarding the work done

$\vec{F}$ along these two paths.

0%
Work done on x-axis is zero
0%
Work done on x-axis is less than on y-axis
0%
Work done on x-axis is more than on y-axis but not zero
0%
Data insufficient

Explanation

since force is applied in -ve y-axis and angle between x & y axis is

${ 90 }^{ o }$
and

$dW=\vec{F}\cdot \vec{ds}$

$W=0$ on x-axis
Alternatively, since

$y=0$ along the path from

$x=0$ ( origin) to

$x=3$ ,

$\vec{F}=\alpha xy^2 \hat{j}=0$

$\Rightarrow W=\vec{F}\cdot \vec{ds}=0$
Similarly,

$W=0$ along y-axis.

Energy stored in the bonds of molecules is a form of:

0%
heat energy
0%
potential energy'
0%
chemical energy
0%
light energy

The work done by the varying force in changing the angular displacement from 0 to

$\theta$ is

0%
$Wh$
0%
$FL\,\sin \theta$
0%
$Fh$
0%
$\dfrac{1}{2}FL\,\sin \theta$

Explanation

Because the process is quasi static, work is equal to increase of potential energy.

$Work =mgh= Wh$ .

A cutting tool under microprocessor control has several forces acting on it. One force is

$\vec{F}=-\alpha xy^2\hat{j}$ , a force in the negative y-direction whose magnitude depend on the position of the tool. The constant is

$\alpha = 2.50\ N/m^3$ . Consider the displacement of the tool from the origin to the point

$x = 3.00 \,m, y = 3.00 \,m$ . Calculate the work done on the tool by

$\vec{F}$ if the tool is first moved out along the x-axis to the point

$x = 3.00m, \:y= 0m$ and then moved parallel to the y-axis to

$x = 3.00m, y = 3.00 \,m$ .

0%
$67.5 J$
0%
$85 J$
0%
$102 J$
0%
$7.5 J$

Explanation

as ,

$dW=f.ds$

$W=\int { \vec{F}.\vec{ds} }$
and force is only along y-axis ,work done along x- axis is zero
work done in moving from

$(3,0)$ to

$(3,3)$ is

$\displaystyle W=\int_0^3 { -\alpha x{ y }^{ 2 }.dy } \cos 180^{\circ}$ and

$x= 3$

$\displaystyle W=\int_0^3 { \alpha \times 3 \times { y }^{ 2 }.dy }$

$\displaystyle W=2.5 \times 3 \times \int _{ 0 }^{ 3 }{ { y }^{ 2 }dy }$

$\displaystyle W=7.5\times\{ \dfrac { { 3 }^{ 3 } }{ 3 } -0\}$

$W=67.5 \ J$

A uniform rod of mass

$m$ and length

$l$ is pivoted at point

$O$ . The rod is initially in vertical position and touching a block of mass

$M$ which is a rest on a horizontal surface. The rod is given a slight jerk and it starts rotating about point

$O$ This causes the block to move forward as shown. The rod loses contact with the block at

$\theta =30^{\circ}$ All surfaces are smooth. Now answer the following questions. The velocity of block when the rod loses contact with the block is

0%
$\displaystyle \frac{3gl}{4}$
0%
$\displaystyle \frac{5gl}{4}$
0%
$\displaystyle \frac{6gl}{4}$
0%
$\displaystyle \frac{7gl}{4}$

Explanation

$\displaystyle \frac{M}{m}=\frac{4}{3} \Rightarrow M=\frac{4}{3} m$

$\displaystyle \frac{\omega l}{2}=v \therefore \omega =\frac{2v}{l}$
Decrease in potential energy of rod

$=$ increase in rotational kinetie energy of rod + translational kinetie energy of block

$\displaystyle \therefore mg\left( \frac{l}{2}-\frac{l}{2}\sin 30^{\circ} \right) =\frac{1}{2} \left( \frac{ml}{3} \right)^{2} \left( \frac{2v}{l} \right)^{2}+\frac{1}{2} \left( \frac{4m}{3} \right) (v^{2})$
Solving this equation. We get

$\displaystyle v=\frac{\sqrt{3 gl}}{4}$

A shown in figure there is a spring block system. Block of mass

$500$ g is pressed against a horizontal spring fixed at one end to compress the spring through

$5.0$ cm. The spring constant is

$500$ N/m. When released, the block moves horizontally till it leaves the spring. Calculate the distance where it will hit the ground

$4$ m below the spring?

0%
$6 m$
0%
$4 m$
0%
$8 m$
0%
$\sqrt { 2 }$ m

Explanation

Mass of the block

$m = 0.5 kg$

When block released, the block moves horizontally with speed

$v$ till it leaves the spring.

By energy conservation, we get

$\displaystyle \dfrac{1}{2} kx^2\, =\, \displaystyle \dfrac{1}{2}mv^2$

where

$x = 0.05 m$ is the compression of the spring,

$k = 500 N/m$ is the spring constant.

$\therefore$ We get

$v^2\, =\, \displaystyle \dfrac{kx^2}{m}\, \quad\, \Rightarrow\, \quad\, v\, =\, \sqrt{\displaystyle \dfrac{kx^2}{m}}$

Time of flight

$t\, =\, \sqrt{\displaystyle \dfrac{2H}{g}}$

where

$H = 4 m$

So, horizontal distance traveled from the free end of the spring

$R = v\, \times\, t$

$\therefore$

$R=\, \sqrt{\displaystyle \dfrac{kx^2}{m}}\, \times\, \sqrt{\displaystyle \dfrac{2H}{g}}$

$R=\, \sqrt {\displaystyle \dfrac{500\, \times\, (0.05)^2}{0.5}}\, \times\, \sqrt{\displaystyle \dfrac{2\, \times\, 4}{10}}\, =\, \sqrt{2}\, m$

So, the block will hit the ground at a horizontal distance of

$\sqrt{2} m$ from the free end of the spring.

A ball of mass '

$m$ ' moves perpendicularly to a wall with a speed

$v$ , strikes it and rebounds with the same speed in the opposite direction. What is the direction and magnitude of the average force acting on the ball due to the wall?

0%
$2 m{v}/{t}$ away from the wall
0%
$3 m{v}/{t}$ away from the wall
0%
$1 m{v}/{t}$ away from the wall
0%
$4 m{v}/{t}$ away from the wall

An athlete in the Olympic games covers a distance of

$100$ m in

$10$ s. His kinetic energy can be estimated to be in the range. (Assume weight = 60kg)

0%
$200J-500J$
0%
$2\times 10^5J-3\times 10^5J$
0%
$20000J-50000J$
0%
$2000J-5000J$

Explanation

Velocity

$V=\dfrac{distance}{time}=\dfrac{100}{10}=10 m/s$

Assuming his mass to be 60kg, his kinetic energy is

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

$= \dfrac{1}{2}\times 60\times 100$

$=60\times50$

$=3000 J$

Hence, Option D

From

$x=0$ to

$x=6$ , the force experienced by an object varies according to the function

$F\left(x\right)=\sqrt{6x-{x}^{2}}$ , as shown above. What is the work done by this force as the object moves from

$x=0$ to

$x=3$ ?
Assume all numbers are given in standard units.

0%
$0 J$
0%
$4.5 J$
0%
$7.1 J$
0%
$9.0 J$
0%
$14.0 J$

Explanation

There is a simpler method to arrive at the solution.

Conceptually, you are integrating force over displacement, so if you are given the graph, the integration is equal to the area under the curve. The plot given is one of a semicircle.

The function

$F\left( x \right) =\sqrt { 6x-{ x }^{ 2 } }$ is one of a semicircle.

It may be easier to see this when it is in the form

$F(x)=\sqrt{9+6x-x^2-9}$ , which corresponds to

$F(x)=3^2-(x-3)^2$ which is the upper half of a circle centered at

$x = 3$ with radius

$3$ .

The area of a circle is given by

$A=\pi r^2$ , and in this case you are looking for the area under half of the top half.

Thus,

$W=\dfrac{1}{4}A=\dfrac{1}{4}\pi (3)^2\approx7.0658\ J$ , that is choice "C".

A moving coin hits another coin and sets it into motion. In this case, energy from the moving coin is _________ to the other coin.

0%
Flown
0%
Taken
0%
Not transferred
0%
Transferred

A helicopter flying in the air has :

0%
Only kinetic energy but not potential energy
0%
Only potential energy but not kinetic energy
0%
Both potential and kinetic energy
0%
Neither kinetic nor potential energy

A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m while applying the force and the ball goes up to 2 in height further, find the magnitude of the force. Consider g =

$10 m/s^2$

0%
16 N
0%
20 N
0%
22 N
0%
180 N

Explanation

Mass ,

$m=0.2 kg$

Total height,

$h=0.2+2=2.2 m$

Work done =Difference in potential energy.

$F.S=mgh$ where S is the distance for ehich the force is applied by hand,

$S=0.2m$

$F=\dfrac{mgh}{S}=\dfrac{0.2\times 10\times 2.2}{0.2}$

$F=22N$

A spherical ball A of mass

$4\ kg$ , moving along a straight line strikes another spherical ball B of mass

$1\ kg$ at rest. After the collision, A and B move with velocities

$v_{1}\ ms^{-1}$ and

$v_{2}\ ms^{-1}$ respectively making angles of

$30^{\circ}$ and

$60^{\circ}$ with respect to the original direction of motion of ball A. The ratio

$\dfrac {v_{1}}{v_{2}}$ will be:

0%
$\dfrac {\sqrt {3}}{4}$
0%
$\dfrac {4}{\sqrt {3}}$
0%
$\dfrac {1}{\sqrt {3}}$
0%
$\sqrt {3}$

Explanation

Since there is no momentum in the line perpendicular to the direction of motion of ball A before collision, after collision also there would be no momentum along that line. So by conservation of momentum along that line we have:

$0=4{ v }_{ 1 }\sin { 30° } -{ v }_{ 2 }\sin { 60° } \Rightarrow \dfrac { { v }_{ 1 } }{ { v }_{ 2 } } =\dfrac { \sqrt { 3 } }{ 4 }$

So, option A is the correct answer.

Under the action of a force,

$2kg$ body moves such that its position

$x$ as a function of time is given by

$x={t}^{3}/3$ where

$x$ is in meters and

$t$ in seconds. The work done by the force in the first two seconds is:

0%
$1.6J$
0%
$16J$
0%
$160J$
0%
$1600J$

A car of mass

$m$ starts moving so that its velocity varies according to the law

$v=\beta \sqrt { s }$ , where

$\beta$ is a constant, and

$s$ is the distance covered. The total work performed by all the forces which are acting on the car during the first

$t$ seconds after the beginning of motion is:

0%
${ m\beta }^{ 4 }{ t }^{ 2 }/8$
0%
${ m\beta }^{ 2 }{ t }^{ 4 }/8$
0%
${ m\beta }^{ 4 }{ t }^{ 2 }/4$
0%
${ m\beta }^{ 2 }{ t }^{ 4 }/4$

Explanation

$v=\beta \sqrt s$

$\dfrac{ds}{dt}=\beta \sqrt s$

On integration

$2\sqrt s=\beta t$

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

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

On differentiation with respect to t

$a=\dfrac{\beta^2}{2}$

Work =F.s=mas=

$\dfrac{m\beta^4 t^2}{8}$

Work done in time

$t$ on a body of mass

$m$ which is accelerated from rest to a speed

$v$ in time as a function of time

$t$ is given by

0%
$\cfrac { 1 }{ 2 } m\cfrac { v }{ { t }_{ 1 } } { t }^{ 2 }\quad$
0%
$m\cfrac { v }{ { t }_{ 1 } } { t }^{ 2 }$
0%
$\cfrac { 1 }{ 2 } m{ \left( \cfrac { v }{ { t }_{ 1 } } \right) }^{ 2 }{ t }^{ 2 }$
0%
$\cfrac { 1 }{ 2 } m\cfrac { v }{ { t }_{ 1 }^{ 2 } } { t }^{ 2 }$

Explanation

Ans : We know

$W = F \times s$

where,

$F = ma$

By using Kinematics,

$s = \dfrac{1}{2}at^{2}$ , if u =0

So,

$W = (ma).(\dfrac{1}{2}at^{2})$

$W = \dfrac{1}{2}ma^{2}t^{2}$ where

$a = v/t_{1}$

$\boxed{W = \dfrac{1}{2}m(\dfrac{v}{t_{1}})^{2}t^{2}}$

Which of the following potential energy curves possibly describe the elastic of two billiard balls?
Here

$r$ is the distance between centres of the balls.

Explanation

The potential energy of a system of two masses varies inversely as the distance

$(r)$ between them.
i.e.

$V(r) \propto \dfrac {1}{r}$
When the two billiard balls touch each other, potential energy becomes zero i.e. at

$r = R + R = 2R, V(r) = 0$
Out of the given graphs, only curve

$(c)$ satisfies these conditions. Therefore, all other curves cannot possibly describes the elastic collision of two billiard balls.

Two small particles of equal masses start moving in opposite directions from a point

$A$ in a horizontal circular orbit. Their tangential velocities are

$v$ and

$2v$ respectively, as shown in the figure. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at

$A$ , these two particles will again reach the point

$A$ ?

0%
$4$
0%
$3$
0%
$2$
0%
$1$

Explanation

$\textbf{Hint:}$ Collision is elastic and mass is same. In elastic collision, the velocity of the particles will interchange.

$\textbf{Step 1:Calculation of angle at which collision occure}$

At time

$t$ , particle will collide in such manner,

$R\theta=v\,t$ ................i

and

$R\left( 2\pi-\theta \right)=2v\,t$ ................. ii

From equation i and ii, we get

$\dfrac{\theta}{\left(2\pi -\theta\right)}=\dfrac{v\,t}{2v\,t}$

$2\theta=2\pi -\theta$

So,

$\theta = \dfrac{2\pi}{3}$

$\theta=120^o$

$\textbf{Step 2:Number of collision}$

As a result, the particle with speed

$v$ will traverse

$120^o$ before colliding for the first time.

The velocities will then swap, and the particle with speed v will now have speed

$2\,v$ , and so on. Before the second impact, the particle with speed v will travel at an angle of

$120^o$ . The pace will be altered once more.

They will arrive at point A immediately before the third collision following two collisions.

$\textbf{Option C is correct.}$

By applying a force

$\vec{F} = (3xy - 5z)\hat{j} + 4z\hat{k}$ a particle is moved along the path

$y=x^{2}$ from point

$(0,0,0)$ to point

$(2,4,0)$ . The work done by the

$F$ on the particle is

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

The bob

$A$ of a pendulum of mass

$m$ released from horizontal to the vertical hits another bob

$B$ of the same mass at rest on a table as shown in figure. If the length of the pendulum is

$1\ m$ , what is the speed with which bob

$B$ starts moving. (Neglect the size of the bobs and assume the collision to be elastic) (Take

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

0%
$4.47\ ms^{-1}$
0%
$5.47\ ms^{-1}$
0%
$6.47\ ms^{-1}$
0%
$3.47\ ms^{-1}$

Explanation

As the collision is elastic and two balls have the same mass, therefore ball

$A$ transfers its entire momentum to the ball

$B$ and ball

$A$ does not rise at all. After a collision, ball

$A$ comes to rest and ball

$B$ moves with speed of ball

$A$ .

$\therefore$ The speed with which bob

$B$ starts moving is

$v = \sqrt {2gh}$

$= \sqrt {2\times 10\ ms^{-2} \times 1m} = \sqrt {20}\ ms^{-1}$

$= 4.47\ ms^{-1}$

Speed of

$C$ just after collison is

0%
$2 m/s$
0%
$2 \sqrt{2} m/s$
0%
$5 m/s$
0%
$(\sqrt{2} - 1) m/s$

Explanation

R.E.F image

Let I be the impulse generated

on collision

As ball A and B are connected

by strings, there will no vertical

components of velocity after

collision

So, their Gorizontal velocities are

equal to 3 m/s

Then,

$\dfrac{J}{\sqrt{2}} = (2)(3)-2(0)$

$J = 2(3)\sqrt{2}$ kgm/s

The impulse of

$\sqrt{2}J$ is acted

upon the ball i on collision, so

find velocity of c is,

$\sqrt{2}J = (1) v_{f} = (1)v_{f}-(1)(-10)$

$\sqrt{2}(2)(3)(\sqrt{2}) = v_{f}+10$

$12-10 = v_{f}$

$\Rightarrow 2 m/s = v_{f}$

Option

$-A$ is correct

1107072_981768_ans_e12f1a522f8f4432955f9d1232017bff.jpg

A man is standing on a plank which is placed on a smooth horizontal surface. There is sufficient friction between the feet of man and plank. Now man starts running over plank, correct statement is/are

0%
Work done by friction on the man with respect to the ground is negative.
0%
Work done by friction on the man with respect to the ground is positive.
0%
Work done by friction on the plank with respect to the ground is positive.
0%
Work done by friction on the man with respect to the plank is zero.

Two bodies

$A$ and

$B$ have masses

$20\ kg$ and

$5\ kg$ respectively. If they acquire the same kinetic energy. Find the ratio of thier velocities.

0%
$\dfrac {1}{2}$
0%
$2$
0%
$\dfrac {2}{5}$
0%
$\dfrac {5}{6}$

Explanation

Given:

$m_{A} = 20 kg$

$m_{B} = 5 kg$

Kinetic Energy of both bodies A and B are the same.

Thus,

$\dfrac{1}{2} m_{A} V_{A}^{2} = \dfrac{1}{2} m_{B} V_{B}^{2} \\$

$\dfrac{V_{A}^{2}}{V_{B}^{2}} = \dfrac{m_{B}}{m_{A}} \\$

$\dfrac{V_{A}^{2}}{V_{B}^{2}} = \dfrac{5}{20}\\$

$\dfrac{V_{A}}{V_{B}} = \dfrac{1}{2}$

So, option A is correct.

A body is acted upon by a force which is proportional to the distance covered. If distance covered is represented by s, then work done by the force will be proportional to.

0%
s
0%
$s^2$
0%
$\sqrt{s}$
0%
None of the above

Explanation

Given that-

$F\propto s$

$\implies F=ks$ where

$k=$ proportionality constant

Now, we know that, Work done

$W=F.s$

$\implies W=ks^2$

$\implies F\propto s^2$

Answer-(B)

A particle of mass

$m$ is moving horizontally with a constant velocity

$v$ towards a rigid wall that is moving in opposite direction with a constant speed

$u$ . Assuming elastic impact between the particle and wall the work done by the wall in reflecting the particle is equal to:

0%
$\left( \dfrac{1}{2} \right) m{ \left( u+v \right) }^{ 2 }\quad \quad$
0%
$\left( \dfrac{1}{2} \right)4 mv\left( { u }+{ v } \right)$
0%
$\dfrac{1}{2}muv$
0%
None of these

Explanation

The correct answer is option (B).

Hint: Make a pictorial representation and then calculate the speed.

Step 1: Make a representation of the given data.

Let speed of the wall be

$v_w=u$

Let the speed after collision be

$v'$

According to the above diagram,

$\dfrac{v'-v_w}{-v-v_w}=-e$

$\dfrac{v'-u}{-v-u}=-1$

$v'=v+2u$

Step 2: Calculate the work done

$W=\dfrac{1}{2}(mv'^2)-\dfrac{1}{2}mv^2$

$W=\dfrac{1}{2}m(v+2u)^2-\dfrac{1}{2}mv^2$

$W=\dfrac{1}{2}(2u)(2v+2u)$

$W=\dfrac{1}{2}4mu(u+v)$

Hence, the work is $W=\dfrac{1}{2}4mu(u+v)$

The correct answer is option (B).

A particle is taken from A to point B under the influence of a force field. Now it is taken from B to A and it is observed that the work done in taking the particle from A to B is not equal to the work done in taking it from B to A. If

${W}_{nc}$ and

${W}_{C}$ are the work done by non-conservation and conservative force field respectively and

$\Delta U$ and

$\Delta K$ , be the change in P.E and K.E then

0%
${ W }_{ nc }-\Delta U=\Delta K$
0%
${ W }_{ C }=-\Delta U$
0%
${ W }_{ nc }+{ W }_{ C }=\Delta K$
0%
${ W }_{ nc }=-\Delta U=-\Delta K$

Two infinitely large sheets having charge densities

$\sigma_1$ and

$\sigma_2$ respectively

$(\sigma_1 > \sigma_2)$ are placed near each other separated by distance '

$d$ '. A small change '

$q$ ' is placed in between two plates such that there is no effect on charge distribution on plates. Now this charge is moved at an angle of

$45^o$ with the horizontal towards plate having charge density

$\sigma_2$ by distance '

$a$ '

$(a << d)$ . Find the work done by electric field in the process.

0%
$\dfrac{qa(\sigma_1 - \sigma_2)}{5\sqrt{2}\in_0}$
0%
$\dfrac{qa(\sigma_1 - \sigma_2)}{2\sqrt{2}\in_0}$
0%
$\dfrac{qa(\sigma_1 - \sigma_2)}{3\sqrt{2}\in_0}$
0%
$\dfrac{qa(\sigma_1 - \sigma_2)}{4\sqrt{2}\in_0}$

Explanation

${ E }_{ 1 }=\dfrac { { \sigma }_{ 1 } }{ { 2E }_{ 0 } }$

${ E }_{ 2 }=\dfrac { { \sigma }_{ 2 } }{ { 2E }_{ 0 } }$

Net electric field

$=\left( \dfrac { { \sigma }_{ 1 }-{ \sigma }_{ 2 } }{ { 2E }_{ 0 } } \right)$

Force on charge

$=q\left( \dfrac { { \sigma }_{ 1 }-{ \sigma }_{ 2 } }{ { 2E }_{ 0 } } \right)$

Work done

$=$ force

$\times$ displacement in the direction of force

$W=\dfrac { q\left( { \sigma }_{ 1 }-{ \sigma }_{ 2 } \right) }{ { 2E }_{ 0 } } \times a\cos{ 45 }^{ 0 }$

$W=\dfrac { qa\left( { \sigma }_{ 1 }-{ \sigma }_{ 2 } \right) }{ 2\sqrt { 2 } { E }_{ 0 } }$

1107158_1013640_ans_2292d61476654935ab27b9aa7a60f490.png

The position

$x$ of a particle moving along

$x-$ axis at time

$(t)$ is given by the equation

$t=\sqrt x+2$ , where

$x$ is in metres and

$t$ in seconds. Find the work done by the force in first four seconds.

0%
$Zero$
0%
$2\ J$
0%
$4\ J$
0%
$8\ J$

Explanation

$x=(t-2)^2=t^2-4t+4$

$v=2t-4$

$a=2$

F=m.a.x

m=1kg

$2\times (4^2-4\times 4+4)=8J$

The graph below represents the relation between displacement

$x$ and force

$F$ . The work done in displacing an object from

$x=8\ m$ to

$x=16\ m$ is approximately.

0%
$25\ J$
0%
$40\ J$
0%
$8\ J$
0%
$16\ J$

The velocity of a body of mass

$20 kg$ decreases from

$20 m/s$ to

$5 m/s$ in a distance of

$100 m$ . Force on the body is :

0%
$-27.5 N$
0%
$-47.5 N$
0%
$-37.5 N$
0%
$-67.5 N$

Explanation

The equation of motion is given as,

${v^2} = {u^2} + 2as$

${\left( 5 \right)^2} = {\left( {20} \right)^2} + 2 \times a \times 100$

$a = - 1.875\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$

The force is given as,

$F = ma$

$F = 20 \times \left( { - 1.875} \right)$

$F = - 37.5\;{\rm{N}}$

Thus, the force on the body is $- 37.5\;{\rm{N}}$ .

What is the amount of work done in raising a glass of water weighing

$0.5 kg$ through a height of

$50 cm ? (g = 10 m/s ^2$ )

0%
$1 J$
0%
$0.4 J$
0%
$0.20 J$
0%
$2.5 J$

Explanation

The amount of the work done in raising the glass of the water through a height is calculated by the amount of potential energy

$W = mgh$

By substituting the value in the above equation we get

$W = 0.5 \times 10 \times 0.5$

Hence the work done in raising the glass is ${\rm{2}}{\rm{.5J}}$

When a capillary tube of radius r is immersed in a liquid of density

$\rho$ , the liquid rises to a height h in it. If m is the mass of the liquid in the capillary tube, the P.E. of this mass of the liquid in the tube is :

0%
$\cfrac{mgh}{4}$
0%
$\cfrac{mgh}{2}$
0%
mgh
0%
2mgh

Explanation

$M = \pi r^2 h \rho$

Now

Surface tension =

$\dfrac{h\rho gr}{2} = \dfrac{h'\rho gr'}{2}$

Where h' is the height to which water rised in the second tube and r' its radius

Since

$r' = 2r, h' = \dfrac{h}{2}$

$M' = \pi r'^{2}h' \rho$

$= \pi(2r)^2 \dfrac{h}{2} \rho = 2\pi r^{2} h \rho = 2m$

So, P.E. = 2Mgh

Hence (D) option is correct

An object of mass

$M_1$ moving horizontally with speed u collides elastically with another object of mass

$M_2$ at rest. Select correct statement.

0%
The momentum of system is conserved only in direction PQ.
0%
Momentum of $M_1$ is conserved in direction perpendicular to SR.
0%
Momentum of $M_2$ is conserved in direction perpendicular to CR.
0%
All of these

The P.E. of a certain spring certain when stretched from natural length through a distance

$0.3 m$ is

$10 J$ . The amount of work in joule that must be done on this spring to stretch it through an additional distance

$0.15 m$ will be :

0%
$10 J$
0%
$20 J$
0%
$7.5 J$
0%
$12.5 J$

Explanation

The initial potential energy of the spring is given as,

${U_i} = \dfrac{1}{2}k{x^2}$

$10 = \dfrac{1}{2}k{\left( {0.3} \right)^2}$

$k = \dfrac{{20}}{{0.09}}$

The final potential energy of the spring is given as,

${U_f} = \dfrac{1}{2}k{x_1}^2$

$= \dfrac{1}{2} \times \dfrac{{20}}{{0.09}}{\left( {0.45} \right)^2}$

$= 22.5\;{\rm{J}}$

The amount of work done is given as,

$W = {U_f} - {U_i}$

$= 22.5 - 10$

$= 12.5\;{\rm{J}}$

Thus, the amount of work done is $12.5\;{\rm{J}}$ .

If increase in linear momentum of a body is 50%, then change in its kinetic energy is

0%
25%
0%
125%
0%
150%
0%
50%

A ball of radius r moving with a speed v collides elastically with another identical stationary ball. The impact parameter for the collision is b as shown in figure.

0%
The balls must scatter at right angles for $0 < b \le 2r.$
0%
For a head on collision b must be zero, and for an oblique collision $$0
0%
After collisions, ball-1 will comes to rest and ball-2 move at an angle $\sin^{-1} (b/r)$ below the x-axis
0%
After collision ball-1 and 2 will move at angles $cos^{-1} (b/r)$ and $$\sin^{-1} (b/2r) above and below the x-axis respectively.

In the figure shown initially spring is in uninstructed state & blocks are at rest. Now

$100N$ force is appiled on block

$A$ and

$B$ as shown in figure. After some time velocity of

$'A'$ becomes

$2m/s$ and that of

$'B'\ 4m/s$ and block

$A$ displaced by amount

$10\ cm$ and spring is streched by amount

$30\ cm$ . then find the work done by sprig force on

$A$ .

0%
$9/3\ J$
0%
$-6\ J$
0%
$6\ J$
0%
$-2 J$

Find the value of

$u$

0%
$\sqrt{\dfrac{9}{2}Rg}$
0%
$\sqrt{3Rg}$
0%
$\sqrt{\dfrac{7}{2}Rg}$
0%
$\sqrt{7Rg}$

Two steel spheres approach each other head on with the same speed and collide elastically. After the collision one of the sphere's of radius r comes to rest, the radius of the other sphere is

0%
$\frac{r}{{{{\left( 3 \right)}^{\frac{1}{3}}}}}$
0%
$\frac{r}{3}$
0%
$\frac{r}{9}$
0%
${\left( 3 \right)^{\frac{1}{2}}}r$

A ball of mass 10 g is projected with an initial velocity of 10 m/s, comes back withd a velocity of 5 m/s at the point of projection. Find the work done by air resistance.(Neglect buoyancy force due to air)

0%
0.375 J
0%
0.275 J
0%
-0.375 J
0%
Zero

CBSE Questions for Class 11 Medical Physics Work, Energy And Power Quiz 12 - 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 12 - MCQExams.com

0:0:1

Practice Class 11 Medical Physics Quiz Questions and Answers

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