MCQ Questions for Class 11 Medical Physics Work, Energy And Power Quiz 2

What is potential energy?

0%
Energy of an object due to its position or arrangement in a system
0%
Energy of an object due to its nature or arrangement in a system
0%
Energy of an object due to its shape or arrangement in a system
0%
None

State whether the given statement is True or False :

The potential energy of a free falling object decreases progressively.

0%
True
0%
False

A body of mass 0.1 kg is dropped from a height of 10 m at a place wheres

$g= 10\, m\, s^{-2}$ . Its KE just before its strikes the ground is:

0%
1 J
0%
1.04 J
0%
3.5 J
0%
10 J

Explanation

Here, by conservation of energy, the potential energy of the body at height

$h$ is converted into kinetic energy before its strikes the ground.

Thus,

$mgh=KE$

$KE=0.1(10)10=10 J$

A ball with mass m and speed

$V_0$ hit a wall and rebounds back with same speed.
Calculate the change in the object's kinetic energy.

0%
$-mv_0 ^2$
0%
$- \frac{1}{2}mv_0 ^2$
0%
Zero
0%
$\frac{1}{2}mv_0 ^2$
0%
$mv_0 ^2$

Explanation

The speed of the ball remains the same

$v_0$ before and after the collision with wall.

Thus the kinetic energy remains

$\dfrac{1}{2}mv_0^2$ .

Since kinetic energy is a scalar quantity, the change in it is zero.

Mechanical energy is the form of which two energies:

0%
Potential and Kinetic
0%
Thermal and Nuclear
0%
Potential and Sound
0%
Light and Sound

What is kinetic energy?

0%
Energy possessed by a body by the virtue of its motion
0%
Energy possessed by a body by the virtue of its shape
0%
Energy possessed by a body by the virtue of its size
0%
None

Explanation

Correct Option: A
Explanation: The energy which a body possesses by virtue of being in motion is called as kinetic energy. If we want to stop a moving body then we must do some work on the body to stop it which means that a moving body possesses energy because of motion.

If an object of mass

$m$ moving with the velocity

$v$ , The kinetic energy is defined as,

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

Hence, Option A is correct.

Refer to the given figures and select the correct option regarding these, when they are operating.

0%
Object I converts mechanical energy into kinetic energy whereas object II converts electrical energy into potential energy
0%
Object II converts electrical energy into kinetic, heat and sound energy
0%
Object I converts potential energy into kinetic energy
0%
Both B and C

A force of

$16N$ is distributed uniformly on one surface of a cube of edge

$8cm$ . The pressure on this surface is

0%
$3500Pa$
0%
$2500Pa$
0%
$4500Pa$
0%
$5500Pa$

Explanation

$F=16N$

$A=8\times 8\times { 10 }^{ -4 }{ m }^{ 2 }$

$P=\cfrac { F }{ A } =\cfrac { 16 }{ 64\times { 10 }^{ -4 } } =\cfrac { 1000 }{ 4 } =2500Pa$

If amount of heat supplied is Q, work done is W and change in internal energy is

$mC_V dT$ , then relation among them is

$(C_V$ gram specific heat)

0%
$mC_V dT = Q + W$
0%
$Q = W + mC_V dT$
0%
$Q + mC_V dT = W$
0%
None of these

Explanation

Given,

Heat supplied

$=Q$

Work done

$=W$

and change in internal energy

$mCvdT$

$A/Q$ to first law of thermodynamics,

Change in internal energy,

$dv=dQ-dW.......(i)$

putting the above values in equation

$(i)$ we get,

$mCvdT=Q-W$

$Q=W+mCvdT$

Hence, Option

$B$ is correct.

A child builds a tower from three blocks. The blocks are uniform cubes of side

$2cm$ . The blocks are initially lying on the same horizontal surface and each block has a mass of

$0.1kg$ . The work done by the child is (in J)

0%
$4$
0%
$0.04$
0%
$6$
0%
$0.06$

Explanation

Total work done is change in kinetic energy plus change in potential energy. Since, change in kinetic energy is zero. So total work done by child is change in potential energy of blocks.

Assuming center of mass of blocks initially is at 1cm level and value of g=10

Change in height of center of mass for-

Block 1- Zero

Block2- 1 cm to 3cm= 2cm

Block 3- 1cm to 5cm= 4cm

Potential energy is-

$E=mg\triangle h\\ \\ E\quad (block1)=0.1\times 10\times 0=0J\\ E\quad (block2)=0.1\times 10\times (\frac { 2 }{ 100 } )=0.02J\\ E\quad (block3)=0.1\times 10\times (\frac { 4 }{ 100 } )=0.04J\\ Adding\quad all=\quad 0+0.02+0.04=\quad 0.06J$

A bread gives a boy of mass

$40kg$ an energy of

$21kJ$ . If the efficiency is

$28$ % then the height can be climbed by him using this energy is:

0%
$5 cm$
0%
$10 cm$
0%
$15 cm$
0%
$20 cm$

Explanation

the total energy dilvered by boy with efficiency is

$TE=E*\eta$

$TE=21*10^3*0.28=5880J$

The total height climbed by boy is

$TE=mgh$

$5880=40*9.8*h$

$h=15m$

In the phenomenon of work done by variable forces, the forces:

0%
remain constant
0%
don't remain constant
0%
increase
0%
decrease

Work-energy theorem is valid in the presence of

0%
External forces only
0%
Internal forces onlhy
0%
Conservative forces only
0%
All type of forces

The velocity of the block after covering

$3m$ distance under the influence of 12 N force :

0%
$12m/s$
0%
$6m/s$
0%
$3.5m/s$
0%
$3m/s$

Explanation

Work of

$F=\triangle K$

$f\times d =\triangle K$

$12 \times 3=\dfrac{1}{2}m[v_g^2-v_i^2]$

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

$v^2=\dfrac{2\times 12\times 3}{6}=12$

$V^2=\sqrt{12}$

$V=3.5 m/s$

A particle moves from

$X = 0\$ to

$X = 2\ m$ on X-axis under the effect of

${ F } (x) = ({ 4x }^{ 3 } - { 3x }^{ 2 } + 2x + 5)\widehat { i }$ Newton. The work done on the particle is

0%
$22\ Joule$
0%
$36\ Joule$
0%
$46\ Joule$
0%
None of these

Explanation

Work done by a force =

$\int Fdx=x^4-x^3+x^2+5x|^2_0=22 \ Joule$

Which of the following statements are correct

0%
Collision does not require physical contact
0%
Collision between sub atomic particles is elastic
0%
Collision between macroscopic bodies is generally inelastic
0%
None of these

A force F =

$-\frac{K}{X^2} (X \neq 0)$ acts on a particle in x-direction. The work done by this force in displacing the particle from x = +a to x = +2a is. (Where k is a positive constant)

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

Explanation

Given that ,

$F= - \dfrac K{x^2}$ , Where

$x$ is the position of particle .

Work done by this force ,

$W=\int _a^{2a} F.dx=\int _a^{2a} -\dfrac K{x^2} dx= [\dfrac Kx ]_a ^{2a}= \dfrac K{2a}- \dfrac Ka = -\dfrac K {2a}$

$\vec{F}=(5\hat{i}+3\hat{j}+2\hat{k})\ N$ is applied over a particle with displaces it from its origin to the point

$\vec{r}=(2\hat{i}-\hat{j})m$ . The work done on the particle is joule is

0%
$-7\ J$
0%
$+7\ J$
0%
$+10\ J$
0%
$13\ J$

Explanation

Work done by a force =

$\int \vec{F}.\vec{dr}=(5\hat{i}+3\hat{j}+2\hat{k}).(2\hat{i}-\hat{j})=7 \ Joule$

If the resultant force on a system of particles is non-zero, then:

0%
The linear momentum of the system must increase
0%
The velocity of the centre of mass of the system must change.
0%
The distance of the centre of mass may remain constant from a fixed point.
0%
Kinetic energy of all particles must either increase simultaneously or decrease simultaneously.

Explanation

It is inertial frame, it means that the frame of reference does not have an acceleration, and so does not have a psudo force

$\sum F = 0,=>\large\dfrac{dp}{dt}=0$

$p$ is constant

kinetic energy and potential energy does not remain constant

Hence a is the correct answer.

Choose one more correct alternatives.
For work to be performed , energy must be _____

0%
transferred from one place to another
0%
concentrated
0%
transformed from one type to another
0%
destroyed

A ball is at rest at the top of a hill. It rolls down the hill. At the bottom of the hill the ball hits a wall and stops.
Which energy changes occur?

0%
Gravitational potential energy $\rightarrow$ internal energy $\rightarrow$ kinetic energy
0%
Gravitational potential energy $\rightarrow$ kinetic energy $\rightarrow$ internal energy
0%
Kinetic energy $\rightarrow$ gravitational potential energy $\rightarrow$ internal energy
0%
Kinetic energy $\rightarrow$ internal energy $\rightarrow$ gravitational potential energy

A ball of mass

$'m'$ moves towards a wall with a velocity

$'u'$ , the direction of motion making an angle

$\theta$ with the surface of the wall and rebounds with the same period. The change in momentum of the ball during the collision is:

0%
$2mu\ sin$ $\theta$ towards the wall
0%
$2mu \ sin$ $\theta$ away from the wall
0%
$2mu\ cos$ $\theta$ towards the wall
0%
$2mu\ cos$ $\theta$ away from the wall

Explanation

As the ball hits the wall at an angle and rebounds, there is a change in the momentum only along the direction perpendicular to the wall. The change along the wall is zero. In diagram

$\vec p_1$ and

$\vec p_2$ is the momentum before and after the collision with wall.
Since , the angle with the wall is

$\theta$ , the change is only in

$sin\theta$ component, which is across

$\vec F_{wall}$ .
Change in momentum

$=m[usin\theta -(-usin\theta )]=2mu\ sin\theta$ away from the wall, as the final velocity is away from the wall.

A car of mass 400 kg travelling at 72 kmph crashes a truck of mass 4000 kg and travelling at 9 kmph in the same direction. The car bounces back with a speed of 18 kmph. The speed of the truck after the impact is

0%
9 kmph
0%
18 kmph
0%
27 kmph
0%
36 kmph

Explanation

Using momentum conservation,

$\displaystyle m_{car} \displaystyle u _{car} + \displaystyle m_{ truck} \displaystyle u_{ truck}= \displaystyle m_{ car} \displaystyle \displaystyle v _{car} + \displaystyle m_{truck} \displaystyle v_{truck}$

$400 \times 72 + 4000 \times 9 = -18 \times 400 + 4000 \times v$

$v = 18 \ kmph$

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 (treat

$C$ as a constant):

0%
$\dfrac {C^{2}}{x^{2}_{1}}$
0%
$Cx^{2}_{1}$
0%
$\dfrac {Cx^{2}_{1}}{2}$
0%
$\dfrac {C}{x^{2}_{1}}$

Explanation

Work Done

$\displaystyle =\int_{0}^{x_1} \vec{F}.d\vec{x}$

$\displaystyle=\int_{0}^{x_1}Cx.dx$

$= \dfrac {Cx^{2}_{1}}{2}$

A perfectly elastic ball

$p_{1}$ of mass 'm' moving with velocity v collides elastically with three exactly similar balls

$p_{2}$ ,

$p_{3}$ ,

$p_{4}$ lying on a smooth table. Velocities of the four balls after collision are

0%
0,0,0,0
0%
v,v,v,v
0%
v,v,v,0
0%
0,0,0,v

Explanation

$\textbf{Hint}$ : Use law of conservation of momentum

$\textbf{Step}$ :

Here when

$P_{1}$ and

$P_{2}$ will collide, their momentum will be exchanged because the collision is elastic and the two bodies are similar.

When

$P_{2}$ and

$P_{3}$ collide same happens.

So, momentum of

$P_{1}$ is transferred to

$P_{4}$

So,

$P_{1}$ ,

$P_{2}$ ,

$P_{3}$ will have zero velocity and

$P_{4}$ has a velocity of

$V$

$\textbf{Step 2}: \textbf{Calculation}$ :

Let us take two balls. Let the first ball has a mass of 'm' and velocity 'v'. Let the second ball has a mass of 'm' and velocity is

$0$ that is it is at rest.

For perfectly elastic collision coefficient of restitution is

$e=1$

Total momentum before collision (Initial momentum)=Total momentum after collision(Final momentum)

$P_i=P_f$

$mV=mV_1+mV_2$

$\implies V_1+V_2=V$

$(1)$

We know coefficient of restitution

$e=1$

So,

$e= \dfrac{V_{separation}}{V_{approach}}$

$1=\dfrac{V_1-V_2}{V}$

$V_1-V_2=V$

$(2)$

Solving equation

$(1)$ and

$(2)$ we get

$2V_1=2V$

$\implies V_1=V$

$V_2=0$

$P_{1}$ ,

$P_{2}$ ,

$P_{3}$ will have zero velocity and

$P_{4}$ has a velocity of

$V$

Thus option D is correct.

A ball of mass M moving with a velocity V collides head on elastically with another of same mass but moving with a velocity v in the opposite direction. After collision,

0%
the velocities are exchanged between the two balls
0%
both the balls come to rest
0%
both of them move at right angles to the original line of motion
0%
one ball comes to rest and another ball travels back with velocity 2V

Explanation

Using conservation of linear momentum we have:

$M_1u_1-M_2u_2=-M_1v_1+M_2v_2$

$MV-Mu_2=-Mv+Mv_2$

$V-u_2=-v+v_2$ .............(i)

and as for a perfect elastic collision we have

$e=1$ , thus we get

$v_2+v=u_2+V$ .................(ii)

Using these two equations we get:

$u_2=v$ and

$v_2=V$

Thus, the velocities are exchanged between the two balls.

A heavy steel ball of mass greater than 1 kg moving with a speed of 2m/ s collides head on with a stationary ping pong ball of mass less than 0.1 g. The collision is elastic. After the collision the ping pong ball moves approximately with a speed

0%
$2 m / s$
0%
$4 m/ s$
0%
$2\times10^{4}m / s$
0%
$2\times10^{3}m / s$

Explanation

Since the body is much heavy these won't be much change in velocity
&

$e = 1$

i.e.,

$\dfrac{v-2}{0-2} = -1$

$\Rightarrow v = 4 m/s$

Which of the following physical quantities is different from others?

0%
Work
0%
Kinetic energy
0%
Force
0%
Potential energy

Explanation

Kinetic energy is the energy possessed by the body by virtue of its motion and potential energy is the energy possessed by the virtue of its position and shape. Energy is required to do work.

Hence work, kinetic energy, potential energy all can be measured using same unit, i.e joule.

So, force is the physical quantity here which is different from others as given by product of mass and acceleration. S.I unit of force is

$N$ (newton).

A man of mass

$60 kg$ climbs up a

$20 m$ long staircase on the top of a building

$10 m$ high. What is the work done by him?

$(Take g = 10 m s^{-2})$

0%
$12 kJ$
0%
$6 kJ$
0%
$3 kJ$
0%
$18 kJ$

Explanation

$=$ 60 kg
Length of staircase

$=$ 20 m
Height of staircase

$=$ 10 m
Workdone

$=$ mgh

$=$ mg (length + height of stair case)

$=60 \times 10 \times (20+10)$

$600 \times 30 = 18 kJ$

An empty truck moving on a straight road with a certain velocity can be stopped over a distance

$s$ by applying the brakes. If the truck is loaded so that its mass now is one and half times that of the empty truck and is moving with the same velocity, it can be stopped by the brakes in a distance of (assume that the same braking force is acting in the two cases)

0%
$\dfrac{s}{2}$
0%
$2 s$
0%
$\dfrac{3s}{2}$
0%
$\dfrac{2s}{3}$

Explanation

The use of break is to reduce the kinetic energy of a car using the frictional force between the tires of the car and the road. The empty car of mass m has a kinetic energy

$=1/2 \times m \times v^2$ and the work done by the frictional force (F) is =

$F\times s$ , where

$s$ is the distance traveled by the truck before it can be stopped.

So,

$1/2 \times m \times v^2$ =

$F\times s$

or,

$F= \dfrac{m \times v^2}{2 \times s}$ ........................ (1)

Now when the truck is loaded then let the mass of it is M. And the distance traveled by the truck before it can be stopped is S.

$1/2 \times M \times v^2$ =

$F\times S$ .............(2)

Using (1), we will get ,

$S=3/2 \times s$ .

A heavier body moving with certain velocity collides head on elastically with a lighter body at rest. Then

0%
smaller body continues to be in the same state of rest
0%
smaller body starts to move in the same direction with same velocity as that of bigger body
0%
the smaller body starts to move with twice the velocity of the bigger body in the same direction
0%
the bigger body comes to rest

Explanation

$\textbf{Hint}$ : Apply Law of conservation of momentum

$\textbf{Step 1:Assumptions}$

Let us assume that there are two bodies. Let the first body has a mass of 'M' and initial velocity=

$V$ and final velocity=

$V_1$ (after collision).

Let the second body has a mass of 'm' and initial velocity=0 ; final velocity=

$V_2$ (after collision)

$\textbf{Step 2:Apply Law of conservation of Linear Momentum}$

We know according to the law of conservation of momentum

Initial momentum=Final Momentum

$MV+m(0)=MV_1+mV_2$

$(1)$

$\textbf{Step3:Coefficient of Restitution}$

We know for elastic collision e=1

$e=1=\dfrac{V_2-V_1}{V-0}$

$V=V_2-V_1$

$V_1=V_2-V$

$(2)$

$\textbf{Step 4:Solve above equations to obtain answer}$

Substitute

$V_1=V_2-V$ in

$(1)$

we get

$MV=M(V_1)+mV_2$

$MV=M(V_2-V)+mV_2$

$MV=MV_2-MV+mV_2$

$2MV=MV_2+mV_2$

$V_2=\dfrac{2VM}{M+m}$

$V_2=2V$ if

$M>>m$

Thus option C is correct

A rain drop of mass (1/10) gram falls vertically at constant speed under the influence of the forces of gravity and viscous drag. In falling through 100 m, the work done by gravity is

0%
0.98 J
0%
0.098 J
0%
9.8 J
0%
98 J

Explanation

work done = mgh

m = mass of drop = 0.1 g = 0.00001kg

g =9.8m/s

h = 100m

work done =

$0.00001 \times 9.8 \times 100$

= 0.098 J

A body of mass

$1$

$kg$ is made to travel with a uniform acceleration of

$30\ cm/s^{2}$ over a distance of

$2$

$m$ , the work done is:

0%
$6 J$
0%
$60 J$
0%
$0.6 J$
0%
$0.3 J$

Explanation

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

$v^{2}- u ^{2} = 2\times 30\times 10^{-2}\times 2$

$= 1.2$

$W.D = \Delta (K.E)$

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

$= \dfrac{1}{2}\times 1\times (1.2)$

$= 0.6J$

lf the two bodies moving at right angles collide and their initial momenta are

$\vec{\mathrm{P}}_{1}$ and

$\vec{\mathrm{P}}_{2}$ , their resultant momentum after collison is :

0%
$\vec{\mathrm{P}_{1}}-\vec{\mathrm{P}}_{2}$
0%
$\vec{\mathrm{P}}_{1}\sim\vec{\mathrm{P}}_{2}$
0%
$\sqrt{\vec{\mathrm{P}_{1}}^{2}+\vec{\mathrm{P}}_{2}^{2}}$
0%
$\sqrt{\vec{\mathrm{P}_{1}}^{2}-\vec{\mathrm{P}}_{2}^{2}}$

Explanation

By momentum conservation,
Resultant momentum after collision = Resultant momentum before collision =

$\sqrt{\vec {P_1}^2 + \vec {P_2}^2}$

A ball moving with a speed of 2.2 m/sec strikes an identical stationary ball. After collision the first ball moves at 1.1 m/sec at

$60^{0}$ with the original line of motion. The magnitude and direction of the ball after collision is

0%
$5 m/sec,90^{0}$
0%
$2 m/sec,60^{0}$
0%
$\sqrt{39(1.1)}m/sec,30^{0}$
0%
$10m/sec,60^{0}$

Explanation

Conserving momentum in X - direction

$m \times 2.2 = 1.1 \times m \times cos 60 + v\:cos\:\theta \times m$

$2.2 = \dfrac{1.1}{2} + v\:cos\:\theta$

$\dfrac{3.3}{2} = v\:cos\:\theta$ ........................(1)

Similarly in Y - direction

$v\:sin\:\theta = 1.1 sin \;60$

$v\:sin\:\theta = \dfrac{1.1\sqrt{3}}{2}$ ........................(2)

from (1) & (2)

$v = \sqrt{3.9(1.1)}m/s$

$\theta = 30^{0}$

A 50 gm ball collides with another ball of mass 150gm, moving in its direction of motion, After collision the two balls move at a an angle

$30^{0}$ with their initial direction. Ratio of their velocities after collision is

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

Explanation

The momentum of the system is conserved in every direction. Since the initial vertical momentum is zero, the final vertical momentum is also zero.
Therefore,

$50v_1 \times sin\,30^{\circ} = 150v_2 \times sin\,30^{\circ}$

$\dfrac{v_1}{v_2} = \dfrac{3}{1}$ .

The total work done on a particle is equal to the change in its kinetic energy.

This statement is true for which of the condition?

0%
always
0%
only if the forces acting on the body are conservative
0%
only if the forces acting on the body are gravitational
0%
only if the forces acting on the body are elastic.

Explanation

From Newton's second law, it can be shown that work on a free, rigid body, is equal to the change in kinetic energy of the velocity and rotation of that body,

$W = \Delta KE$

The work of forces generated by a potential function is known as potential energy and the forces are said to be conservative. Therefore work on an object that is merely displaced in a conservative force field, without change in velocity or rotation, is equal to minus the change of potential energy of the object,

$W = -\Delta PE$

These formulas demonstrate that work is the energy associated with the action of a force, hence the total work done on a particle is equal to the change in its kinetic energy always.

A car is accelerated on a leveled road and attains a velocity 4 times of 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

An 800 g ball is pulled up a slope as shown in the diagram. Calculate the potential energy it gains.

0%
1.96 J
0%
1.568 J
0%
7.84 J
0%
1.225 J

A body of mass 5 kg moving with a speed of

$3 ms^{-1}$ collides head on with a body of mass 3 kg moving in the opposite direction at a speed of

$2 ms^{-1}$ . The first body stops after the collision. Find the final velocity of the second body.

0%
$3 ms^{-1}$
0%
$5 ms^{-1}$
0%
$-9 ms^{-1}$
0%
$30 ms^{-1}$

Explanation

$\underline {Given:}$

$m_1=5kg$

$m_2=3kg$

$u_1=3ms^{-1}$

$u_2=-2ms^{-1}$

Before collision, the 2nd body is moving backwards. Hence its velocity is negative.

$v_1=0ms^{-1}$

After collision, the 1st body stops moving. So its velocity is zero.

$To$

$find:$

$v_2=$

$?$

$\underline {Solution:}$

$m_1 u_1+m_2 u_2$

$=$

$m_1 v_1+m_2 v_2$

Substituting the values, we get,

$(5\times3)$

$+$

$(3\times(-2))$

$=$

$5\times 0$

$+$

$3\times v_2$

$\therefore15$

$-$

$6$

$=$

$3\times v_2$

$\therefore$

$9$

$=$

$3 v_2$

$\therefore$

$v_2$

$=$

$9$

$\div$

$3$

$=$

$3$

$\therefore$

$v_2$

$=$

$3ms^{-1}$

Since we got a positive value for

$v_2,$ it means that the 2nd body moves forwards after collision.

Velocity being a vector quantity, the direction is also important.

After collision, the 2nd body moves with a velocity

$3ms^{-1}$ in the forward direction.

A body of mass 'm' moving with certain velocity collides with another identical body at rest. If the collision is perfectly elastic and after the collision, if both the bodies move, they can move

0%
in the same direction
0%
in opposite direction
0%
in perpendicular direction
0%
making $45^o$ to each other

A ball strikes a horizontal floor at an angle

$45^{0}$ The value of 'e' between the ball and the floor is 1/Find the fraction of K.E. loss during the collision

0%
$\dfrac{3}{4}$
0%
$\dfrac{5}{8}$
0%
$\dfrac{3}{8}$
0%
$\dfrac{7}{8}$

Explanation

$x$ component of velocity will remain same & for

$y$ component

$(V_y) \:final - 0 = -e.V_y$ (velocity in y direction)

so,

$K.E.$ lost

$=\dfrac{\dfrac{1}{2}m\left ( \dfrac{v}{\sqrt{2}} \right )^{2}-\dfrac{1}{2}m\left ( \dfrac{ev}{\sqrt{2}} \right )^{2}}{2 \times \dfrac{1}{2}m\left ( \dfrac{v}{\sqrt{2}} \right )^{2}}$

Here, the horizontal component of velocity willn't undergo any change.

$=\dfrac{1-e^{2}}{2}$

$=\dfrac{1-0.25}{2} = \dfrac{0.75}{2} = \dfrac{3}{8}$

A wagon of mass 10 tons moving at a speed of 12 kmph collides with another wagon of mass 8 tons moving on the same track in the same direction at a speed of 10 kmph. If the speed of the first wagon decreases to 8 kmph. Find the speed of the other after collision

0%
18 kmph
0%
25 kmph
0%
5 kmph
0%
15 kmph

Explanation

Conserving momemtum,

$10 \times 12 + 8 \times10 = 10 \times 8 + 8 \times v$

$200 - 80 = 8 \times v$

$\Rightarrow v = 15 \ kmph$

A particle of mass m has a velocity

$-v_{0}\hat{i}$ . while a second particle of same mass has a velocity

$v_{0}j$ . After the particles collide, first particle is found to have a velocity

$\dfrac{-1}{2}v_{0}\hat{i}$ then the velocity of other particle is

0%
$\dfrac{-1}{2}v_{0}\hat{i}+v_{0}\hat{j}$
0%
$\dfrac{1}{2}v_{0}\hat{i}+v_{0}\hat{j}$
0%
$v_{0}\hat{i}+v_{0}\hat{j}$
0%
$-v_{0}\hat{i}+v_{0}\hat{j}$

Explanation

Using momentum conservation
Final momentum

$=$ Initial momentum

$-\dfrac{m}{2}v_{0}\hat{i} + m(v_{x}\hat{i} + v_{y}\hat{j}) = -mv_{0}\hat{i} + mv_{0}\hat{j}$

Comparing

$v_{x} = -\dfrac{1}{2}v_{0}\hat{i}$

$v_{y} = v_{0}\hat{j}$

A 6 kg mass travelling at

$2.5$ ms

$^{-1}$ collides head on with a stationary 4 kg mass. After the collision the 6 kg mass travels in its original direction with a speed of

$1$ ms

$^{-1}$ . The final velocity of 4 kg mass

0%
$1$ ms $^{-1}$
0%
$2.25$ ms $^{-1}$
0%
$2$ ms $^{-1}$
0%
$0$ ms $^{-1}$

Explanation

Use momentum conservation,

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$
But

$m_1=6$ kg

$, m_2=4$ kg,

$u_1=2.5$ ms

$^{-1}, u_2=0$ ms

$^{-1}, v_1=1$ ms

$^{-1}, v_2=?$

$\Rightarrow 6 \times 2.5 +4\times (0) = 6 \times 1 + 4 \times v_2$

$\Rightarrow 9 = 4v \Rightarrow v = 2.25$ ms

$^{-1}$ .

A coconut fruit hanging high in a palm tree has ......... owing to its location.

0%
Free energy
0%
Kinetic energy
0%
Activation energy
0%
Potential energy

The decrease in the PE of a ball of mass 20 kg which falls from a height of 25 cm is 98 J.

0%
True
0%
False

Explanation

$\Delta PE = mg (h_2 - h_1)$
Given

$h_2 = 25 cm = 0.25 m$

$h_1 =0$

$\therefore \Delta PE = 20 \times 9.8 \times 25 \times 10^{-2}$

$=196 \times 0.25 = 49 J$
But given

$\Delta PE = 98 J$
Statement is false.

When velocity of a moving object is doubled its:

0%
acceleration is doubled
0%
momentum becomes four times more
0%
kinetic energy is increased to four times
0%
potential energy is increased

Explanation

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

$K.E \propto v^2$

$New\ K.E = \dfrac{1}{2}m(2v)^2 = 4 \times K.E$

$\therefore$ The K.E. of a body will increase to 4 times if its velocity doubles.

Water stored in a dam possesses:

0%
no energy
0%
electrical energy
0%
kinetic energy
0%
potential energy

Two satellites of earth,

$S_{1}$ and

$S_{2}$ , are moving in
the same orbit. The mass of

$S_{1}$ is four times the
mass of

$S_{2}$ . Which one of the following statements is true ?

0%
The kinetic energies of the two satellites are

equal
0%
The time period of $S_{1}$ is four times that of $S_{2}$
0%
The potential energies of earth and satellite in

the two cases are equal
0%
$S_{1}$ and $S_{2}$ are moving with the same speed

Explanation

$K.E.=\frac{GMm}{2r}\Rightarrow Kinetic energies are unequal$

$T=\frac{2\pi r^{3/2}}{\sqrt{GM}}\Rightarrow Time period are equal$

$P.E.=-\frac{GMm}{r}\Rightarrow Potential energies are unequal$

$V=\sqrt{\frac{GM}{r}}\Rightarrow Orbital speeds are equal$

CBSE Questions for Class 11 Medical Physics Work, Energy And Power Quiz 2 - 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 2 - MCQExams.com

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Practice Class 11 Medical Physics Quiz Questions and Answers

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