MCQ Questions for Class 11 Medical Physics Work, Energy And Power Quiz 4

A force

$F=(2+x)$ acts on a particle in x-direction where

$F$ is in newton and

$x$ in meter. Find the work done by this force during a displacement from

$x=1.0$ m to

$x=2.0\ m$ .

0%
$3.5 \ J$
0%
$7 \ J$
0%
$4.5 \ J$
0%
$5.5 \ J$

Explanation

As the force is variable, we shall find the work done in a small displacement from

$x$ to

$x+dx$ and then integrate it to find the total work. The work done in this small displacement is

$\displaystyle dW= F\:dx= \left ( 2+x \right )dx$

Thus,

$\displaystyle W= \int_{1.0}^{2.0}dW= \int_{1.0}^{2.0}\left ( 2+x \right )dx$

$\displaystyle = \left [ 2x+\frac{x^{2}}{2} \right ]_{1.0}^{2.0}= 3.5J$

A force

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

$\displaystyle \vec{S}= \left ( 2t^{2}\hat{i}-5\hat{j} \right )m.$ Work done by this force in 2 second is:

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

Explanation

$W = \int \bar F.\bar {ds}$

$\bar ds = 4t\hat i dt$

$dW = (3t\hat i + 5\hat j).(4t\hat i)dt$

$W(t=2)=\int 12t^2dt = 4t^3 = 4(2^3) = 32J$

Equal net forces act on two different blocks A and B of masses m and 4m respectively. For same displacement, identify the correct statement.

0%
Their kinetic energies are in the ratio $\displaystyle \frac{K_{A}}{K_{B}}= \frac{1}{4}$
0%
Their speeds are in the ratio $\displaystyle \frac{v_{A}}{v_{B}}= \frac{1}{1}$
0%
Work done on the blocks are in the ratio $\displaystyle \frac{W_{A}}{W_{B}}= \frac{1}{1}$
0%
All of the above

Explanation

Work done is equal to F.s
Hence for equal force and displacement amount of work done will be equal irrespective of the weight of the bodies.

$\therefore \dfrac{W_A}{W_B}=\dfrac{1}{1}$

A block of mass

$M$ is hanging over a smooth and light pulley through a light string. The other end of the string is pulled by a constant force

$F$ . The kinetic energy of the block increases by

$40$

$J$ in

$1$

$s$ . The tension in the string is:

0%
$F$
0%
$Mg$
0%
$F+Mg$
0%
$F-Mg$

Explanation

Since a force

$F$ is pulling the string, the tension in the string is

$F$ .
However the forces acting on the block are tension

$=F$ upwards and weight

$=Mg$ downwards and thus has a net acceleration upwards.

A particle of mass

$m$ moves on a straight line with its velocity varying with the distance travelled according to the equation

$\displaystyle v= \alpha \sqrt{x},$ where a is a constant. Find the total work done by all the forces during a displacement from

$x=0$ to

$x=d$ .

0%
$\displaystyle \frac{1}{2}ma ^{2}b$
0%
$\displaystyle \frac{1}{4}ma ^{2}b$
0%
$ma ^{2}b$
0%
$2\ ma ^{2}b$

Explanation

$v=\alpha\sqrt{x}$

$\implies \dfrac{dx}{dt}=\alpha\sqrt{x}$

$\implies 2\sqrt{x}=\alpha t$

$\implies x=\dfrac{\alpha^2 t^2}{4}$

Work done by forces

$=$

$\int_0^b Fdx=_0^b\int madx$

$=_0^b\int m\dfrac{d^2x}{dt^2}dx$

$=_0^b\int m(\dfrac{2\alpha^2}{4})dx$

$=\dfrac{1}{2}m\alpha^2 b$

A player kicks a ball of mass 250 g at the centre of a field. The ball leaves his foot with a speed of 10 m/s. Find the work done by the player on the ball.

0%
$10.5 \:J$
0%
$12.5 \:J$
0%
$1.5 \:J$
0%
$2.5 \:J$

Explanation

Given, initial speed of ball,

$u=0$ ;
final speed of ball,

$v =10 \:m/s$ ;
mass of the ball,

$m = 250 \:g =(250/1000) \:kg$

$= 0.25 \:kg$
Work done by the player on the ball,

$W =change \: in \: kinetic \: energy \: (E_k) \: of \: the \: ball$

$\displaystyle W = \dfrac{1}{2}mv^2- \dfrac{1}{2}mu^2$

$\displaystyle =\dfrac{1}{2}(0.25)(10)^2-\dfrac{1}{2}(0.25)(0)^2=12.5 \:J$

A block is constrained to move along x-axis under a force F=-2x. Here, F is in newton and x in metre. Find the work done by this force when the block is displaced from x=2 m to x=-4 m.

0%
-4 J
0%
-8 J
0%
-12 J
0%
-16 J

Explanation

Work done by a body in moving an object from

$x_1$ to

$x_2$

$=\int_{x_1}^{x_2}Fdx$

$=\int_{x_1}^{x_2}(-2x)dx$

$=x_1^2-x_2^2$

$=-12J$

Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of

$60 \:kmh^{-1}$

0%
$-208333 \:J$
0%
$208333 \:J$
0%
$-209333 \:J$
0%
$-207333 \:J$

Explanation

Given,

$mass \: of \: car, \: m = 1500 \: kg$ ;

$velocity \: of \: car, \: v =60 \:kmh^{-1}=60\times \dfrac{5}{18}=\dfrac{50}{3}m/s$

$Work \: done = Change \: in \: kinetic \: energy \: of \: the \: car$

$\displaystyle W=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2=\dfrac{1}{2}m\left (v^2-u^2 \right )$

$\displaystyle W=\dfrac{1}{2}(1500)[(0)^2-(50/3)^2]$

$\displaystyle W=\dfrac{1}{2}\times 1500\times \dfrac{2500}{9}= -208333 \:J$

Certain force acting on a 20 kg mass changes its velocity from

$5 \:ms^{-1}$ to

$2 \:ms^{-1}$ . Calculate the work done by the force.

0%
$-210 \:J$ .
0%
$210 \:J$ .
0%
$-105 \:J$ .
0%
$420 \:J$ .

Explanation

Given,

$mass,\: 'm = 20 \: kg$ ;

$initial \: velocity, \: u =5 \:ms^{-1}$ ;

$final \: velocity, \: v = 2 \:ms^{-1}$

$Work \: done \: by \: the force = change \: in \: kinetic \: energy$

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

$W =\dfrac{1}{2}(20)[(2)^2-(5)^2] = -210 \:J$ .

An object of mass, m is moving with a constant velocity v. How much work should be done on the object in order to bring the object to rest ?

0%
$-\dfrac{1}{2}mv^2$
0%
$\dfrac{1}{2}mv^2$
0%
$-mv^2$
0%
$-\dfrac{1}{2}mv$

Explanation

$\textbf{Hint: Apply Work-enegy theorem}$

Correct option: A

Explanation for correct option:

$\textbf{Step1: Work-energy theorem}$

According to the Work-Energy theorem,
The net work done on an object is equal to the change in its Kinetic Energy.

$\therefore W = \Delta K$

$\textbf{Step2: Find work done}$

$\Delta K =$ Final Kinetic energy - Initial Kinetic energy

Initial kinetic energy will be zero as the object comes to rest.

$\Delta K = 0 - \dfrac{1}{2}m{v^2} = - \dfrac{1}{2}m{v^2}$

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

Hence, option A is correct.

A block is constrained to move along x-axis under a force

$\displaystyle F= \frac{4}{x^{2}}\left ( x\neq 0 \right ).$ Here, F is in newton and x in metre. Find the work done by this force when the block is displaced from x=4 m to x=2 m.

0%
-1 J
0%
-2 J
0%
-3 J
0%
-4 J

Explanation

Work done in moving an object from

$x_1$ to

$x_2$

$=\int_{x_1}^{x_2}Fdx$

$=\int_{x_1}^{x_2}\dfrac{4}{x^2}dx$

$=4(\dfrac{1}{x_1}-\dfrac{1}{x_2})$

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

$=-1J$

A block of mass 30 kg is being brought down by a chain. If the block acquires a speed of 40 cm/s in dropping down 2 m. Find the work done by the chain during the process.

$\displaystyle \left ( g= 10 m/s^{2} \right )$

0%
-297.6 J
0%
-375 J
0%
-597.6 J
0%
-674.6 J

Explanation

Forces acting on the block are of chain and gravity. By work energy theorem

${ W }_{ c }+{ W }_{ g }=\dfrac { 1 }{ 2 } m{ v }^{ 2 }-\dfrac { 1 }{ 2 } m{ u }^{ 2 }\\ { W }_{ c }+mgh=\dfrac { 1 }{ 2 } m{ v }^{ 2 }-0\\ \Rightarrow { W }_{ c }=\dfrac { 1 }{ 2 } \times 30\times { 0.4 }^{ 2 }-30\times 10\times 2=-597.6J$

A bullet of mass 20 g moving with a velocity of 500 m/s, strikes a tree and goes out from the other side with a velocity of 400 m/s. Calculate the work done by the bullet (in joules) in passing through the tree.

0%
$900 \:J$
0%
$800 \:J$
0%
$950 \:J$
0%
$500 \:J$

Explanation

Given, mass of the bullet,

$m = 20 \:g = 0.02 \:kg$ ;
initial velocity of the bullet,

$u = 500 \:m/s$ ;
final velocity of the bullet,

$v = 400 \:m/s$
Work done by the tree against the bullet,

$\displaystyle W_{tree}=\dfrac{1}{2}mv^2-\dfrac{1}{2}mv^2=\dfrac{1}{2}m(v^2-u^2)$

$\displaystyle =\dfrac{1}{2}(0.02)[(400)^2 -(500)^2]= -900 \:J$

Work done by the bullet,

$W_{bullet}=-W_{tree}$
or

$W_{bullet} = -(-900 \:J)= 900 \:J$

Sania, a high-board diver of mass 50 kg is diving from a height of 30 m into a pool (see figure given). What is the potential energy of Sania at point A ?

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

0%
$5000 J$
0%
$10000 J$
0%
$15000 J$
0%
$20000 J$

Explanation

Potential energy is energy stored in an object. This energy has the potential to do work. Gravity gives potential energy to an object. This potential energy is a result of gravity pulling downwards. The gravitational constant,

$g$ , is the acceleration of an object due to gravity. This acceleration is about

$9.8$ meters per second on earth. The formula for potential energy due to gravity is

$PE = mgh$ . As the object gets closer to the ground, its potential energy decreases while its kinetic energy increases.

In this case, the mass is

$50 kg$ , height is

$30 m$ and the gravitational acceleration

$g=10 m/{ s }^{ 2 }$ .

$PE=50\times 30\times 10\quad =\quad 15000\quad J$ .

Hence, the potential energy of Sania at point A is 15,000 J.

Energy associated with moving objects is called:

0%
Potential energy
0%
Kinetic energy
0%
No energy
0%
Work

Flowing water possess same energy as :

0%
Water in a water tank
0%
A stationery object
0%
Flowing air
0%
None of these

Water stored in a dam primarily possesses

0%
Heat energy
0%
Electrical energy
0%
Kinetic energy
0%
Potential energy

The gravitational potential energy possessed by a body is by virtue of its _________.

0%
position
0%
velocity
0%
acceleration
0%
temperature

A car is accelerated on a levelled 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 of initial
0%
becomes 4 times of initial
0%
becomes 16 times of initial

Which of the following is an example of an object with Kinetic Energy?

0%
Watch spring
0%
Bent bow
0%
Stone at a height
0%
Car moving along the road

The moving striker of the carom board will possess------ energy

0%
Kinetic
0%
Potential
0%
Solar
0%
Electric

Potential energy of a body depends on its:

0%
motion
0%
colour
0%
mass
0%
none of these

Explanation

We know that potential energy is the energy possesd by the virtue of its position.
So,

$F = mgh$
Where,

$m =$ mass
So, From above equation we can deduce that potential energy depends on the mass of the substance.

Which one of the following possesses potential energy?

0%
Moving vehicle on the road
0%
A running athlete
0%
Stone on the road
0%
A stretched rubber band

An overhead tank having some water possesses .......... energy.

0%
Kinetic
0%
Potential
0%
Thermal
0%
Electrical

Explanation

We know that Potential energy is the energy possessed by a body by the virtue of its position.

$\therefore$ An overhead tank having some water possesses Potential energy, as it is at a height.

Energy possessed by a body by virtue of its motion is :

0%
Potential energy
0%
Kinetic energy
0%
Chemical energy
0%
Electrical energy

Explanation

Kinetic energy is defined as the energy possessed by a body by virtue of its motion. and it is equal to

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

$m$ and

$v$ are mass and Velocity of moving body.

A student sitting at the top of a tree has ............ than the student who is sitting on the ground.

0%
less potential energy
0%
more potential energy
0%
more kinetic energy
0%
more gravitational energy

Explanation

Potential energy =

$mgh$

$m\rightarrow mass$

$g\rightarrow acceleration\ due\ to\ gravity$

$h\rightarrow height\ from\ ground$

P.E. is directly proportional to the height from the ground. Hence a student sitting at the top of a tree has more potential energy than the student who is sitting on the ground.

The energy directly related to the speed of a moving body and its mass is:

0%
Kinetic
0%
Potential
0%
Solar
0%
Electric

Explanation

The kinetic energy of a body is the energy by virtue of its motion.

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

where

$m$ is mass and

$v$ is velocity.

Hence, Kinetic energy is directly related to the speed of a moving body and its mass.

Energy possessed by a body by virtue of its position at a height is :

0%
Kinetic energy
0%
Chemical energy
0%
Potential energy
0%
Magnetic energy

Explanation

The potential energy of a body is defined as energy possessed by a body by virtue of its position at a height.

$P.E = mgh$

$m$ : mass of the object

$g$ : gravitational acceleration

$h$ : height gained by object

A heavy object has ---- potential energy than a lighter one

0%
Less
0%
More
0%
Same
0%
None of these

Explanation

$P.E= mgh$

The amount of potential energy an object possesses depends on its height and mass. The heavier the object and the higher from the ground, the more potential energy it holds.

Two weights of 5 kg and 10 kg are placed on a horizontal table of height 1.5 m. Which will have more potential energy?

0%
5 kg
0%
10 kg
0%
Both will have equal energy
0%
None of the above

Explanation

We know that , P.E

$= mgh$
So, It is directly proportional to height and mass.

Since both the weights are at the same height, so the weight with a larger mass will have more potential energy.

Since

$10$ kg object has a larger mass than a

$5$ kg,

so, the potential energy of a

$10$ kg mass will be greater.

A solar cell converts solar energy into:

0%
Chemical energy
0%
Light energy
0%
Electrical energy
0%
Muscular energy

From the following statements find the one that is NOT correct:

0%
A flying bird has both potential and kinetic energy.
0%
A ball rolling on the ground has kinetic energy but no mechanical energy.
0%
A stretched rubber band has potential energy
0%
Chemical energy can be converted into electrical energy

The form of energy that wood can be categorized into is:

0%
chemical energy
0%
electrical energy
0%
potential energy
0%
muscular energy

An atom bomb converts __________ into heat and sound energy.

0%
electrical energy
0%
heat energy
0%
mechanical energy
0%
atomic energy

Law of conservation of energy states that:

0%
energy cannot be destroyed but can be created and transformed from one form to another
0%
energy exists in only one form
0%
energy exists in many forms but it cannot be transformed
0%
energy can neither be created nor be destroyed but can be transformed from one form to another

The form of energy present in a wound spring is:

0%
chemical
0%
heat energy
0%
magnetic
0%
mechanical

A wound spring has:

0%
chemical energy
0%
magnetic energy
0%
mechanical energy
0%
muscular energy

A ball of mass 3 kg collides with a wall with velocity

$10\:m/sec$ at an angle of

$30^{\circ}$ with the wall and after collision reflects at the same angle with the same speed. The change in momentum of ball in MKS unit is-

0%
20
0%
30
0%
15
0%
45

Explanation

$\Delta p=2mv\:\cos 60^{\circ}=mv= 2 \times 3 \times 10 \times \frac{1}{2}=30\: kg \:m/s$

Two identical 5 kg blocks are moving with same speed of

$2\:m/s$ towards each other along a friction-less horizontal surface. The two blocks collide, stick together and come to rest. Consider two blocks as a system, the work done on the system by the external forces will be:

0%
$20$ Joule
0%
$-20$ Joule
0%
$0$ Joule
0%
None of these

A position dependent force

$F=7-2x+3x^{2}$ N acts on a small object of mass 2kg to displace it from

$x=0$ to

$x=5m$ . The work done in joules is-

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

Explanation

$W=\int F\mathrm{d} x$

$\displaystyle \int_{0}^{5}(7-2x+3x^{2})\mathrm{d} x=\left ( 7x-\frac{2x^{2}}{2}+\frac{3x^{3}}{3} \right )_{0}^{5}=135J$

A box of mass

$1\ kg$ is pulled on a horizontal plane of length

$1\ m$ by a force of

$8\ N$ then it is raised vertically to a height of

$2$ m, the net work done is:

0%
28 J
0%
8 J
0%
18 J
0%
none of the above

Explanation

$W=mgh+FS=1(10)(2)+8\times 1=20+8=28J$

A particle of mass

$m_{1}$ moving with a velocity of

$5m/s$ collides head on with a stationary particle of mass

$m_{2}$ . After collision both the particle move with a common velocity of

$4m/s$ , then the value

$m_ {1}/m_{2}$ is:

0%
4:1
0%
2:1
0%
1:8
0%
1:2

Explanation

Conservation of Momentum principle,
Initial momentum is

$M_i=5m_1$
Final momentum is

$M_f=4(m_1+m_2)$
By the above stated principle

$M_i=M_f=5m_1=4(m_1+m_2)$

$\implies m_1=4m_2$

$\therefore m_1:m_2=4:1$

A force of

$10$ N acts on a body of

$2$ kg mass for a distance of

$1$ m. The kinetic energy received by the body is:

0%
20 J
0%
10 J
0%
5 J
0%
2.5 J

Explanation

$\displaystyle W=FS=E_{K}=\frac{1}{2}mv^{2}=10\times 1=10J$

A time dependent force F = 10 t is applied on 10 kg block. Find out the work done by F in 2 seconds.

0%
40 J
0%
20 J
0%
120 J
0%
60 J

Explanation

$10\, \times\, \displaystyle \dfrac{dv}{dt}\, =\, 10\, t$

$\int_0^v \ dv =\int_{0}^{t} tdt$

$\Rightarrow\, v\, =\, \displaystyle \dfrac{t^2}{2}$ ..........(1)

$dW\, =\, \vec{F}\, .\, \vec{ds}$

$dW\, =\, 10 t. dx$

$dW\, =\, 10 t. v. dt$ .......... (2)

$\because\, dx\, =\, vdt$

from (1) & (2)

$dW\, =\, 10\, t\, .\, \left ( \displaystyle \dfrac{t^2}{2} \right )dt\, \quad\, dW\, =\, 5t^3dt$

$W\, =\, \displaystyle \dfrac{5}{4}\, [t^4]_0^2\, =\, 20\, J$

Force shown acts for

$2$ seconds. Find out work done by force

$F$ on

$10$ kg in

$3$ seconds.

0%
30 J
0%
20 J
0%
50 J
0%
60 J

Explanation

Work done,

$W=Fd$
Displacement

$d$ is given by,

$d=\dfrac{1}{2}a{t}^{2}$

$F=ma$

$10=10a$ ,

$a=1m/s^2$

$d=\dfrac{1}{2}(1){(2)}^{2}=2m$

$W=10\times2=20J$

A force acts on a

$3$ gm particle in such a way that the position of the particle as a function of time is given by

$x=3t-4t^2+t^3$ , where x is in meters and t is in seconds. The work done during the first

$4$ second is:

0%
$384 mJ$
0%
$168 mJ$
0%
$528 mJ$
0%
$541 mJ$

Explanation

Mass of the particle,

$m=0.003kg$

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

$\therefore a=-8+6t$

Thus force acting on the particle is

$ma$

Thus the work done on the particle

$=\int Fdx$

$=\int_0^4 m(-8+6t)(3-8t+3t^2)dt$

$=528mJ$ -m represents mass here.

Two bodies P and Q of equal masses are kept at heights

$x$ and

$4x$ respectively. What will be the ratio of their potential energies?

0%
$1:8$
0%
$4:1$
0%
$1:4$
0%
$8:1$

Explanation

Potential energy

$P=mgh$
Given,

$h_1=x ; h_2=4x$

Since, the masses are same,
then

$\dfrac{P_1}{P_2}=\dfrac{h_1}{h_2} = \dfrac{x}{4x}=1:4$

What is work done in holding a body of mass

$20\ kg$ at a height of

$2\ m$ above the ground?

$(g = 10\ m/s^2)$

0%
$40\ J$
0%
$400\ J$
0%
$10\ J$
0%
Zero

Explanation

A body of mass

$20\ kg$ is held at a height of

$2\ m$ above the ground means there is no displacement because of which there is no change in its potential energy. Hence work done is zero.

The diagrams represent the potential energy U as a function of the inter-atomic distance r. Which diagram corresponds to stable molecules found in nature.

Explanation

The potential energy function U(r) of a stable of inter-atomic distance r is given by:

$U(r)= -\dfrac{A}{(r^n)}+\dfrac{B}{r^m}$

Hence the correct opton is A

1018387_293763_ans_9293df1f372f4e3590654295a8740972.png

The work done by the frictional force on a surface in drawing a circle of radius r on the surface by a pencil of negligible mass with a normal pressing force N (coefficient of friction

$\mu_k$ ) is :

0%
$4\pi r^{2} \mu_K N$
0%
$-2\pi r^{2} \mu_K N$
0%
$-2\pi r \mu_K N$
0%
Zero

CBSE Questions for Class 11 Medical Physics Work, Energy And Power Quiz 4 - 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 4 - MCQExams.com

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Practice Class 11 Medical Physics Quiz Questions and Answers

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