MCQ Questions for Class 9 Physics Work And Energy Quiz 11

When a boy is sleeping, he is doing work.

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True
0%
False

$4$ bulbs rated

$100$ W each, operate for

$6$ hours per day. What is the cost of the energy consumed in

$30$ days at the rate of Rs.

$5$ /kWh?

0%
Rs. $360$
0%
Rs. $90$
0%
Rs. $120$
0%
Rs. $400$

Explanation

$\displaystyle E= Power(in kW)\times time=\frac{4\times 100\times 6}{1000}$ kWh

$=2.4$ kWh
Consumed in

$30$ days

$=30\times 2.4=72$
Total cost

$=72\times 5=360$ Rs.

Energy can neither be created nor be destroyed, but it can be changed from one form to another. This law is known as:

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potential energy
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kinetic energy
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conservation principle
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conservation of energy

Which of the following is not an example of potential energy?

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A vibrating pendulum at its maximum displacement from the mean position
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A body at rest at some height from the ground
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A wound clock-spring
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A vibrating pendulum when it is just passing through the mean position

In which of the following cases work is said to be done?

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A man pushing a roller and displacing it
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A boy sleeping
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Girl writing in Exam
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All of these

Explanation

In the first case, the man is pushing a roller and it is moving so work is said to be done

and in the other two both are sitting they are not moving so no work is said to be done.

$\text{ work } = \text{ force} \times \text{displacement}$

The diagram below shows the path taken by a ball when Sundram kicks it. The potential energy of the ball is highest at ______________

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P
0%
Q
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R
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S

What is the formula for electric power?

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$P=I^2 Rt$
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$P = \frac{E} {t}$
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P = VI $\times$ t
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P = VQ

Explanation

Formula for Electric Power is

$P = E/t$

where

$E$ = Energy in

$J$ and

$t$ is the time in second.

When the speed of a body is doubled, its kinetic energy becomes

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Double
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Half
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Quadruple
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One-fourth

Explanation

The kinetic energy of a body is given by:

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

Hence, Kinetic energy depends on the velocity as:

$KE\propto v^2$

So, if we double the velocity, then

$KE$ becomes four times.

Hence, option C is correct.

Choose whether true or false:-
Kinetic energy is zero at the start of the free-fall of an object.

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True
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False

A ball bounces to

$80$ % of its original height. What fraction of its potential energy is lost in each bounce?

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$0.20$
0%
$0.60$
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$0.40$
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$1$

Explanation

Given,

$h'=\dfrac{80h}{100}$

where,

$h=$ original height

The kinetic energy of a ball striking the ground is equal to the potential energy of a ball at height

$h$ ,

$K=mgh$

The kinetic energy due to rebounce is given by

$K_r=mgh'$

Loss of kinetic energy,

$\Delta K=K-K_r$

$\Delta K=mgh-mgh'$

$\Delta K=mg(h-\dfrac{80h}{100})$

$\Delta K=\dfrac{20}{100}mgh=0.20mgh$

Fractional loss,

$\dfrac{\Delta K}{K}=\dfrac{0.20mgh}{mgh}=0.20$

The correct option is A.

Which of the following statements is incorrect?

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Kinetic energy may be zero, positive or negative
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Power, energy and work are all scalars
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Potential energy may be zero, positive or negative
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Ballistic pendulum is a device for measuring the speed of bullets

Explanation

The kinetic energy of a body of mass

mm which is moving with velocity

vv is

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

Mass is always positive and if the velocity is either positive or negative, the kinetic energy is always positive due to the square of velocity.

The correct relation between joule and erg is:

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$1\ J = 10^{-5} erg$
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$1\ J = 10^{5} erg$
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$1\ J = 10^{-7} erg$
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$1\ J = 10^{7} erg$

Explanation

Joule and erg both are units of work done. An erg is the amount of work done by applying a force of one dyne for a distance of one centimeter. In the CGS base units, it will be one gram centimeter-squared per second-squared. Whereas joule is the amount of work done by applying a force of one newton for a distance of one meter.

Thus,

$1 joule = 1 newton \times 1 m\\$

$1 joule = \dfrac{1 kg \times 1 m}{1 s ^{2}} \times 1 m\\$

$1 J = \dfrac{1000 g \times 100 cm}{1 s ^{2}} \times 100 cm \\$

$1 J = 10^{7} \ erg$

Thus option D is correct.

A block of mass

$10kg$ accelerates uniformly from rest to a speed of

$2m/s$ in

$20$ sec. The average power developed in time interval of

$0$ to

$20$ sec is

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$10W$
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$1W$
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$20W$
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$2W$

A force

$(4\hat {i} + \hat {j} - 2\hat {k})N$ acting on a body maintains its velocity at

$(2\hat {i} + 2\hat {j} + 3\hat {k})ms^{-1}$ . The power exerted is:

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$4\ W$
0%
$5\ W$
0%
$2\ W$
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$8\ W$

Explanation

Force,

$\vec {F} = (4\hat {i} + \hat {j} - 2\hat {k})N$
Velocity,

$\vec {v} = (2\hat {i} + 2\hat {j} + 3\hat {k})ms^{-1}$
Power,

$P = \vec {F} \cdot \vec {v} = (4\hat {i} + \hat {j} - 2\hat {k})\cdot (2\hat {i} + 2\hat {j} + 3\hat {k})$

$= (8 + 2 - 6)W = 4\ W$ .

A raindrop of mass

$1\ g$ falling from a height of

$1\ km$ hits the ground with a speed of

$50\ ms^{-1}$ . Which of the following statements is correct?
(Take

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

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The loss of potential energy of the drop is $10\ J$
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The gain in kinetic energy of the drop is $1.25\ J$
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The gain in kinetic energy of the drop is not equal to the loss of potential energy of the drop
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All of these

Explanation

$Mass$ =

$m$ =

$1g$ =

$\frac{1}{1000}kg$

$Height$ =

$h$ =

$1km$ =

$1000m$

$Speed=50m/s$

$g=10m/s^2$

The formula of potential energy,

$mgh=\cfrac{1}{1000}\times10\times1000 = 10J$

When it hit the ground this potential energy is lost fully. The loss of potential energy of the drop is

$10J$

Calculate work done on object where applied force is

$10N$ and distance is

$10m$ :

0%
$10\ J$
0%
$50\ J$
0%
$100\ J$
0%
$150\ J$

Explanation

Given data

$f=10N$

$d=10m$

Work done by applied force is

$W=Force \times distance$

$W = Fd$

$W= 10\ N\times 10\ m$

$W = 100\ J$ .

A block of mass

$2\ kg$ initially at rest moves under the action of an applied horizontal force of

$6\ N$ on a rough horizontal surface. The coefficient of friction between block and surface is

$0.1$ . The work done by the applied force in

$10\ s$ is (Take

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

0%
$200\ J$
0%
$-200\ J$
0%
$600\ J$
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$-600\ J$

Explanation

The various forces acting on the block is as shown in the figure.
Here,

$m = 2\ kg, \mu = 0.1, F = 6\ N, g = 10\ ms^{-2}$

Force of friction,

$f = \mu N = 0.1\times 2\ kg\times 10\ ms^{-2} = 2\ N$

Net force with which the block moves

$F' = F - f = 6N - 2N = 4N$

Net acceleration with which the block moves

$a = \dfrac {F'}{m} = \dfrac {4N}{2kg} = 2\ ms^{-2}$

Distance travelled by the block in

$10\ s$ is

$d = \dfrac {1}{2}at^{2} = \dfrac {1}{2}\times 2\ ms^{-2}(10)^{2} = 100\ m(\therefore u = 0)$

As the applied force and displacement are in the same direction therefore angle between the applied force and the displacement is

$\theta = 0^{\circ}$ .

Hence, work done by the applied force,

$W_{F} = Fd\cos \theta = (6N) (100m) \cos 0^{\circ} = 600\ J$ .

Mohan uses one television of

$100$ W for

$10$ hrs. How much energy is consumed by Mohan?

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$E=100 kWh$
0%
$E=0.1 kWh$
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$E=1 kWh$
0%
$E=10 kWh$

Explanation

Energy consumption is given by

$E=P\times t$ ,where P= power and t= time

Given power

$P=100\ W=0.1 KW$ and time

$t=10 hour$

$E=1 kWh$

When a men walks on a horizontal surface then work done by

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friciton is zero
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contact force is non-zero
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gravity is non-zero
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None of these

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

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

Explanation

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

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

$\implies v =4\ m/s$

Work done in lifting a body is calculated by

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Mass of the body $\times$ vertical distance moved
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Force acting on a body $\times$ vertical distance moved
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Weight acting on the body $\times$ vertical distance moved
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None of the above.

If the mass of the moving object is decreased 1/4 of its mass and its velocity is increased to twice its previous velocity, what will be the kinetic energy of the object from the following?

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1/2 of the previous kinetic energy
0%
4 times of previous kinetic energy
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Kinetic energy will remain constant
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2 times of the previous kinetic energy

Explanation

We know,

Kinetic energy,

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

where,

$m=mass$ and

$v=velocity$

If the mass of the moving object is decreased

$1/4$ of its mass and its velocity is increased to twice its previous velocity,

$m'=\dfrac m4$

$v'=2v$

$KE' = \dfrac{1}{2}m'v'^{2}$

$KE' = \dfrac{1}{2}\left(\dfrac m4\times (2v)^2\right)$

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

$KE'=KE$

Which of the following bodies has the largest kinetic energy?

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

Explanation

We know that, the kinetic energy is

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

Checking all options given -

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Potential energy
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nuclear energy
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kinetic energy
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chemical energy

Let us suppose there are two balls of an equal mass shape and size. We apply an equal force on both. Then work done by the force will be:

0%
Same
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Zero
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Different
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None of the above

Explanation

Given,

Two balls of equal mass, shape, and size.

Mass of each ball,

$m$

Force applied on each ball,

$F$

We know,

Work done,

$W=F\times s$

where,

$s=\text{displacement}$

As both the balls are identical, the distance traveled by each ball will be the same.

$\therefore$ Work done by the force will be the same.

A man carries a load on his head through a distance of 5 m. The maximum amount of work is done when he

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Movies it over an inclined plane
0%
Movies it over a horizontal surface
0%
Lift it vertically upwards
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All of the above

An object is thrown vertically upwards. As it rise, its total energy

0%
Decreases
0%
Increases
0%
Remains constant
0%
None

Explanation

Total Energy = Kinetic Energy + Potential Energy

As object move upward is velocity will decrease due to deceleration by gravity.

So its kinetic energy = $\dfrac{1}{2}m{{v}^{2}}$ will decrease as it move upward.

But its potential energy = $mgH$ will increase as it move upward.

Hence, total energy will remain constant.

A body of mass

$m$ kg is lifted by a man to a height of one metre in

$30$ sec.Another man lifts the same mass to the same height in

$60$ sec. The work done by them are in the ratio

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$1 :1$
0%
$2 : 1$
0%
$4 : 1$
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$1 : 2$

Explanation

Work done

$W= F\cdot s$ .....(1)

Force applied to pull the object

$F=$ Wight of the object

$F= W=mg$ .........(2)

From equation 1 and 2

$W=mg\cdot s$ Hight attend by the mass

$=h$ (let)

$W=mgh$

$s=h$

Hence Work done is independent of time taken to lift the mass. so

$W_1:W_2 :: 1:1$ Option A

The

$P.E.$ and

$K.E.$ of a helicopter flying horizontally at a height

$400\ m$ are in the ratio

$5:2$ . The velocity of the helicopter is:

0%
$56\ m/s$
0%
$28\ m/s$
0%
$14\ m/s$
0%
$42\ m/s$

Explanation

Let mass and velocity of the helicopter be

$'m'$ and

$'v'$

So,

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

$PE=mgh$ , where

$[h=400\ m]$

Given ratio,

$\dfrac{PE}{KE}=\dfrac{5}{2}$

$\Rightarrow \dfrac{mgh}{\dfrac{1}{2}mv^2}=\dfrac{5}{2}$

$\Rightarrow v^2=\dfrac{4}{5} gh$ .

$\Rightarrow v=\sqrt{\dfrac{4}{5}\times 10\times 400}$

$\Rightarrow v=40\sqrt{2}\simeq 56.7\ m/s$ .

A staircase has

$40$ steps each of width

$35cm$ and height

$25cm$ .A boy of mass

$20kg$ ascends the staircase.The work done by him is (taken

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

0%
$2\times 10^5J$
0%
$4.8\times 10^3J$
0%
$2\times 10^{3}J$
0%
$4.8 \times 10^5J$

Explanation

$Work done = Force \times displacement$ .

$d=40\times 25cm = 1000cm$

$F=20\times 10 = 200 N$

$W=200 N \times 10 m$

$=2\times 10^3 J$

CBSE Questions for Class 9 Physics Work And Energy Quiz 11 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

CBSE Questions for Class 9 Physics Work And Energy Quiz 11 - MCQExams.com

0:0:2

Practice Class 9 Physics Quiz Questions and Answers

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