MCQ Questions for Class 9 Physics Work And Energy Quiz 3

SI unit of electrical energy is

0%
Joule
0%
Kilowatt hour
0%
KiloJoule
0%
Watt hour

Explanation

SI unit of electrical energy is Joule.

$1 Joule = 1 Volt \times 1 ampere \times 1 second$

The hydroelectric plants do not produce electricity, if the water level in the dam is less than 34 m.

0%
True
0%
False
0%
Ambiguous
0%
Data insufficient

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.

Write true or false for the following statements :
Electric power is also called BOTU.

0%
True
0%
False

A body at rest can't have

0%
Momentum
0%
Kinetic energy
0%
Both $a$ & $b$
0%
None

Explanation

Given, A body is in rest, its mean that

$v=0$

Momentum

$P=mv=m\times 0=0$

Kinetic energy

$K.E=\dfrac12 mv^2=\dfrac12m\times 0^2=0$

Hence, Momentum and Kinetic energy of body will be zero when it is in rest.

There are

$5$ tube-lights each of

$40\ W$ in a house. These are used on an average for

$5\ hrs/day$ . In addition, there is an immersion heater of

$1500\ W$ used on an average for

$1\ hr/day$ . The number of units of electricity that are consumed in a month of

$30\ days$ is:

0%
$25\ units$
0%
$50\ units$
0%
$75\ units$
0%
$100\ units$

Explanation

$E=n_1{ P }_{ 1 }{ t }_{ 1 }+n_2{ P }_{ 2 }{ t }_{ 2 }$

$E= 5 \times 40\times 5\times 30+1 \times 1500\times 30$

$=200\times 150+45000$

$E=75\ kWh$

$\Rightarrow E = 75\ units$

Water stored in a dam possesses:

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

The electrical energy in kilowatt hours consumed in operating ten

$50\ W$ bulbs for

$10\ hrs/day$ in a month of

$30\ days$ is

0%
$1500$
0%
$15000$
0%
$15$
0%
$150$

Explanation

Let

$n$ be the number of bulbs,

$n= 10$

$P$ be the power of each bulb,

$P= 50\ W$
total time to burn the bulb,

$t =10 \times 30 = 300\ hr$

Now the energy

$E$ , consumed in operating the bulbs is:
Energy consume by each built in 30 days

$E=Pt=50\times 300$

Energy consume by 10 bulbs in 30 days

$E=nPt=10 \times 50 \times 300\ Wh$

$E=150\ kWh$ (Option D)

A pump delivers water at the rate of

$2400\ litres$ in

$6\ minutes$ over a head of

$12\ m$ . If

$g = 10 m/s^{2}$ , then its power is:

0%
$200\ W$
0%
$800\ W$
0%
$240\ W$
0%
$24\ kW$

Explanation

Given,

Volume of water delivers

$V=2400\ lit$

Mass of water deliver

$V\times d_w=2400\times 1=2.4\times 10^3\ kg$

Time consumed

$t=6\ mint=6\times 60\ sec$

Power of a pump is given by,

$P = \dfrac{mg h}{t}=\dfrac{2.4\times 10^{3}\times 10\times 12}{6\times 60}$

$= 800 \ W$

A man pushes a wall and fails to displace it. He does :

0%
Negative work
0%
Positive but not maximum work
0%
Maximum work
0%
No work at all

Explanation

$\textbf{Hint}$ : We know work done in displacing the object is equal to the product of the force applied on the object and the distance traveled by it in the direction of force applied. Let us apply this concept to solve the problem.

$\textbf{Step 1}$ :

We know

$Work \quad done(W)=F.S$

This is the work done by the man when he pushes the wall.

Given that man pushes a wall and fails to displace it so displacement is zero i.e..,

$S=0$

$\textbf{Step 2}$ :

$Work \quad done(W)=F.0=0$

So work done is zero.

Thus option (D) is correct.

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

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 house is fitted with ten lamps of each

$60\ W$ . Each lamp burns for

$5\ hrs$ a day on an average. The cost of consumption in a month of

$30\ days$ at

$80$ rupees per unit is:

0%
$Rs. 152$
0%
$Rs. 252$
0%
$Rs. 352$
0%
$Rs. 452$

Explanation

Total energy spent is given by:

$E = nE'$

where

$n:$ number of bulbs

$E':$ Energy spent by each bulb

$E' = Pt$

$E' = 60 \times 5 \times 30\ Wh$

$E = 60 \times 5 \times 30 \times 10$

$= 9000 \times 10$

$= 90 KWh$

Cost

$= 90 \times 2.8$

$= 28 \times 9$

$= 252 Rs$

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

A girl is carrying a school bag of

$3\ kg$ mass on her back and moves

$200\ m$ on a leveled road. The work done against the gravitational force will be (

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

0%
$6\times 10^{3}\ J$
0%
$6\ J$
0%
$0.6\ J$
0%
$Zero$
0%
None of these

Explanation

The amount of work done is expressed in the equation:

$W = F.d cos\theta$ . In this case, the total work done is zero as the displacement is perpendicular to the applied force.

A girl of mass 40 kg climbs 50 stairs of average height 20 cm each in 50 s. Find the power of the girl. $(g = 10 m\:s^{-2})$

0%
$50\ W$
0%
$20\: W$
0%
$80\ W$
0%
$500\: W$

A 220 V, 100 W bulb is connected to a 110 V source. Calculate the power consumed by the bulb.

0%
10 W
0%
15 W
0%
20 W
0%
25 W

Explanation

$P=100W$

$V=220V$

We know

$P=\dfrac{V^2}{R}$

$R=\dfrac{V^2}{P}$ =

$\dfrac{220^2}{100}$ =

$484\Omega$

Supply=

$V=110V$

$P=\dfrac{V^2}{R}$

$P=\dfrac{110^2}{484}$ =

$25W$

If an electric iron of

$1200 \ W$ is used for

$30 \ minutes$ everyday, find electric energy consumed in the month of April.

0%
$12\ kWh$
0%
$18 \ kWh$
0%
$36\ kWh$
0%
$6\ kWh$

Explanation

$Power \ \ P= 1200 W = \dfrac{1200}{1000}=1.2 kW$

$time \ \ t = 30 \ min =\dfrac{30}{60}=0.5 h$

Number of days in the month of april =

$30$

Hence, Total time for which iron is used is given by (in hrs) =

$0.5 \times 30$

$Energy \ \ E =Power\times Total \ time$

$= 1.2\times 0.5\times 30$

$= 18 kW h$

Answer question numbers 72 and 73 based on the following information.
Solar surface radiates energy uniformly at a rate of

$4 \times 10^{26}W$ . This energy spreads or distributes uniformly and normally outwards.

Considering Earth and the Sun to be spherical object, the amount of radiant energy received by the Earth per second is nearly

0%
$1360 \:W.$
0%
$1.75 \times 10^{17}\:W$
0%
$3.5 \times 10^{17}\:W$
0%
$7 \times 10^{17}\:W$

A girl having a mass of

$35\ kg$ sits on a trolley of mass

$5\ kg$ . The trolley is given an initial velocity of

$4\ ms^{-1}$ by applying a force. The trolley comes to rest after traversing a distance of

$16\ m$ . How much work is done on the trolley?

0%
$0\ J$
0%
$320\ J$
0%
$120\ J$
0%
$250\ J$

Explanation

Given,

Mass of the girl,

$m_g=35\ kg$

Mass of the trolley,

$m_t=5\ kg$

Iniial velociy,

$u=4\ ms^{-1}$

Final velocity,

$v=0$

Displacement,

$s=16\ m$

Total mass,

$m=m_g+m_t$

$m=35+5 =40\ kg$

Using Newton's second equation of motion,

$s=ut+\dfrac 12u^2$

$\implies a=\dfrac{v^{2}-u^{2}}{2s}$

$a=-\dfrac{0^2-4^2}{2\times 16}$

$a=-0.5\ ms^{-2}$

Force,

$F=ma$

$F=40\times (-0.5)$

$F=-20\ N$

Magnitude of force,

$F=20\ N$

Work done on the trolley,

$W= (20 \times 16)\ J$

$W= 320\ J$

In a hydro power plant:

0%
Potential energy possessed by stored water is converted into electricity
0%
Kinetic energy possessed by stored water is converted into potential energy
0%
Water is heated to produce electricity
0%
Water is converted into steam to produce electricity

$1kWh=$ _________?

0%
$3600000\ J$
0%
$10000\ J$
0%
$4.2\ J$
0%
$25000\ J$

Explanation

Kilowatt hour is the unit of energy to measure the amount of electricity used in an hour

$1\ kW$ in

$1\ hr$ .

Hence,

$1\ kWh = 1 kW \times 1\ hr$

$= 10^3 W \times 3600\ s$

$= 3600000\ J$

How much work they do in just holding it ?

0%
250 J
0%
2500 J
0%
0 J
0%
625 J

Explanation

The work done by a constant force of magnitude

$F$ on a point that moves a displacement (not distance)

$s$ in the direction of the force is given by:

$W = F\times s$
As the men are just holding the box the value of

$s$ is

$0$ .
Hence, the work done is $W=Fs = F\times 0 = 0$ .

Kilowatt is the unit of electrical _______ but kilowatt-hour is the unit of electrical _______.

0%
Power, Energy
0%
Power, Electric potential
0%
Power, Current
0%
Energy, Power

Explanation

In SI system, unit of power is Watt (W). Kilowatt is equal to one thousand (

$10^3$ ) watts. The kilowatt-hour is a unit of energy and is equal to 1,000 watt-hours. If the energy is being transmitted or used at a constant rate (power) over a period of time, the total energy in kilowatt-hours is the product of the power in kilowatts and the time in hours.

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

Kilowatt-hour is the unit of :

0%
potential difference.
0%
electric power.
0%
electrical energy.
0%
charge.

Explanation

Kilowatt-hour $(KWh)$ is unit of Electrical Energy.

We know that,

$\text{Power} \times \text{time} = \text{Energy}$

$\Rightarrow\ \ \text{Kilo-Watt} \times \text{Hour} = \text{Energy}$

$1 KWh$ is electrical energy consumed when an electrical equipment of power $1\ KiloWatt$ is operated for $1\ Hour$ .
It is also used as commertial unit of energy in households/Buildings/Factories etc where 1 Unit of electricity is equivalent of 1KW-h of energy.

Hence, Option C is correct.

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.

An electric kettle is rated as

$2.5\ kW, \ 250\ V$ . Find the cost of running the kettle for two hours at

$60\ \text{paisa}/\text{unit}$ .

0%
$Rs\ 6$
0%
$Rs\ 3$
0%
$Rs\ 9$
0%
$Rs\ 1$

Explanation

$Rate=0.60 Rs/kWh$

$Energy=Power \times time$

Energy consumed,

$E =2.5 \times 2 = 5\ kWh$

Cost,

$C = 5 \times 0.60 = Rs\ 3$

An air conditioner is rated

$240\ V, 1.5\ kW$ . The air conditioner is switched on

$8\ hrs$ each day. What is electrical energy consumed in

$30\ days$ ?

0%
$2.88\ kWh$
0%
$360\ kWh$
0%
$120\ kWh$
0%
$240\ kWh$

Explanation

Energy consumed is given by:

$E= Pt$

$E$ : Energy consumed

$P$ : Power

$t$ : Time

$\Rightarrow E = 1.5\ kW \times 30 \times 8\ hrs$

$\Rightarrow E= 360\ kWh$

Unit of work in MKS system is

0%
Calorie
0%
Newton
0%
Joule
0%
Watt

Explanation

MKS system: A unit system of measurement based on the metric system and having the meter of length, the kilogram of mass, and the second of time as its fundamental units.

Work = Force

$\times$ displacement

MKS unit system,

Unit of work = N-m or joule

Option C

CBSE Questions for Class 9 Physics Work And Energy Quiz 3 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

CBSE Questions for Class 9 Physics Work And Energy Quiz 3 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

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