MCQ Questions for Class 9 Physics Work And Energy Quiz 7

The unit for the measurement of power is:

0%
$Watt$
0%
$Kilowatt$
0%
$Kilowatt-hour$
0%
$Horsepower$

Explanation

Power is represented by energy per second.

$Power=\quad \dfrac { Energy }{ time } =\quad \dfrac { Joule }{ Second } =\quad Watt$

The

$kilowatt$ is the higher unit of power.

$Kilowatt-hour$ is a commercial unit of energy.

$Horsepower$ is also a unit of power.

You lift a suitcase from the floor and keep it on the table. The work done by you on the suitcase depends on

0%
the path taken by the suitcase
0%
your weight
0%
the weight of the suitcase
0%
the time taken by you in doing so

Explanation

The work done by a person in lifting an object is stored as its potential energy

$mgh$ .

Hence, work done depends on the weight of the object

$mg$ .

Thus option C is correct.

A body at rest may have

0%
speed
0%
velocity
0%
energy
0%
momentum

The work done in lifting 1 kg mass to a height of 9.8 m is about

0%
1 J
0%
9.8 J
0%
$(9.8)^2 J$
0%
$\displaystyle \frac{1}{(9.8)^2} J$

Explanation

The work done in lifting an object is stored as its potential energy.

Mass of object

$m = 1$ kg

Height upto which it is lifted

$h = 9.8$ m

Hence work done

$W= m g h = 1 \times 9.8 \times 9.8 = (9.8)^2 J$

Thus option C is correct.

Work done by centripetal force in revolving a satellite around the earth is

0%
Zero
0%
Unity
0%
Infinity
0%
Nothing can be decided

If a body moves along a frictionless horizontal surface its weight:

0%
Does work
0%
Does not work
0%
Work partially
0%
None of these

Explanation

Work done by the force of gravity on an object (weight) depends on vertical displacement.

Vertical displacement,

$h = 0$

Mass,

$m$

Acceleration due to gravity,

$g$

Work done,

$W = mgh$

$W=mg\times 0$

$W = 0\ J$

$\therefore$ If a body moves along a frictionless horizontal surface its weight does not work.

$kg m^2s^{-2}$ represents the unit of :

0%
kinetic energy only
0%
work done only
0%
potential energy only
0%
all the above

Explanation

Work done is same as energy which is measured in joules. 1 joule is same as 1 newton expanded over 1 meter which is force multiplied by distance.

Force is defined as mass into acceleration. Hence if we accelerate 1kg mass by 1m/s

$^{2}$ in a straight line for a distance of 1 meter, we have indirectly expanded 1 joule of energy .

Also acceleration is distance per second squared.

Thus we have (mass

$\ast$ acceleration

$\ast$ distance) which is Kg m

$^{2}$ S

$^{-2}$ .

A wound watch spring has ______ energy.

0%
Mechanical
0%
Kinetic
0%
Potential
0%
Kinetic and potential

Asha lifts a doll from the floor and places it on a table. If the weight of the doll is known, what else does one need to know in order to calculate the work Asha has done on the doll?

0%
the time required
0%
the height of the table
0%
the power she could deliver
0%
cost of the doll or the table

If the speed of a motor car becomes six times, then the kinetic energy becomes

0%
6 times
0%
36 times
0%
12 times
0%
24 times

Explanation

Kinetic energy, K.E =

$\dfrac{1}{2}$ mv

$^{2}$

If speed becomes 6 times, then

New K.E =

$\dfrac{1}{2}$ m (6v)

$^{2}$ =

$\dfrac{1}{2}$ m x 36 v

$^{2}$ = 36 x (

$\dfrac{1}{2}$ mv

$^{2}$ ) = 36 x K.E

Hence, kinetic energy becomes 36 times.

The flowing water of a river primarily possesses ______ energy.

0%
gravitational
0%
potential
0%
electrical
0%
kinetic

When a body moves with a constant speed along a circle, then which of the following is correct :

0%
no force acts on it
0%
its velocity remains constant
0%
no work is done on it
0%
no acceleration is produced in it

Two bodies of equal weight are kept at heights of h and 1.5 h respectively. The ratio of their P.E. will be :

0%
3: 2
0%
2: 3
0%
1: 1
0%
none of these

Explanation

$\underline{\text{First body:}}$

Mass of the body =

$m$

Height at which the body is kept =

$h$

Potential energy of the body

$U_1=mgh$

$\underline{\text{Second body:}}$

Mass of the body =

$m$

Height at which the body is kept =

$1.5h$

Potential energy of the body

$U_2=1.5mgh$

The ratio of potential energy is

$U_1:U_2=1:1.5=\dfrac{2}{3}=2:3$

A body rolling down a hill has :

0%
K.E. only
0%
P.E. only
0%
neither K.E. nor P.E.
0%
both K.E. and P.E

The kinetic energy of a body is increased by increasing its:

0%
Mass
0%
Weight
0%
Velocity
0%
Potential energy

Explanation

Kinetic Energy, $KE=\dfrac{1}{2}mv^{2}$

(where, $m=$ mass of the body and $v=$ velocity)

$KE\propto mv^{2}$

Kinetic energy can be increased by either increasing mass or velocity.

Kilowatt hour (kWh) represents the unit of :

0%
power
0%
impulse
0%
momentum
0%
energy

Explanation

1 kW h = 1 kW ×1 h = 1000 W × 3600 s = 3600000 J = 3.6

$\ast$ 10

$^{6}$ J

The energy used in households, industries and commercial establishments are usually expressed in kilowatt hour.

The unit of power is:

0%
watt per second
0%
joule
0%
kilo joule
0%
watt

Explanation

Unit of energy is Joule (J).

Power is energy per unit time J/s. Called watt.

So option A not unit of power as unit of power is watt so watt per second can not be unit of power.

Option B is unit of energy.

Option C is also unit of energy. =

$10^3$ J.

Option D is unit of power.

So only correct option is option D.

A ball of mass

$200 g$ falls from a height of

$5 m$ . What is its K.E. when it just reaches the ground ?

0%
$9.8 J$
0%
$98 J$
0%
$980 J$
0%
None of these

Explanation

Given,

$M=200\ g=.2\ kg,\ \ \ \ \ \ h=5m$

From conservation of energy,

Gain in kinetic energy

$K.E =$ loss in

$P.E$

initial P.E

$=$ mgh

initial K.E

$=0$

final P.E

$=0$

$K.E_f-K.E_i=P.E_i-P.E_f$

$K.E_f-0=0.2 \times 9.8\times 5-0\\K.E_f =9.8\quad J$

The number of joules contained in

$1 kWh$ is

0%
$36 \times {10}^{2}$
0%
$36 \times {10}^{3}$
0%
$36 \times {10}^{4}$
0%
$3.6 \times {10}^{6}$

Explanation

1 KWh = 1000

$\ast$ joule/sec

$\ast$ 60

$\ast$ 60 sec

1KWh = 1000

$\ast$ 60

$\ast$ 60joue/sec

$\ast$ sec

1KWh = 1000

$\ast$ 60

$\ast$ 60 joule

1KWh = 36,00,000 joule = 3.6

$\ast$ 10

$^{6}$

An object of mass

$1 kg$ has a P.E. of

$1 J$ relative to the ground when it is at a height of

0%
$0.1 m$
0%
$1 m$
0%
$9.8 m$
0%
$32 m$

Explanation

$P.E$ is given by

$=mgh$

where

$h$ is the height,

$m$ is mass and

$g$ is acceleration due to gravity

$1J=1\times 10\times h\\ \\$

$h=0.1m$

A raised hammer possesses :

0%
K.E.
0%
P.E.
0%
electrical energy
0%
sound energy

A body at rest can have:

0%
speed
0%
energy
0%
momentum
0%
velocity

Work done upon a body is

0%
a vector quantity
0%
a scalar quantity
0%
always positive
0%
always negative

The mass of an object

$P$ is double the mass of

$Q$ . If both move with the same velocity, then the ratio of K.E. of

$P$ to

$Q$ is

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

Explanation

$M_P =2 \times M_Q$

$V_P = V_Q$

$K.E_P: K.E_Q = ?$

$\implies \dfrac{1}{2}M_PV_P^2 : \dfrac{1}{2}M_QV_Q^2$

Since

$V_P$ =

$V_Q$

$\implies M_P : M_Q$

$\implies 2 \times M_Q : M_Q$

$\implies 2 : 1$

Mechanical work done is equal to (symbol have their usual meanings)

0%
$W=\frac { F }{ S }$
0%
$W=FS$
0%
$W=F+S$
0%
$W=F-S$

A person walks

$100$ m, holding a suitcase of weight

$100$ N. The work done by the applied force is ____ J.

0%
$10000$
0%
$100$
0%
zero
0%
$5000$

Explanation

In the given scenario, Direction of force on the suitcase is vertically upward and this force will make

$90^\circ$ angle with displacement vector.

So there is no motion in the direction of force applied. So work done by girl on the suitcase will be zero.

Option C

2106240_398513_ans_5cf91d65aab64715b74ef98f9ab456aa.png

The product of force and displacement equals:

0%
work
0%
power
0%
momentum
0%
none of these

(i) What is the work done by the porter when he climbs up a height of 10 m

$\displaystyle (g\quad = 10\ {ms} ^{-2})$ ?

0%
$5kJ^2$
0%
$50kJ$
0%
$100kJ^2$
0%
$5kJ$

Explanation

The work done by the potter is defined as the product of the force and the displacement.

Work done = force

$\times$ displacement

$m \times g \times 10$ ( Since force = mass

$\times$ gravity)

$50 \times 10 \times 10$

$5KJ$

A force of

$5\ N$ causes a displacement of

$10\ m$ in its direction. Calculate the work done by the force.

0%
$15\ J$
0%
$25\ J$
0%
$50\ J$
0%
$200\ J$

Explanation

Given,

Force applied,

$F=5\ N$

Displacement,

$d=10\ m$

We know,

Work done,

$W=F\times d$

$W=5\times 10$

$W=50\ J$

What is the work done by a force of 2 N in displacing a body by 2 m in the direction of the force?

0%
$4J^2$
0%
$6J$
0%
$4J$
0%
$48J$

Explanation

Work done is given by

$Work \quad done=Force \times Displacement$

$Force=2N$

$Displacement=2m$

$Work\quad done$ =

$2 \times 2$ =

$4J$

CBSE Questions for Class 9 Physics Work And Energy Quiz 7 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

CBSE Questions for Class 9 Physics Work And Energy Quiz 7 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

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