not defined at a position in space

MCQ Questions for Class 9 Physics Work And Energy Quiz 14

Some children were trying to find out which of three bulbs was brightest. Which one of these gives the best start toward finding the answer?

0%
"One bulb looks the brightest to me, so I already know the answer."
0%
"It would help if we had a way to measure the brightness of a light bulb."
0%
"All the bulbs look bright to me, so there cannot be an answer."
0%
"We can take a vote and each person will vote for the bulb he or she thinks is the brightest ."

A man carries a heavy box on his head on a horizontal plane from one place to another. In this case, he does

0%
Maximum work
0%
No work
0%
Negative work
0%
Minimum work

Explanation

$\underline{\text{A man carrying a box on his head:}}$

In this case, the force acting on the man is the weight of the box, which is acting downwards.

The displacement of the man is along the horizontal direction.

Therefore, the force and the displacement are perpendicular to each other. So there is no work involved in this process.

When a body does some work its energy decreases.

0%
True
0%
False

A car weighing 600 kg traveling with

$72\, km/h$ stops at a distance of

$60\ m$ decelerating uniformly. What is the work done by brakes in joules?

0%
$1.2 \times 10^5\ J$
0%
$2.4 \times 10^5\ J$
0%
$3.2 \times 10^5\ J$
0%
$1.4 \times 10^5\ J$

Explanation

Given,

Initial velocity,

$u=72\ km/h$

$u=\dfrac{72\times 1000}{3600}\ m/s=20\ m/s$

Final velocity,

$v=0\ m/s$

Displacement

$s=60\ m$

Using Newton's third equation of motion,

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

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

$\implies a=\dfrac{0^2-20^2}{2\times 60}$

$\implies a=\dfrac{400}{120}$

$a= -\dfrac{10}{3}\ m/s^2$

Magnitude of acceleration,

$a=\dfrac{-10}{3}\ m/s^2$

Force

$F=ma=600\times (10/3)$

$F=2000\ N$

Work done,

$W=F \times S$

$W=2000\times 60$

$W=120000 J$

1 kWh is equal to

0%
$3.6 \times 10^6 MJ$
0%
$3.6 \times 10^5 MJ$
0%
$3.6 \times 10^2 MJ$
0%
$3.6 MJ$

Explanation

1 kilowatt hour is the energy produced by 1 kilowatt power source in 1 hour.

$1kWh=1kW\times 1hour=1000\times 3600 W.s$

$\implies 1kWh=3.6\times 10^6J$

$\implies 1kWh=3.6MJ$

Answer-(D)

Energy cannot be measured in.

0%
$Js^{-1}$
0%
$Ws$
0%
$kWh$
0%
$erg$

Explanation

Energy can be measured in Ws (i.e watt second), kWh and erg.

$Js^{-1}$ is the unit of power

$P=\dfrac{E}{t}$ , energy can not be measured in that.

In which of the following, work is being done?

0%
Man sitting on a bench
0%
Person Standing with a basket of fruit on the head.
0%
Climbing a tree to pluck.
0%
Pushing a wheelbarrow of bricks.

Sunil rolls a marble ball down a frictionless inclined plane as shown above. What kind of energies are possessed by the rolling marble ball?

0%
(i) and (ii) only
0%
(ii) and (iii) only
0%
(iii) and (i) only
0%
(i), (ii) and (iii)

Work done in time

$t$ on a body of mass

$m$ which is accelerated from rest to a speed

$v$ in time as a function of time

$t$ is given by

0%
$\cfrac { 1 }{ 2 } m\cfrac { v }{ { t }_{ 1 } } { t }^{ 2 }\quad$
0%
$m\cfrac { v }{ { t }_{ 1 } } { t }^{ 2 }$
0%
$\cfrac { 1 }{ 2 } m{ \left( \cfrac { v }{ { t }_{ 1 } } \right) }^{ 2 }{ t }^{ 2 }$
0%
$\cfrac { 1 }{ 2 } m\cfrac { v }{ { t }_{ 1 }^{ 2 } } { t }^{ 2 }$

Explanation

Ans : We know

$W = F \times s$

where,

$F = ma$

By using Kinematics,

$s = \dfrac{1}{2}at^{2}$ , if u =0

So,

$W = (ma).(\dfrac{1}{2}at^{2})$

$W = \dfrac{1}{2}ma^{2}t^{2}$ where

$a = v/t_{1}$

$\boxed{W = \dfrac{1}{2}m(\dfrac{v}{t_{1}})^{2}t^{2}}$

A weight lifter lifts a weight off the ground and holds it up then:

0%
Work is done in lifting as well as holding the weight
0%
No work is done in both lifting and holding the weight
0%
Work is done in lifting the weight but no work is required to hold it up
0%
No work is done in lifting the weight but work is required to be done in holding it up

A particle is taken from A to point B under the influence of a force field. Now it is taken from B to A and it is observed that the work done in taking the particle from A to B is not equal to the work done in taking it from B to A. If

${W}_{nc}$ and

${W}_{C}$ are the work done by non-conservation and conservative force field respectively and

$\Delta U$ and

$\Delta K$ , be the change in P.E and K.E then

0%
${ W }_{ nc }-\Delta U=\Delta K$
0%
${ W }_{ C }=-\Delta U$
0%
${ W }_{ nc }+{ W }_{ C }=\Delta K$
0%
${ W }_{ nc }=-\Delta U=-\Delta K$

An athlete in the Olympic games covers a distance of

$100$ m in

$10$ s. His kinetic energy can be estimated to be in the range. (Assume weight = 60kg)

0%
$200J-500J$
0%
$2\times 10^5J-3\times 10^5J$
0%
$20000J-50000J$
0%
$2000J-5000J$

Explanation

Velocity

$V=\dfrac{distance}{time}=\dfrac{100}{10}=10 m/s$

Assuming his mass to be 60kg, his kinetic energy is

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

$= \dfrac{1}{2}\times 60\times 100$

$=60\times50$

$=3000 J$

Hence, Option D

Work is

0%
not defined at an instant
0%
not defined at a position in space
0%
independent of force acting
0%
a vector quantity

Explanation

Work done is defined as.

$Work=Force\times displacement$

So work clearly depends on force so option C is wrong.

Work is a scalar quantity.

Work is not defined at an instant but it is defined for the whole duration of displacement.And it is also not defined for any particular position in space.

A man

${M}_{1}$ of mass

$80kg$ runs up a staircase in

$15s$ . Another man

${M}_{2}$ also of mass

$80kg$ runs up the stair case in

$20s$ . The ratio of the power developed by them will be:

0%
$1$
0%
$4/3$
0%
$16/9$
0%
None of the above

Explanation

Let the height of the staircase is

$h$ .

$\underline{\text{Power of the first man:}}$

Work done by the man =

$mgh=80\times g\times h$

Time taken by the man to ascend the staircase is

$15\ s$

Power of the man

$P_1=\dfrac{mgh}{t}=\dfrac {80\times g\times h}{15}\ W$

$\underline{\text{Power of the second man:}}$

Work done by the man =

$mgh=80\times g\times h$

Time taken by the man to ascend the staircase is

$20\ s$

Power of the man

$P_2=\dfrac{mgh}{t}=\dfrac {80\times g\times h}{20}\ W$

Thus,

$\dfrac{P_1}{P_2}=\dfrac{20}{15}=\dfrac{4}{3}$

Two bodies

$A$ and

$B$ have masses

$20\ kg$ and

$5\ kg$ respectively. If they acquire the same kinetic energy. Find the ratio of thier velocities.

0%
$\dfrac {1}{2}$
0%
$2$
0%
$\dfrac {2}{5}$
0%
$\dfrac {5}{6}$

Explanation

Given:

$m_{A} = 20 kg$

$m_{B} = 5 kg$

Kinetic Energy of both bodies A and B are the same.

Thus,

$\dfrac{1}{2} m_{A} V_{A}^{2} = \dfrac{1}{2} m_{B} V_{B}^{2} \\$

$\dfrac{V_{A}^{2}}{V_{B}^{2}} = \dfrac{m_{B}}{m_{A}} \\$

$\dfrac{V_{A}^{2}}{V_{B}^{2}} = \dfrac{5}{20}\\$

$\dfrac{V_{A}}{V_{B}} = \dfrac{1}{2}$

So, option A is correct.

If increase in linear momentum of a body is 50%, then change in its kinetic energy is

0%
25%
0%
125%
0%
150%
0%
50%

A man displaces a block by

$5\ cm$ on a rough surface

$(\mu=0.5)$ by applying a force

$50\ N$ acting at

$37^{o}$ to the horizontal. The work done by the applied force is

0%
$200\ J$
0%
$100\ J$
0%
$150\ J$
0%
$50\ J$

A block of mass

$10kg$ is moving in x-direction with a constant speed of

$10m/sec$ . It is subjected to a force

$F=-0.1x$ Joules/meter during its travel from

$x=20$ meters to

$x=30$ meters. Its final kinetic energy will be-

0%
$475$ joules
0%
$450$ joules
0%
$275$ joules
0%
$250$ joules

The potential energy of a particle varies with position

$x$ according to the reaction

$U(x) = 2x^{4} - 27x$ the point

$x = 3/2$ is point of

0%
Unstable equilibrium
0%
Stable equilibrium
0%
Neutral equilibrium
0%
None of these

Explanation

Hint: Use the formula $F=-\dfrac{dU}{dx}$

A small solid sphere of mass

$m$ is released from a point

$A$ at a height

$h$ above the bottom of a rough track as shown in the figure. If the sphere rolls down the track without slipping, its rotational kinetic energy when it comes to the bottom of track is

0%
$mgh$
0%
$\cfrac{10}{7}mgh$
0%
$\cfrac{5}{7}mgh$
0%
$\cfrac{2}{7}mgh$

A worker lifts a

$20.0 kg$ bucket of concrete from the ground up to the top of a

$20.0 m$ tall building. What is the amount of work that the worker did in lifting the bucket?

0%
$3.92 kJ$
0%
$400 J$
0%
$560 kJ$
0%
$4.08 J$

Explanation

Given:

The mass of the bucket $m=20\ kg$

The height of the building $h=20\ m$

$\text{Work done by the worker} = \text{Potential energy of bucket at the top of the building}$

$\therefore W = mgh$

$\Rightarrow W = 20 \times 9.8 \times 20$

$\Rightarrow W = 3920\;{\rm{J}}$

Thus, the required work done is $3.92\;k{\rm{J}}$

A solid sphere of mass

$4kg$ and radius

$1m$ is rotating about the given axis

$xx'$ with angular velocity

$10rad/s$ shown in figure. The

${K.E}_{rotation}$ is given by

0%
$180J$
0%
$200J$
0%
$240J$
0%
$280J$

The average kinetic energy of an idea gas per molecule at

$25^{o}C$ , will be

0%
$6.1\times10^{-20}J$
0%
$6.1\times10^{-21}J$
0%
$6.1\times10^{-22}J$
0%
$6.1\times10^{-23}J$

Three equal resistances each of

$10\Omega$ are connected. The maximum power consumed by each resistor is

$20W$ . the maximum power consumed by the system is

0%
60 W
0%
30 W
0%
15 W
0%
13.3 W

The mass of object X is

$M_1$ and that of object Y is

$M_2$ . Keeping their kinetic energy constant, if the velocity of object Y is doubled the velocity of object X, what will be the relation between their masses?

0%
$M_1=2M_2$
0%
$4M_1=M_2$
0%
$M_1=4M_2$
0%
$2M_1=M_2$

Explanation

Mass of object X =

$M_1$ , Velocity of object X =

$v_1$

Mass of object Y =

$M_2$ , Velocity of object Y =

$v_2$

Also,

$v_2$ = 2

$v_1$

K.E is constant.

K.E $_1$ = K.E $_2$

$\dfrac{1}{2}$ $M_1$ $v_1^2$ = $\dfrac{1}{2}$ $M_2$ $v_2^2$

$\dfrac{1}{2}$

$M_1$

$v_1^2$ =

$\dfrac{1}{2}$

$M_2$

$(2v_1)^2$

$\dfrac{1}{2}$

$M_1$

$v_1^2$ =

$\dfrac{1}{2}$

$M_2$

$4\ v_1^2$

$M_1$ = 4

$M_2$

A car weighing 1 ton is moving twice as fast as another car weighing 2 ton. The kinetic energy of the one-ton car is

0%
less than that of the two-ton car is
0%
some as that of the two-ton car is
0%
more than that of the two-ton car is
0%
impossible to compare with that of the two-ton car unless the height of each

Explanation

$K_1=\dfrac{1}{2}(1)(2v)^2=2v^2$

$K_2=\dfrac{1}{2}(2)(v)^2=v^2$

$\implies K_1 > K_2$

The work function of tungsten is

$4.50eV$ . The wavelength of fastest electron emitted when light whose photon energy is

$5.50eV$ falls on tungsten surface, is

0%
$12.27\mathring {A}$
0%
$0.286\mathring {A}$
0%
$12400\mathring{A}$
0%
$1.227\mathring{A}$

Two riffles fire the same number of bullets in a given interval of time. The second fires bullets of mass twice that fired by the first and with a velocity that is half that of the first. The ratio of their powers is?

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

Explanation

$\textbf {Hint :}$ Use

$Power=\dfrac{Energy}{time}$

$\textbf{Step 1: Calculating power of first riffle}$

Let the mass of bullet be

$m$ and velocity be

$v$

Power of first riffle

$P_{1}$

$\dfrac{Energy}{time}=\dfrac{\frac{1}{2}mv^{2}}{t}........(1)$

$\textbf{Step 2: Calculating power of second riffle}$

Given mass becomes

$2m$ and velocity becomes

$\dfrac{v}{2}$

Power of second riffle

$P_2$

$=\dfrac { \frac { 1 }{ 2 } \times 2m\times \dfrac { v^{ 2 } }{ 4 } }{ t } ........(2)$

$\textbf{Step 3: Ratio}$

Equating equation (1) and (2)

$\Rightarrow \dfrac { P_{ 1 } }{ P_{ 2 } } =2:1$

${\textbf{Correct option: D}}$

Refer above figure.

An electric bulb 'P' rated 220 V, 60 W and, another identical bulb 'Q' is connected across the mains as shown here.
The power consumed is _ _ _ _ _

0%
20W
0%
40W
0%
30W
0%
60W

Explanation

Given: Rating of

$P-220V,60W$

Also, resistance of P and Q are identical

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

$\boxed{R=\dfrac{V^2}{P}=\dfrac{220\times 220}{60}}$ ...............(1)

Now, total resistance

$R+R=2R$

$=\dfrac{220\times 220}{60}\times 2$

$V=IR_{net}$

$I=current$

$\therefore$ Current =

$\dfrac{V}{R_{net}}=\dfrac{220\times 60}{220\times 220\times 2}=\dfrac{30}{220}$

$\Rightarrow$ Current (I)

$=\dfrac{3}{22}$

$\therefore Power=I^2R_{net}=\dfrac{3}{22}\times \dfrac{220\times 220\times 2}{60}\times \dfrac{3}{22}=30W$

Option C is correct.

2085506_1263347_ans_b92ec463885f432b9044c771c74d3b59.png

A pump of

$200W$ power is lifting

$2kg$ water from an average depth of

$10m$ in one second. Velocity of water delivered by the pump is :

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

0%
$10m/s$
0%
$2m/s$
0%
$4 m/s$
0%
$1 m/s$

Explanation

acceleration due to gravity

$g = 10m/sec^2$

height

$h = 10\ meters$

mass

$= 2kg$

Potential energy

$= mgh = 2\times 10 \times 10 = 200J$

as power

$= \dfrac{work \ done}{time}$

when power of motor

$= 200w$

$200 = 200/t$

$\Rightarrow t = 1sec$

here displacement of water= height

$= 10m$

time = 1sec

Hence,

Velocity

$V= \dfrac{Displacement}{Time}$

$V= \dfrac{10}{1}$

$V= 10m/sec$

CBSE Questions for Class 9 Physics Work And Energy Quiz 14 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

CBSE Questions for Class 9 Physics Work And Energy Quiz 14 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

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