Work And Energy Test

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Work And Energy Test - 9

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Question 1
1 / -0

A body is moved through a distance of 5 m in the following different ways. Maximum work is done against gravity when the body is

A
pushed over an inclined plane

B
raised vertically upwards

C
pushed over smooth rollers

D
pushed on a plane horizontal surface

Solution

Maximum work is done against gravity when the body is raised vertically upwards. This is because the displacement against the force of gravity is the maximum in this case.
Question 2
1 / -0

If the momentum of a body increases by 20%, the increase in its K.E. will be

A
20%

B
44%

C
60%

D
80%

Solution

Let initial velocity = v₁
Then, final velocity v₂ = v₁ +
= 1.2 v₁
This is because mass is constant and therefore, increase in momentum increases the velocity.
Hence,
Initial kinetic energy KE₁ =
Final kinetic energy KE₂ =
=
= (1.44)
Change =
=
=
Question 3
1 / -0

Work done by centripetal force during circular motion is

A
(mv²/r)

B
zero

C
mv²/r²

D

Solution

Work done by centripetal force during circular motion is zero. This can be seen from the figure below.

As F is at right angle to the direction of motion, therefore,
W = Fd cos
= Fd cos90° = Fd(0)
= 0
Question 4
1 / -0

What is the change in potential energy as a body climbs a distance h and descends a distance h/2? (Assuming g being constant)

A
mgh

B
mgh/2

C
2mgh

D
mgh/4

Solution

Initial height = h
PE₁ = mgh
Final height =
PE₂ =

=
Or | =
Question 5
1 / -0

A work of 2 milli Joules changes the speed of a moving body of mass 0.1 gram. The change in speed

A
is 0.1 m/s

B
is 0.2 m/s

C
is 0.3 m/s

D
depends on the original speed

Solution

The change in speed will be calculated only if the initial speed is known.
This can be understood from the formula given as,
W =
2 10^-3 J = -
Or 2 10^-3 J =
Therefore, for calculation of change in speed we need both v₁ and v₂ which depends upon the original speed.
Question 6
1 / -0

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?

A
Required time

B
Height of the table

C
Mass of the doll

D
Cost of the doll

Solution

Asha lifts a doll from the floor and places it on a table. Therefore, she does work against gravity.
Now weight of the doll is known, therefore, for calculating work the height of table is required.
As we know, =
= mgh
We know mg, thus h is to be known.
Question 7
1 / -0

Two masses of 1 g and 4 g are moving with KE in the ratio 4 : 1. What is the ratio of their linear momenta?

A
1 : 1

B
1 : 2

C
4 : 1

D
6 : 1

Solution

Mass of bodies = 1 g and 4 g
Let velocity of bodies be v_A and v_B

Ratio of momentum = m_Av_A : m_Bv_B
=
= 1 : 1
Question 8
1 / -0

A water fall is 84 m high. Assuming that half the KE of the falling water gets converted to heat, the rise in temperature of water is

A
0.0098°C

B
0.098°C

C
0.98°C

D
9.8°C

Solution

Height of waterfall = 84 m
Assuming that half the KE is converted to heat,
Let us consider mass of water to be 1 kg

= 823.2 J (Half of it is used for raising the temperature)
Therefore, if c is specific heat and is rise in temperature, then
= = 411.6 J
= 98.31 cal
1000 1 T = 98.31 cal
T = 0.098° C
Question 9
1 / -0

A ball is dropped vertically from a height d above the ground. It hits the ground and bounces up vertically to a height d/2. Neglecting subsequent motion and air resistance, its velocity v varies with the height h above the ground as

A

B

C

D

Solution
Question 10
1 / -0

There is a hemispherical bowl of radius 3m. A block of mass 10 kg slides from the rim of the bowl to the bottom. The velocity of the block at the bottom will be

A

B

C

D

Solution

Initial P.E. of the block = mgh
= 10(g) (3)
= 30 g J

Initial K.E. = 0
Final K.E. =
Now, =
= mgh
= 30 g
Therefore, V = m/s
Question 11
1 / -0

a wheelbarrow is an example of a lever of

A
1^st order

B
2^nd order

C
3^rdorder

D
none of the above

Solution

A second-class lever or a 2nd order lever is a stick where the fulcrum is at one end of the stick, you push on the other end, and the weight is in the middle of the stick. Some common second-class levers are doors, staplers, wheelbarrows, and can openers. Thus, a wheelbarrow is a 2nd order lever.
Question 12
1 / -0

Which of the following does not represent a simple machine?

A
Lever

B
Pulley

C
Gear

D
Steam engine

Solution

Steam engine is not a simple machine as compared to a lever, pulley and a gear.
Question 13
1 / -0

What is the power of a pump, which can pump 500 kg of water from a depth of 40 m? (Take g as 10 m/s².)

A
20 Kw

B
2 Kw

C
0.2 Kw

D
200 Kw

Solution

Power = = J/S
Work done =
Work done = mgh
= (500) (10) (40)
= 2 10⁵ J
Power = = 200 kW
Question 14
1 / -0

From a waterfall, water is pouring down at the rate of 100 kg per second on the blades of a turbine. If the height of the fall be 200 m, the power delivered to the turbine is (approximately) equal to

A
200 kW

B
1 W

C
1 kW

D
200 W

Solution

Power = (Here, W is work and T is time)
W = = 100 10 200
= 2 10⁵ J
Question 15
1 / -0

An engine develops 10 kW of power. How much time will it take to lift a mass of 400 kg to a height of 80 m? (g = 10 m/s²)

A
34 s

B
35 s

C
32 s

D
10 s

Solution

P = 10 kW
h = 80 m
m = 400 kg
Work done = mgh
= (400) (10) (80)
= 32 10⁴ J
P = 10 10³ W
P = 10 10³ =
t = 32 seconds
Question 16
1 / -0

A tape-recorder records sound in the form of

A
electrical energy

B
magnetic field on the tape

C
variable resistance on the tape

D
sound waves held on the tape

Solution

A tape recorder records sound in the form of magnetic field on the tape.
Question 17
1 / -0

A moving body need not have

A
potential energy

B
kinetic energy

C
momentum

D
velocity

Solution

A moving body has a speed or a velocity with respect to another stationary object.
Hence, it has 1) KE =
2) Momentum = mv
3) Velocity
The only thing it might not have is potential energy.
Question 18
1 / -0

Water is falling from the top of a high water fall. On falling,

A
it evaporates

B
it freezes

C
there is no change in its condition

D
it gets slightly warmed up

Solution

Water falling from the top of a waterfall gets slightly warmed up because some of the KE possessed by it is converted to heat energy upon striking the ground.
Question 19
1 / -0

A spring, which is initially in its unstretched condition, is first stretched by a length X and then again by a further length X. The work done in the first case is W₁ and that in the second case is W₂. Which of the following relations between W₁ and W₂ is correct?

A
W₂ = W₁

B
W₂ = 2W₁

C
W₂ = 3W₁

D
W₂ = 4W₁

Solution

W₁ is the work done when a spring is stretched by a length X.
W₁ = (K is the spring constant)
W₂ is the work done when a spring is stretched by another length X.
W₂ =
=
Therefore,
W₂ =
W₂ = 3W₁
Question 20
1 / -0

A mass of 2 kg falls from a certain height onto a spring of force constant 625 N/m. If the total distance traveled by the mass, from its point of release to its final position, is 40 cm, then the spring is compressed by

A
15.8 cm

B
158 cm

C
12.8 cm

D
128 cm

Solution

Mass of body, m = 2 kg
Spring constant, k = 625 N/m
Let the spring be compressed by a length x and the distance traveled by the block before hitting the spring be h.
Decrease in the gravitational potential energy of the mass = Increase in the elastic potential energy of the spring

Given, h + x = 40 cm

Or, x = 0.158 m or 15.8 cm.