Work And Energy Test

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Work And Energy Test - 8

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Question 1
1 / -0

One kWh is equal to

A
3.6 ´ 10⁶ J

B
3.6 ´ 10⁵ J

C
3.6 ´ 10⁴ J

D
3.6 ´ 10³ J

Solution

One kWh = 3.6 10⁶ J
= 10³
= 10³ 3600
= 3.6 10⁶ J
Question 2
1 / -0

A body covers a distance of 10 m along a straight line under the action of a force of 5 N. If the work done is 25J, then the angle that the force makes with the direction of motion of the body is

A
0°

B
30°

C
45°

D
90°

Solution

W = (dot product)
or W = Fd cos
F = 5 N
d = 10 m
W = J
= 5(10)(cos )
= cos
Therefore, = cos^-1
cos^–1= 45°
Question 3
1 / -0

The graph given below represents the relation between displacement x and force F. The work done in displacing an object from x = 8 m to x = 14 m is (approximately)

A
30 J

B
40 J

C
8 J

D
16 J

Solution

Let us consider the given graph:

The work done is given by,

The given graphs is drawn between force (F) and displacement (d). Hence, area under the graph represents the work done.

Hence, the work done in displacing an object from x = 8 m to x = 14 m is (approximately),
Question 4
1 / -0

If the momentum of a body increases by 20%, then percentage increase in kinetic energy is

A
40%

B
44%

C
48%

D
56%

Solution
Question 5
1 / -0

Two bodies of mass 2m and 3m are approaching each other at velocities 4v and 5v, respectively. After a while, their velocities become 6v and 7v, respectively. What is the total change in kinetic energy?

A

B

C

D

Solution

The mass of bodies:
m₁ = 2m, m₂ = 3m

Case-1:
Velocities of the bodies,
v₁ = 4v, v₂= 5v

The total kinetic energy of bodies:

Case-2:
Velocities of the bodies,
v₁ = 6v, v₂ = 7v
The total kinetic energy of bodies:

Total change in kinetic energy:
Question 6
1 / -0

Two bodies of equal weights are kept at heights h and 1.5 h respectively. The ratio of their potential energy is

A
3 : 2

B
1 : 1

C
2 : 3

D
3 : 4

Solution

Two bodies of equal weights are kept at h and 1.5 h, respectively.
P.E. of weight at height h = mg(h)
PE of weight at height 1.5 h = (mg) (1.5 h)
P.E. = or 2: 3
Question 7
1 / -0

Two bodies of masses m_A and m_B have equal KE. The ratio of their momenta is

A
m_A : m_B

B
m_B : m_A

C

D

Solution

Let the velocities be V_A and V_B.
(KE)_A = (KE)_B

Momenta = P_A : P_B
m_A v_A : m_B v_B
=
=
Question 8
1 / -0

A truck and a car moving with the same K.E. are brought to rest by the application of brakes which provides equal retarding forces. Which of them will come to rest in a shorter distance?

A
The truck

B
The car

C
Both will travel the same distance before coming to rest

D
The distance of travel will depend on the horsepower of the vehicles

Solution

As the truck and the car have the same initial kinetic energies, the brakes have to do the same amount of work to expend this energy away, mainly in the form of heat. As the retarding force on both is the same, and also the work is force multiplied by the distance, both will travel the same distance.
Question 9
1 / -0

A body falls freely under gravity from height h. Its gravitational potential energy and kinetic energy vs time graph are shown in the figure. The correct variation is represented by

A

B

C

D

Solution

Option 3 is the correction representation.
Question 10
1 / -0

The PE of a simple pendulum is the highest, when it is

A
at the turning points of the oscillation

B
at the equilibrium

C
in between the above two cases

D
at any position, it has always a fixed value

Solution

Potential energy is maximum at the ends or turning points of the oscillation.
Question 11
1 / -0

Which of the following devices does not come under the category of lever?

A
Crowbar

B
A pair of scissors

C
Broom

D
Staircase

Solution

Staircase does not come under the category of lever as it does not have a bar and a fulcrum.
Question 12
1 / -0

Find the mechanical advantage of a machine, which has a velocity ratio as 3.2 and efficiency as 75%.

A
2.4

B
3.6

C
2.8

D
3.9

Solution

Mechanical advantage is simply the product of velocity ratio and the efficiency.
Thus, for the given case, mechanical advantage = 3.2 x 0.75 = 2.4.
Thus, option (1) is correct.
Question 13
1 / -0

Water at Bhakra Dam is falling through 200 m. What will be the amount of energy available from the fall of 40 metric tonnes of water? (g = 10 m/s²)

A
8 10⁶ joules

B
4 10⁶ joules

C
8 10⁷ joules

D
4 10⁷ joules

Solution

m = 40 10³ kg
h = 200 m
g = 10 m/s²
P.E. = mgh
= 40 10³ 200 10
= 8 10⁷ joules
Question 14
1 / -0

Scalar product of force with velocity is called

A
work

B
power

C
energy

D
momentum

Solution

Scalar product of force with velocity is given as,
F.V. = F.
= = power
Hence, it is instantaneous power.
Question 15
1 / -0

A vehicle needs an engine of 7500 watts to keep it moving with a constant velocity of 30 m/s on a horizontal surface. The force resisting the motion is

A
375 dyne

B
250 N

C
1,50,000 dyne

D
1,50,000 N

Solution

P = 7500 W
v = 30 m/s
P = F v
7500 W = F x 30 m/s
F = N = 250 N
Question 16
1 / -0

The temperature of water at the bottom of a waterfall is higher than that of the water at the top because

A
falling water absorbs heat from the sun

B
kinetic energy of the falling water is converted into heat

C
water at the bottom has greater potential energy

D
the rocks on the river bed give out heat

Solution

As the water falls through a height, its PE is converted into KE.
Now, kinetic energy on falling is converted into heat.
Question 17
1 / -0

Sound energy is converted into electrical energy in a

A
loudspeaker

B
microphone

C
sonometer

D
earphone

Solution

Microphone converts sound energy into electrical energy.
Question 18
1 / -0

When a spring is wound, a certain amount of PE is stored in it. If this wound spring is dissolved in acid, the stored energy

A
is completely lost

B
appears in the form of electromagnetic waves

C
appears in the form of heat and raising the temperature of the acid

D
appears in the form of KE by splashing acid drops

Solution

If the wound spring is dissolved in acid then the stored energy appears in the form of heat and it raises the temperature of the acid.
Question 19
1 / -0

The spring of the winding knob of a watch has

A
mechanical energy

B
only kinetic energy

C
only potential energy

D
kinetic or potential energy

Solution

The spring has stored potential energy which is converted into kinetic energy. Therefore, we can say that the spring of the winding knob of a watch has only potential energy.
Question 20
1 / -0

The PE stored in a long spring stretched by x cm is U. If the spring is stretched by 4x cm, its PE will be

A
16 U

B
3 U

C
U/3

D
U/9

Solution

P.E. =
where, is the extension
Then, U =
Also,
So, U_f = 16 U or 16 times the original extension