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# MCQ : Work And Energy - 2

## 20 Questions MCQ Test Science & Technology for UPSC CSE | MCQ : Work And Energy - 2

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This mock test of MCQ : Work And Energy - 2 for Class 9 helps you for every Class 9 entrance exam. This contains 20 Multiple Choice Questions for Class 9 MCQ : Work And Energy - 2 (mcq) to study with solutions a complete question bank. The solved questions answers in this MCQ : Work And Energy - 2 quiz give you a good mix of easy questions and tough questions. Class 9 students definitely take this MCQ : Work And Energy - 2 exercise for a better result in the exam. You can find other MCQ : Work And Energy - 2 extra questions, long questions & short questions for Class 9 on EduRev as well by searching above.
QUESTION: 1

### On tripling the speed of motion of a body, the change in K.E. is:

Solution:

K.E. ∝ v2
So when v → 3v, K.E. → 9K.E.

QUESTION: 2

### The moon revolves around the earth due to the gravitational force (F) of the earth on the moon. The work done by the gravitational force is (r = radius of the circular orbit of moon):

Solution:
• The work done by the moon when it revolves around the earth is zero because here displacement of the moon and gravitational force of earth are perpendicular to each other.
• Work done (W) = F x d x cosƟ. In the case of the revolution of the moon around the earth, Ɵ is 90°. Thus, W = F x d x cos90° = F x d x 0 = 0.
• So, the work done by the moon when it revolves around the earth is Zero.
QUESTION: 3

### An electric bulb of 60 W is used for 5 hours a day. The cost of electricity involved in a month of 30 days at Rs. 3.00 per unit is (in Rs):

Solution:

Cost of Electricity = P*t*cost per kW
=0.06 kW x (5 x 30) x 3= Rs. 27

QUESTION: 4

A body is moved through a distance of 3 m in the following different ways. In which case is the maximum work done?

Solution:

Work done = Mgh
Work will be maximum when the change in height is maximum, thus when it is lifted vertically upward.

QUESTION: 5

A: 50 kg man climbing a ‘slant length of 5 m along a 30° incline.
B: 25 kg man running with 2 m/s speed.
C: A force of 5 N acting on an object moving with 5 m/s speed for 5 min.
If the energies in A, B and C are EA, EB and EC respectively, then:

Solution:

Ea = mv2/2 = 50 x 10 x 5sin30º = 250j
Eb = mv2/2 = 25 x 4/2 = 50j
Ec  = fvt = 5 x 25 x 60 = 7500j
Therefore, Eb < E< Ec

QUESTION: 6

Two bodies of equal weight are kept at heights of h and 1.5 h, respectively. The ratio of their potential energies is:

Solution:

Potential energy is mathematically defined as, P.E = mgh
Potential energy of body 1 = P.E1 = mgh1
Potential energy of body 2 = P.E2 = mgh2
Since weight of two bodies are equal, therefore, P.E1 / P. E2 =  h / 1.5 h = 2:3

QUESTION: 7

If a force of F newton moves a body with constant speed v, the power delivered by it is:

Solution:

Given: Velocity = v & Force = F
We know that work done is equal to the product of force and displacement.
W = F*d
Dividing by t on both sides then we get:
W/t = F*d/t
d/t = v ( Speed = Distance/ Time)
W/t = P (Power)
Therefore, P = F*v

QUESTION: 8

In the SI system, the unit of potential energy is:

Solution:

Potential energy is energy stored in matter. Joule is the SI unit of energy.

QUESTION: 9

A mass ‘m’ falls from a height ‘h’ any point on its path the total energy is:

Solution:
• Energy at start will be = Potential Energy = mgh
• According to the law of conservation of energy, it's total energy will be the same at all points i.e. = mgh.
QUESTION: 10

Newton-metre is the unit of:

Solution:

Work = Force * Displacement (N-m)

QUESTION: 11

If the bulbs of 60 W and 40 W are connected in series to a 220 V source the bulb which glows brighter is:

Solution:
• Two bulbs of 40 W and 60 W are connected in series to an external potential difference.
•  When two bulbs are connected in series, the current I will be the same. But the resistance varies. Higher the resistance brighter the glowing.
• P = V2 / R
⇒ P, R are inversely proportional to each other.
Hence, P40 < P60
⇒ R40 > R60
• So, the resistance of 40 W bulb is higher than the 60 W bulb. So 40 W bulb will dissipate more power and glow brighter.
QUESTION: 12

A porter with a suitcase on his head is climbing up a flight of stairs with a uniform speed. The work done by the ‘weight of the suitcase’ on the suitcase is:

Solution:

Force and displacement are in the same direction.
So, the work done is positive.

QUESTION: 13

A boy holds a mass on his stretched hand. Then:

Solution:
QUESTION: 14

Two bodies of unequal masses are dropped from a cliff. At any instant, they have equal:

Solution:

They will have the same acceleration due to gravity because it is independent of mass.

QUESTION: 15

Two masses m and 2m are dropped from a certain height ‘h’. Then on reaching the ground:

Solution:
• Since the free-falling acceleration is not dependent on mass, the final velocities of both masses are the same.
The kinetic energy = KE = mv2/2
• Hence, the kinetic energy of the heavier mass will be more.
QUESTION: 16

A person A does 500 J of work in 10 minutes and another person B does 600 J of work in 20 minutes. Let the power delivered by A and B be PA and PB respectively. Then,

Solution:

Power = P = Energy/Time​ = Work/Time
PA​= 500/10*60 ​= 0.83J/s
PB= 600/20*60 ​= 0.5J/s
P> PB

QUESTION: 17

In a collision______.

Solution:
QUESTION: 18

In the case of negative work, the angle between the force and displacement is:

Solution:
• Work done by any force is the product of the component of the force in the direction of displacement and the magnitude of displacement.
W = F.d = Fd cosθ
• When  W =−ve, it means cosθ = −ve
Therefore, θ = 180º
QUESTION: 19

One unit of electricity is consumed by,
A: P = 40 W bulb used for t = 25 hours
B: P = 20 W bulb used for t = 50 hours
Both A and B are true as, energy E in kWh is best related as:

Solution:
• In the given question, P is given in Watt and we have to calculate E in kWh. So,  we have to convert watt into kW. For this, we divide Pt by 1000.
• Therefore, E = Pt / 1000  is the correct answer.
QUESTION: 20

The work done on an object does not depend upon the______.

Solution:
• Work done by any force is the product of the force in the direction of displacement and the magnitude of displacement. The work done depends upon the force and displacement.
• W = F.d, when force and displacement are perpendicular to each other work done is zero.
• Hence work done also depends upon the angle between Force and Displacement but not on the velocity of the moving object.