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Chapter Test: Work. Power & Energy - 1 - JEE MCQ


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30 Questions MCQ Test Physics for JEE Main & Advanced - Chapter Test: Work. Power & Energy - 1

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Chapter Test: Work. Power & Energy - 1 - Question 1


Detailed Solution for Chapter Test: Work. Power & Energy - 1 - Question 1


Chapter Test: Work. Power & Energy - 1 - Question 2

Power applied to a particle varies with time as P = [3t2 – 2t + 1] watts. Where t is time in seconds. Then the change in kinetic energy of particle between time t = 2s to t = 4s is –

Detailed Solution for Chapter Test: Work. Power & Energy - 1 - Question 2


Chapter Test: Work. Power & Energy - 1 - Question 3


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Chapter Test: Work. Power & Energy - 1 - Question 4

By burning 1 g of coal, the energy produced is 2 kcal. Then for 1 kWh, the quantity of coal so required will be nearly -

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Chapter Test: Work. Power & Energy - 1 - Question 5

A uniform chain of length L and mass M is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If g is acceleration due to gravity, the work required to pull the hanging part on to the table is –

Detailed Solution for Chapter Test: Work. Power & Energy - 1 - Question 5

Mass of hanging portion is M/3 (one-third) and centre of mass c, is at a distance h = L/6 below the table top.Therefore, the required work done is,

Chapter Test: Work. Power & Energy - 1 - Question 6


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Chapter Test: Work. Power & Energy - 1 - Question 7

Water from a stream is falling on the blades of a turbine at the rate of 100 kg/sec. If the height of the stream is 100 m, then the power delivered to the turbine is -

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Chapter Test: Work. Power & Energy - 1 - Question 8


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Chapter Test: Work. Power & Energy - 1 - Question 9

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Chapter Test: Work. Power & Energy - 1 - Question 10


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Chapter Test: Work. Power & Energy - 1 - Question 11


Detailed Solution for Chapter Test: Work. Power & Energy - 1 - Question 11

The work done by man is negative of magnitude of decrease in potential energy of chain.


Chapter Test: Work. Power & Energy - 1 - Question 12

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Chapter Test: Work. Power & Energy - 1 - Question 13


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Chapter Test: Work. Power & Energy - 1 - Question 14


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Chapter Test: Work. Power & Energy - 1 - Question 15


Detailed Solution for Chapter Test: Work. Power & Energy - 1 - Question 15

As W = ΔK

Force is along negative x-axis and displacement is along + x-axis

∴ W = negative

 Hence

ΔK = negative

Chapter Test: Work. Power & Energy - 1 - Question 16


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Chapter Test: Work. Power & Energy - 1 - Question 17

A particle moves in a straight line with its retardation proportional to its displacement 'x'. Change in kinetic energy is proportional to –

Detailed Solution for Chapter Test: Work. Power & Energy - 1 - Question 17


Chapter Test: Work. Power & Energy - 1 - Question 18

Detailed Solution for Chapter Test: Work. Power & Energy - 1 - Question 18

Given:

  • Mass of the ball, m=20 kg

  • Initial height, h1=100 m

  • Intermediate height, h2=30 m

  • Final height, hf=20 m

  • Acceleration due to gravity, g=10 m/s2

The ball starts at rest at height h1​. Its initial kinetic energy is zero, and its initial potential energy is:

PEinitial = mgh= 20 × 10 × 100 = 20000 J

At the final position (height h= 20 m), the potential energy is:

PEfinal = mgh= 20 × 10 × 20 = 4000 J

The loss in potential energy is converted into kinetic energy:

KEfinal = PEinitial − PEfinal = 20000 − 4000 = 16000 J

The kinetic energy is given by:

KE=1/2mv2

Substitute K E=16000 J and m=20 kg:

16000 = 1/2 × 20 × v2

16000 = 10v2

v2=1600 = 40m/s

Chapter Test: Work. Power & Energy - 1 - Question 19

A particle moves along the x-axis from x = 0 to x = 5m under the influence of a force given by F = 7 – 2x + 3x2. The work done in the process is-

Detailed Solution for Chapter Test: Work. Power & Energy - 1 - Question 19


Chapter Test: Work. Power & Energy - 1 - Question 20


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Chapter Test: Work. Power & Energy - 1 - Question 21


Detailed Solution for Chapter Test: Work. Power & Energy - 1 - Question 21


Chapter Test: Work. Power & Energy - 1 - Question 22

A particle experiences a variable force in a horizontal x-y plane. Assume distance in meters and force in newton. If the particle moves from point (1, 2) to point (2, 3) in the x-y plane; then kinetic energy changes by

Detailed Solution for Chapter Test: Work. Power & Energy - 1 - Question 22

The change in kinetic energy of a particle moving in a horizontal plane can be determined using the work-energy theorem. The theorem states that the work done by a force on a particle equals the change in its kinetic energy. We Know that, 

Chapter Test: Work. Power & Energy - 1 - Question 23

Two cylindrical vessels of equal cross-sectional area 16 cm2 contain water up to heights 100 cm and 150 cm, respectively. The vessels are interconnected so that the water levels in them become equal. The work done by the force of gravity during the process is: [Take density of water = 103 kg/m3 and g = 10 ms-2]

Detailed Solution for Chapter Test: Work. Power & Energy - 1 - Question 23


Chapter Test: Work. Power & Energy - 1 - Question 24

A block of mass m attached to a massless spring is performing oscillatory motion of amplitude 'A' on a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through its equilibrium point, the amplitude of oscillation for the remaining system become fA. The value of f is

Detailed Solution for Chapter Test: Work. Power & Energy - 1 - Question 24

Vmax = A'ω' = Aω


Chapter Test: Work. Power & Energy - 1 - Question 25

A particle of mass 1 kg is hanging from a spring of force constant 100 Nm-1. The mass is pulled slightly downward and released, so that it executes free simple harmonic motion with time period T. The time when the kinetic energy and potential energy of the system will become equal is The value of x is _______. (in integers)

Detailed Solution for Chapter Test: Work. Power & Energy - 1 - Question 25

The time when the kinetic energy and potential energy of the system become equal can be determined as follows:

The condition for equal kinetic and potential energy is given by:
Kinetic Energy (KE) = Potential Energy (PE)

⇒  A2 = 2x2

The displacement X is given by: X = A sin(ωt)

Solving these equations, 

A2 = 2x2

X = A sinωt

x = 8 sec

Chapter Test: Work. Power & Energy - 1 - Question 26

In a spring gun having spring constant 100 N/m, a small ball 'B' of mass 100 g is put in its barrel (as shown in figure) by compressing the spring through 0.05 m. There should be a box placed at a distance 'd' on the ground so that the ball falls in it. If the ball leaves the gun horizontally at a height of 2 m above the ground, the value of d is _______________ m.
(g = 10 m/s2) (in integers)

Detailed Solution for Chapter Test: Work. Power & Energy - 1 - Question 26


Chapter Test: Work. Power & Energy - 1 - Question 27

Two particles of masses m and 4m have linear momenta in the ratio of 2:1. What is the ratio of their kinetic energies?

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Chapter Test: Work. Power & Energy - 1 - Question 28

A particle of mass m has half the kinetic energy of another particle of mass m/2. If the speed of the heavier particle is increased by 2 ms−1, its new kinetic energy equals the original kinetic energy of the lighter particle. The ratio of the original speeds of the lighter and heavier particles is

Detailed Solution for Chapter Test: Work. Power & Energy - 1 - Question 28

Chapter Test: Work. Power & Energy - 1 - Question 29

A body of mass m accelerates uniformly from rest to velocity v1 in time interval T1. The instantaneous power delivered to the body as a function of time t is:

Detailed Solution for Chapter Test: Work. Power & Energy - 1 - Question 29

Let a be the constant acceleration
Using 1st equation of motion

At any instant t, the velocity v of the body

Chapter Test: Work. Power & Energy - 1 - Question 30

The distance x moved by a body of mass 0.5 kg by a force varies with time t as x = 3t2 + 4t + 5 where x is expressed in metre and t in second. What is the work done by the force in the first 2 seconds?

Detailed Solution for Chapter Test: Work. Power & Energy - 1 - Question 30


By work - energy theorem
Total work done = change in kinetic energ

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