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Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) - NEET MCQ


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10 Questions MCQ Test - Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26)

Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) for NEET 2024 is part of NEET preparation. The Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) questions and answers have been prepared according to the NEET exam syllabus.The Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) below.
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Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) - Question 1

Two masses of 1 g and 4g are moving with equal kinetic energy. The ratio of the magnitudes of their momenta is

Detailed Solution for Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) - Question 1


Hence, the Correct Answer is option A

Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) - Question 2

At time t = 0 s particle starts moving along the x-axis. If its kinetic energy increases uniformly with time ‘t’, the net force acting on it must be proportional to

Detailed Solution for Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) - Question 2

Given, Kinetic energy increases uniformly with time
K.E ∝ t
Kinetic energy can be written as KE= ½ mvso we can write
½ mv∝ t


Therefore, v ∝ √t
According to Newton's second law, force (F) is inversely proportional to velocity (F ∝ 1/v). 
F ∝ 1/v
Therefore,
F ∝ 1/√t

Hence, Correct Answer is Option B

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Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) - Question 3

Time rate at which work is done by a force is:

Detailed Solution for Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) - Question 3

We know that Work done, W = F.s
where F is force and s is displacement due to that force.
Thus rate of work done is: 

dW/dt = d(F.s)/dt

So as F is constant we get, 
dW/dt = F.d(s)/dt = F.v = P (Power)

Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) - Question 4

There are two bodies X and Y with equal kinetic energy but different masses m and 4m respectively. The ratio of their linear momentum is:

Detailed Solution for Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) - Question 4

X and Y have equal kinetic energy but their masses are m and 4m respectively.
► 1/2 m1v12 = 1/2 m2v22  
► mv12 = 4m * v22  
► v1 : v2 = 2 : 1
Hence the ratio of their linear momentum is:
m1v1 : m2v2 = m * 2v : 4m * v = 1 : 2

Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) - Question 5

 Find the potential energy stored in a ball of mass 5 kg placed at a height of 3 m above the ground.

Detailed Solution for Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) - Question 5

m = 5 kg,
h = 3 m,
g = 9.81 m/s-2

We know that,
Potential energy = mgh

= 5 * 9.81 * 3 = 147.15 J

Hence, the correct answer is Option C.

Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) - Question 6

Output of a truck is 4500 J and its efficiency is 50%, the input energy provided to the truck is:

Detailed Solution for Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) - Question 6

► η = work output / heat input​
► η = 50% = 50/100 = 1/2
► 1/2 = 4500 / Heat input
► Heat input = 9000 J

Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) - Question 7

By how much does kinetic energy increase if the momentum is increased by 20%?

Detailed Solution for Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) - Question 7

The kinetic energy is given by: 

KE= p2/2m

So, ΔKE = 2pΔp​ / 2m = pΔp / m​
ΔKE / KE ​= (pΔp/m)​ * (2m/p2)​ = 2Δp / p
Since the momentum p increases by 20%, so the final momentum becomes 1.2p.
Hence, KEfinal​ = (1.2p)/ 2m​ = 1.44p2 / 2m​ = 1.44KE
So, % change in KE = 44%

Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) - Question 8

When a body slides against a rough horizontal surface, the work done by friction is:

Detailed Solution for Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) - Question 8
  • If a force acting on a body has a component in the opposite direction of displacement, the work done is negative.
  • So when a body slides against a rough horizontal surface, its displacement is opposite to that of the force of friction. The work done by the friction is negative.
Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) - Question 9

 A machine gun fires 60 bullets per minute, with a velocity of 700 m/s. If each bullet has a mass of 50g, find the power developed by the gun.

Detailed Solution for Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) - Question 9

Mass of the bullets = 60×50 = 3000g = 3kg
v = 700m/s t = 1min = 60s
Power = W/t = (Kinetic energy)/t = 12250W.

Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) - Question 10

 If a force acts perpendicular to the direction of motion of a body, what is the amount of work done?

Detailed Solution for Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) - Question 10

If a force acts perpendicular to the direction of a body, the amount of work done is zero because there is no displacement in the direction of a force.

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