NEET Exam  >  NEET Tests  >  Daily Test for NEET Preparation  >  Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) - NEET MCQ

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


Test Description

10 Questions MCQ Test Daily Test for NEET Preparation - 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 Daily Test for NEET Preparation 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.
Solutions of Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) questions in English are available as part of our Daily Test for NEET Preparation for NEET & Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) solutions in Hindi for Daily Test for NEET Preparation course. Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free. Attempt Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) | 10 questions in 20 minutes | Mock test for NEET preparation | Free important questions MCQ to study Daily Test for NEET Preparation for NEET Exam | Download free PDF with solutions
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

1 Crore+ students have signed up on EduRev. Have you? Download the App
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.

12 docs|366 tests
Information about Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) Page
In this test you can find the Exam questions for Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26) solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Introduction and Work Energy Theorem & Kinetic Energy (June 26), EduRev gives you an ample number of Online tests for practice

Top Courses for NEET

Download as PDF

Top Courses for NEET