Test: Newton's Third Law of Motion (NCERT) - NEET MCQ

# Test: Newton's Third Law of Motion (NCERT) - NEET MCQ

Test Description

## 15 Questions MCQ Test NCERTs at Fingertips: Textbooks, Tests & Solutions - Test: Newton's Third Law of Motion (NCERT)

Test: Newton's Third Law of Motion (NCERT) for NEET 2024 is part of NCERTs at Fingertips: Textbooks, Tests & Solutions preparation. The Test: Newton's Third Law of Motion (NCERT) questions and answers have been prepared according to the NEET exam syllabus.The Test: Newton's Third Law of Motion (NCERT) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Newton's Third Law of Motion (NCERT) below.
Solutions of Test: Newton's Third Law of Motion (NCERT) questions in English are available as part of our NCERTs at Fingertips: Textbooks, Tests & Solutions for NEET & Test: Newton's Third Law of Motion (NCERT) solutions in Hindi for NCERTs at Fingertips: Textbooks, Tests & Solutions course. Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free. Attempt Test: Newton's Third Law of Motion (NCERT) | 15 questions in 15 minutes | Mock test for NEET preparation | Free important questions MCQ to study NCERTs at Fingertips: Textbooks, Tests & Solutions for NEET Exam | Download free PDF with solutions
Test: Newton's Third Law of Motion (NCERT) - Question 1

### We can derive Newtons

Detailed Solution for Test: Newton's Third Law of Motion (NCERT) - Question 1

We can derive Newton’s third and first laws from the second law.

The correct option is C third and first laws from the second law
We can derive Newton's third and first laws from second law. If force on particle is zero, then according to Newton's second law

F=ma

0=ma

a=0 (Since mass cannot be zero)
So, this shows that the body moves with constant velocity or remains at rest if no external force on the system. This is the statement of first law itself.

Suppose an isolated system of two bodies A and B. Suppose A exerts force F1 on B and B exerts F2 on A. Now, rate of change of momentum of A is dp1/dt and that of B is dp2/dt. Therefore, from second law,

F1=dp1/dt …(i)

F2=dp2/dt …(ii)

Adding (��) and (����), we get

F1+F2=dp1/dt+dp2/dt

Since the system is isolated, so the total change in momentum or rate of change of momentum dp1/dt+dp2/dt should be zero. Thus, F1+F2=0 or F1=−F2.This is the statement of third law.

Test: Newton's Third Law of Motion (NCERT) - Question 2

### Which of the following statements is not true regarding the Newton's third law of motion?

Detailed Solution for Test: Newton's Third Law of Motion (NCERT) - Question 2

Action and reaction act on different bodies not on the same body.

 1 Crore+ students have signed up on EduRev. Have you?
Test: Newton's Third Law of Motion (NCERT) - Question 3

### The driver of a car suddenly sees a broad wall in front of him. He should

Detailed Solution for Test: Newton's Third Law of Motion (NCERT) - Question 3

By the retarding force the car can be stopped in a less distance if the driver apply brakes. This retarding force is actually a friction force.

Test: Newton's Third Law of Motion (NCERT) - Question 4

Which of the following statements is incorrect?

Detailed Solution for Test: Newton's Third Law of Motion (NCERT) - Question 4

It is easier to pull lawn mower than to push it . All other statements are true.

Test: Newton's Third Law of Motion (NCERT) - Question 5

Match Column I with Column II

Detailed Solution for Test: Newton's Third Law of Motion (NCERT) - Question 5

Newton's first law defines force. Newton's second law gives us a measure of force. Impulse gives us the effect of force. Recoiling of gun is accounted for by Newton's 3rd law.

Test: Newton's Third Law of Motion (NCERT) - Question 6

A rocket is going upwards with accelerated motion. A man sitting in it feels his weight increased 5 times his own weight. If the mass of the rocket including that of the man is 1.0 x 104 kg, how much force is being applied by rocket engine?(Take g = 10 m s-2).

Detailed Solution for Test: Newton's Third Law of Motion (NCERT) - Question 6

As the weight of man increased by 5 times, so acceleration of  the rocket a = 5g = 5 x 10 = 50 m s-2
Force applied by rocket engine is
F = ma = 1.0 x 104 x 50 = 5 x 105 N

Test: Newton's Third Law of Motion (NCERT) - Question 7

Ten one rupee coins are put on top of each other on a table. Each coin has a mass m. The reaction of the 6th coin (counted from the bottom) on the 7th coin is

Detailed Solution for Test: Newton's Third Law of Motion (NCERT) - Question 7

The sixth coin is under the weight of four coins baove it. Hence,
Reaction of the 6th coin on the 7th coin = Force on the 6th coin due to 7th coin
= 4mg

Test: Newton's Third Law of Motion (NCERT) - Question 8

A cork of mass 10 g is floating on water The net force acting on this cork is

Detailed Solution for Test: Newton's Third Law of Motion (NCERT) - Question 8

When the cork is floating, its weight is balanced by the upthrust. Therefore, net force on the cork is zero.

Test: Newton's Third Law of Motion (NCERT) - Question 9

A stone of mass 1 kg is lying on the floor of a train which is accelerating with 1 m s-2. The net force acting on the stone is

Detailed Solution for Test: Newton's Third Law of Motion (NCERT) - Question 9

Here, mass of the stone, m = 1 kg
As the stone is lying on the floor of the train, its acceleration is same as that of train.
∴ Force acting on the stone F = ma = (1kg)(1 m s-2)
= 1 N

Test: Newton's Third Law of Motion (NCERT) - Question 10

A stream of water flowing horizontally with a speed of 15 m s-1 gushes out of a tube of cross sectional area 10-2 m2 and hit a vertical wall normally. Assuming that it does not rebound from the wall, the force exerted on the wall by the impact of waler is

Detailed Solution for Test: Newton's Third Law of Motion (NCERT) - Question 10

Here, v = 15 m s-1
Area of cross section, A = 10-2 m2
Density of water, p = 103 kg m3
Mass of water hitting the wall per second
= p x A x v
=103 kg m3 x 10-2 mx 15 m s-1
= 150 kg s-1
Force exerted on the wall = Momentum loss of water per second
= 150 kg s-1 xg m3 x 10-2 mx 15 m s-1
= 150 kg s-1
Force exerted on the wall = Momentum loss of water per second
= 150 kg s-1 x 15 m s-1 = 2250 N
= 2.25 x 1000 N = 2250 N
= 2.25 x 103 N

Test: Newton's Third Law of Motion (NCERT) - Question 11

A steam of water flowing horizontally with a speed of 25 m s-1 gushes out of a tube of cross sectional area 10-3 m2, and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water?

Detailed Solution for Test: Newton's Third Law of Motion (NCERT) - Question 11

Here, u = 25 m s-1, v = 0, t = 1, A = 10-3 m2
Density of water p = 1000 kgm-3
Mass of water gushed out per second, m
= volume x density / time = area x distance x density / time
= Area x velocity x density
= 10-3 x 25 x 1000 = 25 kg
F =  = 625 N
F' = F = 625N

Test: Newton's Third Law of Motion (NCERT) - Question 12

A rocket with a lift-off mass 2 x 104 kg is blasted upwards with an initial acceleration of 5 m s-2. The initial thrust of the blast is (Take g = 10 m s-2)

Detailed Solution for Test: Newton's Third Law of Motion (NCERT) - Question 12

Here, m = 2 × 104kg
Initial acceleration, a = 5ms-2
Initial thrust = upthrust required to impart acceleration a + upthrust required to overcome gravitational pull
∴ F  = m(a + g) = (2 × 104kg)(5 + 10)ms−2
= 3 × 105N

Test: Newton's Third Law of Motion (NCERT) - Question 13

A ball of mass m strikes a rigid wall with speed u at an angle of 30o and get reflected with the same speed and at the same angle as shown in the figure. If the ball is in contact with the wall for time t, then the force acting on the wall is

Detailed Solution for Test: Newton's Third Law of Motion (NCERT) - Question 13

Initial momentum of the ball is

Final momentum of the ball is
∴ Change in momentum,

Impulse = Change in momentum =
As impulse and force are in the same direction, therefore, force on the ball due to the wall is normal to the wall along the negative x-axis. Using Newton’s 3rd law of motion the force on the wall due to the ball is normal to the wall along the positive x− direction.
∴ F =

Test: Newton's Third Law of Motion (NCERT) - Question 14

A rocket of initial mass 6000 kg ejects gases at a constant rate of 16 kgs-1 with constant relative speed of11 km s-1. What is the acceleration of the rocket one minute after the blast?

Detailed Solution for Test: Newton's Third Law of Motion (NCERT) - Question 14

Acceleration of the rocket at any instant t is

Here, M = 6000kg, dm/dt = 16kgs−1
Vr = 11kms-1 = 11km s−1 t = 1 min = 60s
∴
≈ 35 m s−2

Test: Newton's Third Law of Motion (NCERT) - Question 15

Two billiard balls A and B. each of mass 50 g and moving in opposite directions with speed of 5 m s-1 each, collide and rebound with the same speed. The impulse imparted to each ball is

Detailed Solution for Test: Newton's Third Law of Motion (NCERT) - Question 15

Initial momentum of ball A = (0.05kg)(5ms(-1)) = 0.25kgms−1
As the speed is reversed on collision,
Final momentum of the ball
A = (0.05kg)(−5ms−1) = −0.25kgms−1
Impulse imparted to the ball A = Change in momentum of ball A
= Final momentum - Initial momentum
= −0.25kg ms−−0.25kg ms−1 = −0.5kg ms−1
Similarly,
Initial momentum of ball B = (0.05kg)(−5ms−1) = −0.25kgms−1
Final momentum of ball B = (0.05kg)(5ms−1) = = +0.25kgms−1
Impulse imparted to ball B = (0.25kgms−1) − (−0.25kgms−1) =  0.5kgms−1
Impulse imparted to each ball is 0.5kgms−1 in magnitude. The two impulses are opposite in direction.

## NCERTs at Fingertips: Textbooks, Tests & Solutions

304 docs|275 tests
Information about Test: Newton's Third Law of Motion (NCERT) Page
In this test you can find the Exam questions for Test: Newton's Third Law of Motion (NCERT) solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Newton's Third Law of Motion (NCERT), EduRev gives you an ample number of Online tests for practice

## NCERTs at Fingertips: Textbooks, Tests & Solutions

304 docs|275 tests