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Test: Newton's laws of Motion (June 2) - JEE MCQ


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10 Questions MCQ Test Daily Test for JEE Preparation - Test: Newton's laws of Motion (June 2)

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Test: Newton's laws of Motion (June 2) - Question 1

A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. The speed is then increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks.

Detailed Solution for Test: Newton's laws of Motion (June 2) - Question 1
  • The stone is moving in a circular path with constant speed, this is uniform circular motion.
  • In uniform circular motion, at any instant velocity (magnitude speed) acts tangentially to the circular path.
  • Hence, when string is cut the stone will fly off tangentially to circular path with velocity of magnitude speed.

Test: Newton's laws of Motion (June 2) - Question 2

A man of mass 70 kg stands on a weighing scale in a lift which is moving upwards with a uniform speed of 10 m s−1, what would be the reading on the scale?

Detailed Solution for Test: Newton's laws of Motion (June 2) - Question 2

Mass of the man, m = 70 kg
Acceleration, a = 0
Using Newton’s second law of motion, We can write the equation of motion as, 
R – mg = ma
∴ R = mg = 70 × 10 = 700 N
∴ the weighing scale = 700 / g = 700 / 10 = 70 kg

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Test: Newton's laws of Motion (June 2) - Question 3

Newton’s third law states that when two bodies interact.

Detailed Solution for Test: Newton's laws of Motion (June 2) - Question 3
  • The third law states that all forces between two objects exist in equal magnitude and opposite direction. 
  • If one object A exerts a force FA on a second object B, then B simultaneously exerts a force FB on A, and the two forces are equal in magnitude and opposite in direction, FA = −FB
  • Newton's third Law:
    Engineering Car Crash Safety with Newton's Third Law | Lesson Plan
Test: Newton's laws of Motion (June 2) - Question 4

Force is required

Detailed Solution for Test: Newton's laws of Motion (June 2) - Question 4
  • Force is required to start a stationary object and to stop a moving object due to inertia. 
  • Inertia is a property of matter by which it continues in its existing state of rest or uniform motion in a straight line, unless that state is changed by an external force.
  • This is also called law of inertia or newton's first law of motion.
  • Newton's first Law:
Test: Newton's laws of Motion (June 2) - Question 5

According to Newton's third law of motion, the action and reaction forces are

Detailed Solution for Test: Newton's laws of Motion (June 2) - Question 5
  • When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.
  • To every action there is always opposed an equal reaction or the mutual actions of two bodies upon each other are always equal and opposite in direction.
Test: Newton's laws of Motion (June 2) - Question 6

According to first law of motion

Detailed Solution for Test: Newton's laws of Motion (June 2) - Question 6

If net force act on a body then acceleration of the body will also be zero. Hence velocity will not be changed i.e. it continues in its existing state of rest or uniform motion in a straight line.

Test: Newton's laws of Motion (June 2) - Question 7

Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg lying on the floor of a train which is accelerating with 1 ms−2, the stone being at rest relative to the train. Neglect air resistance.

Detailed Solution for Test: Newton's laws of Motion (June 2) - Question 7

Weight of the stone is balanced by the reaction of the floor. The only acceleration is provided by the horizontal motion of the train.

a = 1ms-2

Force in horizontal direction 

F = ma = 0.1 x 1 = 0.1N

Test: Newton's laws of Motion (June 2) - Question 8

A man of mass 70 kg stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 10 ms−2 what would be the reading on the scale?

Detailed Solution for Test: Newton's laws of Motion (June 2) - Question 8

According to Newton's second Law of motion,
Apparent weight, R = m(a+g) 

= 70(10+10) 

= 1400N 

mass = R/g

= 1400/10 

= 140Kg

Test: Newton's laws of Motion (June 2) - Question 9

A helicopter of mass 1000 kg rises with a vertical acceleration of 15 ms−2. The crew and the passengers weigh 300 kg. Give the magnitude and direction of the force on the floor by the crew and passengers,

Detailed Solution for Test: Newton's laws of Motion (June 2) - Question 9
  • The moving helicopter is a non-inertial frame and hence we do need to apply pseudo force onto the body so as to solve equations with Newton's laws. 
  • The pseudo force needed to be applied, f = ma = 300 x 15 N = 4500N
  • Thus the net downward force due to the passengers is 7500N.
Test: Newton's laws of Motion (June 2) - Question 10

Detailed Solution for Test: Newton's laws of Motion (June 2) - Question 10

Given:

We have to  find the least force required to make the body slide.

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