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Test: Circular Motion (NCERT) - NEET MCQ


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10 Questions MCQ Test Physics Class 11 - Test: Circular Motion (NCERT)

Test: Circular Motion (NCERT) for NEET 2024 is part of Physics Class 11 preparation. The Test: Circular Motion (NCERT) questions and answers have been prepared according to the NEET exam syllabus.The Test: Circular Motion (NCERT) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Circular Motion (NCERT) below.
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Test: Circular Motion (NCERT) - Question 1

A cyclist bends while taking turn to

Detailed Solution for Test: Circular Motion (NCERT) - Question 1

Tunning means motion on a curved path, which requires centripetal force. Bending of cyclist with respect to vertical direction provides the necessary centripetal force

Test: Circular Motion (NCERT) - Question 2

A motor cyclist rides around the well with a round vertical wall and does not fall down while riding because

Detailed Solution for Test: Circular Motion (NCERT) - Question 2

The component of normal force balances the weight of bike rider. Thus, if the wall is perfectly vertical, only frictional force of the wall balances the weight.

Test: Circular Motion (NCERT) - Question 3

One end of a string of length I is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v, the net force on the particle directed towards the centre is (where T is the tension in the string)

Detailed Solution for Test: Circular Motion (NCERT) - Question 3

The net force on the particle directed towards the center T.

Test: Circular Motion (NCERT) - Question 4

The mass of a bicycle rider along with the bicycle is 100 kg. He wants to cross over a circular turn of radius 100 m with a speed o f 10 m s-1. If the coefficient of friction between the tyres and the road is 0.6, the frictional force required by the rider to cross the turn, is

Detailed Solution for Test: Circular Motion (NCERT) - Question 4

Centripetal force = 
= 100 N
Required frictional force to cross the turn,
= μmg = 0.6 × 100 × 10 = 600 N
As the frictional force is greater than the centripetal force, so the rider will be able to cross the turn.

Test: Circular Motion (NCERT) - Question 5

A stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are
T1 and v1 denotes the tension and speed at the lowest point. T2 and v2 denotes corresponding values at the highest point.

Detailed Solution for Test: Circular Motion (NCERT) - Question 5

At the lowest point, both mg and T2 act downwards and T1 upwards so that net force = mg − T1.
At heights point, both mg and T2 act downwards so that net force = mg + T2

Test: Circular Motion (NCERT) - Question 6

A small object placed on a rotating horizontal turn table just slips when it is placed at a distance 4 cm from the axis of rotation. If the angular velocity of the turn-table is doubled, the object slips when its distance from the axis of rotation is 

Detailed Solution for Test: Circular Motion (NCERT) - Question 6

The object will slip if centripetal force ≥ force of friction
mrω2 ≥ μmg
2 ≥ μg
2 ≥  constant, or 

Test: Circular Motion (NCERT) - Question 7

A particle is moving on a circular path of 10 m radius. At any instant of time, its speed is 5 m s-1 and the speed is increasing at a rate of 2 m s-1. The magnitude of net acceleration at this instant is 

Detailed Solution for Test: Circular Motion (NCERT) - Question 7

Here, r = 10m, v = 5ms−1, at = 2ms−2,

The net acceleration is
a = 

= 3.2 m s-2

Test: Circular Motion (NCERT) - Question 8

The coefficient of frictionbetween the tyres and the road is 0.1. The maximum speed with which a cyclist can take a  circular turn of radius 3m without skidding it.(Take g =  m s-2)

Detailed Solution for Test: Circular Motion (NCERT) - Question 8

Here, r = 3m, μ = 0.1, g = 10ms−2
The maximum speed with which a cyclist can take a turn without skidding is

Test: Circular Motion (NCERT) - Question 9

A stone of mass 5 kg is tied to a string of length 10 m is whirled round in a horizontal circle. Wha is the maximum speed with which the stone can be whirled around if the string can witbstand  a maximum tension of 200 N.

Detailed Solution for Test: Circular Motion (NCERT) - Question 9

Here, m = 5kg , r = 10m, Tmax = 200N
As Tmax
∴ 
∴  
= 400
⇒  vmax =20 m s−1

Test: Circular Motion (NCERT) - Question 10

In the question number 8, the maximum permissible speed to avoid slipping is

Detailed Solution for Test: Circular Motion (NCERT) - Question 10

The maximum permissible speed is given by v(max) = 

= 38.6 m s-1

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