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Test: Friction (June 6) - JEE MCQ


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10 Questions MCQ Test - Test: Friction (June 6)

Test: Friction (June 6) for JEE 2024 is part of JEE preparation. The Test: Friction (June 6) questions and answers have been prepared according to the JEE exam syllabus.The Test: Friction (June 6) MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Friction (June 6) below.
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Test: Friction (June 6) - Question 1

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: Friction (June 6) - Question 1

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

Test: Friction (June 6) - 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: Friction (June 6) - 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.

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Test: Friction (June 6) - 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: Friction (June 6) - Question 3

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

Test: Friction (June 6) - Question 4

A cyclist bends while taking turn to

Detailed Solution for Test: Friction (June 6) - Question 4

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: Friction (June 6) - 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: Friction (June 6) - 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: Friction (June 6) - Question 6

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: Friction (June 6) - Question 6

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

Test: Friction (June 6) - Question 7

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: Friction (June 6) - Question 7

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

Test: Friction (June 6) - Question 8

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

Detailed Solution for Test: Friction (June 6) - Question 8

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

= 38.6 m s-1

Test: Friction (June 6) - Question 9

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: Friction (June 6) - Question 9

Here, r = 10m, v = 5ms−1, at = 2ms−2,

The net acceleration is
a = 

= 3.2 m s-2

Test: Friction (June 6) - Question 10

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: Friction (June 6) - Question 10

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.

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