Exams › JEE Main › Physics

A block of mass m rests on a surface whose vertical cross-section is described by y = x³/6. If the coefficient of friction is 0.5, what is the greatest height above the ground at which the block can be placed so that it does not slide?

1. 1/6 m
2. 2/3 m
3. 1/3 m
4. 1/2 m

Correct answer: 1/6 m

Solution

Slipping is on the verge when dy/dx = mu. Here dy/dx = x^2/2 = 0.5 gives x = 1, so the maximum height is y = x^3/6 = 1/6 m.

Related JEE Main Physics questions

• Which of the following has the same SI unit as impulse?
• A monkey slides downward from a tree branch with uniform acceleration. If the maximum tension the branch can withstand is 75% of the monkey’s weight, what is the least downward acceleration that will allow the monkey to descend without snapping the branch?
• A 1000 kg car is travelling at 30 m/s. The brakes are applied until it stops. If the friction force between the tyres and the road is 5000 N, how long will the car take to come to rest?
• A rocket initially ejects gas from its rear at a mass rate of 0.1 kg s⁻¹. If the exhaust speed relative to the rocket is 50 m s⁻¹ and the rocket’s mass is 2 kg, what is its acceleration in m s⁻²?
• A block of mass M rests on a rough level floor with coefficient of friction μ. A person applies a horizontal pull, but the block remains at rest. If the total contact force exerted by the floor on the block is F, then
• Which of the following types of motion on a frictionless plane surface occurs without any force acting on the body?

⚔️ Practice JEE Main Physics free + battle 1v1 →