Exams › JEE Main › Physics

A small bead of mass m rests on a circular hoop of radius sqrt(2) m at the position where the radius to the bead makes 45 deg with the vertical axis. The hoop rotates about a fixed vertical axis through its centre O. The coefficient of friction between bead and hoop is mu = 0.5. Find the maximum angular speed (in rad/s) of the hoop for which the bead does not slip relative to the hoop. (g = 10 m/s²)

1. sqrt(30)
2. sqrt(5)
3. sqrt(10)
4. sqrt(15)

Correct answer: sqrt(30)

Solution

With theta = 45 deg, r = 1 m, mu = 0.5, the limiting condition gives omega² = 10*(sin45 + 0.5*cos45)/(1*(cos45 - 0.5*sin45)) = 30, so omega(max) = sqrt(30) rad/s.

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 →