JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Laws of Motion

A block of mass

m

is placed on a surface having vertical cross section given by

y = x^{2} /4

. If coefficient of friction is 0.5, the maximum height above the ground at which block can be placed without slipping is:

(A) $1/4$ m✓ Correct
(B) $1/2$ m
(C) $1/6$ m
(D) $1/3$ m

Answer: (A)

Solution

A block resting on a curved surface without slipping requires the surface slope not to exceed the friction angle:

tan θ \leq μ

For the curve

y = \frac{x ^{2}}{4}

, the slope at any point is

tan θ = \frac{d y}{d x} = \frac{x}{2}

The block can be placed up to the point where

tan θ = μ = 0.5

\frac{x}{2} = 0.5 ⟹ x = 1 m

The corresponding height is

y = \frac{x ^{2}}{4} = \frac{1}{4} m

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Single correct option

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