JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Rotational Mechanics

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A solid sphere and a hollow sphere of the same mass and of same radius are rolled on an inclined plane. Let the time taken to reach the bottom by the solid sphere and the hollow sphere be

t_{1}

and

t_{2}

, respectively, then

(A) $t_{1} < t_{2}$ ✓ Correct
(B) $t_{1} = t_{2}$
(C) $t_{1} = 2 t_{2}$
(D) $t_{1} > t_{2}$

Answer: (A)

Solution

For a body of moment of inertia

I_{c m} = k m R^{2}

rolling without slipping down an incline of angle

θ

, the acceleration of the centre of mass is

a_{c m} = \frac{g sin θ}{1 + k} .

For a solid sphere,

k = \frac{2}{5}

a_{1} = \frac{g sin θ}{1 + \frac{2}{5}} = \frac{5 g sin θ}{7} .

For a hollow (thin-shell) sphere,

k = \frac{2}{3}

a_{2} = \frac{g sin θ}{1 + \frac{2}{3}} = \frac{3 g sin θ}{5} .

Comparing,

\frac{5}{7} \approx 0.714

is greater than

\frac{3}{5} = 0.6

, so

a_{1} > a_{2}

: the solid sphere accelerates faster. Since the time to cover a fixed length

ℓ

starting from rest is

t = 2 ℓ / a

, a larger acceleration means a shorter time. Hence

t_{1} < t_{2}

: the solid sphere reaches the bottom first.

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Rotational Mechanics

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