JEE Main 2025 Jan 24 Shift 2, Physics Q9: Rolling Motion

JEE Main2025Jan 24, Shift 2Physics

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
$t_{1} < t_{2}$
B
$t_{1} = t_{2}$
C
$t_{1} = 2 t_{2}$
D
$t_{1} > t_{2}$

Solution

For an object rolling down a plane of inclination $θ$ without slipping, the centre-of-mass acceleration is

$a_{c m} = \frac{g sin θ}{1 + \frac{I _{c m}}{M R ^{2}}}$

For the two bodies:

Solid sphere: $I = \frac{2}{5} M R^{2} \Rightarrow a_{1} = \frac{g sin θ}{1 + 2/5} = \frac{5 g sin θ}{7}$
Hollow sphere: $I = \frac{2}{3} M R^{2} \Rightarrow a_{2} = \frac{g sin θ}{1 + 2/3} = \frac{3 g sin θ}{5}$

Since $\frac{5}{7} \approx 0.714 > \frac{3}{5} = 0.6$ , we have $a_{1} > a_{2}$ .

Both bodies cover the same length $ℓ$ starting from rest, so $t = \frac{2 ℓ}{a}$ is smaller for the larger acceleration. Hence $t_{1} < t_{2}$ .

Correct option: (1)

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JEE Main 2025 Jan 24 Shift 2, Physics Q9: Rolling Motion

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