JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Rotational Mechanics

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A, B and C are disc, solid sphere and spherical shell respectively with same radii and masses. These masses are placed as shown in figure.

The moment of inertia of the given system about PQ is

\frac{x}{15} I

, where

I

is the moment of inertia of the disc about its diameter. The value of x is ____.

Answer: 199

Solution

The axis

P Q

is a diameter of the disc, so its contribution is

I_{A} = \frac{M R ^{2}}{4} = I

. The solid sphere and the spherical shell each have their centre a perpendicular distance

R

from

P Q

, so by the parallel-axis theorem

I_{B} = \frac{2}{5} M R^{2} + M R^{2} = \frac{7}{5} M R^{2}, I_{C} = \frac{2}{3} M R^{2} + M R^{2} = \frac{5}{3} M R^{2} .

Adding, over a common denominator of

60

I_{P Q} = \frac{M R ^{2}}{4} + \frac{7 M R ^{2}}{5} + \frac{5 M R ^{2}}{3} = \frac{15 + 84 + 100}{60} M R^{2} = \frac{199}{60} M R^{2} .

Since

I = \frac{M R ^{2}}{4}

, i.e.

M R^{2} = 4 I

, this is

\frac{199}{60} \times 4 I = \frac{199}{15} I

, so

x = 199

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Integer answer

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JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Rotational Mechanics

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