JEE Main 2025 Apr 7 Shift 2, Physics Q25: Moment of Inertia

JEE Main2025Apr 7, Shift 2Physics

M and R be the mass and radius of a disc. A small disc of radius R/3 is removed from the bigger disc as shown in figure. The moment of inertia of remaining part of bigger disc about an axis AB passing through the centre O and perpendicular to the plane of disc is $\frac{4}{x} M R^{2}$ . The value of x is _______.

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Solution

Treat the disc with cavity as the full disc minus the removed small disc.

Moment of inertia of the full disc about axis AB through O: $I_{1} = \frac{M R ^{2}}{2}$ .

Mass of the removed small disc: since it is made of the same material with area ratio $(R /3)^{2} / R^{2} = 1/9$ , its mass is $m = \frac{M}{9}$ .

Moment of inertia of the removed disc about its own centre O' is $\frac{m ( R /3 ) ^{2}}{2} = \frac{M R ^{2}}{162}$ . Using the parallel axis theorem to shift to axis AB through O (a distance $\frac{2 R}{3}$ away): $I_{2} = \frac{M R ^{2}}{162} + \frac{M}{9} (\frac{2 R}{3})^{2} = \frac{M R ^{2}}{162} + \frac{4 M R ^{2}}{81} = \frac{M R ^{2} + 8 M R ^{2}}{162} = \frac{M R ^{2}}{18}$ .

Remaining moment of inertia: $I = I_{1} - I_{2} = \frac{M R ^{2}}{2} - \frac{M R ^{2}}{18} = \frac{9 M R ^{2} - M R ^{2}}{18} = \frac{8 M R ^{2}}{18} = \frac{4 M R ^{2}}{9}$ .

Comparing with $\frac{4 M R ^{2}}{x}$ , we get $x = 9$ .

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JEE Main 2025 Apr 7 Shift 2, Physics Q25: Moment of Inertia

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