Practice portal › Rotational Motion › Moment of Inertia

A sphere of radius R is cut from a larger solid sphere of radius 2R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is:

Asked in NEET 2025 · Composite and cavity bodies

Figure: Composite and cavity bodies
Answer: (3) 7/(57)

Step-by-step solution

Let the whole sphere (radius 2R) have mass M. The small sphere has 1/8 of the volume, so mass M/8.

Whole sphere about its diameter: I=2/5M(2R)²=8/5MR².

Small sphere: its centre is R from the Y-axis. Iₛ=2/5·/M8R²+/M8R²=7/(40)MR².

Rest: 8/5MR²-7/(40)MR²=(57)/(40)MR².

Ratio: (7/40)/(57/40)=7/(57).

Why the other options are wrong

More Moment of Inertia questionsAll Moment of Inertia questions →

Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.

Practise this with a timer