A sphere of radius R is cut from a larger solid sphere or 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:

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Question

This question was previously asked in

NEET 2025 Official Paper (Held On: 04 May, 2025)

Answer 1

Correct option is: (3) 7 / 57

For a larger solid sphere about diameter Y-axis:

I_whole = (2 / 5) × M × (2R)² = (8 / 5) × M × R²

1 (10)

Density of sphere is uniform:

M / V_whole = M_smaller / V_smaller

M / ((4/3)π(2R)³) = M′ / ((4/3)πR³)

⇒ M′ = M / 8

Using parallel axis theorem for smaller sphere:

I′ = I_cm + M′ × R² = (2 / 5) × (M / 8) × R² + (M / 8) × R² = (7 / 40) × M × R²

Ratio:

Ratio = I_smaller / I_remaining = I′ / (I_whole − I′)

= ((7 / 40) × M × R²) / (((8 / 5) − (7 / 40)) × M × R²)

= 7 / 57