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Physics · PYQ NEET UG 2025

हिन्दी

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 :

  1. \( \frac{7}{8} \)
  2. \( \frac{7}{40} \)
  3. \( \frac{7}{57} \)
  4. \( \frac{7}{64} \)

Correct answer · Option 2

\( \frac{7}{40} \)

Explanation

Using mass proportional to volume and parallel axis theorem for the cavity, mass of large sphere \( M = 8m \), moment of inertia of small sphere \( I_s = \frac{2}{5}mR^2 + mR^2 = \frac{7}{5}mR^2 \). Rest part \( I_{rest} = I_{large} - I_{s} = \frac{32}{5}mR^2 - \frac{7}{5}mR^2 = \frac{25}{5}mR^2 \). Ratio equals \( \frac{7}{40} \).
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