NEET 2025 Physics Moment of Inertia MCQ Question

The correct option is (4) 108 days.

Explanation: The Sun's rotation (spin about its own axis) is considered here, and the problem asks for the new period of rotation (time for one rotation, often called "period of revolution" around its centre in some questions) when the Sun expands to twice its present radius, with no external torque. Since no external torque acts ("without any external influence"), angular momentum is conserved.

For a sphere of uniform density, the moment of inertia about its axis is I = (2/5) M R² (where M = mass, R = radius) Mass M remains constant (expansion without loss of mass). New radius R' = 2R New moment of inertia I' = (2/5) M (2R)² = (2/5) M (4 R²) = 4 × [(2/5) M R²] = 4 I Angular momentum L = I ω = constant (where ω = angular velocity = 2π / T, and T = period of rotation)

So, I ω = I' ω' I (2π / T) = I' (2π / T') Cancel 2π: I / T = I' / T' T' / T = I' / I = 4 Original T = 27 days New period T' = 4 × 27 = 108 days Thus, the period becomes 108 days. (Note: The actual Sun is not uniform density, and its equatorial rotation period is ~25 days while polar is longer, but the problem assumes uniform density sphere and uses 27 days as given.)

Correct answer: (4) 108 days