If the radius of earth is reduced to three-fourth of its present value without change in its mass then value of duration of the day of earth will be ________ hours 30 minutes.

Solution

Physics sir · 3 steps

No outside torque acts on the shrinking earth, so Iω stays fixed and the day scales as R².

1. Conserve angular momentum. Shrinking the earth involves only internal forces, so no external torque acts and L = Iω is constant. Treat the earth as a solid sphere: I = 2/5MR², with M unchanged.

   2/5MR² · 2π/T = constant ⇒ T ∝ R²

2. Put in numbers. With R' = 3/4R, T' = 24 × (3/4)² = 24 × 9/16 = 13.5 hours.
3. So. 13.5 hours is 13 hours 30 minutes, so the blank is 13.

The common slip: 18. Taking the period as proportional to R instead of R² gives 24 × 3/4 = 18 hours, not 13 hours 30 minutes.

The idea to keep. When a spinning body changes shape with no external torque, Iω is conserved. For a sphere I ∝ R², so the period of rotation goes as R²: a smaller earth spins faster.

Answer from the NTA final answer key. Solution by PrepWiser.