A solid sphere of mass 500 g and radius 5 cm is rotated about one of its diameter with angular speed of 10 rad s⁻¹. If the moment of inertia of the sphere about its tangent is x × 10⁻² times its angular momentum about the diameter. Then the value of x will be ___________.

Solution

Physics sir · 3 steps

Shift the axis to the tangent with the parallel axis theorem, then divide by Iω.

1. About the diameter. For a solid sphere, I_d = 2/5MR². With M = 0.5 kg and R = 0.05 m, I_d = 2/5 × 0.5 × 0.0025 = 5 × 10⁻⁴ kg m². So the angular momentum is L = I_d ω = 5 × 10⁻³ kg m² s⁻¹.
2. About the tangent. A tangent is parallel to a diameter at a distance R. The parallel axis theorem gives Iₜ = 2/5MR² + MR² = 7/5MR² = 1.75 × 10⁻³ kg m².

   Iₜ/L = (7/5MR²)/(2/5MR² ω) = 7/2ω = 7/20 = 0.35

3. So. Iₜ = 0.35 L = 35 × 10⁻² L, so x = 35.

The common slip: 10. Using 2/5MR² for the tangent as well, without the parallel axis shift, gives a ratio of 1/ω = 0.1, so 10 instead of 35.

The idea to keep. A tangent to a sphere is a diameter shifted by R, so add MR². When a question compares two quantities of the same body, write the ratio first. Mass and radius often cancel.

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