Solid sphere A is rotating about an axis PQ. If the radius of the sphere is 5 cm then its radius of gyration about PQ will be √x cm. The value of x is ________.

Solution

Physics sir · 3 steps

The axis PQ misses the centre by 10 cm, so shift the axis with the parallel-axis theorem first.

1. About the centre. For a solid sphere about an axis through its centre, I_(cm) = 2/5MR². With R = 5 cm, this is 2/5M × 25 = 10M cm².
2. Shift to PQ. The figure shows PQ parallel to a diameter, at d = 10 cm from the centre. The parallel-axis theorem adds Md² = 100M cm².

   Mk² = I_(PQ) = 2/5MR² + Md² = 10M + 100M = 110M

3. So. k² = 110 cm², so k = √110 cm and x = 110.

The common slip: 10. Using only the centre-axis value 2/5R² gives 10. The axis PQ is 10 cm away from the centre, so the d² = 100 term must be added.

The idea to keep. Radius of gyration comes from I = Mk². For any off-centre axis, get I about the centre first, add Md², then divide by M to read off k².

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