Question

The radius of gyration of a solid sphere of mass \(5 \, \text{kg}\) about \(XY\) is \(5 \, \text{m}\) as shown in figure.
The radius of the sphere is \(\frac{5x}{\sqrt{7}} \, \text{m}\), then the value of \(x\) is:

• A. \(5\)
• B. \(\sqrt{2}\)
• C. \(\sqrt{3}\)
• D. \(\sqrt{5}\)

Answer 1

The Correct Option is D

The moment of inertia I_xy is equal to I_CM + MR², which simplifies to $\frac{2}{5}$MR² + MR², resulting in $\frac{7}{5}$MR². Further calculation gives $\frac{7}{5}$ × 5R² = 7R², denoted as equation (1).

The moment of inertia I_xy is also equal to MK², which is 5 × 5², denoted as equation (2).

Equating the results from (1) and (2), we have 5 × 5² = 7 × R².

This implies R = $\sqrt{\frac{5}{7}} \times 5$, which is given as $\frac{5x}{\sqrt{7}}$.

Therefore, x = √5.