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:
The radius of the sphere is \(\frac{5x}{\sqrt{7}} \, \text{m}\), then the value of \(x\) is:
- \(5\)
- \(\sqrt{2}\)
- \(\sqrt{3}\)
- \(\sqrt{5}\)
The Correct Option is D
Solution and Explanation
The moment of inertia Ixy is equal to ICM + 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 Ixy 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.
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