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
Ixy = ICM + MR2 = $\frac{2}{5}$MR2 + MR2 = $\frac{7}{5}$MR2 = $\frac{7}{5}$ × 5R2 = 7R2 ...(1)
Ixy = MK2 = 5 × 52 ...(2)
5 × 52 = 7 × R2 [From (1) and (2)]
$\implies R = \sqrt{\frac{5}{7}} \times 5 = \frac{5x}{\sqrt{7}}$ (Given)
x = √5
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