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

Answer: (D)

$\sqrt{5}$

Answer 2

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