The radius of gyration of a solid sphere of mass 5 kg about XY is 5 m as shown in figure. The radius of the sphere is (5x)/(√(7)) m, then the value of x is: (a) 5 (b) √(2) (c) √(3) (d) √(5)

Answer: (d). XY is a tangent: I = 2/5MR² + MR² = 7/5MR² = MK². 7/5R² = 25 ⇒ R = (5√(5))/(√(7)) m Comparing with (5x)/(√(7)), x = √(5). So the answer is (d) √(5).

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Moment of Inertia & Radius of Gyration

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:

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Answer: (d)

XY is a tangent: $I = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2 = MK^2$ .
$\frac{7}{5}R^2 = 25 \Rightarrow R = \frac{5\sqrt{5}}{\sqrt{7}}\text{ m}$
Comparing with $\frac{5x}{\sqrt{7}}$ , $x = \sqrt{5}$ .
So the answer is (d) $\sqrt{5}$ .

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