Question

Solid sphere $A$ is rotating about an axis $PQ$ If the radius of the sphere is $5 cm$ then its radius of gyration about $PQ$ will be $\sqrt{x} cm$ The value of $x$ is __

Answer 1

Correct Answer: 110

The radius of gyration \(k\) is related to the moment of inertia \(I\) of the sphere and its mass \(M\) by: \[ I = M \cdot k^2 \quad \Rightarrow \quad k = \sqrt{\frac{I}{M}}. \]
1.Moment of inertia of the sphere about the axis \(PQ\): The moment of inertia of a solid sphere about an axis passing through a point at a distance \(d\) from its center is given by the parallel axis theorem: \[ I = I_{\text{center}} + M \cdot d^2, \] where: - \(I_{\text{center}} = \frac{2}{5} M R^2\) is the moment of inertia about the center, - \(d = 10 \, \text{cm}\) is the distance from the center to the axis \(PQ\), - \(R = 5 \, \text{cm}\) is the radius of the sphere. Substituting: \[ I = \frac{2}{5} M R^2 + M d^2 = \frac{2}{5} M (5)^2 + M (10)^2. \] Simplify: \[ I = \frac{2}{5} M \cdot 25 + M \cdot 100 = \frac{50}{5} M + 100 M = 10 M + 100 M = 110 M. \]
2.Radius of gyration (\(k\)): The radius of gyration is: \[ k = \sqrt{\frac{I}{M}} = \sqrt{\frac{110 M}{M}} = \sqrt{110} \, \text{cm}. \]
3.Relating to \(\sqrt{x}\): From the problem, \(k = \sqrt{x} \, \text{cm}\). Thus: \[ x = 110. \]