Question

A solid sphere of radius \(10\) cm is rotating about an axis which is at a distance \(15\) cm from its centre. The radius of gyration about this axis is \( \sqrt{n} \) cm. Find the value of \( n \).

Hint:

Radius of gyration directly reflects how mass is distributed about the axis—parallel axis theorem is essential here.

Answer 1

Correct Answer: 265

Concept: The radius of gyration \(k\) about an axis is defined by: \[ I = Mk^2 \] For a rigid body rotating about an axis not passing through its centre, the moment of inertia is calculated using the {parallel axis theorem}. For a solid sphere: \[ I_{\text{cm}}=\frac{2}{5}MR^2 \]
Step 1: Apply the parallel axis theorem Distance of the axis from centre: \[ d=15\text{ cm} \] \[ I = I_{\text{cm}} + Md^2 = \frac{2}{5}MR^2 + Md^2 \]
Step 2: Substitute given values \[ R=10\text{ cm} \] \[ I = M\left(\frac{2}{5}\times10^2 + 15^2\right) = M(40+225) = 265M \]
Step 3: Find the radius of gyration \[ Mk^2=265M \Rightarrow k^2=265 \] \[ k=\sqrt{265}\text{ cm} \]
Step 4: Compare with given form \[ k=\sqrt{n}\Rightarrow n=265 \] Final Answer: \[ \boxed{n=265} \]