Question:
Solid sphere is rotating about an axis as shown in figure. If the radius of the sphere is 5 cm, then radius of gyration about that axis will be \(\sqrt{x}\) cm. The value of x is
Solid sphere is rotating about an axis as shown in figure. If the radius of the sphere is 5 cm, then radius of gyration about that axis will be \(\sqrt{x}\) cm. The value of x is
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Use the parallel axis theorem with the axis 10 cm from the centre.
Updated On: Oct 1, 2026
- \(100\)
- \(110\)
- \(120\)
- \(140\)
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The Correct Option is B
Solution and Explanation
Step 1: Read the figure
The figure shows a solid sphere of radius \(5\ \text{cm}\) whose centre is \(10\ \text{cm}\) from the axis of rotation.
Step 2: Moment of inertia
About the centre, \(I_{cm} = \frac25MR^2 = \frac25M(25) = 10M\) (in cm squared). By the parallel axis theorem, \(I = 10M + M(10)^2 = 110M\).
Step 3: Radius of gyration
\(I = Mk^2\) gives \(k^2 = 110\ \text{cm}^2\), so \(k = \sqrt{110}\ \text{cm}\). Thus \(x = 110\), option (B).
Final Answer:
The value of x is 110. \[ \boxed{\text{(B)}\ 110} \]
The figure shows a solid sphere of radius \(5\ \text{cm}\) whose centre is \(10\ \text{cm}\) from the axis of rotation.
Step 2: Moment of inertia
About the centre, \(I_{cm} = \frac25MR^2 = \frac25M(25) = 10M\) (in cm squared). By the parallel axis theorem, \(I = 10M + M(10)^2 = 110M\).
Step 3: Radius of gyration
\(I = Mk^2\) gives \(k^2 = 110\ \text{cm}^2\), so \(k = \sqrt{110}\ \text{cm}\). Thus \(x = 110\), option (B).
Final Answer:
The value of x is 110. \[ \boxed{\text{(B)}\ 110} \]
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