The moment of inertia of a semicircular ring about an axis, passing through the center and perpendicular to the plane of the ring, is \((\frac{1}{x})MR^2\), where R is the radius and M is the mass of the semicircular ring. The value of x will be:
The moment of inertia of a semicircular ring about an axis, passing through the center and perpendicular to the plane of the ring, is \((\frac{1}{x})MR^2\), where R is the radius and M is the mass of the semicircular ring. The value of x will be:
Correct Answer: 2
Solution and Explanation
Step 1: Formula for moment of inertia - Moment of inertia \[ I = R^2 \, dm \] where \[ dm = \frac{M}{\pi R}. \] For a semicircular ring, \[ I = \frac{1}{2} M R^2. \]
Step 2: Compare with given expression - Given \[ I = \frac{1}{x} M R^2. \] Equating \[ \frac{1}{2} M R^2 = \frac{1}{x} M R^2, \] we get \[ x = 2. \]
Final Answer: The value of x is 2.
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