A uniform rod of mass m is bent into a semicircle of radius R. The moment of inertia of the rod about an axis passing through P and perpendicular to plane of semicircular ring is

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${mR}^{2}$

$\frac{2}{3} {mR}^{2}$

$(\frac{5}{π}) {mR}^{2}$

$2 {mR}^{2}$

answer is B.

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Detailed Solution

$I = 2 {mR}^{2} (Full ring)$
Semicircular ring = mR²
By parallel axis theorem about P point
$\begin{array}{l} I_{P} = I_{semicircular} + {mR}^{2} \\ = {mR}^{2} + {mR}^{2} \\ = 2 {mR}^{2} \end{array}$

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