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If D and d are external and internal diameters of a circular shaft respectively, its polar moment of inertia is
$$\frac{\pi(D^{4}-d^{4})}{2}$$
$$\frac{\pi(D^{4}-d^{4})}{4}$$
$$\frac{\pi(D^{4}-d^{4})}{64}$$
$$\frac{\pi(D^{4}-d^{4})}{32}$$
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