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Q.

Moment of inertia of a thin rod of mass M and length L about an axis perpendicular to the rod at a distance $\frac{L}{4}$ from end is

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a

$\frac{7 M L^{2}}{48}$

b

$\frac{7 M L^{2}}{24}$

c

$\frac{M L^{2}}{6}$

d

$\frac{M L^{2}}{12}$

answer is D.

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Detailed Solution

$I = I_{C} + M r^{2}$

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