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Q.

Moment of inertia of a rod of mass m and length l about its one end is I. If one-fourth of its length is cut away, then moment of inertia of the remaining rod about its one end will be

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a

$\frac{9}{16} I$

b

$\frac{3}{4} I$

c

$\frac{I}{16}$

d

$\frac{27}{64} I$

answer is C.

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Detailed Solution

$I = m l^{2} / 3, I^{'} = \frac{(3 m / 4) (3 l / 4)^{2}}{3}$

$= \frac{27}{67} (\frac{m l^{2}}{3}) = \frac{27}{64} I$

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