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Four identical rods each of mass ‘m’ and length $l$ make a square frame in the xy – plane as shown in the figure. The M.I. about the Z –axis is

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$\frac{5 m l^{2}}{3}$

$\frac{10 m l^{2}}{3}$

$\frac{m l^{2}}{3}$

$4 m l^{2}$

answer is B.

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Detailed Solution

$I = 3_{1} + I_{2} + I_{3} + I_{4}$
$= \frac{m l^{2}}{3} + \frac{m l^{2}}{3} + (I_{c m} + \frac{m 5 l^{2}}{4}) 2$
$= (\frac{2}{3} + \frac{2}{12} + \frac{10}{4}) m l^{2} = \frac{10}{3} m l^{2}$

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