$(a)\;\large\frac{10}{3} \rho A l^{3} \qquad(b)\;\large\frac{8}{3} \rho A l^{3}\qquad(c)\;\large\frac{4}{3} \rho A l^{3}\qquad(d)\;\large\frac{16}{3} \rho A l^{3}$

Thermodynamics

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Answer : $\;\large\frac{10}{3} \rho A l^{3}$

Explanation :

Finally volume remains same

So $\;(l)2A=(l^{'})A$

$l^{'}=2l$

Moment of inertia of Rod about an axis passing through center and perpendicular to Rod is $\;\large\frac{mL^2}{12}\;$ Wher m is the mass of Rod

L is the length of Rod

Hence

$m=(\rho) 4lA$

Moment of inertia = $\;\large\frac{m (4l)^2}{12}$

$=\large\frac{\rho 4 l A 16l^2}{12}$

$= \large\frac{16}{3}\;\rho Al^3$

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