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2 Rods of length l and 2l , Area 2A and A are joined end to end . Now by hammering we change the area of rod from 2A to A . Find moment of inertia about the joint in final condition . It is given that both the rods have some density$\;\rho$

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

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Answer : $\;\large\frac{10}{3} \rho A l^{3}$
Explanation :
Finally volume remains same
So $\;(l)2A=(l^{'})A$
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
$m=(\rho) 4lA$
Moment of inertia = $\;\large\frac{m (4l)^2}{12}$
$=\large\frac{\rho 4 l A 16l^2}{12}$
$= \large\frac{16}{3}\;\rho Al^3$
answered Mar 15, 2014 by yamini.v
edited Mar 15, 2014 by yamini.v

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