# The linear mass density of a thin rod AB of length L varies from A to B as $\lambda (x)=\lambda _{0}\left ( 1+\frac{x}{L} \right ),$ where x is the distance from A. If M is the mass of the rod then its moment of inertia about an axis passing through A and perpendicular to the rod is :   Option: 1   Option: 2 Option: 3   Option: 4

\begin{aligned} &I=\int r^{2} d m=\int x^{2} \lambda d x\\ &I=\int_{0}^{L} x^{2} \lambda_{0}\left(1+\frac{x}{L}\right) d x\\ &I=\lambda_{0} \int_{0}^{1}\left(x^{2}+\frac{x^{3}}{L}\right) d x\\ &I=\lambda\left[\frac{L^{3}}{3}+\frac{L^{3}}{4}\right]\\ &I=\frac{7 L^{3} \lambda_{0}}{12} \dots(i)\\ &M=\int_{0}^{L} \lambda d x=\int_{0}^{L} \lambda_{0}\left(1+\frac{x}{L}\right) d x\\ &M=\lambda_{0}\left(L+\frac{L}{2}\right)=\lambda_{0} \frac{3 L}{2}\\ &\frac{2}{3} M=\left(\lambda_{0} L\right) \dots(ii) \end{aligned}

\begin{aligned} &\text { From (i) \& (ii) }\\ &I=\frac{7}{12}\left(\frac{2}{3} M\right) L^{2}=\frac{7 M L^{2}}{18} \end{aligned}

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