**Question:**

The linear mass density of a thin rod $A B$ of length $L$ varies

from A to B as $\lambda(x)=\lambda_{0}\left(1+\frac{x}{\text { L }}\right)$, where $x$ is the distance

from $\mathrm{A}$. If $\mathrm{M}$ is the mass of the rod then its moment of inertia about an axis passing through $\mathrm{A}$ and perpendicular to the rod is :

Correct Option: , 2

**Solution:**

Mass of the small element of the rod

$d m=\lambda \cdot d x$

Moment of inertia of small element,

$d I=d m \cdot x^{2}=\lambda_{0}\left(1+\frac{x}{L}\right) \cdot x^{2} d x$

Moment of inertia of the complete rod can be obtained by integration

$I=\lambda_{0} \int_{0}^{L}\left(x^{2}+\frac{x^{3}}{L}\right) d x$

$=\lambda_{0}\left|\frac{x^{3}}{3}+\frac{x^{4}}{4 L}\right|_{0}^{L}=\lambda_{0}\left[\frac{L^{3}}{3}+\frac{L^{3}}{4}\right]$

$\Rightarrow I=\frac{7 \lambda_{0} L^{3}}{12}$ ....(1)

Mass of the thin rod,

$M=\int_{0}^{L} \lambda d x=\int_{0}^{L} \lambda_{0}\left(1+\frac{x}{L}\right) d x=\frac{3 \lambda_{0} L}{2}$

$\therefore \lambda_{0}=\frac{2 M}{3 L}$

$\therefore I=\frac{7}{12}\left(\frac{2 M}{3 L}\right) L^{3} \Rightarrow I=\frac{7}{18} M L^{2}$

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