Question 64 of 280

The linear mass density of a thin rod $AB$ of length $L$ varies from $A$ to $B$ as $\lambda \left(x\right)={\lambda }_{0}\left(1+\dfrac{x}{L}\right)$, 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 :

A$\dfrac{5}{12}{ML}^{2}$B$\dfrac{7}{18}{ML}^{2}$C$\dfrac{2}{5}{ML}^{2}$D$\dfrac{3}{7}{ML}^{2}$

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Answer: (B) $\dfrac{7}{18}{ML}^{2}$

$dm=\lambda .dv$
$dI=dm.{x}^{2}$
$={\lambda }_{0}\left(1+\dfrac{x}{L}\right).{x}^{2}dx$
$I={\lambda }_{0}{\int }_{0}^{1}\left({x}^{2}+\dfrac{{x}^{3}}{L}\right)dx$
$={\lambda }_{0}{\left|\dfrac{{x}^{3}}{3}+\dfrac{{x}_{4}}{4L}\right|}_{0}^{L}$
$={\lambda }_{0}\left[\dfrac{{L}^{3}}{3}+\dfrac{{L}_{3}}{4}\right]$
$I=\dfrac{7{\lambda }_{0}{L}^{3}}{12}$
$m=\int _{0}^{L}\lambda dx=\dfrac{3{\lambda }_{0}L}{2}$
$\dfrac{I}{m}=\dfrac{7{L}^{2}}{12\times \dfrac{3}{2}}=\dfrac{7{L}^{2}}{18}$
$\Rightarrow$ $I=\dfrac{7}{18}{mL}^{2}$