Q 11-06-178JEE MainJEE Main 2020 (6 Sep, Shift 2)Medium
The linear mass density of a thin rod AB of length $L$ varies from A to B as $\lambda(x) = \lambda_0\left(1 + \dfrac xL\right)$, 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:
Answer: (B) $\dfrac{7}{18}ML^2$
$$M = \int_0^L \lambda_0\left(1 + \frac xL\right)dx = \frac32\lambda_0L$$
$$I = \int_0^L \lambda_0\left(1 + \frac xL\right)x^2\,dx = \lambda_0\left(\frac{L^3}{3} + \frac{L^3}{4}\right) = \frac{7}{12}\lambda_0L^3$$
With $\lambda_0 = \dfrac{2M}{3L}$: $I = \dfrac{7}{12}\cdot\dfrac{2M}{3L}\cdot L^3 = \dfrac{7}{18}ML^2$.
Solution by Sreeraj P, M.Sc Physics