The linear mass density of a thin rod AB of length L varies from A to… | JEE Main 2020 (6 Sep, Shift 2)

Q154JEE Main 2020 · 6 SEP · SHIFT 2Difficult

Moment of Inertia & Radius of Gyration

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:

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Answer: (c)

$M = \int_0^L \lambda_0\left(1+\frac{x}{L}\right)dx = \frac{3}{2}\lambda_0 L$
$I_A = \int_0^L \lambda_0\left(1+\frac{x}{L}\right)x^2 dx = \lambda_0\left(\frac{L^3}{3}+\frac{L^3}{4}\right) = \frac{7}{12}\lambda_0 L^3$
Put $\lambda_0 = \frac{2M}{3L}$ : $I_A = \frac{7}{12}\cdot\frac{2M}{3L}L^3 = \frac{7}{18}ML^2$
So the answer is (c) $\frac{7}{18}ML^2$ .

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