Q70 / 75 · JEE Main 2020 (06 Sep Shift 2) · Physics

The linear mass density of a thin rod $AB$ of length $L$ varies from $A$ to $B$ as $λ (x) = λ_{0} (1 + \frac{x}{L})$ , 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 :

Solution

$dm = λ . dv$

$dI = dm . x^{2}$

$= λ_{0} (1 + \frac{x}{L}) . x^{2} dx$

$I = λ_{0} \int_{0}^{1} (x^{2} + \frac{x^{3}}{L}) dx$

$= λ_{0} {|\frac{x^{3}}{3} + \frac{x_{4}}{4 L}|}_{0}^{L}$

$= λ_{0} [\frac{L^{3}}{3} + \frac{L_{3}}{4}]$

$I = \frac{7 λ_{0} L^{3}}{12}$

$m = \int_{0}^{L} λdx = \frac{3 λ_{0} L}{2}$

$\frac{I}{m} = \frac{7 L^{2}}{12 \times \frac{3}{2}} = \frac{7 L^{2}}{18}$

$\Rightarrow$ $I = \frac{7}{18} {mL}^{2}$

Hence, the correct answer is option: 'B'.