Q75 / 75 · JEE Main 2026 (21 January Shift 1) · Physics
Two identical thin rods of mass $M \mathrm{~kg}$ and length $L \mathrm{~m}$ are connected as shown in figure. Moment of inertia of the combined rod system about an axis passing through point $P$ and perpendicular to the plane of the rods is $\frac{x}{12} \mathrm{ML}^{2} \mathrm{~kg} \mathrm{~m}^{2}$. The value of $x$ is $\_\_\_\_$.
Correct answer: 17
Solution
For vertical rod with axis at end P: $I_1 = \frac{ML^2}{3}$
For horizontal rod centered at distance L from P:
$I_2 = \int_{-L/2}^{L/2} \frac{M}{L}(x^2 + L^2)dx = \frac{M}{L}\left[\frac{x^3}{3} + L^2x\right]_{-L/2}^{L/2}$
$= \frac{M}{L}\left[\frac{L^3}{12} + L^3\right] = \frac{13ML^2}{12}$
Total: $I = \frac{ML^2}{3} + \frac{13ML^2}{12} = \frac{4ML^2 + 13ML^2}{12} = \frac{17ML^2}{12}$
$x = 17$
For horizontal rod centered at distance L from P:
$I_2 = \int_{-L/2}^{L/2} \frac{M}{L}(x^2 + L^2)dx = \frac{M}{L}\left[\frac{x^3}{3} + L^2x\right]_{-L/2}^{L/2}$
$= \frac{M}{L}\left[\frac{L^3}{12} + L^3\right] = \frac{13ML^2}{12}$
Total: $I = \frac{ML^2}{3} + \frac{13ML^2}{12} = \frac{4ML^2 + 13ML^2}{12} = \frac{17ML^2}{12}$
$x = 17$
Hence, the correct answer is 17.