JEE Main 2026 — Jan 21, Shift 1 — Physics solution

JEE Main2026Jan 21, Shift 1Physics

Two identical thin rods of mass M kg and length L 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 $\frac{x}{12} M L^{2}$ kg m $^{2}$ . The value of x is ______________.

Your answer

Solution

Compute the moment of inertia of each rod about point $P$ , then sum.

Vertical rod (one end at $P$ ): $I_{1} = \frac{M L ^{2}}{3}$ .

Horizontal rod: its centre lies a distance $L$ from $P$ (at the bottom end of the vertical rod). By the parallel-axis theorem, $I_{2} = \frac{M L ^{2}}{12} + M L^{2} = \frac{13 M L ^{2}}{12}$ .

Total: $I = I_{1} + I_{2} = \frac{4 M L ^{2} + 13 M L ^{2}}{12} = \frac{17 M L ^{2}}{12}$ .

Comparing with the given form $\frac{x}{12} M L^{2}$ : $x = 17$ .

← Back to full paper

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →