JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Rotational Mechanics

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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 ______.

Answer: 17

Solution

The vertical rod rotates about its end at

P

, contributing

\frac{M L ^{2}}{3}

. The horizontal rod has its centre a distance

L

from

P

, so by the parallel-axis theorem it contributes

\frac{M L ^{2}}{12} + M L^{2} = \frac{13 M L ^{2}}{12}

. Adding the two,

I = \frac{M L ^{2}}{3} + \frac{13 M L ^{2}}{12} = \frac{17 M L ^{2}}{12}

, so

x = 17

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Integer answer

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