For vertical rod with axis at end P: I1=3ML2
For horizontal rod centered at distance L from P:
I2=∫−L/2L/2LM(x2+L2)dx=LM[3x3+L2x]−L/2L/2
=LM[12L3+L3]=1213ML2
Total: I=3ML2+1213ML2=124ML2+13ML2=1217ML2
x=17
JEE Main 2026 — Physics Mechanics
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 12xML2 kg m2. The value of x is ____.

Held on 21 Jan 2026 · Verified 6 Jul 2026.
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