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JEE Main 202621 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Morning Shift
Subject
Physics
Chapter
Rotational Motion

JEE Main 2026 · 21 January · Morning Shift · Physics · Rotational Motion

Two identical thin rods of mass M ~kg and length L ~m are connected...

Two identical thin rods of mass M kgM \mathrm{~kg} and length L mL \mathrm{~m} are connected as shown in figure. Moment of inertia of the combined rod system about an axis passing through point PP and perpendicular to the plane of the rods is x12ML2 kg m2\frac{x}{12} \mathrm{ML}^{2} \mathrm{~kg} \mathrm{~m}^{2}. The value of xx is ____\_\_\_\_.
Two identical thin rods of mass M ~kg and length L ~m are connected as shown in figure. Moment of inertia of the

Correct answer

0

Step-by-step solution

For vertical rod with axis at end P: I1=ML23I_1 = \frac{ML^2}{3}
For horizontal rod centered at distance L from P:
I2=∫−L/2L/2ML(x2+L2)dx=ML[x33+L2x]−L/2L/2I_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}
=ML[L312+L3]=13ML212= \frac{M}{L}\left[\frac{L^3}{12} + L^3\right] = \frac{13ML^2}{12}
Total: I=ML23+13ML212=4ML2+13ML212=17ML212I = \frac{ML^2}{3} + \frac{13ML^2}{12} = \frac{4ML^2 + 13ML^2}{12} = \frac{17ML^2}{12}
x=17x = 17
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