Physics · Rotational Dynamics

JEE Main 2026 — 21 January, Morning Shift — Question 45

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 $\frac{X}{2} M^{2} \mathrm{~kg} \mathrm{~m}^{2}$ . The value of $x$ is $\_\_\_\_$ .

Answer: 17

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{I}=\frac{\mathrm{ML}^{2}}{3}+\left(\frac{\mathrm{ML}^{2}}{12}+\mathrm{ML}^{2}\right)$ $=\frac{4 \mathrm{ML}^{2}+\mathrm{ML}^{2}+12 \mathrm{ML}^{2}}{12}$ $\mathrm{I}=\frac{17}{12} \mathrm{ML}^{2}$ $\therefore \mathrm{x}=17$ Correct Answer : 17

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 21 January, Morning Shift
Subject: Physics
Chapter: Rotational Dynamics
Topic: Moment of Inertia