JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2026
- Easy
- Numerical
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 \(\frac{x}{2} M L^2 kg m ^2\). The value of \(x\) is _________.

Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) 17
Step-by-step Solution
Detailed explanation
\(I =\frac{ ML ^2}{3}+\left(\frac{ ML ^2}{12}+ ML ^2\right)\)
\(=\frac{4 ML ^2+ ML ^2+12 ML ^2}{12}\)

\(I =\frac{17}{12} ML ^2\)
\(\therefore \quad x =17\)
\(=\frac{4 ML ^2+ ML ^2+12 ML ^2}{12}\)

\(I =\frac{17}{12} ML ^2\)
\(\therefore \quad x =17\)
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