JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2024
- Hard
- Numerical
Two identical spheres each of mass \(2 \mathrm{~kg}\) and radius \(50 \mathrm{~cm}\) are fixed at the ends of a light rod so that the separation between the centers is \(150 \mathrm{~cm}\). Then, moment of inertia of the system about an axis perpendicular to the rod and passing through its middle point is \(\frac{x}{20} \mathrm{~kg} \mathrm{~m}^2\), where the value of \(\mathrm{x}\) is _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(53\)
Step-by-step Solution
Detailed explanation
\(\mathrm{I}=\left(\frac{2}{5} \mathrm{mR}^2+\mathrm{md}^2\right) \times 2\)
\(\mathrm{I}=2\left(\frac{2}{5} \times 2 \times\left(\frac{1}{2}\right)^2+2 \times\left(\frac{3}{4}\right)^2\right)=\frac{53}{20} \mathrm{~kg}-\mathrm{m}^2\)
\(\mathrm{X}=53\)
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