JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2023
- Medium
- Numerical
Two identical solid spheres each of mass \(2\,kg\) and radii \(10\,cm\) are fixed at the ends of a light rod. The separation between the centres of the spheres is \(40\,cm\). The moment of inertia of the system about an axis perpendicular to the rod passing through its middle point is \(........... \times 10^{-3}\,kg - m ^2\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(176\)
Step-by-step Solution
Detailed explanation
\(I =2\left( I _{ cm }+ md ^2\right)\)
\(=2\left(\frac{2}{5} mr ^2+ md ^2\right)\)
\(=\frac{4}{5} \times 2 \times(0.1)^2+2(2)(0.20)^2\)
\(=\frac{8}{5} \times 10^{-2}+16 \times 10^{-2}\)
\(=(1.6+16) \times 10^{-2}\)
\(=17.6 \times 10^{-2}\)
\(I =176 \times 10^{-3}\,kg\,m ^2\)
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