Rotational Motion Question 10
Question 10 - 2024 (31 Jan Shift 2)
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
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Answer: (53)
Solution:
$\mathrm{I}=\left(\frac{2}{5} \mathrm{mR}^{2}+\mathrm{md}^{2}\right) \times 2$
$\mathrm{I}=2\left(\frac{2}{5} \times 2 \times\left(\frac{1}{2}\right)^{2}+2 \times\left(\frac{3}{4}\right)^{2}\right)=\frac{53}{20} \mathrm{~kg}-\mathrm{m}^{2}$
$\mathrm{X}=53$
