Question 30 of 280

Two identical spheres each of mass $2kg$ and radius $50cm$ are fixed at the ends of a light rod so that the separation between the centers is $150cm$. Then, moment of inertia of the system about an axis perpendicular to the rod and passing through its middle point is $\dfrac{x}{20}kg{m}^{2}$, where the value of $x$ is

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Answer: 53

Using parallel axis theorem, we can write total moment of inertia as, $I=\left(\dfrac{2}{5}m{R}^{2}+m{d}^{2}\right)\times 2$
$\Rightarrow I=2\left(\dfrac{2}{5}\times 2\times {\left(\dfrac{1}{2}\right)}^{2}+2\times {\left(\dfrac{3}{4}\right)}^{2}\right) =\dfrac{53}{20}kg{m}^{2}$
Therefore, $x=53$.