Two identical spheres each of mass 2 kg and radius 50 cm are fixed at the ends of a light rod so that the separation between the centers is 150 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 x is $\_\_\_\_$ .
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$\begin{aligned} & I=\left(\frac{2}{5} m R^2+m d^2\right) \times 2 \\ & I=2\left(\frac{2}{5} \times 2 \times\left(\frac{1}{2}\right)^2+2 \times\left(\frac{3}{4}\right)^2\right)=\frac{53}{20} \mathrm{~kg}-\mathrm{m}^2 \\ & X=53\end{aligned}$