Two identical spheres, each of mass 2 ,kg and | vedclass

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(D) The moment of inertia of a solid sphere about an axis passing through its center is $I_{cm} = \frac{2}{5} mR^2$.Using the parallel axis theorem, t

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$53$

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(D) The moment of inertia of a solid sphere about an axis passing through its center is $I_{cm} = \frac{2}{5} mR^2$.Using the parallel axis theorem, the moment of inertia of one sphere about an axis passing through the midpoint of the rod (at a distance $d = 75 \,cm = 0.75 \,m = \frac{3}{4} \,m$ from the center of the sphere) is:$I_{sphere} = I_{cm} + md^2 = \frac{2}{5} mR^2 + md^2$.Given $m = 2 \,kg$, $R = 50 \,cm = 0.5 \,m = \frac{1}{2} \,m$, and $d = 0.75 \,m = \frac{3}{4} \,m$.$I_{sphere} = \frac{2}{5} 	imes 2 	imes (\frac{1}{2})^2 + 2 	imes (\frac{3}{4})^2 = \frac{4}{5} 	imes \frac{1}{4} + 2 	imes \frac{9}{16} = \frac{1}{5} + \frac{9}{8} = \frac{8 + 45}{40} = \frac{53}{40} \,kg \,m^2$.Since there are two identical spheres, the total moment of inertia of the system is:$I_{total} = 2 	imes I_{sphere} = 2 	imes \frac{53}{40} = \frac{53}{20} \,kg \,m^2$.Comparing this with $\frac{x}{20} \,kg \,m^2$, we get $x = 53$.

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