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

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(D) The moment of inertia of a solid sphere about its diameter is $I_{cm} = \frac{2}{5}mr^2$. Using the parallel axis theorem,the moment of inertia of

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$176$

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(D) The moment of inertia of a solid sphere about its diameter 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 $d = 20\,cm = 0.2\,m$ from the center of the sphere) is $I = I_{cm} + md^2$. Since there are two identical spheres,the total moment of inertia is $I_{total} = 2(I_{cm} + md^2) = 2\left(\frac{2}{5}mr^2 + md^2ight)$. Given $m = 2\,kg$,$r = 10\,cm = 0.1\,m$,and $d = 20\,cm = 0.2\,m$: $I_{total} = 2 \left[ \frac{2}{5} 	imes 2 	imes (0.1)^2 + 2 	imes (0.2)^2 ight]$ $I_{total} = 2 \left[ \frac{4}{5} 	imes 0.01 + 2 	imes 0.04 ight]$ $I_{total} = 2 \left[ 0.8 	imes 0.01 + 0.08 ight]$ $I_{total} = 2 \left[ 0.008 + 0.08 ight] = 2 	imes 0.088 = 0.176\,kg\,m^2$. Converting to the required form: $0.176\,kg\,m^2 = 176 	imes 10^{-3}\,kg\,m^2$.

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