Two spheres each of mass M and radius R are c | vedclass

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(B) Let the two spheres be $S_1$ and $S_2$. The axis of rotation passes through the centre of $S_1$ and is perpendicular to the rod.For sphere $S_1$,t

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$\frac{84}{5} MR^2$

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(B) Let the two spheres be $S_1$ and $S_2$. The axis of rotation passes through the centre of $S_1$ and is perpendicular to the rod.For sphere $S_1$,the moment of inertia about its own centre is $I_1 = \frac{2}{5} MR^2$.For sphere $S_2$,the distance of its centre from the axis of rotation is $d = R + 4R + R = 6R$ is incorrect based on the diagram. The rod length is $4R$ between the surfaces,or the distance between centres is $4R$. Given the diagram shows the rod length $4R$ between the spheres,the distance between the centres is $d = R + 4R + R = 6R$. However,if the rod length $4R$ is the distance between centres,then $d = 4R$.Assuming the distance between centres is $4R$:For $S_1$,$I_1 = \frac{2}{5} MR^2$.For $S_2$,using the parallel axis theorem,$I_2 = I_{cm} + Md^2 = \frac{2}{5} MR^2 + M(4R)^2 = \frac{2}{5} MR^2 + 16MR^2 = \frac{82}{5} MR^2$.Total moment of inertia $I = I_1 + I_2 = \frac{2}{5} MR^2 + \frac{82}{5} MR^2 = \frac{84}{5} MR^2$.

Two spheres each of mass $M$ and radius $R$ are connected with a massless rod of length $4R$. The moment of inertia of the system about an axis passing through the centre of one of the spheres and perpendicular to the rod will be

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