Three solid spheres each of mass $M$ and radius $R$ are arranged as shown in the figure. The moment of inertia of the system about YY' will be
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Always separate composite bodies into individual elements relative to the target axis. Recognizing that the lower spheres are shifted by exactly $d = R$ allows you to immediately assign them a value of $\frac{7}{5}MR^2$ using the standard parallel axis shift, turning the calculation into simple fraction addition.
Step 1: Describe the arrangement. Three identical solid spheres, each of mass $M$ and radius $R$, are stacked so the axis $YY'$ passes through the centre of the top sphere, while the two lower spheres touch it on either side. Step 2: Recall a sphere's moment of inertia. About a diameter through its own centre, a solid sphere has $I_{cm} = \dfrac{2}{5}MR^2$. Step 3: Handle the top sphere. The axis goes through its centre, so $I_1 = \dfrac{2}{5}MR^2$. Step 4: Use the parallel axis theorem for the lower spheres. Each lower sphere's centre is a distance $R$ from $YY'$, so $I = \dfrac{2}{5}MR^2 + MR^2 = \dfrac{7}{5}MR^2$. Step 5: Account for both lower spheres. By symmetry each contributes $\dfrac{7}{5}MR^2$, giving $\dfrac{7}{5}MR^2 + \dfrac{7}{5}MR^2$. Step 6: Add all three. $I = \dfrac{2}{5}MR^2 + \dfrac{7}{5}MR^2 + \dfrac{7}{5}MR^2 = \dfrac{16}{5}MR^2$, which is option (1). \[ \boxed{I = \frac{16}{5}MR^2} \]
