The moment of inertia \( I \) of each sphere about the central axis (using the parallel axis theorem) is:
\[ I_{\text{total}} = 2 \left( I_{\text{sphere}} + md^2 \right). \]For a solid sphere:
\[ I_{\text{sphere}} = \frac{2}{5}mR^2 = \frac{2}{5} \times 2 \times (0.5)^2 = 0.2 \, \text{kg m}^2. \]Distance \( d \) from the center of each sphere to the midpoint of the rod is \( 0.75 \, \text{m} \).
So,
\[ I_{\text{total}} = 2 \left( 0.2 + 2 \times (0.75)^2 \right) = 2 \left( 0.2 + 1.125 \right) = \frac{53}{20} \, \text{kg m}^2. \]Thus, \( x = 53 \).
A sphere of radius R is cut from a larger solid sphere of radius 2R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is :