JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Rotational Mechanics

Each solid sphere has a moment of inertia $\frac{2}{5}m{\left(\frac{d}{2}\right)}^2=\frac{m{d}^2}{10}$ about its own centre. The three centres form an equilateral triangle of side $d$, so the centroid lies a distance $\frac{d}{\sqrt{3}}$ from every centre. For the axis through the centroid perpendicular to the plane, the parallel-axis theorem gives $I_0=3\left[\frac{m{d}^2}{10}+m{\left(\frac{d}{\sqrt{3}}\right)}^2\right]=3\left[\frac{m{d}^2}{10}+\frac{m{d}^2}{3}\right]=\frac{13m{d}^2}{10}$. About an axis through the centre of one sphere, that sphere offers only its own $\frac{m{d}^2}{10}$, while the other two sit a distance $d$ away and each contribute $\frac{m{d}^2}{10}+m{d}^2$, so $I_A=\frac{m{d}^2}{10}+2\cdot \frac{11m{d}^2}{10}=\frac{23m{d}^2}{10}$. Dividing the two, $\frac{I_0}{I_A}=\frac{13}{23}$.