JEE Main · 2018 · 08 Apr 2018 · Offline · Rotational Mechanics

Build the moment of inertia about the central axis first and then shift it to P. The central disc contributes $\frac{M{R}^2}{2}$. Each of the six surrounding discs has its own centre a distance $2R$ from the central axis, so by the parallel-axis theorem each contributes $\frac{M{R}^2}{2}+M(2R{)}^2=\frac{9M{R}^2}{2}$, and the six with the centre give $I_{cm}=\frac{M{R}^2}{2}+6\times \frac{9M{R}^2}{2}=\frac{55M{R}^2}{2}$. Point P lies $3R$ from the centre of the whole arrangement, so carrying the full mass $7M$ there gives $I_P=\frac{55M{R}^2}{2}+7M(3R{)}^2=\frac{55M{R}^2}{2}+63M{R}^2=\frac{181M{R}^2}{2}$.