JEE Main 2025 Apr 7 Shift 1, Physics Q21: Moment of Inertia

Let the common mass be $M$ and the common radius $R$. The moment of inertia of the disc about a diameter is

$I=\frac{M{R}^2}{4}$

Contribution from each body about axis PQ (use parallel-axis theorem, with the perpendicular distance from each body's centre to PQ equal to $R$ for B and C by geometry; for A the axis passes through its centre along a diameter):

• Disc A (axis is its own diameter): $I_A=\frac{M{R}^2}{4}$
• Solid sphere B: about a diameter $\frac{2}{5}M{R}^2$; parallel-axis shift by $R$ adds $M{R}^2$: $I_B=\frac{2}{5}M{R}^2+M{R}^2=\frac{7}{5}M{R}^2$
• Spherical shell C: about a diameter $\frac{2}{3}M{R}^2$; parallel-axis shift by $R$ adds $M{R}^2$: $I_C=\frac{2}{3}M{R}^2+M{R}^2=\frac{5}{3}M{R}^2$

Total:

$I_{PQ}=\frac{M{R}^2}{4}+\frac{7}{5}M{R}^2+\frac{5}{3}M{R}^2$

Bring to common denominator 60:

$I_{PQ}=\frac{15+84+100}{60}M{R}^2=\frac{199}{60}M{R}^2$

Express as a multiple of $I=M{R}^2/4$, i.e. $M{R}^2=4I$:

$I_{PQ}=\frac{199}{60}\cdot 4I=\frac{199}{15}I$

Matching $\frac{x}{15}I$: $x=199$.

$\therefore$ Answer = 199.