The figure shows two solid discs with radius R and r respectively. If mass per unit area is same for both, what is the ratio of MI of bigger disc around axis AB (Which is ⊥ to the plane of the disc and passing through its centre) of MI of smaller disc around one of its diameters lying on its plane? Given 'M' is the mass of the larger disc. (MI stands for moment of inertia)

Solution

Physics sir · 4 steps

Same mass per unit area means the small disc is lighter: find its mass first, then compare.

1. Mass of small disc. With surface density σ, the big disc has M = σ π R² and the small disc has m = σ π r². Dividing gives m = Mr²/R².
2. Bigger disc. The axis AB is perpendicular to the plane and passes through the centre, so I₁ = 1/2MR².
3. Smaller disc. About a diameter, the perpendicular axis theorem gives half the central value, so I₂ = 1/4mr² = 1/4Mr⁴/R².

   I₁/I₂ = (1/2MR²)/(1/4Mr⁴/R²) = 2R⁴/r⁴

4. So. The ratio is 2R⁴ : r⁴, option (d).

The idea to keep. When two bodies share the same density, write each mass in terms of its size before comparing. For a disc, mass goes as r², so its moment of inertia goes as r⁴.

Answer from the NTA final answer key. Solution by PrepWiser.