The figure shows two solid disc 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) to 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)

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2R² : r²

2R⁴ : R⁴

R² : r²

2R⁴ : r⁴

answer is A.

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Detailed Solution

Let σ is both discs' mass per unit area.
mass of the smaller disc, $m = σ ({πr}^{2})$
and the mass of the larger disc, $M = σ (π R^{2})$

$\frac{{MI}_{big}}{{MI}_{small}} = \frac{\frac{1}{2} {MR}^{2}}{\frac{1}{4} {mr}^{2}} = \frac{2 {MR}^{2}}{{mr}^{2}}$

$\frac{{MI}_{big}}{{MI}_{small}} = \frac{2 ({σπR}^{2}) R^{2}}{({σπr}^{2}) r^{2}} = \frac{2 R^{4}}{r^{4}}$

Hence the correct answer is $2 R^{4} : r^{4} .$

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