JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Rotational Mechanics

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) 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)

(A) $R^{2} : r^{2}$
(B) $2 r^{4} : R^{4}$
(C) $2 R^{4} : r^{4}$ ✓ Correct
(D) $2 R^{2} : r^{2}$

Answer: (C)

Solution

About the perpendicular axis through its centre the bigger disc has

I_{1} = \frac{M R ^{2}}{2}

. Equal mass per unit area

σ

gives

M = σ π R^{2}

and

m = σ π r^{2}

, so

m = M \frac{r ^{2}}{R ^{2}}

. About a diameter a disc has half the perpendicular axis value, so for the smaller disc

I_{2} = \frac{m r ^{2}}{4} = \frac{M r ^{4}}{4 R ^{2}}

Dividing,

\frac{I _{1}}{I _{2}} = \frac{M R ^{2}}{2} \cdot \frac{4 R ^{2}}{M r ^{4}} = \frac{2 R ^{4}}{r ^{4}}

The two moments of inertia are in the ratio

2 R^{4} : r^{4}

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Single correct option

Practise more: all JEE Main PYQs on Rotational Mechanics →

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Rotational Mechanics

Solution

More JEE Main questions on Rotational Mechanics