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 \bot 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)

Rotational Motion PYQ — JEE Main 2021 · 27 Jul · Shift 1 · Q55 with Solution

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Correct answer: D

Let surface mass density be $\sigma$ .

Then, $M_{\text{big}} = \sigma \pi R^2 = M$

For the smaller disc, $m_{\text{small}} = \sigma \pi r^2$

So, $m_{\text{small}} = M\frac{r^2}{R^2}$

Moment of inertia of bigger disc about axis through centre perpendicular to plane

For a solid disc, $I_{\text{big}} = \frac{1}{2}MR^2$

For a solid disc, MI about a diameter is $I_{\text{diameter}} = \frac{1}{4}mr^2$

Thus for the smaller disc,

= \frac{1}{4}M\frac{r^4}{R^2}$$ 5. **Required ratio** $$I_{\text{big}} : I_{\text{small}} = \frac{1}{2}MR^2 : \frac{1}{4}M\frac{r^4}{R^2}$$ Cancel $M$: $$= \frac{1}{2}R^2 : \frac{1}{4}\frac{r^4}{R^2}$$ Multiply both terms by $4R^2$: $$= 2R^4 : r^4$$ 6. **Final answer** $$\boxed{2R^4 : r^4}$$ So the correct option is **D**.