JEE Main 2021 — Rotational Motion Question with Solution

According to question, a circular disc reaches from top to bottom of an inclined plane of length L. This can be shown as


When the disc slips down the inclined plane, it takes time t₁. Therefore, in this case its acceleration, a₁ = g sin$$\theta$$

$$\because$$ s = ut₁ + $$\frac{1}{2}$$a₁t$${}_1^2$$

$$\Rightarrow$$ s = $$\frac{1}{2}$$ g sin$$\theta$$t$${}_1^2$$ .... (i)

And when the disc rolls down the inclined plane, it takes time t₂. Therefore in this case, its acceleration,

$$a_2=\frac{g\sin \theta }{1+\frac{K^2}{R^2}}=\frac{g\sin \theta }{1+\frac{1}{2}}=\frac{2}{3}g\sin \theta$$ [$$\because$$ for disc, $$\frac{K^2}{R^2}=\frac{1}{2}$$]

$$\because$$ s = ut₂ + $$\frac{1}{2}$$ a₂t$${}_2^2$$ = $$\frac{1}{2}$$ . $$\frac{2}{3}$$ g sin$$\theta$$ t$${}_2^2$$ .... (ii)

On dividing Eq. (i) by Eq. (ii), we get

$$\frac{t_2}{t_1}=\sqrt{\frac{3}{2}}$$ .... (iii)

According to question, value of $$\frac{t_2}{t_1}$$ is $$\sqrt{\frac{3}{x}}$$.

Comparing it with Eq. (iii), we get x = 2