JEE Main 2024 — Circular Motion Question with Solution

\begin{aligned} & \left|\vec{a}_c\right|=\left|\vec{a}_t\right| \\ & \frac{v^2}{r}=\frac{d v}{d t} \\ & \Rightarrow \int_\limits4^v \frac{d v}{v^2}=\int_\limits0^t \frac{d t}{r} \\ & \Rightarrow\left[\frac{-1}{v}\right]_4^v=\frac{t}{r} \\ & \Rightarrow \frac{-1}{v}+\frac{1}{4}=2 t \end{aligned}

$$\begin{array}{rl}& \Rightarrow \mathrm{v}=\frac{4}{1-8\mathrm{t}}=\frac{\mathrm{ds}}{\mathrm{dt}}\\ & 4{\int }_0^{\mathrm{t}}\frac{\mathrm{dt}}{1-8\mathrm{t}}={\int }_0^{\mathrm{s}}\mathrm{ds}\\ & (\mathrm{r}=0.5\mathrm{m}\\ & \mathrm{s}=2\pi \mathrm{r}=\pi )\\ & 4\times \frac{[\ell \mathrm{n}(1-8\mathrm{t}){]}_0^{\mathrm{t}}}{-8}=\pi \\ & \ell \mathrm{n}(1-8\mathrm{t})=-2\pi \\ & 1-8\mathrm{t}={\mathrm{e}}^{-2\pi }\\ & \mathrm{t}=\left(1-{\mathrm{e}}^{-2\pi }\right)\frac{1}{8}\mathrm{s}\end{array}$$

So, $$\alpha =8$$