JEE Main 2024 — 29 January, Shift 2 — Question 57

$\left|\vec{a}_{C}\right|=\left|\vec{a}_{t}\right|$ $\frac{v^{2}}{r}=\frac{d v}{d t}$

$\Rightarrow \int_{4}^{\mathrm{v}} \frac{\mathrm{dv}}{\mathrm{v}^{2}}=\int_{0}^{\mathrm{t}} \frac{\mathrm{dt}}{\mathrm{r}}$ $\Rightarrow\left[\frac{-1}{\mathrm{v}}\right]_{4}^{\mathrm{v}}=\frac{\mathrm{t}}{\mathrm{r}}$

$\Rightarrow \frac{-1}{\mathrm{v}}+\frac{1}{4}=2 \mathrm{t}$

$\Rightarrow \mathrm{v}=\frac{4}{1-8 \mathrm{t}}=\frac{\mathrm{ds}}{\mathrm{dt}}$

$4 \int_{0}^{\mathrm{t}} \frac{\mathrm{dt}}{1-8 \mathrm{t}}=\int_{0}^{\mathrm{s}} \mathrm{ds}$

$(\mathrm{r}=0.5 \mathrm{~m}$ $\mathrm{s}=2 \pi \mathrm{r}=\pi)$

$4 \times \frac{[\ln (1-8 \mathrm{t})]_{0}^{\mathrm{t}}}{-8}=\pi$

$\ln (1-8 \mathrm{t})=-2 \pi$ $1-8 \mathrm{t}=\mathrm{e}^{-2 \pi}$

$\mathrm{t}=\left(1-\mathrm{e}^{-2 \pi}\right) \frac{1}{8} \mathrm{~s}$ So, $\alpha=8$