A particle is moving in a circle of radius $$50$$ cm in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at $$t = 0$$ is $$4 \text{ m s}^{-1}$$, the time taken to complete the first revolution will be $$\frac{1}{\alpha}\left[1 - e^{-2\pi}\right]$$ s, where $$\alpha =$$ ______.

We need to find $$\alpha$$ where the time for the first revolution is $$\frac{1}{\alpha}(1 - e^{-2\pi})$$ s.

Considering radius $$R = 0.5$$ m and initial speed $$v_0 = 4$$ m/s, and using $$a_t = a_n$$, we have $$\frac{dv}{dt} = \frac{v^2}{R}\,. $$

Separating variables gives $$\frac{dv}{v^2} = \frac{dt}{R}\,,$$ and integrating yields $$-\frac{1}{v} = \frac{t}{R} + C\,. $$ At $$t = 0$$, $$C = -\frac{1}{v_0} = -\frac{1}{4}$$, so $$v = \frac{4}{1 - 8t}\,. $$

The distance for one revolution is $$s = 2\pi R = \pi$$ and also $$s = \int_0^T v\,dt = \int_0^T \frac{4}{1-8t}\,dt = -\frac{1}{2}\ln(1-8T)\,. $$ Setting this equal to $$\pi$$ gives $$-\frac{1}{2}\ln(1-8T) = \pi\,, $$ so $$1-8T = e^{-2\pi}$$ and $$T = \frac{1 - e^{-2\pi}}{8} = \frac{1}{8}(1 - e^{-2\pi})\,. $$

Comparing this result with the given expression $$\frac{1}{\alpha}(1-e^{-2\pi})$$ shows that $$\alpha = 8\,. $$

Option: 8