A planet takes $$200$$ days to complete one revolution around the Sun. If the distance of the planet from Sun is reduced to one fourth of the original distance, how many days will it take to complete one revolution?

We need to find the new orbital period when the distance from the Sun is reduced to one-fourth.

According to Kepler’s Third Law, $$T^2 \propto r^3$$, where $$T$$ is the orbital period and $$r$$ is the orbital radius.

Applying this to the initial and new conditions, we get $$\frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3} = \left(\frac{r_2}{r_1}\right)^3 = \left(\frac{1}{4}\right)^3 = \frac{1}{64}$$, which implies $$\frac{T_2}{T_1} = \frac{1}{8}$$.

Since the original period $$T_1$$ is 200 days, the new period is $$T_2 = \frac{T_1}{8} = \frac{200}{8} = 25$$ days.

The correct answer is Option (1): 25 days.