The distance of Neptune and Saturn from sun are nearly 10¹³ and 10¹² meters respectively. Assuming that they move in circular orbits, their periodic times will be in the ratio

The correct option is 10√(10).

To find the ratio of the periodic times of Neptune and Saturn, we can use Kepler's Third Law of planetary motion. This law states that the square of the periodic time ($T$) of a planet in a circular orbit is directly proportional to the cube of its orbital radius ($r$) from the Sun:

$T^2\propto r^3$

Let the orbital radius and periodic time of Neptune be $r_1$ and $T_1$, and those of Saturn be $r_2$ and $T_2$ respectively. From the proportionality, we can write the ratio of their periodic times as:

${\left(\frac{T_1}{T_2}\right)}^2={\left(\frac{r_1}{r_2}\right)}^3$

We are given the following values for the distances from the Sun:
- Distance of Neptune, $r_1={10}^{13}\text{m}$
- Distance of Saturn, $r_2={10}^{12}\text{m}$

Substituting these values into the ratio formula:

${\left(\frac{T_1}{T_2}\right)}^2={\left(\frac{{10}^{13}}{{10}^{12}}\right)}^3$

Simplify the fraction inside the parentheses:

$\frac{{10}^{13}}{{10}^{12}}=10$

Substitute this back into the equation:

${\left(\frac{T_1}{T_2}\right)}^2={10}^3=1000$

Taking the square root of both sides gives the ratio of their periodic times:

$\frac{T_1}{T_2}=\sqrt{1000}$

$\frac{T_1}{T_2}=\sqrt{100\times 10}=10\sqrt{10}$

Thus, the ratio of the periodic times of Neptune and Saturn is $10\sqrt{10}$.