When we increase the distance between the notched wheel and the mirror, it will take longer time for the light to come back. To compensate for this change, $textbf{the rotational speed of the wheel must be decreased}$ in order to catch up for the returning light.

The time taken by the light to return back si

$$
t=frac{2d}{c}
$$

Here we have $d=10.5 mathrm{km}=10.5times 10^3$ m. and velocity of light $c=3times 10^8$ m/s. Hence

$$
t=frac{2times (10.5times 10^3 m)}{(3times 10^8 m/s)}=7times 10^{-5} s
$$

$$
7times 10^{-5} mathrm{s}
$$