Q 11-07-087JEE MainJEE Main 2023 (24 Jan, Shift 2)Easy
If the distance of the earth from Sun is $1.5\times10^6\ \text{km}$, then the distance of an imaginary planet from Sun, if its period of revolution is $2.83$ years, is
Answer: (C) $3\times10^6\ \text{km}$
Kepler's third law: $r\propto T^{2/3}$. With $2.83\approx2\sqrt2=2^{3/2}$:
$$r=1.5\times10^6\times\left(2^{3/2}\right)^{2/3}=1.5\times10^6\times2=3\times10^6\ \text{km}$$
Solution by Sreeraj P, M.Sc Physics