Q 11-01-094JEE MainJEE Main 2024 (4 Apr, Shift 2)Easy
Applying the principle of homogeneity of dimensions, determine which one is correct, where $T$ is time period, $G$ is the gravitational constant, $M$ is mass and $r$ is the radius of the orbit.
Answer: (C) $T^2 = \dfrac{4\pi^2r^3}{GM}$
$[GM] = [\text{M}^{-1}\text{L}^3\text{T}^{-2}][\text{M}] = [\text{L}^3\text{T}^{-2}]$.
So $\dfrac{r^3}{GM}$ has dimensions $\dfrac{[\text{L}^3]}{[\text{L}^3\text{T}^{-2}]} = [\text{T}^2]$, matching $T^2$. The other options fail this check.
Solution by Sreeraj P, M.Sc Physics