1. Write dimensions of each quantity

We use:

• Time period: $[T] = T$
• Radius: $[r] = L$
• Mass: $[M] = M$
• Gravitational constant: from Newton's law, $F = \frac{G m_1 m_2}{r^2}$ so $[G] = \frac{[F][r]^2}{[M]^2} = \frac{(M L T^{-2})L^2}{M^2} = M^{-1}L^3T^{-2}$

Thus, $[G] = M^{-1}L^3T^{-2}$

We need the RHS to have dimensions of $T^2$.

2. Check option A

$T^2 = \frac{4\pi^2 r}{GM^2}$ Since $4\pi^2$ is dimensionless, $\left[\frac{r}{GM^2}\right] = \frac{L}{(M^{-1}L^3T^{-2})(M^2)}$ $= \frac{L}{ML^3T^{-2}} = M^{-1}L^{-2}T^2$ This is not equal to $T^2$.

So, A is incorrect.

3. Check option B

$T^2 = 4\pi^2 r^3$ $[r^3] = L^3$ This is not $T^2$.

So, B is incorrect.

4. Check option C

$T^2 = \frac{4\pi^2 r^3}{GM}$ Now, $\left[\frac{r^3}{GM}\right] = \frac{L^3}{(M^{-1}L^3T^{-2})(M)}$ $= \frac{L^3}{L^3T^{-2}} = T^2$ This matches the dimension of $T^2$.

So, C is correct.

5. Check option D

$T^2 = \frac{4\pi^2 r^2}{GM}$ $\left[\frac{r^2}{GM}\right] = \frac{L^2}{(M^{-1}L^3T^{-2})(M)}$ $= \frac{L^2}{L^3T^{-2}} = L^{-1}T^2$ This is not $T^2$.

So, D is incorrect.

6. Final answer

The only dimensionally correct relation is $T^2 = \frac{4\pi^2 r^3}{GM}$ So the correct option is C.