Units and Measurements question for JEE Main, with solution

Answer: (A) $\sqrt{\dfrac{hG}{c^3}}$

Let length $= c^a G^b h^p$:
$$[L] = [LT^{-1}]^a\,[M^{-1}L^3T^{-2}]^b\,[ML^2T^{-1}]^p$$

M: $-b + p = 0$
L: $a + 3b + 2p = 1$
T: $-a - 2b - p = 0$

From M, $p = b$. Then T gives $a = -3b$, and L gives $-3b + 3b + 2b = 1 \Rightarrow b = \tfrac12$.

So $p = \tfrac12$ and $a = -\tfrac32$:
$$\text{Length} = \sqrt{\frac{hG}{c^3}}$$
This is the Planck length, about $10^{-35}\ \text{m}$.

Solution by Sreeraj P, M.Sc Physics