Applications of Dimensional Analysis question from JEE Main 2019 (8 Apr, Shift 2)

Answer: (a)

1. $[S] = MT^{-2}$, $[I] = ML^2$, $[h] = ML^2T^{-1}$; $p = MLT^{-1} = S^aI^bh^c$.
2. M: $a + b + c = 1$; L: $2b + 2c = 1$; T: $-2a - c = -1$.
3. First two give $a = \frac{1}{2}$; then $c = 0$ and $b = \frac{1}{2}$.
4. So the answer is (a) $S^{1/2}I^{1/2}h^0$.