Q 11-01-220JEE MainJEE Main 2019 (8 Apr, Shift 2)Medium
If surface tension ($S$), moment of inertia ($I$) and Planck's constant ($h$), were to be taken as the fundamental units, the dimensional formula for linear momentum would be
Answer: (A) $S^{1/2}I^{1/2}h^0$
Let $p = S^aI^bh^c$ with $[S] = MT^{-2}$, $[I] = ML^2$, $[h] = ML^2T^{-1}$, $[p] = MLT^{-1}$.
- $M$: $a + b + c = 1$
- $L$: $2b + 2c = 1$
- $T$: $-2a - c = -1$
From the first two, $a = \frac12$; then the third gives $c = 0$ and so $b = \frac12$.
$$p = S^{1/2}I^{1/2}h^0$$
Solution by Sreeraj P, M.Sc Physics