Q 11-01-233JEE MainJEE Main 2017 (8 Apr)Medium
Time $(T)$, velocity $(C)$ and angular momentum $(h)$ are chosen as fundamental quantities instead of mass, length and time. In terms of these, the dimensions of mass would be:
Answer: (A) $[M] = [T^{-1}C^{-2}h]$
Let $M = T^aC^bh^c$ with $[C] = LT^{-1}$ and $[h] = ML^2T^{-1}$:
$$M^1L^0T^0 = T^a(LT^{-1})^b(ML^2T^{-1})^c = M^cL^{b+2c}T^{a-b-c}$$
- $M$: $c = 1$
- $L$: $b + 2c = 0 \Rightarrow b = -2$
- $T$: $a - b - c = 0 \Rightarrow a = -1$
So $[M] = [T^{-1}C^{-2}h]$.
Solution by Sreeraj P, M.Sc Physics