JEE Main On 8 April 2017 Question 12
Question: Time (T), velocity (3) 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 - [JEE Online 08-04-2017]
Options:
A) $ [M]=[{T^{-1}}{C^{-2}}h] $
B) $ [M]=[T{C^{-2}}h] $
C) $ [M]=[{T^{-1}}{C^{-2}}{h^{-1}}] $
D) $ [M]=[{T^{-1}}C^{2}h] $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ M\propto T^{x}v^{y}h^{z} $ $ M’L^{0}T^{0}={{(T’)}^{x}}{{(L^{1}{T^{-1}})}^{y}}{{(M^{1}L^{2}{T^{-1}})}^{z}} $ $ M^{1}L^{0}T^{0}=M^{z}{L^{y+2z}}+{T^{x-y-z}} $ $ z=1 $ $ y+2z=0 $ $ x-y-z=0 $ $ y=-2 $ $ x+2-1=0 $ $ x=-1 $ $ M\Rightarrow {T^{-1}}{C^{-2}}h^{1} $
