Physics And Measurement Question 252
Question: What are the dimensions of A/B in the relation $ F=A\sqrt{x}+Bt^{2} $ , where F is the force, x is the distance and t is time?
Options:
A) $ ML^{2}{{T}^{-2}} $
B) $ {{L}^{-1/2}}T^{2} $
C) $ {{L}^{-1/2}}{{T}^{-1}} $
D) $ L{{T}^{-2}} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ [F]=[A]\times {{[x]}^{1/2}}=[B]{{[t]}^{2}} $
$ \Rightarrow \frac{[A]}{[B]}=\frac{[t{{[}^{2}}}{{{[x]}^{1/2}}}=[{{L}^{-1/2}}T^{2}] $
