Physics And Measurement Question 81
Question: The velocity of a freely falling body changes as $ g^{p}h^{q} $ where g is acceleration due to gravity and $ h $ is the height. The values of $ p $ and $ q $ are
[NCERT 1983; EAMCET 1994]
Options:
A) $ 1,\frac{1}{2} $
B) $ M^{0}L^{2}{{T}^{-2}} $
C) $ \frac{1}{2},1 $
D) $ 1,1 $
Show Answer
Answer:
Correct Answer: B
Solution:
$ v\propto g^{p}h^{q} $ (given)
By substituting the dimension of each quantity and comparing the powers in both sides we get $ [L{{T}^{-1}}]={{[L{{T}^{-2}}]}^{p}}{{[L]}^{q}} $
$ \Rightarrow $
$ p+q=1,-2p=-1,\therefore p=\frac{1}{2},q=\frac{1}{2} $
