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 $

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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} $



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