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Question: The velocity of a freely falling body changes as g<sup>p</sup>h<sup>q</sup> where g is acceleration ...

The velocity of a freely falling body changes as gphq where g is acceleration due to gravity and h is the height. The values of p and q are

A

1,121,\frac{1}{2}

B

12,12\frac{1}{2},\frac{1}{2}

C

12,1\frac{1}{2},1

D

1, 1

Answer

12,12\frac{1}{2},\frac{1}{2}

Explanation

Solution

[V]=[g]p[h]q\lbrack V\rbrack = \lbrack g\rbrack^{p}\lbrack h\rbrack^{q}

[M0L1T−1]−[LT−2]p[M0K1T0]q=M0LP+qT−2p\lbrack M^{0}L^{1}T^{- 1}\rbrack - \lbrack LT^{- 2}\rbrack^{p}\lbrack M^{0}K^{1}T^{0}\rbrack^{q} = M^{0}L^{P + q}T^{- 2p}

⇒ 1 = p + q and – 1 = –2P

**∴ **p=12,q=12p = \frac{1}{2},q = \frac{1}{2}