The velocity of a body falling freely under acceleration due to gravity g varies as `g^(p)h^(q)` where h is decrease in height of the body. What are the values of p and q?

Options

A. `(1)/(2),1`
B. `(1)/(2),(1)/(2)`
C. `1,(1)/(2)`
D. `1,1`

Answer

b

Explanation

The dimensional equation of v is `[v]=[L^(1)T^(-1)]`
The dimensional equation of g is `[g]=[L^(1)M^(0)T^(-2)]`.
The velocity v is given by `v=g^(p)h^(q)`
`therefore [L^(1)T^(-1)]=[L^(1)T^(-2)]^(p)[L^(1)]^(q)`
`therefore L^(1)T^(-1)=L^(p)T^(-2p)L^(q)=L^(p+q)T^(-2p)`
Equating the powers of L and T on both sides, we get
`p+q=1and-2p=-1 therefore p=(1)/(2)`
and `(1)/(2)+q=1 therefore q=(1)/(2)`
Thus, `vpropg^(1//2)h^(1//2)orvpropsqrt(gh)`