The velocity of the wave `v` depends upon the wave length, density of water `'d'` and acceleration due to gravity `g` Then :

Options

A. `v^(2)=Klambda^(-1)g^(-1)d^(-1)`
B. `v^(2)=Klambda^(1)g^(1)d^(1)`
C. `v^(2)=Klambda^(1)g^(1)d^(0)`
D. `v^(2)=Klambda^(3)g^(-1)d^(-1)`

Answer

C

Explanation

Here `v=Klambda^(a)g^(b)d^(c )`
`M^(0)L^(1)T^(-1)=[L^(1)]^(a)xx[L^(1)T^(-2)]^(b)xx[ML^(-3)]^(c )`
`=M^(c )L^(a+b-3c)T^(-2b)`
Thus `c=0,-2b=-1` or `b=(1)/(2)`
`a+b-3c=1` or `a=1-(1)/(2)=(1)/(2)`
Hence `v=Klambda^(1/2)g^(1/2)d^(0)`
`v^(2)=Klambdag`
Hence correct choice is `(c )`.