Derive the relationship between the velocity v of water which depends

1.	Derive the relationship between the velocity v of water which depends on the density of water rho and the acceleration due to gravity g
Answer» CORRECT QUESTION: DERIVE the relationship between the velocity v of water which depends on wavelength, density of water RHO and the acceleration due to gravity g. ANSWER: $\sf v = <klux>K</klux>\ \sqrt{\lambda\ g}$ GIVEN: Velocity of water which depends on wavelength, density of water and the acceleration due to gravity. TO FIND: Relationship between wavelength, density and gravity. EXPLANATION: $\boldsymbol{[ v ] = [ M^0 \ L \ T^{-1}]}$ $\boldsymbol{[\lambda] = [M^0\ L^1\ T^0]}$ $\boldsymbol{[\rho] = [M^1\ L^{-3}\ T^0]}$ $\boldsymbol{[g] =[M^0\ L \ T^{-2}]}$ $\sf v\propto \lambda^a \ \rho^b \ g^c$ $\sf v = k\ \lambda^a \ \rho^b \ g^c$ K is the constant of proportionality and has no unit and dimensions. $\boldsymbol{ [v]= [ \lambda]^a \ [\rho]^b \ [g]^c}$ $\boldsymbol{[\lambda]^{a} = [M^0\ L^1\ T^0]^{a}}$ $\boldsymbol{[\lambda]^{a} = [ L^a]}$ $\boldsymbol{[\rho] ^{b}= [M^1\ L^{-3}\ T^0]^{b}}$ $\boldsymbol{[\rho] ^{b}= [M^b\ L^{-3b}]}$ $\boldsymbol{[g]^{c} =[M^0\ L \ T^{-2}]^{c}}$ $\boldsymbol{[g]^{c} =[ L^{c} \ T^{-2c}]}$ $\boldsymbol{[ \lambda]^a \ [\rho]^b \ [g]^c = [ L^a] [M^b\ L^{-3b}][ L^{c} \ T^{-2c}]}$ Add the power for same bases. $\boldsymbol{[ \lambda]^a \ [\rho]^b \ [g]^c =[M^b \ L^{a - 3b+c} \ T^{-2c}]}$ $\boldsymbol{[ M^0 \ L \ T^{-1}]=[M^b \ L^{a - 3b+c} \ T^{-2c}]}$ Equate the powers for same bases. $\sf M^0 = M^b$ $\sf b = 0$ $\sf{T^{-1}=T^{-2c}}$ $\sf{-1= -2c}$ $\sf{c = \dfrac{1}{2} }$ $\sf L = L^{a - 3b+c}$ $\sf a - 3b +{ \dfrac{1}{2} } = 1$ $\sf a -0 = { \dfrac{1}{2} }$ $\sf a = \dfrac{1}{2}$ $\large v = k\ \lambda^{ \frac{1}{2} } \ \rho^0 \ g^{ \frac{1}{2} }$ $\sf v = k\ \sqrt{\lambda\ g}$