Question

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 1

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:

• $\sf v = k\ \sqrt{\lambda\ g}$

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.

$\sf M^0 = M^b$

$\sf b = 0$

$\sf{T^{-1}=T^{-2c}}$

$\sf{-1= -2c}$

$\sf{c = \dfrac{1}{2} }$

$\sf L = L^{a - 3b+c}$

$\sf a - 3b +{ \dfrac{1}{2} } = 1$

$\sf a -0 = { \dfrac{1}{2} }$

$\sf a = \dfrac{1}{2}$

$\large v = k\ \lambda^{ \frac{1}{2} } \ \rho^0 \ g^{ \frac{1}{2} }$

$\sf v = k\ \sqrt{\lambda\ g}$