Question

The velocity v of water waves depends on the wavelength λ, density of water ρ and acceleration due to gravity g. The relationship between these quantities will be – 

1. \(v \propto \sqrt {\lambda g} \)
2. \(v \propto \sqrt {\lambda/ g}\)
3. \(v \propto \left( {\frac{1}{{\sqrt {\lambda g} }}} \right)\)
4. None 

Answer 1

Correct Answer - Option 1 : \(v \propto \sqrt {\lambda g} \)

Let \(v\; = \;K{\lambda ^a}{\rho ^b}{g^c}\)

Where K = a dimensionless constant.

Dimensions of various quantities are, \(\left[ v \right]\; = \;L{T^{ - 1}},\;\left[ \lambda \right]\; = \;L,\;\left[ \rho \right]\; = \;M{L^{ - 3}},\;\left[ g \right]\; = \;L{T^{ - 2}}\)

Substituting these dimensions in equations we get,

\(\begin{array}{l} \left[ {L{T^{ - 1}}} \right]\; = \;{\left[ L \right]^a}\;{\left[ {M{L^{ - 3}}} \right]^b}\;{\left[ {L{T^{ - 2}}} \right]^c}\\ {M^0}{L^1}{T^{ - 1}}\; = \;{M^b}{L^{a - 3b + c\;}}{T^{ - 2c}} \end{array}\)

Equating the powers of M,L and T on both sides,

b = 0, a – 3b + c = 1, - 2c = -1

By solving, a = ½, b = 0, c = ½

\(So,\;v\; = \;K{\lambda ^{\frac{1}{2}}}{\rho ^0}{g^{\frac{1}{2}}}\; = \;K\sqrt {\lambda g} \)