Question

The velocity of water waves may depend on their wavelength λ, the density of water ρ and the acceleration due to gravity g. The method of dimensional analysis gives the relation between these quantities as : (where K is a dimensionless constant)

• A. $v^{2} = K\lambda^{- 1}g^{- 1}\rho^{- 1}$
• B. v² = K λ g
• C. v² = K λ g ρ
• D. v² = K λ³ g^–1ρ^–1

Answer 1

Answer: (B)

v² = K λ g

Answer 2

$v \propto \lambda^{a}\rho^{b}g^{c}$

$\lbrack M^{0}\text{ }\text{L}^{1}\text{ T}\text{ }^{\text{-1}}\rbrack\ = \ \lbrack M^{0}\text{ }\text{L}^{1}\text{ }\text{T}^{0}\rbrack^{a}\ \lbrack M^{1}\text{ }\text{L}^{\text{-3}}\text{ }\text{T}^{0}\rbrack^{b}\ \lbrack M^{0}\text{ }\text{L}^{1}\text{ T}\text{ }^{\text{-2}}\rbrack^{c}$ $\left[ M ^ { 0 } L ^ { 1 } T ^ { - 1 } \right] = M ^ { b } L ^ { a - 3 b + c } T ^ { - 2 C }$ ∴ b = 0 a = 1/2

a – 3b + c = 1 ⇒ b = 0

– 2c = – 1 c = ½

$v \propto \lambda^{1/2}\rho^{0}g^{1/2}$

or $v^{2}\ = \text{ K }\lambda\ g$where K is constant.