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Physics Question on Surface tension

The speed vv of ripples on the surface of water depends on surface tension σ \sigma density p and wavelength λ\lambda. The square of speed v is proportional to

A

σρ λ\frac{ \sigma }{\rho \, \lambda }

B

ρσ λ \frac{ \rho }{ \sigma \, \lambda }

C

λσ ρ\frac{ \lambda }{ \sigma \, \rho }

D

ρλσ\rho \lambda \sigma

Answer

σρ λ\frac{ \sigma }{\rho \, \lambda }

Explanation

Solution

Let v ∝ σa ρb λc\propto \, \sigma^a \, \rho^b \, \lambda^c Equating dimensions on both sides [M0LT−1]∝[MT−2]a [ML−3]b [L]c[ M^0 L T^{ - 1} ] \propto [ MT^{ - 2} ]^a \, [ ML^{ - 3} ]^b \, [ L]^c ∝[ML]a+b[L]−3b+c[T]−2a\propto [ ML]^{ a + b } [ L ]^{ - 3b + c } [ T ] ^{ - 2a } Equating the powers of M, L, T on both sides, we get a + b = 0 - 3b + c = 1 -2a = - 1 Solving, we get a = 12, b=12, c=12\frac{1}{2} , \, b = \frac{1}{2}, \, c = \frac{1}{2} ∴v∝σ1/2 ρ−1/2 λ−1/2\therefore v \propto \sigma^{1/2} \, \rho^{ - 1/2 } \, \lambda^{ - 1 / 2} ∴v2∝σρλ\therefore v^2 \propto \frac{ \sigma }{ \rho \lambda}