The speed of ripples (v) on a water surface d | vedclass

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Question

(A) The speed of ripples $v$ depends on surface tension $\sigma$,density $\rho$,and wavelength $\lambda$. We can write the relation as $v = k \sigma^a

Vedclass · Accepted Answer

$\frac{\sigma}{\rho \lambda}$

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(A) The speed of ripples $v$ depends on surface tension $\sigma$,density $\rho$,and wavelength $\lambda$. We can write the relation as $v = k \sigma^a \rho^b \lambda^c$,where $k$ is a dimensionless constant. The dimensional formula for speed $v$ is $[M^0 L T^{-1}]$. The dimensional formula for surface tension $\sigma$ is $[M L^0 T^{-2}]$. The dimensional formula for density $\rho$ is $[M L^{-3} T^0]$. The dimensional formula for wavelength $\lambda$ is $[M^0 L T^0]$. Substituting these into the equation: $[M^0 L T^{-1}] = [M L^0 T^{-2}]^a [M L^{-3} T^0]^b [L]^c$ $[M^0 L T^{-1}] = [M]^{a+b} [L]^{-3b+c} [T]^{-2a}$ Equating the powers of $M, L,$ and $T$ on both sides: $a + b = 0$ $-3b + c = 1$ $-2a = -1$ From $-2a = -1$,we get $a = 1/2$. Substituting $a = 1/2$ into $a + b = 0$,we get $b = -1/2$. Substituting $b = -1/2$ into $-3b + c = 1$,we get $-3(-1/2) + c = 1 \Rightarrow 1.5 + c = 1 \Rightarrow c = -0.5 = -1/2$. Thus,$v \propto \sigma^{1/2} \rho^{-1/2} \lambda^{-1/2}$. Therefore,$v \propto \sqrt{\frac{\sigma}{\rho \lambda}}$. Squaring both sides,$v^2 \propto \frac{\sigma}{\rho \lambda}$.

The speed of ripples $(v)$ on a water surface depends on surface tension $(\sigma)$,density $(\rho)$,and wavelength $(\lambda)$. Then the square of speed $(v^2)$ is proportional to

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