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(C) The dimension of velocity $v$ is $[M^0 L^1 T^{-1}]$.The dimension of wavelength $\lambda$ is $[M^0 L^1 T^0]$.The dimension of acceleration due to

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$v^2 \propto g \lambda$

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(C) The dimension of velocity $v$ is $[M^0 L^1 T^{-1}]$.The dimension of wavelength $\lambda$ is $[M^0 L^1 T^0]$.The dimension of acceleration due to gravity $g$ is $[M^0 L^1 T^{-2}]$.The dimension of density $\rho$ is $[M^1 L^{-3} T^0]$.Let the relation be $v = k \lambda^a g^b \rho^c$,where $k$ is a dimensionless constant.Equating dimensions on both sides:$[L T^{-1}] = [L]^a [L T^{-2}]^b [M L^{-3}]^c$$[M^0 L^1 T^{-1}] = [M^c L^{a+b-3c} T^{-2b}]$Comparing the powers of $M, L,$ and $T$:For $M$: $c = 0$For $T$: $-2b = -1 \implies b = 0.5$For $L$: $a + b - 3c = 1 \implies a + 0.5 - 0 = 1 \implies a = 0.5$Substituting the values,$v \propto \lambda^{0.5} g^{0.5} \rho^0$.Therefore,$v \propto \sqrt{g \lambda}$,which implies $v^2 \propto g \lambda$.

The velocity of water waves $v$ may depend upon their wavelength $\lambda$,the density of water $\rho$,and the acceleration due to gravity $g$. The method of dimensions gives the relation between these quantities as:

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