A) \[{{v}^{2}}rg\]
B) \[{{v}^{2}}\propto g\lambda \rho \]
C) \[{{v}^{2}}\propto g\lambda \]
D) \[{{v}^{2}}\propto {{g}^{-1}}{{\lambda }^{-3}}\]
Correct Answer: C
Solution :
Let \[{{v}^{x}}=k{{g}^{y}}{{\lambda }^{z}}{{\rho }^{\delta }}.\]Now by substituting the dimensions of each quantities and equating the powers of M, L and T we get \[\delta =0\] and \[x=2,\,y=1,\,z=1\] .
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