Q. The velocity $v$ of water waves may depend on 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
A) $v^{2} \propto \lambda^{- 1}\rho^{- 1}$
B) $v^{2} \propto g\lambda$
C) $v^{2} \propto g\lambda\rho$
D) $g^{- 1} \propto \lambda^{3}$
Explanation
Velocity $v = k\lambda^{a}\rho^{b}g^{c} \Longrightarrow \left\lbrack M^{0}LT^{- 1} \right\rbrack = \left\lbrack L^{a} \right\rbrack\left\lbrack M^{b}L^{- 3b} \right\rbrack\lbrack L^{c}T^{- 2c}\rbrack$ Or $\left\lbrack M^{0}LT^{- 1} \right\rbrack = \lbrack M^{b}L^{a - 3b + c}T^{- 2c}\rbrack$ Equating powers of $M,L$ and $T,$ we get $- 2c = - 1$ Again, $a - 3b + c = 1,b = 0,c = \frac{1}{2}$ $\therefore\ v = k\lambda^{1/2}\rho^{0}g^{1/2}$ or $v^{2} \propto g\lambda$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.