Question

the velocity V of water waves may depend on their wavelength lemda density of water p and the acceleration due to gravity g find out relation between these quantity by dimension methode​

Answer 1

To find:

Relationship between velocity of water wave, wavelength, density and acceleration due to gravity by dimensional analysis.

Calculation:

Let Velocity be related as follows:

$\sf{ \therefore \: v \propto { \lambda}^{x} \times { \rho}^{y} \times {g}^{z} }$

$\sf{ = > \: \bigg \{ L{T}^{ - 1} \bigg\} \propto { \bigg \{L \bigg \}}^{x} \times { \bigg \{M{L}^{ - 3} \bigg \}}^{y} \times { \bigg \{L{T}^{ - 2} \bigg \} }^{z} }$

$\sf{ = > \: \bigg \{ L{T}^{ - 1} \bigg\} \propto \bigg \{{M}^{y}{L}^{(x - 3y + z)} {T}^{ - 2z} \bigg \}}$

Comparing both sides, we get :

$\sf{1) \: y = 0 }\\ \sf{ 2)x - 3y + z = 1 }\\ \sf{3) - 2z = -1}$

Solving these equations, we get:

$\sf{ \therefore \: x = \dfrac{1}{2} ,y = 0 ,z = \dfrac{1}{2} }$

Putting these values:

$\sf{ \therefore \: v \propto { \lambda}^{ \frac{1}{2} } \times { \rho}^{0} \times {g}^{ \frac{1}{2} } }$

$\sf{ = > \: v = k \bigg( { \lambda}^{ \frac{1}{2} } \times { \rho}^{0} \times {g}^{ \frac{1}{2} } \bigg)}$

[Here "k" is dimension-less constant]

So, final answer is:

$\boxed{ \bf{ v = k \bigg( { \lambda}^{ \frac{1}{2} } \times { \rho}^{0} \times {g}^{ \frac{1}{2} } \bigg)}}$