Physics, asked by prashantmaan212, 6 months ago

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​

Answers

Answered by nirman95
4

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)}}

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