Question

The velocity (v) of water waves may depend upon their wavelength (λ) , the density of water (ρ) and the acceleration due to gravity (g). Find the relation between these quantities by the method of dimensions.
with explanation and method..​

Answer 1

Answer:

Relation between the quantities is:

$\;\;\;\bigstar\;\;\bf{v=\sqrt{\lambda\;g}}$

Explanation:

Given that,

velocity (v) of water waves depend upon wavelength (λ), density of water (ρ), and acceleration due to gravity (g).

• We need to find the relation between these quantities.

Now, As we know

dimensions of v: $\rm{[ M^0L^1T^{-1}]}$

dimensions of λ: $\rm{[M^0L^1T^0]}$

dimensions of ρ: $\rm{[M^1 L^{-3} T^0]}$

dimensions of g: $\rm{[M^0 L^1 T^{-2}]}$

Let, v is related to λ, ρ and g in the way

$\to \rm{v \propto {\lambda}^x \; {\rho}^y \;{g}^z}$

then, solving dimensionally

$\to \rm{ [M^0L^1T^{-1}] = [ M^0 L^1T^0 ]^x\; [M^1L^{-3}T^0]^y\;[M^0L^1T^{-2}]^z}$

$\to \rm{ [M^0L^1T^{-1}] = [ L^1 ]^x\; [M^1L^{-3}]^y\;[L^1T^{-2}]^z}$

$\to \rm{ [M^0L^1T^{-1}] = [ L^x ]\; [M^yL^{-3y}]\;[L^zT^{-2z}]}$

$\to \rm{ [M^0L^1T^{-1}] = [ M^y \;L^{x-3y+z} \;T^{-2z} ]}$

Using the rule of homogeneity of dimensions

We will get,

$\bigstar\boxed{\rm{\;y=0\;}}$

Also,

→ -2 z = -1

$\bigstar\boxed{\rm{\;\;z = \dfrac{1}{2}\;\;}}$

And

→ x - 3y + z = 1

substituting values of y and z we will get,

$\bigstar\boxed{\rm{\;\;x=\dfrac{1}{2}\;\;}}$.

Therefore,

Correct relation between the quantities is

$\boxed{\boxed{\rm{v \propto \rho^0 \lambda^{\frac{1}{2}} \; g^{\frac{1}{2}}\propto \sqrt{\lambda\;g}}}}$.

Answer 2

$\huge\red\checkmark$ Dimension of Velocity = $\bf{[M^0\:L^1\:T^{-1}]}$

$\huge\red\checkmark$ Dimension of wavelength = $\bf{[M^0\:L^1\:T^0]}$

$\huge\red\checkmark$ Dimension of g = $\bf{[M^0\:L^1\:T^{-2}]}$

$\huge{\color{aqua}\checkmark}$ Dimension of $\red{\rho}$ = $\bf{[M^1\:L^{-3}\:T^0]}$

Let,

• $\bf\pink{v\:\propto\:\lambda^{a}\:g^{b}\:\rho^{c}\:}$

☯︎ Using dimensional method,

$\bf{:\implies\:[M^0\:L^1\:T^{-1}]\:=\:[L^1]^{a}\:[L^1\:T^{-2}]^{b}\:[M^1\:L^{-3}]^{c}\:}$

$\bf{:\implies\:[M^0\:L^1\:T^{-1}]\:=\:[L^a]\:[L^b\:T^{-2b}]\:[M^c\:L^{-3c}]\:}$

$\bf{:\implies\:[M^0\:L^1\:T^{-1}]\:=\:\Big[M^c\:L^{(a\:+\:b\:-\:3c)}\:T^{-2b}\Big]\:}$

✈︎ From L.H.S & R.H.S, we get

$\bf{1)\:c\:=\:0\:}$

$\bf{2)\:a\:+\:b\:-\:3c\:=\:1\:}$

$\bf{3)\:-2b\:=\:-1\:}$

$\bf{:\implies\:b\:=\:\dfrac{1}{2}\:}$

➪ Now putting the value of c & b in condition (2), we get

$\bf{2)\:a\:+\:b\:-\:3c\:=\:1\:}$

$\bf{:\implies\:a\:+\:\dfrac{1}{2}\:-\:3\times{0}\:=\:1\:}$

$\bf{:\implies\:a\:+\:\dfrac{1}{2}\:-\:0\:=\:1\:}$

$\bf{:\implies\:a\:=\:1\:-\:\dfrac{1}{2}\:}$

$\bf{:\implies\:a\:=\:\dfrac{2\:-\:1}{2}\:}$

$\bf{:\implies\:a\:=\:\dfrac{1}{2}\:}$

$\huge\red\therefore$ $\bf{v\:\propto\:\:\lambda^{1/2}\:g^{1/2}\:\rho^{0}\:}$

$\bf\green{:\implies\:v\:\propto\:\:\sqrt{\lambda\:g}\:}$

$\bf\blue{:\implies\:v^2\:\propto\:\:{\lambda\:g}\:}$