Question

solve correct answer​

Answer 1

$\huge\red\checkmark$ 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

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

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

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

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

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

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

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

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

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

Answer 2

Step-by-step explanation:

the correct solution is given below