i)Velocity potential function for two dimensional flow(∅)=x(2y-1)

Velocity, u=$\frac{∂∅}{∂x}$

Therefore, u= $\frac{(∂(x(2y-1))}{∂x}$

= (2y-1)+(x×(2(0)-0))

=(2y-1) (ans)

Velocity, v=$\frac{(∂∅)}{∂y}$

Therefore, v=$\frac{(∂(x(2y-1)))}{∂y}$

= x×(2-1)

= x. (ans)

ii) Stream function $ψ$=f(x,y) In velocity we can write u=$\frac{∂ψ}{∂y}$ , v=$-\frac{∂ψ}{∂x}$ ∴ (2y-1)= $\frac{∂ψ}{∂y}$ (2y-1) ∂y= ∂ψ Now integrating above equation $∫(2y-1)∂y=∫∂ψ$ $2\frac{y^2}{2}+ y+C(x)=ψ$ $y^{2}+y+C(x)=ψ$…………… (1) Where C(x) is an integral constant and may be a function of x Differentiating the above expression with x, we get $\frac{∂ψ}{∂x} = \frac{(∂C(x))}{∂x}$…………………(2) But, from relationship between velocity potential and stream function we have $\frac{∂ψ}{∂y}=\frac{(∂∅)}{∂x}$ $\frac{∂ψ}{∂y}$=2y-1 $\frac{∂ψ}{∂x}=-\frac{(∂∅)}{∂y}$ $\frac{∂ψ}{∂x}4$=-2x Substitute these in equation 2. $\frac{(∂C(x))}{∂x}$=-2x Integrating above equation with respect to x C(x)=$-x^2$ From equation 1 $ψ=y^2+y-x^2$

This is the stream function required for given potential function

Weight of stone in air=245N

Mass of stone =$\frac{245}{9.81}$=24.97Kg

Volume=$\frac{mass}{(density \ of \ water)}$=$\frac{24.97}{1000}=.02497 m^3$ ….(ans)

$ρ_{fluid}$=s×1000

Mass of stone in water= $\frac{168}{9.81}=17.25kg$

$ρ_{fluid}$=$\frac{mass}{volume}=\frac{17.25}{.02497}$=685.823

Specific gravity=$ρ_{fluid}{1000}$=$\frac{685.823}{1000}$=.6858 ……..(ans)