Question

Consider a water tank shown in the figure. It has one wall at x = L and can be taken to be very wide in the z direction. When filled with a liquid of surface tension S and density ρ, the liquid surface makes angle θ(θ << 1) with the x-axis at x = L. If y(x) is the height of the surface then the equation for y(x) is:

(take θ(x) = sin θ(x) = tan θ(x) = `dy/dx`, g is the acceleration due to gravity)

पर्याय

• `(d^2y)/(dx^2) = (rho g)/S x`

• `(d^2 y)/(dx^2) = (rho g)/S y`

• `(d^2 y)/(dx^2) = sqrt((rho g)/S)`

• `dy/dx = sqrt((rho g)/S) x`

Answer 1

`bb((d^2 y)/(dx^2) = (rho g)/S y)`

Explanation:

From Pascal’s law, PA = PB = Po (atmospheric pressure)

P_B = P_o = P_C − yρg

⇒ PC = P_o − yρg

`P_o  - S/r`

S = surface tension

r = radius of the meniscus at C

∴ yρg = `S/r`    ...(i)

r = `([1 + (dy/dx)^2]^(3/2))/((d^y)/(d x^2))`

Slope = `dy/dx` = tan θ

Since θ is very small, `dy/dx` is also very small and hence rejected.

Hence, r = `1/((d^2 y)/(d x^2))`    ...(ii)

Putting r in Eq. (i), we get

yρg = `S (d^2 y)/(d x^2)`

`(d^2 y)/(d x^2) = (rho g)/S y`