Question

A U tube manometer has water poured into the left side and oil (density is 790 kg/m^3) poured into to the right side. The water in left side is measured to be 70 cm high, while the right side contains both oil and water with oil height 4 times as high as water height. Determine the height of oil and height of water in right side of U tube.

Answer 1

1) Height of oil in right limb = 67.31 cm

2) Height of water in the right limb = 16.83 cm

Step-by-step explanation:

The U-tube manometer is shown in the attached figure

Foe equilibrium The pressure at the bottom of the U tube should be same

Let the height of the water in the left limb of the manometer be
`h_L`

Thus the pressure at the bottom is found using the equation of pressure statics as

`P_(bottom)=P_(atm)+\rho _(water)* g* h_(L).............(i)`

Similarly for the liquid in the right limb the pressure at the bottom is the sum of the oil column and the water column

Thus we can write

`P_(bottom)=P_(atm)+\rho _(water)* g* h_{}+\rho _(oil)* g* 4h_{}...........(ii)`

Equating the equations 'i' and 'ii' we get

`P_(atm)+\rho _(water)* g* h_(L)=P_(atm)+\rho _(water)* g* h_{}+\rho _(oil)* g* 4h_{}\\\\\rho _(water)* g* 0.7=\rho _(oil)* 4h* g+\rho _(water)* h* g\\\\\therefore h_{}=(0.7* \rho _(water))/(4* \rho _(oil)+\rho _(water))\\\\h_{}=(0.7* 1000)/(4* 790+1000)=16.83cm`

Thus the height of oil is
`4* 16.83=67.31cm` andthe height of water in the right limb is 16.83 cm.