Question

A U-tube manometer with both ends open, contains 0.35 m of oil on its left limb with an interface with water below it. If the water level on the right limb is 0.28 m above the interface, what is the SG of oil?

Answer 1

0.8

Step-by-step explanation:

Given:

`h_(oil)` = 0.35m

`h_(water)` = 0.28m

Equating the pressure in the manometer at both ends

we have

Pressure at the left limb = Pressure at right limb

`\rho_(oil) gh_(oil)=\rho_(water) gh_(water)`

substituting the values in the above equation, we get

`\rho_(oil)* g* 0.35=\rho_(water)* g* 0.28`

`(\rho_(oil))/(\rho_(water)) =( 0.28)/(0.35)`

`(\rho_(oil))/(\rho_(water)) =0.8`

we know that specific gravity is defined as the ratio of the density of the fluid with respect to the density of water

thus, SG of oil = 0.8