Water and gasoline surface of relative density 1 and 0.6 are open to the atmosphere at the same elevation. Find the height h of the third liquid of relative density 1.6 in the right leg as shown in

Explanation

`Here, rho_w =1 xx 1000 kg//m^3 , rho_(gas) = 0.6 xx 10000 kg//m^3`
`rho_(liquid) = 1.6 xx 1000 kg//m^3` the pressure due to left side = pressure due to right side `1.5 rho_w g = (2.5 - h) rho_(gas) xx g + (h -1)rho_(liquid) xx g`
or `1.5 xx 1000 xx g = (2.5 - h) xx 600 g+ (h -1) xx 1600 xx g` on solving h = 1.6 m