One arm of a U-shaped tube (open at both ends) contains water, and the other alcohol. If the two fluids meet at exactly the bottom of the U, and the alcohol is at a height of 16.0 cm, at what height will the water be?

The height at which alcohol is placed is \(h = 16\;{\rm{cm}}\).

In this problem, the fluid is stationary because of which the pressure at the place where the alcohol and water meet must be equivalent.

The relation from Bernoulli’s equation is given by,

\(\begin{array}{c}{P_1} = {P_2}\\{P_0} + {\rho _1}gh = {P_0} + {\rho _2}gh'\\{\rho _1}h = {\rho _2}h'\\h' = \left( {\frac{{{\rho _1}h}}{{{\rho _2}}}} \right)\end{array}\)

Here, \(g\) is the gravitational acceleration, \({P_0}\) is the atmospheric pressure, \({P_1}\) and \({P_2}\) are the pressure of alcohol and water, \({\rho _1}\) and \({\rho _2}\) are the density of alcohol and water and \(h'\) is the height of the water.

On plugging the values in the above relation.

\(\begin{array}{l}h' = \left( {\frac{{\left( {789\;{\rm{kg/}}{{\rm{m}}^3}} \right)\left( {16\;{\rm{cm}} \times \frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)}}{{\left( {1000\;{\rm{kg/}}{{\rm{m}}^3}} \right)}}} \right)\\h' = 0.12\;{\rm{m}}\end{array}\)

Thus, the height at which water is placed is \(0.12\;{\rm{m}}\).