Answer to Question #153752 in Classical Mechanics for Hussain Darboe

Question

A u-tube contains some mercury of relative density 13.6 kerosine of relative density 0.8 is poured into one arm until the height of mercury in that arm is reduced by 1.5cm. Calculate the height of kerosine? Water is then poured into the other arm until the mercury level in both arms is the same. Calculate the height of water?

Answer 1

Answer

The pressure increases when going down in a liquid, so we will add P=ρgh when going down the tube, and subtract ρgh when moving upward in the tube.

Let’s write the pressure equation as we move from point A to B.

Remember that when we get to the bottom of the water column, we can shortcut across to the right hand tube since we are in the same liquid (mercury) and at the same horizontal level (same pressure):

$P_A+(ρgh)_{water}−(ρgh)_{mercury}=P_B$

PA=PB=0

Therefore

(gauge pressure) since they are exposed to atmospheric pressure

∴ $(103kg/m3)(9.81m/s2)(0.136m)−(13600kg/m3)(9.81m/s2)(2y)=0$

So we can write

Height of kerosine

y=0.005m or 5mm