Answer to Question #252228 in Mechanics | Relativity for Hussain Darboe

Question

A U-tube contains some mercury of a relative density 13.6

(a) Kerosene of relative density 0.8 is poured into one arm of the U-tube until the height of the mercury in that arm is reduced by 1.5cm. Calculate the height of kerosene?

(b) Water is now poured into the arm of the U-tube until the level of the mercury in both arms are again the same. Calculate the height of water column.

Answer 1

(a) Pressure increases when moving down in a liquid, so we will add P=ρgh when moving down the tube, and subtract ρgh when moving upward in the tube. 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). The pressure must be equal in both arms:

$$\rho_kgh_k=\rho_mg\Delta h_m,\\\space\\ h_k=\Delta h_m\frac{\rho_m}{\rho_k}=25.5\text{ cm}.$$

(b) When the levels of mercury are the same, we can ignore its presence and weights forces created by the kerosene and water only:

$$\rho_kgh_k=\rho gh,\\\space\\ h=h_k\frac{\rho_k}{\rho}=20.4\text{ cm}.$$