Question

A thread of mercury of length 16 cm is used to trap some air in a capillary tube of uniform cross – sectional area and closed at one end. When the tube is held vertically, with the closed end at the bottom, the length of the trapped air column is 30 cm. calculate the length of the air column when the tube is held,
1. Horizontally
2. Vertically with the open end underneath. [Atmospheric pressure = 76 cm of mercury

Answer 1

When the tube is placed vertically with closed end down, the pressure of the trapped air withstands the pressure of the mercury $h$ and of the air above mercury $h_0$:

$$p_1=\rho g h+\rho g h_0.$$

When the tube is placed horizontally, the trapped air only withstands the atmospheric pressure, temperature does not change, so

$$p_1V_1=p_2V_2,\\p_2=p_0,\\ V_2=V_1\frac{p_1}{p_2}.$$

Since the tube is of uniform cross-sectional area, we can replace volumes with heights because

$$V_1,V_2,...V_n\leftrightarrow Ah_1,Ah_2,...,Ah_n.$$

Therefore:

$$h_2=h_1\frac{\rho g h+\rho g h_0}{\rho g h_0}=\\\space\\ =h_1\frac{h+h_0}{h_0}=30\frac{16+76}{76}=36\text{ cm}.$$

When we flip the tube and the closed end looks up, the trapped air pressure withstands 16 cm of mercury "pulling" down minus the atmospheric pressure "pushing" the mercury upward:

$$p_1h_1=p_3h_3,\\ h_3=h_1\frac{p_1}{p_3},\\p_3=p_0-p_m=\rho g h-\rho gh_m,\\ h_3=h_1\frac{h+h_0}{h_0-h}=30\frac{16+76}{76-16}=46\text{ cm}.$$