Answer to Question #113274 in Physics for Eze Chisom

Question

A certain mass of air is trapped into the sealed end of a narrow tube of uniform bore by a thread of mercury 15cm long. When the tube is held vertically with the closed end at the bottom, the air column is 10cm long, but when it is inverted,the air column is 15cm long. calculate the atmospheric pressure and the length of the air column when the tube is held horizontally

Answer 1

$$P_1=(p+15), V_1=10A\\P_2=(p-15), V_1=15A\\ (p+15)10=(p-15)15$$

The atmospheric pressure

$$p=75\text{ cm of Hg}=750\text{ mm of Hg}$$

$$(p+15)10=(p+0)l\to (75+15)10=75l$$

$$l=12\ cm$$