Question

A thin tube of uniform cross section has trapped air columns of lengths 46 and 44.5 cm above and below a middle pellet of mercury 5 cm in length (see figure) when held at an angle of 60° to the vertical. When placed horizontally, the columns are equal in length. The temperature is 27°C. The pressure of air columns, in cm Hg, when the tube is horizontal, is

• A. 76.0 cm
• B. 71.7 cm
• C. 73.2 cm
• D. 75.4 cm

Answer 1

Answer: (D)

75.4 cm

Answer 2

Let $P_1$ and $P_2$ be the pressures of the air columns above and below the mercury pellet, respectively, when the tube is inclined. Let $l_1 = 46$ cm and $l_2 = 44.5$ cm be their lengths. The mercury pellet has length $L_{Hg} = 5$ cm. The vertical height difference $h$ across the mercury pellet is $h = L_{Hg} \cos(60^\circ) = 5 \times \frac{1}{2} = 2.5$ cm. Thus, the pressure relation is $P_2 = P_1 + h = P_1 + 2.5$ cm Hg.

When the tube is placed horizontally, the air columns are equal in length, say $l'$. The total length of trapped air is $l_1 + l_2 = 46 + 44.5 = 90.5$ cm. When horizontal, the total length of air is $2l'$. Thus, $2l' = 90.5$ cm, so $l' = 45.25$ cm. Let $P$ be the pressure of the air columns when the tube is horizontal. Since the columns are equal in length and the tube is horizontal, their pressures are equal, $P_1' = P_2' = P$.

By Boyle's Law ($PV = \text{constant}$ at constant temperature): $P_1 V_1 = P_1' V_1' \implies P_1 (46A) = P (45.25A) \implies 46 P_1 = 45.25 P$ $P_2 V_2 = P_2' V_2' \implies P_2 (44.5A) = P (45.25A) \implies 44.5 P_2 = 45.25 P$ From these, we get $46 P_1 = 44.5 P_2$.

We have a system of equations:

1. $P_2 = P_1 + 2.5$
2. $46 P_1 = 44.5 P_2$

Substitute (1) into (2): $46 P_1 = 44.5 (P_1 + 2.5)$ $46 P_1 = 44.5 P_1 + 111.25$ $1.5 P_1 = 111.25$ $P_1 = \frac{111.25}{1.5} = \frac{445}{6}$ cm Hg.

Now, we find $P$ using $46 P_1 = 45.25 P$: $P = \frac{46 P_1}{45.25} = \frac{46 \times \frac{445}{6}}{45.25} = \frac{20470/6}{45.25} \approx 75.396$ cm Hg.