Problem: Determine the length of the air column when the capillary tube is held vertically with the open end up, given that a 10 cm column of air is trapped by a column of Hg 4 cm long in a capillary tube of uniform box.

Solution:

Given: Length of the column of air, L1 = 10 cm; Length of the column of Hg, L2 = 4 cm

To find: Length of the air column when the capillary tube is held vertically with the open end up, L3.

Assumptions: The density of mercury, ρHg = 13.6 g/cm³; Acceleration due to gravity, g = 9.8 m/s²; Atmospheric pressure, P0 = 1 atm.

Formula:

Pressure due to a column of liquid, P = ρgh

where, ρ = density of liquid, g = acceleration due to gravity, h = height of the column of liquid.

Total pressure, Ptotal = P0 + ρHgh

where, P0 = atmospheric pressure, ρHg = density of mercury, g = acceleration due to gravity, h = height of the column of mercury.

Calculation:

The pressure at the bottom of the capillary tube when held horizontally is given by:

Ptotal = P0 + ρHg * L2 * g
= 1 atm + 13.6 g/cm³ * 4 cm * 9.8 m/s²
= 1 atm + 532.48 Pa
= 1.00053248 atm

The pressure at the top of the air column is equal to the pressure at the bottom of the capillary tube when held horizontally, i.e., P1 = Ptotal.

The pressure at the top of the air column when the capillary tube is held vertically with the open end up is given by:

P2 = P0 + ρHg * L3 * g + ρair * g * L3

where, ρair = density of air and L3 = length of the air column.

Since the pressure at the top of the air column is equal to the pressure at the bottom of the capillary tube when held horizontally, we have:

P2 = P1
P0 + ρHg * L3 * g + ρair * g * L3 = P0 + ρHg * L2 * g
ρair * g * L3 = ρHg * (L2 - L3) * g
L3 = L2 - (ρHg / ρair) * L3

Substituting the values, we get:

L3 = 4 cm - (13.6 g/cm³ / 1.2 g/cm³) * L3
L3 = 4 cm - 45.33 cm * L3
46.33 cm * L3 = 4 cm
L3 = 0.086 cm

Answer: The length of the air column when the capillary tube is held vertically with the open end up is 0.086 cm.