A U-tube manometer contains mercury. The open end is exposed to atmospheric pressure of 75 cmHg. A column of water 68 cm high exerts pressure. The difference in mercury levels is 30 cm. Calculate the

The atmospheric pressure is given as $75 \text{ cmHg}$.

$$h_{atm} = 75 \text{ cm} = 0.75 \text{ m}$$

The height of the water column is $68 \text{ cm}$.

$$h_{water} = 68 \text{ cm} = 0.68 \text{ m}$$

The difference in the mercury levels in the U-tube is $30 \text{ cm}$.

$$h_{Hg\_diff} = 30 \text{ cm} = 0.30 \text{ m}$$

The formula for pressure due to a fluid column is $P = \rho g h$.
Given density of mercury ($\rho_{Hg}$) = $13600 \text{ kg/m}^3$.
Acceleration due to gravity ($g$) = $9.8 \text{ m/s}^2$.

$$P_{atm} = \rho_{Hg} g h_{atm}$$ $$P_{atm} = 13600 \text{ kg/m}^3 \times 9.8 \text{ m/s}^2 \times 0.75 \text{ m}$$ $$P_{atm} = 99960 \text{ Pa}$$

Density of water ($\rho_{water}$) = $1000 \text{ kg/m}^3$.

$$P_{water} = \rho_{water} g h_{water}$$ $$P_{water} = 1000 \text{ kg/m}^3 \times 9.8 \text{ m/s}^2 \times 0.68 \text{ m}$$ $$P_{water} = 6664 \text{ Pa}$$

$$P_{Hg\_diff} = \rho_{Hg} g h_{Hg\_diff}$$ $$P_{Hg\_diff} = 13600 \text{ kg/m}^3 \times 9.8 \text{ m/s}^2 \times 0.30 \text{ m}$$ $$P_{Hg\_diff} = 39984 \text{ Pa}$$

By balancing the pressure at the horizontal level of the lower mercury surface (indicated by the dashed line):
The pressure on the left side (open to atmosphere) is $P_{atm}$.
The pressure on the right side (containing trapped air, water, and mercury above the dashed line) is the sum of the trapped air pressure, the pressure due to the water column, and the pressure due to the mercury column difference.

$$P_{atm} = P_{air} + P_{water} + P_{Hg\_diff}$$

Rearrange to solve for $P_{air}$:

$$P_{air} = P_{atm} - P_{water} - P_{Hg\_diff}$$ $$P_{air} = 99960 \text{ Pa} - 6664 \text{ Pa} - 39984 \text{ Pa}$$ $$P_{air} = 53312 \text{ Pa}$$

The pressure of the trapped air is:

$$\textbf{53312 Pa}$$