A gas is supplied at atmospheric pressure of 74 cmHg. A manometer containing water shows a difference of 10 cm between the two levels. Calculate the gas pressure.

The atmospheric pressure is given as $74 \text{ cmHg}$. We need to convert this to meters of mercury and then to Pascals using the formula $P = \rho gh$.
We'll use the density of mercury ($\rho_{Hg} = 13600 \text{ kg/m}^3$) and acceleration due to gravity ($g = 10 \text{ N/kg} = 10 \text{ m/s}^2$) from the context of similar problems.

$$P_{atm} = \rho_{Hg} \cdot g \cdot h_{Hg}$$ $$P_{atm} = 13600 \text{ kg/m}^3 \times 10 \text{ m/s}^2 \times (74 \text{ cm} \times \frac{1 \text{ m}}{100 \text{ cm}})$$ $$P_{atm} = 13600 \text{ kg/m}^3 \times 10 \text{ m/s}^2 \times 0.74 \text{ m}$$ $$P_{atm} = 100640 \text{ Pa}$$

From the figure, the water level on the gas supply side is at $5 \text{ cm}$, and on the open side, it is at $15 \text{ cm}$. The difference in height ($h_w$) of the water column is:

$$h_w = 15 \text{ cm} - 5 \text{ cm} = 10 \text{ cm}$$

Convert this to meters: $h_w = 0.10 \text{ m}$.
The pressure difference ($\Delta P$) due to this water column is:

$$\Delta P = \rho_w \cdot g \cdot h_w$$

Using the density of water ($\rho_w = 1000 \text{ kg/m}^3$) and $g = 10 \text{ m/s}^2$:

$$\Delta P = 1000 \text{ kg/m}^3 \times 10 \text{ m/s}^2 \times 0.10 \text{ m}$$ $$\Delta P = 1000 \text{ Pa}$$

Since the water level is lower on the gas supply side, the gas pressure is higher than the atmospheric pressure. Therefore, the gas pressure ($P_{gas}$) is the sum of the atmospheric pressure and the pressure difference due to the water column.

$$P_{gas} = P_{atm} + \Delta P$$ $$P_{gas} = 100640 \text{ Pa} + 1000 \text{ Pa}$$ $$P_{gas} = 101640 \text{ Pa}$$

The gas pressure in SI units is $\text{101640 Pa}$.