Calculate the pressure exerted by the mercury column.

The pressure due to a column of fluid is given by the formula $P = \rho g h$.
Given:
Density of mercury ($\rho$) = $14,000 \text{ kg/m}^3$
Acceleration due to gravity ($g$) = $10 \text{ N/kg} = 10 \text{ m/s}^2$
Difference in height ($h$) = $72 \text{ m}$

$$P_{column} = \rho g h$$ $$P_{column} = 14,000 \text{ kg/m}^3 \times 10 \text{ m/s}^2 \times 72 \text{ m}$$ $$P_{column} = 10,080,000 \text{ Pa}$$

$$1 \text{ kPa} = 1000 \text{ Pa}$$ $$P_{column} = \frac{10,080,000}{1000} \text{ kPa}$$ $$P_{column} = 10,080 \text{ kPa}$$

From the diagram, the mercury level in the arm connected to the gas supply (left side) is lower than the mercury level in the arm open to the atmosphere (right side). This indicates that the pressure of the gas supply is greater than the atmospheric pressure.

$$P_{gas} = P_{atm} + P_{column}$$

Given atmospheric pressure ($P_{atm}$) = $100 \text{ kPa}$.

$$P_{gas} = 100 \text{ kPa} + 10,080 \text{ kPa}$$ $$P_{gas} = 10,180 \text{ kPa}$$

The pressure of the gas supply is $\boxed{\text{10,180 kPa}}$.

10.

• Effect on stability: His stability will increase.
• Explanation: Spreading his feet further apart increases the base area (or base of support) of the athlete. For an object to be stable, its center of gravity must fall within its base of support. A larger base area means the center of gravity can shift a greater distance horizontally before it falls outside the base, making the athlete more stable and less likely to topple over, especially when lifting heavy weights.