A test vehicle contains a U-tube manometer for measuring differences of air pressure. The manometer is so mounted that, when the vehicle is on level ground, the plane of the U is vertical and in the fore-and-aft direction. The arms of the \(\mathrm{U}\) are \(60 \mathrm{~mm}\) apart, and contain alcohol of relative density \(0.79 .\) When the vehicle is accelerated forwards down an incline at \(20^{\circ}\) to the horizontal at \(2 \mathrm{~m} \cdot \mathrm{s}^{-2}\) the difference in alcohol levels (measured parallel to the arms of the \(\mathrm{U}\) ) is \(73 \mathrm{~mm}\), that nearer the front of the vehicle being the higher. What is the difference of air pressure to which this reading corresponds?

Pressure difference in a U-tube manometer is the main concept to understand in this exercise. A U-tube manometer measures the difference in pressure between two points by the height difference (\text{Δ}h) of a liquid column in its arms. The pressure difference is influenced by the liquid's density and gravity. A basic formula to represent the pressure difference is: \( \text{ΔP} = ρgh \), where \( ρ \) is the liquid's density, \( g \) is the acceleration due to gravity, and \( h \) is the height difference. When the manometer is exposed to additional forces, such as acceleration, these forces must be accounted for to compute the accurate pressure difference.

Relative density, also known as specific gravity, is the ratio of the density of a substance to the density of a reference substance, typically water for liquids. In this exercise, the relative density of alcohol is given as 0.79, which means its density is 79% that of water. The density of alcohol can be calculated by multiplying its relative density by the density of water: \( \rho_\text{alcohol} = 0.79 \times 1000 \text{ kg/m}^3 = 790 \text{ kg/m}^3 \). This density is crucial for calculating pressure differences using the manometer.

The vehicle in the problem is accelerating, which affects the fluid in the U-tube manometer. When a vehicle accelerates, the fluid experiences an additional pressure component due to this acceleration. Acceleration causes the fluid to redistribute, creating a height difference in the manometer arms not solely due to gravity. This effect is captured by adding an acceleration term in the pressure difference formula. Thus, the total pressure difference considering gravity and acceleration is given by: \( \text{ΔP} = ρgh + ρa \frac{L}{2} \), where \( a \) is the acceleration and \( L \) is the distance between the arms of the U-tube.

The total pressure difference measured by a U-tube manometer in an accelerating vehicle is the sum of gravitational and accelerative components. The gravitational component is given by \( P_g = \rho gh \), and the accelerative component is given by \( P_a = \rho a \frac{L}{2} \). In our exercise, the gravitational contribution given by the height difference in alcohol levels was: \( P_g = 790 \text{ kg/m}^3 \times 9.81 \text{ m/s}^2 \times 0.073 \text{ m} = 564.753 \text{ Pa} \). The acceleration contribution was: \( P_a = 790 \text{ kg/m}^3 \times 2 \text{ m/s}^2 \times 0.03 \text{ m} = 47.4 \text{ Pa} \). Summing these gives the total pressure difference: \( \text{ΔP} = P_g + P_a = 564.753 \text{ Pa} + 47.4 \text{ Pa} = 612.153 \text{ Pa} \).