Density and pressure — practice questions

1. (a) Calculate the gauge pressure: The gauge pressure at a certain depth in a fluid is given by the formula $p = \rho g h$.

• $\rho = 1030 \text{ kg m}^{-3}$
• $g = 9.81 \text{ m s}^{-2}$
• $h = 250 \text{ m}$ $p = (1030) \times (9.81) \times (250) = 2,526,275 \text{ Pa}$. Rounding to two significant figures, $p = 2.5 \times 10^6 \text{ Pa}$ or 2.5 MPa.

1. (b) Calculate the force on the window: First, calculate the area of the circular window using $A = \pi r^2$.

• $r = 0.20 \text{ m}$ $A = \pi \times (0.20)^2 = 0.12566... \text{ m}^2$.

Next, calculate the force using the formula $F = pA$. We use the unrounded pressure value for better accuracy. $F = (2,526,275 \text{ Pa}) \times (0.12566... \text{ m}^2) = 317,468... \text{ N}$. Rounding to two significant figures, $F = 3.2 \times 10^5 \text{ N}$.

1. Calculate the volume of the wood: $V = 0.10 \text{ m} \times 0.10 \text{ m} \times 0.10 \text{ m} = 0.0010 \text{ m}^3$.
2. (a) Calculate the density of the wood: $\rho_{wood} = \frac{m}{V} = \frac{0.80 \text{ kg}}{0.0010 \text{ m}^3} = 800 \text{ kg m}^{-3}$.
3. (b) Calculate the upthrust (buoyant force): Using Archimedes' Principle, $F_B = \rho_{fluid} g V_{displaced}$. Since the wood is fully submerged, $V_{displaced} = V_{wood}$. $F_B = 1000 \text{ kg m}^{-3} \times 9.81 \text{ m s}^{-2} \times 0.0010 \text{ m}^3 = 9.81 \text{ N}$.
4. (c) Calculate the weight of the wood: $W_{wood} = m g = 0.80 \text{ kg} \times 9.81 \text{ m s}^{-2} = 7.848 \text{ N}$.
5. Calculate the net force: The net force is the difference between the upward force (upthrust) and the downward force (weight). Net force = Upthrust - Weight = $9.81 \text{ N} - 7.848 \text{ N} = 1.962 \text{ N}$ (upwards). Since the net force is upwards, the block will accelerate towards the surface when released.