Thrust, Pressure, and Buoyancy — Long Answer Questions
Medium Level (Application & Explanation)
Q1. Differentiate between thrust and pressure. Give one real-life example of each and explain how they are related but distinct.
Answer:
Thrust is a vector quantity — it is a force that pushes or pulls an object in a specific direction. For example, a rocket engine produces thrust to push the rocket upward.
Pressure is a scalar defined as force per unit area, acting perpendicular to a surface: Pressure = Force / Area. For example, when a thumbtack pierces a wall, its small pointed area produces high pressure, so the force applied penetrates the wall.
Relationship and difference: Both involve force, but thrust is the actual force (with direction), while pressure describes how that force is distributed over an area. A large thrust spread over a large area gives lower pressure; the same thrust concentrated on a smaller area gives higher pressure. Thus, two identical thrusts can produce very different effects depending on the contact area.
Q2. Explain how pressure in a fluid changes with depth and give one practical consequence of this behavior.
Answer:
In a fluid at rest, pressure increases with depth according to the relation p = p₀ + ρgh, where p₀ is the pressure at the surface (often atmospheric pressure), ρ is the fluid density, g is acceleration due to gravity, and h is depth below the surface.
This means each extra metre of depth adds the same extra pressure equal to ρg per metre. Pressure acts in all directions at a point, not just downward.
Practical consequence: The force on the base of a dam depends on the water height, not on the total amount of water. This is why dams must be thicker at the bottom — because pressure increases with depth, producing greater force on lower parts of the wall. Another example: divers feel increasing pressure as they descend; equipment must be designed to tolerate higher pressures at depth.
Q3. A school bag weighs 8 kg. Compare the pressure on the shoulder when the bag hangs from a thin string strap of width 1 cm versus a wide strap of width 5 cm. Explain why the thin strap is uncomfortable.
Answer:
Weight of bag = 8 kg. Force due to weight ≈ W = mg = 8 × 9.8 ≈ 78.4 N.
Assume strap contacts shoulder over a length of 10 cm (0.10 m) along the shoulder.
Thin strap: width = 1 cm = 0.01 m → contact area A₁ = 0.01 × 0.10 = 0.001 m². Pressure P₁ = 78.4 / 0.001 = 78,400 Pa.
Wide strap: width = 5 cm = 0.05 m → A₂ = 0.05 × 0.10 = 0.005 m². Pressure P₂ = 78.4 / 0.005 = 15,680 Pa.
The thin strap produces nearly five times larger pressure, concentrating the same force on a much smaller area. This high pressure presses into the skin and muscle, causing discomfort and pain. The wider strap spreads force over larger area, reducing pressure and making it comfortable.
Q4. Using Archimedes’ Principle, explain why a large iron ship floats while a small iron nail sinks in water.
Answer:
Archimedes’ Principle states that the buoyant force on an object immersed in a fluid equals the weight of fluid displaced by the object.
For the iron nail, its volume is small, so when placed in water it displaces only a small amount of water. The buoyant force is less than the nail’s weight, so the nail sinks.
For an iron ship, although made of iron (which is dense), the overall shape encloses air so the total volume of the ship is very large. This large volume displaces a large mass of water, producing a buoyant force equal to the ship’s weight. The ship’s average (bulk) density — total mass divided by total volume (including air) — is less than the density of water, so it floats.
Thus, flotation depends on average density and displaced fluid weight, not just the material of the object.
Q5. Why do fluids exert pressure equally in all directions? Describe one experimental observation that supports this fact.
Answer:
In a fluid at rest, molecules move randomly and collide with each other and with container walls. These collisions produce force per unit area that has no preferred direction, so pressure at a point is the same in all directions.
Experimental observation: A tin can with holes experiment — if you poke holes at the same level around a sealed, water-filled can and then let water flow out, the jets are of equal strength and height in every direction. This shows that at the same depth inside the can, the pressure is equal in all directions and pushes water out equally.
Consequence: Devices like the hydraulic press and U-tube manometers rely on the fact that pressure at the same depth in a continuous fluid is equal horizontally, enabling predictable force transmission.
High Complexity (Analytical & Scenario-Based)
Q6. Two cubes of equal volume are fully submerged in water — one is made of steel (density ≈ 7800 kg/m³) and the other of wood (density ≈ 800 kg/m³). Compare the buoyant forces on the two cubes and explain which cube is more likely to float when released in water.
Answer:
When fully submerged, the buoyant force on an object equals the weight of the fluid displaced: F_b = ρ_water × V × g. Since both cubes have the same volume V and water density ρ_water is the same, both experience the same buoyant force.
However, whether they float depends on the object’s weight (mg) relative to buoyant force. The steel cube’s mass is much larger (m_steel = ρ_steel × V) than the wooden cube’s (m_wood = ρ_wood × V). Steel cube’s weight = ρ_steel V g; wooden cube’s weight = ρ_wood V g.
Because ρ_steel ≫ ρ_water, the steel cube’s weight exceeds the buoyant force, so it sinks. For wood, ρ_wood < ρ_water, so its weight is less than buoyant force; it will float (or stay partially submerged) until buoyant force equals its weight.
Key point: Buoyant force depends on displaced fluid volume, not on the object's material; flotation depends on object’s density relative to fluid.
Q7. An iceberg floats in seawater. If the density of ice is 917 kg/m³ and that of seawater is 1025 kg/m³, what fraction of the iceberg remains submerged? Explain the reasoning using Archimedes’ principle.
Answer:
For a floating object, buoyant force = weight of the object. Let the iceberg's total volume be V and the submerged volume be V_sub.
Buoyant force = ρ_water × V_sub × g. Weight = ρ_ice × V × g. Equate: ρ_water × V_sub × g = ρ_ice × V × g. Cancelling g and rearranging gives V_sub / V = ρ_ice / ρ_water.
Substitute values: V_sub / V = 917 / 1025 ≈ 0.894, i.e. about 89.4% of the iceberg’s volume is submerged and roughly 10.6% sticks above water.
This result shows how density ratio directly determines the submerged fraction: denser fluids support a larger submerged fraction for the same object. It also explains why most of an iceberg is under water, making it dangerous for ships.
Q8. A U-tube contains two immiscible liquids: liquid A (density ρ_A = 800 kg/m³) on the left and liquid B (density ρ_B = 1000 kg/m³) on the right. If the height of liquid A column above the common interface is 0.20 m, find the height of liquid B above the interface. Show the hydrostatic reasoning.
Answer:
At the common horizontal level (the interface), the pressure from left column must equal the pressure from right column because liquids are in contact. Using hydrostatic pressure p = ρgh, set pressures equal at the interface: ρ_A g h_A = ρ_B g h_B. g cancels.
So the column of the denser liquid B will have a smaller height (0.16 m) than that of the lighter liquid A (0.20 m) to produce the same pressure at the interface. This demonstrates the principle that pressure at the same depth in connected fluids is equal, and height adjusts inversely with density.
Q9. A diver descends to a depth of 30 m in seawater (density ≈ 1025 kg/m³). Calculate the absolute pressure experienced by the diver and explain the physiological implications of increased pressure at depth.
Answer:
Hydrostatic pressure at depth h is p = p_atm + ρgh. Take atmospheric pressure p_atm ≈ 1.01 × 10^5 Pa, ρ = 1025 kg/m³, g = 9.8 m/s², h = 30 m. Compute ρgh = 1025 × 9.8 × 30 ≈ 301,650 Pa.
Absolute pressure p ≈ 101,000 + 301,650 = 402,650 Pa ≈ 4.0 × 10^5 Pa, which is about 4.0 atm (since 1 atm ≈ 1.01 × 10^5 Pa).
Physiological implications: At greater pressures the partial pressures of gases in the lungs and blood increase. This can lead to issues like nitrogen narcosis and increased oxygen partial pressure risks at depth. Ears and sinuses must equalize pressure to avoid pain or injury. When ascending, divers must ascend slowly to avoid decompression sickness (the formation of gas bubbles in tissues). Thus, pressure effects are both mechanical and physiological.
Q10. A cargo ship displaces 2.0 × 10^6 kg of water when unloaded. After loading cargo of mass 5.0 × 10^4 kg, how much additional volume of water must be displaced, and by how much will the ship sink if its water-plane area (area of hull at waterline) is 500 m²? (Take g = 9.8 m/s², density of water = 1000 kg/m³.) Explain the steps and physical meaning.
Answer:
Additional mass added: Δm = 5.0 × 10^4 kg. To float, ship must displace an additional Δm of water. Since water density ρ = 1000 kg/m³, additional displaced volume ΔV = Δm / ρ = (5.0 × 10^4) / 1000 = 50 m³.
If the ship’s water-plane area A = 500 m², the increase in draft (sink amount) Δh = ΔV / A = 50 / 500 = 0.10 m (10 cm).
Physical meaning: The ship floats by displacing a volume of water whose weight equals the ship’s total wei...
