Archimedes and Buoyancy — Long Answer Questions (Class 9 Physics)
Medium Level (Application & Explanation)
Q1. Explain Archimedes' Principle and describe how Activity 9.6 (cork and iron nail in water) demonstrates this principle.
Answer:
- Archimedes' Principle states that an object immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced by the object. This force acts opposite to the object's weight.
- In Activity 9.6, the cork floats because it displaces a volume of water whose weight equals the cork’s weight before the cork is fully submerged. The upward buoyant force balances the cork’s weight, so it floats.
- The iron nail sinks because, for the nail’s small volume, the weight of displaced water is less than the nail’s weight; the buoyant force cannot balance its weight, so it sinks.
- Thus the activity shows how volume displaced and fluid density decide floating or sinking, directly illustrating Archimedes' Principle.
Q2. Using the idea of density and buoyant force, explain clearly why some objects float while others sink. Use simple examples.
Answer:
- Density is mass per unit volume. An object floats if its average density is less than the fluid's density; it sinks if its density is greater than the fluid's density.
- When an object is placed in a fluid, it displaces a certain volume of fluid. The fluid pushes upward with a buoyant force equal to the weight of that displaced fluid.
- Example 1: A wooden block has lower density than water, so it displaces enough water to generate an upward buoyant force equal to its weight—so it floats.
- Example 2: A small iron piece has higher density than water; even when fully submerged it displaces less water weight than its own weight, so it sinks.
- Shape matters only in how much fluid is displaced for a given mass; a hollow steel ship can float because its average density (including air) is lower than water.
Q3. Describe how a hydrometer and a lactometer work. Explain the principle and how readings tell us about liquid density or milk purity.
Answer:
- Both instruments work on buoyancy: a weighted, floatable stem rests in a liquid; it sinks until the weight of displaced liquid equals the instrument's weight. The depth depends on the liquid’s density.
- A hydrometer has scale marks on the stem. In a denser liquid, the hydrometer floats higher (less stem submerged); in a less dense liquid it sinks more. The scale directly gives the liquid’s density or specific gravity.
- A lactometer is a type of hydrometer calibrated for milk. If milk is watered, its density decreases and the lactometer sinks deeper, showing lower reading—indicating impurity or dilution.
- Thus, both instruments use Archimedes' Principle: deeper sinking = lower density, shallower sinking = higher density.
Q4. Explain the role of displaced volume and hull shape in designing a ship so it floats safely even when loaded.
Answer:
- A ship floats because it displaces water whose weight equals the ship’s total weight (hull + cargo + fuel + people). The designer ensures this by making the hull enclose a large volume so the ship’s average density is less than water.
- Hull shape affects stability and how much water is displaced for a given load. A wide, roomy hull displaces more water and gives higher buoyant force at small depths, allowing heavy loads without sinking.
- For safety, designers use freeboard (height of hull above water) and reserve buoyancy to handle waves and extra load. The metacentric height and center of buoyancy positions ensure the ship returns upright after tilting.
- Good design balances displacement, stability, and resistance to move efficiently while staying afloat under expected loads.
Q5. Design a simple experiment to measure the buoyant force acting on a solid object and explain how you would calculate it from measurements.
Answer:
- Use a spring balance, a beaker of water, and the solid object. First, measure the object’s weight in air (W_air) using the spring balance. Record the reading.
- Next, suspend the object fully in water (without touching beaker sides) and measure the apparent weight in water (W_water) from the balance. The buoyant force (B) equals the difference: B = W_air − W_water.
- To check Archimedes' Principle, measure the volume of displaced water by collecting overflow or using a graduated cylinder when object is submerged; calculate weight of displaced water = volume × water density × g. The result should match B (allowing small experimental error).
- This method gives a clear numerical verification of buoyant force using simple measurements.
High Complexity (Analytical & Scenario-Based)
Q6. An object of mass 600 g floats with 60% of its volume submerged in water. Calculate the object's density and explain the reasoning. (Show conceptually, no complex algebra expected.)
Answer:
- When the object floats, the weight of the object equals the weight of water displaced. Let the object's volume be V. The mass is 0.600 kg, so its weight is 0.600 g-force units (conceptually). The displaced water mass equals the object mass when floating.
- If 60% of volume is submerged, then displaced water mass = water density × (0.60 V). Since water density is 1 g/cm³ (or 1000 kg/m³), the object’s density must be equal to the average density of the displaced portion relative to the whole volume. So object density = displaced mass / total volume = (water density × 0.60 V) / V = 0.60 × water density.
- Therefore the object’s density = 0.60 × 1 g/cm³ = 0.60 g/cm³ (or 600 kg/m³). It floats because its density is less than water’s density.
Q7. You are asked to check if a gold crown is pure gold without damaging it. Describe how you would use Archimedes' Principle (the classical “Eureka” method) and explain the steps and reasoning.
Answer:
- First, measure the crown’s mass in air using a balance. Record as m_air.
- Next, suspend the crown fully submerged in water and measure its apparent mass in water (or the balance reading while submerged). The buoyant force equals the difference m_air − m_water; this equals the mass of water displaced.
- Calculate the volume of the crown = mass of displaced water / water density. Knowing the crown’s mass and volume allows calculation of its density = mass / volume.
- Compare the calculated density with pure gold’s known density (~19.3 g/cm³). If the crown’s density is significantly less, it is not pure (mixed with lighter metals). This method is non-destructive and uses Archimedes' Principle directly.
Q8. A cargo ship lists (tilts) after loading heavy goods on one side. Using the concepts of center of gravity and center of buoyancy, explain why this happens and how moving cargo can restore stability.
Answer:
- A floating body’s equilibrium depends on the relative positions of center of gravity (G) and center of buoyancy (B). G is the point where weight acts downward; B is where buoyant force (upward) acts through the centroid of displaced fluid.
- When heavy cargo is placed to one side, G shifts toward that side, making a torque that tilts the ship. As the hull tilts, the shape of displaced volume changes and B moves, but if G moves beyond the point where restoring torque exists, the ship lists.
- To restore stability, move cargo so G shifts back toward the centerline, producing a restoring moment because B will lie to create an upward force that rights the ship. Designers also use ballast tanks to adjust G and trim.
- Thus balancing weight distribution controls G and keeps B positioned to produce restoring torque and maintain upright stability.
Q9. Compare the buoyant force on the same object when submerged in fresh water, sea water, and oil. If the object displaces 2.0 L of each fluid, explain qualitatively and compute relative buoyant forces using typical densities.
Answer:
- Buoyant force equals weight of displaced fluid = (displaced volume) × (fluid density) × g. For 2.0 L (2.0 × 10−3 m³), use typical densities: fresh water ≈ 1000 kg/m³, sea water ≈ 1025 kg/m³, oil ≈ 900 kg/m³.
- Relative buoyant forces (proportional to density) will be: fresh water: 1000; sea water: 1025; oil: 900. So sea water gives the largest buoyant force, oil the smallest.
- Numerically, buoyant force in fresh water ≈ 2.0 × 10−3 × 1000 × g = 2.0 g-newtons units (i.e., about 19.6 N if g = 9.8 m/s²). Sea water ≈ 2.05 g-newtons (≈20.1 N). Oil ≈ 1.8 g-newtons (≈17.6 N).
- So an object floats better (less submerged) in sea water than fresh water, and floats least well in oil, because fluid density controls buoyant force.
Q10. Explain how a submarine dives and resurfaces using ballast tanks. Include the role of displaced volume, buoyant force, and how controlled changes let the submarine remain at a chosen depth.
Answer:
- A submarine controls its buoyancy using ballast tanks that can be filled with air or water. On the surface the tanks are filled with air, giving the submarine a large volume of trapped air, so its average density is lower than water and it floats.
- To dive, the submarine floods ballast tanks with water, increasing its overall mass while not greatly increasing volume. This increases average density so the weight becomes greater than buoyant force, and it sinks.
- To resurface, compressed air forces water out of the tanks, reducing mass and restoring buoyant force greater than weight, causing it to rise.
- To remain at a chosen depth (neutral buoyancy), the submarine adjusts the quantity of water/air so weight equals buoyant force exactly; small changes in tanks or trim planes maintain depth. This is a direct application of Archimedes' Principle and controlled displacement.
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