Upthrust in Fluids, Archimedes' Principle and Floatation - Questions & Answers
Explained, Step by Step
EXERCISE-5(A)(A) MULTIPLE CHOICE TYPE :
1. If an empty can is pushed into water, we feel :
(a) a downward force
(b) an upward force
(c) it is easy to push the can into water
(d) both (a) and (c)
Answer: (b) an upward force
Step 1: When an object is placed in a liquid, the liquid exerts an upward force.
Step 2: This upward force is called upthrust or buoyant force.
Step 3: Hence, we feel an upward force opposing our push.
Step 1: When an object is placed in a liquid, the liquid exerts an upward force.
Step 2: This upward force is called upthrust or buoyant force.
Step 3: Hence, we feel an upward force opposing our push.
2. The effect of upthrust is that the weight of a body immersed in a liquid appears to be :
(a) more than its actual weight
(b) the same as its weight outside the liquid
(c) less than its actual weight
(d) None of these
Answer: (c) less than its actual weight
Step 1: The true weight pulls the body vertically downwards.
Step 2: The upthrust from the liquid pushes vertically upwards.
Step 3: The net downward force (apparent weight) is less than the true weight.
Step 1: The true weight pulls the body vertically downwards.
Step 2: The upthrust from the liquid pushes vertically upwards.
Step 3: The net downward force (apparent weight) is less than the true weight.
3. Which of the following are the correct characteristic properties of upthrust ?
(i) the smaller the volume of a body submerged in a liquid, the smaller is the upthrust
(ii) the upthrust acts on the body in an upward direction at the centre of buoyancy
(iii) for the same volume of a body inside a fluid, the less the density of the fluid, greater will be the upthrust
(a) (i) & (ii)
(b) (i) & (iii)
(c) (ii) & (iii)
(d) (i), (ii) & (iii)
Answer: (a) (i) & (ii)
Step 1: Upthrust increases proportionally with the submerged volume of the body (making (i) correct).
Step 2: Upthrust acts upwards at the centre of buoyancy (making (ii) correct).
Step 3: Upthrust is greater when fluid density is greater, so statement (iii) is incorrect.
Step 1: Upthrust increases proportionally with the submerged volume of the body (making (i) correct).
Step 2: Upthrust acts upwards at the centre of buoyancy (making (ii) correct).
Step 3: Upthrust is greater when fluid density is greater, so statement (iii) is incorrect.
4. The correct relation for upthrust on a solid of volume V immersed in a liquid of density ρ is :
(a) F_B = V/g
(b) F_B = ρg/V
(c) F_B = Vρg
(d) F_B = Vg/ρ
Answer: (c) F_B = Vρg
Step 1: Archimedes principle says upthrust equals the weight of the displaced liquid.
Step 2: Mass of displaced liquid = Volume (V) × Density (ρ).
Step 3: Weight = Mass × gravity (g) = V × ρ × g.
Step 1: Archimedes principle says upthrust equals the weight of the displaced liquid.
Step 2: Mass of displaced liquid = Volume (V) × Density (ρ).
Step 3: Weight = Mass × gravity (g) = V × ρ × g.
5. Archimedes' principle applies on :
(a) solids and liquids
(b) liquids and gases
(c) solids and gases
(d) solids, liquids and gases
Answer: (b) liquids and gases
Step 1: Archimedes' principle applies universally to all fluids.
Step 2: In physics, both liquids and gases are classified as fluids because they can flow.
Step 1: Archimedes' principle applies universally to all fluids.
Step 2: In physics, both liquids and gases are classified as fluids because they can flow.
6. If a body is completely immersed in a liquid, the volume of the liquid displaced will be ............... its own volume and the upthrust will be ...............
(a) less than, minimum
(b) equal to, minimum
(c) more than, maximum
(d) equal to, maximum
Answer: (d) equal to, maximum
Step 1: A completely immersed body takes up space exactly equal to its own total volume, displacing that exact amount of liquid.
Step 2: Since the displaced volume has reached its maximum possible amount, the upthrust is also maximum.
Step 1: A completely immersed body takes up space exactly equal to its own total volume, displacing that exact amount of liquid.
Step 2: Since the displaced volume has reached its maximum possible amount, the upthrust is also maximum.
7. The apparent loss in weight is equal to the upthrust on the body verifies :
(a) Pascal's law
(b) Newton's third law
(c) Archimedes' principle
(d) Newton's second law
Answer: (c) Archimedes' principle
Step 1: Archimedes' principle directly defines that the buoyant force (upthrust) equals the weight of the displaced fluid.
Step 2: This upward force directly subtracts from the true weight, causing the apparent loss in weight.
Step 1: Archimedes' principle directly defines that the buoyant force (upthrust) equals the weight of the displaced fluid.
Step 2: This upward force directly subtracts from the true weight, causing the apparent loss in weight.
8. A body weighed 'W' in air by a sensitive spring balance. It will weigh ............ in vacuum.
(a) the same
(b) slightly more
(c) slightly less
(d) zero
Answer: (b) slightly more
Step 1: Air is a fluid and exerts a small upthrust on the body, reducing its measured weight 'W' slightly.
Step 2: In a vacuum, there is no air and therefore no upthrust.
Step 3: So, the true weight recorded in a vacuum is slightly more than what was measured in air.
Step 1: Air is a fluid and exerts a small upthrust on the body, reducing its measured weight 'W' slightly.
Step 2: In a vacuum, there is no air and therefore no upthrust.
Step 3: So, the true weight recorded in a vacuum is slightly more than what was measured in air.
9. Floating of a cork on the surface of water indicates that :
(a) the density of water is more than the density of cork.
(b) the density of water is less than the density of cork.
(c) the density of water is equal to the density of cork.
(d) the density of water plays no role in the floatation of cork.
Answer: (a) the density of water is more than the density of cork.
Step 1: An object floats when its density is less than the density of the liquid it is placed in.
Step 2: Since the cork floats, it must be lighter (less dense) than the water.
Step 1: An object floats when its density is less than the density of the liquid it is placed in.
Step 2: Since the cork floats, it must be lighter (less dense) than the water.
10. Sinking of an iron nail in water implies that :
(a) the density of nail is more than the density of water
(b) the density of nail is less than the density of water
(c) the density of nail is equal to the density of water
(d) the sinking of nail does not depend upon the density of water
Answer: (a) the density of nail is more than the density of water
Step 1: An object sinks when its density is greater than the density of the surrounding liquid.
Step 2: Since the iron nail sinks, its density is higher than water's density.
Step 1: An object sinks when its density is greater than the density of the surrounding liquid.
Step 2: Since the iron nail sinks, its density is higher than water's density.
11. Bodies of density .............. than that of the liquid sink in it, while bodies of average density equal to or .............. than that of the liquid float on it.
(a) smaller, greater
(b) greater, greater
(c) greater, smaller
(d) smaller, smaller
Answer: (c) greater, smaller
Step 1: Sinking occurs when the object's density is greater than the liquid's density.
Step 2: Floating occurs when the object's density is equal to or smaller than the liquid's density.
Step 1: Sinking occurs when the object's density is greater than the liquid's density.
Step 2: Floating occurs when the object's density is equal to or smaller than the liquid's density.
12. A body will experience minimum upthrust when it is completely immersed in :
(a) turpentine
(b) water
(c) glycerine
(d) mercury
Answer: (a) turpentine
Step 1: Upthrust depends directly on the density of the liquid.
Step 2: From known tables, turpentine has a lower density than water, glycerine, and mercury.
Step 3: The liquid with the lowest density will provide the minimum upthrust.
Step 1: Upthrust depends directly on the density of the liquid.
Step 2: From known tables, turpentine has a lower density than water, glycerine, and mercury.
Step 3: The liquid with the lowest density will provide the minimum upthrust.
13. A body of density ρ sinks in a liquid of density ρ_L. The densities ρ and ρ_L are related as :
(a) ρ = ρ_L
(b) ρ < ρ_L
(c) ρ > ρ_L
(d) nothing can be said.
Answer: (c) ρ > ρ_L
Step 1: A body sinks only when its weight is greater than the maximum upthrust.
Step 2: This physically means the body's material density (ρ) must be greater than the liquid's density (ρ_L).
Step 1: A body sinks only when its weight is greater than the maximum upthrust.
Step 2: This physically means the body's material density (ρ) must be greater than the liquid's density (ρ_L).
(B) VERY SHORT ANSWER TYPE :
1. In what direction and at what point does the buoyant force on a body due to a liquid, act ?
Answer: Upwards, at the centre of buoyancy.
Step 1: The buoyant force opposes gravity, meaning it acts in the upward vertical direction.
Step 2: It acts through a specific point called the centre of buoyancy, which is the centre of gravity of the displaced liquid.
Step 1: The buoyant force opposes gravity, meaning it acts in the upward vertical direction.
Step 2: It acts through a specific point called the centre of buoyancy, which is the centre of gravity of the displaced liquid.
2. Define upthrust and state its S.I. unit.
Answer: Upthrust is the upward force exerted on a body by the fluid in which it is submerged. Its S.I. unit is newton (N).
Step 1: State the definition relating to the upward push from a fluid.
Step 2: State the standard scientific force unit, the newton.
Step 1: State the definition relating to the upward push from a fluid.
Step 2: State the standard scientific force unit, the newton.
3. Why is a force needed to keep a block of wood inside water ?
Answer: Because the upthrust due to water on the fully submerged block is more than its weight.
Step 1: Wood has a lower density than water.
Step 2: When pushed completely underwater, the upward upthrust force is much stronger than the downward weight.
Step 3: To stop it from rising, a downward external force is required to balance it.
Step 1: Wood has a lower density than water.
Step 2: When pushed completely underwater, the upward upthrust force is much stronger than the downward weight.
Step 3: To stop it from rising, a downward external force is required to balance it.
4. A body experiences an upthrust F₁ in river water and F₂ in sea water when dipped up to the same level. Which is more F₁ or F₂ ? Give reason.
Answer: F₂ is more than F₁.
Step 1: Upthrust depends directly on the density of the liquid.
Step 2: Sea water contains salts, making it denser than fresh river water.
Step 3: Higher density means greater upthrust, therefore F₂ > F₁.
Step 1: Upthrust depends directly on the density of the liquid.
Step 2: Sea water contains salts, making it denser than fresh river water.
Step 3: Higher density means greater upthrust, therefore F₂ > F₁.
5. A body of volume V and density ρ is kept completely immersed in a liquid of density ρ_L. If g is the acceleration due to gravity, write expressions for the following : (i) the weight of the body, (ii) the upthrust on the body, (iii) the apparent weight of the body in liquid, (iv) the loss in weight of the body.
Answer:
Step 1: Weight is Mass × g. Mass is Volume × object Density. So, Weight = Vρg. (Answer i)
Step 2: Upthrust is the weight of the displaced liquid. So, Upthrust = Vρ_Lg. (Answer ii)
Step 3: Apparent weight is True Weight - Upthrust. So, Apparent weight = Vρg - Vρ_Lg = V(ρ - ρ_L)g. (Answer iii)
Step 4: Loss in weight is equal to the upthrust. So, Loss in weight = Vρ_Lg. (Answer iv)
Step 1: Weight is Mass × g. Mass is Volume × object Density. So, Weight = Vρg. (Answer i)
Step 2: Upthrust is the weight of the displaced liquid. So, Upthrust = Vρ_Lg. (Answer ii)
Step 3: Apparent weight is True Weight - Upthrust. So, Apparent weight = Vρg - Vρ_Lg = V(ρ - ρ_L)g. (Answer iii)
Step 4: Loss in weight is equal to the upthrust. So, Loss in weight = Vρ_Lg. (Answer iv)
6. Complete the following sentences :
(a) Two balls, one of iron and the other of aluminium experience the same upthrust when dipped completely in water if ...............
(b) An empty tin container with its mouth closed has an average density equal to that of a liquid. The container is taken 2 m below the surface of that liquid and is left there. Then the container will ...............
(c) A piece of wood is held under water. The upthrust on it will be ............... the weight of the wood piece.
Answer:
Step 1: For (a), upthrust depends only on the submerged volume. To have the same upthrust, they must have equal volumes.
Step 2: For (b), if average density exactly equals liquid density, upthrust exactly balances weight. It will remain at the same position.
Step 3: For (c), wood is lighter than water, so upthrust when completely submerged is more than its weight.
Step 1: For (a), upthrust depends only on the submerged volume. To have the same upthrust, they must have equal volumes.
Step 2: For (b), if average density exactly equals liquid density, upthrust exactly balances weight. It will remain at the same position.
Step 3: For (c), wood is lighter than water, so upthrust when completely submerged is more than its weight.
7. A sphere of iron and another of wood of the same radius are held under water. Compare the upthrust on the two spheres.
[Hint : Both have equal volume inside water.]
Answer: 1 : 1
Step 1: Both spheres have the same radius, which means they have the exact same volume.
Step 2: Both are submerged in the exact same liquid (water).
Step 3: Since volume and liquid density are identical, the upthrust is identical, giving a 1:1 ratio.
Step 1: Both spheres have the same radius, which means they have the exact same volume.
Step 2: Both are submerged in the exact same liquid (water).
Step 3: Since volume and liquid density are identical, the upthrust is identical, giving a 1:1 ratio.
8. A body of density ρ is immersed in a liquid of density ρ_L. State condition when the body will (i) float, (ii) sink, in liquid.
Answer:
Step 1: A body floats if its density is equal to or less than the liquid. (i) Float condition: ρ ≤ ρ_L.
Step 2: A body sinks if its density is greater than the liquid. (ii) Sink condition: ρ > ρ_L.
Step 1: A body floats if its density is equal to or less than the liquid. (i) Float condition: ρ ≤ ρ_L.
Step 2: A body sinks if its density is greater than the liquid. (ii) Sink condition: ρ > ρ_L.
9. State Archimedes' principle.
Answer: Archimedes' principle states that when a body is immersed partially or completely in a liquid, it experiences an upthrust, which is equal to the weight of the liquid displaced by it.
Step 1: Mention the prerequisite condition: an object partially or completely immersed in a fluid.
Step 2: Mention the result: an upward force (upthrust) acts on it.
Step 3: State the magnitude: it perfectly equals the weight of the displaced fluid.
Step 1: Mention the prerequisite condition: an object partially or completely immersed in a fluid.
Step 2: Mention the result: an upward force (upthrust) acts on it.
Step 3: State the magnitude: it perfectly equals the weight of the displaced fluid.
(C) SHORT ANSWER TYPE :
1. What is meant by the term buoyancy ?
Answer: The property of a liquid to exert an upward force on a body immersed in it, is called buoyancy.
Step 1: Note that liquids push up on objects.
Step 2: Define this physical property as buoyancy.
Step 1: Note that liquids push up on objects.
Step 2: Define this physical property as buoyancy.
2. What is the cause of upthrust ? At which point it can be considered to act ?
Answer: It is caused by the pressure difference across the submerged body. It acts at the centre of buoyancy.
Step 1: Liquid pressure increases with depth.
Step 2: Therefore, upward pressure on the bottom is greater than downward pressure on the top.
Step 3: This pressure difference creates a net upward force.
Step 4: It acts centrally through the centre of buoyancy.
Step 1: Liquid pressure increases with depth.
Step 2: Therefore, upward pressure on the bottom is greater than downward pressure on the top.
Step 3: This pressure difference creates a net upward force.
Step 4: It acts centrally through the centre of buoyancy.
3. State three characteristic properties of upthrust.
Answer:
Step 1: Upthrust increases as the submerged volume of the body increases.
Step 2: Upthrust is greater when the density of the fluid is greater.
Step 3: Upthrust acts in the upward direction through the centre of buoyancy.
Step 1: Upthrust increases as the submerged volume of the body increases.
Step 2: Upthrust is greater when the density of the fluid is greater.
Step 3: Upthrust acts in the upward direction through the centre of buoyancy.
4. A piece of wood if left under water, comes to the surface. Explain the reason.
Answer:
Step 1: The density of wood is lower than the density of water.
Step 2: This means the upthrust acting on the fully submerged wood is greater than its weight.
Step 3: This net upward force quickly pushes the wood up to the surface.
Step 1: The density of wood is lower than the density of water.
Step 2: This means the upthrust acting on the fully submerged wood is greater than its weight.
Step 3: This net upward force quickly pushes the wood up to the surface.
5. Will a body weigh more in air or in vacuum when weighed with a spring balance ? Give a reason for your answer.
Answer: It will weigh more in vacuum.
Step 1: Air is a fluid and it exerts a tiny upward upthrust on all objects.
Step 2: This upthrust makes the object's apparent weight slightly less in air.
Step 3: In a vacuum, there is no air upthrust, so it shows its true, heavier weight.
Step 1: Air is a fluid and it exerts a tiny upward upthrust on all objects.
Step 2: This upthrust makes the object's apparent weight slightly less in air.
Step 3: In a vacuum, there is no air upthrust, so it shows its true, heavier weight.
6. A metal solid cylinder tied to a thread is hanging from the hook of a spring balance. The cylinder is gradually immersed into water contained in a jar. What changes do you expect in the readings of spring balance ? Explain your answer.
Answer:
Step 1: As the cylinder enters the water, it begins to experience an upward upthrust.
Step 2: As it goes deeper, the submerged volume increases, so the upthrust increases.
Step 3: The upthrust reduces the apparent weight, so the spring balance reading will gradually decrease until it is completely submerged.
Step 1: As the cylinder enters the water, it begins to experience an upward upthrust.
Step 2: As it goes deeper, the submerged volume increases, so the upthrust increases.
Step 3: The upthrust reduces the apparent weight, so the spring balance reading will gradually decrease until it is completely submerged.
7. A body dipped into a liquid experiences an upthrust. State two factors on which upthrust on the body depends.
Answer:
Step 1: It depends on the volume of the body submerged inside the liquid.
Step 2: It depends on the density of the liquid itself.
Step 1: It depends on the volume of the body submerged inside the liquid.
Step 2: It depends on the density of the liquid itself.
8. How is the upthrust related to the volume of the body submerged in a liquid ?
Answer:
Step 1: Upthrust is directly proportional to the submerged volume.
Step 2: As the submerged volume increases, the upward upthrust force increases correspondingly.
Step 1: Upthrust is directly proportional to the submerged volume.
Step 2: As the submerged volume increases, the upward upthrust force increases correspondingly.
9. A bunch of feathers and a stone of the same mass are released simultaneously in air. Which will fall faster and why ? How will your observation be different if they are released simultaneously in vacuum ?
Answer:
Step 1: The stone will fall faster.
Step 2: The bunch of feathers has a much larger volume, so it experiences a greater upthrust and resistance from the air.
Step 3: In a vacuum, there is no air, meaning no upthrust or air resistance.
Step 4: Therefore, both would fall simultaneously in a vacuum.
Step 1: The stone will fall faster.
Step 2: The bunch of feathers has a much larger volume, so it experiences a greater upthrust and resistance from the air.
Step 3: In a vacuum, there is no air, meaning no upthrust or air resistance.
Step 4: Therefore, both would fall simultaneously in a vacuum.
10. A small block of wood is held completely immersed in (i) water, (ii) glycerine and then released. In each case, what do you observe ? Explain the difference in your observation in the two cases.
Answer:
Step 1: The wood block will float to the top in both liquids.
Step 2: However, glycerine has a higher density than water.
Step 3: Because glycerine is denser, a smaller fraction of the wood needs to be submerged to balance its weight.
Step 4: Therefore, it floats with less volume submerged in glycerine than in water.
Step 1: The wood block will float to the top in both liquids.
Step 2: However, glycerine has a higher density than water.
Step 3: Because glycerine is denser, a smaller fraction of the wood needs to be submerged to balance its weight.
Step 4: Therefore, it floats with less volume submerged in glycerine than in water.
11. A sphere of iron and another of wood, both of same radius are placed on the surface of water. State which of the two will sink ? Give reason to your answer.
Answer: The sphere of iron will sink.
Step 1: Iron has a density much higher than water.
Step 2: Because its density is higher, its downward weight is greater than the maximum upward upthrust water can provide, pulling it down.
Step 1: Iron has a density much higher than water.
Step 2: Because its density is higher, its downward weight is greater than the maximum upward upthrust water can provide, pulling it down.
12. How does the density of material of a body determine whether it will float or sink in water ?
Answer:
Step 1: Compare the object's material density to water's density.
Step 2: If the object's density is greater than water's, it sinks.
Step 3: If the object's density is equal to or less than water's, it floats.
Step 1: Compare the object's material density to water's density.
Step 2: If the object's density is greater than water's, it sinks.
Step 3: If the object's density is equal to or less than water's, it floats.
13. It is easier to lift a heavy stone under water than in air. Explain.
Answer:
Step 1: While underwater, the stone displaces water and experiences a strong upward upthrust.
Step 2: This upthrust acts against the downward force of gravity (weight).
Step 3: This reduces the apparent weight of the stone, making it feel significantly lighter to lift.
Step 1: While underwater, the stone displaces water and experiences a strong upward upthrust.
Step 2: This upthrust acts against the downward force of gravity (weight).
Step 3: This reduces the apparent weight of the stone, making it feel significantly lighter to lift.
(D) LONG ANSWER TYPE :
1. What do you understand by the term upthrust of a fluid ? Describe an experiment to show its existence.
Answer:
Step 1: Upthrust is the net upward force exerted by any fluid on an object immersed in it.
Step 2: Take an empty plastic bottle and close it with a tight stopper.
Step 3: Try to press the bottle down into a bucket filled with water.
Step 4: You will physically feel a distinct upward force pushing back against your hand. This demonstrates the existence of upthrust.
Step 1: Upthrust is the net upward force exerted by any fluid on an object immersed in it.
Step 2: Take an empty plastic bottle and close it with a tight stopper.
Step 3: Try to press the bottle down into a bucket filled with water.
Step 4: You will physically feel a distinct upward force pushing back against your hand. This demonstrates the existence of upthrust.
2. Describe an experiment to show that a body immersed in a liquid appears lighter than it really is.
Answer:
Step 1: Tie a heavy stone or metal block to the hook of a spring balance and record its weight in the air.
Step 2: Slowly lower the stone into a beaker containing water until it is fully immersed.
Step 3: Read the new weight on the spring balance.
Step 4: The reading will be noticeably lower, proving the body appears lighter in the liquid.
Step 1: Tie a heavy stone or metal block to the hook of a spring balance and record its weight in the air.
Step 2: Slowly lower the stone into a beaker containing water until it is fully immersed.
Step 3: Read the new weight on the spring balance.
Step 4: The reading will be noticeably lower, proving the body appears lighter in the liquid.
3. A body held completely immersed inside a liquid experiences two forces : (i) F₁, the force due to gravity and (ii) F₂, the buoyant force. Draw a diagram showing the direction of these forces acting on the body and state the conditions when the body will float or sink.
Answer:
Step 1: (Visualizing Diagram: A rectangular body inside liquid. F₁ is an arrow pointing straight down. F₂ is an arrow pointing straight up.)
Step 2: Float Condition: The body will float if F₁ is less than or equal to the maximum F₂.
Step 3: Sink Condition: The body will sink if F₁ is greater than the maximum F₂.
Step 1: (Visualizing Diagram: A rectangular body inside liquid. F₁ is an arrow pointing straight down. F₂ is an arrow pointing straight up.)
Step 2: Float Condition: The body will float if F₁ is less than or equal to the maximum F₂.
Step 3: Sink Condition: The body will sink if F₁ is greater than the maximum F₂.
4. Prove that the loss in weight of a body when immersed wholly or partially in a liquid is equal to the buoyant force (or upthrust) and this loss is because of the difference in pressure exerted by liquid on the upper and lower surfaces of the submerged part of the body.
Answer:
Step 1: Consider a solid cylinder of area A immersed in liquid of density ρ. The upper surface is at depth h₁ and lower surface at depth h₂.
Step 2: Downward pressure on the upper surface = h₁ρg. Thus, downward force F₁ = h₁ρgA.
Step 3: Upward pressure on the lower surface = h₂ρg. Thus, upward force F₂ = h₂ρgA.
Step 4: The net upward force (Upthrust F_B) = F₂ - F₁ = (h₂ - h₁)ρgA.
Step 5: The term (h₂ - h₁)A is the volume V of the submerged cylinder. Thus, F_B = Vρg.
Step 6: Vρg is exactly the weight of the displaced liquid. This net upward force reduces the apparent weight, proving the principle.
Step 1: Consider a solid cylinder of area A immersed in liquid of density ρ. The upper surface is at depth h₁ and lower surface at depth h₂.
Step 2: Downward pressure on the upper surface = h₁ρg. Thus, downward force F₁ = h₁ρgA.
Step 3: Upward pressure on the lower surface = h₂ρg. Thus, upward force F₂ = h₂ρgA.
Step 4: The net upward force (Upthrust F_B) = F₂ - F₁ = (h₂ - h₁)ρgA.
Step 5: The term (h₂ - h₁)A is the volume V of the submerged cylinder. Thus, F_B = Vρg.
Step 6: Vρg is exactly the weight of the displaced liquid. This net upward force reduces the apparent weight, proving the principle.
5. Describe an experiment to verify Archimedes' principle.
Answer:
Step 1: Suspend a solid cylinder from the left arm of a physical balance. Balance it with weights on the right arm.
Step 2: Place a eureka can filled with water up to its spout underneath the cylinder.
Step 3: Lower the cylinder entirely into the eureka can and catch the displaced water in a separate beaker.
Step 4: The balance will tip because the cylinder lost apparent weight due to upthrust.
Step 5: Take the beaker of displaced water and weigh the displaced liquid.
Step 6: The loss in weight shown by the balance will exactly equal the weight of the water collected in the beaker, verifying the principle.
Step 1: Suspend a solid cylinder from the left arm of a physical balance. Balance it with weights on the right arm.
Step 2: Place a eureka can filled with water up to its spout underneath the cylinder.
Step 3: Lower the cylinder entirely into the eureka can and catch the displaced water in a separate beaker.
Step 4: The balance will tip because the cylinder lost apparent weight due to upthrust.
Step 5: Take the beaker of displaced water and weigh the displaced liquid.
Step 6: The loss in weight shown by the balance will exactly equal the weight of the water collected in the beaker, verifying the principle.
(E) NUMERICALS :
1. A body of volume 100 cm³ weighs 5 kgf in air. It is completely immersed in a liquid of density 1.8 × 10³ kg m⁻³. Find : (i) the upthrust due to liquid and (ii) the weight of the body in liquid.
Answer:
Step 1: Convert the volume from cm³ to m³: V = 100 / 1,000,000 = 10⁻⁴ m³.
Step 2: Note the liquid density: ρ = 1.8 × 10³ kg/m³ = 1800 kg/m³.
Step 3: Calculate mass of liquid displaced = Volume × Density = 10⁻⁴ × 1800 = 0.18 kg.
Step 4: Therefore, the upthrust = 0.18 kgf. (Answer i)
Step 5: Calculate apparent weight = True weight - Upthrust.
Step 6: Weight in liquid = 5 kgf - 0.18 kgf = 4.82 kgf. (Answer ii)
Step 1: Convert the volume from cm³ to m³: V = 100 / 1,000,000 = 10⁻⁴ m³.
Step 2: Note the liquid density: ρ = 1.8 × 10³ kg/m³ = 1800 kg/m³.
Step 3: Calculate mass of liquid displaced = Volume × Density = 10⁻⁴ × 1800 = 0.18 kg.
Step 4: Therefore, the upthrust = 0.18 kgf. (Answer i)
Step 5: Calculate apparent weight = True weight - Upthrust.
Step 6: Weight in liquid = 5 kgf - 0.18 kgf = 4.82 kgf. (Answer ii)
2. A body weighs 450 gf in air and 310 gf when completely immersed in water. Find : (i) the volume of the body, (ii) the loss in weight of the body, and (iii) the upthrust on the body. State the assumption made in part (i).
Answer:
Step 1: Calculate the loss in weight: 450 gf - 310 gf = 140 gf. (Answer ii)
Step 2: By Archimedes' principle, loss in weight is the upthrust. So, upthrust = 140 gf. (Answer iii)
Step 3: Upthrust = Volume × Density of water. Assume water density is 1.0 g/cm³.
Step 4: 140 gf = Volume × 1.0 g/cm³. Thus, Volume = 140 cm³. (Answer i)
Step 5: Assumption: We assumed the density of water is exactly 1.0 g cm⁻³.
Step 1: Calculate the loss in weight: 450 gf - 310 gf = 140 gf. (Answer ii)
Step 2: By Archimedes' principle, loss in weight is the upthrust. So, upthrust = 140 gf. (Answer iii)
Step 3: Upthrust = Volume × Density of water. Assume water density is 1.0 g/cm³.
Step 4: 140 gf = Volume × 1.0 g/cm³. Thus, Volume = 140 cm³. (Answer i)
Step 5: Assumption: We assumed the density of water is exactly 1.0 g cm⁻³.
3. You are provided with a hollow iron ball A of volume 15 cm³ and mass 12 g and a solid iron ball B of mass 12 g. Both are placed on the surface of water contained in a large tub. (a) Find upthrust on each ball. (b) Which ball will sink ? Give reason for your answer. (Density of iron = 8.0 g cm⁻³)
Answer:
Step 1: For ball A, maximum possible upthrust = Volume (15 cm³) × Density of water (1 g/cm³) = 15 gf.
Step 2: Since weight of A (12 gf) is less than max upthrust (15 gf), it floats. While floating, upthrust = weight. So upthrust on A = 12 gf.
Step 3: For ball B, calculate its volume = Mass / Density of iron = 12 g / 8.0 g/cm³ = 1.5 cm³.
Step 4: Maximum upthrust on B = Volume (1.5 cm³) × water density (1 g/cm³) = 1.5 gf.
Step 5: Since B sinks, the upthrust on B = 1.5 gf. (Answer a)
Step 6: Ball B will sink. Reason: The max upthrust on it (1.5 gf) is much less than its true weight (12 gf). (Answer b)
Step 1: For ball A, maximum possible upthrust = Volume (15 cm³) × Density of water (1 g/cm³) = 15 gf.
Step 2: Since weight of A (12 gf) is less than max upthrust (15 gf), it floats. While floating, upthrust = weight. So upthrust on A = 12 gf.
Step 3: For ball B, calculate its volume = Mass / Density of iron = 12 g / 8.0 g/cm³ = 1.5 cm³.
Step 4: Maximum upthrust on B = Volume (1.5 cm³) × water density (1 g/cm³) = 1.5 gf.
Step 5: Since B sinks, the upthrust on B = 1.5 gf. (Answer a)
Step 6: Ball B will sink. Reason: The max upthrust on it (1.5 gf) is much less than its true weight (12 gf). (Answer b)
4. A solid of density 5000 kg m⁻³ weighs 0.5 kgf in air. It is completely immersed in water of density 1000 kg m⁻³. Calculate the apparent weight of the solid in water.
Answer:
Step 1: Find the mass of the solid: Mass = 0.5 kg.
Step 2: Find the volume of the solid: Volume = Mass / Density = 0.5 / 5000 = 10⁻⁴ m³.
Step 3: Calculate maximum upthrust in water: Upthrust = Volume × Density of water = 10⁻⁴ m³ × 1000 kg/m³ = 0.1 kg.
Step 4: So, the upthrust force = 0.1 kgf.
Step 5: Apparent weight = True weight - Upthrust = 0.5 kgf - 0.1 kgf = 0.4 kgf.
Step 1: Find the mass of the solid: Mass = 0.5 kg.
Step 2: Find the volume of the solid: Volume = Mass / Density = 0.5 / 5000 = 10⁻⁴ m³.
Step 3: Calculate maximum upthrust in water: Upthrust = Volume × Density of water = 10⁻⁴ m³ × 1000 kg/m³ = 0.1 kg.
Step 4: So, the upthrust force = 0.1 kgf.
Step 5: Apparent weight = True weight - Upthrust = 0.5 kgf - 0.1 kgf = 0.4 kgf.
5. Two spheres A and B, each of volume 100 cm³ are placed on water (density = 1.0 g cm⁻³). The sphere A is made of wood of density 0.3 g cm⁻³ and the sphere B is made of iron of density 8.9 g cm⁻³. (a) Find : (i) the weight of each sphere, and (ii) the upthrust on each sphere. (b) Which sphere will float ? Give reason.
Answer:
Step 1: Weight of A = Volume × Density of A = 100 cm³ × 0.3 g/cm³ = 30 gf.
Step 2: Weight of B = Volume × Density of B = 100 cm³ × 8.9 g/cm³ = 890 gf. (Answer a-i)
Step 3: Max upthrust for full immersion = Volume (100 cm³) × water density (1.0 g/cm³) = 100 gf.
Step 4: Sphere A weighs less than max upthrust (30 < 100), so it floats. Upthrust on floating A equals its weight = 30 gf.
Step 5: Sphere B weighs more than max upthrust (890 > 100), so it sinks. Upthrust on B = maximum upthrust = 100 gf. (Answer a-ii)
Step 6: Sphere A will float. Reason: The density of its wood (0.3 g/cm³) is less than the density of water (1.0 g/cm³).
Step 1: Weight of A = Volume × Density of A = 100 cm³ × 0.3 g/cm³ = 30 gf.
Step 2: Weight of B = Volume × Density of B = 100 cm³ × 8.9 g/cm³ = 890 gf. (Answer a-i)
Step 3: Max upthrust for full immersion = Volume (100 cm³) × water density (1.0 g/cm³) = 100 gf.
Step 4: Sphere A weighs less than max upthrust (30 < 100), so it floats. Upthrust on floating A equals its weight = 30 gf.
Step 5: Sphere B weighs more than max upthrust (890 > 100), so it sinks. Upthrust on B = maximum upthrust = 100 gf. (Answer a-ii)
Step 6: Sphere A will float. Reason: The density of its wood (0.3 g/cm³) is less than the density of water (1.0 g/cm³).
6. The mass of a block made of a certain material is 13.5 kg and its volume is 15 × 10⁻³ m³. (a) Calculate upthrust on the block if it is held fully immersed in water. (b) Will the block float or sink in water when released ? Give reason for your answer. (c) What will be the upthrust on block while floating? Take density of water = 1000 kg m⁻³.
Answer:
Step 1: Calculate max upthrust (fully immersed) = Volume × Density of water = (15 × 10⁻³ m³) × 1000 kg/m³ = 15 kg.
Step 2: So, upthrust when fully immersed = 15 kgf. (Answer a)
Step 3: Compare this to the block's true weight, which is 13.5 kgf.
Step 4: Since max upthrust (15 kgf) is greater than weight (13.5 kgf), the block will float when released. (Answer b)
Step 5: According to floatation principle, upthrust on a floating body perfectly equals its true weight. So, upthrust while floating = 13.5 kgf. (Answer c)
Step 1: Calculate max upthrust (fully immersed) = Volume × Density of water = (15 × 10⁻³ m³) × 1000 kg/m³ = 15 kg.
Step 2: So, upthrust when fully immersed = 15 kgf. (Answer a)
Step 3: Compare this to the block's true weight, which is 13.5 kgf.
Step 4: Since max upthrust (15 kgf) is greater than weight (13.5 kgf), the block will float when released. (Answer b)
Step 5: According to floatation principle, upthrust on a floating body perfectly equals its true weight. So, upthrust while floating = 13.5 kgf. (Answer c)
7. A piece of brass weighs 175 gf in air and 150 gf when fully immersed in water. The density of water is 1.0 g cm⁻³. (i) What is the volume of the brass piece ? (ii) Why does the brass piece weigh less in water ?
Answer:
Step 1: Calculate loss in weight = True weight - Apparent weight = 175 gf - 150 gf = 25 gf.
Step 2: This loss in weight is the upthrust (25 gf).
Step 3: Upthrust equals Volume × Density of water. 25 = Volume × 1.0 g/cm³. Thus, Volume = 25 cm³. (Answer i)
Step 4: It weighs less due to the upward buoyant force (upthrust) applied by the water. (Answer ii)
Step 1: Calculate loss in weight = True weight - Apparent weight = 175 gf - 150 gf = 25 gf.
Step 2: This loss in weight is the upthrust (25 gf).
Step 3: Upthrust equals Volume × Density of water. 25 = Volume × 1.0 g/cm³. Thus, Volume = 25 cm³. (Answer i)
Step 4: It weighs less due to the upward buoyant force (upthrust) applied by the water. (Answer ii)
8. A metal cube of edge 5 cm and density 9.0 g cm⁻³ is suspended by a thread so as to be completely immersed in a liquid of density 1.2 g cm⁻³. Find the tension in thread. (Take g = 10 m s⁻²)
Answer:
Step 1: Calculate volume of the cube: Volume = 5 cm × 5 cm × 5 cm = 125 cm³.
Step 2: Convert volume to SI units: 125 cm³ = 125 × 10⁻⁶ m³.
Step 3: Find mass of the cube: Mass = Volume × Density = 125 cm³ × 9.0 g/cm³ = 1125 g = 1.125 kg.
Step 4: Calculate true weight = Mass × g = 1.125 kg × 10 m/s² = 11.25 N.
Step 5: Calculate upthrust = Volume × Density of liquid × g = (125 × 10⁻⁶ m³) × 1200 kg/m³ × 10 = 1.5 N.
Step 6: Tension in thread equals apparent weight = True weight - Upthrust = 11.25 N - 1.5 N = 9.75 N.
Step 1: Calculate volume of the cube: Volume = 5 cm × 5 cm × 5 cm = 125 cm³.
Step 2: Convert volume to SI units: 125 cm³ = 125 × 10⁻⁶ m³.
Step 3: Find mass of the cube: Mass = Volume × Density = 125 cm³ × 9.0 g/cm³ = 1125 g = 1.125 kg.
Step 4: Calculate true weight = Mass × g = 1.125 kg × 10 m/s² = 11.25 N.
Step 5: Calculate upthrust = Volume × Density of liquid × g = (125 × 10⁻⁶ m³) × 1200 kg/m³ × 10 = 1.5 N.
Step 6: Tension in thread equals apparent weight = True weight - Upthrust = 11.25 N - 1.5 N = 9.75 N.
9. A block of wood is floating on water with its dimensions 50 cm × 50 cm × 50 cm inside water. Calculate the buoyant force acting on the block. Take g = 9.8 N kg⁻¹.
Answer:
Step 1: Calculate the volume that is submerged inside water: V = 50 cm × 50 cm × 50 cm = 125,000 cm³.
Step 2: Convert volume to m³: 125,000 cm³ = 0.125 m³.
Step 3: Find the mass of the displaced water = Submerged volume × Density of water = 0.125 m³ × 1000 kg/m³ = 125 kg.
Step 4: Calculate buoyant force (weight of displaced water) = Mass × g = 125 kg × 9.8 N/kg = 1225 N.
Step 1: Calculate the volume that is submerged inside water: V = 50 cm × 50 cm × 50 cm = 125,000 cm³.
Step 2: Convert volume to m³: 125,000 cm³ = 0.125 m³.
Step 3: Find the mass of the displaced water = Submerged volume × Density of water = 0.125 m³ × 1000 kg/m³ = 125 kg.
Step 4: Calculate buoyant force (weight of displaced water) = Mass × g = 125 kg × 9.8 N/kg = 1225 N.
10. A body of mass 3.5 kg displaces 1000 cm³ of water when fully immersed inside it. Calculate : (i) the volume of body, (ii) the upthrust on body and (iii) the apparent weight of body in water.
Answer:
Step 1: A fully immersed body displaces its own volume. So, Volume of body = 1000 cm³. (Answer i)
Step 2: Find mass of the displaced water = Volume × Density = 1000 cm³ × 1 g/cm³ = 1000 g = 1 kg.
Step 3: Upthrust equals the weight of displaced water. Upthrust = 1 kgf. (Answer ii)
Step 4: Calculate apparent weight = True weight - Upthrust.
Step 5: True weight is 3.5 kgf. Apparent weight = 3.5 kgf - 1 kgf = 2.5 kgf. (Answer iii)
Step 1: A fully immersed body displaces its own volume. So, Volume of body = 1000 cm³. (Answer i)
Step 2: Find mass of the displaced water = Volume × Density = 1000 cm³ × 1 g/cm³ = 1000 g = 1 kg.
Step 3: Upthrust equals the weight of displaced water. Upthrust = 1 kgf. (Answer ii)
Step 4: Calculate apparent weight = True weight - Upthrust.
Step 5: True weight is 3.5 kgf. Apparent weight = 3.5 kgf - 1 kgf = 2.5 kgf. (Answer iii)
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Quick Review Flashcards - Click to flip and test your knowledge!
Question
What is the name of the upward force exerted on a body by the fluid in which it is submerged?
Answer
Upthrust (or buoyant force).
Question
What is the standard symbol used to denote upthrust?
Answer
F_B
Question
The property of a liquid to exert an upward force on a body immersed in it is known as _____.
Answer
Buoyancy
Question
How does the magnitude of the force required to push a can into water change as more of the can is submerged?
Answer
The required force increases as the submerged volume increases.
Question
When a body is completely immersed in a fluid, the upthrust reaches its _____ value.
Answer
Maximum
Question
What happens to a piece of cork pushed into water and then released?
Answer
It rises to the surface and floats.
Question
Define 'Apparent weight' of an object in a fluid.
Answer
The weight an object seems to have when placed in a fluid, which is its real weight minus the upthrust.
Question
What is the mathematical formula for Apparent weight?
Answer
Apparent\ weight = Real\ weight - Upthrust
Question
Why does a bucket of water feel lighter when it is inside a well compared to when it is in the air?
Answer
The bucket experiences an upward force (upthrust) due to the water.
Question
What are the two vertical forces acting on a body submerged in a liquid?
Answer
Its weight (W) acting downwards and the upthrust (F_B) acting upwards.
Question
Under what condition regarding forces W and F'_B will a body sink?
Answer
A body sinks if its weight (W) is greater than the maximum upthrust (F'_B).
Question
If the maximum upthrust (F'_B) is greater than the weight (W) of a body, how will the body behave when released?
Answer
It will float with only a portion of its volume submerged.
Question
What is the net force acting on a floating body?
Answer
Zero
Question
What is the S.I. unit of upthrust?
Answer
Newton (N)
Question
How does the volume of the submerged part of a body affect the upthrust it experiences?
Answer
The larger the submerged volume, the greater the upthrust.
Question
How does fluid density affect the upthrust on a submerged body of the same volume?
Answer
The denser the fluid, the greater the upthrust.
Question
Where does the upthrust act on a body in relation to the displaced fluid?
Answer
Upward at the centre of buoyancy (the centre of gravity of the displaced fluid).
Question
Why does a pebble fall faster through air than a bunch of feathers of the same mass?
Answer
The feathers experience greater upthrust because they have a larger volume.
Question
In a vacuum, how do a bunch of feathers and a pebble fall relative to each other?
Answer
They fall together because there is no upthrust in a vacuum.
Question
Why does a denser liquid exert a greater upthrust for the same submerged volume?
Answer
Upthrust is proportional to the density of the fluid.
Question
What is the primary cause of upthrust in a liquid?
Answer
The difference in liquid pressure between the upper and lower surfaces of the submerged body.
Question
Formula for pressure (P) at depth h in a liquid of density \rho is _____.
Answer
P = h \rho g
Question
Mathematically, upthrust (F_B) is equal to the product of volume (V), fluid density (\rho), and _____.
Answer
Acceleration due to gravity (g)
Question
State Archimedes' principle.
Answer
When a body is immersed partially or completely in a liquid, it experiences an upthrust equal to the weight of the liquid displaced by it.
Question
Does Archimedes' principle apply to gases as well as liquids?
Answer
Yes, it applies to all fluids (liquids and gases).
Question
In the experimental verification of Archimedes' principle, the loss in weight of a solid is equal to the weight of the _____.
Answer
Displaced liquid
Question
A body of density \rho sinks in a liquid of density \rho_L if _____.
Answer
\rho > \rho_L
Question
A body floats just inside the surface of a liquid when its density \rho is _____ the density of the liquid \rho_L.
Answer
Equal to
Question
Define 'Density' of a substance.
Answer
Mass per unit volume of the substance.
Question
What is the S.I. unit of density?
Answer
kg\ m^{-3}
Question
What is the C.G.S. unit of density?
Answer
g\ cm^{-3}
Question
Convert 1\ g\ cm^{-3} to the S.I. unit of density.
Answer
1000\ kg\ m^{-3}
Question
How does an increase in temperature typically affect the density of a substance?
Answer
Density usually decreases as temperature increases (due to expansion).
Question
At what temperature is the density of water at its maximum?
Answer
4^\circ C
Question
Define 'Relative Density' (R.D.).
Answer
The ratio of the density of a substance to the density of water at 4^\circ C.
Question
What is the unit for Relative Density?
Answer
It has no unit (it is a pure ratio).
Question
In the C.G.S. system, how does the numerical value of Relative Density relate to density?
Answer
Relative Density is equal to the numerical value of density in g\ cm^{-3}.
Question
Using weights, what is the formula for the Relative Density of a solid denser than water?
Answer
R.D. = \frac{Weight\ of\ body\ in\ air}{Loss\ in\ weight\ of\ body\ in\ water}
Question
If W_1 is weight in air and W_2 is weight in water, what is the formula for R.D.?
Answer
R.D. = \frac{W_1}{W_1 - W_2}
Question
How can you find the R.D. of a solid that is soluble in water?
Answer
Use a liquid in which the solid is insoluble and multiply the calculated ratio by the R.D. of that liquid.
Question
State the Principle of Floatation.
Answer
The weight of a floating body is equal to the weight of the liquid displaced by its submerged part.
Question
For a floating body, what is the value of the apparent weight?
Answer
Zero
Question
What is the relationship between the submerged volume (v), total volume (V), body density (\rho_s), and liquid density (\rho_L) for a floating body?
Answer
\frac{v}{V} = \frac{\rho_s}{\rho_L}
Question
Why does an iron ship float while an iron nail sinks in water?
Answer
The ship is hollow and contains air, making its average density less than that of water.
Question
A loaded ship is submerged _____ than an unloaded ship.
Answer
More
Question
What is the purpose of the 'Plimsoll line' on a ship?
Answer
It indicates the safe limit for loading the ship in water of different densities.
Question
Why does a ship sink further when it sails from sea water into river water?
Answer
River water is less dense than sea water, so a larger volume must be displaced to balance the ship's weight.
Question
What is used as 'ballast' in an unloaded ship to lower its centre of gravity?
Answer
Sand or stones filled at the bottom.
Question
Why is it easier to swim in sea water than in fresh water?
Answer
Sea water is denser due to dissolved salts, providing a greater upthrust.
Question
How does a submarine dive underwater?
Answer
Its ballast tanks are filled with water to increase its average density beyond that of sea water.
Question
Roughly what percentage of an iceberg remains submerged in sea water?
Answer
Approximately 89.3\% (or about 90\%).
Question
Why are icebergs dangerous for ships?
Answer
Most of the iceberg is submerged (hidden), making it difficult to estimate its size and avoid collision.
Question
What organ in a fish acts like the ballast tank of a submarine?
Answer
The swim bladder.
Question
A balloon filled with hydrogen rises because the upthrust is _____ than the weight of the balloon.
Answer
Greater
Question
Why does a rising balloon eventually stop rising?
Answer
The density of air decreases with altitude, so upthrust decreases until it equals the balloon's weight.
Question
What is the formula for the Relative Density of a liquid using a solid displacement method?
Answer
R.D.\ of\ liquid = \frac{W_1 - W_3}{W_1 - W_2} (where W_1 is weight in air, W_2 in water, and W_3 in the liquid).
Question
If a body floats in a liquid, how does the density of the body compare to the density of the liquid?
Answer
The density of the body is less than or equal to the density of the liquid.
Question
The centre of buoyancy coincides with the centre of gravity of the body only if the body is _____.
Answer
Completely immersed (and uniform).
Question
True or False: The upthrust on a body depends on the mass of the body.
Answer
False (it depends on the volume of the submerged part and the density of the fluid).
Question
What happens to the level of water when a floating piece of ice melts?
Answer
The water level remains unchanged.