Chapter 5 · Solved paper40 marks

Model Test – Solved ഉത്തരങ്ങൾ

Every question of the model paper, in the same order, with the full answer, the marking scheme, an animated figure or working for every numerical and explanation question, and a link to the page where the idea is taught. Try each one again before you open the answer, and mark your own paper honestly.

📐 How the figures are drawn

The weight (ഭാരം) is a red arrow pointing down, the buoyant force (പ്ലവക്ഷമബലം) a green arrow pointing up; equal forces have equal arrows. Water is blue. Every number in the animations is worked out with F_B = W₁ − W₂ = ρ × V × g, g = 9.8 m/s² and 1 kgwt = 9.8 N.

SECTION AAnswer all · 8 × 1 = 8 marks

Liquids and gases together are called .

(1 mark)

💡 Water and air both flow. What name covers both?

✅ Answer: fluids (ദ്രവങ്ങൾ)

Liquids (ദ്രാവകങ്ങൾ) and gases (വാതകങ്ങൾ) can flow, so together they are called fluids. Every fluid – water, oil, even air – exerts a buoyant force on objects in it.

Marking: 1 mark.

📖 Review this on Page 5 · Fluids and key box 1

An object weighs 12 N in air and 9 N when it is fully immersed in water. The buoyant force on it is

(a) 21 N(b) 12 N(c) 9 N(d) 3 N

(1 mark)

💡 Buoyant force = weight in air − weight in water.

✅ Answer: (d) 3 N

Buoyant force = apparent loss of weight (ഭാരനഷ്ടം) = W₁ − W₂ = 12 N − 9 N = 3 N.

Marking: 1 mark for (d).

📖 Review this on Page 5 · Key box 2: F_B = W₁ − W₂

The unit of relative density is

(a) kg/m³(b) g/cm³(c) N(d) it has no unit

(1 mark)

💡 What is (kg/m³) ÷ (kg/m³)?

✅ Answer: (d) no unit

Relative density (ആപേക്ഷികസാന്ദ്രത) = density of the substance ÷ density of water. Both are in kg/m³, so the units cancel – it is a pure number.

Marking: 1 mark for (d).

📖 Review this on Page 9 · Key box 7

Find the relation and fill in: hydrometer : relative density of liquids :: lactometer :

(1 mark)

💡 A lactometer is a hydrometer for which liquid?

✅ Answer: relative density (purity) of milk

A lactometer (ലാക്ടോമീറ്റർ) is a hydrometer (ഹൈഡ്രോമീറ്റർ) made for milk: it measures the relative density of milk and so shows whether water has been added.

Marking: 1 mark (“purity of milk” also accepted).

📖 Review this on Page 9 · Lactometer

A wooden block floats in water. The buoyant force on it is

(a) greater than its weight(b) less than its weight(c) equal to its weight(d) zero

(1 mark)

💡 The block is at rest – the forces are balanced.

✅ Answer: (c) equal to its weight

Principle of floatation (പ്ലവനതത്വം): a floating object displaces its own weight of fluid, so the buoyant force equals its weight. If it were greater, the block would rise further out of the water until they became equal.

Marking: 1 mark for (c).

📖 Review this on Page 8 · Key box 6

A 100 g brass weight is fully immersed in turn in each liquid below. In which is the buoyant force on it the greatest?

(a) kerosene(b) water(c) saturated salt solution(d) glycerine

(1 mark)

💡 Same volume – which liquid is densest?

✅ Answer: (d) glycerine

The volume immersed is the same each time, so the densest liquid gives the greatest buoyant force. Glycerine (ഗ്ലിസറിൻ, 1260 kg/m³) is denser than salt solution (1025), water (1000) and kerosene (810).

Marking: 1 mark for (d).

📖 Review this on Page 6 · Table 5.3, key box 3

The density of mercury is 13 600 kg/m³. Its relative density is .

(1 mark)

💡 Divide by the density of water, 1000 kg/m³.

✅ Answer: 13.6

Relative density = 13 600 ÷ 1000 = 13.6 (no unit): mercury is 13.6 times as dense as water.

Marking: 1 mark (writing a unit: ½ mark).

📖 Review this on Page 9 · Mercury question

The scientist of Syracuse who found the volume of the king’s crown from the water that overflowed when he got into a full bathtub was .

(1 mark)

💡 “Eureka!”

✅ Answer: Archimedes

Archimedes (born 287 BC) realised that the volume of water an object pushes out equals the volume of the object. Knowing the crown’s volume, he found its density and showed that it was not pure gold.

Marking: 1 mark.

📖 Review this on Page 7 · Archimedes box

SECTION BAnswer all · 5 × 2 = 10 marks

An object weighs 0.6 kgwt in air and 0.45 kgwt when it is fully immersed in water. (a) Find the buoyant force on it in kgwt. (b) Express it in newton. (1 kgwt = 9.8 N)

(2 marks)

💡 Loss of weight = buoyant force; then multiply by 9.8.

🎬 Step by step

✅ Answer: (a) 0.15 kgwt (b) 1.47 N

Given: W₁ = 0.6 kgwt (in air); W₂ = 0.45 kgwt (in water); 1 kgwt = 9.8 N

Formula: F_B = W₁ − W₂

Substitution: (a) F_B = 0.6 − 0.45 = 0.15 kgwt; (b) 0.15 × 9.8

Answer: (a) 0.15 kgwt (b) 1.47 N

The apparent loss of weight is the buoyant force; it is also the weight of the water displaced (Archimedes’ principle).

Marking: (a) 1 mark (formula ½, answer ½); (b) 1 mark (answer with unit).

📖 Review this on Page 5 · Measuring buoyant force · Page 1 · kgwt and newton

10.

A ship rises a little out of the water when it sails from a river into the sea. Why?

(2 marks)

💡 Which is denser – river water or sea water? What stays the same?

🎬 Step by step

✅ Answer: sea water is denser, so less of it needs to be displaced

The ship floats in both, so the buoyant force always equals its weight (principle of floatation). Sea water is denser than river water, so a smaller volume of sea water already weighs as much as the ship. Less of the hull needs to be under water – the ship rises.

Marking: 1 mark for “sea water is denser”, 1 mark for “a smaller volume is displaced for the same weight (F_B = W)”.

📖 Review this on Page 6 · Science diary: ship, egg, Dead Sea · Page 12 · Ship lab

11.

A small iron nail sinks in water, but a huge ship made of iron floats. Explain.

(2 marks)

💡 Compare the volume of water each one can push aside.

🎬 Step by step

✅ Answer: the hollow ship displaces water equal to its weight; the nail cannot

The solid nail displaces only its own small volume of water. The weight of that water is far less than the nail’s weight (iron is 7.8 times as dense as water), so the buoyant force is less than its weight – it sinks.

The ship is hollow: its shape pushes aside a huge volume of water. It settles until the weight of the water displaced equals its own weight – it floats. (Like the aluminium foil boat and the crumpled ball, Fig 5.7.)

Marking: 1 mark for the nail (F_B < W), 1 mark for the ship (hollow shape, displaces its own weight of water).

📖 Review this on Page 8 · Foil boat and the ship

12.

The density of coconut oil is 920 kg/m³. (a) Find its relative density. (b) Will coconut oil float on water? Why?

(2 marks)

💡 Divide by 1000; compare the answer with 1.

🎬 Step by step

✅ Answer: (a) 0.92 (b) yes – it is less dense than water

Given: density of coconut oil = 920 kg/m³; density of water = 1000 kg/m³

Formula: relative density = density of the substance ÷ density of water

Substitution: 920 ÷ 1000

Answer: (a) 0.92 (no unit)

(b) Yes. Its relative density is less than 1 – it is less dense than water – so it floats as a layer on top (വെളിച്ചെണ്ണ on water).

Marking: (a) 1 mark; (b) ½ for “yes”, ½ for the reason.

📖 Review this on Page 9 · Relative density · Page 3 · Oil floats on water

13.

A lactometer is placed first in pure milk and then in milk to which water has been added. (a) In which does it sink deeper? (b) Why?

(2 marks)

💡 Less dense liquid → more volume needed to balance the same weight.

🎬 Step by step

✅ Answer: (a) in the watered milk (b) it is less dense

(a) In the milk with water added.

(b) Water (relative density 1.000) makes the milk less dense. The lactometer floats where the weight of the milk it displaces equals its own weight, so in less dense milk it must displace a bigger volume – it sinks deeper and shows a lower reading.

Marking: 1 mark for (a); 1 mark for the reason in (b).

📖 Review this on Page 9 · Lactometer lab · Page 12 · Milk society lab

SECTION CAnswer all · 4 × 3 = 12 marks

14.

A stone of mass 0.5 kg weighs 4.9 N in air. When fully immersed in water, the spring balance reads 3.9 N (figure). (a) Find the buoyant force. (b) Find the volume of the stone. (c) What will the balance read if the stone is fully immersed in saturated salt solution (density 1025 kg/m³)? (g = 9.8 m/s², density of water 1000 kg/m³)

(3 marks)

💡 The loss of weight is the weight of the water displaced; weight of water = ρVg.

🎬 Step by step

✅ Answer: (a) 1.0 N (b) 1.02 × 10⁻⁴ m³ ≈ 102 cm³ (c) about 3.88 N

Given: W₁ = 4.9 N, W₂ = 3.9 N

Formula: F_B = W₁ − W₂

Substitution: 4.9 − 3.9

Answer: (a) F_B = 1.0 N

Given: F_B = 1.0 N = weight of the water displaced; ρ_water = 1000 kg/m³, g = 9.8 m/s²

Formula: F_B = ρ × V × g, so V = F_B ÷ (ρ × g)

Substitution: V = 1.0 ÷ (1000 × 9.8)

Answer: (b) V ≈ 1.02 × 10⁻⁴ m³ ≈ 102 cm³ – the stone is fully under, so this is its volume

(c) In salt solution: F_B = 1025 × 1.02 × 10⁻⁴ × 9.8 ≈ 1.025 N, so the reading = 4.9 − 1.025 ≈ 3.88 N – a little less than in water, because the solution is denser.

Marking: (a) 1 mark; (b) 1 mark (formula ½, answer with unit ½); (c) 1 mark.

📖 Review this on Page 5 · Spring-balance lab · Page 7 · Archimedes’ principle · Page 6 · Density of the liquid

15.

A 100 g brass weight (weight in air 0.98 N) is fully immersed in three liquids P, Q and R. The spring balance readings are shown below.

Liquid	P	Q	R
Reading (N)	0.881	0.826	0.858

(a) Find the buoyant force in each liquid. (b) Which liquid is the densest? Why? (c) Q is glycerine and R is water. Is P kerosene or saturated salt solution? Give a reason.

(3 marks)

💡 Subtract each reading from 0.98 N; more buoyant force = denser liquid.

🎬 Step by step

✅ Answer: (a) 0.099 N, 0.154 N, 0.122 N (b) Q (c) kerosene

(a) F_B = W₁ − W₂: P: 0.98 − 0.881 = 0.099 N; Q: 0.98 − 0.826 = 0.154 N; R: 0.98 − 0.858 = 0.122 N.

(b) Q. The same weight is fully immersed each time (same volume), so the liquid giving the greatest buoyant force is the densest.

(c) Kerosene. The buoyant force in P (0.099 N) is less than in water (0.122 N), so P is less dense than water. Kerosene (810 kg/m³) is less dense than water; saturated salt solution (1025 kg/m³) is denser and would give more than 0.122 N.

Marking: (a) 1 mark (all three); (b) ½ answer + ½ reason; (c) ½ answer + ½ reason.

📖 Review this on Page 6 · Table 5.2 lab · Page 5 · Key box 2

16.

A wooden block of mass 0.6 kg floats in water. (a) What is its weight in newton? (b) What is the buoyant force on it? (c) Find the volume of water it displaces. (g = 9.8 m/s², density of water 1000 kg/m³)

(3 marks)

💡 Floating: weight of water displaced = weight of the block.

🎬 Step by step

✅ Answer: (a) 5.88 N (b) 5.88 N (c) 6 × 10⁻⁴ m³ = 600 cm³

Given: m = 0.6 kg; g = 9.8 m/s²

Formula: W = m × g

Substitution: 0.6 × 9.8

Answer: (a) W = 5.88 N (= 0.6 kgwt)

(b) The block floats, so by the principle of floatation F_B = W = 5.88 N.

Given: weight of water displaced = 5.88 N, so its mass = 0.6 kg; ρ = 1000 kg/m³

Formula: volume = mass ÷ density

Substitution: 0.6 ÷ 1000

Answer: (c) V = 6 × 10⁻⁴ m³ = 600 cm³

Marking: 1 mark for each part.

📖 Review this on Page 8 · Principle of floatation · Page 1 · W = mg · Page 3 · Density = mass ÷ volume

17.

A crown said to be made of pure gold has a mass of 1 kg. When it is lowered into a jar full of water, 60 cm³ of water overflows. (Density of pure gold = 19 300 kg/m³.) (a) What is the volume of the crown? Why? (b) Find the density of the crown. (c) Is the crown pure gold? Explain.

(3 marks)

💡 Overflowing water volume = volume of the crown; then density = m ÷ V.

🎬 Step by step

✅ Answer: (a) 60 cm³ (b) about 16 700 kg/m³ (c) no – it is adulterated

(a) 60 cm³ = 6 × 10⁻⁵ m³. A fully immersed object pushes out a volume of water equal to its own volume (Archimedes’ discovery).

Given: m = 1 kg; V = 6 × 10⁻⁵ m³

Formula: density = mass ÷ volume

Substitution: 1 ÷ (6 × 10⁻⁵)

Answer: (b) ≈ 16 700 kg/m³ (16 667)

(c) No. Its density is less than that of pure gold (19 300 kg/m³), so a lighter metal such as silver has been mixed in. A pure-gold crown of 1 kg would push out only 1 ÷ 19 300 m³ ≈ 51.8 cm³ of water.

Marking: 1 mark for each part (½ answer, ½ reason / unit).

📖 Review this on Page 7 · Archimedes and the crown · Page 3 · Density

SECTION DAnswer all · 2 × 5 = 10 marks

18.

A hydrometer floats in water and in a liquid X as shown in the figure. (a) What is a hydrometer? On what principle does it work? (b) What does it read in pure water? (c) In X it floats higher than in water. Is the relative density of X more or less than 1? (d) The density of kerosene is 810 kg/m³. Find its relative density, and say whether the hydrometer will float higher or sink deeper in kerosene than in water. (e) Why do the values on a hydrometer scale increase downwards?

(5 marks)

💡 Denser liquid → less volume needed → floats higher.

🎬 Step by step

✅ Answer: (a) measures RD of liquids; floatation (b) 1.00 (c) more than 1 (d) 0.81; sinks deeper (e) it floats higher in denser liquids

(a) A hydrometer (ഹൈഡ്രോമീറ്റർ) is an instrument to measure the relative density of liquids. It works on the principle of floatation: it sinks until the weight of the liquid it displaces equals its own weight.

(b) 1.00 (the relative density of water).

(c) More than 1. It floats higher, so it displaces a smaller volume of X to balance the same weight – X is denser than water.

Given: density of kerosene = 810 kg/m³; density of water = 1000 kg/m³

Formula: relative density = density of kerosene ÷ density of water

Substitution: 810 ÷ 1000

Answer: (d) 0.81; kerosene is less dense than water, so the hydrometer sinks deeper in it

(e) In a denser liquid the hydrometer floats higher, so a point lower down on its stem comes to the liquid surface. The marks for larger relative densities are therefore lower on the stem: the values increase downwards.

Marking: 1 mark for each part ((a): ½ use, ½ principle; (d): ½ value, ½ deeper).

📖 Review this on Page 9 · Hydrometer, Fig 5.8 · Page 11 · Build a hydrometer

19.

A house boat on Vembanad lake displaces 3000 kg of water. (a) Find the buoyant force on it in kgwt and in newton. (b) What is the weight of the boat? Which principle tells you this? (c) Four tourists of total mass 240 kg get on. How much more water does the boat displace? Give the mass and the volume. (d) Life jackets are compulsory on such boats. Explain why, using the idea of buoyant force.

(5 marks)

💡 Floating: buoyant force = weight = weight of the water displaced.

🎬 Step by step

✅ Answer: (a) 3000 kgwt = 29 400 N (b) 3000 kgwt – floatation (c) 240 kg, 0.24 m³ (d) more volume, little weight

Given: mass of water displaced = 3000 kg; 1 kgwt = 9.8 N

Formula: F_B = weight of the water displaced (Archimedes’ principle)

Substitution: 3000 kgwt = 3000 × 9.8 N

Answer: (a) F_B = 3000 kgwt = 29 400 N

(b) 3000 kgwt (29 400 N). By the principle of floatation, a floating object’s weight equals the weight of the fluid it displaces.

Given: extra weight = 240 kgwt; density of water = 1000 kg/m³

Formula: extra water displaced has the same weight; volume = mass ÷ density

Substitution: 240 ÷ 1000

Answer: (c) 240 kg more water, 0.24 m³ – the boat sinks a little deeper

(d) A life jacket (ലൈഫ് ജാക്കറ്റ്) is filled with very light foam or air: it adds a large volume but very little weight. A person wearing it displaces more water, so the buoyant force can equal the total weight while the head stays above the water – the person floats and can breathe until help comes.

Marking: (a) 1 (½ kgwt, ½ N); (b) 1 (½ value, ½ principle); (c) 1 (½ mass, ½ volume); (d) 2 (large volume, little weight 1; more water displaced, F_B = weight with head out 1).

📖 Review this on Page 8 · Boat 6000 kg question · Page 11 · Let us assess Q4 · Page 12 · Life-jacket lab

Marks summary

Section	Questions	Marks each	Total	Topics
A	1–8	1	8	fluids, F_B = W₁ − W₂, unit of relative density, lactometer, floating, density of liquids, mercury, Archimedes
B	9–13	2	10	kgwt → N numerical, ship from river to sea, nail and ship, relative density of coconut oil, lactometer in watered milk
C	14–17	3	12	spring balance and volume, three liquids, floating block, Archimedes’ crown
D	18–19	5	10	hydrometer and relative density, house boat and life jackets
Total			40

⏱️ Time management (15 min cool-off + 90 minutes)

Cool-off (15 min) – read the whole paper. Mark every numerical and note what is given in kgwt and what in newton.
Section A – about 10 minutes. Short answers; no unit for relative density.
Section B – about 20 minutes. For Q9 and Q12 write Given, Formula, Substitution, Answer. In explanations name the principle (Archimedes / floatation).
Section C – about 25 minutes. Show every subtraction W₁ − W₂ and every unit (N, m³, cm³, kg/m³).
Section D – about 25 minutes. Answer every part; for “why” parts use the words density, volume displaced, weight of water displaced.
Checking – keep 10 minutes: kgwt ↔ N (× 9.8)? grams changed to kilograms? floating means F_B = W, not greater?

🎉 Finished the chapter?

Count your marks: 32 or more – excellent; 24–31 – good, revise the pages linked under the questions you missed; below 24 – go back through pages 5, 6, 7, 8 and 9 with their labs and do the scored quiz again. Then return to the chapter home and tick off what you have learned.