Chapter 5 · Let us assessTextbook pages 105–106

Let Us Assess & Extended Activities – Solutions വിലയിരുത്താം

All seven end-of-chapter questions – loss of weight in kgwt, two spheres of equal diameter, balloons and an unloading ship, life jackets, the relative density of salt solution, the four spring-balance pictures of Fig 5.10 and a floating 800 N object – followed by the two extended activities തുടർപ്രവർത്തനങ്ങൾ: making a hydrometer and testing milk at a milk society. Try each question in your science diary first, then step through the solution.

📖 What you need (pages 1–10 of this guide)

Question 1 · 0.2 kgwt in air, 0.19 kgwt in water

1. ഒരു വസ്തുവിന്റെ വായുവിലെ ഭാരം 0.2 kgwt ഉം പൂർണ്ണമായും ജലത്തിൽ മുങ്ങിയിരിക്കുമ്പോഴുള്ള ഭാരം 0.19 kgwt ഉം ആണെങ്കിൽ The weight of an object in air is 0.2 kgwt and its weight when fully immersed in water is 0.19 kgwt.

a) വസ്തുവിന് ജലത്തിലുള്ള ഭാരനഷ്ടം എത്രയായിരിക്കും?
b) വസ്തുവിൽ അനുഭവപ്പെടുന്ന പ്ലവക്ഷമബലം എത്ര?
  1. What will be the loss of weight of the object in water?
  2. What is the buoyant force on the object?

Loss of weight = W₁ − W₂, and the buoyant force is equal to that loss. Keep the unit kgwt (or convert with 1 kgwt = 9.8 N).

🎬 Solution in exam layout – Given → Formula → Substitution → Answer
🧪 Try other numbers
Given: W₁ = 0.2 kgwt (in air); W₂ = 0.19 kgwt (fully in water).
Formula: loss of weight = W₁ − W₂; buoyant force = loss of weight.
Substitution: 0.2 kgwt − 0.19 kgwt = 0.01 kgwt.
Answer: (a) loss of weight = 0.01 kgwt (= 0.01 × 9.8 = 0.098 N); (b) buoyant force = 0.01 kgwt (= 0.098 N).
⚠️ Common trap

Writing “0.01 kg”. kgwt (kilogram weight) is a unit of force; kg is a unit of mass. And keep the order W₁ − W₂ (air minus water) – the answer can never be negative.

Question 2 · A copper sphere and an iron sphere of the same diameter

2. ഒരേ വ്യാസമുള്ള ഒരു ചെമ്പുഗോളവും ഇരുമ്പുഗോളവും ജലത്തിനടിയിൽ സ്ഥിതി ചെയ്യുന്നു. (ചെമ്പിന്റെ സാന്ദ്രത 8900 kg/m³, ഇരുമ്പിന്റെ സാന്ദ്രത 7800 kg/m³). രണ്ടിലും അനുഭവപ്പെടുന്ന പ്ലവക്ഷമബലം തുല്യമാണോ? ഉത്തരം സാധൂകരിക്കുക. A copper sphere and an iron sphere of the same diameter are under water (density of copper 8900 kg/m³, density of iron 7800 kg/m³). Is the buoyant force on both equal? Justify your answer.

Same diameter ⇒ same volume. Which two things decide the buoyant force on a fully immersed object? Is the density of the object one of them?

🎬 Solution animation – equal spheres, equal push
🧪 Try it – the equal-volume spheres lab

Change the diameter or the liquid: the two buoyant forces always stay equal to each other. Only the weights (and so the readings) differ.

Yes, the buoyant forces are equal.
Both spheres have the same diameter, so they have the same volume. Fully under water, each displaces the same volume of water, and hence the same weight of water. By Archimedes' principle the buoyant force = weight of water displaced, so the buoyant force is the same on both.
For a fully immersed object the buoyant force depends only on the density of the liquid and the volume of the object. The densities of copper and iron are different, but that only makes their weights different (the copper sphere is heavier) – not the buoyant force. (Example, 4 cm diameter: FB = 0.33 N on each; weights 2.92 N and 2.56 N.)
⚠️ Common trap

“Copper is denser, so it gets more (or less) buoyant force.” The density in the formula FB = ρ V g is the density of the liquid, not of the object. Compare with Table 5.4: there the weights were equal and the volumes different; here the volumes are equal and the weights different.

Question 3 · Give reasons – balloons, and a ship that rises

3. താഴെ കൊടുത്ത പ്രസ്താവനകൾക്ക് കാരണം എഴുതുക. Write the reasons for the following statements.

a) ഹീലിയം വാതകം നിറച്ച കളിപ്പാട്ട ബലൂണുകൾ വായുവിൽ മുകളിലേക്ക് ഉയരുകയും കാർബൺ ഡൈ ഓക്സൈഡ് നിറച്ച ബലൂണുകൾ താഴുകയും ചെയ്യുന്നു.
b) തുറമുഖത്തെത്തി ഭാരം ഇറക്കിക്കഴിയുമ്പോൾ കപ്പൽ ജലത്തിൽ കൂടുതൽ പൊങ്ങുന്നു.
  1. Toy balloons filled with helium gas rise up in air, while balloons filled with carbon dioxide sink.
  2. When a ship reaches the port and its load is unloaded, it rises higher in the water.

(a) Both balloons have the same size, so the air pushes both up equally. Compare their weights: which gas is lighter than air, which heavier? (b) A floating ship always displaces its own weight of water. What happens to that weight when cargo is taken off?

🎬 Solution animation – (a) helium balloon and CO₂ balloon
🧪 Try it – (b) unload the ship

Sample ship (approx.): box-shaped hull 50 m × 10 m, empty ship 500 tonnes. Depth under water = total mass ÷ (density × 500 m²).

(a) Two balloons of the same size displace the same volume of air, so the air pushes both up with the same buoyant force (the weight of that much air). Helium is much less dense than air, so the helium balloon's weight is less than the buoyant force – it rises. Carbon dioxide is denser than air, so the CO₂ balloon's weight is more than the buoyant force – it sinks.
(b) A floating ship displaces water whose weight equals its own weight (principle of floatation). When the cargo is unloaded the ship's weight decreases, so it needs to displace less water – a smaller part of it stays under water and the ship rises higher.

Question 4 · Why are life jackets compulsory in boats?

4. ബോട്ടുകളിൽ ലൈഫ് ജാക്കറ്റ് ഉപയോഗിക്കുന്നത് നിർബന്ധമാണ്. പ്ലവക്ഷമബലത്തിന്റെ അടിസ്ഥാനത്തിൽ വിശദീകരിക്കുക. Using life jackets ലൈഫ് ജാക്കറ്റ് in boats is compulsory. Explain on the basis of buoyant force.

A life jacket is full of foam or air. Does it add much weight? Does it add much volume? Which one decides the buoyant force?

⚠ Always wear a properly fastened life jacket in a boat, a ferry or the Water Metro – even if you can swim. Never play near deep water without an adult.
🎬 Solution animation – with and without a life jacket
The human body is only about as dense as water, so even fully under water the buoyant force on a person is about equal to (often a little less than) their weight – the head, nose and mouth do not stay above water, and a tired or frightened person can drown.
A life jacket is filled with very light material (foam or air). It adds a large volume but very little weight. The person + jacket can now displace much more water, so the buoyant force becomes equal to their total weight while a good part of the body – the head and shoulders – is still above the water. The person floats safely (principle of floatation). That is why life jackets are compulsory in boats.
🔢 Sample numbers (approx.)

Person 50 kg, body volume 49.5 L: weight 490 N; buoyant force even when fully under = 485 N – less than the weight. Life jacket 0.5 kg, volume 10 L: total weight 495 N; buoyant force possible up to 583 N. The person + jacket floats with only 50.5 L under water – about 9 L (the head and shoulders) stays above the surface.

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Question 5 · Relative density of saturated salt solution

5. പൂരിത ഉപ്പുലായനിയുടെ സാന്ദ്രത 1025 kg/m³ ആണ്. എങ്കിൽ ആപേക്ഷിക സാന്ദ്രത എത്രയായിരിക്കും? The density of saturated salt solution is 1025 kg/m³. What will be its relative density?

Relative density = density of the substance ÷ density of water (1000 kg/m³).

🎬 Solution in exam layout – Given → Formula → Substitution → Answer
Given: density of saturated salt solution = 1025 kg/m³; density of water = 1000 kg/m³.
Formula: relative density = density of the substance ÷ density of water.
Substitution: = 1025 kg/m³ ÷ 1000 kg/m³.
Answer: relative density = 1.025 (no unit). A hydrometer placed in it would read 1.025.

Question 6 · The same object, four ways (Fig 5.10)

6. ഒരേ വസ്തു സ്പ്രിങ് ബാലൻസ് ഉപയോഗിച്ച് വായുവിലും വ്യത്യസ്ത രീതികളിൽ ഒരു ദ്രാവകത്തിലും താഴ്ത്തി വച്ചിരിക്കുന്നത് ചിത്രീകരിച്ചിരിക്കുന്നു. The figure shows the same object hung from a spring balance in air and lowered into a liquid in different ways: Fig 1 – in air, 50 N; Fig 2 – partly immersed, 40 N; Fig 3 – fully immersed just below the surface, 30 N; Fig 4 – fully immersed, deeper, __ N.

a) ചിത്രം 2, 3 എന്നിവയിൽ പ്ലവക്ഷമബലത്തിൽ വ്യത്യാസം അനുഭവപ്പെടുന്നതിന് കാരണമെന്ത്?
b) നാലാമത്തെ ചിത്രത്തിൽ സ്പ്രിങ് ബാലൻസിലെ റീഡിങ് എത്രയാണ്? എന്തുകൊണ്ട്?
  1. What is the reason for the difference in buoyant force in Fig 2 and Fig 3?
  2. What is the spring-balance reading in the fourth figure? Why?

Work out the buoyant force in each picture: 50 − 40 and 50 − 30. What changed between Fig 2 and Fig 3? Between Fig 3 and Fig 4, does the volume under the liquid change at all?

🎬 Solution animation – Fig 5.10 redrawn
🧪 Try it – lower the block slowly (depth slider)

Sample block (consistent with Fig 5.10): a cube of side 12.7 cm (volume about 2.04 L, so FB = 20 N when fully in water), weight 50 N.

Buoyant force in each figure: Fig 2: 50 − 40 = 10 N; Fig 3: 50 − 30 = 20 N.
(a) In Fig 2 the object is only partly immersed, so it displaces less liquid; in Fig 3 it is fully immersed and displaces more liquid (its whole volume). The buoyant force equals the weight of liquid displaced, so it is greater in Fig 3 (20 N) than in Fig 2 (10 N).
(b) The reading in Fig 4 is 30 N. Once the object is fully immersed, the volume of liquid it displaces is its own whole volume – it does not change as the object goes deeper. So the buoyant force stays 20 N and the reading stays 50 N − 20 N = 30 N.
⚠️ Common trap

“Deeper means more pressure, so more buoyant force – the reading must be less than 30 N.” The pressure on the top and the bottom faces both increase with depth by the same amount, so their difference – the buoyant force – stays the same (page 4).

Question 7 · An 800 N object floats

7. വായുവിൽ 800 N ഭാരമുള്ള ഒരു വസ്തു ജലത്തിൽ പൊങ്ങിക്കിടക്കുന്നു. ആ വസ്തുവിന്റെ ജലത്തിലെ ഭാരം എത്രയായിരിക്കും? വസ്തു ആദേശം ചെയ്ത ജലത്തിന്റെ ഭാരം എത്ര? An object weighing 800 N in air floats in water. What will be its weight in water? What is the weight of the water displaced by the object?

Floating means the water holds up the whole weight: buoyant force = weight. What would a spring balance read then?

🎬 Solution in exam layout – Given → Formula → Substitution → Answer
🧪 Try other numbers
Given: weight in air W₁ = 800 N; the object floats in water.
Formula: floating ⇒ buoyant force = weight (principle of floatation); weight in water W₂ = W₁ − FB; weight of displaced water = FB (Archimedes).
Substitution: FB = 800 N; W₂ = 800 N − 800 N.
Answer: weight in water = 0 N (a spring balance would read zero); weight of displaced water = 800 N. (That is 800 ÷ 9.8 ≈ 81.6 kg, i.e. about 81.6 L of water.)
⚠️ Common trap

“Weight in water 0 N” does not mean gravity has stopped – the object still weighs 800 N. It means the water is holding up all of it, so nothing is left for the spring balance.

Extended activity 1 · Make your own hydrometer

1. ഒരു ടെസ്റ്റ്ട്യൂബ്, ലെഡ്ഷോട്ട്, മെഴുക്, അങ്കനം ചെയ്ത പേപ്പർ എന്നിവ ഉപയോഗിച്ച് ഒരു ഹൈഡ്രോമീറ്റർ നിർമ്മിച്ച് ശുദ്ധമായ ജലത്തിൽ താഴ്ത്തി ആപേക്ഷിക സാന്ദ്രത അടയാളപ്പെടുത്തുക. അതിനുശേഷം വിവിധ ദ്രാവകങ്ങളുടെ ആപേക്ഷികസാന്ദ്രത ജലത്തിന്റേതുമായി താരതമ്യം ചെയ്യുക. Make a hydrometer ഹൈഡ്രോമീറ്റർ using a test tube, lead shot, wax and a graduated paper strip. Put it in pure water and mark the relative density. Then compare the relative densities of different liquids with that of water.

🛠️ Build card
  1. Ballast: put a few lead shots (or small steel nuts / clean sand) in a dry test tube so that it floats upright in water with about half of it under water.
  2. Seal: with an adult's help, pour a little melted wax over the shots so they cannot move.
  3. Scale: slip a strip of graph paper inside the tube, numbers facing out.
  4. Mark 1.00: float the tube in pure water in a tall jar; mark “1.00” where the water surface meets the scale.
  5. More marks: use the lab below (or liquids of known density from Table 5.3) to find where 0.8, 0.9, 1.1 and 1.2 go. Notice the marks get closer together lower down, and values increase downwards.
  6. Test: float it in kerosene, coconut oil, salt water, glycerine … read the mark at the surface and record it in your science diary.
⚠ Lead is poisonous: handle lead shot only with an adult, never put it in your mouth, and wash your hands afterwards (steel nuts or sand work too). Melt wax only with an adult. Don't taste any of the liquids.
🧪 Virtual calibration lab – where do the marks go?

The tube floats when its weight equals the weight of liquid displaced: m g = ρ A h g, so the depth under the surface is h = m / (ρ A) (A = cross-section area; with m in g, ρ in g/cm³ = relative density, A in cm², h comes out in cm). The tube is 15 cm long; the rounded bottom is ignored.

💡 Why this matters

Because h = m/(ρA), doubling the density halves the depth – the scale is not evenly spaced. Real hydrometers use a heavy bulb and a thin stem (small A) so that small changes in density move the reading a lot – that is how a lactometer can tell 1.030 from 1.025.

Extended activity 2 · A lactometer at the milk society

2. ഒരു ലാക്ടോമീറ്റർ നിർമ്മിക്കുക. നിങ്ങളുടെ സമീപത്തെ പാൽസൊസൈറ്റിയിൽ എത്തുന്ന ഏതാനും പാൽപാത്രങ്ങളിലെ പാലിന്റെ ആപേക്ഷികസാന്ദ്രത കണ്ടെത്തി സൊസൈറ്റിയിലെ പാൽ വിലസൂചിക ചാർട്ടുമായി താരതമ്യം ചെയ്യുക. Make a lactometer. Find the relative density of the milk in a few milk cans brought to your nearby milk society ക്ഷീരസംഘം and compare it with the society's milk price chart.

🥛 Guidance card
  1. Make it: a lactometer is a hydrometer with a very thin stem, so its scale spreads out between 1.015 and 1.040. Your test-tube hydrometer can be used: calibrate it in water (1.000) and in a salt solution of known density.
  2. At the society: ask the secretary or the milk tester for permission and watch how they test each can. Many societies write the lactometer reading as a whole number: “28” means a relative density of 1.028.
  3. Record: for each sample note the lactometer reading, the relative density, the fat % and the price paid per litre.
  4. Compare: pure cow's milk is usually about 1.028–1.032 (approx.). The price chart is based on fat % and SNF % (solids-not-fat, worked out from the lactometer reading and fat %) – higher values, higher price.
  5. Think: a low lactometer reading suggests added water; but skimmed milk with water can read normal, which is why fat % is also measured.
SampleLactometer readingRelative densityFat %Price per litre (from the chart)Remark
1
2
3

Copy this table into your science diary and fill it at the society.

🧪 Sample checker

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