Goyal Brothers Solutions for ICSE Class 9 Physics Chapter 5 Upthrust And Archimedes Principle

Practice Problem 1:

Question 1. A solid of density \( 2700 \text{ kg m}^{-3} \) and of volume \( 0.0015 \text{ m}^3 \) is completely immersed in alcohol of density \( 800 \text{ kg m}^{-3} \). Calculate :
1. Weight of solid in SI system.
2. Upthrust on solid in SI system.
3. Apparent weight of solid in alcohol.
4. Will the apparent weight of solid be less or more, if it is immersed completely in brine solution? Give a reason, \( [g = 10 \text{ m s}^{-2}] \)

Answer: Density of solid \( (\rho) = 2700 \text{ kg m}^{-3} \)
Volume of solid \( (V) = 0.0015 \text{ m}^3 \)
Density of alcohol \( (\rho') = 800 \text{ kg m}^{-3} \)

1. Mass of solid \( = m = V \times \rho \)
\( \implies m = 0.0015 \times 2700 = 4.05 \text{ kg} \)
Weight of solid \( = mg = 4.05 \times 10 = 40.5 \text{ N} \)

2. Volume of alcohol displaced = Volume of solid
\( \implies V = 0.0015 \text{ m}^3 \)
Mass of alcohol displaced \( = m' = V \times \rho' \)
\( \implies m' = 0.0015 \times 800 = 1.2 \text{ kg} \)
Upthrust = Weight of alcohol displaced
\( \implies \text{Upthrust} = m'g = 1.2 \times 10 = 12 \text{ N} \)

3. Apparent weight of solid in alcohol = Actual weight of solid - Upthrust
\( \implies \text{Apparent weight} = 40.5 - 12 = 28.5 \text{ N} \)

4. When a solid is immersed completely in brine solution, then upthrust acts on it in upward direction, as a result, its apparent weight of solid will be less than actual weight of solid because the density of brine solution is more than the density of alcohol, providing more upthrust.

In simple words: First, we find the real weight by multiplying mass and gravity. Then, we find the "upward push" (upthrust) by seeing how much alcohol the solid pushes aside. The apparent weight is just the real weight minus that upward push.

📝 Teacher's Note: Remind students that the volume of liquid displaced is always equal to the volume of the solid if it is completely submerged. Use this as the starting point for Archimedes' calculations.

🎯 Exam Tip: Always state the units clearly (Newtons for weight and upthrust, kg for mass) to avoid losing half-marks on numerical problems.

Question 2. A stone of density \( 3000 \text{ kg m}^{-3} \) is lying submerged in water of density \( 1000 \text{ kg m}^{-3} \). If the mass of stone in air is \( 150 \text{ kg} \), calculate the force required to lift the stone. \( [g = 10 \text{ m s}^{-2}] \)

Answer: Density of stone \( (\rho) = 3000 \text{ kg m}^{-3} \)
Density of water \( (\rho') = 1000 \text{ kg m}^{-3} \)
Mass of stone \( = m = 150 \text{ kg} \)
Acceleration due to gravity \( = g = 10 \text{ m s}^{-2} \)
Volume of stone \( = V = \frac{m}{\rho} \)
\( \implies V = \frac{150}{3000} = \frac{1}{20} = 0.05 \text{ m}^3 \)
Actual weight of stone \( = mg = 150 \times 10 = 1500 \text{ N} \)
Volume of water displaced = Volume of stone \( = 0.05 \text{ m}^3 \)
Mass of water displaced \( = m' = V \times \rho' \)
\( \implies m' = 0.05 \times 1000 = 50 \text{ kg} \)
Upthrust \( = m'g = 50 \times 10 = 500 \text{ N} \)
Force required to lift the stone = Actual weight of stone - upthrust
\( \implies \text{Force} = 1500 - 500 = 1000 \text{ N} \)

In simple words: Lifting a stone underwater is easier because the water helps push it up. This upward help is the "upthrust." The force you need to apply is just the leftover weight after water does its part.

📝 Teacher's Note: Use the analogy of a friend helping you lift a heavy box—the total effort needed is reduced by the amount of help provided.

🎯 Exam Tip: The "force required to lift" is exactly equal to the "apparent weight" of the object in that fluid.

Question 3. A solid of area of cross-section \( 0.004 \text{ m}^2 \) and length \( 0.60 \text{ m} \) is completely immersed in water of density \( 1000 \text{ kg m}^{-3} \). Calculate :
1. Wt of solid in SI system
2. Upthrust acting on the solid in SI system.
3. Apparent weight of solid in water.
4. Apparent weight of solid in brine solution of density \( 1050 \text{ kg m}^{-3} \).
\( [\text{Take } g = 10 \text{ N/kg; Density of solid } = 7200 \text{ kg m}^{-3}] \)

Answer: Area of cross-section of solid \( = A = 0.004 \text{ m}^2 \)
Length of the solid \( = l = 0.60 \text{ m} \)
Density of water \( (\rho') = 1000 \text{ kg m}^{-3} \)
Acceleration due to gravity \( = g = 10 \text{ m s}^{-2} \)
Density of solid \( (\rho) = 7200 \text{ kg m}^{-3} \)

1. Volume of solid \( = V = A \times l \)
\( \implies V = 0.004 \times 0.60 = 0.0024 \text{ m}^3 \)
Mass of solid \( = m = V \times \rho \)
\( \implies m = 0.0024 \times 7200 = 17.28 \text{ kg} \)
Weight of the solid \( = mg = 17.28 \times 10 = 172.8 \text{ N} \)

2. Volume of water displaced = Volume of solid \( = 0.0024 \text{ m}^3 \)
Mass of water displaced \( = m' = V \times \rho' \)
\( \implies m' = 0.0024 \times 1000 = 2.4 \text{ kg} \)
Upthrust = Weight of water displaced \( = m'g = 2.4 \times 10 = 24 \text{ N} \)

3. Apparent weight of solid in water = Actual weight of solid - upthrust
\( \implies \text{Apparent weight} = 172.8 - 24 = 148.8 \text{ N} \)

4. Density of brine solution \( (\rho_b) = 1050 \text{ kg m}^{-3} \)
Mass of brine solution displaced \( = m_b = V \times \rho_b \)
\( \implies m_b = 0.0024 \times 1050 = 2.52 \text{ kg} \)
Upthrust acting on solid in brine solution = Weight of brine solution displaced \( = m_b g \)
\( \implies \text{Upthrust} = 2.52 \times 10 = 25.2 \text{ N} \)
Apparent weight of solid in brine solution = Actual weight - Upthrust
\( \implies \text{Apparent weight} = 172.8 - 25.2 = 147.6 \text{ N} \)

In simple words: This problem shows that the saltier (denser) the water, the harder it pushes up on the object. That's why things feel even lighter in salt water than in fresh water.

📝 Teacher's Note: This question is excellent for comparing fresh water and brine. Highlight that higher density of liquid results in higher upthrust and lower apparent weight.

🎯 Exam Tip: When a question gives length and area, first find the volume using \( V = \text{Area} \times \text{Length} \) before using Archimedes' principle formulas.

Practice Problem 2:

Question 1. A solid of density \( 7600 \text{ kg m}^{-3} \) is found to weigh 0.950 kgf in air. If 4/5 volume of solid is completely immersed in a solution of density \( 900 \text{ kg m}^{-3} \), find the apparent weight of solid in liquid.

Answer: Weight of solid in air = 0.950 kgf
\( \therefore \text{Mass of solid in air } (m) = 0.950 \text{ kg} \)
Density of solid \( (\rho) = 7600 \text{ kg m}^{-3} \)
Volume of solid \( (V) = \frac{m}{\rho} \)
\( \implies V = \frac{0.950}{7600} = 0.000125 \text{ m}^3 \)
Density of solution \( (\rho') = 900 \text{ kg m}^{-3} \)
\( \because \frac{4}{5} \text{ volume of solid is completely immersed in the given solution} \)
\( \therefore \text{Volume of solution displaced } (V') = \frac{4}{5} \times \text{Volume of solid} \)
\( \implies V' = \frac{4}{5} \times 0.000125 = 0.0001 \text{ m}^3 \)
Mass of solution displaced \( = m' = V' \times \rho' \)
\( \implies m' = 0.0001 \times 900 = 0.09 \text{ kg} \)
Upthrust = Weight of solution displaced \( = m'g \)
\( \implies \text{Upthrust} = 0.09 \text{ kgf} \)
Apparent weight of solid in liquid = Actual weight - Upthrust
\( \implies \text{Apparent weight} = 0.950 - 0.09 = 0.860 \text{ kgf} \)

In simple words: Even if only part of an object is underwater, it still gets pushed up. Here, 80% (or 4/5) of the object was submerged, so we only calculated the upward push for that part.

📝 Teacher's Note: Clarify the unit 'kgf' (kilogram-force). It is a convenient unit where 1 kg mass weighs 1 kgf. This simplifies calculations as \( g \) is not numerically required until you want the answer in Newtons.

🎯 Exam Tip: Watch for fractions! When a solid is partially immersed, use only the fraction of the volume that is *inside* the liquid for upthrust calculations.

Question 2. A glass cylinder of length \( 12 \times 10^{-2} \text{ m} \) and area of cross-section \( 5 \times 10^{-4} \text{ m}^2 \) has a density of \( 2500 \text{ kg m}^{-3} \). It is immersed in a liquid of density \( 1500 \text{ kg m}^{-3} \), such that 3/8 of its length is above liquid. Find the apparent weight of glass cylinder in newtons.

Answer: Length of glass cylinder \( (l) = 12 \times 10^{-2} \text{ m} \)
Area of cross-section \( (A) = 5 \times 10^{-4} \text{ m}^2 \)
Volume of glass cylinder \( (V) = A \times l \)
\( \implies V = (5 \times 10^{-4}) \times (12 \times 10^{-2}) = 60 \times 10^{-6} = 0.00006 \text{ m}^3 \)
Acceleration due to gravity \( (g) = 9.8 \text{ m/s}^2 \)
Density of glass cylinder \( (\rho) = 2500 \text{ kg m}^{-3} \)
Density of liquid \( (\rho') = 1500 \text{ kg m}^{-3} \)
\( \because \frac{3}{8} \text{ length of glass cylinder is above the liquid} \)
\( \therefore \text{Length of glass cylinder inside the liquid} = 1 - \frac{3}{8} = \frac{5}{8} l \)
\( \therefore \text{Volume of liquid displaced by glass cylinder } (V') = \frac{5}{8} \times \text{Volume of glass cylinder} \)
\( \implies V' = \frac{5}{8} \times 0.00006 = 0.0000375 \text{ m}^3 \)
Mass of glass cylinder \( = m = V \times \rho \)
\( \implies m = 0.00006 \times 2500 = 0.15 \text{ kg} \)
Weight of glass cylinder \( = mg = 0.15 \times 10 = 1.5 \text{ N} \)
Mass of liquid displaced \( = m' = V' \times \rho' \)
\( \implies m' = 0.0000375 \times 1500 = 0.05625 \text{ kg} \)
Upthrust = Weight of liquid displaced \( = m'g = 0.05625 \times 10 = 0.5625 \text{ N} \)
Apparent weight of glass cylinder in liquid = Actual weight - Upthrust
\( \implies \text{Apparent weight} = 1.5 - 0.5625 = 0.9375 \text{ N} \)

In simple words: The cylinder has a total height, but only 5/8 of it is "doing the pushing" in the liquid. We calculate the real weight and the upward push of that submerged part to find how heavy it feels.

📝 Teacher's Note: Help students with scientific notation calculations. Practicing powers of 10 is essential for these units.

🎯 Exam Tip: "3/8 length above" means "5/8 length below." Always use the portion *below* the surface for upthrust.

Practice Problems 3:

Question 1. A solid weighs 0.08 kgf in air and 0.065 kgf in water. Find
(1) R.D. of solid
(2) Density of solid in SI system. \( [\text{Density of water } = 1000 \text{ kg m}^{-3}] \)

Answer: Weight of solid in air \( (W_{\text{air}}) = 0.08 \text{ kgf} \)
Weight of solid in water \( (W_{\text{water}}) = 0.065 \text{ kgf} \)
Density of water \( = 1000 \text{ kg m}^{-3} \)

(1) Relative density (R.D.) of solid = \( \frac{\text{Weight of solid in air}}{\text{Weight of solid in air} - \text{Weight of solid in water}} \)
\( \implies \text{R.D.} = \frac{0.08}{0.08 - 0.065} \)
\( \implies \text{R.D.} = \frac{0.08}{0.015} = 5.3333 \)

(2) R.D. of solid \( = \frac{\text{Density of solid}}{\text{Density of water}} \)
\( \implies 5.3333 = \frac{\text{Density of solid}}{1000} \)
\( \implies \text{Density of solid} = 5.3333 \times 1000 = 5333.3 \text{ kg m}^{-3} \)

In simple words: Relative Density is a comparison of how heavy an object is compared to water. We find how much weight the object "lost" in water to see how it compares.

📝 Teacher's Note: Explain that Relative Density has no units because it is a ratio of two similar quantities. This is a very common conceptual question.

🎯 Exam Tip: The denominator "Weight in air - Weight in water" is the loss in weight, which is equal to the upthrust.

Question 2. A solid of R.D. = 2.5 is found to weigh 0.120 kgf in water. Find the wt. of solid in air.

Answer: Relative density of solid \( (\text{R.D.}) = 2.5 \)
Weight of solid in water \( (W') = 0.120 \text{ kgf} \)
Weight of solid in air \( (W) = ? \)
\( \text{R.D.} = \frac{\text{Weight of solid in air}}{\text{Weight of solid in air} - \text{Weight of solid in water}} \)
\( \implies \text{R.D.} = \frac{W}{W - W'} \)
\( \implies 2.5 = \frac{W}{W - 0.120} \)
\( \implies 2.5W - 2.5 \times 0.120 = W \)
\( \implies 2.5W - 0.3 = W \)
\( \implies 2.5W - W = 0.3 \)
\( \implies 1.5W = 0.3 \)
\( \implies W = \frac{0.3}{1.5} = 0.20 \text{ kgf} \)
So, weight of solid in air = 0.20 kgf.

In simple words: We know the object is 2.5 times denser than water. Using algebra, we work backwards from its underwater weight to find out what it weighs in the open air.

📝 Teacher's Note: Practice rearranging this specific R.D. formula as it involves basic linear equations which many students find tricky.

🎯 Exam Tip: If the R.D. is greater than 1, the weight in air *must* be greater than the weight in water. Use this to check your final answer.

Question 3. A solid of R.D. 4.2 is found to weigh 0.200 kgf in air. Find its apparent weight in water.

Answer: Relative density of solid \( (\text{R.D.}) = 4.2 \)
Weight of solid in air \( (W) = 0.200 \text{ kgf} \)
\( \text{R.D.} = \frac{\text{Weight of solid in air}}{\text{Weight of solid in air} - \text{Weight of solid in water}} \)
Also, \( \text{Weight of solid in air} - \text{Weight of solid in water} = \text{Upthrust} \)
\( \implies \text{R.D.} = \frac{\text{Weight of solid in air}}{\text{Upthrust}} \)
\( \implies 4.2 = \frac{0.200}{\text{Upthrust}} \)
\( \implies \text{Upthrust} = \frac{0.200}{4.2} = 0.0476 \text{ kgf} \)
So, apparent weight of solid in water = Weight of solid in air - Upthrust
\( \implies \text{Apparent weight} = 0.200 - 0.0476 = 0.1524 \approx 0.15 \text{ kgf} \)

In simple words: We use the density comparison to find the upward "push" (upthrust) water gives the object. Subtracting that push from the original weight gives the new, lighter weight.

📝 Teacher's Note: This is a variation of the R.D. formula. Showing that \( \text{R.D.} = \text{Weight in air} / \text{Upthrust} \) is a very useful shortcut for students.

🎯 Exam Tip: For small decimals, keep up to 3 or 4 decimal places during calculations to ensure the final rounded answer is accurate.

Practice Problems 4:

Question 1. A sinker is found to weigh 56.7 gf in water. When the sinker is tied to a cork of weight 6 gf, the combination is found to weigh 40.5 gf in water. Calculate R.D. of cork.

Answer: Weight of sinker in water = 56.7 gf
Weight of cork = 6 gf
Weight of sinker in water + Weight of cork in air = \( 56.7 + 6 = 62.7 \text{ gf} \) …(1)
Weight of cork in water + Weight of sinker in water = 40.5 gf …(2)
Subtract eq. (1) from eq. (2)
Weight of cork in air - Weight of cork in water \( = 62.7 - 40.5 = 22.2 \text{ gf} \)
\( \therefore \text{R.D. of cork} = \frac{\text{Weight of cork in air}}{\text{Weight of cork in air} - \text{Weight of cork in water}} \)
\( \implies \text{R.D.} = \frac{6}{22.2} = 0.27 \)

In simple words: A cork floats, so we have to tie it to a heavy "sinker" to force it underwater. By measuring how much the total weight changes when both are underwater, we find the volume of the cork.

📝 Teacher's Note: This is the "Sinker Method" used for finding the R.D. of solids less dense than water. Explain that the difference in readings (22.2 gf) represents the massive upthrust acting on the light cork.

🎯 Exam Tip: Notice that for a floating object like cork, the R.D. must be less than 1. If your result is above 1, check your logic.

Question 2. A solid lighter than water is found to weigh 7.5 gf in air. When tied to a sinker the combination is found to weigh. If the sinker alone weighs 72.5 gf in water, find R.D. of solid.

Answer: Weight of solid in air = 7.5 gf
Weight of sinker in water = 72.5 gf
Weight of sinker in water + Weight of solid in air = \( 72.5 + 7.5 = 80.0 \text{ gf} \) …(1)
Weight of solid in water + Weight of sinker in water = 62.5 gf …(2)
Subtract eq. (1) from eq. (2)
Weight of solid in air - Weight of solid in water \( = 80 - 62.5 = 17.5 \text{ gf} \).
\( \therefore \text{R.D. of solid} = \frac{\text{Weight of solid in air}}{\text{Weight of solid in air} - \text{Weight of solid in water}} \)
\( \implies \text{R.D.} = \frac{7.5}{17.5} = 0.428 \)

In simple words: Just like the cork problem, we use a heavy partner to pull a light object underwater. The massive drop in the scale reading shows how strongly water wants to push the light object back up.

📝 Teacher's Note: Emphasize that the denominator "Weight in air - Weight in water" for the solid is purely the upthrust on that solid alone.

🎯 Exam Tip: The "weight of solid in water" for a floating object is technically a negative value, which is why the "weight in air - weight in water" result (17.5) is larger than the weight in air (7.5).

Practice Problems 5:

Question 1. An aluminium cube of side 5 cm and RD. 2.7 is suspended by a thread in alcohol of relative density 0.80. Find the tension in thread.

Answer: Side of an aluminium cube \( = l = 5 \text{ cm} \)
Volume of aluminium cube \( (V) = l^3 = (5)^3 = 125 \text{ cm}^3 \)
Relative density of aluminium \( (\text{R.D.}) = 2.7 \)
Relative density of alcohol \( (\text{R.D.}) = 0.80 \)
Density of water \( = 1 \text{ g cm}^{-3} \)
\( \text{R.D. of aluminium} = \frac{\text{Density of aluminium}}{\text{Density of water}} \)
\( \implies 2.7 = \frac{\text{Density of aluminium}}{1} \)
Density of aluminium \( (\rho) = 2.7 \text{ g cm}^{-3} \)
Mass of aluminium \( = V \times \rho \)
\( \implies m = 125 \times 2.7 = 337.5 \text{ g} \)
Weight of aluminium cube acting downwards = 337.5 gf
Volume of alcohol displaced = Volume of cube \( = V = 125 \text{ cm}^3 \)
\( \text{Now R.D. of alcohol} = \frac{\text{Density of alcohol}}{\text{Density of water}} \)
\( \implies 0.80 = \frac{\rho_{\text{alcohol}}}{1} \)
\( \implies \rho_{\text{alcohol}} = 0.80 \text{ g cm}^{-3} \)
Upthrust due to alcohol \( = V \times \rho_{\text{alcohol}} \times g \)
\( \implies \text{Upthrust} = 125 \times 0.80 \times g = 100 \text{ gf} \)
So tension in thread = Weight of aluminium cube - Upthrust
\( \implies \text{Tension} = 337.5 - 100 = 237.5 \text{ gf} \)

In simple words: The thread doesn't have to carry the full weight of the aluminium because the alcohol helps by pushing it up. We calculate the real weight and the upward push of alcohol; the thread only has to hold the remaining "tension."

📝 Teacher's Note: Tension in the string is always equal to the apparent weight of the object in the liquid. Draw a free body diagram with weight acting down and upthrust + tension acting up.

🎯 Exam Tip: When \( g \) is included in the density product \( (V \rho g) \), the unit becomes 'gf' (gram-force). This is a standard laboratory unit in many physics curricula.

Question 2. A cube of lead of side 8 cm and R.D. 10.6 is suspended from the hook of a spring balance. Find the reading of spring balance. The cube is now completely immersed in sugar solution of R.D. 1.4. Calculate the new reading of spring balance.

Answer: Length of side of cube \( (l) = 8 \text{ cm} \)
Volume of cube \( (V) = (8)^3 = 512 \text{ cm}^3 \)
Relative density of lead cube \( (\text{R.D.}) = 10.6 \)
Relative density of sugar solution \( (\text{R.D.}) = 1.4 \)
Density of water \( = 1 \text{ g cm}^{-3} \)
\( \text{R.D. of lead} = \frac{\text{Density of lead}}{\text{Density of water}} \)
\( \implies 10.6 = \frac{\rho_{\text{lead}}}{1} \)
\( \rho_{\text{lead}} = 10.6 \text{ g cm}^{-3} \)
Mass of lead \( = m = V \times \rho_{\text{lead}} \)
\( \implies m = 512 \times 10.6 = 5427.2 \text{ g} \)
Weight of lead cube \( = mg = 5427.2 \text{ gf} \)
Volume of sugar solution displaced = Volume of lead cube \( = V = 512 \text{ cm}^3 \)
\( \text{R.D. of sugar solution} = \frac{\text{Density of sugar solution}}{\text{Density of water}} \)
\( \implies 1.4 = \frac{\rho_{\text{sugar}}}{1} \)
\( \rho_{\text{sugar}} = 1.4 \text{ g cm}^{-3} \)
Upthrust due to sugar solution \( = V \times \rho_{\text{sugar}} \times g = 512 \times 1.4 \times g \)
\( \implies \text{Upthrust} = 716.8 \text{ gf} \)
Now reading of spring balance = Actual weight - Upthrust
\( \implies \text{Reading} = 5427.2 - 716.8 = 4710.4 \text{ gf} \)

In simple words: The spring balance initially feels the full weight of the lead. When lowered into the sugar solution, the sticky liquid pushes up on the cube, making the spring stretch less and giving a lower reading on the scale.

📝 Teacher's Note: Use the term "sugar solution" to introduce students to the idea that R.D. of solutions can be varied. Higher concentration of sugar means higher R.D. and more upthrust.

🎯 Exam Tip: Always state the "Initial Reading" (weight in air) and "New Reading" (weight in liquid) separately as requested by the two-part question.

Practice Problems 1:

Question 1. A hollow cylinder of copper of length 25 cm and area of cross-section \( 15 \text{ cm}^2 \), floats in water with 3/5 of its length inside water. Calculate :
(1) apparent density of hollow copper cylinder.
(2) wt. of cylinder.
(3) extra force required to completely submerge it in water.

Answer: (1) Length of hollow cylinder of copper \( = h_{cu} = l = 25 \text{ cm} \)
Length of hollow cylinder of copper inside water \( = h_{\text{water}} = \frac{3}{5} l \)
\( \implies h_{\text{water}} = \frac{3}{5} \times 25 = 15 \text{ cm} \)
Area of cross-section of hollow cylinder of copper \( = 15 \text{ cm}^2 \)
Density of water \( = \rho_{\text{water}} = 1 \text{ g cm}^{-3} \)
Apparent density of hollow copper cylinder \( = \rho_{cu} = ? \)
By law of flotation :
\( h_{cu} \times \rho_{cu} = h_{\text{water}} \times \rho_{\text{water}} \)
\( \implies 25 \times \rho_{cu} = 15 \times 1 \)
\( \implies \rho_{cu} = \frac{15}{25} = 0.6 \text{ g cm}^{-3} \)

(2) Volume of cylinder \( = V = A \times h_{cu} = 15 \times 25 = 375 \text{ cm}^3 \)
Mass of hollow cylinder of copper \( = V \times \rho_{cu} \)
\( \implies m = 375 \times 0.6 = 225 \text{ g} \)
Weight of hollow cylinder of copper \( = mg = 225 \text{ gf} \)

(3) Total upthrust when hollow copper cylinder is completely immersed in water \( = V \rho_{\text{water}} g \)
\( \implies \text{Total Upthrust} = 375 \times 1 \times g = 375 \text{ gf} \)
Extra force required to submerge complete the cylinder in water = Upthrust - down thrust (actual weight)
\( \implies \text{Force} = 375 - 225 = 150 \text{ gf} \)

In simple words: The cylinder floats because its average density (including the air inside) is less than water. To force it all the way down, you have to push with enough power to overcome the remaining upward push of the water.

📝 Teacher's Note: The "apparent density" of a hollow object is its total mass divided by its total volume. This explains why steel ships (high density) can float—their *average* density is low due to air spaces.

🎯 Exam Tip: The "Law of Flotation" is \( \frac{\text{immersed volume}}{\text{total volume}} = \frac{\text{density of solid}}{\text{density of liquid}} \). In cylinders of uniform area, this simplifies to the ratio of heights.

Question 2. A cork cut in the form of a cylinder floats in alcohol of density \( 0.8 \text{ g cm}^{-3} \), such that 3/7 of its length is outside alcohol. If the total length of cylinder is 35 cm and area of cross-section \( 25 \text{ cm}^2 \), calculate :
(1) density of cork
(2) wt. of cork
(3) extra force required to submerge it in alcohol

Answer: (1) Density of alcohol \( = \rho_{\text{alcohol}} = 0.8 \text{ g cm}^{-3} \)
Total length of cork cylinder \( = h_{\text{cork}} = 35 \text{ cm} \)
Area of cross-section of cork cylinder \( = A = 25 \text{ cm}^2 \)
\( \because \frac{3}{7} \text{ of the length of cork cylinder is outside the alcohol} \)
\( \therefore \text{Length of cork cylinder inside alcohol} = l \times (1 - \frac{3}{7}) \)
\( \implies h_{\text{alcohol}} = \frac{4}{7} \times 35 = 20 \text{ cm} \)
Density of water \( = \rho_{\text{water}} = 1 \text{ g cm}^{-3} \)
By law of flotation :
\( h_{\text{cork}} \times \rho_{\text{cork}} = h_{\text{alcohol}} \times \rho_{\text{alcohol}} \)
\( \implies 35 \times \rho_{\text{cork}} = 20 \times 0.8 \)
\( \implies \rho_{\text{cork}} = \frac{20 \times 0.8}{35} = 0.457 \text{ g cm}^{-3} \)

(2) Volume of cork \( = V = A \times h_{\text{cork}} = 25 \times 35 = 875 \text{ cm}^3 \)
Mass of cork \( = m = V \times \rho_{\text{cork}} \)
\( \implies m = 875 \times 0.457 = 399.8 \text{ g} \approx 400 \text{ g} \)
Wt. of cork \( = mg = 400 \times g = 400 \text{ gf} \)

(3) Total upthrust when cork is completely immersed in water \( = V \rho_{\text{alcohol}} \times g \)
\( \implies \text{Upthrust} = 875 \times 0.8 \times g = 700 \text{ g} = 700 \text{ gf} \)
Extra force required to submerge complete the cork in alcohol = Upthrust - down thrust
\( \implies \text{Force} = 700 - 400 = 300 \text{ gf} \)

In simple words: This is just like the copper cylinder problem but with alcohol. We find the cork's density first, then its weight, and finally the extra "push" needed to dunk it completely under the surface.

📝 Teacher's Note: Watch for the "outside" versus "inside" length. Upthrust is only generated by the part *inside* the liquid.

🎯 Exam Tip: Use "By law of flotation" as a heading for your work to show the examiner you are using the correct physical principle.

Practice Problems 2:

Question 1. A cylinder made of copper and aluminium floats in mercury of density \( 13.6 \text{ g cm}^{-3} \), such that 0.26th part of it is below mercury. Find the density of solid.

Answer: Density of mercury \( = \rho_{\text{Hg}} = 13.6 \text{ g cm}^{-3} \)
Density of solid cylinder \( = \rho_{\text{solid}} = ? \)
0.26th part of the cylinder is below mercury
Let \( V_{\text{solid}} = \) Volume of solid cylinder
Volume of mercury displaced by immersed part of the solid cylinder \( = V_{\text{Hg}} = 0.26 V_{\text{solid}} \)
By law of flotation :
Weight of the solid cylinder = Weight of mercury displaced by immersed part of solid cylinder
\( V_{\text{solid}} \times \rho_{\text{solid}} \times g = V_{\text{Hg}} \times \rho_{\text{Hg}} \times g \)
\( \implies V_{\text{solid}} \times \rho_{\text{solid}} = 0.26 V_{\text{solid}} \times 13.6 \)
\( \implies \rho_{\text{solid}} = 0.26 \times 13.6 = 3.536 \text{ g cm}^{-3} \)
So, density of solid \( = 3.536 \text{ g cm}^{-3} \)

In simple words: Mercury is so heavy that only about a quarter of this metal cylinder needs to go underwater to support its whole weight. We multiply that fraction by mercury's density to find the object's density.

📝 Teacher's Note: This is a simple application of the flotation ratio \( \rho_s / \rho_l = V_{\text{sub}} / V_{\text{total}} \). It shows why even heavy metals float on mercury.

🎯 Exam Tip: The volumes cancel out from both sides, so you don't need to know the actual size of the cylinder to solve this.

Question 2. An iceberg floats in sea water of density \( 1.17 \text{ g cm}^{-3} \), such that 2/9 of its volume is above sea water. Find the density of iceberg.

Answer: Density of sea water \( = \rho_w = 1.17 \text{ g cm}^{-3} \)
Density of solid iceberg \( = \rho_i = ? \)
\( \because \frac{2}{9} \text{th volume of iceberg is above the sea water} \)
\( \therefore \text{Volume of iceberg inside water } (V_i) = (1 - \frac{2}{9}) V \)
Where \( V = \) Total volume of iceberg
\( \implies V_i = \frac{7}{9} V \)
\( \implies \text{Volume of sea water displaced by immersed part of the iceberg} = V_w = \frac{7}{9} V \)
By law of flotation :
Weight of iceberg = Weight of sea water displaced by iceberg
\( V \times \rho_i \times g = V_w \times \rho_w \times g \)
\( \implies V \times \rho_i = \frac{7}{9} V \times 1.17 \)
\( \implies \rho_i = \frac{7}{9} \times 1.17 = 0.91 \text{ g cm}^{-3} \)
\( \implies \text{Density of iceberg} = 0.91 \text{ g cm}^{-3} \)

In simple words: Most of an iceberg is hidden underwater (7/9ths of it!). We use the density of sea water and the hidden fraction to find out the density of the ice itself.

📝 Teacher's Note: This is the classic "tip of the iceberg" problem. It's a great real-world example of flotation laws.

🎯 Exam Tip: Be careful to subtract the fraction "above" from 1 to get the fraction "below" before starting the calculation.

Practice Problems 3:

Question 1. A wooden block floats in alcohol with 3/8 of its length above alcohol. If it is made to float in water, what fraction of its length is above water? Density of alcohol is \( 0.80 \text{ g cm}^{-3} \).

Answer: Let length of wooden block \( = x \)
\( \therefore \text{Length of the block above alcohol} = \frac{3}{8} x \)
Length of the block below alcohol \( = x - \frac{3}{8} x = \frac{5}{8} x \)
Density of water \( = \rho_w = 1 \text{ g cm}^{-3} \)
Density of alcohol \( = \rho_{\text{alcohol}} = 0.80 \text{ g cm}^{-3} \)
By the law of flotation :
\( h_{\text{block}} \times \rho_{\text{block}} = h_{\text{alcohol}} \times \rho_{\text{alcohol}} \)
\( \implies x \times \rho_{\text{block}} = \frac{5x}{8} \times 0.80 \)
\( \implies \rho_{\text{block}} = \frac{5}{8} \times 0.80 = 0.5 \text{ g cm}^{-3} \)
When block is floating in water :
By law of flotation :
\( h_{\text{block}} \times \rho_{\text{block}} = h_{\text{water}} \times \rho_w \)
\( \implies x \times 0.5 = h_{\text{water}} \times 1 \)
\( \implies h_{\text{water}} = 0.5x \text{ or } \frac{1}{2} x \)
So, length of block below water \( = \frac{1}{2} x \)
Length of block above water \( = x - \frac{1}{2} x = \frac{x}{2} \)
Fraction of the block above the water \( = \frac{x/2}{x} = \frac{1}{2} \text{ part} \)
\( \implies \frac{1}{2} \text{th part of wooden block is above the water.} \)

In simple words: First we use alcohol to find out the wood's density. Then, we use that density to see how it would sit in water. Since water is heavier than alcohol, it pushes the wood higher up.

📝 Teacher's Note: This is a two-step flotation problem. Step 1 finds the solid's density, Step 2 finds its new submerged volume in a different liquid.

🎯 Exam Tip: Always solve for the density of the solid first in problems involving two different liquids.

Question 2. A hollow metal cylinder of length 10 cm floats in alcohol of density \( 0.80 \text{ g cm}^{-3} \), with 1 cm of its length above it. What length of cylinder will be above copper sulphate solution of density \( 1.25 \text{ g cm}^{-3} \)?

Answer: Length of hollow metal cylinder \( = x = 10 \text{ cm} \)
1 cm length of cylinder is above the alcohol
\( \therefore \text{Length of the cylinder below alcohol} = (10-1) = 9 \text{ cm} \)
Density of alcohol \( = \rho_{\text{alcohol}} = 0.80 \text{ g cm}^{-3} \)
Density of copper sulphate solution \( = \rho_{\text{CuSO4}} = 1.25 \text{ g cm}^{-3} \)
When block floats in alcohol By the law of flotation :
\( h_{\text{block}} \times \rho_{\text{block}} = h_{\text{alcohol}} \times \rho_{\text{alcohol}} \)
\( \implies 10 \times \rho_{\text{block}} = 9 \times 0.80 \)
\( \implies \rho_{\text{block}} = \frac{9 \times 0.80}{10} = 0.72 \text{ g cm}^{-3} \)
When block floats in copper sulphate solution :
By law of flotation :
\( h_{\text{block}} \times \rho_{\text{block}} = h_{\text{CuSO4}} \times \rho_{\text{CuSO4}} \)
\( \implies 10 \times 0.72 = h_{\text{CuSO4}} \times 1.25 \)
\( \implies h_{\text{CuSO4}} = \frac{7.2}{1.25} = 5.76 \text{ cm} \)
Length of metal block below copper sulphate solution = 5.76 cm
So, length of metal block above copper sulphate solution \( = 10 – 5.76 = 4.24 \text{ cm} \)

In simple words: The CuSO4 solution is much denser and "thicker" than alcohol, so it pushes the cylinder much higher out of the liquid (4.24 cm out instead of just 1 cm).

📝 Teacher's Note: Remind students that the denser the liquid, the less volume needs to be submerged to balance the weight of the object.

🎯 Exam Tip: Be sure to finish the problem! The calculation often finds the part *below* the surface, but the question asks for the part *above*.

Practice Problems 4:

Question 1. What fraction of an iceberg of density \( 910 \text{ kg m}^{-3} \) will be above the surface of sea water of density \( 1170 \text{ kg m}^{-3} \)?

Answer: Let volume of iceberg \( = V_i = x \)
Volume of iceberg inside sea water \( = V_w \)
Density of iceberg \( = \rho_i = 910 \text{ kg m}^{-3} \)
Density of sea water \( = \rho_w = 1170 \text{ kg m}^{-3} \)
By law of flotation :
Weight of iceberg = Weight of sea water displaced by the iceberg
\( V_i \times \rho_i \times g = V_w \times \rho_w \times g \)
\( \implies x \times 910 = V_w \times 1170 \)
\( \implies V_w = x \times \frac{910}{1170} \)
\( \implies V_w = \frac{7}{9} x = \frac{7}{9} \text{th of total volume} \)
Volume of iceberg inside sea water = Volume of sea water displaced by iceberg \( = \frac{7}{9} x \)
Volume of iceberg above the sea water \( = x - \frac{7}{9} x \)
\( \implies (1 - \frac{7}{9}) x = \frac{9 - 7}{9} x = \frac{2}{9} x \)
Fraction of iceberg above sea water \( = \frac{2}{9} \times \frac{1}{x} = \frac{2}{9} \text{ part} \)
\( \implies \frac{2}{9} \text{th part of iceberg is above the sea water.} \)

In simple words: This shows the math behind why icebergs are dangerous—almost 80% (7/9) of the giant block of ice is hidden under the water.

📝 Teacher's Note: Ensure students understand that "fraction above" = \( 1 - (\text{density of solid} / \text{density of liquid}) \).

🎯 Exam Tip: Simplifying the fraction \( 910 / 1170 \) into \( 7 / 9 \) immediately makes the rest of the subtraction much easier.

Question 2. What fraction of metal of density \( 3400 \text{ kg m}^{-3} \) will be above the surface of mercury of density \( 13600 \text{ kg m}^{-3} \), while floating in mercury?

Answer: Density of metal \( (\rho_m) = 3400 \text{ kg m}^{-3} \)
Density of mercury \( (\rho_{\text{Hg}}) = 13600 \text{ kg m}^{-3} \)
Let volume of metal \( = x \)
and volume of metal inside mercury \( = y \)
By law of flotation :
Weight of mercury displaced by metal = wt. of metal
\( V_{\text{Hg}} \times \rho_{\text{Hg}} \times g = V_{\text{metal}} \times \rho_{\text{metal}} \times g \)
\( \implies y \times 13600 = x \times 3400 \)
\( \implies y = \frac{3400}{13600} x = \frac{1}{4} x \)
Volume of metal inside mercury = Volume of mercury displaced \( = \frac{1}{4} \times \text{Volume of metal} \)
Volume of metal above the surface of mercury \( = x - y \)
\( \implies = x - \frac{1}{4} x = (1 - \frac{1}{4}) x = \frac{4-1}{4} x = \frac{3}{4} x \)
Fraction of metal above surface of mercury \( = \frac{3}{4} \times \frac{1}{x} = \frac{3}{4} \)
\( \implies \frac{3}{4} \text{th part of metal lies above the surface of mercury.} \)

In simple words: Mercury is so incredibly heavy that it can hold up a metal block while only touching the bottom 25% of it. Most of the metal (75%) stays dry above the surface.

📝 Teacher's Note: Mercury has such high density that it can float even heavy metals like iron or copper easily.

🎯 Exam Tip: Always define your variables (\( x \) for total volume, \( y \) for submerged volume) at the start of your answer.

Practice Problems 5:

Question 1. A balloon of volume \( 1000 \text{ m}^3 \) is filled with a mixture of hydrogen and helium of density \( 0.32 \text{ kg m}^{-3} \). If the fabric of balloon weighs 40 kgf and the density of cold air is \( 1.32 \text{ kg m}^{-3} \), find the tension in the tope, which is holding the balloon to ground.

Answer: Volume of balloon \( (V) = 1000 \text{ m}^3 \)
Density of mixture of hydrogen and helium \( (\rho) = 0.32 \text{ kg m}^{-3} \)
Density weight of empty balloon \( = 40 \text{ kgf} \)
Density of cold air \( (\rho') = 1.32 \text{ kg m}^{-3} \)
Volume of balloon = Volume of mixture of hydrogen and helium gas = Volume of cold air displaced by balloon \( = V = 1000 \text{ m}^3 \)
Weight of mixture of hydrogen and helium gas in balloon \( = V \rho g = 1000 \times 0.32 \times g = 320 \text{ kgf} \)
Down thrust = Weight of empty balloon + Weight of mixture of hydrogen and helium gas
\( \implies \text{Down thrust} = 40 + 320 = 360 \text{ kgf} \)
Upthrust = Weight of cold air displaced by balloon \( = V \rho' g \)
\( \implies \text{Upthrust} = 1000 \times 1.32 \times g = 1320 \text{ kgf} \)
Tension in the rope = Upthrust - downthrust
\( \implies \text{Tension} = 1320 - 360 = 960 \text{ kgf} \)

In simple words: The air pushes the balloon up with a force of 1320. The balloon's own weight and gas pull it down with 360. The rope has to fight the difference (960) to keep the balloon from flying away.

📝 Teacher's Note: Explain that "down thrust" is the total weight acting downwards (fabric + gas). The "upthrust" is the weight of the air that the balloon replaces.

🎯 Exam Tip: The tension is positive because the upthrust is greater than the total weight. If the weight was greater, the tension would be zero as the balloon wouldn't even lift off.

Question 2. A balloon of volume \( 800 \text{ cm}^3 \) is filled with hydrogen gas of density \( 9 \times 10^{-5} \text{ g cm}^{-3} \). If the empty balloon weighs 0.3 gf and density of air is \( 1.3 \times 10^{-3} \text{ g cm}^{-3} \), calculate the lifting power of balloon.

Answer: Volume of balloon \( (V) = 800 \text{ cm}^3 \)
Density of hydrogen gas \( (\rho_H) = 9 \times 10^{-5} \text{ g cm}^{-3} \)
Weight of empty balloon \( = 0.3 \text{ gf} \)
Density of air \( (\rho_a) = 1.3 \times 10^{-3} \text{ g cm}^{-3} \)
Weight of hydrogen gas in balloon \( = V \rho_H g = 800 \times 9 \times 10^{-5} \times g = 72 \times 10^{-3} \text{ gf} = 0.072 \text{ gf} \)
Volume of balloon = Volume of hydrogen gas in the balloon = Volume of air displaced by balloon.
Downthrust = Wt. of empty balloon + wt. of hydrogen gas in balloon
\( \implies \text{Downthrust} = 0.3 + 0.072 = 0.372 \text{ gf} \)
Upthrust = Wt. of air displaced by balloon \( = V \rho_a g = 800 \times 1.3 \times 10^{-3} \times g = 1.04 \text{ gf} \)
Lifting power of balloon = Upthrust - Down thrust
\( \implies \text{Lifting power} = 1.04 - 0.372 = 0.668 \text{ gf} \)

In simple words: Lifting power is just how much "extra" upward force the balloon has after lifting its own skin and gas. It's the maximum weight it could carry up into the sky.

📝 Teacher's Note: This uses CGS units. Be very careful with the powers of 10 (\( 10^{-5} \) and \( 10^{-3} \)).

🎯 Exam Tip: "Lifting power" is the same as the "net upward force" or "net upthrust."

Question 3. A balloon of volume \( 120 \text{ m}^3 \) is filled with hot air, of density \( 0.38 \text{ kg m}^{-3} \). If the fabric of balloon weighs 12 kg, such that an additional equipment of wt. \( x \) is attached to it, calculate the magnitude of \( x \). Density of cold air is \( 1.30 \text{ kg m}^{-3} \).

Answer: Volume of balloon \( (V) = 120 \text{ m}^3 \)
Density of hot air \( (\rho_{\text{hot air}}) = 0.38 \text{ kg m}^{-3} \)
Mass of empty balloon = 12 kg \( \implies \) Weight of the empty balloon = 12 kgf
Weight of the additional equipment attached with the balloon \( = x \text{ kgf} \)
Density of cold air \( (\rho_{\text{coldair}}) = 1.30 \text{ kg m}^{-3} \)
Volume of cold air displaced by balloon \( = V = 120 \text{ m}^3 \)
Weight of hot air \( = V \rho_{\text{hotair}} g = 120 \times 0.38 \times g = 45.6 \text{ kgf} \)
Weight of empty balloon + Weight of hot air inside the balloon + Weight of equipment = Downthrust
\( \implies 12 + 45.6 + x = \text{Downthrust} \)
\( \implies \text{Downthrust} = 57.6 + x \)
Upthrust = Weight of cold air displaced by balloon \( = V \rho_{\text{coldair}} g = 120 \times 1.30 \times g = 156 \text{ kgf} \)
By law of flotation :
Downthrust = Upthrust
\( \implies 57.6 + x = 156 \)
\( \implies x = 156 - 57.6 = 98.4 \text{ kgf} \)

In simple words: For the balloon to just sit there floating (equilibrium), the total weight pulling down must exactly equal the air pushing up. We add up all the weights (fabric, gas, and mystery gear) and set them equal to the 156 kgf push from the air.

📝 Teacher's Note: This is an "equilibrium" problem. "Just floats" means upthrust equals downthrust exactly.

🎯 Exam Tip: In hot air balloon problems, don't forget to include the weight of the air *inside* the balloon in the total downward weight.

Practice Problems 6:

Question 1. A test tube weighing 17 gf, floats in alcohol to the level P. When the test tube is made to float in water to the level P, 3 gf of the lead shots are added in it. find the R.D. of alcohol.

Answer: When tube floats in alcohol :
Weight of test tube = 17 gf
By law of flotation :
Weight of alcohol displaced by test tube = Weight of test tube = 17 gf
When tube floats in water :
When test tube is made to float in water to the same level, as in alcohol then 3g lead stones are added in it.
\( \therefore \text{Weight of test tube} = 17 \text{ gf} + 3 \text{ gf} = 20 \text{ gf} \)
Weight of water displaced by test tube = 20 gf
Volume of alcohol displaced = Volume of water displaced (since it floats to the same mark P)
\( \therefore \text{R.D. of alcohol} = \frac{\text{Weight of alcohol displaced}}{\text{Weight of equal volume of water displaced}} \)
\( \implies \text{R.D.} = \frac{17}{20} = 0.85 \)

In simple words: The tube sinks to the same depth in both liquids. Because alcohol is lighter, the tube only needed its own 17g weight to push the alcohol aside. In water, it needed an extra 3g to reach that same depth. Dividing the two weights gives the ratio of their densities.

📝 Teacher's Note: This setup is how a constant immersion hydrometer works. It uses different weights to keep the volume submerged constant.

🎯 Exam Tip: The key phrase is "floats... to the level P." This tells you that the displaced volumes are exactly equal.

Question 2. A test tube loaded with lead shots, weighs 150 gf and floats upto the mark X in water. The test tube is then made to float in alcohol. It is found that 27 gf of lead shots have to be removed, so as to float it to level X. Find R.D. of alcohol.

Answer: When tube floats in water :
Weight of test tube = 150 gf
By law of flotation:
Weight of water displaced = Weight of test tube = 150 gf
When test tube floats in alcohol :
When test tube is made to float in alcohol, then 27 gf of lead shots have to removed, so that it can float upto the same level as in water.
\( \therefore \text{Weight of test tube in alcohol} = 150 - 27 = 123 \text{ gf} \)
By law of flotation:
Weight of alcohol displaced by test tube = Weight of test tube in alcohol = 123 gf
As volume of alcohol displaced = Volume of water displaced
\( \therefore \text{R.D. of alcohol} = \frac{\text{Weight of alcohol displaced}}{\text{Weight of equal volume of water displaced}} \)
\( \implies \text{R.D.} = \frac{123}{150} = 0.82 \)

In simple words: Alcohol is lighter than water, so it can't hold up 150g at the "X" mark. We have to take out 27g of lead to make the tube light enough for the alcohol to support it at that same mark.

📝 Teacher's Note: This is the inverse logic of the previous question. You remove weight for lighter liquids and add weight for denser liquids to maintain the same submerged volume.

🎯 Exam Tip: Always show the subtraction (\( 150 - 27 \)) clearly to get marks for identifying the weight of the displaced liquid.

QUESTIONS BASED ON ICSE EXAMINATIONS

(A) Objective Questions

Question 1. The force experienced by a body when partially or fully immersed in water is called:
(a) aparent weight
(b) upthrust
(c) down thrust
(d) none of these

Answer: (b) upthrust
In simple words: This is the "buoyant force" or the upward push that water gives to anything dipped into it.

📝 Teacher's Note: Help students distinguish between "upthrust" (the force) and "apparent weight" (the resulting feeling of being lighter).

🎯 Exam Tip: "Upthrust" and "Buoyant Force" are interchangeable terms in exams.

Question 2. When a body is floating in a liquid :
(a) The weight of the body is less than the upthrust due to immersed part of the body
(b) The weight of body is more than the upthrust due to the immersed part of the body
(c) The weight of body is equal to the upthrust due to the immersed part of the body
(d) none of the above

Answer: (c) The weight of body is equal to the upthrust due to the immersed part of the body
In simple words: To float perfectly still on the surface, the weight pulling down and the water pushing up must be perfectly balanced.

📝 Teacher's Note: This is the fundamental "Principle of Flotation." If the forces were unequal, the body would either sink or accelerate upwards.

🎯 Exam Tip: Look for the word "equal" in flotation multiple-choice questions.

Question 3. With the increase in the density of the fluid, the upthrust experienced by a body immersed in it :
(a) decreases
(b) increases
(c) remains same
(d) none of these

Answer: (b) increases
In simple words: Denser liquids like salt water or mercury are "stronger" and push up with much more force than lighter liquids like oil.

📝 Teacher's Note: Use the formula \( F = V \rho g \). Since \( F \) is directly proportional to \( \rho \), a higher density means a higher force.

🎯 Exam Tip: This explains why it is easier to swim in the Dead Sea (high density salt water) than in a regular pool.

Question 4. The apparent weight of a body in a fluid is :
(a) equal to weight of fluid displaced
(b) volume of fluid displaced
(c) difference between its weight in air and weight of fluid displaced
(d) none of the options
Answer: (c) difference between its weight in air and weight of fluid displaced
In simple words: When you are in water, you feel lighter because the water is pushing you up. Your "new" weight is your real weight minus the strength of that upward push from the water.

📝 Teacher's Note: Use the term "Loss in Weight" interchangeably with "Weight of fluid displaced" to help students connect Archimedes' Principle to the concept of apparent weight.

🎯 Exam Tip: Remember the formula: \( \text{Apparent Weight} = W_{\text{air}} - \text{Upthrust} \). The upthrust is equal to the weight of the fluid displaced.

Question 5. The phenomenon due to which a solid experiences upward force when immersed in water is called :
(a) floatation
(b) buoyancy
(c) density
(d) none of the options
Answer: (b) buoyancy
In simple words: Buoyancy is like an invisible hand in the water that tries to lift objects up. It is the reason why heavy ships stay afloat and why you feel lighter in a swimming pool.

📝 Teacher's Note: Differentiate between "Buoyant Force" (the force itself) and "Buoyancy" (the property or phenomenon). Students often use these terms loosely.

🎯 Exam Tip: Buoyancy is the property of the *fluid*, while upthrust is the *force* exerted by the fluid. If the question asks for the "phenomenon," the answer is buoyancy.

Question 6. When an object sinks in a liquid, its :
(a) buoyant force is more than the weight of object
(b) buoyant force is less than the weight of object
(c) buoyant force is equal to the weight of the object
(d) none of the options
Answer: (b) buoyant force is less than the weight of object
In simple words: Imagine a tug-of-war between the object's weight pulling it down and the water's push (buoyancy) lifting it up. If the weight is stronger, the object loses and sinks to the bottom.

📝 Teacher's Note: Use a simple diagram of a sinking stone with a long arrow pointing down (weight) and a short arrow pointing up (upthrust) to illustrate this balance of forces.

🎯 Exam Tip: For an object to sink, its density must be greater than the density of the liquid, which results in its weight being greater than the maximum upthrust.

Question 7. The SI unit of density is :
(a) \( \text{g cm}^{-3} \)
(b) \( \text{kg cm}^{-3} \)
(c) \( \text{kg m}^{-3} \)
(d) \( \text{g m}^{-3} \)
Answer: (c) \( \text{kg m}^{-3} \)
In simple words: Density tells us how much "stuff" (mass) is packed into a certain "space" (volume). In the standard international system, we measure mass in kilograms and space in cubic metres.

📝 Teacher's Note: Remind students that \( 1 \text{ g cm}^{-3} = 1000 \text{ kg m}^{-3} \). This conversion is the most frequent source of errors in fluid mechanics numericals.

🎯 Exam Tip: Always look for "kg" and "m" together for SI units. "g" and "cm" are CGS units, which are common but not the standard SI units.

Question 8. When a body is wholly or partially immersed in a liquid, it experiences a buoyant force which is equal to :
(a) volume of liquid displaced by it
(b) weight of liquid displaced by it
(c) both (a) and (b)
(d) none of the options
Answer: (b) weight of liquid displaced by it
In simple words: This is the core of Archimedes' rule: the "upward push" an object gets is exactly equal to the weight of the water it pushed out of the way.

📝 Teacher's Note: This is a definition-based question. Make sure students don't confuse "volume" with "weight." While the volume of the displaced liquid equals the volume of the immersed body, the *force* equals the *weight*.

🎯 Exam Tip: This is the literal statement of Archimedes' Principle. Memorizing this phrase perfectly helps with both MCQs and descriptive questions.

Question 9. The ratio between the mass of a substance and the mass of an equal volume of water at 4°C is called :
(a) relative density
(b) density
(c) weight
(d) pressure
Answer: (a) relative density
In simple words: Relative density is a way to say "how many times heavier is this compared to water?" It's like a scale where water is 1. If something is 2, it's twice as heavy as water.

📝 Teacher's Note: Explain that we use 4°C because that is when water is at its maximum density. This provides a precise and unchanging standard for comparison.

🎯 Exam Tip: Relative density is a ratio, so it has NO units. If you see a numerical answer for R.D. with a unit like "kg/m³," it is incorrect.

Question 10. A body has density \( 9.6 \text{ g cm}^{-3} \). Its density in SI system is :
(a) \( 96 \text{ kg m}^{-3} \)
(b) \( 960 \text{ kg m}^{-3} \)
(c) \( 9600 \text{ kg m}^{-3} \)
(d) \( 96,000 \text{ kg m}^{-3} \)
Answer: (c) 9600 \( \text{kg m}^{-3} \)
In simple words: To change from grams/cm to the bigger kilogram/metre units, you just multiply by 1000. It's like changing from cents to dollars, but in the world of density.

📝 Teacher's Note: Show the derivation: \( 1 \text{ g/cm}^3 = \frac{10^{-3} \text{ kg}}{(10^{-2} \text{ m})^3} = \frac{10^{-3}}{10^{-6}} \text{ kg/m}^3 = 10^3 \text{ kg/m}^3 \).

🎯 Exam Tip: Shortcut: \( \text{Density in SI} = \text{Density in CGS} \times 1000 \). Use this for quick checks in MCQs.

(B) Subjective Questions

Question 1. A wooden block floats in water with two third of its volume submerged.
(1) Calculate density of wood.
(2) When the same block is placed in oil, three quarter of its volume is immersed in oil. Calculate the density of oil.
Answer:
(1) Let total vol. of wood \( = V \)
Vol. of wood submerged \( v' = \frac{2}{3} V \)
By law of flotation:
\( \frac{d_s}{d_w} = \frac{v'}{V} = \frac{\frac{2}{3} V}{V} = \frac{2}{3} \)
\( \implies d_s = \frac{2}{3} d_w = \frac{2}{3} \times 1000 = 667 \text{ kg m}^{-3} \)
but \( d_w = 1000 \text{ kg m}^{-3} \)
Density of wood \( d_s = 667 \text{ kg m}^{-3} \)

(2) Now \( \frac{d_s}{d_L} = \frac{v'}{V} \)
\( \implies \frac{667}{d_L} = \frac{3}{4} \frac{V}{V} = \frac{3}{4} \)
\( \implies \frac{2000}{3 d_L} = \frac{3}{4} \)
\( \implies d_L = \frac{2000 \times 4}{3 \times 3} = \frac{8000}{9} = 889 \text{ kg m}^{-3} \)
\( \therefore \text{Density of oil} = 889 \text{ kg m}^{-3} \)
In simple words: A block floats better in heavier liquids. Since it sinks less in water than in oil, we know water is heavier. We use the fraction submerged to find the exact density of the wood and the oil.

📝 Teacher's Note: The "Law of Flotation" formula \( \frac{\text{Density of Solid}}{\text{Density of Liquid}} = \frac{\text{Volume Submerged}}{\text{Total Volume}} \) is the most direct way to solve this. Ensure students understand that this only applies to floating bodies.

🎯 Exam Tip: When given fractions like "two third" or "three quarter," always use the formula \( \rho_s = \text{fraction} \times \rho_L \). It saves time and prevents algebraic errors.

Question 2. A metal cube of 5 cm edge and relative density 9 is suspended by a thread so as to be completely immersed in a liquid of relative density 1.2. Find the tension in the thread.
Answer:
Volume of metal cube \( = (\text{side})^3 = 5^3 = 125 \text{ cm}^3 \)
Density of cube \( = 9 \text{ g cm}^{-3} \) (since R.D. is 9)
Weight of cube acting downward \( = mg = V \times d \)
\( F_1 \downarrow = 125 \times 9 = 1125 \text{ gf} \)
Density of liquid \( d_L = 1.2 \text{ g cm}^{-3} \)
\( \therefore \) Upthrust due to liquid in the upward direction:
\( F_2 \uparrow = V \times d_L = 125 \times 1.2 = 150.0 \text{ gf} \)
Tension in the string = Net downward force \( = F_1 - F_2 \)
\( = 1125 - 150 = 975 \text{ gf} \)
In SI units, Tension \( = 9.75 \text{ N} \) (approx, if \( g = 10 \text{ m/s}^2 \)).
In simple words: The metal block is heavy, but the liquid pushes it up a little bit. The string doesn't have to carry the whole weight, just what is left over after the liquid helps.

📝 Teacher's Note: Remind students that Tension is essentially the "Apparent Weight" of the object. Using the unit 'gf' makes the calculations simpler for beginners than using Newtons directly.

🎯 Exam Tip: If the question provides Relative Density (R.D.), the density in \( \text{g/cm}^3 \) is numerically the same as the R.D. value. Use this shortcut to skip a conversion step.

Question 3. A weather forecasting plastic balloon of volume \( 15 \text{ m}^3 \) contains hydrogen of density \( 0.09 \text{ kg m}^{-3} \). The volume of equipment carried by the balloon is negligible compared to its own volume. The mass of the empty balloon is 7.15 kg. The balloon is floating in air of density \( 1.3 \text{ kg m}^{-3} \).
(1) Calculate the mass of hydrogen in balloon.
(2) Calculate the mass of hydrogen and the balloon.
(3) If the mass of equipment is x kg, write down the total mass of hydrogen, the balloon and the equipment.
(4) Calculate the mass of air displaced by balloon.
(5) Using the law of floatation, calculate the mass of equipment.
Answer:
Volume of Hydrogen \( V = 15 \text{ m}^3 \)
Density of hydrogen \( d = 0.09 \text{ kg m}^{-3} \)
(1) Mass of hydrogen in balloon \( = V \times d = 15 \times 0.09 = 1.35 \text{ kg} \)
(2) Mass of empty balloon \( = 7.15 \text{ kg} \)
Mass of hydrogen and balloon \( = 1.35 + 7.15 = 8.50 \text{ kg} \)
(3) Mass of equipment \( = x \text{ kg} \)
Total mass of hydrogen + Balloon + Equipment \( = (8.50 + x) \text{ kg} \)
(4) Density of air \( = 1.3 \text{ kg m}^{-3} \)
Mass of air displaced by balloon \( = V \times d = 15 \times 1.3 = 19.5 \text{ kg} \)
(5) According to law of floatation:
Total downward weight = UPTHRUST (Weight of air displaced)
\( 8.5 + x = 19.5 \)
\( \implies x = 19.5 - 8.5 \)
Mass of equipment \( x = 11 \text{ kg} \)
In simple words: To float in the air, the balloon's total weight (balloon + gas + gear) must be balanced by the weight of the air it pushed aside. We found out how much air was pushed aside and then worked out how much heavy gear the balloon could carry to stay level.

📝 Teacher's Note: This is an excellent real-world application of flotation. Explain that the air provides upthrust to balloons just as water provides upthrust to ships.

🎯 Exam Tip: In balloon problems, "Total Downward Weight" must include the gas inside, the balloon fabric, and any cargo/equipment.

Question 4. (a) State the principle of floatation. (b) The mass of a block made of certain material is 1.35 kg and its volume is \( 1.5 \times 10^{-3} \text{ m}^3 \).
1. Find the density of block.
2. Will this block float or sink? Give reasons for your answer.
Answer:
(a) PRINCIPLE OF FLOATATION : “When a solid is floating in a fluid, the weight of whole solid acting vertically downward at its CENTRE OF GRAVITY, is equal to the weight of fluid displaced by the IMMERSED part of solid acting upward, at its CENTRE OF BUOYANCY or at the centre of the BULK OF LIQUID displaced.”
OR
“The weight of a floating body is equal to the weight of the liquid displaced by its SUBMERGED part.”

(b) Mass of block \( = m = 1.35 \text{ kg} \)
Volume of block \( = V = 1.5 \times 10^{-3} \text{ m}^3 \)
(1) \( \text{Density of block} = \frac{m}{V} = \frac{1.35}{1.5 \times 10^{-3}} = 900 \text{ kg m}^{-3} \)
(2) Density of block (\( 900 \text{ kg m}^{-3} \)) is less than density of water (\( 1000 \text{ kg m}^{-3} \)).
\( \therefore \) Block will float in water.
In simple words: To float, an object's weight must be balanced by the weight of the water it pushes aside. Since this block is lighter than the same amount of water, it floats easily.

📝 Teacher's Note: Help students with the scientific notation in part (1). \( 10^{-3} \) in the denominator moves to the numerator as \( 10^3 \) (1000).

🎯 Exam Tip: When explaining why something floats, always compare its density to the density of water (\( 1000 \text{ kg m}^{-3} \)).

Question 5. (a) State Archimedes’ Principle. (b) A block of mass 7 kg and volume \( 0.07 \text{ m}^3 \) floats in a liquid of density \( 140 \text{ kg/m}^3 \). Calculate :
1. Volume of block above the surface of liquid.
2. Density of block.
Answer:
(a) ARCHIMEDES’ PRINCIPLE : “Whenever a body is immersed in a liquid (fluid), wholly or partially, it loses weight equal to the weight of liquid displaced by it.”

(b) Mass of block \( = m = 7 \text{ kg} \)
Volume of block \( = V = 0.07 \text{ m}^3 \)
Density of liquid \( = \rho_l = 140 \text{ kg m}^{-3} \)
Let \( V' = \) Volume of block immersed in the liquid
By law of flotation:
Weight of block = Weight of liquid displaced by the immersed part
\( V \rho_s g = V' \rho_l g \)
\( V' = \frac{\rho_s}{\rho_l} V \)
(2) \( \text{Density of block } = \rho_s = \frac{m}{V} = \frac{7}{0.07} = 100 \text{ kg m}^{-3} \)
From (1):
\( V' = \frac{100}{140} \times 0.07 = \frac{5}{7} \times \frac{7}{100} \)
\( \implies V' = \frac{1}{20} = 0.05 \text{ m}^3 \)
(1) Volume of block above the surface of liquid \( = V_{\text{total}} - V' \)
\( \implies \text{Volume above} = 0.07 - 0.05 = 0.02 \text{ m}^3 \)
In simple words: First we find the block's own density (100). Since it is lighter than the liquid (140), it floats. We calculate how much sits underwater (0.05) and subtract that from the total to see how much peeks out above (0.02).

📝 Teacher's Note: This is a sophisticated flotation problem. Use it to check if students understand that "submerged volume" and "total volume" are different for floating bodies.

🎯 Exam Tip: The volume *above* the liquid is often what catches students out. Always calculate submerged volume first, then subtract from total volume.

Question 6. (a) A body whose volume is \( 100 \text{ cm}^3 \) weighs 1 kgf in air. Find its weight in water. (b) Why is it easier to swim in sea water than in river water?
Answer:
(a) Volume of body \( = V = 100 \text{ cm}^3 = 10^{-4} \text{ m}^3 \)
Weight of body in air \( = 1 \text{ kgf} \)
Density of water \( = \rho_w = 1000 \text{ kg m}^{-3} \)
We know volume of water displaced = Volume of body \( = V = 100 \text{ cm}^3 \)
Upthrust = Weight of water displaced by body \( = V \rho_w g \)
In CGS: \( \text{Upthrust} = 100 \text{ cm}^3 \times 1 \text{ g cm}^{-3} \times g = 100 \text{ gf} = 0.1 \text{ kgf} \)
Weight of body in water = Weight of body in air - Upthrust
\( = 1 - 0.1 = 0.90 \text{ kgf} \)

(b) Sea water has dissolved salts, which makes its density higher than that of river water. According to Archimedes' principle, a denser liquid provides more upthrust. With a smaller portion of the man’s body submerged in sea water, the weight of sea water displaced is already equal to the total weight of the body. While to displace the same weight of river water, a larger portion of the body will have to be submerged. Hence, it is easier for a man to swim in sea water.
In simple words: Water pushes up on you. Salt water is "stronger" and pushes up harder than fresh water, so it holds you up more easily.

📝 Teacher's Note: Use the sea water vs. river water example to explain the role of density in buoyancy. Mention the Dead Sea as an extreme example where you can float without trying.

🎯 Exam Tip: For part (b), use the key phrase: "Sea water is denser, so it provides more upthrust for the same volume displaced."

Question 7. Why does a ship made of iron not sink in water, while an iron nail sinks in it?
Answer:
Density of iron is more than density of water, therefore the weight of an iron nail is more than the weight of water displaced by it and the nail SINKS. While the shape of an iron ship is made in such a way that it displaces a WEIGHT OF WATER equal to its own weight. Secondly, the ship is HOLLOW and THE EMPTY SPACE contains AIR which makes the AVERAGE DENSITY OF THE SHIP LESS THAN THAT OF WATER and hence the ship floats on water.
In simple words: A nail is a solid chunk of heavy metal, so it sinks. A ship is a giant hollow bowl filled with air. This makes the ship act like it is very light for its size, so it floats.

📝 Teacher's Note: This is the most classic "paradox" in buoyancy. Focus on "Average Density"—the combination of the iron hull and the huge amount of air inside.

🎯 Exam Tip: Use the keyword "Hollow" and "Average Density" to explain why large vessels float despite being made of dense materials.

Question 8. A solid of density \( 5000 \text{ kg m}^{-3} \) weighs 0.5 kgf in air. It is completely immersed in a liquid of density \( 800 \text{ kg m}^{-3} \). Calculate the apparent weight of the solid in liquid.
Answer:
Density of solid \( d_s = 5000 \text{ kg m}^{-3} \)
Weight of body in air \( = 0.5 \text{ kgf} \)
\( \implies \text{Mass } m = 0.5 \text{ kg} \)
Volume of solid \( V = \frac{m}{d_s} = \frac{0.5}{5000} = \frac{1}{10000} \text{ m}^3 \)
Vol. of liquid displaced \( = V = \frac{1}{10000} \text{ m}^3 \)
Density of liquid \( = 800 \text{ kg m}^{-3} \)
Mass of liquid displaced \( = V \times D = \frac{1}{10000} \times 800 = \frac{8}{100} \text{ kg} \)
Weight of displaced liquid (Upthrust) \( = 0.08 \text{ kgf} \)
Apparent weight of the solid in liquid = Actual Weight - Upthrust
\( = 0.5 - 0.08 = 0.42 \text{ kgf} \)
In simple words: The object is dense and heavy. We find how much space it takes up, then calculate the weight of the liquid that would fill that same space. Subtracting that liquid weight from the object's weight gives us its "underwater" weight.

📝 Teacher's Note: Part (2) of the OCR text mentioned a density of 800 for the liquid being zero weight—this is a typo in the original source material. If solid density > liquid density, it will always have a non-zero apparent weight and will sink.

🎯 Exam Tip: Calculate volume first, then use that volume to find the upthrust (\( V \times \rho_{\text{liquid}} \)). This two-step process is the safest way to solve these.

Question 9. (a) A body dipped in a liquid experiences an upthrust. State the factors on which the upthrust depends. (b) While floating, is the weight of body greater than, equal to or less than upthrust?
Answer:
(a) Factors on which upthrust depends are:
1. Volume of body immersed in fluid: Upthrust is maximum when the body is completely immersed in the fluid.
2. Density of the fluid: Upthrust \( \propto \) density of fluid. Larger the density of the fluid, larger will be the upthrust acting on the body.

(b) When the body floats, then the weight of the body is equal to the upthrust acting on the body.
In simple words: Upthrust depends on how big the object is and how heavy the liquid is. When things float, it's because the "push up" and "pull down" are exactly equal.

📝 Teacher's Note: Clarify that upthrust does NOT depend on the mass of the solid, only on the volume submerged. A 1kg iron block and a 1kg wood block (different volumes) will experience different upthrusts.

🎯 Exam Tip: Listing these two factors is a common 2-mark question. Be specific about "Volume of *immersed* part."

Question 10. A sinker is first weighed alone under water. It is then tied to a cork and again weighed under water. In which of the two cases weight under water is less and why?
Answer:
The weight of the sinker, when tied to a cork, under water is less than when it is weighed alone under water. Because the cork displaces a large amount of water relative to its own weight, a large upthrust acts on the combination (especially on the cork), reducing the overall reading on the scale.
In simple words: Adding a cork is like tying a balloon to the sinker. Even though the balloon has its own small weight, it pushes up so hard that it makes the whole package feel much lighter.

📝 Teacher's Note: This demonstrates that the *net* force determines the reading. The upthrust on the cork is much greater than the weight of the cork, creating an "upward pull" on the sinker.

🎯 Exam Tip: Explain that cork has very low density, so it provides a huge upthrust for a very small addition of weight.

Question 11. A solid weighs 105 kgf in air. When completely immersed in water, it displaces \( 30,000 \text{ cm}^3 \) of water. Calculate relative density of solid.
Answer:
Weight of solid in air \( = 105 \text{ kgf} \)
Volume of solid = Volume of water displaced \( = 30000 \text{ cm}^3 \)
\( \implies V = 30000 \times 10^{-6} \text{ m}^3 = 0.03 \text{ m}^3 \)
\( \rho_w = \text{Density of water} = 1000 \text{ kg m}^{-3} \)
Weight of water displaced by solid \( = V \rho_w g = 0.03 \times 1000 \times g = 30 \text{ kgf} \)
\( \text{R.D. of solid} = \frac{\text{Weight of solid in air}}{\text{Weight of water displaced by solid}} \)
\( \implies \text{R.D.} = \frac{105}{30} = 3.5 \)
In simple words: We know the block weighs 105 kg. We found out that the water it pushed away weighs 30 kg. Dividing 105 by 30 tells us the block is 3.5 times denser than water.

📝 Teacher's Note: Remind students that \( 1 \text{ litre} = 1000 \text{ cm}^3 \). So \( 30,000 \text{ cm}^3 \) is 30 litres, which weighs exactly 30 kgf.

🎯 Exam Tip: Using the definition \( \text{R.D.} = \frac{W_{\text{air}}}{\text{Upthrust}} \) is the most efficient way to solve this.

Question 12. A test tube loaded with lead shots weighs 25 gf and floats upto the mark X in water. When the test tube is made to float in brine solution, it needs 5 gf more of lead shots to float upto level X. Find the relative density of brine solution.
Answer:
When test tube floats in water :
Weight of test tube \( = 25 \text{ gf} \)
By law of flotation: Weight of water displaced = Weight of test tube \( = 25 \text{ gf} \)
When test tube floats in brine solution, it needs 5 gf more of lead shots to float upto same level X as in water.
Weight of test tube in brine \( = 25 + 5 = 30 \text{ gf} \)
By law of flotation : Weight of brine solution displaced = Weight of test tube \( = 30 \text{ gf} \)
As the test tube floats to the same level X, the volume of brine displaced is equal to the volume of water displaced.
\( \therefore \text{R.D. of brine solution} = \frac{\text{Weight of brine solution displaced}}{\text{Weight of equal volume of water displaced}} \)
\( \implies \text{R.D.} = \frac{30}{25} = 1.2 \)
In simple words: In saltier water, the tube floats higher. To sink it back down to the same "X" mark, we have to add more weight. The ratio of the total weight in brine to the total weight in water gives us the relative density.

📝 Teacher's Note: This problem explains the logic of a constant-volume hydrometer. By adding weights, we ensure the volume submerged stays constant, making density comparison easy.

🎯 Exam Tip: Identify that "floats to level X" means the volume displaced is constant in both cases. This allows you to use the weights directly in the R.D. formula.

Question 13. A wooden block is weighed with iron, such that combination just floats in water at room temperature. State your observations when :
(1) water is heated above room temperature
(2) water is cooled below 4°C. Give reasons to your answers in (1) and (2).
Answer:
(1) When water is heated, its density decreases. As a result, the upthrust decreases and the wooden block weighed with iron will sink more.
(2) Density of water is maximum at 4°C. When water is cooled below 4°C, its density decreases (anomalous expansion). Consequently, the upthrust acting on the block decreases. So, the wooden block weighed with iron sinks more than earlier.
In simple words: Water is at its strongest (densest) at 4°C. If you make it hotter OR colder than 4°C, it becomes "thinner" and less able to push objects up, so the block sinks deeper.

📝 Teacher's Note: This question tests the "Anomalous Expansion of Water." Water is unique because it becomes less dense both above and below 4°C.

🎯 Exam Tip: Mention "Anomalous expansion" and "Maximum density at 4°C" to score full marks for the reasoning part.

Question 14. A rubber ball floats in water with 2/7 of its volume above the surface of water. Calculate the average relative density of rubber ball.
Answer:
Let volume of rubber ball \( = V \)
Volume of rubber ball above the water surface \( = \frac{2}{7} V \)
Volume of rubber ball below the water surface \( = V - \frac{2}{7} V = \frac{5}{7} V \)
\( \implies \text{Volume of water displaced by the immersed part} = \frac{5}{7} V \)
By law of flotation :
Weight of rubber ball = Weight of water displaced
\( V \times \rho_{\text{ball}} \times g = \text{Volume submerged} \times \rho_w \times g \)
\( \implies V \times \rho = \frac{5}{7} V \times \rho_w \)
\( \implies \frac{\rho}{\rho_w} = \frac{5}{7} = 0.71 \)
\( \therefore \text{Average relative density of rubber ball} = 0.71 \)
In simple words: Since 5/7ths of the ball is underwater, the ball must be 5/7ths as heavy as water. We turn that fraction into a decimal to find the relative density.

📝 Teacher's Note: This is a very common type of question. The R.D. of a floating object is simply the fraction of its volume that is *submerged*.

🎯 Exam Tip: Shortcut: \( \text{R.D.} = \frac{\text{Submerged Volume}}{\text{Total Volume}} \). Just remember to use the fraction *under* the water, not the fraction *above*.

Question 15. A cube of ice whose side is 4.0 cm is allowed to melt. The volume of water formed is found to be \( 58.24 \text{ cm}^3 \). Find the density of ice.
Answer:
Side of ice cube \( = l = 4 \text{ cm} \)
Volume of ice cube \( V = l^3 = 4^3 = 64 \text{ cm}^3 \)
Volume of water formed \( V_w = 58.24 \text{ cm}^3 \)
Density of water \( \rho_w = 1 \text{ g cm}^{-3} \)
When ice melts, its mass remains constant.
\( \text{Mass of ice} = \text{Mass of water formed} \)
\( \text{Volume of ice} \times \text{Density of ice} = \text{Volume of water} \times \text{Density of water} \)
\( 64 \times \rho_i = 58.24 \times 1 \)
\( \rho_i = \frac{58.24}{64} = 0.91 \text{ g cm}^{-3} \)
In simple words: Melting doesn't change how much "stuff" is there, just the space it takes up. We use the amount of water created to figure out the mass, and then see how dense the original ice block must have been.

📝 Teacher's Note: This problem emphasizes "Conservation of Mass." Ice is less dense than water, which is why the same mass occupies a larger volume (\( 64 \text{ cm}^3 \)) as ice than it does as liquid water (\( 58.24 \text{ cm}^3 \)).

🎯 Exam Tip: Always state that "Mass remains constant during the change of state" as the basis for your equation.

Question 16. A jeweller claims to make ornaments of pure gold of relative density 19.3. A customer buys from him a bangle of weight 25.25 gf. The customer then weighs the bangle under water and finds its weight 23.075 gf. With the help of suitable calculations explain whether the bangle is of pure gold or not.
Answer:
R.D of pure gold \( = 19.3 \)
Weight of bangle in air \( = 25.25 \text{ gf} \)
Weight of bangle in water \( = 23.075 \text{ gf} \)
\( \text{R.D. of bangle} = \frac{\text{weight of bangle in air}}{\text{wt. in air} - \text{wt. of bangle in water}} \)
\( \implies \text{R.D. of bangle} = \frac{25.25}{25.25 - 23.075} = \frac{25.25}{2.175} \)
\( \implies \text{R.D. of bangle} = 11.61 \)
Since the calculated R.D. (11.61) is much lower than that of pure gold (19.3), the bangle is NOT PURE.
In simple words: Pure gold is very dense. We calculated the density of the bangle and found it's almost half as dense as it should be. This means the jeweller mixed in cheaper, lighter metals.

📝 Teacher's Note: This is a classic application of Archimedes' Principle—the very one used to detect the fraud in King Hiero's crown!

🎯 Exam Tip: Always conclude your numerical answer with a clear sentence answering the "Yes/No" part of the question.

Question 17. (a) When a piece of ice floating in water melts, the level of water inside the glass remains same. Explain. (b) An inflated balloon is placed inside a big glass jar which is connected to an evacuating pump. What will you observe when the evacuating pump starts working? Give a reason for your answer.
Answer:
(a) A piece of ice displaces an amount of water equal to its own weight. When the ice cube melts, its volume decreases and the resulting water exactly occupies the volume of water which was displaced by the submerged part of the ice. As a result, the level of water inside the glass remains the same.
(b) When the evacuating pump starts working, the air pressure inside the glass jar reduces. As the pressure inside the balloon is now much higher than the pressure outside the balloon, the balloon will expand and eventually burst.
In simple words: (a) Melting ice doesn't overflow a glass because the "hole" the ice made in the water is the exact same size as the water the ice turns into. (b) Without the outside air squeezing it, the balloon's own inside air pushes out so hard that it pops.

📝 Teacher's Note: For (a), explain that water expands when it freezes, so it takes up more space as ice. When it melts, it "shrinks" back into the exact space it displaced. For (b), this demonstrates that atmospheric pressure is a squeezing force.

🎯 Exam Tip: For (a), the key is that "Weight of ice = Weight of water formed." For (b), the reason is the "Pressure difference."

Question 18. (a) A trawler is fully loaded in sea water to maximum capacity. What will happen to this trawler, if moved to river water? Explain your answer. (b) A body of mass 50 g is floating in water. What is the apparent weight of body in water? Explain your answer.
Answer:
(a) The density of sea water is higher than the density of river water. Therefore, river water provides less upthrust to the trawler compared to sea water for the same submerged volume. When a trawler fully loaded in sea water is moved to river water, it needs to sink deeper to displace more water to balance its weight. Since it was already at "maximum capacity," it will sink.
(b) Mass of body \( = 50 \text{ g} \).
For any floating body, the weight of the body in air is exactly balanced by the upthrust.
\( \text{Apparent weight} = \text{Weight in air} - \text{Upthrust} \)
Since \( \text{Weight} = \text{Upthrust} \), the result is zero.
\( \implies \text{Apparent weight of the floating body} = 0 \)
In simple words: (a) Salt water is stronger at holding things up than river water. A ship that is "just okay" in the ocean might be "too heavy" for a river. (b) Floating is like weightlessness—the water is carrying all of your weight, so you feel like you weigh nothing.

📝 Teacher's Note: The "apparent weight of a floating body is zero" is a fundamental law. Students often want to give a small positive number—remind them that if it was not zero, the object would be moving up or down.

🎯 Exam Tip: "Apparent weight = 0" is the universal answer for any body that is floating at rest on a surface.

Question 19. A body of mass ‘m’ is floating in a liquid of density ‘p’ (1) what is the apparent weight of body? (2) what is the loss of weight of body?
Answer:
Mass of body \( = m \)
Weight in air \( = mg \)
(1) When a body floats in a liquid, the weight of the body is equal to the upthrust.
\( \text{Apparent weight} = \text{Weight in air} - \text{Upthrust} = mg - mg = 0 \).
(2) Loss in weight of the body is equal to the upthrust, which for a floating body is equal to the actual weight of the body (\( mg \)).
In simple words: (1) If it floats, it feels weightless (0). (2) It "lost" its entire weight to the water's upward push.

📝 Teacher's Note: This is a more generalized version of Question 18(b). Use it to solidify the concept of equilibrium in flotation.

🎯 Exam Tip: For the "loss in weight," you can write either "equal to upthrust" or "equal to the weight of the body in air." Both are correct for a floating object.

Question 20. A block of wood of volume \( 25 \text{ cm}^3 \) floats in water with \( 20 \text{ cm}^3 \) of its volume immersed in water. Calculate : (1) density of wood (2) the weight of block of wood.
Answer:
Total Volume of wooden block \( V = 25 \text{ cm}^3 \)
Volume of wooden block immersed in water \( v' = 20 \text{ cm}^3 \)
Density of water \( \rho_w = 1 \text{ g cm}^{-3} \)
(1) By law of flotation:
\( V \times \rho_{\text{wood}} = v' \times \rho_w \)
\( 25 \times \rho_{\text{wood}} = 20 \times 1 \)
\( \rho_{\text{wood}} = \frac{20}{25} = 0.8 \text{ g cm}^{-3} \)
(2) Weight of wooden block \( = V \times \rho_{\text{wood}} \times g \)
\( = 25 \times 0.8 \times g = 20 \text{ gf} \)
In simple words: 80% of the block is underwater, so its density is 80% of water's density (0.8). Its total weight is then easily found by multiplying that density by its size.

📝 Teacher's Note: Remind students that they can also calculate weight using the upthrust: \( v' \times \rho_w \times g = 20 \times 1 \times g = 20 \text{ gf} \). This confirms the law of flotation.

🎯 Exam Tip: Using CGS units (\( \text{g/cm}^3 \)) for wood and water problems is often much faster and less prone to zero-counting errors than using SI units.

Question 21. A solid body weighs 2.10 N in air. Its relative density is 8.4. How much will the body weigh if placed (1) in water, (2) in liquid of relative density 1.2?
Answer:
Weight of solid body in air \( = 2.10 \text{ N} \)
R.D. of solid \( = 8.4 \)
\( \text{R.D. of solid} = \frac{\text{Weight of solid in air}}{\text{Weight of water displaced by body}} \)
\( \implies 8.4 = \frac{2.10}{\text{Weight of water displaced}} \)
\( \implies \text{Weight of water displaced by body} = \frac{2.10}{8.4} = 0.25 \text{ N} \)
(1) Weight of body in water = Weight in air - Weight of water displaced
\( = 2.10 - 0.25 = 1.85 \text{ N} \)
(2) Upthrust due to water \( = 0.25 \text{ N} \)
Upthrust due to liquid \( = \text{Upthrust due to water} \times \text{R.D. of liquid} \)
\( = 0.25 \times 1.2 = 0.30 \text{ N} \)
Weight of body in liquid = Weight of body in air - Upthrust due to liquid
\( = 2.10 - 0.30 = 1.80 \text{ N} \)
In simple words: Water pushes up with 0.25 N of force. The other liquid is thicker (RD 1.2) and pushes up harder with 0.30 N. We subtract these pushes from the original air weight to find the two "apparent" weights.

📝 Teacher's Note: This provides a powerful shortcut: \( \text{Upthrust in any liquid} = \text{Upthrust in water} \times \text{R.D. of that liquid} \). This saves the step of calculating volume.

🎯 Exam Tip: When working with Newtons (N), don't confuse them with grams-force (gf). The math works the same way, but keep your units consistent throughout the whole answer.