Pressure & Floatation is a small but scoring chapter in AMU. 1-2 questions come every year — mostly on pressure calculation (thrust/area), Archimedes' principle, and why things float or sink. The best part? The numericals are straightforward formula-plugging. This expanded guide covers atmospheric pressure, barometer, buoyancy, Pascal's law, 15+ PYQs, 15+ practice problems, and common mistakes. Let's lock this in.
1) Thrust — The Perpendicular Force
In physics, the force acting perpendicularly (normally) on a surface is called thrust. Thrust is a vector quantity — it has both magnitude and direction, and its direction is always perpendicular to the surface. The SI unit of thrust is the Newton (N).
🟡 Key distinction: Thrust is the force itself. It depends only on how much force is applied, NOT on the area over which it acts. For example, if you push a wall with 50 N force, the thrust is 50 N regardless of whether you use your palm or fingertip.
2) Pressure — Thrust per Unit Area
Pressure is defined as the thrust (normal force) acting per unit area. While thrust tells you the total perpendicular force, pressure tells you how concentrated that force is. A small force on a tiny area can produce enormous pressure!
Pressure (P) = Thrust (F) / Area (A)
P = F / A
SI unit:
Pascal (Pa) = N/m²CGS unit:
dyne/cm²Conversion:
1 Pa = 1 N/m²1 atm = 1.013 × 10⁵ Pa1 bar = 10⁵ Pa
2.1 Factors Affecting Pressure
- Force (F) ↑ → Pressure (P) ↑ (directly proportional — if you push harder, pressure increases)
- Area (A) ↓ → Pressure (P) ↑ (inversely proportional — smaller area means more concentrated force)
🟡 Real-life examples for AMU:
• Sharp knife cuts better — thin edge = small area → high pressure
• School bag with wide straps — larger area on shoulders → less pressure → more comfortable
• Nails with pointed tips — small area → high pressure → easily penetrates walls
• Tank / Bulldozer tracks — very wide tracks → large area → less pressure → don't sink into mud
• Drawing pins (thumbtacks) — sharp pin end (small area) goes into board; flat thumb end (large area) doesn't hurt thumb
• Skiing / Snowshoes — large surface area → less pressure on snow → don't sink
• High-heeled shoes vs flat shoes — heels have tiny area → high pressure on ground (can dent wooden floors!)
• Sharp knife cuts better — thin edge = small area → high pressure
• School bag with wide straps — larger area on shoulders → less pressure → more comfortable
• Nails with pointed tips — small area → high pressure → easily penetrates walls
• Tank / Bulldozer tracks — very wide tracks → large area → less pressure → don't sink into mud
• Drawing pins (thumbtacks) — sharp pin end (small area) goes into board; flat thumb end (large area) doesn't hurt thumb
• Skiing / Snowshoes — large surface area → less pressure on snow → don't sink
• High-heeled shoes vs flat shoes — heels have tiny area → high pressure on ground (can dent wooden floors!)
2.2 Numericals on Pressure
Numerical 1: A force of 200 N acts on an area of 0.5 m². Calculate pressure.
Solution: P = F/A = 200/0.5 = 400 Pa
Solution: P = F/A = 200/0.5 = 400 Pa
Numerical 2: A woman of mass 50 kg wears high-heel shoes. Each heel has area 1 cm². Find pressure exerted by one heel on ground (g = 10 m/s²).
Solution: Force = mg = 50 × 10 = 500 N (assuming weight equally distributed, but mostly on heels). Area = 1 cm² = 1 × 10⁻⁴ m². P = 500 / 10⁻⁴ = 5 × 10⁶ Pa = 5 MPa (That's enormous!)
Solution: Force = mg = 50 × 10 = 500 N (assuming weight equally distributed, but mostly on heels). Area = 1 cm² = 1 × 10⁻⁴ m². P = 500 / 10⁻⁴ = 5 × 10⁶ Pa = 5 MPa (That's enormous!)
Numerical 3: A rectangular block of mass 10 kg has dimensions 50 cm × 20 cm × 10 cm. Find pressure when placed on different faces (g = 10).
Solution: Force = 10 × 10 = 100 N.
Face A (50×20): A = 0.5 × 0.2 = 0.1 m², P = 100/0.1 = 1000 Pa
Face B (50×10): A = 0.5 × 0.1 = 0.05 m², P = 100/0.05 = 2000 Pa
Face C (20×10): A = 0.2 × 0.1 = 0.02 m², P = 100/0.02 = 5000 Pa
→ Smallest face gives maximum pressure!
Solution: Force = 10 × 10 = 100 N.
Face A (50×20): A = 0.5 × 0.2 = 0.1 m², P = 100/0.1 = 1000 Pa
Face B (50×10): A = 0.5 × 0.1 = 0.05 m², P = 100/0.05 = 2000 Pa
Face C (20×10): A = 0.2 × 0.1 = 0.02 m², P = 100/0.02 = 5000 Pa
→ Smallest face gives maximum pressure!
🔥 AMU PYQ: The SI unit of pressure is:
(A) N (B) N/m (C) N/m² (D) N/m³
Ans: C (Pascal = N/m²)
🔥 AMU PYQ: A force of 100 N acts on an area of 0.5 m². Pressure is:
(A) 50 Pa (B) 100 Pa (C) 150 Pa (D) 200 Pa
Ans: D (P = 100/0.5 = 200 Pa)
(A) N (B) N/m (C) N/m² (D) N/m³
Ans: C (Pascal = N/m²)
🔥 AMU PYQ: A force of 100 N acts on an area of 0.5 m². Pressure is:
(A) 50 Pa (B) 100 Pa (C) 150 Pa (D) 200 Pa
Ans: D (P = 100/0.5 = 200 Pa)
3) Pressure in Fluids — Liquids and Gases
Fluids (liquids and gases) also exert pressure. The pressure inside a fluid has some unique characteristics:
3.1 Characteristics of Fluid Pressure
- Fluid pressure acts in all directions (not just downward — it pushes on all surfaces of a container)
- Pressure increases with depth — the deeper you go, the more the weight of fluid above you
- Pressure depends on the density of the fluid — denser fluids exert more pressure at the same depth
- At any given depth, pressure is the same in all directions (Pascal's law for static fluids)
3.2 Formula for Liquid Pressure
Pressure due to a liquid column: P = h ρ g
h = height (depth) of the liquid column (m)ρ = density of the liquid (kg/m³)g = acceleration due to gravity (m/s²)
3.3 Derivation of P = hρg
Consider a liquid column of height h and cross-sectional area A. The volume of this column = A × h. Mass = ρ × Volume = ρAh. Weight = mg = ρAhg. This weight acts on the base area A. So, Pressure = Force / Area = (ρAhg) / A = hρg.
🟡 Important observations:
• Liquid pressure depends ONLY on h, ρ, and g — NOT on the area of the container or the total volume of liquid
• A thin tube and a wide tank with same depth have SAME pressure at bottom
• This is why dams are built thicker at the bottom — pressure is highest there
• Water tanks on rooftops have to be elevated — the height creates the pressure needed for taps
• Liquid pressure depends ONLY on h, ρ, and g — NOT on the area of the container or the total volume of liquid
• A thin tube and a wide tank with same depth have SAME pressure at bottom
• This is why dams are built thicker at the bottom — pressure is highest there
• Water tanks on rooftops have to be elevated — the height creates the pressure needed for taps
Numerical — Dam design:
Calculate pressure at a depth of 50 m in a lake (ρ = 1000 kg/m³, g = 10 m/s²).
Solution: P = hρg = 50 × 1000 × 10 = 5 × 10⁵ Pa
At 100 m depth: P = 100 × 1000 × 10 = 10⁶ Pa
→ Dam must be thicker at 100 m depth because pressure doubles!
Calculate pressure at a depth of 50 m in a lake (ρ = 1000 kg/m³, g = 10 m/s²).
Solution: P = hρg = 50 × 1000 × 10 = 5 × 10⁵ Pa
At 100 m depth: P = 100 × 1000 × 10 = 10⁶ Pa
→ Dam must be thicker at 100 m depth because pressure doubles!
🔥 AMU PYQ: Pressure at a point inside a liquid depends on:
(A) Depth only (B) Density only (C) Depth and density (D) Area of container
Ans: C (P = hρg — depends on depth, density, and g)
🔥 AMU PYQ: Pressure at a depth of 10 m in water is (density = 1000 kg/m³, g = 10 m/s²):
(A) 10⁴ Pa (B) 10⁵ Pa (C) 10⁶ Pa (D) 10³ Pa
Ans: B (P = 10 × 1000 × 10 = 10⁵ Pa)
(A) Depth only (B) Density only (C) Depth and density (D) Area of container
Ans: C (P = hρg — depends on depth, density, and g)
🔥 AMU PYQ: Pressure at a depth of 10 m in water is (density = 1000 kg/m³, g = 10 m/s²):
(A) 10⁴ Pa (B) 10⁵ Pa (C) 10⁶ Pa (D) 10³ Pa
Ans: B (P = 10 × 1000 × 10 = 10⁵ Pa)
4) Atmospheric Pressure
The blanket of air surrounding the Earth exerts pressure on all surfaces. This is called atmospheric pressure. Even though air seems weightless, the entire column of air from the Earth's surface to the top of the atmosphere has significant weight.
Value of atmospheric pressure (at sea level):
1 atm = 1.013 × 10⁵ Pa1 atm = 1013 millibar (mbar)1 atm = 76 cm of Hg (760 mm of Hg)1 atm = 760 torr1 atm = 1 bar (approximately: 1 bar = 10⁵ Pa)1 atm ≈ 10.3 m of water column
4.1 Importance of Atmospheric Pressure
- Why don't we get crushed? Because the pressure inside our body (blood pressure, internal fluids) balances the external atmospheric pressure. Our bodies have evolved to exist under this pressure.
- Drinking straw — When you suck, you reduce pressure inside the straw; atmospheric pressure pushes the liquid up
- Vacuum seal — Removing air creates a pressure difference that holds lids tight
- Suction cups — Work because of atmospheric pressure pushing against the cup (not "sucking")
- Syringes — Pulling the plunger reduces pressure inside; atmospheric pressure pushes fluid in
4.2 Variation of Pressure with Altitude
As we go to higher altitudes (mountains), the atmospheric pressure decreases because there is less air above us.
- Mountaineers carry oxygen cylinders — At high altitudes, low atmospheric pressure means less oxygen available per breath. Even though the percentage of oxygen is still ~21%, the lower total pressure means fewer oxygen molecules enter the lungs.
- Why do ears "pop" on airplanes? — As altitude changes rapidly, the pressure outside the eardrum changes faster than the pressure inside the middle ear. The eardrum bulges until pressure equalizes (often through a yawn or swallow), causing the "pop" sensation.
- Airplane cabins are pressurized — They maintain a cabin pressure equivalent to about 8000 ft altitude even when flying at 35,000 ft.
- Water boils at lower temperature on mountains — Lower atmospheric pressure means water boils at < 100°C, making it harder to cook food.
4.3 Measurement of Atmospheric Pressure — Torricelli's Barometer
Evangelista Torricelli (1643) invented the mercury barometer to measure atmospheric pressure.
Construction and Working:
1. A long glass tube (~1 m) is filled completely with mercury
2. The open end is covered, inverted, and placed in a trough of mercury
3. The mercury level in the tube drops to about 76 cm above the trough level
4. The space above the mercury column is a vacuum (Torricellian vacuum)
5. The height of the mercury column (h) is a measure of atmospheric pressure
Why 76 cm? P_atm = hρg = 0.76 × 13600 × 9.8 = 1.013 × 10⁵ Pa
If water is used instead of mercury: h = P/(ρg) = 1.013×10⁵/(1000×9.8) ≈ 10.3 m — impractical for a barometer!
1. A long glass tube (~1 m) is filled completely with mercury
2. The open end is covered, inverted, and placed in a trough of mercury
3. The mercury level in the tube drops to about 76 cm above the trough level
4. The space above the mercury column is a vacuum (Torricellian vacuum)
5. The height of the mercury column (h) is a measure of atmospheric pressure
Why 76 cm? P_atm = hρg = 0.76 × 13600 × 9.8 = 1.013 × 10⁵ Pa
If water is used instead of mercury: h = P/(ρg) = 1.013×10⁵/(1000×9.8) ≈ 10.3 m — impractical for a barometer!
🟡 Why mercury?
• Mercury is very dense (13.6 g/cm³) → the column height is manageable (76 cm)
• Water would require a 10.3 m tall tube — impractical!
• Mercury does not stick to glass and has a shiny surface for easy reading
• Mercury is very dense (13.6 g/cm³) → the column height is manageable (76 cm)
• Water would require a 10.3 m tall tube — impractical!
• Mercury does not stick to glass and has a shiny surface for easy reading
4.4 Manometer
A manometer is a U-shaped tube containing a liquid (usually mercury or water) used to measure the pressure of gases. One end is open to the atmosphere, the other is connected to the gas source. The difference in liquid levels gives the gauge pressure of the gas.
🔥 AMU PYQ: The value of 1 atmospheric pressure is:
(A) 1.013 × 10⁴ Pa (B) 1.013 × 10⁵ Pa (C) 1.013 × 10⁶ Pa (D) 1.013 × 10³ Pa
Ans: B
🔥 AMU PYQ: At higher altitudes, atmospheric pressure:
(A) Increases (B) Decreases (C) Remains same (D) Becomes zero
Ans: B (decreases with altitude)
🔥 AMU PYQ: Why does a drinking straw work?
Ans: Sucking reduces pressure inside the straw; atmospheric pressure pushes the liquid up into the straw.
(A) 1.013 × 10⁴ Pa (B) 1.013 × 10⁵ Pa (C) 1.013 × 10⁶ Pa (D) 1.013 × 10³ Pa
Ans: B
🔥 AMU PYQ: At higher altitudes, atmospheric pressure:
(A) Increases (B) Decreases (C) Remains same (D) Becomes zero
Ans: B (decreases with altitude)
🔥 AMU PYQ: Why does a drinking straw work?
Ans: Sucking reduces pressure inside the straw; atmospheric pressure pushes the liquid up into the straw.
5) Buoyancy — The Upward Push
Have you ever noticed that it's easier to lift a stone underwater than in air? Or that a piece of wood feels lighter when partially submerged? This is because fluids exert an upward force on objects immersed in them. This force is called buoyant force or upthrust.
Buoyant force (F_b) = Weight of fluid displaced by the objectF_b = ρ_fluid × V_immersed × gρ_fluid = density of the fluid (kg/m³)V_immersed = volume of the object that is submerged (m³)g = acceleration due to gravity (m/s²)
5.1 Cause of Buoyancy — Pressure Difference
When an object is immersed in a fluid, the fluid pressure on its bottom surface is greater than the pressure on its top surface (because pressure increases with depth, and the bottom is deeper than the top). This pressure difference creates a net upward force — the buoyant force.
🟡 Simple explanation:
Bottom of object is deeper → higher pressure pushing up
Top of object is shallower → lower pressure pushing down
Net result: UPWARD force = Buoyant force
Bottom of object is deeper → higher pressure pushing up
Top of object is shallower → lower pressure pushing down
Net result: UPWARD force = Buoyant force
5.2 Why Do Objects Float or Sink?
Three forces compete: the weight of the object (downward) and the buoyant force (upward).
| Condition | Net Force | What happens? | Example |
|---|---|---|---|
| Weight < Buoyant force | Net upward | Object floats (rises to surface) | Cork, wood, ice in water |
| Weight = Buoyant force | Net = 0 | Object floats at any level (neutral buoyancy) | Fish, submarine at constant depth |
| Weight > Buoyant force | Net downward | Object sinks | Iron nail, stone in water |
In terms of density: If the density of the object is less than the density of the fluid → it floats. If the object's density is greater than the fluid's density → it sinks. Equal densities → neutral buoyancy.
🟡 Trick to remember:
• Floating: Weight = Buoyant force → ρ_body × V_body × g = ρ_fluid × V_immersed × g
• Fraction immersed = ρ_body / ρ_fluid
• For ice floating in water: ρ_ice = 0.9 g/cm³, ρ_water = 1 g/cm³ → fraction immersed = 0.9/1 = 0.9 (90% underwater)
• Apparent weight in liquid = True weight − Buoyant force
• Floating: Weight = Buoyant force → ρ_body × V_body × g = ρ_fluid × V_immersed × g
• Fraction immersed = ρ_body / ρ_fluid
• For ice floating in water: ρ_ice = 0.9 g/cm³, ρ_water = 1 g/cm³ → fraction immersed = 0.9/1 = 0.9 (90% underwater)
• Apparent weight in liquid = True weight − Buoyant force
🔥 AMU PYQ: Which of the following will sink in water?
(A) Ice (B) Wood (C) Iron (D) Oil
Ans: C (iron density > water density)
🔥 AMU PYQ: The apparent loss of weight of a body immersed in water is equal to:
(A) Weight of body (B) Weight of water displaced (C) Volume of body (D) Density of body
Ans: B
(A) Ice (B) Wood (C) Iron (D) Oil
Ans: C (iron density > water density)
🔥 AMU PYQ: The apparent loss of weight of a body immersed in water is equal to:
(A) Weight of body (B) Weight of water displaced (C) Volume of body (D) Density of body
Ans: B
6) Archimedes' Principle ⭐ (Most Important)
This is THE most important topic from this chapter for AMU. At least 1 question every year!
Statement:
"When a body is immersed fully or partially in a fluid, it experiences an upward buoyant force that is equal to the weight of the fluid displaced by the body."
"When a body is immersed fully or partially in a fluid, it experiences an upward buoyant force that is equal to the weight of the fluid displaced by the body."
6.1 Understanding Archimedes' Principle
Imagine a solid object being lowered into water. As it goes in, it displaces (pushes aside) some water. The volume of water displaced is exactly equal to the volume of the object that is immersed. The weight of this displaced water equals the buoyant force.
🟡 Key deductions from Archimedes' principle:
1. Apparent loss of weight = Weight of fluid displaced = Buoyant force
2. Apparent weight = True weight − Buoyant force
3. If an object weighs 80 gf in air and 64 gf in water → apparent loss = 16 gf = weight of water displaced
4. Volume of water displaced = mass of water displaced / density of water
1. Apparent loss of weight = Weight of fluid displaced = Buoyant force
2. Apparent weight = True weight − Buoyant force
3. If an object weighs 80 gf in air and 64 gf in water → apparent loss = 16 gf = weight of water displaced
4. Volume of water displaced = mass of water displaced / density of water
6.2 Verification of Archimedes' Principle
- Take a solid object and suspend it from a spring balance. Note its weight in air (W₁).
- Immerse the object fully in water. Note the apparent weight (W₂).
- Collect the water displaced by the object in a beaker. Measure its weight (W_d).
- Observation: W₁ − W₂ = W_d (Loss in weight = Weight of displaced water)
6.3 Applications of Archimedes' Principle (AMU Favourites)
| Device / Application | How it works |
|---|---|
| Hydrometer | Measures relative density (specific gravity) of liquids. Floats at different levels in different liquids — the denser the liquid, the less it sinks. Used to check purity of liquids (e.g., battery acid, milk). |
| Lactometer | Tests purity of milk. Pure milk has a specific density. If water is added, density decreases → lactometer sinks more. |
| Ships | Made of steel (density ~7800 kg/m³, which is more than water)! Yet they float because they are hollow. The hollow shape displaces a huge volume of water → buoyant force equals the ship's weight. Average density of the ship (total mass / total volume) is less than water. |
| Submarines | Can float or sink at will. Ballast tanks are filled with water to increase weight → sink. Compressed air pushes water out → weight decreases → rise. |
| Hot Air Balloons | Hot air is less dense than cool air → buoyant force (weight of cool air displaced) exceeds weight of balloon → it rises. |
| Icebergs | About 90% of an iceberg is under water (because density of ice = 0.9 g/cm³, water = 1 g/cm³). That's why "tip of the iceberg" means only a small part is visible! |
🔥 AMU PYQ: Archimedes' principle states that the buoyant force on an object is equal to the:
(A) Weight of the object (B) Weight of fluid displaced (C) Volume of object (D) Mass of object
Ans: B
🔥 AMU PYQ: The working of a hydrometer is based on:
(A) Newton's law (B) Archimedes' principle (C) Pascal's law (D) Bernoulli's theorem
Ans: B
🔥 AMU PYQ: A submarine sinks when its ballast tanks are filled with water because:
(A) Weight increases (B) Buoyancy decreases (C) Volume decreases (D) Density decreases
Ans: A (weight increases → weight > buoyant force → sinks)
(A) Weight of the object (B) Weight of fluid displaced (C) Volume of object (D) Mass of object
Ans: B
🔥 AMU PYQ: The working of a hydrometer is based on:
(A) Newton's law (B) Archimedes' principle (C) Pascal's law (D) Bernoulli's theorem
Ans: B
🔥 AMU PYQ: A submarine sinks when its ballast tanks are filled with water because:
(A) Weight increases (B) Buoyancy decreases (C) Volume decreases (D) Density decreases
Ans: A (weight increases → weight > buoyant force → sinks)
7) Density and Relative Density
7.1 Density (ρ)
Density is the measure of how much mass is packed into a given volume. It is a characteristic property of a substance.
Density (ρ) = Mass (m) / Volume (V)SI unit:
kg/m³CGS unit:
g/cm³Conversion:
1 g/cm³ = 1000 kg/m³Density of water = 1000 kg/m³ = 1 g/cm³Density of mercury = 13600 kg/m³ = 13.6 g/cm³
7.2 Relative Density (Specific Gravity)
Relative density (R.D.) is the ratio of the density of a substance to the density of a reference substance (usually water at 4°C). Since it's a ratio of two same units, it has NO units — it's a pure number.
R.D. = Density of substance / Density of water at 4°CR.D. = Weight of substance in air / Loss of weight in waterR.D. = Mass of substance / Mass of equal volume of waterRelative density has NO unit — it's a pure ratio.
🟡 Key facts about R.D.:
• R.D. > 1 → Substance sinks in water (denser than water)
• R.D. < 1 → Substance floats on water (less dense than water)
• R.D. = 1 → Same density as water; floats at any level
• R.D. of water = 1 (by definition)
• R.D. of mercury = 13.6
• R.D. of ice = 0.9 (ice floats — 90% under water)
• R.D. of aluminium = 2.7, iron = 7.8, gold = 19.3
• R.D. > 1 → Substance sinks in water (denser than water)
• R.D. < 1 → Substance floats on water (less dense than water)
• R.D. = 1 → Same density as water; floats at any level
• R.D. of water = 1 (by definition)
• R.D. of mercury = 13.6
• R.D. of ice = 0.9 (ice floats — 90% under water)
• R.D. of aluminium = 2.7, iron = 7.8, gold = 19.3
Numerical — Finding R.D. by weight loss method:
A solid weighs 120 gf in air and 90 gf in water. Find: (a) loss of weight in water (b) R.D.
Solution:
(a) Loss of weight = 120 − 90 = 30 gf
(b) R.D. = W_air / loss of weight = 120 / 30 = 4
(c) Density = R.D. × density of water = 4 × 1000 = 4000 kg/m³
A solid weighs 120 gf in air and 90 gf in water. Find: (a) loss of weight in water (b) R.D.
Solution:
(a) Loss of weight = 120 − 90 = 30 gf
(b) R.D. = W_air / loss of weight = 120 / 30 = 4
(c) Density = R.D. × density of water = 4 × 1000 = 4000 kg/m³
Numerical — Fraction submerged for a floating body:
A block of wood of density 600 kg/m³ floats in water (density = 1000 kg/m³). What fraction of its volume is above water?
Solution:
For floating: Weight = Buoyant force
ρ_body × V_body × g = ρ_water × V_immersed × g
V_immersed / V_body = ρ_body / ρ_water = 600 / 1000 = 0.6 = 60%
So 60% is under water, 40% is above water
A block of wood of density 600 kg/m³ floats in water (density = 1000 kg/m³). What fraction of its volume is above water?
Solution:
For floating: Weight = Buoyant force
ρ_body × V_body × g = ρ_water × V_immersed × g
V_immersed / V_body = ρ_body / ρ_water = 600 / 1000 = 0.6 = 60%
So 60% is under water, 40% is above water
🔥 AMU PYQ: Relative density of a substance is:
(A) Has unit kg/m³ (B) Has unit g/cm³ (C) Is unitless (D) Has unit N/m²
Ans: C (unitless — it's a ratio)
🔥 AMU PYQ: The relative density of a substance is greater than 1. What does it signify?
Ans: The substance is denser than water → it will sink in water.
🔥 AMU PYQ: The relative density of mercury is:
(A) 1 (B) 3.6 (C) 13.6 (D) 1000
Ans: C (13.6 times denser than water)
(A) Has unit kg/m³ (B) Has unit g/cm³ (C) Is unitless (D) Has unit N/m²
Ans: C (unitless — it's a ratio)
🔥 AMU PYQ: The relative density of a substance is greater than 1. What does it signify?
Ans: The substance is denser than water → it will sink in water.
🔥 AMU PYQ: The relative density of mercury is:
(A) 1 (B) 3.6 (C) 13.6 (D) 1000
Ans: C (13.6 times denser than water)
8) Floatation — Law of Floatation
The Law of Floatation is derived from Archimedes' principle: A body floats in a liquid if its weight is equal to the weight of the liquid displaced by the immersed part of the body.
8.1 Conditions for Floatation
| Density Comparison | Behaviour | Examples |
|---|---|---|
| ρ_body < ρ_liquid | Floats (partially immersed) | Wood, ice, oil in water |
| ρ_body = ρ_liquid | Floats at any level (neutral) | Fish (using swim bladder), submarine at constant depth |
| ρ_body > ρ_liquid | Sinks | Iron, stone, coin in water |
8.2 Why Does an Iron Ship Float but an Iron Nail Sinks?
This is a classic question from AMU! An iron nail is solid — its density (7800 kg/m³) is much greater than water, so it sinks. But a ship is hollow — it contains a lot of air inside. The total volume of the ship (iron + air) is very large, but the mass is relatively low. So the average density of the ship (total mass / total volume) is less than water, making it float. The ship displaces enough water to create a buoyant force equal to its total weight.
🟡 Remember: It's not about the material's density alone — it's about the average density of the whole object. A hollow iron ball floats; a solid iron ball sinks!
8.3 Plimsoll Line (Load Line)
Ships have a marking on their side called the Plimsoll line or load line. It indicates the maximum safe depth to which a ship can be loaded in different types of water (fresh water, sea water, different seasons). If the ship is overloaded beyond this line, it's unsafe and may sink.
9) Pascal's Law — Transmission of Pressure
Statement: Pressure applied to an enclosed fluid at rest is transmitted equally and undiminished to every point of the fluid and to the walls of the containing vessel.
Mathematical form:
Where F₁ is the small input force on area A₁, producing pressure P = F₁/A₁, and F₂ is the large output force on area A₂.
F₁ / A₁ = F₂ / A₂Where F₁ is the small input force on area A₁, producing pressure P = F₁/A₁, and F₂ is the large output force on area A₂.
F₂ = F₁ × (A₂ / A₁)
9.1 Applications of Pascal's Law
| Application | Working Principle |
|---|---|
| Hydraulic Lift / Jack | A small force on a small piston creates pressure in the fluid. This pressure is transmitted to a large piston, producing a much larger force. Used to lift cars in garages. |
| Hydraulic Brakes | When the brake pedal is pressed, a small piston applies force on brake fluid. The pressure is transmitted equally to all wheels, applying braking force on each wheel. |
| Hydraulic Press | Used for compressing materials (like cotton bales, metals). Same principle — small force → large force multiplication. |
Numerical — Hydraulic Lift:
In a hydraulic lift, the small piston has area 0.01 m² and the large piston has area 0.5 m². If a force of 100 N is applied on the small piston, what force is exerted on the large piston?
Solution:
F₂ = F₁ × (A₂ / A₁) = 100 × (0.5 / 0.01) = 100 × 50 = 5000 N
So a 100 N force can lift a 5000 N (500 kg) load! That's mechanical advantage of 50×!
In a hydraulic lift, the small piston has area 0.01 m² and the large piston has area 0.5 m². If a force of 100 N is applied on the small piston, what force is exerted on the large piston?
Solution:
F₂ = F₁ × (A₂ / A₁) = 100 × (0.5 / 0.01) = 100 × 50 = 5000 N
So a 100 N force can lift a 5000 N (500 kg) load! That's mechanical advantage of 50×!
10) PYQ Battle — AMU Past Questions on Pressure & Floatation
Q1. A wooden block of mass 10 kg has dimensions 50 cm × 20 cm × 10 cm. What is the pressure on the table when the 20 cm × 10 cm face is in contact? (g = 10 m/s²)
Ans: F = mg = 100 N. Area = 200 cm² = 0.02 m². P = 100/0.02 = 5000 Pa
Ans: F = mg = 100 N. Area = 200 cm² = 0.02 m². P = 100/0.02 = 5000 Pa
Q2. Pressure at a depth of 10 m in water is (density = 1000 kg/m³, g = 10 m/s²):
(A) 10⁴ Pa (B) 10⁵ Pa (C) 10⁶ Pa (D) 10³ Pa
Ans: B (P = 10 × 1000 × 10 = 10⁵ Pa)
(A) 10⁴ Pa (B) 10⁵ Pa (C) 10⁶ Pa (D) 10³ Pa
Ans: B (P = 10 × 1000 × 10 = 10⁵ Pa)
Q3. A solid weighs 80 g in air and 64 g in water. The relative density of the solid is:
Ans: Loss = 16 g. R.D. = 80/16 = 5
Ans: Loss = 16 g. R.D. = 80/16 = 5
Q4. Why do nails have pointed tips?
(A) To increase area (B) To decrease pressure (C) To increase pressure (D) To look good
Ans: C (small area → high pressure)
(A) To increase area (B) To decrease pressure (C) To increase pressure (D) To look good
Ans: C (small area → high pressure)
Q5. The working of a hydrometer is based on:
(A) Newton's law (B) Archimedes' principle (C) Pascal's law (D) Bernoulli's theorem
Ans: B
(A) Newton's law (B) Archimedes' principle (C) Pascal's law (D) Bernoulli's theorem
Ans: B
Q6. A force of 100 N acts on an area of 0.5 m². Pressure is:
(A) 50 Pa (B) 100 Pa (C) 150 Pa (D) 200 Pa
Ans: D (P = 100/0.5 = 200 Pa)
(A) 50 Pa (B) 100 Pa (C) 150 Pa (D) 200 Pa
Ans: D (P = 100/0.5 = 200 Pa)
Q7. Which of the following will sink in water?
(A) Ice (B) Wood (C) Iron (D) Oil
Ans: C (density > water)
(A) Ice (B) Wood (C) Iron (D) Oil
Ans: C (density > water)
Q8. The SI unit of density is:
(A) kg/m³ (B) kg/m² (C) g/cm³ (D) N/m³
Ans: A
(A) kg/m³ (B) kg/m² (C) g/cm³ (D) N/m³
Ans: A
Q9. A submarine sinks when its ballast tanks are filled with water because:
(A) Weight increases (B) Buoyancy decreases (C) Volume decreases (D) Density decreases
Ans: A (weight increases → weight > buoyant force)
(A) Weight increases (B) Buoyancy decreases (C) Volume decreases (D) Density decreases
Ans: A (weight increases → weight > buoyant force)
Q10. The relative density of mercury is:
(A) 1 (B) 3.6 (C) 13.6 (D) 1000
Ans: C (13.6)
(A) 1 (B) 3.6 (C) 13.6 (D) 1000
Ans: C (13.6)
Q11. Pascal's law is used in:
(A) Hydraulic lift (B) Lactometer (C) Barometer (D) Hydrometer
Ans: A
(A) Hydraulic lift (B) Lactometer (C) Barometer (D) Hydrometer
Ans: A
Q12. The value of atmospheric pressure is about:
(A) 10³ Pa (B) 10⁴ Pa (C) 10⁵ Pa (D) 10⁶ Pa
Ans: C (1.013 × 10⁵ Pa)
(A) 10³ Pa (B) 10⁴ Pa (C) 10⁵ Pa (D) 10⁶ Pa
Ans: C (1.013 × 10⁵ Pa)
Q13. A body weighs 60 N in air and 45 N when fully immersed in water. The buoyant force is:
(A) 60 N (B) 45 N (C) 15 N (D) 105 N
Ans: C (60 − 45 = 15 N)
(A) 60 N (B) 45 N (C) 15 N (D) 105 N
Ans: C (60 − 45 = 15 N)
Q14. Why is mercury preferred over water in a barometer?
(A) Mercury is cheaper (B) Mercury is denser → shorter column (C) Mercury is colorless (D) Mercury expands more
Ans: B (density 13.6 → 76 cm column; water would need 10.3 m)
(A) Mercury is cheaper (B) Mercury is denser → shorter column (C) Mercury is colorless (D) Mercury expands more
Ans: B (density 13.6 → 76 cm column; water would need 10.3 m)
Q15. An iron nail sinks but an iron ship floats because:
(A) Ship is heavier (B) Ship has larger volume → lower average density (C) Ship uses engines (D) Nail is smaller
Ans: B
(A) Ship is heavier (B) Ship has larger volume → lower average density (C) Ship uses engines (D) Nail is smaller
Ans: B
Q16. The Plimsoll line on a ship indicates:
(A) Speed limit (B) Maximum safe loading depth (C) Fuel capacity (D) Age of ship
Ans: B
(A) Speed limit (B) Maximum safe loading depth (C) Fuel capacity (D) Age of ship
Ans: B
11) Practice Problems — 15+ Numericals with Solutions
Practice 1: A force of 250 N acts on an area of 25 cm². Find pressure in Pa.
Solution: Area = 25 cm² = 25 × 10⁻⁴ m² = 0.0025 m². P = 250/0.0025 = 1,00,000 Pa = 10⁵ Pa
Solution: Area = 25 cm² = 25 × 10⁻⁴ m² = 0.0025 m². P = 250/0.0025 = 1,00,000 Pa = 10⁵ Pa
Practice 2: At what depth in water is the pressure equal to 2 atm? (1 atm = 10⁵ Pa, ρ = 1000 kg/m³, g = 10 m/s²)
Solution: P_total = P_atm + hρg. 2 × 10⁵ = 10⁵ + h × 1000 × 10. h × 10⁴ = 10⁵. h = 10 m
Solution: P_total = P_atm + hρg. 2 × 10⁵ = 10⁵ + h × 1000 × 10. h × 10⁴ = 10⁵. h = 10 m
Practice 3: A 60 kg person stands on one foot. If foot area = 150 cm², find pressure on ground (g = 10).
Solution: F = 60 × 10 = 600 N. A = 150 × 10⁻⁴ = 0.015 m². P = 600/0.015 = 40,000 Pa
Solution: F = 60 × 10 = 600 N. A = 150 × 10⁻⁴ = 0.015 m². P = 600/0.015 = 40,000 Pa
Practice 4: A block of density 800 kg/m³ floats in water. What fraction of its volume is immersed?
Solution: V_immersed/V = ρ_body/ρ_water = 800/1000 = 0.8 = 80%
Solution: V_immersed/V = ρ_body/ρ_water = 800/1000 = 0.8 = 80%
Practice 5: A body weighs 150 N in air and 120 N in water. Find volume of body (g = 10, ρ_water = 1000 kg/m³).
Solution: Buoyant force = 150 − 120 = 30 N. F_b = ρVg → 30 = 1000 × V × 10 → V = 30/10000 = 0.003 m³ = 3000 cm³
Solution: Buoyant force = 150 − 120 = 30 N. F_b = ρVg → 30 = 1000 × V × 10 → V = 30/10000 = 0.003 m³ = 3000 cm³
Practice 6: Find R.D. of a substance if a 200 g solid (in air) weighs 150 g in water.
Solution: Loss = 200 − 150 = 50 g. R.D. = 200/50 = 4
Solution: Loss = 200 − 150 = 50 g. R.D. = 200/50 = 4
Practice 7: A body of mass 2 kg and volume 2500 cm³ is immersed in water. Will it float or sink? (ρ_water = 1000 kg/m³)
Solution: ρ_body = 2 / 0.0025 = 800 kg/m³. ρ_body < ρ_water → Floats
Solution: ρ_body = 2 / 0.0025 = 800 kg/m³. ρ_body < ρ_water → Floats
Practice 8: A hydraulic press has pistons of areas 0.02 m² and 0.8 m². A force of 50 N on the small piston produces what force on the large piston?
Solution: F₂ = 50 × (0.8/0.02) = 50 × 40 = 2000 N
Solution: F₂ = 50 × (0.8/0.02) = 50 × 40 = 2000 N
Practice 9: Calculate the pressure at the bottom of a 30 m deep well filled with water (g = 10).
Solution: P = hρg = 30 × 1000 × 10 = 3 × 10⁵ Pa
Solution: P = hρg = 30 × 1000 × 10 = 3 × 10⁵ Pa
Practice 10: A cube of side 10 cm and mass 800 g is placed on a table. Find pressure exerted (g = 10).
Solution: F = 0.8 × 10 = 8 N. Area = 0.1 × 0.1 = 0.01 m². P = 8/0.01 = 800 Pa
Solution: F = 0.8 × 10 = 8 N. Area = 0.1 × 0.1 = 0.01 m². P = 8/0.01 = 800 Pa
Practice 11: A piece of ice (density = 900 kg/m³) floats in water. Find the fraction of volume above water level.
Solution: Fraction immersed = 900/1000 = 0.9. Fraction above = 1 − 0.9 = 0.1 = 10%
Solution: Fraction immersed = 900/1000 = 0.9. Fraction above = 1 − 0.9 = 0.1 = 10%
Practice 12: A body floats with 2/5th of its volume above the surface of water. Find its density.
Solution: V_immersed/V = 1 − 2/5 = 3/5 = 0.6. ρ_body = ρ_water × 0.6 = 1000 × 0.6 = 600 kg/m³
Solution: V_immersed/V = 1 − 2/5 = 3/5 = 0.6. ρ_body = ρ_water × 0.6 = 1000 × 0.6 = 600 kg/m³
Practice 13: The density of gold is 19.3 g/cm³. Find its R.D. and mass of gold block of volume 100 cm³.
Solution: R.D. = 19.3 (same number, no unit). Mass = 19.3 × 100 = 1930 g = 1.93 kg
Solution: R.D. = 19.3 (same number, no unit). Mass = 19.3 × 100 = 1930 g = 1.93 kg
Practice 14: A solid of R.D. 2.5 weighs 125 gf in air. Find its weight in water.
Solution: R.D. = W_air/(W_air − W_water). 2.5 = 125/(125 − W_w). 125 − W_w = 50. W_w = 125 − 50 = 75 gf
Solution: R.D. = W_air/(W_air − W_water). 2.5 = 125/(125 − W_w). 125 − W_w = 50. W_w = 125 − 50 = 75 gf
Practice 15: A barometer reads 74 cm of Hg. Find the atmospheric pressure in Pa. (ρ_Hg = 13600 kg/m³, g = 10)
Solution: P = hρg = 0.74 × 13600 × 10 = 1,00,640 Pa ≈ 1.006 × 10⁵ Pa
Solution: P = hρg = 0.74 × 13600 × 10 = 1,00,640 Pa ≈ 1.006 × 10⁵ Pa
Practice 16: The density of seawater is 1025 kg/m³. Find pressure at 100 m depth in sea (g = 10).
Solution: P = hρg = 100 × 1025 × 10 = 1.025 × 10⁶ Pa ≈ 10.25 atm
Solution: P = hρg = 100 × 1025 × 10 = 1.025 × 10⁶ Pa ≈ 10.25 atm
12) Common Mistakes & Tricks
⚠️ Top common mistakes in AMU exam:
Mistake 1: Area unit conversion
→ Students forget to convert cm² to m². Remember: 1 m² = 10,000 cm² (NOT 100). So 100 cm² = 0.01 m².
Mistake 2: Atmospheric pressure in liquid pressure problems
→ P_total at depth h = P_atm + hρg. Many students forget to add atmospheric pressure when asked for "total pressure". But if asked for "pressure due to water column", just use hρg.
Mistake 3: Relative density confusion
→ R.D. has NO units. It's a pure number. Don't write kg/m³ or g/cm³ for R.D.
Mistake 4: Buoyant force direction
→ Buoyant force is ALWAYS upward, regardless of the shape or orientation of the object.
Mistake 5: Floating object weight vs buoyant force
→ For a floating body, Weight = Buoyant force. For a sinking body, Weight > Buoyant force.
Mistake 6: kg-wt vs kg mass
→ In RD problems using spring balance, the readings are in gf (gram-force) or N. 1 kg-wt = 10 N (approximately).
Mistake 1: Area unit conversion
→ Students forget to convert cm² to m². Remember: 1 m² = 10,000 cm² (NOT 100). So 100 cm² = 0.01 m².
Mistake 2: Atmospheric pressure in liquid pressure problems
→ P_total at depth h = P_atm + hρg. Many students forget to add atmospheric pressure when asked for "total pressure". But if asked for "pressure due to water column", just use hρg.
Mistake 3: Relative density confusion
→ R.D. has NO units. It's a pure number. Don't write kg/m³ or g/cm³ for R.D.
Mistake 4: Buoyant force direction
→ Buoyant force is ALWAYS upward, regardless of the shape or orientation of the object.
Mistake 5: Floating object weight vs buoyant force
→ For a floating body, Weight = Buoyant force. For a sinking body, Weight > Buoyant force.
Mistake 6: kg-wt vs kg mass
→ In RD problems using spring balance, the readings are in gf (gram-force) or N. 1 kg-wt = 10 N (approximately).
✅ Tricks to score full marks:
• Always write: P = F/A, then substitute with correct units
• 1 m² = 10⁴ cm², so to convert cm² → m², divide by 10000
• Density of water = 1000 kg/m³ = 1 g/cm³ — memorize FOREVER
• For floatation: fraction immersed = ρ_body / ρ_fluid
• For RD: R.D. = W_air / (W_air − W_liquid)
• In hydraulic lift: F₂ = F₁ × (A₂/A₁) — force multiplies by area ratio
• 1 atm pressure supports 76 cm Hg column or about 10.3 m water column
• Apparent weight = True weight − Buoyant force
• Always write: P = F/A, then substitute with correct units
• 1 m² = 10⁴ cm², so to convert cm² → m², divide by 10000
• Density of water = 1000 kg/m³ = 1 g/cm³ — memorize FOREVER
• For floatation: fraction immersed = ρ_body / ρ_fluid
• For RD: R.D. = W_air / (W_air − W_liquid)
• In hydraulic lift: F₂ = F₁ × (A₂/A₁) — force multiplies by area ratio
• 1 atm pressure supports 76 cm Hg column or about 10.3 m water column
• Apparent weight = True weight − Buoyant force
13) Important Formulae — At a Glance
| Quantity | Formula | SI Unit |
|---|---|---|
| Thrust (Force) | F | N |
| Pressure | P = F/A | Pa (N/m²) |
| Liquid pressure | P = hρg | Pa |
| Buoyant force | F_b = ρ V g | N |
| Density | ρ = m/V | kg/m³ |
| Relative Density | R.D. = ρ_substance / ρ_water | No unit |
| R.D. using weights | R.D. = W_air / (W_air − W_liquid) | No unit |
| Apparent weight | W_app = W_true − F_b | N |
| Fraction immersed (floating) | V_immersed / V_total = ρ_body / ρ_fluid | — |
| Pascal's law | F₁/A₁ = F₂/A₂ | — |
| 1 Pa | = 1 N/m² | — |
| 1 atm | = 1.013 × 10⁵ Pa = 76 cm Hg = 760 torr | — |
| 1 bar | = 10⁵ Pa | — |
| Density of water | 1000 kg/m³ = 1 g/cm³ | — |
| Density of Hg | 13600 kg/m³ = 13.6 g/cm³ | — |
14) Quick Revision — Power-packed One-liners
- Thrust = perpendicular force (N). Pressure = thrust/area (Pa)
- P = F/A. Small area → large pressure (nails, knives, heels)
- Liquid pressure: P = hρg. Depends only on depth, density, and g. NOT on container shape or area.
- Pressure in a fluid acts equally in all directions at the same depth.
- Dams are thicker at the bottom because pressure increases with depth.
- Atmospheric pressure: 1 atm = 1.013 × 10⁵ Pa = 76 cm Hg = 760 torr = 1.013 bar
- Atmospheric pressure decreases with altitude (mountaineers need oxygen).
- Torricelli's barometer measures atmospheric pressure using a mercury column.
- Buoyant force exists due to pressure difference between top and bottom of immersed object.
- Buoyant force = weight of fluid displaced = ρ V g
- Archimedes' principle: Apparent loss of weight = weight of fluid displaced
- Object floats if ρ_body < ρ_liquid; sinks if ρ_body > ρ_liquid
- For floating: Weight = Buoyant force. Fraction immersed = ρ_body/ρ_fluid.
- Iron ship floats because hollow shape → large volume → low average density.
- Relative Density (R.D.) = density of substance / density of water. No units!
- R.D. = W_air / (W_air − W_water). R.D. of water = 1, mercury = 13.6, ice = 0.9.
- Hydrometer / Lactometer → Archimedes' principle. Measures density/purity.
- Pascal's law: F₁/A₁ = F₂/A₂ (hydraulic lift, brakes, press).
- Plimsoll line: max safe loading depth on a ship.
- 1 Pa = 1 N/m², 1 bar = 10⁵ Pa, 1 atm = 1.013 × 10⁵ Pa.
✅ Pressure & Floatation — COMPLETE. This chapter is all about P = F/A, Archimedes' principle, and the floatation condition. Master these three and you're set for AMU! Next up: Sound & Waves!
15) Solved Examples from NCERT/Exemplar
Example 1: A block of wood floats in water with 2/3 of its volume submerged. Find its density. (ρ_water = 1000 kg/m³)
Solution: For floating: ρ_body × V = ρ_water × V_immersed → ρ_body = ρ_water × (V_immersed/V) = 1000 × (2/3) ≈ 667 kg/m³
Solution: For floating: ρ_body × V = ρ_water × V_immersed → ρ_body = ρ_water × (V_immersed/V) = 1000 × (2/3) ≈ 667 kg/m³
Example 2: A hydraulic lift has two pistons of areas 0.01 m² and 0.5 m². What force on the small piston can lift a 2000 kg car? (g = 10)
Solution: Load on large piston = 2000 × 10 = 20000 N. F₁/A₁ = F₂/A₂ → F₁/0.01 = 20000/0.5 → F₁ = 20000 × 0.01 / 0.5 = 400 N
Solution: Load on large piston = 2000 × 10 = 20000 N. F₁/A₁ = F₂/A₂ → F₁/0.01 = 20000/0.5 → F₁ = 20000 × 0.01 / 0.5 = 400 N
Example 3: The density of a metal is 7.8 g/cm³. Find its R.D. and the mass of a 50 cm³ block.
Solution: R.D. = density of metal / density of water = 7.8/1 = 7.8. Mass = density × volume = 7.8 × 50 = 390 g
Solution: R.D. = density of metal / density of water = 7.8/1 = 7.8. Mass = density × volume = 7.8 × 50 = 390 g
Example 4: A body of mass 2 kg and volume 0.002 m³ is fully immersed in water. Find the buoyant force and apparent weight. (ρ_water = 1000 kg/m³, g = 10)
Solution: F_b = ρVg = 1000 × 0.002 × 10 = 20 N. True weight = mg = 2 × 10 = 20 N. Apparent weight = 20 − 20 = 0 N (body is neutrally buoyant — floats at any level!)
Solution: F_b = ρVg = 1000 × 0.002 × 10 = 20 N. True weight = mg = 2 × 10 = 20 N. Apparent weight = 20 − 20 = 0 N (body is neutrally buoyant — floats at any level!)
Example 5: At a depth of 20 m in a lake, find (a) pressure due to water alone (b) total pressure. (ρ = 1000 kg/m³, g = 10, P_atm = 10⁵ Pa)
Solution: (a) P_water = hρg = 20 × 1000 × 10 = 2 × 10⁵ Pa. (b) P_total = P_atm + P_water = 10⁵ + 2×10⁵ = 3 × 10⁵ Pa
Solution: (a) P_water = hρg = 20 × 1000 × 10 = 2 × 10⁵ Pa. (b) P_total = P_atm + P_water = 10⁵ + 2×10⁵ = 3 × 10⁵ Pa
16) Bonus — AMU Exam Day Tips
- If a question gives dimensions in cm, convert to m before using P = F/A (1 m² = 10⁴ cm²)
- If a body "floats" → Weight = Buoyant force → use ρ_body × V_total = ρ_fluid × V_immersed
- If a body is "fully immersed" → V_immersed = V_body
- Read the question carefully: "pressure due to liquid" means hρg; "total pressure at depth" means P_atm + hρg
- For hydraulic lift questions, identify which area is the input (small) and which is output (large)
- In barometer questions, the mercury column height is the measure of atmospheric pressure
- If a ship sails from river to sea, it rises (because seawater is denser → more buoyant force for same immersed volume)
- An egg floats in salty water but sinks in fresh water — because salty water has higher density
- Dead Sea is so salty that people float effortlessly — high density of water → high buoyant force
- Always write the formula first, then substitute values with units — this gets partial credit!