Chapter 9 of 12

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CBSE Class 9 Science★ 8–10 marks · CBSE Class 9 board (Unit III — Motion, Force and Work)

Gravitation

The same force that pulls an apple to the ground keeps the moon in orbit. Class 9 Gravitation covers Newton's universal law, free fall and g, mass vs weight, thrust and pressure, buoyancy, Archimedes' principle, relative density, and 25+ exam-style problems with full step-by-step solutions.

⏱ 40 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1State Newton's universal law of gravitation and identify each variable's units
2Compute acceleration due to gravity g from G, M and R; explain its variation with altitude/latitude
3Distinguish mass and weight; convert kg to N using g; predict how weight changes on Moon or other planets
4Define thrust and pressure; compute pressure given force and area
5State Archimedes' principle and apply it to predict floating vs sinking
6Define relative density and use it to compare materials
7Solve free-fall and projectile-style numericals with g = 9.8 or 10 m/s²

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Why this chapter matters

Gravitation extends the F = ma framework to a force that acts at a distance, across the whole universe. The thrust/pressure/buoyancy half of the chapter is the foundation of fluid mechanics — used in ships, dams, plumbing, blood pressure, weather systems.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Motion (Class 9 Ch7)

Equations of motion for free fall under g.

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Force and Laws of Motion (Ch8)

F = ma, weight as force = mg.

Gravitation — Class 9 (CBSE)

When Newton saw the apple fall, he didn't ask "why does it fall?" Falling was obvious. He asked, "Does the same force that pulls the apple also reach the moon?" The answer turned out to be yes — and that single insight unified terrestrial and celestial physics for the first time. This chapter is about how gravity works on Earth, how it relates to weight, and how the same physics explains why heavy ships float.

1. The story — from an apple to a universal law

In 1666, with the plague closing Cambridge University, 23-year-old Isaac Newton retreated to his family farm. There he watched an apple fall from a tree (or so the famous story goes), and made the conceptual leap:

"If gravity reaches the top of a tree, why not the top of a mountain? And if the top of a mountain, why not the moon?"

He went on to formulate Newton's universal law of gravitation, published in 1687: every particle of matter in the universe attracts every other particle with a force proportional to the product of their masses and inversely proportional to the square of the distance between them.

This was the FIRST time a single law explained both terrestrial physics (falling apples) and celestial physics (orbiting planets). Before Newton, these were thought to obey different rules. After Newton, the universe became unified.

2. Newton's universal law of gravitation

For two point masses $m_{1}$ and $m_{2}$ separated by distance $r$ :

$F = G \frac{m _{1} m _{2}}{r ^{2}}$

Where:

$F$ = gravitational force between them (newtons, N).
$G$ = gravitational constant = $6.674 \times 1 0^{- 11} N\cdotm^{2} / kg^{2}$ . Universal and unchanging.
$m_{1}, m_{2}$ = the two masses (kg).
$r$ = distance between centres (m).

Three important properties of gravity

Always attractive — never repulsive (unlike electricity).
Acts at a distance — no contact needed; works across the vacuum of space.
Universal — the same law applies to atoms, balls, planets and galaxies.

Why we don't feel gravity from objects around us

Because $G$ is incredibly small. Two 70 kg people standing 1 m apart attract each other with force:

$F = (6.674 \times 1 0^{- 11}) \cdot \frac{70 \times 70}{1 ^{2}} \approx 3.3 \times 1 0^{- 7} N$

That's 0.00000033 newtons — totally imperceptible. It takes a massive body (like the Earth, $6 \times 1 0^{24} kg$ ) to produce a noticeable force.

3. Free fall and the acceleration due to gravity (g)

Drop an object near Earth's surface. It falls toward the Earth — that's free fall. The acceleration it experiences is $g$ .

From Newton's universal law applied to an object near Earth's surface:

$F = G \frac{M m}{R ^{2}}$

Where $M$ is Earth's mass ( $\approx 6 \times 1 0^{24}$ kg) and $R$ is Earth's radius ( $\approx 6.4 \times 1 0^{6}$ m).

But by Newton's 2nd law, $F = m g$ (weight). So:

$g = \frac{GM}{R ^{2}} = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 6 \times 1 0 ^{24} )}{( 6.4 \times 1 0 ^{6} ) ^{2}}$

Calculating: $g \approx 9.8 m/s^{2}$ .

That's why a free-falling object accelerates at $9.8 m/s^{2}$ regardless of its mass — the mass cancels in $F = m g = GM m / R^{2}$ .

What varies the value of g?

Altitude: $g$ decreases as you go higher. Top of Mount Everest: $g \approx 9.77 m/s^{2}$ .
Depth: $g$ also decreases as you go deeper into Earth (mass below shrinks). Centre of Earth: $g = 0$ .
Latitude: Earth bulges at the equator due to spin, so $R$ is larger at the equator → $g$ is slightly smaller. Equator: $g \approx 9.78$ . Poles: $g \approx 9.83$ .

For Class 9 problems, $g$ is taken as either $9.8 m/s^{2}$ or $10 m/s^{2}$ (whichever simplifies calculations).

Galileo's experiment revisited

Aristotle: heavy objects fall faster. Galileo (~1590) dropped a heavy iron ball and a light wooden ball from the Leaning Tower of Pisa — both hit the ground at the same time.

Modern proof: on the Moon (no air resistance), an astronaut dropped a hammer and a feather. They fell side by side and hit the lunar surface together. Beautifully confirmed by Apollo 15 astronaut David Scott in 1971.

The reason: $g$ is mass-independent. All objects experience the same acceleration in free fall (assuming negligible air resistance).

4. Mass vs weight

The most-confused distinction in physics. Let's be clear.

Mass ( $m$ ) — the amount of matter in a body. Scalar. SI unit: kilogram (kg). Universal — doesn't change with location.

Weight ( $W$ ) — the force exerted by gravity on a body. Vector (directed down). SI unit: newton (N).

$W = m g$

How they differ

Feature	Mass	Weight
Type	Scalar	Vector
Unit	kg	N
Depends on location?	No	Yes (depends on g)
Measured by	Pan balance (compares masses)	Spring balance (measures force)
On the Moon	Same	About 1/6 of Earth's

"I weigh 60 kg"

In everyday language, you say "I weigh 60 kg," but you're actually stating your mass. Your weight on Earth is 60 × 9.8 = 588 N. On the Moon: 60 × 1.6 = 96 N. Your mass is the same in both places.

On the Moon, you'd feel "lighter" because the gravitational force on you is smaller. Your inertia (resistance to motion) is the same.

5. Thrust and pressure

Thrust = force acting perpendicular to a surface. SI unit: newton (N).

Pressure = thrust per unit area.

$P = \frac{F}{A}$

Where $F$ is force (in N) and $A$ is area (in m²). SI unit of pressure: pascal (Pa) = N/m².

Why pressure matters

Two examples with the same force but different pressure:

A blunt knife (large contact area) doesn't cut bread easily.
A sharp knife (small contact area) cuts through the same bread — same force, same arm, same cutting motion. The difference is higher pressure due to smaller area.

This is why:

Knives are sharpened (reduce area → increase pressure).
Camels have broad feet (large area → low pressure → don't sink in sand).
A drawing pin's tip is sharp (high pressure on the wall) while its head is broad (low pressure on your finger).
Trucks have wider tyres than cars (distribute weight over more area → less damage to roads).

Pressure in fluids

Liquids and gases exert pressure on the walls of their container AND on any object placed in them.

Pascal's principle: pressure applied to an enclosed fluid is transmitted equally in all directions. Basis of hydraulic systems (car brakes, lifts, hydraulic presses).

Pressure due to a column of fluid:

$P = ρ g h$

Where $ρ$ is fluid density (kg/m³), $g$ is gravity, $h$ is height. So pressure increases linearly with depth.

6. Buoyancy and Archimedes' principle

When an object is partially or fully immersed in a fluid, the fluid exerts an upward force on it called buoyant force or upthrust. This is buoyancy.

Why buoyancy exists

A submerged object experiences pressure from all sides. The pressure at the bottom is higher than at the top (deeper fluid → more pressure). The pressure difference produces a net upward force = buoyant force.

Archimedes' principle

The buoyant force on an object equals the weight of the fluid the object displaces.

$F_{b} = weight of displaced fluid = ρ_{fluid} \cdot V_{submerged} \cdot g$

Three cases when an object is placed in a fluid:

Object sinks: object's density > fluid's density. Buoyant force < object's weight. Net force downward → sinks.
Object floats: object's density ≤ fluid's density. Buoyant force = object's weight. Net force zero → floats in equilibrium.
Object floats partly submerged: object adjusts how much it submerges until displaced fluid's weight equals its own weight.

Why a ship floats and a coin sinks

The coin: small volume, high density (much higher than water). It displaces only a tiny volume of water; the displaced water's weight is much less than the coin's weight. Coin sinks.

The ship: hollow shape means LARGE volume despite the steel's high density. The ship displaces a huge volume of water; the displaced water's weight equals the ship's total weight. Ship floats.

Relative density

Relative density (specific gravity) = ratio of an object's density to water's density.

$Relative density = \frac{Density of substance}{Density of water} = \frac{ρ _{substance}}{1000 kg/m ^{3}}$

If relative density < 1: floats in water. If > 1: sinks.

Common values:

Cork: ≈ 0.2 (floats).
Wood: ≈ 0.7 (floats).
Ice: ≈ 0.92 (floats — but only just; 92% submerged).
Water: 1 (reference).
Aluminium: 2.7.
Iron: 7.8.
Lead: 11.3.
Gold: 19.3.

This is also why icebergs are dangerous to ships — only ~8% of the ice is above water; ~92% is hidden below.

7. Worked example — buoyancy in water

A solid block of density $800 kg/m^{3}$ and volume $0.5 m^{3}$ is placed in water. Will it sink or float? If it floats, what fraction is submerged?

Step 1 — Density check.

Block's density: 800 kg/m³.
Water's density: 1000 kg/m³.
Block's density < water's density → floats.

Step 2 — Find fraction submerged.

The block floats in equilibrium → buoyant force = weight of block.

$ρ_{water} \cdot V_{submerged} \cdot g = ρ_{block} \cdot V_{total} \cdot g$

Cancel $g$ : $ρ_{water} \cdot V_{submerged} = ρ_{block} \cdot V_{total}$ $\frac{V _{submerged}}{V _{total}} = \frac{ρ _{block}}{ρ _{water}} = \frac{800}{1000} = 0.8$

Answer: 80% of the block is submerged. The remaining 20% is above water.

This is a general result: fraction submerged = ratio of densities.

8. Closing thought

The universal law of gravitation is one of those rare equations that connects everything:

Why apples fall.
Why the moon doesn't fall but orbits.
Why ocean tides exist (moon's gravity pulling on the oceans).
Why galaxies form (gravity over cosmic time pulling matter together).
Why GPS satellites need clock corrections (gravity slows time, slightly).

The same equation gives you everyday weight and buoyancy via the simple chain $F = GM m / R^{2} = m g$ → $W = m g$ → buoyant force = weight of displaced fluid.

Three centuries on, this law is so well-tested that NASA still uses Newton's gravity (not Einstein's general relativity) to compute spacecraft trajectories. Only at extreme conditions (very strong fields or very high speeds) do Einstein's corrections matter. Newton built the physics that put humans on the moon.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Newton's universal gravitation

F = G × m₁m₂ / r²

G = 6.674 × 10⁻¹¹ N·m²/kg². Universal constant.

Acceleration due to gravity

g = G × M / R²

M = Earth mass, R = Earth radius. g ≈ 9.8 m/s² at surface.

Weight

W = m × g

Force of gravity on a mass. SI unit: newton. Different on different planets.

Equations of motion (free fall)

v = u + gt; h = ut + ½gt²; v² = u² + 2gh

Replace 'a' with 'g'. Sign convention: take downward positive (or specify your axis).

Pressure

P = F / A

SI unit: pascal (Pa) = N/m². Higher P = same force on smaller area.

Pressure in fluids

P = ρ × g × h

Depth-dependent pressure. ρ = density, h = depth.

Density

ρ = m / V

Mass per unit volume. SI unit: kg/m³.

Relative density (specific gravity)

R.D. = density of substance / density of water (= 1000 kg/m³)

Dimensionless. < 1 floats, > 1 sinks.

Archimedes' principle

Buoyant force = ρ_fluid × V_submerged × g

Force = weight of fluid displaced. Always upward.

Fraction submerged (floating)

= ρ_object / ρ_fluid

Used for ice on water (~92% submerged), icebergs, etc.

⚠️

Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Saying gravity gets weaker as masses get bigger

✓ Opposite: F = Gm₁m₂/r². Bigger masses → BIGGER force. The force gets weaker with distance (1/r²), not with mass.

WATCH OUT

✗ Confusing mass and weight in numerical problems

✓ If a quantity has units of kg → mass. If newtons → weight. Always check units first.

WATCH OUT

✗ Saying weight on the moon is 6 times less because mass is less

✓ Mass is the SAME on the moon. Weight is less because g_moon ≈ 1.6 m/s² (vs Earth's 9.8 ≈ 6×). Same body, different gravity, different weight.

WATCH OUT

✗ Forgetting that buoyant force depends on VOLUME submerged, not mass

✓ F_b = ρ_fluid × V_submerged × g. A balloon of any mass — what matters is V_submerged. Object density determines whether it sinks (then V_submerged = V_total) or floats (V_submerged < V_total).

WATCH OUT

✗ Using pressure formula P = ρgh for solids

✓ P = ρgh is for FLUIDS only. For solid objects on a surface, use P = F/A = (mg)/A.

WATCH OUT

✗ Saying objects feel 'lighter' in water because gravity is weaker

✓ Gravity is the same — the body experiences an UPWARD buoyant force from the water that partially cancels its weight. Apparent weight = actual weight − buoyant force.

WATCH OUT

✗ Confusing density and relative density

✓ Density has units (kg/m³). Relative density is dimensionless. R.D. of iron = 7.8 means iron's density is 7.8 × that of water = 7800 kg/m³.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Section 9.1 (in-text)

Universal law of gravitation; variation of g

Questions

Section 9.2 (in-text)

Free fall; g; mass vs weight; numericals

Questions

Section 9.3 (in-text)

Thrust and pressure; pressure in fluids; Pascal's principle

Questions

Section 9.4 (in-text)

Buoyancy; Archimedes' principle; floating/sinking

Questions

Section 9.5 (in-text)

Relative density; experimental determination

Questions

End-of-chapter

Mixed: gravitational force, free-fall, weight calculations, buoyancy

Questions

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Definition

State Newton's universal law of gravitation.

▶ Show solution

Step 1 — Write the formal statement.
  Every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Step 2 — Symbolic form.
  F = G × m₁ × m₂ / r².
  G = 6.674 × 10⁻¹¹ N·m²/kg², the universal gravitational constant.

✦ Answer: As above. F ∝ m₁m₂ / r².

Q2EASY· Weight

Find the weight of a 60 kg person on (a) Earth and (b) the Moon. (g_Earth = 10 m/s², g_Moon = 1.6 m/s²)

▶ Show solution

Step 1 — W = mg.
  W_Earth = 60 × 10 = 600 N.
  W_Moon = 60 × 1.6 = 96 N.

✦ Answer: (a) 600 N on Earth, (b) 96 N on the Moon.

Note: Mass is the same (60 kg) in both places. Only weight changes because g is different.

Q3EASY· Pressure

A box of weight 200 N is placed on a table. The base of the box has area 0.5 m². What pressure does the box exert on the table?

▶ Show solution

Step 1 — Use P = F/A.
  F = weight = 200 N. A = 0.5 m².

Step 2 — Substitute.
  P = 200 / 0.5 = 400 Pa.

✦ Answer: 400 Pa.

Q4EASY· Free fall

A ball is dropped from a height of 20 m. Find the time it takes to reach the ground. (g = 10 m/s²)

▶ Show solution

Step 1 — Use h = ut + ½gt² with u = 0.
  20 = 0 + ½ × 10 × t²
  20 = 5t²
  t² = 4
  t = 2 s.

✦ Answer: 2 seconds.

Q5EASY· Buoyancy

Why does an iron nail sink in water but a much heavier ship made of iron floats?

▶ Show solution

Step 1 — Key principle: floating depends on DENSITY (or volume relative to mass), not on weight alone.

Step 2 — Iron nail.
  Nail is solid iron. Density of iron (~ 7800 kg/m³) > density of water (1000 kg/m³). Buoyant force < weight → SINKS.

Step 3 — Iron ship.
  Ship is HOLLOW. Its overall (average) density (mass / total volume including the air pocket) is LESS than water's density. By Archimedes' principle, the ship displaces a huge volume of water; weight of displaced water = ship's total weight → FLOATS.

✦ Answer: A nail's solid-iron density > water's. A ship's hollow design gives a much LOWER average density (mass/volume including the interior air), allowing buoyant force to equal its weight. Hence ship floats; nail sinks.

Q6MEDIUM· F = GmM/r²

Two bodies of masses 100 kg and 200 kg are placed 10 m apart. Find the gravitational force between them. (G = 6.67 × 10⁻¹¹ N·m²/kg²)

▶ Show solution

Step 1 — Use F = G × m₁m₂ / r².

Step 2 — Substitute.
  F = (6.67 × 10⁻¹¹) × (100 × 200) / (10²)
   = (6.67 × 10⁻¹¹) × 20,000 / 100
   = (6.67 × 10⁻¹¹) × 200
   = 1.334 × 10⁻⁸ N.

✦ Answer: ≈ 1.33 × 10⁻⁸ N (about 0.000000013 N).

That's incredibly small — which is why we don't feel gravity from each other in a room.

Q7MEDIUM· Throw up

A stone is thrown vertically upward with initial velocity 20 m/s. Find (a) the maximum height reached, (b) the time to reach the maximum height. (g = 10 m/s²)

▶ Show solution

Step 1 — Take upward as positive. u = +20, a = −10 m/s² (gravity opposes upward motion). At max height, v = 0.

Step 2 (a) — Use v² = u² + 2as.
  0 = (20)² − 2 × 10 × h
  h = 400/20 = 20 m.

Step 3 (b) — Use v = u + at.
  0 = 20 − 10t → t = 2 s.

✦ Answer: (a) Maximum height = 20 m. (b) Time to max height = 2 s.

Q8MEDIUM· Pressure depth

Find the pressure exerted by sea water (density 1030 kg/m³) at a depth of 100 m. (g = 10 m/s²)

▶ Show solution

Step 1 — Use P = ρgh.

Step 2 — Substitute.
  P = 1030 × 10 × 100
   = 1,030,000 Pa
   ≈ 1.03 × 10⁶ Pa
   ≈ 1030 kPa
   ≈ 10 atm (atmospheric pressures).

Step 3 — Note: this is JUST the water pressure at that depth. The total pressure at that depth is water pressure + atmospheric pressure (~1 atm) = ~ 11 atm.

✦ Answer: P ≈ 1.03 × 10⁶ Pa (≈ 10 atm of water pressure).

This is why deep-sea submarines need extremely thick hulls — at 100 m depth, the pressure on each square metre of hull is ~ 10 tons of force.

Q9MEDIUM· Relative density

An object has mass 200 g and volume 50 cm³. Find its density and relative density. State whether it will sink or float in water.

▶ Show solution

Step 1 — Density = mass/volume = 200 g / 50 cm³ = 4 g/cm³.
  In SI: 4 g/cm³ = 4000 kg/m³.

Step 2 — Relative density = density of object / density of water.
  = 4000 / 1000 = 4. (Equivalently, 4 g/cm³ / 1 g/cm³ = 4.)

Step 3 — Verdict.
  Density (4000) > density of water (1000) → R.D. > 1 → SINKS.

✦ Answer: Density = 4 g/cm³ = 4000 kg/m³. Relative density = 4. The object will SINK in water.

Q10MEDIUM· g variation

Why is the value of g less at the equator than at the poles?

▶ Show solution

Step 1 — Two effects make g smaller at the equator.

(i) Earth's shape:
  Earth is not a perfect sphere — it bulges at the equator (oblate spheroid). The equatorial radius (R_eq ≈ 6378 km) is LARGER than the polar radius (R_pole ≈ 6357 km).
  Since g = GM/R², a larger R gives smaller g at the equator.

(ii) Earth's rotation:
  The earth spins on its axis. At the equator, this spin creates a 'centrifugal effect' that effectively subtracts from g (this is a non-inertial-frame effect; in the rotating frame of the Earth, you feel slightly less weight at the equator).
  At the poles, you're on the axis of rotation — no centrifugal effect.

Step 2 — Combined: g_pole ≈ 9.83 m/s², g_equator ≈ 9.78 m/s² (about 0.5% difference).

✦ Answer: (i) Earth is oblate — larger radius at the equator → g = GM/R² gives smaller g there. (ii) Earth's rotation produces a centrifugal effect at the equator that effectively reduces g, while the poles have no such effect.

Q11HARD· Buoyancy

A cubical wooden block of side 10 cm and density 600 kg/m³ floats in water. (a) What fraction of the block is submerged? (b) What is the depth of the submerged part below the water surface?

▶ Show solution

Step 1 — Find total volume.
  V_total = (0.1 m)³ = 10⁻³ m³.

Step 2 (a) — Fraction submerged.
  By Archimedes: fraction submerged = ρ_block / ρ_water = 600 / 1000 = 0.6.
  So 60 % of the block is under water.

Step 3 (b) — Depth of the submerged part.
  Submerged volume = 0.6 × V_total = 0.6 × 10⁻³ = 6 × 10⁻⁴ m³.
  Since the cross-section (top view) of the block is 0.1 × 0.1 = 0.01 m², the depth h is given by:
  V_submerged = A × h ⇒ 6 × 10⁻⁴ = 0.01 × h ⇒ h = 0.06 m = 6 cm.

✦ Answer: (a) 60 % submerged; (b) Submerged depth = 6 cm.

Q12HARD· Numerical

Calculate the force of gravitational attraction between Earth (mass 6 × 10²⁴ kg) and a person of mass 60 kg standing on Earth's surface. (G = 6.67 × 10⁻¹¹, R_Earth = 6.4 × 10⁶ m). Show that the result is the person's weight on Earth.

▶ Show solution

Step 1 — Use F = G × Mm/R².

Step 2 — Substitute.
  F = (6.67 × 10⁻¹¹) × (6 × 10²⁴ × 60) / (6.4 × 10⁶)²
   = (6.67 × 10⁻¹¹) × (3.6 × 10²⁶) / (4.096 × 10¹³)
   = (6.67 × 10⁻¹¹) × 8.79 × 10¹²
   ≈ 586 N.

Step 3 — Compare to weight from W = mg.
  W = 60 × 9.8 = 588 N. ✓ (within rounding error of 586 N).

Step 4 — Conclusion.
  The two methods give the same answer because g IS defined as GM/R². So mg = m × GM/R² = GMm/R² is the force of gravity — by construction.

✦ Answer: F ≈ 586 N, which equals W = mg = 588 N. This proves that weight = gravitational pull of Earth on the body.

Q13HARD· Free fall

A body is thrown vertically upward from a tower of height 25 m with velocity 15 m/s. Find the total time the body remains in the air before hitting the ground. (g = 10 m/s²)

▶ Show solution

Step 1 — Pick up as positive. Body starts at top of tower (set this as origin). u = +15, a = −10. Ground is at h = −25 (below origin).

Step 2 — Use h = ut + ½at².
  −25 = 15t + ½(−10)t²
  −25 = 15t − 5t²
  5t² − 15t − 25 = 0
  t² − 3t − 5 = 0.

Step 3 — Solve quadratic.
  t = (3 ± √(9 + 20))/2 = (3 ± √29)/2.
  Take the positive root: t = (3 + 5.385)/2 ≈ 4.19 s.
  (The negative root would correspond to time before the throw — unphysical.)

✦ Answer: Total time ≈ 4.19 s.

Sanity check: time to max height = u/g = 15/10 = 1.5 s. So body goes up for 1.5 s, then falls. From max height, body falls (25 + ½ × 10 × 1.5²) = 25 + 11.25 = 36.25 m. Fall time: √(2 × 36.25/10) = √7.25 ≈ 2.69 s. Total = 1.5 + 2.69 ≈ 4.19 s. ✓

Q14HARD· HOTS

A ship floats with 5 m of its hull below the water surface in fresh water. When it sails into sea water, the hull is now only 4.85 m below the surface. Why? (density_seawater = 1030 kg/m³, density_freshwater = 1000 kg/m³)

▶ Show solution

Step 1 — Use Archimedes: weight of ship = weight of displaced water.
  In fresh water: W = ρ_fw × V_fw × g, where V_fw = submerged volume in fresh water.
  In sea water: W = ρ_sw × V_sw × g.

Step 2 — Same ship → same weight in both cases.
  ρ_fw × V_fw = ρ_sw × V_sw
  1000 × V_fw = 1030 × V_sw
  V_sw / V_fw = 1000/1030 ≈ 0.971.

Step 3 — Volumes scale linearly with depth (constant cross-section).
  Depth in sea water = 0.971 × Depth in fresh water = 0.971 × 5 ≈ 4.85 m. ✓

Step 4 — Physical interpretation.
  Sea water is DENSER than fresh water. Same displaced volume in sea water has more mass → more upward buoyant force per unit volume. So the ship needs to displace LESS volume to balance its weight → sits HIGHER (less depth submerged) in sea water.

✦ Answer: Sea water (denser than fresh water) gives more buoyant force per unit volume of displacement. The ship needs to displace less volume to support its weight → sits ~3% higher in sea water. Hence depth submerged decreases from 5.00 m to 4.85 m.

Q15HARD· Pressure

Why is it easier to swim in sea water than in fresh water?

▶ Show solution

Step 1 — Density comparison.
  Sea water density ≈ 1030 kg/m³. Fresh water density ≈ 1000 kg/m³. Sea water is ~3% denser.

Step 2 — Apply Archimedes' principle.
  Buoyant force on a swimmer = ρ_water × V_submerged × g.
  Greater density → greater buoyant force for the same submerged volume.
  Sea water gives ~3% more buoyant force than fresh water for the same swimmer.

Step 3 — What this means for the swimmer.
  The swimmer's apparent weight (actual weight − buoyant force) is reduced more in sea water.
  Less downward 'effective weight' means less energy is needed to stay afloat → easier to swim.

Step 4 — Extreme example.
  The Dead Sea (Israel/Jordan) has water density ≈ 1240 kg/m³ (extremely salty). Buoyant force is so high that people FLOAT on the surface with much of their body above water — they can't even fully submerge themselves without effort.

✦ Answer: Sea water is denser than fresh water → larger buoyant force on the swimmer's body → smaller effective weight → swimmer doesn't have to expend as much energy to stay afloat. Easiest case: Dead Sea, where people automatically float without swimming.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Newton's law of gravitation: F = Gm₁m₂/r². G = 6.674 × 10⁻¹¹ N·m²/kg² (universal constant).
•Acceleration due to gravity: g = GM/R² ≈ 9.8 m/s² at Earth's surface.
•g varies: lower at the equator (Earth bulges + spin effect); lower at altitude or depth.
•Mass = matter content (kg, scalar). Weight = gravitational force (N, vector). W = mg.
•Free-fall equations: replace 'a' with 'g' in motion equations.
•Thrust = force perpendicular to surface. Pressure P = F/A. Unit: Pa = N/m².
•Pressure in a fluid at depth h: P = ρgh.
•Pascal's principle: pressure applied to a confined fluid is transmitted equally in all directions. Basis of hydraulics.
•Archimedes' principle: buoyant force = weight of fluid displaced.
•Fraction submerged when floating = ρ_object / ρ_fluid.
•Relative density (specific gravity) = ρ_substance / ρ_water (dimensionless). < 1 floats, > 1 sinks.

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Questions students ask

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Mass is always positive (no 'negative mass' in classical physics). Two positive masses always attract. Electric charge can be + or −, so like charges repel and unlike charges attract — but mass has no analogous duality.

It IS falling — but it has enough sideways velocity that as it 'falls' toward Earth, it also moves sideways enough to keep missing. Result: a stable orbit. Newton showed that orbiting is just 'continuously falling without ever hitting the ground.'

Feeling heavy = weight = mg. Heavier mass → bigger weight (more force on your hand). But the ACCELERATION when dropped (g) is the same. Holding still requires you to support the weight; dropping doesn't.

P = F/A. Same force F divided by smaller A gives a bigger P. That's why a sharp knife cuts: the same arm force concentrated into a tiny cross-section creates immense pressure at the edge.

No — solid ice is just slightly less dense than liquid water (density 920 vs 1000 kg/m³ — about 8 % less). That's why 92 % of an iceberg is underwater. Water expanding when it freezes is unusual; most substances become denser as solids.

Yes — but only your APPARENT weight (what a spring scale reads). In a downward-accelerating elevator, your scale reads less because the floor presses up on you with less force. Your actual mass and the actual gravitational force are unchanged. (In free fall, scale reads zero — 'weightlessness'.)

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