Gravitation

www.akankshaclasses.com
CLASS IX Science ~8 marks/year Ch 10 of 15
Gravitation

Class 9 · Science · NCERT chapter notes · Akanksha Classes

Snapshot
  • Universal Law of Gravitation: every object in the universe attracts every other object with a force $F = \dfrac{Gm_1m_2}{r^2}$; $G = 6.674\times10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$.
  • Acceleration due to gravity: $g = 9.8\ \text{m s}^{-2}$ at Earth's surface — derived from the Universal Law as $g = GM/R^2$.
  • Free fall: any object falling under gravity alone (air resistance ignored) accelerates at $g$ downward; all objects fall at the same rate in vacuum.
  • Weight $= mg$ — a force (vector, newtons); mass is a scalar, constant everywhere in the universe.
  • Weight on Moon $= \dfrac{1}{6}\times$ weight on Earth, because $g_\text{Moon}\approx\dfrac{g_\text{Earth}}{6}$.
  • Pressure $= F/A$ (Pa); fluids exert pressure in all directions; $P = h\rho g$ at depth $h$.
  • Archimedes' Principle: buoyant force = weight of fluid displaced by the immersed body.
  • Relative density $= \dfrac{\text{density of substance}}{\text{density of water}}$ — dimensionless; $<1$ floats, $>1$ sinks.
  • Board weightage: ~8 marks/year — typically one derivation ($g$ from universal law), one weight/mass numerical, one Archimedes' / buoyancy problem.
Detailed Notes

1. Universal Law of Gravitation

Isaac Newton observed that the same force pulling an apple to the ground also keeps the Moon in orbit around Earth. In 1687 he stated:

Every object in the universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The force acts along the line joining the centres of the two bodies.

Mathematically, the gravitational force between two point masses $m_1$ and $m_2$ separated by a distance $r$ is:

$$F = G\,\frac{m_1\,m_2}{r^2}$$

where $G = 6.674\times10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$ is the Universal Gravitational Constant.

  • $F$ is always attractive — it acts along the line joining the two bodies.
  • $G$ is the same everywhere in the universe; it was first measured by Henry Cavendish in 1798 using a torsion balance.
  • If $r$ doubles, $F$ becomes $\dfrac{1}{4}$ of its original value (inverse-square relationship).
  • If either mass doubles, $F$ doubles (directly proportional to each mass).
  • $G$ is not the same as $g$: $G$ is universal and constant; $g$ varies with location.
  • Newton's third law applies: Earth pulls the apple downward with force $F$, and the apple pulls Earth upward with the same force $F$ — but Earth's acceleration is negligible because of its enormous mass.
NCERT Example 10.1 — Force between Earth and Moon

Given: mass of Earth $M_E = 6\times10^{24}\ \text{kg}$, mass of Moon $M_M = 7.4\times10^{22}\ \text{kg}$, distance $r = 3.84\times10^{8}\ \text{m}$, $G = 6.7\times10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$.

$$F = \frac{G\,M_E\,M_M}{r^2} = \frac{6.7\times10^{-11}\times6\times10^{24}\times7.4\times10^{22}}{(3.84\times10^{8})^2}$$

Numerator: $6.7\times6\times7.4 = 297.48$; powers: $10^{-11+24+22}=10^{35}$. So numerator $= 297.48\times10^{35}$.

Denominator: $(3.84)^2\times10^{16} = 14.7456\times10^{16}$.

$$F = \frac{297.48\times10^{35}}{14.7456\times10^{16}} \approx 20.18\times10^{19} \approx \boxed{2.02\times10^{20}\ \text{N}}$$

2. Importance of the Universal Law of Gravitation

The Universal Law unified what seemed like completely unrelated phenomena:

  • It explained the force that binds us to Earth and prevents the atmosphere from escaping into space.
  • It explained the motion of the Moon around Earth and of planets around the Sun (Kepler's laws are consequences of it).
  • It explained tides — the periodic rise and fall of sea level caused by the differential gravitational pull of the Moon (and to a lesser extent, the Sun) on different parts of Earth.
  • It predicted the existence of Neptune — astronomers noticed Uranus was being pulled off course and used Newton's law to calculate where an unseen planet must be. Neptune was found exactly there in 1846.
  • It is used to calculate satellite trajectories, launch velocities, and orbital periods.
  • It explains why objects fall to Earth and not away from it.

The law applies from two dust particles to two galaxies — making it genuinely universal.

3. Free Fall and Acceleration Due to Gravity

When an object falls toward Earth under the influence of gravity alone (no air resistance, no support), it is said to be in free fall.

Galileo's discovery: all freely falling objects near Earth's surface accelerate at the same rate, regardless of their mass. A hammer and a feather fall at the same rate in a vacuum — as NASA demonstrated on the Moon in 1971.

This acceleration is called the acceleration due to gravity, denoted $g$:

$$g = 9.8\ \text{m s}^{-2}\ \text{(at Earth's surface, directed downward)}$$

Key points about free fall:

  • The direction of $g$ is always toward the centre of Earth (i.e., downward).
  • $g$ is always positive in the downward direction; when an object moves upward, $g$ acts opposite to its motion, causing deceleration.
  • In air, a feather falls more slowly than a stone because of air resistance — but in a vacuum, both fall identically.
  • $g$ varies slightly with location: slightly less at the equator (farther from Earth's centre) and slightly more at the poles; it decreases with altitude and depth.
  • For NCERT problems, use $g = 9.8\ \text{m s}^{-2}$ unless told to use $g = 10\ \text{m s}^{-2}$.

4. Calculating $g$ from the Universal Law of Gravitation

Consider an object of mass $m$ on Earth's surface. Earth has mass $M$ and radius $R$. By the Universal Law, Earth pulls the object with:

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

By Newton's second law, $F = ma$. For free fall, $a = g$, so $F = mg$. Equating:

$$mg = G\,\frac{Mm}{R^2} \quad\Longrightarrow\quad \boxed{g = \frac{GM}{R^2}}$$

Substituting known values:

  • $G = 6.7\times10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$
  • $M = 6\times10^{24}\ \text{kg}$ (mass of Earth)
  • $R = 6.4\times10^{6}\ \text{m}$ (radius of Earth)

$$g = \frac{6.7\times10^{-11}\times6\times10^{24}}{(6.4\times10^{6})^2} = \frac{40.2\times10^{13}}{40.96\times10^{12}} = \frac{402\times10^{12}}{40.96\times10^{12}} \approx 9.8\ \text{m s}^{-2}$$

Important conclusions from $g = GM/R^2$:

  • $g$ does not depend on the mass of the falling object — it depends only on Earth's mass and radius.
  • At height $h$ above the surface: $g_h = \dfrac{GM}{(R+h)^2}$ — $g$ decreases with altitude.
  • On different planets, $g$ varies: Moon $\approx 1.63\ \text{m s}^{-2}$, Mars $\approx 3.7\ \text{m s}^{-2}$, Jupiter $\approx 24.8\ \text{m s}^{-2}$.

5. Equations of Motion Applied to Free Fall

The three kinematic equations apply to free fall with $a = g = 9.8\ \text{m s}^{-2}$. Let $u$ = initial velocity, $v$ = final velocity, $s$ = displacement, $t$ = time:

Falling downward (take downward as positive):

1.  $v = u + g\,t$

2.  $s = u\,t + \dfrac{1}{2}g\,t^2$

3.  $v^2 = u^2 + 2g\,s$

Thrown upward (take upward as positive, so $a = -g$):

Replace $g$ with $-g$ in all three equations.

Key results for upward throw with initial speed $u_0$:

  • Maximum height: $H = \dfrac{u_0^2}{2g}$
  • Time to reach top: $t_\text{up} = \dfrac{u_0}{g}$
  • Total time of flight: $T = \dfrac{2u_0}{g}$ (time up = time down)
  • Speed when it returns to the launch point = $u_0$ (same magnitude, opposite direction)
NCERT Example 10.2 — Object dropped from a cliff

An object is dropped from a cliff of height $19.6\ \text{m}$. Find (a) the time to reach the ground and (b) the velocity on impact. ($g = 9.8\ \text{m s}^{-2}$)

Given: $u = 0$ (dropped), $s = 19.6\ \text{m}$, $g = 9.8\ \text{m s}^{-2}$.

(a) Using $s = ut + \dfrac{1}{2}gt^2$:

$19.6 = 0 + \dfrac{1}{2}(9.8)t^2 = 4.9\,t^2 \Rightarrow t^2 = \dfrac{19.6}{4.9} = 4 \Rightarrow \boxed{t = 2\ \text{s}}$.

(b) Using $v = u + gt = 0 + 9.8\times2 = \boxed{19.6\ \text{m s}^{-1}}$.

Check with $v^2 = u^2 + 2gs = 0 + 2(9.8)(19.6) = 384.16 \Rightarrow v = 19.6\ \text{m s}^{-1}$. Consistent.

NCERT Example 10.3 — Object thrown vertically upward

An object is thrown vertically upward with initial velocity $19.6\ \text{m s}^{-1}$. Find (a) the maximum height reached and (b) the time taken to return to the starting point.

Take upward as positive; $u = +19.6\ \text{m s}^{-1}$, $g = -9.8\ \text{m s}^{-2}$.

(a) At maximum height, $v = 0$. $v^2 = u^2 + 2gs$:

$0 = (19.6)^2 + 2(-9.8)s \Rightarrow 19.6\,s = \dfrac{(19.6)^2}{2} = \dfrac{384.16}{2} = 192.08 \Rightarrow \boxed{s = 19.6\ \text{m}}$.

(b) Time to reach top: $0 = 19.6 - 9.8\,t \Rightarrow t = 2\ \text{s}$. Total time = $2\times2 = \boxed{4\ \text{s}}$.

6. Mass vs Weight

Students often confuse mass and weight. They are fundamentally different quantities:

Property Mass Weight
DefinitionQuantity of matter in a bodyForce of gravity acting on the body
Formula— (intrinsic)$W = m\,g$
Quantity typeScalarVector (directed toward Earth's centre)
SI Unitkilogram (kg)newton (N)
Varies?Constant everywhereVaries with $g$ (changes on Moon, at altitude, etc.)
Measured byPhysical balance (beam balance)Spring balance
$$W = m \times g$$

Example: weight of a $10\ \text{kg}$ object on Earth $= 10\times9.8 = 98\ \text{N}$.

Why a beam balance works anywhere: a beam balance compares the gravitational pull on two masses — since both experience the same $g$ at any location, the balance reading (mass) is correct even on the Moon. A spring balance measures the actual force $W = mg$, so it reads differently on the Moon.

Weightlessness in space: an astronaut in orbit is in continuous free fall toward Earth — the spacecraft falls at the same rate, so there is no contact force between them. The astronaut feels "weightless" even though gravity still acts strongly (it is what keeps them in orbit).

7. Weight on the Moon

The Moon's mass is about $\dfrac{1}{100}$ of Earth's mass, and its radius is about $\dfrac{1}{4}$ of Earth's radius. Applying $g = GM/R^2$:

$$g_\text{Moon} = \frac{G(M_E/100)}{(R_E/4)^2} = \frac{G\,M_E\times16}{100\,R_E^2} = \frac{16}{100}\,g_E \approx \frac{1}{6}\,g_E$$

So $g_\text{Moon} \approx 1.63\ \text{m s}^{-2}$, which is $\dfrac{1}{6}$ of $g_\text{Earth} = 9.8\ \text{m s}^{-2}$.

$$W_\text{Moon} = m\,g_\text{Moon} = m\times\frac{g_\text{Earth}}{6} = \frac{W_\text{Earth}}{6}$$
NCERT Example 10.4 — Weight of a man on the Moon

A man weighs $600\ \text{N}$ on Earth. (a) What is his mass? (b) What is his weight on the Moon?

(a) $m = \dfrac{W_E}{g_E} = \dfrac{600}{10} = \boxed{60\ \text{kg}}$ (using $g\approx10\ \text{m s}^{-2}$).

(b) $W_\text{Moon} = m\,g_\text{Moon} = 60\times\dfrac{10}{6} = 60\times1.67 \approx \boxed{100\ \text{N}}$.

Or directly: $W_\text{Moon} = \dfrac{W_E}{6} = \dfrac{600}{6} = 100\ \text{N}$. His mass remains $60\ \text{kg}$ on the Moon.

On the Moon, you can jump about 6 times higher with the same muscular effort, because $g$ is only $1/6$ of Earth's.

8. Thrust and Pressure

Thrust is the force acting perpendicular (normal) to a surface. Its SI unit is the newton (N).

Pressure is the thrust per unit area on which it acts:

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

SI unit: pascal (Pa). $1\ \text{Pa} = 1\ \text{N m}^{-2}$. Named after Blaise Pascal (1623–1662).

Why area matters — everyday examples:

  • A knife cuts because all force concentrates on a tiny edge — extremely high pressure on the object being cut.
  • A nail is pointed so that a small hammer blow creates very high pressure to penetrate wood.
  • School bags with wide straps distribute the weight over a larger area, reducing pressure on shoulders.
  • Army tanks have broad caterpillar tracks — to spread their enormous weight over a large area and avoid sinking into soft ground (same principle as snow shoes).
  • Camel's feet are padded and large — they spread weight over a big area to avoid sinking into sand.
  • A person wearing stiletto heels exerts more pressure on the floor than an elephant, because the heel area is tiny.
NCERT Example 10.5 — Pressure from a wooden block

A block of wood measures $0.4\ \text{m}\times0.2\ \text{m}\times0.1\ \text{m}$ and has mass $8\ \text{kg}$. Find the pressure on the table when resting on (a) its largest face, (b) its smallest face. ($g = 9.8\ \text{m s}^{-2}$)

Thrust (weight) $= 8\times9.8 = 78.4\ \text{N}$ in both cases.

(a) Largest face $= 0.4\times0.2 = 0.08\ \text{m}^2$. $P = \dfrac{78.4}{0.08} = \boxed{980\ \text{Pa}}$.

(b) Smallest face $= 0.2\times0.1 = 0.02\ \text{m}^2$. $P = \dfrac{78.4}{0.02} = \boxed{3920\ \text{Pa}}$.

The pressure on the smallest face is $4\times$ that on the largest face — same weight, $1/4$ the area.

9. Pressure in Fluids

A fluid (liquid or gas) exerts pressure on any surface in contact with it. Unlike solids, fluids exert pressure in all directions — upward, downward, and sideways.

Pressure at depth $h$ in a liquid of density $\rho$:

$$P = h\,\rho\,g$$

This arises because the column of liquid of height $h$ and cross-sectional area $A$ has weight $= \rho\,A\,h\,g$, so pressure $= \text{weight}/A = h\rho g$.

Key facts about fluid pressure:

  • Pressure at a given depth is the same in all horizontal directions at that depth (Pascal's principle).
  • Pressure increases with depth — this is why submarine hulls must be very thick and why deep-sea fish would be crushed at the surface.
  • Pressure depends only on depth, not on the shape or cross-section of the container (hydrostatic paradox).
  • Atmospheric pressure at sea level $\approx 1.013\times10^5\ \text{Pa} = 1\ \text{atm}$ — equivalent to the weight of a $10\ \text{m}$ tall column of water above every square metre.
  • We do not feel atmospheric pressure crushing us because the pressure of fluids inside our bodies (blood, tissue fluid) pushes outward equally.

10. Buoyancy and the Buoyant Force

Push a rubber ball into water — you feel an upward force pushing it back. This upward force exerted by a fluid on an immersed object is called the buoyant force or upthrust.

Why does buoyancy arise? A submerged object experiences higher fluid pressure on its lower face (greater depth) than on its upper face (lesser depth). The net upward force from this pressure difference is the buoyant force.

Effect on apparent weight: A body weighed in air shows weight $W_1$; the same body fully submerged in water shows lower reading $W_2$. The difference $W_1 - W_2$ equals the buoyant force — this "apparent loss of weight" is the upthrust.

When does an object float or sink?

  • Buoyant force $>$ weight: object is pushed up and floats (partially submerged, with enough volume submerged to make buoyant force equal weight at equilibrium).
  • Buoyant force $=$ weight: object is suspended in the fluid — neutral buoyancy (floats fully submerged without sinking or rising).
  • Buoyant force $<$ weight: object sinks to the bottom.

Equivalently — an object floats if its average density is less than the fluid density; it sinks if greater.

  • Wood (density $\approx 600\ \text{kg m}^{-3}$) floats in water ($1000\ \text{kg m}^{-3}$).
  • Iron ($7800\ \text{kg m}^{-3}$) sinks in water — but a hollow steel ship floats because its average density (including the enclosed air) is less than water.
  • Ice ($917\ \text{kg m}^{-3}$) floats in water — about $\dfrac{1}{11}$ of an iceberg protrudes above the surface.

11. Archimedes' Principle

Statement:

When a body is wholly or partially immersed in a fluid, it experiences an upward buoyant force (upthrust) equal to the weight of the fluid displaced by it.
$$F_b = \rho_\text{fluid}\times V_\text{displaced}\times g = \text{weight of fluid displaced}$$

where $V_\text{displaced}$ is the volume of fluid displaced (equal to the submerged volume of the object).

Historical note: Archimedes (287–212 BCE) discovered this principle while stepping into a bath — he noticed the water level rose by an amount equal to the volume of his body that was submerged. He used this to prove that the King of Syracuse's crown was not pure gold, without melting it.

Applications of Archimedes' Principle:

  • Ships and submarines: ships are shaped so that the volume of water displaced produces a buoyant force equal to the ship's total weight. Submarines pump water into or out of ballast tanks to change their average density and dive or surface.
  • Hydrometers: the depth to which a hydrometer sinks indicates the fluid density — used to test battery acid concentration and fuel purity.
  • Lactometers: check milk purity — pure milk has a standard density; diluted milk is less dense, so the lactometer sinks deeper.
  • Hot-air balloons: buoyant force of air supports the balloon when average density (balloon + hot air) is less than surrounding cooler air.
  • Finding density of irregular solids: weigh in air ($W_1$), weigh fully submerged in water ($W_2$). Volume displaced $= (W_1-W_2)/(g\,\rho_w)$. Density of solid $= \dfrac{W_1}{W_1-W_2}\times\rho_\text{water}$.
  • Life jackets increase the volume of a person in water, displacing more water and increasing the buoyant force to support the person afloat.
NCERT — Buoyant force calculation

A solid cube of side $0.5\ \text{m}$ is fully immersed in water (density $= 1000\ \text{kg m}^{-3}$). Find the buoyant force acting on it. ($g = 9.8\ \text{m s}^{-2}$)

Volume of cube $= (0.5)^3 = 0.125\ \text{m}^3$. Volume displaced $= 0.125\ \text{m}^3$.

$F_b = \rho\,V\,g = 1000\times0.125\times9.8 = \boxed{1225\ \text{N}}$.

NCERT Activity — Apparent weight in water

A stone weighs $5\ \text{N}$ in air. When fully submerged in water, a spring balance reads $3.5\ \text{N}$. Find the buoyant force and the volume of the stone. ($\rho_\text{water} = 1000\ \text{kg m}^{-3}$)

Buoyant force $= W_\text{air} - W_\text{water} = 5 - 3.5 = \boxed{1.5\ \text{N}}$.

$F_b = \rho_w\,V\,g \Rightarrow V = \dfrac{1.5}{1000\times9.8} = \dfrac{1.5}{9800} \approx 1.53\times10^{-4}\ \text{m}^3 \approx 153\ \text{cm}^3$.

12. Relative Density (Specific Gravity)

The density of a substance relative to the density of water at $4^\circ\text{C}\ (1000\ \text{kg m}^{-3})$ is called its relative density or specific gravity:

$$\text{Relative Density} = \frac{\text{Density of the substance}}{\text{Density of water at } 4^\circ\text{C}}$$

Since it is a ratio of quantities with the same units, relative density is dimensionless (has no unit).

Relative densities of common substances:

Substance Relative Density Floats in water?
Ice0.917Yes
Wood (oak)0.6–0.9Yes
Iron / Steel7.87No
Gold19.3No
Mercury13.6No (iron floats on mercury!)

Quick rule: relative density $< 1$ — floats in water; relative density $> 1$ — sinks in water.

Calculating relative density using Archimedes' Principle:

Weigh a solid in air ($W_1$) and fully in water ($W_2$):

$$\text{Relative Density} = \frac{W_1}{W_1 - W_2}$$

This is the operating principle of the hydrometer. The denominator $W_1 - W_2$ is the apparent loss of weight = buoyant force = weight of water displaced = $\rho_w\,V\,g$.

NCERT Example — Relative density from spring balance readings

A body weighs $500\ \text{N}$ in air and $400\ \text{N}$ when fully submerged in water. Find (a) the buoyant force and (b) the relative density of the body.

(a) Buoyant force $= 500 - 400 = \boxed{100\ \text{N}}$.

(b) Relative density $= \dfrac{500}{500-400} = \dfrac{500}{100} = \boxed{5}$.

The body is $5$ times as dense as water.

13. Common Mistakes and Exam Tips

  • $G$ vs $g$: $G = 6.674\times10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$ is universal; $g \approx 9.8\ \text{m s}^{-2}$ is specific to Earth's surface. Never swap their values.
  • Mass is not weight: "I weigh 50 kg" is scientifically incorrect — 50 kg is mass; weight $= 50\times9.8 = 490\ \text{N}$.
  • Free fall vs thrown: any object under gravity alone is in free fall — a stone thrown horizontally is also in free fall vertically.
  • Sign convention: in free-fall problems, state which direction you take as positive — examiners deduct marks if this is missing.
  • Buoyant force $\neq$ pressure: buoyant force is a net upward force from the pressure difference between top and bottom surfaces of the submerged object.
  • Displaced vs surrounding fluid: buoyant force equals the weight of the displaced fluid, not all the fluid in the container.
  • Beam balance vs spring balance: beam balance measures mass (works on Moon), spring balance measures weight (reads differently on Moon).
  • Relative density has no unit — do not write "kg/m³" or "g/cm³" for relative density.
Practice MCQs
1. The value of the Universal Gravitational Constant $G$ is:
  1. $9.8\ \text{N m}^2\ \text{kg}^{-2}$
  2. $6.674\times10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$
  3. $6.674\times10^{11}\ \text{N m}^2\ \text{kg}^{-2}$
  4. $9.8\times10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$
Answer: (B) $G = 6.674\times10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$ — first measured by Henry Cavendish in 1798. Option (A) is $g$, not $G$.
2. The gravitational force between two masses is $F$. If the distance between them is doubled and each mass is also doubled, the new force is:
  1. $F$
  2. $2F$
  3. $F/2$
  4. $4F$
Answer: (A) $F' = G(2m_1)(2m_2)/(2r)^2 = G\cdot4m_1m_2/(4r^2) = Gm_1m_2/r^2 = F$. The two effects cancel exactly.
3. A stone is dropped from rest. After $3$ seconds its velocity is: ($g = 10\ \text{m s}^{-2}$)
  1. $10\ \text{m s}^{-1}$
  2. $20\ \text{m s}^{-1}$
  3. $30\ \text{m s}^{-1}$
  4. $45\ \text{m s}^{-1}$
Answer: (C) $v = u + gt = 0 + 10\times3 = 30\ \text{m s}^{-1}$.
4. Weight of an object on the Moon compared to Earth:
  1. same
  2. six times more
  3. one-sixth
  4. depends on the object's mass
Answer: (C) $W_\text{Moon} = W_\text{Earth}/6$ because $g_\text{Moon} \approx g_\text{Earth}/6$. Mass is the same everywhere.
5. The SI unit of pressure is:
  1. Newton
  2. Pascal
  3. $\text{N m}^2$
  4. $\text{kg m}^{-3}$
Answer: (B) Pascal (Pa) $= 1\ \text{N m}^{-2}$, named after Blaise Pascal.
6. When a body is fully immersed in water, the buoyant force acting on it equals:
  1. weight of the body
  2. weight of water displaced by the body
  3. weight of all water in the container
  4. weight of water above the body
Answer: (B) By Archimedes' Principle — buoyant force = weight of fluid displaced.
7. An iron ball and a wooden ball of the same volume are dropped from the same height in a vacuum. Which reaches the ground first?
  1. Iron ball
  2. Wooden ball
  3. Both simultaneously
  4. Whichever is denser
Answer: (C) In vacuum, $g$ is the same for all objects regardless of mass — both fall at identical rate (Galileo's result).
8. A substance has relative density $0.75$. It will:
  1. sink in water
  2. float in water with $75\%$ of volume submerged
  3. float in water with $25\%$ of volume submerged
  4. sink in all liquids
Answer: (B) Fraction submerged $= \rho_\text{object}/\rho_\text{water} = 0.75$, so $75\%$ is under water and $25\%$ is above.
9. At height $h = R$ (Earth's radius) above Earth's surface, the acceleration due to gravity becomes:
  1. $g/2$
  2. $g/4$
  3. $g/6$
  4. $g/8$
Answer: (B) $g' = \dfrac{GM}{(R+R)^2} = \dfrac{GM}{4R^2} = \dfrac{g}{4}$.
10. Which instrument uses Archimedes' Principle to measure the purity of milk?
  1. Thermometer
  2. Hydrometer
  3. Lactometer
  4. Barometer
Answer: (C) A lactometer floats to a different depth depending on the density (and hence purity) of milk — based on Archimedes' Principle.
11. A body weighs $500\ \text{N}$ in air and $300\ \text{N}$ when fully submerged in water. The relative density of the body is:
  1. $2.5$
  2. $3$
  3. $1.67$
  4. $5$
Answer: (A) Relative density $= \dfrac{W_\text{air}}{W_\text{air}-W_\text{water}} = \dfrac{500}{500-300} = \dfrac{500}{200} = 2.5$.
12. A hammer drives a nail into wood more easily when the nail is pointed because:
  1. the mass of the nail is small
  2. the same force acts over a tiny area, giving very high pressure
  3. nails are made of harder material than wood
  4. gravity pulls the nail downward
Answer: (B) $P = F/A$ — the pointed tip has a very small area, so the same hammer force produces very high pressure on the wood.
Assertion–Reason
A: Mass of a body remains constant everywhere in the universe.    R: Mass is the amount of matter in a body, independent of gravity.
Answer: Both A and R are true, and R is the correct explanation of A. Mass is an intrinsic property; $g$ affects weight ($W=mg$) but not mass. A beam balance therefore gives the same reading on the Moon as on Earth.
A: A steel ship floats, but a solid steel ball of the same mass sinks.    R: The average density of the ship (including the air inside) is less than the density of water.
Answer: Both A and R are true, and R correctly explains A. Buoyancy depends on the average density of the entire object — the hollow shape of the ship allows it to displace enough water to support its weight.
Previous-Year Questions
PYQ 1. State the Universal Law of Gravitation. Write its mathematical expression and state the SI unit of the Universal Gravitational Constant. (CBSE Board, 3 marks)
Answer: Statement: every object in the universe attracts every other object with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Expression: $F = Gm_1m_2/r^2$. SI unit of $G$: $\text{N m}^2\ \text{kg}^{-2}$.
PYQ 2. Derive an expression for the acceleration due to gravity $g$ at Earth's surface using the Universal Law of Gravitation. How does $g$ change with altitude? (CBSE Board, 3 marks)
Answer: Force on mass $m$: $F = GMm/R^2$. Also $F = mg$. Equating: $g = GM/R^2$. At altitude $h$: $g_h = GM/(R+h)^2 < g$, so $g$ decreases with altitude.
PYQ 3. An object is dropped from a height of $80\ \text{m}$. Taking $g = 10\ \text{m s}^{-2}$, find (a) the time to reach the ground and (b) the velocity just before impact. (CBSE Board, 3 marks)
Answer: $u=0$, $s=80\ \text{m}$, $g=10\ \text{m s}^{-2}$. (a) $80 = \tfrac{1}{2}(10)t^2 \Rightarrow t^2=16 \Rightarrow \boxed{t=4\ \text{s}}$. (b) $v = 0+10\times4 = \boxed{40\ \text{m s}^{-1}}$.
PYQ 4. State Archimedes' Principle. A solid weighs $250\ \text{N}$ in air and $200\ \text{N}$ in water. Find (a) the buoyant force, (b) the volume of the solid, and (c) its relative density. ($g = 10\ \text{m s}^{-2}$, $\rho_w = 1000\ \text{kg m}^{-3}$) (CBSE Board, 5 marks)
Archimedes' Principle: when a body is wholly or partially immersed in a fluid it experiences an upward force equal to the weight of the fluid displaced. (a) $F_b = 250-200 = \boxed{50\ \text{N}}$. (b) $F_b = \rho_w\,V\,g \Rightarrow V = 50/(1000\times10) = \boxed{5\times10^{-3}\ \text{m}^3}$. (c) Relative density $= 250/50 = \boxed{5}$.
PYQ 5. Differentiate between mass and weight. An astronaut weighs $630\ \text{N}$ on Earth. What will be his (a) weight and (b) mass on the Moon? ($g_E = 9.8\ \text{m s}^{-2}$) (CBSE Board, 3 marks)
Mass vs Weight: mass = quantity of matter, scalar, constant, in kg; weight = gravitational force, vector, in N, varies with $g$. (a) $W_\text{Moon} = 630/6 = \boxed{105\ \text{N}}$. (b) Mass $= W_E/g_E = 630/9.8 \approx \boxed{64.3\ \text{kg}}$ — same on Moon.
Want personal coaching in Dwarka?
Book a free demo class
More Class 9 Science chapters