Non-Rationalised Science NCERT Notes and Solutions (Class 6th to 10th)
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Non-Rationalised Science NCERT Notes and Solutions (Class 11th)
Physics		Chemistry			Biology
Non-Rationalised Science NCERT Notes and Solutions (Class 12th)
Physics		Chemistry			Biology

Class 9th Chapters
1. Matter In Our Surroundings	2. Is Matter Around Us Pure?	3. Atoms And Molecules
4. Structure Of The Atom	5. The Fundamental Unit Of Life	6. Tissues
7. Diversity In Living Organisms	8. Motion	9. Force And Laws Of Motion
10. Gravitation	11. Work And Energy	12. Sound
13. Why Do We Fall Ill?	14. Natural Resources	15. Improvement In Food Resources

Chapter 10: Gravitation

In earlier chapters, we explored the concepts of motion and force. We learned that an unbalanced force is required to change an object's speed or direction. We frequently observe that objects fall towards the Earth when dropped, and celestial bodies like planets orbit the Sun and the Moon orbits the Earth.

Isaac Newton recognised that the same fundamental force is responsible for all these phenomena: the force of gravitation. This force explains why objects fall to Earth, why the Moon orbits the Earth, and why planets orbit the Sun.

This chapter introduces the universal law of gravitation, discusses motion under Earth's gravitational force (free fall), explores the concepts of mass and weight (including variations with location), and examines the effects of force on area (thrust and pressure), leading to the principle of buoyancy and Archimedes' principle.

Gravitation

The phenomenon of objects falling towards the Earth (like an apple falling from a tree) and the Moon orbiting the Earth prompted Newton to propose that the same type of force is at play. He reasoned that the Earth must be attracting the Moon, continuously causing it to change direction and effectively "fall" towards the Earth instead of moving in a straight line. Similarly, the Sun attracts the planets, keeping them in orbit.

The force that causes an object to move in a circular path, continuously changing its direction towards the centre, is called the centripetal force (centre-seeking force).

A stone tied to a thread being whirled in a circle. Arrow shows the direction of motion if the thread is released (tangential).

The Moon's motion around the Earth is due to the gravitational force exerted by the Earth, which provides the necessary centripetal force. If this force were absent, the Moon would move off in a straight line (tangent to its orbit) according to Newton's First Law of Motion.

According to Newton's Third Law of Motion, if the Earth attracts an object (like an apple or the Moon), that object also attracts the Earth with an equal and opposite force. However, we do not observe the Earth moving significantly towards the apple or the Moon. This is explained by Newton's Second Law ($a=F/m$). Since the Earth's mass is vastly greater than the mass of an apple or the Moon, the acceleration of the Earth due to these forces is negligible.

Based on these observations and reasoning, Newton concluded that every object in the universe attracts every other object. This force of attraction is called the gravitational force.

Universal Law Of Gravitation

Newton formulated the quantitative law describing this gravitational force:

Statement: Every object in the universe attracts every other object with a force which 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 objects.

Diagram showing two objects, A and B, with masses M and m, separated by distance d. Arrows indicate mutual gravitational force F along the line joining their centres.

Let $M$ and $m$ be the masses of two objects, and $d$ be the distance between their centres. Let $F$ be the force of attraction between them.

According to the law:

$F \propto M \times m$ (Directly proportional to the product of masses)
$F \propto \frac{1}{d^2}$ (Inversely proportional to the square of the distance)

Combining these proportionalities:

$F \propto \frac{M \times m}{d^2}$

Introducing a constant of proportionality, $G$:

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

This is the mathematical expression of the Universal Law of Gravitation.

The constant $G$ is called the Universal Gravitation Constant. It is called universal because its value is believed to be the same for any two objects anywhere in the universe.

The value of $G$ was experimentally determined by Henry Cavendish. The accepted value is $G = 6.673 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$.

The SI unit of $G$ is N m² kg⁻². This can be derived from the formula $G = \frac{F d^2}{M m}$.

The gravitational force is generally very weak unless the masses involved are extremely large, like those of planets or stars. This is why we don't feel the gravitational attraction between ourselves and nearby objects like furniture or other people, even though it exists.

Newton's derivation of the inverse-square rule: Newton used Kepler's Third Law of planetary motion ($r^3/T^2 = \text{constant}$, where $r$ is orbital radius and $T$ is orbital period) along with the centripetal force formula ($F \propto v^2/r$) to show that the gravitational force keeping planets in orbit must be inversely proportional to the square of the distance from the Sun ($F \propto 1/r^2$).

Diagram illustrating Kepler's first and second laws of planetary motion: elliptical orbit with sun at focus, and equal areas swept in equal times.

Kepler's Laws:

Planetary orbits are ellipses with the Sun at one focus.
A line segment joining a planet and the Sun sweeps out equal areas in equal intervals of time (planets move faster when closer to the Sun).
The square of the orbital period ($T^2$) is proportional to the cube of the mean distance from the Sun ($r^3$) ($r^3/T^2 =$ constant).

Example 10.1. The mass of the earth is 6 × 10²⁴ kg and that of the moon is 7.4 × 10²² kg. If the distance between the earth and the moon is 3.84×10⁵ km, calculate the force exerted by the earth on the moon. (Take G = 6.7 × 10⁻¹¹ N m² kg⁻²)

Answer:

Given:

Mass of the Earth, $M = 6 \times 10^{24}$ kg
Mass of the Moon, $m = 7.4 \times 10^{22}$ kg
Distance between Earth and Moon, $d = 3.84 \times 10^5$ km
Universal Gravitation Constant, $G = 6.7 \times 10^{-11}$ N m² kg⁻²

Convert distance to metres:

$d = 3.84 \times 10^5 \text{ km} = 3.84 \times 10^5 \times 10^3 \text{ m} = 3.84 \times 10^8 \text{ m}$.

Using the Universal Law of Gravitation formula, $F = G \frac{M m}{d^2}$:

$F = (6.7 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}) \times \frac{(6 \times 10^{24} \text{ kg}) \times (7.4 \times 10^{22} \text{ kg})}{(3.84 \times 10^8 \text{ m})^2}$

$F = (6.7 \times 10^{-11}) \times \frac{(6 \times 7.4 \times 10^{24+22})}{(3.84)^2 \times 10^{8 \times 2}} \text{ N}$

$F = (6.7 \times 10^{-11}) \times \frac{(44.4 \times 10^{46})}{14.7456 \times 10^{16}} \text{ N}$

$F \approx 6.7 \times \frac{44.4}{14.7456} \times 10^{-11 + 46 - 16} \text{ N}$

$F \approx 6.7 \times 3.011 \times 10^{19} \text{ N}$

$F \approx 20.17 \times 10^{19} \text{ N} = 2.017 \times 10^{20} \text{ N}$.

The force exerted by the Earth on the Moon is approximately 2.02 $\times 10^{20}$ N.

Importance Of The Universal Law Of Gravitation

The Universal Law of Gravitation is highly significant as it provided a unified explanation for several seemingly unrelated phenomena:

The force that keeps us grounded on Earth.
The motion of the Moon around the Earth.
The motion of planets around the Sun.
The occurrence of tides in oceans and seas due to the gravitational forces exerted by the Moon and the Sun.

Free Fall

We observe that an object dropped from a height falls towards the Earth. This is due to the Earth's gravitational force. When an object falls towards the Earth under the influence of only the gravitational force (neglecting air resistance), it is said to be in free fall.

During free fall, an object's direction of motion remains unchanged (towards the Earth), but its velocity changes in magnitude due to the Earth's attraction. This change in velocity implies acceleration.

The acceleration experienced by an object in free fall due to the Earth's gravitational force is called the acceleration due to gravity. It is denoted by the symbol $g$. The unit of $g$ is the same as acceleration, m/s² or m s⁻².

According to Newton's Second Law ($F = ma$), the gravitational force ($F$) acting on an object of mass $m$ near the Earth's surface causing acceleration $g$ is:

$F = m \times g$

Also, from the Universal Law of Gravitation, the force between the Earth (mass $M$) and an object (mass $m$) at a distance $d$ from the Earth's centre is $F = G \frac{M m}{d^2}$.

For an object on or near the Earth's surface, the distance $d$ is approximately equal to the Earth's radius, $R$. So, $F = G \frac{M m}{R^2}$.

Equating the two expressions for force:

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

The mass of the object $m$ cancels out from both sides.

$$ g = G \frac{M}{R^2} $$

This shows that the acceleration due to gravity ($g$) is independent of the mass of the falling object. This means that all objects, regardless of their mass, should fall towards the Earth at the same rate in the absence of air resistance.

To Calculate The Value Of G

We can calculate the approximate value of $g$ on the surface of the Earth using the values of $G$, the mass of the Earth ($M$), and the radius of the Earth ($R$).

Given values:

$G = 6.673 \times 10^{-11}$ N m² kg⁻²
Mass of Earth, $M = 6 \times 10^{24}$ kg
Radius of Earth, $R = 6.4 \times 10^6$ m

Using the formula $g = G \frac{M}{R^2}$:

$g = (6.673 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}) \times \frac{6 \times 10^{24} \text{ kg}}{(6.4 \times 10^6 \text{ m})^2}$

$g = (6.673 \times 10^{-11}) \times \frac{6 \times 10^{24}}{(6.4)^2 \times 10^{12}} \text{ m s}^{-2}$

$g \approx 6.673 \times \frac{6}{40.96} \times 10^{-11 + 24 - 12} \text{ m s}^{-2}$

$g \approx 6.673 \times 0.1465 \times 10^1 \text{ m s}^{-2}$

$g \approx 0.978 \times 10^1 \text{ m s}^{-2} = 9.78 \text{ m s}^{-2}$.

The calculated value of $g$ is approximately 9.8 m s⁻². For most calculations near the Earth's surface, $g$ is taken as a constant with this value.

Note that the Earth is not a perfect sphere; its radius is slightly larger at the equator and smaller at the poles. According to the formula $g = G \frac{M}{R^2}$, the value of $g$ is slightly greater at the poles than at the equator because $R$ is smaller at the poles.

Motion Of Objects Under The Influence Of Gravitational Force Of The Earth

Since $g$ is nearly constant for objects near the Earth's surface, the motion of objects under free fall is a form of uniformly accelerated motion.

The equations of motion (Chapter 8) for uniformly accelerated motion can be applied to objects in free fall by replacing the acceleration $a$ with the acceleration due to gravity $g$.

$v = u + gt$
$s = ut + \frac{1}{2}gt^2$
$v^2 = u^2 + 2gs$

When applying these equations, it's important to use sign conventions:

If the motion is downwards (in the direction of gravity), $g$ is taken as positive (+g).
If the motion is upwards (against the direction of gravity), $g$ is taken as negative (-g).
Initial velocity ($u$), final velocity ($v$), and displacement ($s$) should also be assigned appropriate signs based on the chosen direction convention. For motion under gravity, it is common to take the upward direction as positive and the downward direction as negative, or vice versa, consistently throughout the calculation. If we take downward as positive, $g$ is positive. If we take upward as positive, $g$ is negative.

Example 10.2. A car falls off a ledge and drops to the ground in 0.5 s. Let g = 10 m s⁻² (for simplifying the calculations).

(i) What is its speed on striking the ground?

(ii) What is its average speed during the 0.5 s?

(iii) How high is the ledge from the ground?

Answer:

Given: Time of fall, $t = 0.5$ s. Initial velocity, $u = 0$ m s⁻¹ (falls from rest). Acceleration due to gravity, $g = 10$ m s⁻². Since the car is falling downwards, we take $g$ as positive, $a = +10$ m s⁻².

(i) Speed on striking the ground (final velocity, $v$): Use $v = u + at$.

$v = 0 \text{ m s}^{-1} + (10 \text{ m s}^{-2}) \times (0.5 \text{ s}) = 5 \text{ m s}^{-1}$.

The speed on striking the ground is 5 m s⁻¹.

(ii) Average speed: For uniform acceleration along a straight line, average speed is $\frac{u+v}{2}$.

Average speed = $\frac{0 \text{ m s}^{-1} + 5 \text{ m s}^{-1}}{2} = \frac{5}{2} \text{ m s}^{-1} = 2.5 \text{ m s}^{-1}$.

The average speed during the 0.5 s is 2.5 m s⁻¹.

(iii) Height of the ledge from the ground (distance fallen, $s$): Use $s = ut + \frac{1}{2}at^2$.

$s = (0 \text{ m s}^{-1}) \times (0.5 \text{ s}) + \frac{1}{2} \times (10 \text{ m s}^{-2}) \times (0.5 \text{ s})^2$

$s = 0 + \frac{1}{2} \times 10 \text{ m s}^{-2} \times 0.25 \text{ s}^2 = 5 \times 0.25 \text{ m} = 1.25 \text{ m}$.

The height of the ledge from the ground is 1.25 m.

Example 10.3. An object is thrown vertically upwards and rises to a height of 10 m. Calculate (i) the velocity with which the object was thrown upwards and (ii) the time taken by the object to reach the highest point.

Answer:

Given: Distance risen, $s = 10$ m. At the highest point, the final velocity is $v = 0$ m s⁻¹. Acceleration due to gravity, $g = 9.8$ m s⁻². Since the object is moving upwards (opposite to gravity), we take $a = -9.8$ m s⁻².

(i) To find the initial velocity ($u$) with which it was thrown: Use $v^2 = u^2 + 2as$.

$(0 \text{ m s}^{-1})^2 = u^2 + 2 \times (-9.8 \text{ m s}^{-2}) \times (10 \text{ m})$

$0 = u^2 - 196 \text{ m}^2\text{s}^{-2}$

$u^2 = 196 \text{ m}^2\text{s}^{-2}$

$u = \sqrt{196} \text{ m s}^{-1} = 14 \text{ m s}^{-1}$. (Taking the positive root as velocity is upwards initially).

The object was thrown upwards with a velocity of 14 m s⁻¹.

(ii) To find the time ($t$) taken to reach the highest point: Use $v = u + at$.

$0 \text{ m s}^{-1} = 14 \text{ m s}^{-1} + (-9.8 \text{ m s}^{-2}) \times t$

$-14 \text{ m s}^{-1} = -9.8 \text{ m s}^{-2} \times t$

$t = \frac{-14 \text{ m s}^{-1}}{-9.8 \text{ m s}^{-2}} = \frac{14}{9.8} \text{ s} \approx 1.43 \text{ s}$.

The time taken to reach the highest point is approximately 1.43 s.

Mass

As learned in the previous chapter, mass is a measure of an object's inertia. It represents the amount of matter in an object.

The mass of an object is a fundamental property and remains constant regardless of its location in the universe (Earth, Moon, space). Mass does not change from place to place.

The SI unit of mass is the kilogram (kg).

Weight

While mass is a measure of inertia and is constant, weight is the force with which a celestial body (like the Earth or Moon) attracts an object.

The weight ($W$) of an object is the force ($F$) exerted on it due to gravity. From Newton's Second Law ($F=ma$), we can write the weight of an object of mass $m$ on the Earth as:

$$ W = F = m \times g $$

where $g$ is the acceleration due to gravity at that location.

Weight is a force, acting vertically downwards towards the centre of the Earth. It has both magnitude and direction (it is a vector quantity). The SI unit of weight is the same as that of force, which is Newton (N).

Since the value of $g$ varies from place to place (e.g., slightly different at poles and equator, decreases with altitude, significantly different on other celestial bodies), the weight of an object is not constant and changes with location. However, the mass of the object remains the same.

At a given location where $g$ is constant, weight is directly proportional to mass ($W \propto m$). This is why weighing machines, which actually measure weight (force), are calibrated to display mass at that specific location.

Weight Of An Object On The Moon

The Moon also attracts objects towards its surface with a gravitational force. The weight of an object on the Moon ($W_m$) is the force with which the Moon attracts it.

Using the Universal Law of Gravitation, the weight of an object of mass $m$ on the surface of the Moon is given by:

$$ W_m = G \frac{M_m m}{R_m^2} $$

where $M_m$ is the mass of the Moon and $R_m$ is the radius of the Moon.

Similarly, the weight of the same object on the surface of the Earth ($W_e$) is:

$$ W_e = G \frac{M_e m}{R_e^2} $$

where $M_e$ is the mass of the Earth and $R_e$ is the radius of the Earth.

By comparing the values of $M_m, R_m$ with $M_e, R_e$ (from tables like Table 10.1 in the text), we can find the ratio of weight on the Moon to weight on the Earth:

$\frac{W_m}{W_e} = \frac{G \frac{M_m m}{R_m^2}}{G \frac{M_e m}{R_e^2}} = \frac{M_m}{R_m^2} \times \frac{R_e^2}{M_e}$

Substituting the approximate values:

$\frac{W_m}{W_e} = \frac{7.36 \times 10^{22} \text{ kg}}{(1.74 \times 10^6 \text{ m})^2} \times \frac{(6.37 \times 10^6 \text{ m})^2}{5.98 \times 10^{24} \text{ kg}}$

$\frac{W_m}{W_e} \approx \frac{7.36 \times 10^{22}}{3.028 \times 10^{12}} \times \frac{40.577 \times 10^{12}}{5.98 \times 10^{24}}$

$\frac{W_m}{W_e} \approx (2.43 \times 10^{10}) \times (6.78 \times 10^{-12})$

$\frac{W_m}{W_e} \approx 1.65 \times 10^{-2} \times 10^{10-12} \approx 0.0165 \times 10^{-2}$

Using the calculated values of $g$ on Earth ($g_e \approx 9.8$ m s⁻²) and Moon ($g_m \approx 1.62$ m s⁻²), the ratio is $\frac{g_m}{g_e} \approx \frac{1.62}{9.8} \approx 0.165 \approx \frac{1}{6}$.

So, the weight of an object on the Moon is approximately one-sixth ($1/6$) of its weight on the Earth.

$$ W_m = \frac{1}{6} W_e $$

Example 10.4. Mass of an object is 10 kg. What is its weight on the earth?

Answer:

Given: Mass, $m = 10$ kg. Acceleration due to gravity on Earth, $g = 9.8$ m s⁻².

Weight on Earth, $W_e = m \times g$

$W_e = 10 \text{ kg} \times 9.8 \text{ m s}^{-2} = 98 \text{ N}$.

The weight of the object on the Earth is 98 N.

Example 10.5. An object weighs 10 N when measured on the surface of the earth. What would be its weight when measured on the surface of the moon?

Answer:

Given: Weight on Earth, $W_e = 10$ N.

The weight of an object on the Moon is approximately $\frac{1}{6}$ of its weight on the Earth.

Weight on Moon, $W_m = \frac{1}{6} \times W_e$

$W_m = \frac{1}{6} \times 10 \text{ N} = \frac{10}{6} \text{ N} \approx 1.67 \text{ N}$.

The weight of the object on the surface of the Moon would be approximately 1.67 N.

Thrust And Pressure

The effect of a force depends not only on its magnitude but also on the area over which it is applied. This distinction is important in understanding phenomena like why sharp knives cut better or why wider tyres are used on heavy vehicles.

Thrust: The force acting on an object perpendicular to a surface is called thrust. When you press a drawing pin into a board, the force exerted by your thumb on the head of the pin (which is the weight of the block in Example 10.6) is thrust. This thrust acts perpendicularly on the surface of the board through the pin's tip.

Pressure: The effect of thrust is greater when it acts on a smaller area. Pressure is defined as the thrust per unit area.

$$ \text{Pressure} = \frac{\text{Thrust}}{\text{Area}} $$

The SI unit of pressure is Newton per square metre (N/m² or N m⁻²). This unit is also called pascal (Pa), in honour of scientist Blaise Pascal. 1 Pa = 1 N/m².

A given thrust produces a larger pressure on a smaller area and a smaller pressure on a larger area. This explains why:

Cutting tools like knives and scissors have sharp edges (small area to exert large pressure).
Needles and nails have pointed tips.
Building foundations are wide (large area to distribute weight and reduce pressure on the ground).
Camels can walk easily in deserts (broad feet increase contact area, reducing pressure on the sand).
Heavy vehicles have wide tyres or multiple wheels.

Example 10.6. A block of wood is kept on a tabletop. The mass of wooden block is 5 kg and its dimensions are 40 cm × 20 cm × 10 cm. Find the pressure exerted by the wooden block on the table top if it is made to lie on the table top with its sides of dimensions (a) 20 cm × 10 cm and (b) 40 cm × 20 cm.

Answer:

Given: Mass of the block, $m = 5$ kg. Dimensions = 40 cm × 20 cm × 10 cm.

The thrust exerted by the block on the table is its weight. Thrust = $W = m \times g$. Using $g = 9.8$ m s⁻².

Thrust = $5 \text{ kg} \times 9.8 \text{ m s}^{-2} = 49 \text{ N}$.

(a) Block lies on the side with dimensions 20 cm × 10 cm.

Area = $20 \text{ cm} \times 10 \text{ cm} = 200 \text{ cm}^2$. Convert to m²: $200 \text{ cm}^2 = 200 \times (10^{-2} \text{ m})^2 = 200 \times 10^{-4} \text{ m}^2 = 0.02 \text{ m}^2$.

Pressure = $\frac{\text{Thrust}}{\text{Area}} = \frac{49 \text{ N}}{0.02 \text{ m}^2} = \frac{49}{2/100} \text{ Pa} = 49 \times \frac{100}{2} \text{ Pa} = 49 \times 50 \text{ Pa} = 2450 \text{ Pa}$.

The pressure exerted is 2450 Pa.

(b) Block lies on the side with dimensions 40 cm × 20 cm.

Area = $40 \text{ cm} \times 20 \text{ cm} = 800 \text{ cm}^2$. Convert to m²: $800 \text{ cm}^2 = 800 \times 10^{-4} \text{ m}^2 = 0.08 \text{ m}^2$.

Pressure = $\frac{\text{Thrust}}{\text{Area}} = \frac{49 \text{ N}}{0.08 \text{ m}^2} = \frac{49}{8/100} \text{ Pa} = 49 \times \frac{100}{8} \text{ Pa} = 49 \times 12.5 \text{ Pa} = 612.5 \text{ Pa}$.

The pressure exerted is 612.5 Pa.

As expected, the larger area results in less pressure for the same thrust.

Pressure In Fluids

Fluids (liquids and gases) also have weight and exert pressure. Fluids exert pressure not only on the base of the container but also on the walls. Pressure exerted by a confined fluid is transmitted equally in all directions.

Buoyancy

When an object is immersed in a fluid (like water), we feel an upward force on the object. This upward force is called the buoyant force or upthrust.

Experimental Observation: Pushing an empty bottle into a bucket of water requires increasing downward force as it is pushed deeper. Releasing it causes it to bounce back to the surface. This demonstrates that water exerts an upward force (buoyant force) opposing the downward force of gravity (weight of the bottle).

Diagram showing a bottle being pushed into water. Upward arrow indicates buoyant force.

The magnitude of the buoyant force depends on the density of the fluid. Denser fluids exert a greater buoyant force.

Why Objects Float Or Sink When Placed On The Surface Of Water?

Whether an object floats or sinks when placed on the surface of a liquid depends on the relative magnitudes of two forces acting on it: the downward force of gravity (its weight) and the upward buoyant force exerted by the liquid.

Consider an object placed on the surface of water:

If the downward force (weight) is greater than the upward buoyant force, the net force is downwards, and the object sinks.
If the downward force (weight) is equal to the upward buoyant force, the net force is zero, and the object floats fully submerged (equilibrium at any depth).
If the downward force (weight) is less than the maximum possible upward buoyant force (which occurs when the object is fully submerged), the net force is upwards, and the object rises until it displaces just enough liquid for the buoyant force to equal its weight. The object then floats partially submerged.

This also relates to the density of the object compared to the density of the liquid.

Based on the concept of buoyant force (and Archimedes' principle, discussed next):

An object with a density less than the density of the liquid will float.
An object with a density equal to the density of the liquid will float fully submerged.
An object with a density greater than the density of the liquid will sink.

Diagram showing an iron nail sinking and a cork floating on water

Example: An iron nail sinks in water because the density of iron is greater than the density of water, so its weight is greater than the buoyant force even when fully submerged. A cork floats because the density of cork is less than the density of water; it displaces a weight of water equal to its own weight while only partially submerged.

Archimedes’ Principle

The magnitude of the buoyant force experienced by an object immersed in a fluid is described by Archimedes' Principle.

Experimental Observation: Suspending a stone from a spring balance and lowering it into water shows that the reading on the balance decreases as the stone is immersed. This indicates an upward buoyant force from the water opposing the weight. The decrease in the reading is equal to the magnitude of the buoyant force.

Diagram showing a stone suspended from a spring balance, (a) in air and (b) partially/fully immersed in water, showing reduced reading.

Statement of Archimedes’ Principle: When a body is immersed fully or partially in a fluid, it experiences an upward force (buoyant force) that is equal to the weight of the fluid displaced by it.

The buoyant force is equal to the weight of the volume of fluid that the object pushes aside or displaces. If an object is fully submerged, it displaces a volume of fluid equal to its own volume. If it is partially submerged (floating), it displaces a volume of fluid such that the weight of that displaced fluid is equal to the object's weight.

Archimedes, a Greek scientist, discovered this principle. His principle has numerous applications, including:

Designing ships and submarines.
Lactometers (instruments for measuring the purity of milk based on its density).
Hydrometers (instruments for measuring the density of liquids).

Relative Density

The density of a substance is a fundamental property, defined as its mass per unit volume.

Density = Mass / Volume ($\rho = m/V$)

The SI unit of density is kilogram per cubic metre (kg m⁻³). The density of a pure substance is constant under specific conditions and can be used to determine its purity.

Sometimes, it is convenient to compare the density of a substance to the density of a reference substance, usually water. This gives the relative density.

Relative Density: The ratio of the density of a substance to the density of water.

$$ \text{Relative Density} = \frac{\text{Density of a substance}}{\text{Density of water}} $$

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

The density of water is approximately 1000 kg m⁻³ or 1 g cm⁻³.

Example 10.7. Relative density of silver is 10.8. The density of water is 10³ kg m⁻³. What is the density of silver in SI unit?

Answer:

Given:

Relative density of silver = 10.8
Density of water = 10³ kg m⁻³

Using the formula for relative density:

Relative Density of silver = $\frac{\text{Density of silver}}{\text{Density of water}}$

Density of silver = Relative Density of silver $\times$ Density of water

Density of silver = $10.8 \times (10^3 \text{ kg m}^{-3})$

Density of silver = $10800 \text{ kg m}^{-3}$.

The density of silver in SI unit is 10800 kg m⁻³.

Intext Questions

Page No. 134

Question 1. State the universal law of gravitation.

Answer:

Question 2. Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.

Answer:

Page No. 136

Question 1. What do you mean by free fall?

Answer:

Question 2. What do you mean by acceleration due to gravity?

Answer:

Page No. 138

Question 1. What are the differences between the mass of an object and its weight?

Answer:

Question 2. Why is the weight of an object on the moon $\frac{1}{6}$th its weight on the earth?

Answer:

Page No. 141

Question 1. Why is it difficult to hold a school bag having a strap made of a thin and strong string?

Answer:

Question 2. What do you mean by buoyancy?

Answer:

Question 3. Why does an object float or sink when placed on the surface of water?

Answer:

Page No. 142

Question 1. You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?

Answer:

Question 2. You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than other. Can you say which one is heavier and why?

Answer:

Exercises

Question 1. How does the force of gravitation between two objects change when the distance between them is reduced to half ?

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Question 2. Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?

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Question 3. What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is $6 \times 10^{24}$ kg and radius of the earth is $6.4 \times 10^6$ m.)

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Question 4. The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why?

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Question 5. If the moon attracts the earth, why does the earth not move towards the moon?

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Question 6. What happens to the force between two objects, if

(i) the mass of one object is doubled?

(ii) the distance between the objects is doubled and tripled?

(iii) the masses of both objects are doubled?

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Question 7. What is the importance of universal law of gravitation?

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Question 8. What is the acceleration of free fall?

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Question 9. What do we call the gravitational force between the earth and an object?

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Question 10. Amit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why? [Hint: The value of g is greater at the poles than at the equator.]

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Question 11. Why will a sheet of paper fall slower than one that is crumpled into a ball?

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Question 12. Gravitational force on the surface of the moon is only $\frac{1}{6}$ as strong as gravitational force on the earth. What is the weight in newtons of a 10 kg object on the moon and on the earth?

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Question 13. A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate

(i) the maximum height to which it rises,

(ii) the total time it takes to return to the surface of the earth.

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Question 14. A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.

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Question 15. A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 $m/s^2$, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

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Question 16. Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = $6 \times 10^{24}$ kg and of the Sun = $2 \times 10^{30}$ kg. The average distance between the two is $1.5 \times 10^{11}$ m.

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Question 17. A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.

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Question 18. A ball thrown up vertically returns to the thrower after 6 s. Find

(a) the velocity with which it was thrown up,

(b) the maximum height it reaches, and

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Question 19. In what direction does the buoyant force on an object immersed in a liquid act?

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Question 20. Why does a block of plastic released under water come up to the surface of water?

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Question 21. The volume of 50 g of a substance is 20 $cm^3$. If the density of water is 1 $g cm^{-3}$, will the substance float or sink?

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Question 22. The volume of a 500 g sealed packet is 350 $cm^3$. Will the packet float or sink in water if the density of water is 1 $g cm^{-3}$? What will be the mass of the water displaced by this packet?

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