What is the difference between work and energy?

Work is energy transfer due to displacement by force; energy is the ability to perform work.

What is the principle behind a battery?

It stores chemical energy and converts it to electrical energy.

How does energy flow in an ecosystem?

Energy flows from the sun to producers, then to consumers via food chain.

What determines the power rating of an appliance?

The rate at which the appliance consumes energy per unit time.

Which energy change occurs when a ball falls?

Potential energy converts to kinetic energy.

State the importance of the law of conservation of energy in technology.

It guides efficient energy usage and device design.

Can energy be destroyed according to science?

No, according to the law of conservation of energy.

Gravitation — NCERT Solutions | Class 9 Science

📘 Concept & Theory Concept Behind the Question ›

This question is based on Newton's Universal Law of Gravitation, which states that every two objects in the universe attract each other with a force that is:

Directly proportional to the product of their masses.
Inversely proportional to the square of the distance between their centres.

Mathematically,

\[ F=G\frac{m_1m_2}{r^2} \]

where

\(F\) = gravitational force
\(G\) = universal gravitational constant
\(m_1,m_2\) = masses of the two objects
\(r\) = distance between their centres

Since the force depends on the square of the distance, even a small change in distance causes a significant change in gravitational force.

🗺️ Solution Roadmap Step-by-step Plan ›

Write Newton's law of gravitation.
Assume the initial distance between the objects is \(R\).
Reduce the distance to \(\dfrac{R}{2}\).
Calculate the new gravitational force.
Compare the initial and final forces.
Draw the required conclusion.

📊 Graph / Figure Graph / Figure ›

✏️ Solution Complete Solution ›

Step-by-step Solution · 16 steps

According to Newton's Universal Law of Gravitation,
\[F=G\frac{m_1m_2}{r^2}\]
Let the initial distance between the two objects be\[ R\]
Therefore, the initial gravitational force is\[F_i=G\frac{m_1m_2}{R^2}\]
When the distance is reduced to half,\[r=\frac{R}{2}\]
Therefore, the new gravitational force becomes\[F_f=G\frac{m_1m_2}{\left(\frac{R}{2}\right)^2}\]
First, simplify the denominator:\[\left(\frac{R}{2}\right)^2=\frac{R^2}{4}\]
Substitute this into the equation:, \[ F_f= G\frac{m_1m_2}{\frac{R^2}{4}} \]
Dividing by a fraction is equivalent to multiplying by its reciprocal.
\[ \begin{aligned} F_f&=Gm_1m_2\times\frac{4}{R^2}\\ &=4G\frac{m_1m_2}{R^2} \end{aligned} \]
Since \[F_i=G\frac{m_1m_2}{R^2}\]
therefore,\[F_f=4F_i\]
Alternatively, comparing the initial and final forces,
\[ \begin{aligned} \frac{F_i}{F_f} &= \frac{G\dfrac{m_1m_2}{R^2}} {G\dfrac{m_1m_2}{\left(\dfrac{R}{2}\right)^2}} \\ &= \frac{\left(\dfrac{R}{2}\right)^2}{R^2} \\ &= \frac{R^2/4}{R^2} \\ &= \frac{1}{4} \end{aligned} \]
Hence,\[\boxed{F_f=4F_i}\]
Therefore, when the distance between two objects is reduced to half, the gravitational force becomes four times the original force.

🎯 Exam Significance Exam Significance ›

This is one of the most frequently asked conceptual questions in CBSE board examinations.
It develops a clear understanding of the inverse-square relationship in gravitation.
It helps solve numerical problems involving changes in distance.
The same concept is extensively used in orbital motion, satellite mechanics and planetary motion.
Questions based on changing the distance to half, double, three times or four times are common in school exams as well as competitive examinations.
This concept is useful for Olympiads, NTSE, KVPY-level conceptual practice, JEE Foundation and NEET Foundation courses.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 6 points

Gravitational force varies inversely as the square of the distance between two objects.
Reducing the distance to half increases the force by a factor of four.
Doubling the distance decreases the force to one-fourth.
The masses of the objects remain unchanged in this question; only the distance changes.
Always square the change in distance before calculating the new gravitational force.
Remember the important relation: \[ F\propto\frac{1}{r^2} \]

📘 Concept & Theory Concept Behind the Question ›

This question combines two important concepts:

Newton's Universal Law of Gravitation
Newton's Second Law of Motion

According to the Universal Law of Gravitation, the gravitational force acting on an object is directly proportional to its mass.

\[ F=G\frac{Mm}{R^2} \]

where

\(M\) = mass of the Earth
\(m\) = mass of the object
\(R\) = radius of the Earth
\(G\) = universal gravitational constant

On the other hand, according to Newton's Second Law,

\[ F=ma \]

Although a heavier object experiences a larger gravitational force, it also possesses greater inertia (greater mass), making it equally difficult to accelerate. As a result, the increase in force is exactly balanced by the increase in mass, giving every freely falling object the same acceleration due to gravity.

🗺️ Solution Roadmap Step-by-step Plan ›

Write the gravitational force acting on an object
Apply Newton's Second Law of Motion.
Substitute the gravitational force into the equation.
Simplify the expression by cancelling the mass of the object.
Conclude why all objects fall with the same acceleration.

📊 Graph / Figure Graph / Figure ›

✏️ Solution Complete Solution ›

Step-by-step Solution · 10 steps

Let
- \(M\) = mass of the Earth
- \(m\) = mass of the falling object
- \(R\) = radius of the Earth
The gravitational force acting on the object is\[F=G\frac{Mm}{R^2}\]
According to Newton's Second Law of Motion,\[F=ma\]
Substituting the gravitational force,\[ma=G\frac{Mm}{R^2}\]
Divide both sides by \(m\):\[a=\frac{G\frac{Mm}{R^2}}{m}\]
Cancelling \(m\) from the numerator and denominator,\[a=\frac{GM}{R^2}\]
Since the mass of the object has been cancelled, the acceleration does not depend on whether the object is heavy or light.
This acceleration is called the acceleration due to gravity, represented by \(g\).\[g=\frac{GM}{R^2}\]
Thus,
- A heavier object experiences a larger gravitational force.
- However, it also has proportionally greater inertia because of its larger mass.
- These two effects exactly balance each other.
Therefore, in the absence of air resistance, all objects fall towards the Earth with the same acceleration, irrespective of their masses.

🎯 Exam Significance Exam Significance ›

This is a very common conceptual question in CBSE board examinations.
It tests the understanding of the relationship between gravitational force, inertia and acceleration.
Students often mistakenly believe that heavier objects fall faster; this question removes that misconception.
The derivation of \[ g=\frac{GM}{R^2} \] is frequently used in numerical and conceptual problems.
This concept forms the foundation for studying free fall, satellites and orbital motion in higher classes.
It is also important for Olympiads, NTSE, JEE Foundation and NEET Foundation examinations.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 7 points

Gravitational force is directly proportional to the mass of the object.
Acceleration depends on both force and mass.
The larger force acting on a heavier object is exactly balanced by its greater inertia.
The mass of the falling object cancels during the derivation.
All freely falling objects have the same acceleration due to gravity.
Only air resistance causes objects of different masses to fall at different rates in everyday life.
Remember the important relation: \[ g=\frac{GM}{R^2} \]

📘 Concept & Theory Concept Behind the Question ›

This question is a direct application of Newton's Universal Law of Gravitation. Every object on the Earth's surface is attracted towards the Earth by a gravitational force.

The gravitational force between two objects is given by

\[ F=G\frac{m_1m_2}{R^2} \]

where

\(F\) = gravitational force
\(G=6.67\times10^{-11}\;Nm^2kg^{-2}\) = universal gravitational constant
\(m_1\) = mass of the Earth
\(m_2\) = mass of the object
\(R\) = distance between the centres of the Earth and the object

Since the object is placed on the Earth's surface, the distance between their centres is equal to the radius of the Earth.

This problem also establishes the important fact that the weight of a 1 kg object on Earth is approximately 9.8 N.

🗺️ Solution Roadmap Step-by-step Plan ›

Write the given quantities.
Use Newton's Universal Law of Gravitation.
Substitute the given values carefully.
Simplify the powers of 10 separately.
Perform the numerical calculation step by step.
Write the final answer with the correct SI unit.

📊 Graph / Figure Graph / Figure ›

✏️ Solution Complete Solution ›

Step-by-step Solution · 8 steps

Given
- Mass of the Earth, \[ m_1=6\times10^{24}\;kg \]
- Mass of the object, \[ m_2=1\;kg \]
- Radius of the Earth, \[ R=6.4\times10^6\;m \]
- Universal gravitational constant, \[ G=6.67\times10^{-11}\;Nm^2kg^{-2} \]
Step-by-step Solution
According to Newton's Universal Law of Gravitation,\[F=G\frac{m_1m_2}{R^2}\]
Substituting the given values, \[ F= 6.67\times10^{-11} \times \frac{6\times10^{24}\times1} {(6.4\times10^6)^2} \]
First, square the radius: \[ \begin{aligned} (6.4\times10^6)^2&=6.4^2\times(10^6)^2\\ &=40.96\times10^{12} \end{aligned} \]
Therefore, \[ F= 6.67\times10^{-11} \times \frac{6\times10^{24}} {40.96\times10^{12}} \]
Combine the numerical values and powers of 10 separately. \[ \begin{aligned} F&=\frac{6.67\times6}{40.96}\times10^{-11+24-12}\\ &=\frac{40.02}{40.96}\times10^1\\ &=\frac{400.2}{40.96}\\ &=\approx9.77\;N \end{aligned} \]
Rounding off to one decimal place,\[\boxed{F\approx9.8\;N}\]
Therefore, the gravitational force acting between the Earth and a 1 kg object on its surface is 9.8 N.

🎯 Exam Significance Exam Significance ›

This is a standard numerical problem based on Newton's Universal Law of Gravitation.
It tests the correct substitution of values in scientific notation.
Students must know how to simplify powers of ten without making calculation errors.
This question establishes the relationship between gravitational force and the weight of an object.
The result of approximately 9.8 N for a 1 kg object is frequently used in later chapters involving work, energy, pressure and motion.
Questions of this type are common in CBSE Board examinations, Olympiads, NTSE, JEE Foundation and NEET Foundation.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 7 points

For an object on the Earth's surface, the distance between the object and the Earth's centre equals the Earth's radius.
Always use SI units before substituting values into the formula.
Square both the numerical value and the power of ten while squaring the radius.
Simplify numerical values and powers of ten separately to avoid mistakes.
The gravitational force on a 1 kg object on Earth is approximately \[ 9.8\;N \]
This value is numerically equal to the weight of a 1 kg object on Earth.
Remember the fundamental formula: \[ F=G\frac{m_1m_2}{R^2} \]

📘 Concept & Theory Concept Behind the Question ›

This question is based on the combined application of Newton's Universal Law of Gravitation and Newton's Third Law of Motion.

According to the Universal Law of Gravitation, every two objects in the universe attract each other with a gravitational force given by

\[ F=G\frac{m_1m_2}{r^2} \]

The above equation gives the magnitude of the force between two bodies. Since the masses \(m_1\) and \(m_2\) appear together in the same expression, the force acting on each body has the same magnitude.

According to Newton's Third Law of Motion,

Every action has an equal and opposite reaction.

Therefore, if the Earth pulls the Moon with a certain gravitational force, the Moon also pulls the Earth with an equal force in the opposite direction.

Although the forces are equal, the accelerations produced are different because acceleration depends on the mass of the object.

🗺️ Solution Roadmap Step-by-step Plan ›

Recall Newton's Universal Law of Gravitation.
Apply Newton's Third Law of Motion.
Compare the forces acting on the Earth and the Moon.
Explain why their motions are different despite equal forces.
State the final conclusion.

📊 Graph / Figure Graph / Figure ›

Fig. 1 — Free body diagram

✏️ Solution Complete Solution ›

Step-by-step Solution · 12 steps

According to Newton's Universal Law of Gravitation, the gravitational force between the Earth and the Moon is \[F=G\frac{M_E M_M}{r^2}\]
where
- \(M_E\) = mass of the Earth
- \(M_M\) = mass of the Moon
- \(r\) = distance between their centres
This equation gives a single value for the gravitational force between the two bodies. Therefore, the force exerted by the Earth on the Moon and the force exerted by the Moon on the Earth are equal in magnitude.
According to Newton's Third Law of Motion,
Every action has an equal and opposite reaction.
Hence,\[\boxed{F_{\text{Earth on Moon}}=F_{\text{Moon on Earth}}}\]
The two forces have
- equal magnitude, and
- opposite directions.
However, the Earth and the Moon do not move equally because their masses are very different.
From Newton's Second Law,\[a=\frac{F}{m}\]
Since the Earth has a much larger mass than the Moon, its acceleration produced by the same force is very small. On the other hand, the Moon has a much smaller mass, so it experiences a much larger acceleration.
Consequently, the Moon revolves around the Earth, while the Earth's motion due to the Moon is very small and not easily noticeable.
Therefore, the Earth attracts the Moon with the same gravitational force with which the Moon attracts the Earth. The difference lies only in the acceleration produced because their masses are different.

🎯 Exam Significance Exam Significance ›

This is one of the most important conceptual questions based on Newton's Third Law of Motion.
It helps students distinguish between force and acceleration.
Students often wrongly assume that the Earth exerts a greater force because it is much larger. This question removes that misconception.
The concept is frequently asked in CBSE board examinations as reasoning-based questions.
It forms the foundation for understanding orbital motion, satellites and planetary systems.
Questions based on this concept are common in Olympiads, NTSE, JEE Foundation and NEET Foundation examinations.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 7 points

The Earth and the Moon exert equal gravitational forces on each other.
The forces act in opposite directions according to Newton's Third Law.
The gravitational force between two bodies is given by \[ F=G\frac{m_1m_2}{r^2} \]
Equal force does not imply equal acceleration.
The Earth experiences a very small acceleration because of its enormous mass.
The Moon experiences a much larger acceleration because of its comparatively smaller mass.
Always remember:
- Equal force
- Different acceleration
- Reason: Different masses

📘 Concept & Theory Concept Behind the Question ›

This question is based on the combined application of Newton's Universal Law of Gravitation, Newton's Third Law of Motion and Newton's Second Law of Motion.

According to Newton's Universal Law of Gravitation, every two objects in the universe attract each other with equal gravitational force.

\[ F=G\frac{m_1m_2}{r^2} \]

According to Newton's Third Law of Motion,

Every action has an equal and opposite reaction.

Thus, the Moon attracts the Earth with the same force with which the Earth attracts the Moon.

However, the acceleration produced by a force depends on the mass of the object.

\[ a=\frac{F}{m} \]

Since the Earth's mass is enormously greater than that of the Moon, the same gravitational force produces only a very small acceleration of the Earth. Therefore, the Earth's motion towards the Moon is extremely small and is not noticeable.

🗺️ Solution Roadmap Step-by-step Plan ›

Recall that the Earth and the Moon exert equal gravitational forces on each other.
Apply Newton's Second Law of Motion.
Compare the masses of the Earth and the Moon.
Explain why the Earth's acceleration is very small.
State the final conclusion.

📊 Graph / Figure Graph / Figure ›

✏️ Solution Complete Solution ›

Step-by-step Solution · 14 steps

The Earth and the Moon attract each other with gravitational force according to Newton's Universal Law of Gravitation.
According to Newton's Third Law of Motion,
Every action has an equal and opposite reaction.
Therefore,\[\boxed{F_{\text{Earth on Moon}}=F_{\text{Moon on Earth}}}\]
Thus, the Moon attracts the Earth with the same gravitational force with which the Earth attracts the Moon.
However, the acceleration produced by a force depends upon the mass of the object.
According to Newton's Second Law,\[F=ma\]
Therefore,\[a=\frac{F}{m}\]
The mass of the Earth is about\[6\times10^{24}\;kg\]
whereas the mass of the Moon is only about\[7.35\times10^{22}\;kg\]
Since the Earth's mass is nearly 81 times greater than the Moon's mass, the same gravitational force produces only a very small acceleration of the Earth.
As a result, the Earth's motion towards the Moon is extremely small and cannot be noticed in everyday life.
In reality, the Earth and the Moon do not remain completely stationary. Instead, both revolve around their common centre of mass (barycentre), which lies inside the Earth because the Earth is much more massive than the Moon.
Therefore, although the Moon attracts the Earth with the same gravitational force, the Earth does not appear to move towards the Moon because its enormous mass results in a negligible acceleration

🎯 Exam Significance Exam Significance ›

This is a frequently asked conceptual question in CBSE Board examinations.
It checks the understanding of the difference between force and acceleration.
Students often incorrectly think that the Earth remains completely stationary. In reality, both the Earth and the Moon revolve around their common centre of mass.
The question integrates three important concepts:
- Universal Law of Gravitation
- Newton's Third Law of Motion
- Newton's Second Law of Motion
This concept forms the basis for understanding satellite motion, planetary motion and binary star systems.
Similar reasoning-based questions are common in Olympiads, NTSE, JEE Foundation and NEET Foundation examinations.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 7 points

The Earth and the Moon exert equal gravitational forces on each other.
Equal force does not necessarily produce equal motion.
Acceleration depends on mass according to \[ a=\frac{F}{m} \]
The Earth's enormous mass gives it a very small acceleration.
The Moon experiences a much larger acceleration because it has a much smaller mass.
The Earth and the Moon revolve around their common centre of mass (barycentre).
Remember the concept:
- Equal Force
- Different Mass
- Different Acceleration
- Different Motion

📘 Concept & Theory Concept Behind the Question ›

This question is based on Newton's Universal Law of Gravitation, which states that the gravitational force between two objects depends on their masses and the distance between them.

The mathematical expression is

\[ F=G\frac{m_1m_2}{r^2} \]

From the above equation, we observe that:

The gravitational force is directly proportional to the product of the masses.
The gravitational force is inversely proportional to the square of the distance between the objects.

Therefore,

If the masses increase, the force increases proportionally.
If the distance increases, the force decreases according to the square of the distance.

🗺️ Solution Roadmap Step-by-step Plan ›

Write Newton's Universal Law of Gravitation.
Treat each case independently.
Substitute the changed mass or distance into the formula.
Compare the new force with the original force.
State the result for each case.

📊 Graph / Figure Graph / Figure ›

✏️ Solution Complete Solution ›

Step-by-step Solution · 17 steps

According to Newton's Universal Law of Gravitation,\[F=G\frac{m_1m_2}{r^2}\]
Let the original gravitational force between the two objects be\[F=G\frac{m_1m_2}{r^2}\]
(i) When the mass of one object is doubled
Suppose the mass \(m_1\) becomes \(2m_1\).
The new gravitational force is \[ \begin{aligned} F' &=G\frac{(2m_1)m_2}{r^2} \\ &=2G\frac{m_1m_2}{r^2} \\ &=2F \end{aligned} \]
Therefore, the gravitational force becomes twice the original force.
(ii) When the distance between the objects is doubled
Let the new distance be\[2r\]
The new gravitational force becomes
\[ \begin{aligned} F' &=G\frac{m_1m_2}{(2r)^2} \\ &=G\frac{m_1m_2}{4r^2} \\ &=\frac{F}{4} \end{aligned} \]
Therefore, the gravitational force becomes one-fourth of its original value.
When the distance is tripled
Let the new distance be\[3r\]
The new gravitational force is
\[ \begin{aligned} F' &=G\frac{m_1m_2}{(3r)^2} \\ &=G\frac{m_1m_2}{9r^2} \\ &=\frac{F}{9} \end{aligned} \]
Therefore, the gravitational force becomes one-ninth of the original force.
(iii) When the masses of both objects are doubled
Let the new masses be\[2m_1 \quad \text{and} \quad 2m_2\]
The new gravitational force becomes
\[ \begin{aligned} F' &=G\frac{(2m_1)(2m_2)}{r^2} \\ &=4G\frac{m_1m_2}{r^2} \\ &=4F \end{aligned} \]
Therefore, the gravitational force becomes four times the original force.

💡 Answer Final Answer ›

Change	New Gravitational Force
One mass is doubled	\(2F\)
Distance is doubled	\(\dfrac{F}{4}\)
Distance is tripled	\(\dfrac{F}{9}\)
Both masses are doubled	\(4F\)

🎯 Exam Significance Exam Significance ›

This is one of the most frequently asked conceptual and numerical questions from the Gravitation chapter.
It tests the student's understanding of direct and inverse proportionality.
Students must remember that distance is squared, whereas masses are not.
Questions involving doubling, halving or tripling of masses and distances are common in CBSE Board examinations.
This concept is also important for Olympiads, NTSE, JEE Foundation and NEET Foundation examinations.
It forms the basis for solving higher-level numerical problems involving satellites and planetary motion.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 8 points

The gravitational force is directly proportional to the product of the masses.
The gravitational force is inversely proportional to the square of the distance.
Doubling one mass doubles the gravitational force.
Doubling both masses increases the force four times.
Doubling the distance reduces the force to one-fourth.
Tripling the distance reduces the force to one-ninth.
Always remember the governing equation: \[ F=G\frac{m_1m_2}{r^2} \]
Whenever the distance changes, square the new distance before calculating the force.

📘 Concept & Theory Concept Behind the Question ›

The Universal Law of Gravitation, proposed by Sir Isaac Newton, states that every object in the universe attracts every other object with a force that depends upon their masses and the distance between them.

The mathematical expression for this law is

\[ F=G\frac{m_1m_2}{r^2} \]

This law is called universal because it applies to every object in the universe—from tiny particles to planets, stars and galaxies.

It provides a single explanation for the motion of celestial bodies as well as many natural phenomena observed on the Earth.

🗺️ Solution Roadmap Step-by-step Plan ›

State what the Universal Law of Gravitation explains.
Mention its role in planetary and satellite motion.
Explain its importance in natural phenomena on Earth.
Conclude by highlighting its significance in physics and astronomy.

📊 Graph / Figure Graph / Figure ›

Importance of the universal law of gravitation

💡 Answer Final Answer ›

The Universal Law of Gravitation is one of the most important laws of physics because it explains the gravitational attraction between every pair of objects in the universe.

Its importance can be understood through the following applications:

Explains the motion of planets around the Sun
The gravitational attraction between the Sun and the planets keeps the planets moving in their respective orbits.
Explains the motion of the Moon around the Earth
The Earth's gravitational pull keeps the Moon in its orbit around the Earth.
Helps artificial satellites remain in orbit
Communication, weather forecasting, navigation and remote sensing satellites revolve around the Earth due to gravitational attraction.
Explains the force that binds us to the Earth
Gravity gives weight to objects and prevents everything on the Earth's surface from floating away.
Explains ocean tides
The gravitational pull of the Moon, along with that of the Sun, causes high tides and low tides in the oceans.
Explains natural phenomena on the Earth
Gravity is responsible for rainfall, snowfall, the downward flow of rivers and waterfalls, and the movement of all freely falling objects.
Provides a unified explanation of gravitational forces
The same law explains the motion of celestial bodies as well as everyday events occurring on the Earth, making it one of the fundamental laws of nature.

Thus, the Universal Law of Gravitation provides a common explanation for the motion of planets, moons, satellites and many natural phenomena occurring on the Earth, making it one of the cornerstones of physics and astronomy.

🎯 Exam Significance Exam Significance ›

This is a very important long-answer question frequently asked in CBSE Board examinations.
Students are expected to mention the major applications of the Universal Law of Gravitation rather than merely defining the law.
The applications related to planets, satellites, tides and falling objects are commonly tested in examinations.
This concept forms the foundation for higher studies in mechanics, astronomy, astrophysics and space science.
Questions based on these applications are common in Olympiads, NTSE, JEE Foundation and NEET Foundation examinations.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 9 points

The Universal Law of Gravitation applies to every object in the universe.
It explains why planets revolve around the Sun.
It explains why the Moon revolves around the Earth.
It keeps artificial satellites in stable orbits.
It causes tides in oceans.
It is responsible for the weight of objects and free fall.
It explains rainfall, snowfall, rivers and waterfalls.
Remember the governing equation: \[ F=G\frac{m_1m_2}{r^2} \]
The same law explains both celestial motion and everyday natural phenomena on Earth.

📘 Concept & Theory Concept Behind the Question ›

Whenever an object is dropped from a certain height and falls only under the influence of the Earth's gravitational force (without air resistance), it is said to be in free fall.

During free fall, the object continuously gains speed because the Earth exerts a constant gravitational pull on it.

The acceleration produced by the Earth's gravitational force is called the acceleration due to gravity, represented by the symbol \(g\).

It is given by

\[ g=\frac{GM}{R^2} \]

where

\(G\) = Universal Gravitational Constant
\(M\) = Mass of the Earth
\(R\) = Radius of the Earth

🗺️ Solution Roadmap Step-by-step Plan ›

Define free fall.
Define acceleration of free fall.
Mention its symbol and SI unit.
State its approximate value near the Earth's surface.
Explain its significance.

📊 Graph / Figure Graph / Figure ›

✏️ Solution Complete Solution ›

Step-by-step Solution · 8 steps

Acceleration of free fall is the acceleration acquired by an object when it falls freely towards the Earth under the influence of gravity alone, without any other force such as air resistance.
It is represented by the symbol\[g\]
Near the Earth's surface, its value is approximately\[\boxed{g=9.8\;m/s^2}\]
This means that the velocity of a freely falling object increases by \[9.8\;m/s\]
every second in the downward direction.
The SI unit of acceleration due to gravity is\[m/s^2\]
The acceleration due to gravity depends on the mass and radius of the Earth and is given by\[g=\frac{GM}{R^2}\]
Therefore, acceleration of free fall is the constant acceleration produced by the Earth's gravitational force on a freely falling object, and its average value near the Earth's surface is \(9.8\;m/s^2\).

🎯 Exam Significance Exam Significance ›

This is one of the most fundamental definitions in the Gravitation chapter.
Students should clearly distinguish between acceleration due to gravity (\(g\)) and gravitational force (\(F\)).
The value of \(g\) is frequently used in numerical problems involving free fall and equations of motion.
This concept serves as the foundation for projectile motion, satellite motion and gravitational potential in higher classes.
It is frequently tested in CBSE Board examinations, Olympiads, NTSE, JEE Foundation and NEET Foundation examinations.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 7 points

Free fall occurs when only gravity acts on an object.
The acceleration of free fall is represented by \[ g \]
Near the Earth's surface, \[ g\approx9.8\;m/s^2 \]
Its SI unit is \[ m/s^2 \]
The velocity of a freely falling object increases by approximately \(9.8\;m/s\) every second.
The acceleration due to gravity is directed vertically downward towards the centre of the Earth.
Remember the relation: \[ g=\frac{GM}{R^2} \]

📘 Concept & Theory Concept Behind the Question ›

Every object on or near the Earth's surface is attracted towards the Earth due to its gravitational pull. This attractive force acts on all objects, irrespective of their shape, size or composition.

The gravitational force exerted by the Earth on an object is known as the weight of the object.

Weight depends on two factors:

The mass of the object.
The acceleration due to gravity (\(g\)).

Therefore, although the mass of an object remains constant, its weight changes from one celestial body to another because the value of \(g\) changes.

🗺️ Solution Roadmap Step-by-step Plan ›

Define the gravitational force between the Earth and an object.
State its name.
Write the mathematical expression for weight.
Mention the SI unit of weight.

✏️ Solution Complete Solution ›

Step-by-step Solution · 8 steps

The gravitational force exerted by the Earth on an object is called the weight of the object.
Weight is the force with which the Earth attracts an object towards its centre.
It is represented by the symbol\[W\]
The weight of an object is calculated using the relation\[\boxed{W=mg}\] where
- \(W\) = Weight of the object
- \(m\) = Mass of the object
- \(g\) = Acceleration due to gravity
Since the value of \(g\) near the Earth's surface is approximately\[9.8\;m/s^2\]
the weight of an object depends on both its mass and the local value of gravitational acceleration.
The SI unit of weight is\[\boxed{\text{newton (N)}}\]
Therefore, the gravitational force between the Earth and an object is called the weight of the object.

🎯 Exam Significance Exam Significance ›

This is one of the most important definitions in the Gravitation chapter.
Students should clearly understand the difference between mass and weight.
The formula \[ W=mg \] is frequently used in board examination numericals.
Questions asking for the SI unit, symbol and formula of weight are commonly asked in school examinations.
This concept forms the basis for studying pressure, work, energy and mechanics in higher classes.
It is also important for Olympiads, NTSE, JEE Foundation and NEET Foundation examinations.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 6 points

The gravitational force exerted by the Earth on an object is called its weight.
Weight is a force and is represented by \[ W \]
The mathematical relation is \[ W=mg \]
The SI unit of weight is newton (N).
Mass remains constant, whereas weight changes with the value of \(g\).
An object weighs less on the Moon because the Moon's gravitational acceleration is smaller than that of the Earth.

📘 Concept & Theory Concept Behind the Question ›

This question is based on the concepts of mass, weight and the variation in the acceleration due to gravity (\(g\)) at different places on the Earth.

The weight of an object is given by

\[ W=mg \]

where

\(W\) = weight of the object
\(m\) = mass of the object
\(g\) = acceleration due to gravity

The mass of an object remains constant everywhere, but its weight changes because the value of \(g\) is not the same at all places on the Earth.

The value of \(g\) is slightly greater at the poles than at the equator because:

The Earth is slightly flattened at the poles and bulged at the equator, making the poles closer to the Earth's centre.
The Earth's rotation produces a centrifugal effect, which is maximum at the equator and reduces the effective value of \(g\).

🗺️ Solution Roadmap Step-by-step Plan ›

Recall the relation between weight and gravity.
Compare the values of \(g\) at the poles and the equator.
Explain how the weight changes while the mass remains constant.
Draw the required conclusion.

📊 Graph / Figure Graph / Figure ›

✏️ Solution Complete Solution ›

Step-by-step Solution · 8 steps

No. The friend will not agree with the weight of the gold when it is handed over at the equator.
The weight of an object depends on the acceleration due to gravity and is given by\[W=mg\]
Since the value of \(g\) is greater at the poles than at the equator,\[g_{\text{pole}}>g_{\text{equator}}\]
Therefore,\[W_{\text{pole}}>W_{\text{equator}}\]
Although the mass of the gold remains exactly the same, its weight decreases when it is taken from the poles to the equator because the value of \(g\) becomes smaller.
Hence, if the gold is weighed at the equator using a spring balance (which measures weight), it will appear slightly lighter than it did at the poles.
However, if the gold is measured using a beam balance, there will be no difference because a beam balance compares masses, and the mass remains unchanged everywhere.
Therefore, the friend will not agree with the weight of the gold, because the weight is smaller at the equator due to the lower value of \(g\), although the mass of the gold remains unchanged.

🎯 Exam Significance Exam Significance ›

This is one of the most frequently asked reasoning-based questions from the Gravitation chapter.
It tests the student's understanding of the difference between mass and weight.
Students should remember that the value of \(g\) is maximum at the poles and minimum at the equator.
Questions involving beam balance and spring balance are commonly asked in CBSE Board examinations.
This concept is important for understanding weight variation on different planets and satellites in higher classes.
Similar conceptual questions frequently appear in Olympiads, NTSE, JEE Foundation and NEET Foundation examinations.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 7 points

Weight depends on gravity and is given by \[ W=mg \]
Mass remains constant everywhere.
Weight changes with the value of \(g\).
The value of \(g\) is greater at the poles than at the equator.
An object weighs slightly more at the poles than at the equator.
A spring balance measures weight, whereas a beam balance compares mass.
Always remember:
- Mass → Constant
- Weight → Variable
- \(g_{\text{pole}}>g_{\text{equator}}\)

📘 Concept & Theory Concept Behind the Question ›

This question is based on the concept of air resistance (drag) acting on a falling object.

In the absence of air, all objects fall with the same acceleration due to gravity (\(g\)), irrespective of their masses and shapes.

However, under normal conditions, the atmosphere exerts an upward force called air resistance, which opposes the motion of a falling object.

The magnitude of air resistance depends on several factors, including:

The surface area exposed to the air.
The shape of the object.
The speed of the object.

A larger exposed surface area experiences greater air resistance, causing the object to fall more slowly.

🗺️ Solution Roadmap Step-by-step Plan ›

State the role of air resistance.
Compare the surface areas of a flat sheet and a crumpled paper ball.
Explain how air resistance affects their speeds.
State what would happen in the absence of air.

📊 Graph / Figure Graph / Figure ›

✏️ Solution Complete Solution ›

Step-by-step Solution · 7 steps

A flat sheet of paper has a much larger surface area exposed to the air than a crumpled paper ball.
As the sheet falls, it experiences a greater upward force due to air resistance (drag), which opposes its downward motion.
Because of this larger opposing force, the net downward force acting on the sheet is reduced, causing it to fall more slowly.
In contrast, a crumpled paper ball has a much smaller surface area.
Therefore, it experiences much less air resistance and falls more quickly under the influence of gravity.
If the experiment is performed in a vacuum, where there is no air resistance, both the flat sheet and the crumpled paper ball will fall with the same acceleration due to gravity and reach the ground simultaneously.
Therefore, a sheet of paper falls more slowly than a crumpled paper ball because its larger surface area experiences greater air resistance, which slows down its motion.

🎯 Exam Significance Exam Significance ›

This is a very common conceptual question in CBSE Board examinations.
It tests the student's understanding of the difference between free fall and motion in the presence of air resistance.
Students should remember that gravity acts equally on both objects, but air resistance is different due to their shapes.
Questions based on vacuum experiments are frequently asked in school and competitive examinations.
This concept is important for understanding parachutes, skydiving, aircraft design and terminal velocity in higher classes.
It is also useful for Olympiads, NTSE, JEE Foundation and NEET Foundation examinations.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 7 points

Air resistance opposes the motion of a falling object.
A flat sheet has a larger surface area and therefore experiences greater air resistance.
A crumpled paper ball has a smaller surface area and experiences less air resistance.
Less air resistance allows the crumpled paper ball to fall faster.
In a vacuum, both objects fall with the same acceleration due to gravity.
The difference in falling speed is caused by air resistance, not by gravity.
Remember:

📘 Concept & Theory Concept Behind the Question ›

This question is based on the concepts of mass, weight and acceleration due to gravity.

The weight of an object is the gravitational force exerted on it and is given by

\[ W=mg \]

where

\(W\) = Weight of the object (in newtons)
\(m\) = Mass of the object (in kilograms)
\(g\) = Acceleration due to gravity (in \(m/s^2\))

The mass of an object remains the same everywhere, but its weight changes because the value of \(g\) changes from one celestial body to another.

Since the Moon's gravitational pull is only \(\dfrac{1}{6}\) of the Earth's,

\[ g_{\text{Moon}}=\frac{g_{\text{Earth}}}{6} \]

Therefore, the weight of an object on the Moon is also \(\dfrac{1}{6}\) of its weight on the Earth.

🗺️ Solution Roadmap Step-by-step Plan ›

Write the formula for weight.
Calculate the weight on the Earth.
Find the value of gravitational acceleration on the Moon.
Calculate the weight on the Moon.
Compare the two results.

✏️ Solution Complete Solution ›

Step-by-step Solution · 7 steps

given
- Mass of the object, \[ m=10\;kg \]
- Acceleration due to gravity on the Earth, \[ g_{\text{Earth}}=9.8\;m/s^2 \]
- Acceleration due to gravity on the Moon, \[ g_{\text{Moon}}=\frac{9.8}{6}\;m/s^2 \]
Step 1: Calculate the weight on the Earth
Using the formula, \[ W=mg \] \[ \begin{aligned} W_{\text{Earth}} &=10\times9.8 \ &=98\;N \end{aligned} \]
Therefore,\[\boxed{W_{\text{Earth}}=98\;N}\]
Step 2: Calculate the weight on the Moon
The acceleration due to gravity on the Moon is\[g_{\text{Moon}}=\frac{9.8}{6}\;m/s^2\]
Again using\[W=mg\]
\[ \begin{aligned} W_{\text{Moon}} &=10\times\frac{9.8}{6} \ &=\frac{98}{6} \ &=16.33\;N \end{aligned} \]
Therefore,\[\boxed{W_{\text{Moon}}\approx16.3\;N}\]

🎯 Exam Significance Exam Significance ›

This is a standard numerical question based on the formula \[ W=mg \]
Students must remember that mass remains constant while weight changes from one celestial body to another.
The relation \[ g_{\text{Moon}}=\frac{g_{\text{Earth}}}{6} \] is frequently used in board examination numericals.
This concept is important for solving problems related to planets, satellites and space missions.
Similar questions are commonly asked in CBSE Board examinations, Olympiads, NTSE, JEE Foundation and NEET Foundation examinations.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 6 points

Weight is given by \[ W=mg \]
Mass of an object remains constant everywhere.
Weight depends upon the local value of gravitational acceleration.
The Moon's gravity is approximately \[ \frac{1}{6} \] of the Earth's gravity.
A 10 kg object weighs 98 N on the Earth but only 16.3 N on the Moon.
Always distinguish between mass (kg) and weight (N).

📘 Concept & Theory Concept Behind the Question ›

This question is based on the motion of a body under gravity. When a ball is thrown vertically upwards, the only force acting on it is the Earth's gravitational force.

Since gravity acts downward while the ball initially moves upward, the acceleration is

\[ g=-9.8\;m/s^2 \]

As the ball rises, its velocity gradually decreases until it becomes zero at the maximum height. After that, the ball starts falling back towards the Earth with increasing speed.

The equations of motion used are:

\[ v^2=u^2+2as \] \[ v=u+at \]

Since the motion is symmetrical (neglecting air resistance), the time of ascent equals the time of descent.

🗺️ Solution Roadmap Step-by-step Plan ›

Write the given values.
Use the equation \[ v^2=u^2+2gh \] to calculate the maximum height.
Use \[ v=u+gt \] to calculate the time taken to reach the maximum height.
Double this time to obtain the total time of flight.

📊 Graph / Figure Graph / Figure ›

✏️ Solution Complete Solution ›

Step-by-step Solution · 11 steps

Given
- Initial velocity, \[ u=49\;m/s \]
- Final velocity at maximum height, \[ v=0 \]
- Acceleration due to gravity, \[ g=-9.8\;m/s^2 \]
(i) Maximum height reached
Using the third equation of motion,\[v^2=u^2+2gh\]
Substituting the given values, \[ \begin{aligned} 0^2&=49^2+2(-9.8)h \\ 0&=2401-19.6h \\ 19.6h&=2401 \\ h&=\frac{2401}{19.6} \\ h&=122.5\;m \end{aligned} \]
Therefore,\[\boxed{h=122.5\;m}\]
(ii) Total time taken to return to the Earth
First, calculate the time taken to reach the maximum height.
Using the first equation of motion,\[v=u+gt\]
Substituting the given values, \[ \begin{aligned} 0&=49+(-9.8)t \\ 9.8t&=49 \\ t&=\frac{49}{9.8} \\ t&=5\;s \end{aligned} \]
The motion of the ball is symmetrical. Therefore,
- Time of ascent = 5 s
- Time of descent = 5 s
Hence, \[ \begin{aligned} T&=2t \\ &=2\times5 \\ &=10\;s \end{aligned} \]
Therefore,\[\boxed{T=10\;s}\]
Hence,
- Maximum height reached: \[ \boxed{122.5\;m} \]
- Total time taken to return to the Earth: \[ \boxed{10\;s} \]

🎯 Exam Significance Exam Significance ›

This is one of the most frequently asked numerical questions based on motion under gravity.
Students should remember that the velocity becomes zero at the maximum height.
The acceleration due to gravity is taken as negative during upward motion because it acts opposite to the direction of motion.
Using \[ v=u+gt \] to calculate the time of ascent is simpler and shorter than using the displacement equation.
In the absence of air resistance, the time of ascent equals the time of descent.
This concept is widely used in projectile motion and kinematics in higher classes and is frequently tested in CBSE Board examinations, Olympiads, NTSE, JEE Foundation and NEET Foundation.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 8 points

At the highest point, \[ v=0 \]
Acceleration due to gravity always acts downward.
During upward motion, \[ g=-9.8\;m/s^2 \]
The maximum height is calculated using \[ v^2=u^2+2gh \]
The time of ascent is calculated using \[ v=u+gt \]
Time of ascent = Time of descent.
Total time of flight is \[ T=2t \]
For this problem:
- Maximum Height = 122.5 m
- Total Time = 10 s

📘 Concept & Theory Concept Behind the Question ›

This question is based on the motion of a freely falling object. When a stone is released from rest, it falls vertically downward under the influence of the Earth's gravity.

Since the stone is released (not thrown), its initial velocity is zero.

During free fall:

The only force acting on the stone is gravity.
The acceleration remains constant and is equal to the acceleration due to gravity.
The velocity of the stone continuously increases until it reaches the ground.

The required quantity is the final velocity just before the stone touches the ground. This can be found using the third equation of motion:

\[ v^2=u^2+2gh \]

🗺️ Solution Roadmap Step-by-step Plan ›

Write the given values.
Use the third equation of motion.
Substitute the known values.
Simplify step by step.
Take the positive square root since only the magnitude of velocity is required.

📊 Graph / Figure Graph / Figure ›

✏️ Solution Complete Solution ›

Step-by-step Solution · 6 steps

Given
- Height of the tower, \[ h=19.6\;m \]
- Initial velocity, \[ u=0\;m/s \]
- Acceleration due to gravity, \[ g=9.8\;m/s^2 \]
- Final velocity, \[ v=? \]
Step-by-step Solution
Using the third equation of motion,\[v^2=u^2+2gh\]
Substituting the given values, \[ \begin{aligned} v^2 &=0^2+2\times9.8\times19.6 \\ &=19.6\times19.6 \\ &=384.16 \end{aligned} \]
Taking the square root of both sides, \[ \begin{aligned} v &=\sqrt{384.16} \\ &=\sqrt{19.6\times19.6} \\ &=19.6\;m/s \end{aligned} \]
Since the stone is moving downward, its velocity is directed vertically downward.
Therefore, the final velocity of the stone just before touching the ground is\[\boxed{v=19.6\;m/s\text{ downward}}\]

🎯 Exam Significance Exam Significance ›

This is a standard numerical problem based on free fall and equations of motion.
Students should remember that a body released from rest has \[ u=0 \]
The third equation of motion is the most suitable because time is not given.
Always mention the direction of the final velocity whenever applicable.
This concept is frequently tested in CBSE Board examinations, Olympiads, NTSE, JEE Foundation and NEET Foundation examinations.
It forms the basis for solving projectile motion and vertical motion problems in higher classes.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 6 points

A body released from rest has \[ u=0 \]
During free fall, the acceleration due to gravity is \[ g=9.8\;m/s^2 \]
The third equation of motion is \[ v^2=u^2+2gh \]
The velocity increases continuously during free fall.
The final velocity depends on the height through which the object falls.
For this problem, \[ \boxed{v=19.6\;m/s\text{ downward}} \]

📘 Concept & Theory Concept Behind the Question ›

This question is based on vertical motion under gravity and the concepts of distance and displacement.

When a stone is thrown vertically upward:

Its velocity gradually decreases because gravity acts downward.
At the highest point, its velocity becomes zero.
It then falls back to the point from where it was thrown.

Since gravity acts opposite to the upward motion, the acceleration is taken as

\[ a=-10\;m/s^2 \]

The maximum height can be calculated using the third equation of motion:

\[ v^2=u^2+2ah \]

Remember:

Distance is the total path travelled.
Displacement is the shortest distance between the initial and final positions.

🗺️ Solution Roadmap Step-by-step Plan ›

Write the given values.
Use \[ v^2=u^2+2ah \] to calculate the maximum height.
Determine the total distance travelled.
Find the net displacement after the stone returns to its starting point.

📊 Graph / Figure Graph / Figure ›

✏️ Solution Complete Solution ›

Step-by-step Solution · 10 steps

Given
- Initial velocity, \[ u=40\;m/s \]
- Final velocity at the maximum height, \[ v=0 \]
- Acceleration due to gravity, \[ a=-10\;m/s^2 \]
Step 1: Calculate the maximum height
Using the third equation of motion,\[v^2=u^2+2ah\]
Substituting the given values, \[ \begin{aligned} 0^2 &=40^2+2(-10)h \\ 0 &=1600-20h \\ 20h &=1600 \\ h &=\frac{1600}{20} \\ h &=80\;m \end{aligned} \]
Therefore,\[\boxed{h=80\;m}\]
Step 2: Calculate the total distance travelled
The stone travels
- 80 m upward, and
- 80 m downward.
Hence, \[ \begin{aligned} \text{Total Distance} &=80+80 \\ &=160\;m \end{aligned} \]
Therefore,\[\boxed{\text{Total Distance}=160\;m}\]
Step 3: Calculate the net displacement
After completing the journey, the stone returns to the same point from where it was thrown.
Therefore,\[\boxed{\text{Net Displacement}=0\;m}\]
Hence,
- Maximum height: \[ \boxed{80\;m} \]
- Total distance covered: \[ \boxed{160\;m} \]
- Net displacement: \[ \boxed{0\;m} \]

🎯 Exam Significance Exam Significance ›

This is one of the most important numerical questions based on vertical motion under gravity.
Students should remember that the velocity becomes zero at the maximum height.
Always take the acceleration due to gravity as negative during upward motion.
The distinction between distance and displacement is frequently tested in board examinations.
Students often incorrectly write the displacement as 160 m instead of 0 m because the object returns to its starting point.
This concept is important for projectile motion and is frequently asked in CBSE Board examinations, Olympiads, NTSE, JEE Foundation and NEET Foundation examinations.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 7 points

At the highest point, \[ v=0 \]
During upward motion, \[ a=-g \]
The maximum height is calculated using \[ v^2=u^2+2ah \]
Distance is the total path travelled.
Displacement depends only on the initial and final positions.
If an object returns to its starting point, \[ \boxed{\text{Displacement}=0} \]
For this problem:
- Maximum Height = 80 m
- Total Distance = 160 m
- Net Displacement = 0 m

📘 Concept & Theory Concept Behind the Question ›

This question is a direct application of Newton's Universal Law of Gravitation, which states that every two objects in the universe attract each other with a force that depends upon:

The product of their masses.
The square of the distance between their centres.

The gravitational force is given by

\[ F=G\frac{m_1m_2}{R^2} \]

where

\(F\) = Gravitational force
\(G=6.67\times10^{-11}\;Nm^2kg^{-2}\)
\(m_1\) = Mass of the Earth
\(m_2\) = Mass of the Sun
\(R\) = Distance between their centres

This gravitational force is responsible for keeping the Earth in its orbit around the Sun.

🗺️ Solution Roadmap Step-by-step Plan ›

Write the given values.
Use Newton's Universal Law of Gravitation.
Substitute all the given values carefully.
Simplify the numerical values and powers of ten separately.
Write the final answer with the correct SI unit.

📊 Graph / Figure Graph / Figure ›

✏️ Solution Complete Solution ›

Step-by-step Solution · 8 steps

Given
- Mass of the Earth, \[ m_E=6\times10^{24}\;kg \]
- Mass of the Sun, \[ m_S=2\times10^{30}\;kg \]
- Distance between the Earth and the Sun, \[ R=1.5\times10^{11}\;m \]
- Universal Gravitational Constant, \[ G=6.67\times10^{-11}\;Nm^2kg^{-2} \]
Step-by-step Solution
According to Newton's Universal Law of Gravitation,\[F=G\frac{m_Em_S}{R^2}\]
Substituting the given values, \[ F= 6.67\times10^{-11} \times \frac{6\times10^{24}\times2\times10^{30}} {(1.5\times10^{11})^2} \]
First, square the distance: \[ (1.5\times10^{11})^2 &=1.5^2\times10^{22}\\ &=2.25\times10^{22} \]
Therefore, \[ F= 6.67\times10^{-11} \times \frac{12\times10^{54}} {2.25\times10^{22}} \]
Combine the numerical values separately. \[ \begin{aligned} F &= \frac{6.67\times12}{2.25} \times 10^{-11+54-22} \\ &= \frac{80.04}{2.25} \times10^{21} \\ &= 35.57\times10^{21} \\ &= 3.56\times10^{22}\;N \end{aligned} \]
Therefore,\[\boxed{F\approx3.56\times10^{22}\;N}\]
Hence, the gravitational force between the Earth and the Sun is\[\boxed{3.56\times10^{22}\;N}\]

🎯 Exam Significance Exam Significance ›

This is a standard numerical question based on Newton's Universal Law of Gravitation.
Students should simplify the powers of ten separately from the numerical values.
Always square both the numerical value and the power of ten while squaring the distance.
Writing every intermediate step helps avoid mistakes in scientific notation.
This problem demonstrates the enormous gravitational force that keeps the Earth revolving around the Sun.
Similar numericals are frequently asked in CBSE Board examinations, Olympiads, NTSE, JEE Foundation and NEET Foundation examinations.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 6 points

Newton's Universal Law of Gravitation is \[ F=G\frac{m_1m_2}{R^2} \]
The distance between two bodies must always be measured between their centres.
Square both the numerical value and the power of ten while squaring the distance.
Handle numerical values and exponents separately during calculations.
The gravitational force between the Earth and the Sun is approximately \[ \boxed{3.56\times10^{22}\;N} \]
This enormous force keeps the Earth in a stable orbit around the Sun.

📘 Concept & Theory Concept Behind the Question ›

This problem involves the simultaneous motion of two bodies under gravity. One stone is released from the top of the tower, while the other is projected vertically upward from the ground at the same instant.

Both stones experience the same downward acceleration due to gravity:

\[ g=9.8\;m/s^2 \]

Since both stones have the same acceleration, we write separate equations of motion for each stone and use the fact that they meet at the same point after the same time.

🗺️ Solution Roadmap Step-by-step Plan ›

Assume the stones meet after \(t\) seconds.
Let the first stone fall through a distance \(x\).
Express the upward distance travelled by the second stone as \(100-x\).
Apply the equations of motion to both stones.
Solve for the time of meeting.
Find the position where the stones meet.

📊 Graph / Figure Graph / Figure ›

✏️ Solution Complete Solution ›

Step-by-step Solution · 17 steps

Given
- Height of the tower, \[ H=100\;m \]
- Initial velocity of the upward-thrown stone, \[ u=25\;m/s \]
- Acceleration due to gravity, \[ g=9.8\;m/s^2 \]
Step-by-step Solution
Let the two stones meet after\[t\;s\]
Let the stone dropped from the tower fall through a distance\[x\;m\]
Therefore, the upward-thrown stone travels\[(100-x)\;m\]
Step 1: Equation for the falling stone
Since the stone is released from rest,\[u=0\]
Using\[s=ut+\frac{1}{2}gt^2\]
\[ \begin{aligned} x &=0+\frac{1}{2}gt^2 \\ &=\frac{1}{2}gt^2 \end{aligned} \]
Step 2: Equation for the upward-thrown stone
Using\[s=ut-\frac{1}{2}gt^2\]
\[100-x=25t-\frac{1}{2}gt^2\]
Step 3: Substitute the value of \(x\)
From the first equation,\[x=\frac{1}{2}gt^2\]
Substituting into the second equation, \[ \begin{aligned} 100-\frac{1}{2}gt^2 &=25t-\frac{1}{2}gt^2 \\ 100&=25t \\ t&=\frac{100}{25} \\ &=4\;s \end{aligned} \]
Therefore,\[\boxed{t=4\;s}\]
Step 4: Find the meeting position
Substitute\[t=4\;s\] into \[x=\frac{1}{2}gt^2\]
\[ \begin{aligned} x &=\frac{1}{2}\times9.8\times4^2 \\ &=4.9\times16 \\ &=78.4\;m \end{aligned} \]
Thus, the stone released from the top falls\[\boxed{78.4\;m}\]
Since the tower is 100 m high, the meeting point is\[100-78.4=21.6\;m\] above the ground.
Therefore,
- The two stones meet after \[ \boxed{4\;s} \]
- The meeting point is \[ \boxed{21.6\;m} \] above the ground
  or equivalently \[ \boxed{78.4\;m} \] below the top of the tower.

🎯 Exam Significance Exam Significance ›

This is one of the most important higher-order numerical problems from the chapter on Gravitation and Motion.
It tests the ability to apply equations of motion simultaneously to two moving bodies.
Students should clearly identify the displacement of each object before writing the equations.
Notice that the gravitational terms cancel while solving for the meeting time.
A common mistake is reporting the meeting position only from the top or only from the ground. Mentioning both avoids ambiguity.
Questions of this type are common in CBSE Board examinations, Olympiads, NTSE, JEE Foundation and NEET Foundation examinations.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Both stones experience the same downward acceleration, \[ g=9.8\;m/s^2 \]
For the falling stone, \[ s=\frac{1}{2}gt^2 \]
For the upward-moving stone, \[ s=ut-\frac{1}{2}gt^2 \]
The meeting time is \[ \boxed{4\;s} \]
The stones meet \[ \boxed{21.6\;m} \] above the ground
or \[ \boxed{78.4\;m} \] below the top of the tower.

📘 Concept & Theory Concept Behind the Question ›

This problem is based on the vertical motion of a body under gravity. During its upward journey, the ball slows down because gravity acts downward. At the highest point, its velocity becomes zero, after which it falls back with increasing speed.

Since the ball returns to the thrower after 6 s, the motion is symmetrical. Therefore,

Time of ascent = Time of descent
Total time of flight = 6 s
Time to reach the maximum height = 3 s

The equations of motion used are:

\[ v=u+at \] \[ h=ut+\frac{1}{2}at^2 \] \[ v^2=u^2+2ah \]

🗺️ Solution Roadmap Step-by-step Plan ›

Find the time taken to reach the maximum height.
Calculate the initial velocity using the first equation of motion.
Calculate the maximum height reached.
Find the position of the ball after 4 s using the displacement equation.

📊 Graph / Figure Graph / Figure ›

✏️ Solution Complete Solution ›

Step-by-step Solution · 16 steps

Given
- Total time of flight, \[ T=6\;s \]
- Acceleration due to gravity, \[ a=-9.8\;m/s^2 \]
Since the motion is symmetrical,
\[ \text{Time to reach maximum height} =\frac{T}{2} =\frac{6}{2} =3\;s \]
(a) Velocity with which the ball was thrown
At the maximum height,\[v=0\]
Using the first equation of motion,\[v=u+at\]
\[ \begin{aligned} 0 &=u+(-9.8)(3) \\ 0 &=u-29.4 \\ u &=29.4\;m/s \end{aligned} \]
Therefore,\[\boxed{u=29.4\;m/s}\]
(b) Maximum height reached
Using the equation,\[h=ut+\frac{1}{2}at^2\]
\[ \begin{aligned} h &=29.4(3)+\frac{1}{2}(-9.8)(3)^2 \\ &=88.2-44.1 \\ &=44.1\;m \end{aligned} \]
Therefore,\[\boxed{h=44.1\;m}\]
(c) Position of the ball after 4 s
After 4 s, the ball is already on its downward journey. Instead of calculating the downward distance separately, we directly use the equation of motion with \(t=4\) s.
\[ h=ut+\frac{1}{2}at^2 \] \[ \begin{aligned} h &=29.4(4)+\frac{1}{2}(-9.8)(4)^2 \\ &=117.6-78.4 \\ &=39.2\;m \end{aligned} \]
Therefore, after 4 s, the ball is\[\boxed{39.2\;m}\]
above the point of projection.
Alternatively,
Since the maximum height is 44.1 m, after 4 s (i.e., 1 s after reaching the highest point), the ball has fallen \[\frac{1}{2}\times9.8\times1^2=4.9\;m\]
Hence,\[44.1-4.9=39.2\;m\] which confirms the same result.
Therefore,
- Velocity of projection: \[ \boxed{29.4\;m/s} \]
- Maximum height: \[ \boxed{44.1\;m} \]
- Position after 4 s: \[ \boxed{39.2\;m\text{ above the point of projection}} \]

🎯 Exam Significance Exam Significance ›

This is one of the most important numerical problems based on vertical motion under gravity.
Students should remember that the time of ascent equals the time of descent in the absence of air resistance.
Using \[ v=u+at \] is the shortest method to calculate the initial velocity.
A common mistake is writing the position after 4 s as 4.9 m. Actually, 4.9 m is the distance fallen from the highest point. The required position from the ground is 39.2 m.
Questions of this type are frequently asked in CBSE Board examinations, Olympiads, NTSE, JEE Foundation and NEET Foundation examinations.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 7 points

Total time of flight = 6 s.
Time to reach the highest point = 3 s.
Velocity at the highest point is \[ 0 \]
Maximum height is \[ \boxed{44.1\;m} \]
Velocity of projection is \[ \boxed{29.4\;m/s} \]
After 4 s, the ball is \[ \boxed{39.2\;m} \] above the point of projection.
- Time of Ascent = Time of Descent
- Velocity at Highest Point = 0
- Use the displacement equation whenever the position at any instant is required.

📘 Concept & Theory Concept Behind the Question ›

When an object is partially or completely immersed in a liquid, the liquid exerts a force on the object due to the pressure of the liquid.

The pressure exerted by a liquid increases with depth. Therefore, the pressure acting on the bottom surface of the object is greater than the pressure acting on its top surface.

This difference in pressure produces a resultant upward force called the buoyant force or upthrust.

According to Archimedes' Principle, the buoyant force is equal to the weight of the liquid displaced by the immersed object.

🗺️ Solution Roadmap Step-by-step Plan ›

Define buoyant force.
Explain why it is produced.
State its direction.
Mention its effect on immersed objects.

📊 Graph / Figure Graph / Figure ›

✏️ Solution Complete Solution ›

Step-by-step Solution · 5 steps

The buoyant force acting on an object immersed in a liquid always acts vertically upward, opposite to the direction of gravity.
This upward force arises because the pressure exerted by the liquid on the lower surface of the object is greater than the pressure exerted on its upper surface.
As a result, the object experiences a net upward force known as the buoyant force or upthrust.
The buoyant force opposes the weight of the object and may cause it to float, sink more slowly, or remain suspended in the liquid depending on the relative magnitudes of the buoyant force and the object's weight.
Therefore, the buoyant force on an object immersed in a liquid always acts vertically upward.

🎯 Exam Significance Exam Significance ›

This is one of the most frequently asked conceptual questions from the chapter on Gravitation.
Students should remember that buoyant force is also known as upthrust.
This concept forms the basis of Archimedes' Principle and the law of floatation.
Understanding the direction of buoyant force helps explain why ships float, submarines dive and hot-air balloons rise.
Similar questions are commonly asked in CBSE Board examinations, Olympiads, NTSE, JEE Foundation and NEET Foundation examinations.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 6 points

Buoyant force is the upward force exerted by a liquid on an immersed object.
It always acts vertically upward.
It acts opposite to the weight of the object.
It is produced because liquid pressure increases with depth.
Buoyant force is also called upthrust.
According to Archimedes' Principle, the buoyant force equals the weight of the displaced liquid.

📘 Concept & Theory Concept Behind the Question ›

This question is based on the concepts of buoyant force, density and Archimedes' Principle.

Every object immersed in a liquid experiences an upward force called buoyant force or upthrust. This force acts opposite to the weight of the object.

Whether an object floats or sinks depends upon the comparison between:

The weight of the object.
The buoyant force acting on it.

If the buoyant force is greater than the weight of the object, the object rises until it reaches the surface and floats.

Since most plastics have a density lower than that of water, they experience sufficient buoyant force to rise to the surface.

🗺️ Solution Roadmap Step-by-step Plan ›

State that water exerts a buoyant force on the plastic block.
Compare the buoyant force with the weight of the block.
Relate the phenomenon to the density of plastic and water.
Draw the conclusion.

📊 Graph / Figure Graph / Figure ›

✏️ Solution Complete Solution ›

Step-by-step Solution · 7 steps

A block of plastic released under water comes up to the surface because it experiences an upward buoyant force exerted by the water.
Most plastics have a density lower than that of water. Therefore, the weight of the displaced water is greater than the weight of the plastic block.
According to Archimedes' Principle, the buoyant force acting on the block is equal to the weight of the displaced water.
Since the buoyant force acting on the plastic block is greater than its weight, \[ \boxed{\text{Buoyant Force}>\text{Weight}} \]
the block accelerates upward until it reaches the water surface.
After reaching the surface, the block floats with only a part of it submerged, where the buoyant force becomes equal to its weight. \[ \boxed{\text{Buoyant Force}=\text{Weight}} \]
Therefore, a block of plastic rises to the surface because the buoyant force exerted by water is greater than its weight, owing to its lower density than water.

🎯 Exam Significance Exam Significance ›

This is a frequently asked conceptual question based on Archimedes' Principle.
Students should understand that floating depends on the comparison between buoyant force and weight, not simply on the size of the object.
The concept of density plays an important role in explaining why some objects float while others sink.
This concept forms the basis for understanding ships, submarines, hydrometers and life jackets.
Similar reasoning questions are commonly asked in CBSE Board examinations, Olympiads, NTSE, JEE Foundation and NEET Foundation examinations.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 6 points

Every immersed object experiences an upward buoyant force.
Plastic is generally less dense than water.
If \[ \text{Buoyant Force}>\text{Weight} \] the object rises.
When the object floats, \[ \text{Buoyant Force}=\text{Weight} \]
Floating depends on density and buoyant force.
This phenomenon is explained by Archimedes' Principle.

📘 Concept & Theory Concept Behind the Question ›

This question is based on the concepts of density and floatation.

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

\[ \boxed{\text{Density}=\frac{\text{Mass}}{\text{Volume}}} \]

The behaviour of an object in a liquid depends on the comparison between its density and the density of the liquid.

If the density of the object is less than the density of the liquid, it floats.
If the density of the object is greater than the density of the liquid, it sinks.
If both densities are equal, the object remains suspended in the liquid.

🗺️ Solution Roadmap Step-by-step Plan ›

Calculate the density of the given substance.
Compare it with the density of water.
Use the principle of floatation to determine whether it floats or sinks.

📊 Graph / Figure Graph / Figure ›

✏️ Solution Complete Solution ›

Step-by-step Solution · 9 steps

Given
- Mass of the substance, \[ m=50\;g \]
- Volume of the substance, \[ V=20\;cm^3 \]
- Density of water, \[ \rho_{\text{water}}=1\;g\,cm^{-3} \]
Step 1: Calculate the density of the substance
Using the formula,\[\text{Density}=\frac{\text{Mass}}{\text{Volume}}\]
\[ \begin{aligned} \rho &=\frac{50}{20} \\ &=2.5\;g\,cm^{-3} \end{aligned} \]
Therefore,\[\boxed{\rho_{\text{substance}}=2.5\;g\,cm^{-3}}\]
Step 2: Compare the densities
\[\rho_{\text{substance}}=2.5\;g\,cm^{-3}\] \[\rho_{\text{water}}=1\;g\,cm^{-3}\]
Since,\[2.5>1\]
the density of the substance is greater than the density of water.
Hence, the substance is denser than water and will sink when placed in water.
Therefore,\[\boxed{\text{The substance will sink in water.}}\]

🎯 Exam Significance Exam Significance ›

This is a standard numerical question based on the concept of density.
Students should always calculate the density first before deciding whether an object floats or sinks.
Remember that floating depends on the comparison of densities rather than the mass of the object alone.
Questions involving density and floatation are frequently asked in CBSE Board examinations.
This concept forms the foundation for Archimedes' Principle, buoyancy and applications such as ships, submarines and hydrometers.
Similar questions are also common in Olympiads, NTSE, JEE Foundation and NEET Foundation examinations.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 6 points

Density is given by \[ \boxed{\rho=\frac{m}{V}} \]
Density of the given substance is \[ \boxed{2.5\;g\,cm^{-3}} \]
Density of water is \[ \boxed{1\;g\,cm^{-3}} \]
If \[ \rho_{\text{object}}>\rho_{\text{liquid}} \] the object sinks.
If \[ \rho_{\text{object}}<\rho_{\text{liquid}} \] the object floats.
Since \[ 2.5>1 \] the given substance sinks in water.

📘 Concept & Theory Concept Behind the Question ›

This question is based on the concepts of density, buoyancy and Archimedes' Principle.

The density of an object is given by

\[ \rho=\frac{\text{Mass}}{\text{Volume}} \]

The floating or sinking of an object depends upon the comparison of its density with the density of the liquid.

If \[ \rho_{\text{object}}<\rho_{\text{liquid}} \] the object floats.
If \[ \rho_{\text{object}}>\rho_{\text{liquid}} \] the object sinks.

According to Archimedes' Principle, the mass of water displaced depends upon the volume of water displaced. If an object sinks completely, it displaces a volume of water equal to its own volume.

🗺️ Solution Roadmap Step-by-step Plan ›

Calculate the density of the packet.
Compare it with the density of water.
Decide whether the packet floats or sinks.
Calculate the mass of water displaced using the volume of the packet.

📊 Graph / Figure Graph / Figure ›

✏️ Solution Complete Solution ›

Step-by-step Solution · 11 steps

Given
- Mass of the packet, \[ m=500\;g \]
- Volume of the packet, \[ V=350\;cm^3 \]
- Density of water, \[ \rho_{\text{water}}=1\;g\,cm^{-3} \]
Step 1: Calculate the density of the packet
Using the formula,\[\rho=\frac{\text{Mass}}{\text{Volume}}\]
\[ \begin{aligned} \rho_{\text{packet}} &=\frac{500}{350} \\ &=1.4286 \\ &\approx1.43\;g\,cm^{-3} \end{aligned} \]
Therefore,\[\boxed{\rho_{\text{packet}}\approx1.43\;g\,cm^{-3}}\]
Step 2: Decide whether the packet floats or sinks
Comparing the densities,\[1.43\;g\,cm^{-3}>1\;g\,cm^{-3}\]
Since the density of the packet is greater than the density of water, the packet will sink.\[\boxed{\text{The packet sinks in water.}}\]
Step 3: Calculate the mass of water displaced
Since the packet sinks completely, it displaces water equal to its own volume.\[V=350\;cm^3\]
Using\[\text{Mass}=\text{Density}\times\text{Volume}\]
\[ \begin{aligned} \text{Mass of water displaced} &=1\times350 \\ &=350\;g \end{aligned} \]
Therefore,\[\boxed{\text{Mass of water displaced}=350\;g}\]
Hence,
- The packet sinks in water.
- The mass of water displaced is \[ \boxed{350\;g} \]

🎯 Exam Significance Exam Significance ›

This is a standard application of density and Archimedes' Principle.
Students should compare densities before deciding whether an object floats or sinks.
Remember that a sinking object displaces a volume of liquid equal to its own volume.
A common mistake is to assume that the mass of displaced water is equal to the mass of the object. It actually depends on the volume displaced and the density of the liquid.
Questions based on density, buoyancy and displaced liquid are frequently asked in CBSE Board examinations, Olympiads, NTSE, JEE Foundation and NEET Foundation examinations.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Density is calculated using \[ \rho=\frac{m}{V} \]
The density of the packet is \[ \boxed{1.43\;g\,cm^{-3}} \]
Since \[ 1.43>1 \] the packet sinks in water.
A completely submerged object displaces a volume of water equal to its own volume.
The mass of water displaced is \[ \boxed{350\;g} \]

Gravitation

Chapter Complete!

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Force and Laws of Motion — Learning Resources

Frequently Asked Questions

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Gravitation

Chapter Complete!

Recent posts

Force and Laws of Motion — Learning Resources

Frequently Asked Questions

Q 01 What is the difference between work and energy?

Q 02 What is the principle behind a battery?

Q 03 How does energy flow in an ecosystem?

Q 04 What determines the power rating of an appliance?

Q 05 Example of electrical energy in daily life?

Q 06 Which energy change occurs when a ball falls?

Q 07 State the importance of the law of conservation of energy in technology.

Q 08 Can energy be destroyed according to science?

Q 09 What form of energy is utilized in photosynthesis?

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