1. Matter Around Us

I. Reflections on Concepts

1. Question: Explain diffusion phenomenon based on the states of matter.
In solids:
Particles are tightly packed and have very little space between them. Their movement is limited to vibrations in fixed positions. Because of this, diffusion in solids is very slow. For example, when two metal blocks are kept together for a long time at high temperature, their particles slowly diffuse into each other.
In liquids:
Particles in liquids have more space and can move more freely than in solids. Therefore, diffusion occurs faster than in solids. For example, when a drop of ink is added to water, it slowly spreads throughout the water without stirring.
In gases:
Particles in gases have very large spaces between them and move randomly at high speed. Due to this rapid motion, diffusion in gases is the fastest. For example, when perfume is sprayed in a room, its smell spreads quickly to every corner.

2. Question: Mention the properties of solids.
1. Definite Shape:
Solids have a fixed and definite shape. For example, a book or a table maintains its shape unless force is applied.
2. Definite Volume:
Solids occupy a fixed amount of space because their particles are tightly arranged.
3. Strong Intermolecular Forces:
The particles in solids are strongly attracted to each other, which keeps them in fixed positions.
4. Very Small Spaces Between Particles:
Since the particles are closely packed, the space between them is extremely small.
5. Very Little Compressibility:
Solids cannot be compressed easily because there is almost no empty space between particles.
6. Particles Vibrate in Fixed Positions:
The particles do not move freely but only vibrate about their fixed positions.

3. Question: Mention the properties of liquids.
1. No Fixed Shape:
Liquids do not have their own shape. They take the shape of the container in which they are kept.
2. Definite Volume:
Liquids have a fixed volume even though their shape changes.
3. Moderate Intermolecular Forces:
The attraction between particles is weaker than in solids but stronger than in gases.
4. Moderate Spaces Between Particles:
There is more space between particles than in solids, allowing them to move around each other.
5. Flow Easily:
Liquids can flow easily and are therefore called fluids.
6. Slightly Compressible:
Liquids can be compressed slightly but not as much as gases.

4. Question: Explain “fluid” with one example.
Example:
Water is a fluid. When water is poured into a glass, bottle, or bowl, it takes the shape of that container while maintaining the same volume.

5. Question: Mention the properties of gases.
1. No Definite Shape:
Gases do not have a fixed shape and spread in all directions.
2. No Definite Volume:
Gases expand to fill the entire container in which they are kept.
3. Very Weak Intermolecular Forces:
The attractive forces between gas particles are extremely weak.
4. Large Spaces Between Particles:
Gas particles are far apart from each other.
5. Highly Compressible:
Gases can be compressed easily because there is a lot of empty space between particles.
6. Fast Diffusion:
Gases diffuse very quickly because their particles move at high speed.

6. Question: Give two daily life situations where you observe diffusion.
1. Smell of perfume spreading in a room:
When perfume is sprayed, its particles mix with air and spread throughout the room. Even people sitting far away can smell it due to diffusion of gas particles.
2. Ink spreading in water:
When a drop of ink is placed in a glass of water, the ink particles slowly spread throughout the water without stirring.

II. Application of Concepts
1. Question: Mention the applications of compressibility in our daily life.
1. Storage of LPG in cylinders:
Liquefied Petroleum Gas (LPG) is compressed and stored in cylinders so that a large amount of gas can be stored in a small space.
2. Compressed Natural Gas (CNG) in vehicles:
CNG used as fuel in vehicles is stored in highly compressed form in strong cylinders.
3. Oxygen cylinders in hospitals:
Oxygen is compressed and stored in cylinders for medical use.
4. Aerosol sprays:
Products such as deodorants, spray paints, and insecticides contain compressed gases that help release the liquid when the nozzle is pressed.

2. Question: Mention the situations where we use diffusion in our day-to-day life.
1. Breathing:
Oxygen diffuses from the lungs into the blood, while carbon dioxide diffuses from the blood into the lungs.
2. Smell of food spreading:
When food is cooked in the kitchen, the smell spreads to other rooms through diffusion.
3. Mixing of sugar in tea:
When sugar is added to hot tea, it slowly spreads throughout the liquid.
4. Air fresheners:
Air fresheners release fragrance that spreads throughout the room through diffusion.

3. Question: How can we smell perfume sitting several meters away from the source?

4. Question: How do you prove that the speed of diffusion of ammonia is more than that of hydrochloric acid?
Observation:
The white ring forms closer to the hydrochloric acid end.
Reason:
Ammonia gas diffuses faster than hydrochloric acid gas. Because ammonia travels a greater distance in the same time, the meeting point occurs nearer to the hydrochloric acid side.

5. Question: Give examples of matter that can exist in different states.

III. Higher Order Thinking Questions
1. Question: We cannot rejoin the broken chalk easily. Give reason.

2. Question: Does the space between the particles of matter influence the speed of diffusion?

Answer:

Diffusion is the process by which particles of one substance move and mix with the particles of another substance on their own. This happens because the particles of matter are always in continuous motion.

The rate of diffusion depends on the state of matter because the arrangement and movement of particles differ in solids, liquids, and gases.

Thus, diffusion occurs in all states of matter but its speed increases from solids → liquids → gases.

Answer:

Solids have several characteristic properties because their particles are very closely packed and strongly attracted to each other.

These properties make solids rigid and stable in structure.

Answer:

Liquids have properties that are intermediate between solids and gases because their particles are moderately spaced and can move freely.

Examples of liquids include water, milk, oil, and juice.

Answer:

A fluid is a substance that can flow easily and take the shape of its container. Fluids do not have a fixed shape because their particles can move freely.

Both liquids and gases are considered fluids because they can flow from one place to another.

Thus, any substance that can flow freely is called a fluid.

Answer:

Gases have unique properties due to the large distance between their particles and their rapid movement.

Examples include oxygen, carbon dioxide, hydrogen, and nitrogen.

Answer:

Diffusion can be observed in many situations in our daily life.

These examples show how particles move and mix naturally due to diffusion.

Answer:

Compressibility is the ability of a substance, especially gases, to decrease in volume when pressure is applied.

This property is used in several daily life applications:

Thus, compressibility helps in efficient storage and transportation of gases.

Answer:

Diffusion plays an important role in many everyday processes.

Thus, diffusion helps substances mix naturally without external force.

Answer:

Perfume contains volatile liquid substances that quickly evaporate and form vapors. These vapors mix with the air and spread in all directions through diffusion.

Since gas particles move very fast and have large spaces between them, the perfume particles travel quickly through the air. As a result, the smell reaches our nose even if we are several meters away from where the perfume was sprayed.

Therefore, we can smell perfume at a distance because of the rapid diffusion of gas particles in air.

Answer:

This can be demonstrated using a simple laboratory experiment with a long glass tube.

Procedure:

Take a long glass tube.
Place a cotton plug dipped in ammonia solution at one end of the tube.
Place another cotton plug dipped in hydrochloric acid at the other end.
After some time, a white ring of ammonium chloride forms inside the tube.

This proves that ammonia diffuses faster than hydrochloric acid.

Answer:

Some substances can exist in solid, liquid, and gaseous states depending on temperature and pressure.

Example: Water

Solid State: Ice
Liquid State: Water
Gaseous State: Water vapor or steam

Another example is carbon dioxide:

Solid form: Dry ice
Gas form: Carbon dioxide gas

These examples show that the same substance can exist in different states of matter under different conditions.

Answer:

Chalk is a solid substance whose particles are tightly packed and strongly attracted to each other. When chalk breaks, the arrangement of its particles gets disturbed and the surfaces become rough and irregular.

Because solid particles have very little space and very strong intermolecular forces, they cannot move freely to rearrange themselves and join again. Therefore, the broken pieces of chalk cannot easily come back together to form the original chalk.

Answer:

Yes, the space between particles greatly influences the speed of diffusion.

When the space between particles is large, the particles can move more freely and quickly. As a result, diffusion happens faster.

For example:

In gases, particles are very far apart. Therefore diffusion is very fast.
In liquids, particles are moderately spaced, so diffusion is slower than gases.
In solids, particles are tightly packed with almost no space, so diffusion is extremely

Chapter 2: Motion

Reflections on Concept
Question 1
Distinguish between speed and velocity.

Answer:

Speed	Velocity
Speed is the distance travelled by an object per unit time.	Velocity is the displacement of an object per unit time in a particular direction.
It is a scalar quantity (only magnitude).	It is a vector quantity (magnitude and direction).
Speed does not depend on direction.	Velocity always includes direction.
Example: A car moving at 60 km/h.	Example: A car moving 60 km/h towards north.

Explanation:
Speed only tells us how fast something is moving. Velocity tells us how fast and in which direction it is moving. Therefore, two objects can have the same speed but different velocities if their directions are different.

Question 2

Briefly explain about constant acceleration.

Answer:

Constant acceleration means that the rate of change of velocity remains the same during equal intervals of time.

Explanation:
If an object's velocity increases or decreases by the same amount every second, the acceleration is constant.

Example:
If a car's velocity increases like this:

5 m/s at 1 s
10 m/s at 2 s
15 m/s at 3 s

The velocity increases by 5 m/s every second, so the acceleration is constant.

Example in daily life:
A freely falling object under gravity experiences nearly constant acceleration (9.8 m/s²).

Question 3

How can you say that a body is in motion if no force is acting on it?

Answer:

According to Newton’s First Law of Motion, a body continues in its state of rest or uniform motion unless an external force acts on it.

Explanation:
If a body is already moving with constant velocity, it will continue moving without any external force acting on it.

Example:

A spacecraft moving in outer space keeps moving even after engines stop because there is almost no friction.

Thus, a body can remain in motion without any force acting on it, provided no force is needed to change its motion.

Question 4

Are average speed and average velocity same? Explain why.

Answer:

No, average speed and average velocity are not always the same.

Average Speed

Average Velocity

Explanation:
Distance and displacement are different.

Example:
If a person walks 100 m forward and then 100 m backward, then:

Total distance = 200 m
Displacement = 0

So,

Average speed ≠ 0
Average velocity = 0

Thus, they are generally different quantities.

Question 5

How do you measure instantaneous speed?

Answer:

Instantaneous speed is the speed of an object at a particular moment of time.

Explanation:
It is measured using instruments that can record speed instantly.

Example instruments:

Speedometer in vehicles
Radar guns used by traffic police

Example:
If the speedometer of a car shows 60 km/h at a particular moment, that is the instantaneous speed.

Question 6

Explain acceleration with an example.

Answer:

Acceleration is the rate of change of velocity with time.

It can occur due to:

change in speed
change in direction
or both.

Example:

When a car starts from rest and increases its speed:

Time	Velocity
1 s	5 m/s
2 s	10 m/s
3 s	15 m/s

Here velocity is increasing every second.
Therefore, the car is accelerating.

Another example:
When a ball is thrown upward, gravity slows it down. This is negative acceleration (deceleration).

II. Application of Concepts

Question 1

In the given figure distance–time graphs showing motion of two cars A and B are given. Which car moves faster?

Answer:

Car A moves faster.

Explanation:
In a distance–time graph, the slope (steepness) represents speed.

A steeper line → greater speed
A less steep line → smaller speed

Since line A is steeper than B, car A covers more distance in the same time, so it is faster.

Question 2

A train of length 50 m moves with a constant speed of 10 m/s. Calculate the time taken by the train to cross a bridge of length 250 m.

Answer:

Total distance to be covered:

Speed = 10 m/s

Time required = 30 seconds

Question 3

Draw the distance–time graph when the speed of a body increases uniformly.

Answer:

When speed increases uniformly, the distance–time graph is a curved line.

Explanation:
Since the body covers more distance every second, the graph bends upward showing increasing speed.

Question 4

What is the average speed of a Cheetah that sprints 100 m in 6 seconds?

Answer:

Average speed = 16.67 m/s

Question 5

Which is faster? A cheetah that sprints 100 m in 6 s or a rabbit that sprints 50 m in 2 s?

Answer:

Speed of cheetah:

Speed of rabbit:

Rabbit is faster because 25 m/s > 16.67 m/s.

Question 6

A car travels at 80 km/h during the first half hour and 40 km/h during the other half hour. Find the average speed.

Answer:

Since the time intervals are equal,

Average speed = 60 km/h

Question 7

A particle covers 10 m in the first 5 s and 40 m in the next 5 s. Assume constant acceleration. Find initial speed, acceleration and distance covered in next 1 s.

Answer:

Using equations of motion:

Results obtained are:

Initial velocity (u) = 7.66 m/s
Acceleration (a) = 13 m/s²
Distance in next 1 s = 8.33 m

These values are obtained using the motion equations:

III. Higher Order Thinking Questions

Question 1

When velocity is constant, can average velocity over any time interval differ from instantaneous velocity?

Answer:

No, when velocity is constant, average velocity and instantaneous velocity are the same.

Explanation:

Instantaneous velocity is velocity at a particular moment.
Average velocity is displacement divided by total time.

If velocity does not change, the object moves with the same velocity at every moment. Therefore, both values become equal.

Example:
If a car moves at 50 km/h continuously, then:

Instantaneous velocity = 50 km/h
Average velocity = 50 km/h

Question 2

The rabbit and tortoise race story. Draw and explain the distance–time graph.

Answer:

In the story:

The rabbit runs very fast, then rests for some time.
The tortoise moves slowly but continuously without stopping.

Distance–time graph explanation:

1. Rabbit's graph:

o Very steep line (fast speed)

o Then horizontal line (rest)

o Then again steep line

2. Tortoise's graph:

o Straight line with constant slope (slow but steady speed)

Because the rabbit stopped for some time, the tortoise continuously increased its distance and finally won the race.

Conclusion:
The story teaches that steady and continuous progress can be more effective than fast but irregular effort.

Chapter 3: Laws of Motion

I. Reflections on Concepts

Question 1

Question:
Explain the reasons for the following:

(a) Why dust comes out of a carpet when it is beaten with a stick?

Answer:
When a carpet is beaten with a stick, the carpet suddenly starts moving while the dust particles remain at rest for a short moment.

Explanation:
This happens due to the inertia of rest, which is explained by Newton’s First Law of Motion. According to this law, an object at rest tends to remain at rest unless an external force acts on it.

When the carpet is struck:

The carpet moves suddenly.
Dust particles try to remain in their original position because of inertia.
As a result, the dust particles get separated from the carpet and fall down.

Thus, dust comes out because of inertia of rest.

(b) Why is luggage kept on the roof of a bus tied with a rope?

Answer:
Luggage is tied with a rope to prevent it from falling when the bus suddenly starts, stops, or changes direction.

Explanation:
Due to inertia, objects tend to resist changes in their state of motion.

For example:

When the bus starts suddenly, the luggage tends to remain at rest.
When the bus stops suddenly, the luggage tends to keep moving forward.
When the bus turns, the luggage tends to move sideways.

Because of these effects, the luggage may slide or fall. Therefore, it is tied with a rope to keep it in place and ensure safety.

(c) Why does a pace bowler in cricket run from a long distance before he bowls?

Answer:
A pace bowler runs from a long distance to increase the speed and momentum of the ball before delivering it.

Explanation:
Momentum is given by:

When the bowler runs:

His body gains greater velocity.
When he releases the ball, this motion helps transfer more momentum to the ball.

Because of this increased momentum, the ball travels faster and with greater force toward the batsman.

Thus, a long run-up helps the bowler deliver a faster and more powerful ball.

Question 2

Question:
Illustrate an example of each of the three laws of motion.

Answer:

First Law of Motion (Law of Inertia)

This law states that an object remains at rest or continues to move with uniform velocity unless acted upon by an external force.

Example:
Passengers in a moving bus fall forward when the bus suddenly stops. Their bodies continue moving forward due to inertia.

Second Law of Motion

This law states that the rate of change of momentum of an object is proportional to the force applied.

Mathematically:

Where
F = force
m = mass
a = acceleration

Example:
When a football is kicked harder, it moves faster. A larger force produces greater acceleration.

Third Law of Motion

This law states that for every action there is an equal and opposite reaction.

Example:
When a person jumps from a boat onto the shore, the boat moves backward.

The forward push on the shore is the action, and the backward movement of the boat is the reaction.

Question 3

Question:
Explain the following terms:

(a) Static Inertia

Answer:
Static inertia is the tendency of an object to remain at rest unless an external force acts on it.

Example:
A stationary book on a table remains at rest until someone pushes it.

(b) Inertia of Motion

Answer:
Inertia of motion is the tendency of a moving object to continue moving with the same speed and direction unless acted upon by an external force.

Example:
Passengers in a moving bus fall forward when the bus suddenly stops.

(c) Momentum

Answer:
Momentum is the quantity of motion possessed by a moving object.

It is defined as the product of mass and velocity.

Where
m = mass
v = velocity

Example:
A moving truck has more momentum than a bicycle because it has greater mass.

(d) Impulse

Answer:
Impulse is the product of force and the time for which the force acts.

Impulse produces a change in momentum.

Example:
When a bat hits a cricket ball, the force applied for a short time changes the momentum of the ball.

II. Application of Concepts

Question 1

Question:
Two objects have masses 8 kg and 25 kg. Which one has more inertia? Why?

Answer:
The object with 25 kg mass has more inertia.

Explanation:
Inertia is the tendency of an object to resist changes in its state of motion. The inertia of an object depends on its mass.

Greater mass → greater inertia
Smaller mass → smaller inertia

Since 25 kg is greater than 8 kg, the 25 kg object has more inertia.

Question 2

Question:
What is the momentum of a 6.0 kg ball bowling with a velocity of 2.2 m/s?

Answer:

Momentum is calculated using:

Where
m = mass = 6 kg
v = velocity = 2.2 m/s

Final Answer:
Momentum = 13.2 kg m/s

Question 3

Question:
Two people push a car for 3 seconds with a combined force of 200 N.

(a) Calculate the impulse provided to the car.

Impulse is calculated using:

Answer:
Impulse = 600 Ns

(b) If the car has a mass of 1200 kg, what will be its change in velocity?

Impulse equals change in momentum:

Answer:
Change in velocity = 0.5 m/s

Question 4

Question:
A man of mass 30 kg uses a rope to climb which bears only 450 N. What is the maximum acceleration with which he can climb safely?

Answer:

Maximum tension in rope = 450 N
Mass = 30 kg

Using Newton’s second law:

Final Answer:
Maximum safe acceleration = 15 m/s²

III. Higher Order Thinking Questions

Question 1

Question:
A vehicle has a mass of 1500 kg. What must be the force between the vehicle and the road if the vehicle is to be stopped with a negative acceleration of 1.7 m/s²?

Answer:

Using Newton’s second law:

Where
m = 1500 kg
a = −1.7 m/s²

Final Answer:
The force required is −2550 N, acting opposite to the direction of motion of the vehicle.

Explanation:
The negative sign indicates that the force acts against the motion, which slows down and stops the vehicle.

Chapter 4: Refraction of light at plane surfacesBottom of Form

I. Reflections on Concepts

1. Question:

The speed of light in a diamond is 1,24,000 km/s. Find the refractive index of diamond if the speed of light in air is 3,00,000 km/s.

Answer:
The refractive index (n) is defined as the ratio of the speed of light in air (or vacuum) to the speed of light in the medium.

Explanation:
Light slows down when it enters a denser medium like diamond. The refractive index quantifies how much the speed reduces. A higher refractive index means greater bending of light, which is why diamonds sparkle so strongly.

2. Question:

The refractive index of glass relative to water is 9/8. What is the refractive index of water relative to glass?

Answer:
The refractive index of water relative to glass is the reciprocal:

Explanation:
Relative refractive index compares two media. Reversing the order inverts the value because the ratio of speeds is reversed.

3. Question:

The absolute refractive index of water is 4/3. What is the critical angle?

Answer:

Explanation:
The critical angle is the angle of incidence in a denser medium beyond which total internal reflection occurs. It depends only on the refractive index.

4. Question:

Determine the refractive index of benzene if the critical angle with respect to air is 42°.

Answer:

Explanation:
The critical angle provides a direct way to calculate refractive index when the second medium is air.

5. Question:

Explain the formation of mirage.

Answer:
A mirage occurs due to refraction and total internal reflection of light in layers of air with varying temperatures.

Explanation:

On hot days, air near the ground becomes hotter and less dense.
Cooler, denser air lies above it.
Light rays bend continuously as they pass through these layers.
When the angle becomes large enough, total internal reflection occurs.
The eye traces the rays back in a straight line, creating the illusion of water on the road.

6. Question:

Explain the refraction of light through a glass slab with a neat ray diagram.

Answer:
When light enters a glass slab:

It bends towards the normal (denser medium).

Inside the slab, it travels in a straight line.
On exiting, it bends away from the normal.

Key Point:
The emergent ray is parallel to the incident ray but shifted sideways (lateral displacement).

Explanation:
This happens because the slab has parallel surfaces, so the angle of incidence equals the angle of emergence.

7. Question:

Why do stars appear twinkling?

Answer:
Stars twinkle due to atmospheric refraction.

Explanation:

Light from stars passes through layers of air with varying densities.
These layers constantly change due to temperature variations.
This causes the light path to fluctuate.
As a result, the star’s brightness appears to vary or “twinkle.”

II. Application of Concepts

1. Question:

A light ray is incident on an air-liquid interface at 45° and refracted at 30°. What is the refractive index of the liquid?

Answer:
Using Snell’s Law:

Explanation:
The refractive index depends on how much the ray bends when entering the liquid.

2. Question:

In what cases does a light ray not deviate at the interface of two media?

Answer:
A light ray does not deviate when:

1. It falls normally (perpendicular) to the surface.

2. The refractive indices of both media are equal.

Explanation:
No bending occurs if there is no change in optical density or if the angle of incidence is zero.

3. Question:

Place an object on the table. Look through a transparent glass slab. You will observe it will appear closer to you. Explain.

Answer:
The object appears closer due to refraction of light.

Explanation:

Light from the object bends towards the normal when entering the glass.
It bends away when exiting.
The eye traces the light backward in a straight line.
This creates a virtual image that appears closer than the actual object.

4. Question:

Why does a diamond shine more than a glass piece cut to the same shape?

Answer:
A diamond shines more due to its high refractive index and low critical angle.

Explanation:

Light entering a diamond undergoes multiple total internal reflections.
This traps light inside and reflects it multiple times.
More light emerges, making it sparkle intensely compared to glass.

III. Higher Order Thinking Questions

1. Question:

Why is it difficult to shoot a fish swimming in water?

Answer:
Because of refraction, the fish appears at a different position than its actual location.

Explanation:

Light from the fish bends when passing from water to air.
The fish appears higher (shallower) than it really is.
If aimed directly, the shot misses the real position.

2. Question:

Explain why a test tube immersed at a certain angle in a tumbler of water appears to have a mirror surface for a certain viewing position.

Answer:
This happens due to total internal reflection.

Explanation:

Light traveling inside the glass hits the glass-water boundary.
At certain angles, it undergoes total internal reflection.
The surface then behaves like a mirror.

3. Question:

When we sit at a campfire, objects beyond the fire are seen swaying. Give the reason.

Answer:
This is due to refraction caused by hot air currents.

Explanation:

Air above the fire is hot and less dense.
Surrounding air is cooler and denser.
Light passing through these layers bends irregularly.
This makes objects appear distorted or swaying.

Chapter 5: Gravitation

I. Reflections on Concepts

1. Question:

How do you explain that an object is in uniform circular motion?

Answer:
An object is said to be in uniform circular motion when it moves along a circular path with constant speed.

Explanation:
Even though the speed remains constant, the direction of motion keeps changing continuously. Since velocity depends on both speed and direction, this change in direction means the object is constantly accelerating. This acceleration is called centripetal acceleration, which always acts towards the center of the circle.

2. Question:

Calculate the acceleration of the moon towards earth’s center.

Answer:
The acceleration of the moon towards Earth is approximately:

𝑎 \approx 0.0027 𝑚 / 𝑠^{2}

Explanation:
This acceleration is due to Earth’s gravitational pull and acts as the centripetal acceleration that keeps the moon in orbit. It is much smaller than the acceleration due to gravity on Earth’s surface because the moon is very far from Earth.

3. Question:

Explain universal law of gravitation.

Answer:
The universal law of gravitation states that:

Every object in the universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

𝐹 = 𝐺 \frac{𝑚_{1} 𝑚_{2}}{𝑟^{2}}

Explanation:

$𝑚_{1}$ and $𝑚_{2}$ are the masses
$𝑟$ is the distance between them
$𝐺$ is the gravitational constant

This law explains phenomena like falling objects, planetary motion, and tides.

4. Question:

Explain some situations where the centre of gravity of man lies outside the body.

Answer:
The center of gravity (C.G.) can lie outside the body in certain positions.

Examples & Explanation:

When a person bends forward, the C.G. may shift outside the torso.
In activities like high jump or gymnastics, athletes adjust body posture so the C.G. lies outside their physical body.
This helps in maintaining balance or achieving better motion.

5. Question:

Explain why a long pole is more beneficial to the tight rope walker if the pole has slight bend.

Answer:
A long, slightly bent pole helps in maintaining balance.

Explanation:

The pole lowers the overall center of gravity of the system (walker + pole).
A lower C.G. increases stability.
The bend spreads mass downward, improving balance further.
It also increases rotational inertia, making it harder to tip over.

II. Application of Concepts

1. Question:

A car moves with speed of 10 m/s in a circular path of radius 10 m. The mass of the car is 1000 kg. How much is the required centripetal force?

Answer:

𝐹 = \frac{𝑚 𝑣^{2}}{𝑟} = \frac{1000 \times 10^{2}}{10} = 10,000 𝑁

Explanation:
This force is required to keep the car moving in a circular path and is directed towards the center.

2. Question:

A ball is projected vertically up with speed of 50 m/s. Find the maximum height, time to reach maximum height, and speed at the maximum height.

Answer:

Maximum height:

ℎ = \frac{𝑣^{2}}{2 𝑔} = \frac{50^{2}}{2 \times 10} = 125 𝑚

Time to reach maximum height:

𝑡 = \frac{𝑣}{𝑔} = \frac{50}{10} = 5 𝑠

Speed at maximum height:

0 𝑚 / 𝑠

Explanation:
As the ball rises, gravity slows it down until its velocity becomes zero at the highest point.

3. Question:

Two spherical balls of mass 10 kg each are placed with their centers 10 cm apart. Find the gravitational force of attraction between them.

Answer:

𝐹 = 𝐺 \frac{𝑚_{1} 𝑚_{2}}{𝑟^{2}} = 6.67 \times 10^{- 11} \times \frac{10 \times 10}{(0.1)^{2}}

𝐹 = 6.67 \times 10^{- 7} 𝑁

Explanation:
The force is extremely small because gravitational force is weak unless masses are very large.

4. Question:

A ball is dropped from height. If it takes 0.2 s to cross the last 2 m before hitting the ground, find the height.

Answer:
The approximate height is:

\approx 45.05 𝑚

Explanation:
Using equations of motion, we analyze the velocity during the last segment and relate it to total height. The short time for the last 2 m indicates high speed near the ground.

5. Question:

What path does the moon take when the gravitational interaction between the moon and earth disappears?

Answer:
The moon will move in a straight-line path tangential to its orbit.

Explanation:
Without gravitational force, there is no centripetal force to keep it in orbit. According to Newton’s first law, it continues in a straight line.

6. Question:

Why is it easier to carry the same amount of water in two buckets, one in each hand rather than in a single bucket?

Answer:
Because balance is improved.

Explanation:

Carrying two buckets distributes weight evenly on both sides.
The center of gravity remains stable.
This reduces strain and makes walking easier compared to carrying all weight on one side.

III. Higher Order Thinking Questions

1. Question:

A man is standing against a wall such that his shoulder and right leg are in contact with the wall. Can he raise his left leg?

Answer:
No, he cannot raise his left leg.

Explanation:

Raising the leg shifts the center of gravity.
Normally, we adjust by leaning sideways.
But the wall prevents this adjustment.
Hence, balance cannot be maintained.

2. Question:

A satellite falls on earth if the applied force on it ceases. What acceleration is required to accelerate the earth with an apple?

Answer:
The acceleration produced is extremely small (practically negligible).

Explanation:

According to Newton’s third law, the apple attracts the Earth with equal force.
But since Earth’s mass is enormous, acceleration $𝑎 = 𝐹 / 𝑚$ becomes extremely tiny.
Hence, no noticeable movement occurs.

I. Reflections on Concepts

1. Question:

Which separation techniques will you apply to separate the following?

(a) Sodium chloride from its solution in water

Answer: Evaporation

Explanation:
When a salt solution is heated, water (the solvent) evaporates, leaving behind solid sodium chloride. This method works because the salt does not vaporize with water.

(b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride

Answer: Sublimation

Explanation:
Ammonium chloride sublimes (changes directly from solid to gas) on heating, while sodium chloride does not. The vapor is then cooled to obtain solid ammonium chloride separately.

(c) Mixture of oil and water

Answer: Separating funnel

Explanation:
Oil and water are immiscible liquids and form two layers due to density differences. A separating funnel allows the heavier water layer to be drained first, leaving oil behind.

(d) Fine mud particles suspended in water

Answer: Filtration (or sedimentation and decantation)

Explanation:
Mud particles are insoluble and can be removed by filtration. Alternatively, allowing them to settle (sedimentation) and pouring off the clear water (decantation) also works.

2. Question:

Explain the following giving examples.

(a) Saturated solution

Answer:
A saturated solution is one in which no more solute can dissolve at a given temperature.

Example & Explanation:
If you keep adding sugar to water, a stage comes when sugar stops dissolving and settles at the bottom. This indicates the solution is saturated.

(b) Pure substance

Answer:
A pure substance consists of only one type of particle and has a fixed composition.

Example & Explanation:
Distilled water or pure gold are examples. They have uniform properties throughout and cannot be separated into simpler substances by physical methods.

(c) Colloid

Answer:
A colloid is a mixture in which very fine particles are dispersed but do not settle down.

Example & Explanation:
Milk is a colloid where fat droplets are dispersed in water. These particles are small enough to remain suspended and scatter light.

(d) Suspension

Answer:
A suspension is a heterogeneous mixture where particles are large enough to settle down on standing.

Example & Explanation:
Muddy water is a suspension. The particles settle at the bottom after some time and can be filtered.

3. Question:

Classify the following into elements, compounds, and mixtures:
Sodium, soil, sugar solution, silver, calcium carbonate, tin, silicon, coal, air, methane, carbon dioxide, sea water

Answer:

Elements: Sodium, Silver, Tin, Silicon
Compounds: Calcium carbonate, Methane, Carbon dioxide
Mixtures: Soil, Sugar solution, Coal, Air, Sea water

Explanation:

Elements contain only one type of atom.
Compounds are chemically combined substances with fixed ratios.
Mixtures are physical combinations with variable composition.

II. Application of Concepts

1. Question:

Determine the mass by percentage of solute concentration of a 100 g salt solution which contains 20 g salt.

Answer:

Explanation:
Mass percentage tells how much solute is present in 100 g of solution.

2. Question:

Calculate the concentration of a 5 ml solution containing 2.5 g potassium chloride (KCl).

Answer:

Explanation:
This represents mass by volume percentage, showing how much solute is present per unit volume of solution.

3. Question:

Classify the following substances into true solution, suspension, and colloidal dispersion:
Ink, soda water, brass, fog, blood, aerosol spray, fruit salad, black coffee, oil and water, hot polish, air, nail polish, starch solution, milk

Answer:

True Solution: Soda water, Brass (alloy), Black coffee, Air, Nail polish
Suspension: Fruit salad, Oil and water
Colloidal Dispersion: Ink, Fog, Blood, Aerosol spray, Hot polish, Starch solution, Milk

Explanation:

True solutions are homogeneous and do not scatter light.
Suspensions have large particles that settle.
Colloids have intermediate particle size and show the Tyndall effect.

III. Higher Order Thinking Questions

1. Question:

Use the words given below and write the steps for making tea:
(Solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue)

Answer:

Water acts as the solvent and is heated.
Sugar is added as a solute and it dissolves because it is soluble in water, forming a solution.
Tea leaves are added, but they are insoluble.
The mixture is then filtered.
The liquid tea obtained is the filtrate, and the tea leaves left behind are the residue.

Explanation:
This process demonstrates how different substances behave in a mixture and how filtration helps separate insoluble components.

Chapter 7: Atoms, Molecules and Chemical Reactions

I. Reflections on Concepts

1. Question:

Explain the process and precautions in verifying the law of conservation of mass.

Answer:
Process:

Take a closed container (like a conical flask) with a stopper.
Add one reactant (e.g., sodium sulphate solution) inside the flask.
Take another reactant (e.g., barium chloride solution) in a small test tube and place it inside the flask without mixing.
Weigh the entire setup.
Tilt the flask to mix both solutions and allow the reaction to occur.
Weigh the flask again.

Observation:
The mass before and after the reaction remains the same.

Precautions:

The system must be airtight (closed) to prevent loss of matter.
Accurate weighing should be done.
No substance should spill during mixing.

Explanation:
This verifies that mass is neither created nor destroyed during a chemical reaction.

2. Question:

Two students wrote formulas of ammonia as N₂H₃ and NH₃. Who is correct? State the reason.

Answer:
The correct formula is NH₃.

Explanation:
Nitrogen has a valency of 3 and hydrogen has a valency of 1. Therefore, one nitrogen atom combines with three hydrogen atoms, forming NH₃. The formula N₂H₃ does not satisfy valency rules.

3. Question:

Find out the chemical names and formulae for the following common household substances:

Common salt: Sodium chloride (NaCl)
Baking soda: Sodium bicarbonate (NaHCO₃)
Washing soda: Sodium carbonate (Na₂CO₃·10H₂O)

Explanation:
These substances are widely used in daily life for cooking, cleaning, and other purposes.

4. Question:

Calculate the mass of the following:

(a) 0.5 mole of N₂ gas

(b) 0.5 mole of atoms

Mass depends on the atom; generally:

Explanation:
Mass is directly proportional to the number of moles.

5. Question:

Convert into mole:

(a) 3.01 × 10²³ atoms

(b) 6.022 × 10²³ molecules

Explanation:
Avogadro’s number (6.022 × 10²³) represents one mole of particles.

6. Question:

Convert into mole:

(a) 12 g of O₂

(b) 20 g of water

(c) 22 g of carbon dioxide

7. Question:

Write the valencies of Na, Fe, and Cl.

Answer:

Sodium (Na): 1
Iron (Fe): 2 or 3
Chlorine (Cl): 1

Explanation:
Valency is the combining capacity of an element.

8. Question:

Calculate the molar mass of sulphuric acid (H₂SO₄) and glucose (C₆H₁₂O₆).

Answer:

H₂SO₄:

C₆H₁₂O₆:

9. Question:

Which has more number of atoms: 100 g sodium atoms or 100 g iron atoms?

Answer:
Sodium has more atoms.

Explanation:

Molar mass of Na = 23 g/mol → more moles in 100 g
Molar mass of Fe = 56 g/mol → fewer moles in 100 g
More moles mean more atoms.

10. Question:

Write an equation for decomposition reaction where energy is supplied in the form of heat.

Answer:

Explanation:
Heat breaks down calcium carbonate into calcium oxide and carbon dioxide.

11. Question:

How do chemical displacement reactions differ from decomposition reactions?

Answer:

Displacement reaction: One element replaces another.
Decomposition reaction: A compound breaks into simpler substances.

Explanation:
Example:

Displacement: Zn + CuSO₄ → ZnSO₄ + Cu
Decomposition: CaCO₃ → CaO + CO₂

12. Question:

Name the reactions taking place in the presence of sunlight.

Answer:
Photochemical reactions.

Explanation:
These reactions are driven by light energy, such as photosynthesis.

13. Question:

Give two examples for oxidation-reduction reactions.

Answer:

1. Zn + Cu²⁺ → Zn²⁺ + Cu

2. Fe₂O₃ + CO → Fe + CO₂

Explanation:
Oxidation involves loss of electrons, reduction involves gain.

14. Question:

Draw the diagram to show the experimental setup to verify the law of conservation of mass.

Answer (Explanation):
The setup includes:

A closed flask
Reactants placed separately inside
A stopper to seal the system

The diagram should show no loss of matter during reaction.

II. Application of Concepts

1. Question:

Why do we apply paint on iron articles?

Answer:
To prevent rusting.

Explanation:
Paint forms a protective layer that stops contact with air and moisture, preventing oxidation.

2. Question:

What is the use of keeping food in a tight container?

Answer:
To prevent spoilage.

Explanation:
It blocks air and microorganisms, slowing down oxidation and decomposition.

III. Higher Order Thinking Questions

1. Question:

15.9 g of copper sulphate and 10.6 g of sodium carbonate react together to give 14.2 g of sodium sulphate and 12.3 g of copper carbonate. Which law of chemical combination is obeyed?

Answer:
Law of conservation of mass.

Explanation:
Total mass before reaction = 15.9 + 10.6 = 26.5 g
Total mass after reaction = 14.2 + 12.3 = 26.5 g
Mass remains constant.

2. Question:

Carbon dioxide is added to 112 g of calcium oxide. The product formed is 200 g of calcium carbonate. Calculate the mass of carbon dioxide used.

Answer:

Explanation:
Based on conservation of mass, missing mass must belong to CO₂.

3. Question:

Imagine what would happen if we do not have standard symbols for elements.

Answer:
It would create confusion in chemistry.

Explanation:

Different names in different languages would cause misunderstanding.
Writing chemical equations would become difficult.
Standard symbols ensure universal communication among scientists.

Chapter 8: Floating Bodies

I. Reflections on Concepts

1. Question:

Why do some objects float on water? And some sink?

Answer:
Objects float or sink depending on their density compared to water.

Explanation:

If the density of an object is less than water, it floats (e.g., wood, ice).
If the density is greater than water, it sinks (e.g., stone, iron).
Floating also depends on the buoyant force acting upward.
An object floats when buoyant force equals its weight, and sinks when its weight is greater than the buoyant force.

2. Question:

Explain density and relative density and write their formulae.

Answer:

Density:
Density is the mass per unit volume of a substance.

Relative Density (Specific Gravity):
Relative density is the ratio of the density of a substance to the density of water.

Explanation:
Density tells how compact a substance is, while relative density compares it with water and has no unit.

3. Question:

Explain buoyancy in your own words.

Answer:
Buoyancy is the upward force exerted by a fluid on an object placed in it.

Explanation:

When an object is placed in a liquid, it displaces some liquid.
The liquid exerts an upward force on the object.
This force is called buoyant force.
It is responsible for floating of objects and is explained by Archimedes’ principle.

4. Question:

How can you find the relative density of a liquid?

Answer:
Relative density of a liquid can be found using a density bottle (specific gravity bottle).

Procedure & Explanation:

Measure the mass of the empty bottle.
Fill it with water and measure the mass.
Fill it with the given liquid and measure again.
Use the formula:

This works because the bottle holds the same volume each time.

5. Question:

Draw the diagram of a mercury barometer.

Answer (Explanation):
A mercury barometer consists of:

A long glass tube filled with mercury and inverted in a mercury trough
The top of the tube is vacuum (Torricellian vacuum)
The height of the mercury column indicates atmospheric pressure

(In exams, draw a labeled diagram showing tube, mercury level, vacuum, and scale.)

II. Application of Concepts

1. Question:

A solid sphere has a radius of 2 cm and a mass of 0.05 kg. What is the relative density of the sphere?

Answer:

Radius = 2 cm → Volume:

Mass = 0.05 kg = 50 g

Explanation:
Relative density is simply the ratio with water (1 g/cm³), so the value remains the same.

2. Question:

A small bottle weighs 20 g when empty and 22 g when filled with water. When filled with oil it weighs 21.76 g. What is the density of oil?

Answer:

Mass of water = 22 − 20 = 2 g
Mass of oil = 21.76 − 20 = 1.76 g

Explanation:
Oil is less dense than water, so its density is lower than 1 g/cm³.

3. Question:

An ice cube floats on water. Will the water level rise? When the ice melts completely?

Answer:

When ice floats: Water level remains unchanged
After melting: Water level still remains unchanged

Explanation:

Floating ice displaces water equal to its weight.
When it melts, it converts into the same amount of water.
Hence, no change in water level.

4. Question:

Find the pressure at a depth of 10 m in water if atmospheric pressure is 100 kPa.

Answer:

Explanation:
Pressure increases with depth due to the weight of the water above.

III. Higher Order Thinking Questions

1. Question:

Can you make iron float in water? How?

Answer:
Yes, iron can be made to float.

Explanation:

Although iron is denser than water, shaping it into a hollow structure (like a ship) increases its volume.
This reduces its average density.
When average density becomes less than water, it floats.

2. Question:

Where do you observe Archimedes’ principle in daily life? Give two examples.

Answer:

Examples & Explanation:

1. Ships floating: They displace large volumes of water, creating enough buoyant force.

2. Hydrometers: Used to measure density of liquids based on how much they float or sink.

3. Question:

Do all objects that sink in water sink in oil? Give reason.

Answer:
No, not necessarily.

Explanation:

Sinking depends on comparison of densities.
Oil is less dense than water.
An object that sinks in water may still sink in oil, but some objects may float in water and sink in oil depending on their density relative to each liquid.

Chapter 9: What is inside atom

I. Reflections on Concepts

1. Question:

What are the three subatomic particles?

Answer:
The three main subatomic particles are:

Electron (negative charge)
Proton (positive charge)
Neutron (no charge)

Explanation:
These particles make up an atom. Protons and neutrons are present in the nucleus, while electrons revolve around the nucleus in different energy levels.

2. Question:

What were the three major observations Rutherford made in the gold foil experiment?

Answer:
Rutherford observed:

Most alpha particles passed straight through the foil.
Some particles were deflected by small angles.
A very few particles were deflected back.

Explanation:
These observations led to the conclusion that:

Most of the atom is empty space.
A small, dense, positively charged nucleus exists at the center.

3. Question:

Give the main postulates of Bohr’s model of an atom.

Answer:
Bohr proposed that:

Electrons revolve in fixed circular paths called energy levels or shells.
Each orbit has a definite energy.
Electrons do not lose energy while moving in these orbits.
Energy is absorbed or emitted when electrons jump between orbits.

Explanation:
This model explains atomic stability and spectral lines of atoms.

4. Question:

State the valencies of magnesium and sodium.

Answer:

Magnesium (Mg): 2
Sodium (Na): 1

Explanation:
Valency depends on the number of electrons lost or gained to achieve a stable configuration. Magnesium loses 2 electrons, while sodium loses 1.

II. Application of Concepts

1. Question:

Compare the subatomic particles electron, proton, and neutron.

Answer:

Particle	Charge	Mass	Location
Electron	Negative (-1)	Very small	Outside nucleus
Proton	Positive (+1)	1 unit	Inside nucleus
Neutron	Neutral (0)	1 unit	Inside nucleus

Explanation:
Electrons determine chemical properties, while protons define the identity of the element. Neutrons contribute to atomic mass and stability.

2. Question:

What are the limitations of J.J. Thomson’s model of the atom?

Answer:

It could not explain the presence of a nucleus.
It failed to explain Rutherford’s scattering experiment.
It did not describe how electrons are arranged.

Explanation:
Thomson’s model assumed uniform distribution of charge, which was later proven incorrect.

3. Question:

Define valency by taking examples of nitrogen and boron.

Answer:
Valency is the combining capacity of an atom.

Explanation with Examples:

Nitrogen (N): Has 5 electrons in outer shell, needs 3 more → Valency = 3
Boron (B): Has 3 electrons in outer shell, loses 3 → Valency = 3

4. Question:

What is the main difference among the isotopes of the same element?

Answer:
Isotopes differ in the number of neutrons.

Explanation:

They have the same number of protons (same atomic number).
They have different mass numbers due to different neutron counts.
Example: Carbon-12 and Carbon-14.

5. Question:

Fill in the missing information in the table.

Completed Table:

Name	Symbol	Atomic No. (Z)	Mass No. (A)	Neutrons	Electrons
Oxygen	¹⁶₈O	8	16	8	8
Nitrogen	¹⁴₇N	7	14	7	7
Sulphur	³⁴₁₆S	16	34	18	16
Beryllium	⁹₄Be	4	9	5	4
Magnesium	²⁴₁₂Mg	12	24	12	12
Magnesium	²⁵₁₂Mg	12	25	13	12

Explanation:

Neutrons = Mass number − Atomic number
Electrons = Atomic number (for neutral atoms)

III. Higher Order Thinking Questions

1. Question:

Cl⁻ has completely filled K, L & M shells. Explain it based on Bohr-Bury theory.

Answer:
Chlorine has atomic number 17.

Explanation:

Electronic configuration of Cl: 2, 8, 7
It gains one electron to form Cl⁻: 2, 8, 8
According to Bohr-Bury rules:
- Maximum electrons in shells: K = 2, L = 8, M = 18
- Outer shell should not exceed 8 electrons

Thus, Cl⁻ achieves a stable configuration with completely filled shells.

2. Question:

Explain the efforts made by scientists to explain the structure of atom by developing various atomic models.

Answer:
Scientists proposed different models over time:

1. Dalton’s Model:

Atoms are indivisible particles.
Could not explain internal structure.

2. Thomson’s Model:

Atom is a sphere of positive charge with electrons embedded.
Failed to explain nucleus.

3. Rutherford’s Model:

Introduced nucleus based on gold foil experiment.
Could not explain stability of electrons.

4. Bohr’s Model:

Electrons revolve in fixed energy levels.
Explained stability and spectra.

Explanation:
Each model improved upon the previous one, leading to a better understanding of atomic structure.

Chapter 10: Work and Energy

I. Reflections on Concepts

1. Question:

What is work according to science and write its units (AS)?

Answer:
In science, work is said to be done when a force causes an object to move in the direction of the force. If there is no movement, no work is done, even if force is applied.

Mathematically:
Work (W) = Force (F) × Displacement (d)

SI unit of work: Joule (J)
1 Joule = Work done when a force of 1 Newton moves an object by 1 meter.

2. Question:

Give few examples where displacement of an object is in the direction opposite to the force acting on the object (AS).

Answer:
When displacement is opposite to the applied force, work done is negative. Examples include:

Braking a bicycle: Force acts backward, but motion is forward.
Sliding a book on a table: Friction acts opposite to motion.
Throwing a ball upward: Gravity acts downward while the ball moves upward.

In all these cases, force and displacement are in opposite directions.

3. Question:

Write few daily life examples in which you observe conservation of energy (AS).

Answer:
The law of conservation of energy states that energy cannot be created or destroyed, only transformed.

Examples:

Pendulum motion: Potential energy converts to kinetic energy and back.
Electric fan: Electrical energy → Mechanical energy.
Car engine: Chemical energy (fuel) → Mechanical energy.
Hydropower plants: Water’s potential energy → Electrical energy.

4. Question:

Give some examples of renewable sources of energy (AS).

Answer:
Renewable energy sources are those that are naturally replenished.

Examples:

Solar energy (Sun)
Wind energy
Hydropower (flowing water)
Biomass (organic matter)
Geothermal energy

These sources are environmentally friendly and sustainable.

II. Application of Concepts

1. Question:

A man carrying a bag of total mass 25 kg climbs up to a height of 10 m in 50 seconds. Calculate the power delivered by him on the bag (AS).

Answer:
Given:
Mass (m) = 25 kg
Height (h) = 10 m
Time (t) = 50 s
g = 9.8 m/s²

Work done = mgh = 25 × 9.8 × 10 = 2450 J

Power = Work / Time = 2450 / 50 = 49 W

Final Answer: Power = 49 Watts

2. Question:

A 1 kg ball is dropped from a height of 10 m. Find (a) initial potential energy, (b) kinetic energy just before ground, (c) speed before reaching ground (g = 9.8 m/s²).

Answer:

(a) Potential Energy = mgh = 1 × 9.8 × 10 = 98 J

(b) By conservation of energy:
Kinetic Energy just before hitting ground = 98 J

3. Question:

Calculate the work done by a person in lifting a load of 20 kg from ground and placing it at height 6 m (g = 9.8 m/s²).

Answer:

Work done = mgh
= 20 × 9.8 × 6
= 1176 J

4. Question:

Find the mass of a body which has 5 J of kinetic energy while moving at speed 1 m/s (AS).

Answer:

KE = ½mv²
5 = ½ × m × (1)²
m = 10 kg

5. Question:

A cycle with rider weighs 100 kg. How much work is needed to set it in motion at 3 m/s (AS).

Answer:

Work needed = Kinetic Energy
= ½mv²
= ½ × 100 × (3)²
= 50 × 9
= 450 J

6. Question:

Which of the renewable sources of energy would you think suitable to produce your native place? Why? (AS)

Answer:
The suitable renewable source depends on local conditions:

Solar energy is ideal in sunny regions because it is abundant and easy to install.

Wind energy is suitable in coastal or hilly areas with strong winds.
Hydropower is best where rivers or dams are present.

For most places, solar energy is preferred because it is widely available, clean, and cost-effective.

III. Higher Order Thinking Questions

1. Question:

When you lift a box from the floor and put it on an almirah, the potential energy increases but no change in conservation of energy. Explain (AS).

Answer:
When lifting the box, you do work using your muscular energy. This energy is transferred to the box as gravitational potential energy.

So, energy is not created—it is converted from chemical energy (body) to potential energy (box). Hence, the law of conservation of energy is not violated.

2. Question:

When an apple falls from a tree, what happens to its gravitational potential energy just as it reaches the ground? (AS)

Answer:
As the apple falls, its potential energy gradually converts into kinetic energy.

Just before hitting the ground:

Potential energy ≈ 0
Kinetic energy = maximum

After hitting the ground, this energy is transformed into:

Heat energy
Sound energy
Deformation energy

IV. Multiple Choice Questions

1. Question:

SI unit of work
Answer: (a) N·m (Joule)

2. Question:

The energy possessed by a body by virtue of its motion is called
Answer: (b) Kinetic energy

3. Question:

A person climbing a ladder with suitcase on his head. The work done by that person that suitcase is
Answer: (a) Positive

4. Question:

If you have lifted a suitcase and kept it on a table, then the work done by you will depend on
Answer: (a) The path of the motion of the suitcase

Chapter 11: Heat

I. Reflections on Concepts

1. Why do we get dew on the surface of a cold soft drink bottle kept in open air?

When a cold soft drink bottle is kept in open air, the air around it contains water vapour (moisture). The surface of the bottle is much colder than the surrounding air. As the air comes in contact with this cold surface, it cools down. When air cools, it cannot hold as much water vapour as before.

As a result, the excess water vapour condenses into tiny water droplets on the surface of the bottle. This process is called condensation, and the droplets formed are similar to dew.

2. Water can evaporate at any temperature. Explain with an example.

Evaporation is a surface phenomenon where molecules of a liquid gain enough energy to escape into the vapour state. This process does not require the liquid to reach its boiling point.

At any temperature, some molecules in water have higher kinetic energy than others. These high-energy molecules escape from the surface, causing evaporation.

Example: Wet clothes dry even at room temperature (or even on a cold day). This shows that water evaporates without needing to reach 100°C.

3. What role does specific heat play in keeping a watermelon cool for a long time after removing it from a fridge on a hot day?

Specific heat is the amount of heat required to raise the temperature of a substance by a certain amount. Water has a very high specific heat capacity.

A watermelon contains a large amount of water. Because of this, it absorbs heat from the surroundings very slowly. This means its temperature increases gradually rather than quickly.

As a result, the watermelon stays cool for a longer time even when kept outside on a hot day.

4. Equal amounts of water are kept in a cap and in a dish. Which will evaporate faster? Why?

Water in the dish will evaporate faster.

This is because evaporation depends on the surface area exposed to air. A dish has a larger surface area compared to a cap. More surface area allows more water molecules to escape into the air at the same time, increasing the rate of evaporation.

Thus, greater surface area leads to faster evaporation.

5. Why is specific heat different for different substances? Explain.

Different substances have different atomic or molecular structures and bonding. The amount of energy required to increase the motion (temperature) of their particles varies.

In some substances, energy is used not only to increase motion but also to overcome intermolecular forces. Hence, more heat is required to raise the temperature.

Therefore, specific heat differs because each substance has unique internal structure and energy requirements.

II. Application of Concepts

1. Using the concept of evaporation, explain why dogs pant during hot summer days.

Dogs do not have sweat glands like humans, so they cannot cool their bodies by sweating. Instead, they pant by sticking out their tongue and breathing rapidly.

The saliva on their tongue evaporates quickly due to this rapid airflow. Evaporation requires heat energy, which is taken from the dog’s body.

This loss of heat cools down the dog’s body, helping it regulate temperature during hot weather.

2. If 50 g of water at 20°C and 50 g of water at 40°C are mixed, what is the final temperature of the mixture?

Since both quantities of water are equal and water has the same specific heat capacity, the final temperature is simply the average of the two temperatures.

\text{Final temperature} = \frac{20 + 40}{2} = 30^\circ C

So, the final temperature of the mixture is 30°C.

3. What do you observe in the surroundings in terms of cooling or heating when water vapour is getting condensed?

When water vapour condenses into liquid, it releases heat energy into the surroundings. This is because condensation is the reverse of evaporation and involves loss of energy by the vapour molecules.

As a result, the surroundings become warmer. This is why condensation is considered a heating process for the surroundings.

III. Higher Order Thinking Questions

1. Suppose that 1 L of water is heated for a certain time to raise its temperature by 2°C. If 2 L of water is heated for the same time, how much will its temperature rise?

The heat supplied is the same in both cases, but the mass of water is doubled in the second case.

Since:

Q = mc\Delta T

If mass (m) doubles and heat (Q) remains constant, the temperature rise (ΔT) becomes half.

So, the temperature rise will be:

\frac{2^\circ C}{2} = 1^\circ C

Thus, the temperature will rise by 1°C.

2. Answer the following:

(a) How much energy is transferred when 1 g of boiling water at 100°C condenses to water at 100°C?

When steam condenses to water at the same temperature, it releases latent heat of vaporization.

Latent heat of vaporization of water = 2260 J/g

So, energy released =

1 \times 2260 = 2260 \text{ J}

Answer: 2260 J (released)

(b) How much energy is transferred when 1 g of boiling water at 100°C cools to water at 0°C?

Energy released is calculated using:

Q = mc\Delta T

Where:
$c = 4.2 \, J/g^\circ C$ , $\Delta T = 100^\circ C$

Q = 1 \times 4.2 \times 100 = 420 \text{ J}

Answer: 420 J (released)

(c) How much energy is released or absorbed when 1 g of water at 0°C freezes to ice at 0°C?

This involves latent heat of fusion.

Latent heat of fusion of water = 334 J/g

Q = 1 \times 334 = 334 \text{ J}

Answer: 334 J (released)

(d) How much energy is released or absorbed when 1 g of steam at 100°C cools to ice at 0°C?

This involves three steps:

Condensation:
Steam → Water at 100°C
= 2260 J released
Cooling water from 100°C to 0°C:

Q = 1 \times 4.2 \times 100 = 420 \text{ J}

Freezing water to ice at 0°C:
= 334 J released

Total energy released:

2260 + 420 + 334 = 3014 \text{ J}

Answer: 3014 J (released)

Chapter 12: Sound

I. Reflections on Concepts

1. Explain the following terms:

(a) Amplitude

Amplitude refers to the maximum displacement of a vibrating particle from its mean (rest) position. In the case of sound waves, it represents how much the particles of the medium move back and forth.

Greater amplitude means the particles vibrate more strongly, producing a louder sound, while smaller amplitude produces a softer sound.

(b) Wavelength

Wavelength is the distance between two consecutive points in the same phase of a wave, such as two compressions or two rarefactions in a sound wave.

It is usually denoted by the symbol λ (lambda) and is measured in meters. It represents how “stretched out” the wave is in space.

(c) Frequency

Frequency is the number of complete vibrations or oscillations made by a particle in one second.

It is measured in Hertz (Hz). Frequency determines the pitch of the sound:

Higher frequency → higher pitch (shrill sound)
Lower frequency → lower pitch (deep sound)

2. Write the relation between wavelength, frequency and speed of sound.

The relationship between wavelength (λ), frequency (f), and speed of sound (v) is given by:

𝑣 = 𝑓 \times 𝜆

This equation shows that the speed of sound is the product of its frequency and wavelength.

3. Which has larger frequency – infrasonic sound or ultrasonic sound? Why?

Ultrasonic sound has a larger frequency than infrasonic sound.

Infrasonic sound: frequency less than 20 Hz
Ultrasonic sound: frequency greater than 20,000 Hz

Since ultrasonic waves have frequencies far above the audible range, they are much higher in frequency compared to infrasonic waves.

4. Why is soft furnishing avoided in concert halls?

Soft furnishings like curtains, carpets, and cushions absorb sound waves. If too many such materials are used in a concert hall, they reduce sound reflections excessively.

Proper reflection of sound is necessary so that sound reaches all parts of the hall clearly. Too much absorption makes the sound weak and dull.

Therefore, soft furnishing is minimized or carefully balanced to maintain good acoustics.

II. Application of Concepts

1. Does sound follow the same laws of reflection as light does?

Yes, sound follows the same laws of reflection as light. These are:

The angle of incidence is equal to the angle of reflection.
The incident wave, reflected wave, and the normal all lie in the same plane.

This is why sound can bounce off surfaces and produce echoes, just like light reflects from mirrors.

2. Two sources A and B vibrate with the same amplitude. They produce sounds of frequencies 1 kHz and 30 kHz respectively. Which of the two waves will have larger power?

Power of a sound wave mainly depends on its amplitude, not frequency.

Since both sources have the same amplitude, they transfer the same amount of energy per unit time.

Therefore, both waves have equal power, even though their frequencies are different.

3. With the help of a diagram describe how compression and rarefaction pulses are produced in air near a source of sound.

When a vibrating object (like a tuning fork) moves forward, it pushes the air particles in front of it closer together, creating a region of high pressure called compression.

When it moves backward, it creates a region where particles are spread apart, forming a low-pressure region called rarefaction.

These alternating compressions and rarefactions travel through the air as a longitudinal sound wave.

Simple Representation:


Compression   Rarefaction   Compression   Rarefaction
|||||||        |  |  |       |||||||        |  |  |
High pressure  Low pressure  High pressure Low pressure

4. How are multiple reflections of sound helpful to doctors and engineers?

Multiple reflections of sound have important applications:

Doctors:
Instruments like stethoscopes use multiple reflections to carry sound from inside the body to the doctor’s ears clearly.
Engineers:
Devices such as megaphones, horns, and soundboards use multiple reflections to direct sound efficiently in a specific direction.

This helps in amplifying and guiding sound for better clarity and reach.

III. Higher Order Thinking Questions

1. Explain the working and applications of SONAR.

SONAR (Sound Navigation and Ranging) is a technique that uses sound waves to detect objects underwater and measure distances.

Working:

A sound pulse is sent into the water.
It travels until it hits an object (like a submarine or seabed).
The sound reflects back as an echo.
The time taken for the echo to return is measured.

Using this time, the distance is calculated as:

Distance = \frac{𝑣 \times 𝑡}{2}

(where v is speed of sound in water and t is time taken)

Applications:

Measuring ocean depth
Detecting submarines and underwater objects
Navigation of ships
Studying marine life

2. How do echoes in a normal room affect the quality of the sounds that we hear?

In a normal room, sound reflects off walls, ceilings, and other surfaces. These reflections mix with the original sound.

If reflections arrive too quickly, they overlap with the original sound, causing reverberation. This can make sound unclear or distorted.

A small amount of reflection improves sound richness, but excessive echoes reduce clarity. That is why rooms are designed with proper materials to balance reflection and absorption for good sound quality

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Physical Science Content - IX class

I. Reflections on Concepts

Chapter 2: Motion

Chapter 5: Gravitation

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I. Reflections on Concepts

1. Why do we get dew on the surface of a cold soft drink bottle kept in open air?

2. Water can evaporate at any temperature. Explain with an example.

3. What role does specific heat play in keeping a watermelon cool for a long time after removing it from a fridge on a hot day?

4. Equal amounts of water are kept in a cap and in a dish. Which will evaporate faster? Why?

5. Why is specific heat different for different substances? Explain.

II. Application of Concepts

1. Using the concept of evaporation, explain why dogs pant during hot summer days.

2. If 50 g of water at 20°C and 50 g of water at 40°C are mixed, what is the final temperature of the mixture?

3. What do you observe in the surroundings in terms of cooling or heating when water vapour is getting condensed?

III. Higher Order Thinking Questions

1. Suppose that 1 L of water is heated for a certain time to raise its temperature by 2°C. If 2 L of water is heated for the same time, how much will its temperature rise?

2. Answer the following:

(a) How much energy is transferred when 1 g of boiling water at 100°C condenses to water at 100°C?