Class 9

AP State Syllabus AP Board 9th Class Physical Science Important Questions Chapter 3 Is Matter Pure?

AP State Syllabus 9th Class Physical Science Important Questions 3rd Lesson Is Matter Pure?

9th Class Physical Science 3rd Lesson Is Matter Pure? 1 Mark Important Questions and Answers

Question 1.
There is a mixture with sand and iron filings. Write an activity for the separation of iron filings from sand.
Answer:

1. Take mixture of sand and iron filing in a tub.
2. Take a magnet and move over the mixture.
3. Iron filings are separated by sticking to the magnet.

9th Class Physical Science 3rd Lesson Is Matter Pure? 2 Marks Important Questions and Answers

Question 1.
Frame two questions to understand “Homogeneous mixture”.
Answer:

• Can you see components in homogenous mixture?
• In which mixture components are uniformly distributed, heterogeneous or homogeneous?

Question 2.
How can you use milk to show Tyndall effect?
Answer:

1. Take milk in a glass.
2. Pass light through the milk.
3. Milk particles shine due to scattering of light.
4. This is the Tyndall effect.
5. Milk is a colloidal solution.

9th Class Physical Science 3rd Lesson Is Matter Pure? 4 Marks Important Questions and Answers

Question 1.
Draw the arrangement of apparatus of fractional distillation experiment. What is the advantage of using the fractionating column?
Answer:

1. Fractional distillation column contains glass beads.
2. These glass beads in fractional distillation column provide maximum possible surface area for the vapours to cool and condense repeatedly.

Question 2.
Draw and label the apparatus set up for the separation of a mixture by sublimation.
Answer:

Question 3.
Draw and label the apparatus set up for the separation of a mixture by evaporation.
Answer:

Question 4.
Explain how two miscible liquids can be separated if their boiling points are close to each other.
Answer:

To separate two or more miscible liquids when the difference in their boiling points is less than 25°C, fractional distillation process is used.

Aim :
To separate two miscible liquids (water and acetone) by fractional distillation.

Materials required :
Stand, distillation flask, thermometer, condenser, beaker, acetone and water, one holed rubber cork.

Procedure:

1. Take a mixture of acetone and water in a distillation flask.
2. Fix a thermometer as shown in the figure and clamp to the stand.
3. Attach condenser to the flask.
4. Keep a beaker to collect distillate at the end of condenser.
5. Heat the mixture slowly.
6. Observe the reading of thermometer.
7. The acetone (low boiling point) vapourises and condenses.
8. It can be collected from the condenser outlet.
9. Water remains in the distillation flask. ,

Question 5.
800 ml of solution contains 20 grams of solute. Calculate the concentration in terms of mass by volume percentage of the solution.
Answer:

Question 6.
a) Diagram shows one of the process of separation by mixtures. Based on the diagram answer the following questions.
i) Identify the process involved in the diagram.
Answer:
Sublimation.

ii) Is something missing in the diagram. If so what is that?
Answer:
Stove is missed for heating.

iii) If ‘B’ represent ammonium chloride, then what is ‘A’ represent?
Answer:
Mixture of ammonium chloride and salt.

iv) Give one more example for separation of mixture using above process?

Answer:
Camphor and salt.

Question 7.
Name the instrument used to separate immiscible liquids. Draw a neat diagram of it taking kerosene and water as immiscible liquids.
Answer:
The instrument used to separate immiscible liquids is separating funnel.

9th Class Physical Science 3rd Lesson Is Matter Pure? Important Questions and Answers

9th Class Physical Science 3rd Lesson Is Matter Pure? 1 Mark Important Questions and Answers

Question 1.
What is meant by ‘Pure substance’?
Answer:
In our day to day language, ‘pure’ means something with no adulteration. A substance is said to be pure i.e., homogeneous when the composition doesn’t change, no matter which part of the substance you take for examination.

Question 2.
What is a mixture?
Answer:
A mixture is generally made of two or more components that are not chemically combined.

Question 3.
What is a homogeneous mixture? Give examples.
In a homogeneous mixture the components of the mixture of uniformly distributed throughout it.
Ex : Lemonade, sugar solution, air, etc.

Question 4.
What is a heterogeneous mixture? Give examples.
Answer:
A heterogeneous mixture is a mixture made up of different substances or the same substance in different states which are not uniformly distributed in it.
Ex : Mixture of oil and water; water and naphthalene, etc.

Question 5.
What are the factors affecting rate of dissolving?
Answer:
The factors affecting rate of dissolving are
i) Temperature of solvent,
ii) Size of solute particles,
iii) Stirring of the solution.

Question 6.
When do you say that a solution is dilute solution?
Answer:
If the amount of solute present is little, the solution is said to be dilute.

Question 7.
What do you say that a solution is a concentrated solution?
Answer:
If the amount of solute present is more in a solution, then the solution is said to be a concentrated solution.

Question 8.
Define suspension.
Suspensions are the heterogeneous mixtures of a solid and a liquid in which the solids do not dissolve, like mixtures of soil and water.
Ex : Mixture of sand and water.

Question 9.
Define emulsion and give an example.
Answer:
Emulsion is a mixture consisting of two liquids that do not mix and settle into layers when they left undisturbed.
Ex : Mixture of oil and water.

Question 10.
What is a colloid?
Answer:
Colloids are heterogeneous mixtures in which the particle size is too small to be seen with the naked eye, but big enough to scatter light.
Ex : Milk, Cheese, Ghee, etc.

Question 11.
What is Tyndall effect?
Answer:
Scattering of a beam of light is called Tyndall effect.

Question 12.
What technique do you use to separate the colours?
Answer:
The laboratory technique called chromatography is used for the separation of mixtures into its individual components like inks and dyes.

Question 13.
When do we use fractional distillation method for the separation of miscible liquids?
Answer:
To separate two or more miscible liquids when the difference in their boiling points is less than 25°C, fractional distillation process is used.

Question 14.
What is the definition given by Lavoisier for the ‘element’?
Answer:
According to Lavoisier, an element is a form of matter that cannot be broken down by chemical reactions into simpler substances.

Question 15.
What is the principle involved in separation of immiscible liquids using separation funnel?
Answer:
The underlying principle involved in separation of immiscible liquids using separating funnel is that the immiscible liquids separate out into layers depending on their densities.

Question 16.
What is the use of glass beads in the fractional distillation column?
Answer:
The glass beads in fractional distillation column provide maximum possible surface area for the vapours to cool and condense repeatedly.

Question 17.
How will you separate a mixture containing kerosene and petrol (difference in their boiling points is more then 25°C). Which are miscible with each other?
Answer:
Kerosene and petrol are separated by using distillation process because the difference in boiling point is more than 25°C.

Question 18.
What type of mixtures are separated by the techniques of crystallisation?
Answer:
Salt from water, sugar from water and sodium chloride from its solution in water, etc. are the mixtures to be separated by the techniques of crystallisation.

9th Class Physical Science 3rd Lesson Is Matter Pure? 2 Marks Important Questions and Answers

Question 1.
What is a centrifuge? What are its uses?
Answer:
Centrifuge is a machine used to separate the particles of higher mass and lower mass from a mixture.
Uses :

1. To separate cream from milk.
2. In diagnostic laboratories, to test blood and urine samples.
3. Dryer in a washing machine.

Question 2.
Define a mixture and mention its properties.
Answer:
Mixture :
A mixture is generally made of two or more components that are not chemically combined.

Properties :

1. The substances in a mixture retain their own properties.
2. The substances in a mixture can be physically separated.

Question 3.
What are homogeneous and heterogeneous mixtures? Give examples.
Answer:
Homogeneous mixture:
A homogeneous mixture is a mixture in which the components of the mixture are uniformly distributed throughout it.
Ex :
Lemonade, sugar solution, etc.

Heterogeneous mixture :
A heterogeneous mixture is a mixture made up of different substances or the same substance in different states which are not uniformly distributed in it.
Ex :
Oil and water; Naphthalene and water, etc.

Question 4.
Define solution, solvent and solute.
Answer:
Solution :
The homogeneous mixture of two or more substances is that we cannot separate them by the process of filtration called a solution.

Solvent:
The component of the solution that dissolve the other component in it (usually the component present in larger quantity) is called solvent.

Solute :
The component of the solution that is dissolved in the solvent (usually the component present in lesser quantity) is called solute.

Question 5.
Mention the properties of a solution.
Answer:

1.  The particles of a solution cannot be seen with our naked eye.
2. They do not show Tyndall effect.
3. The solute particles do not settle down when left undisturbed.
4. Solution is a homogeneous mixture.

Question 6.
What are the disperse phase and dispersion medium of a colloidal solution?
Answer:
Disperse phase :
Disperse phase is the substance that present in small proportions and consists of particles of colloidal size (lnm to 100 nm).

Dispersion medium :
Dispersion medium is the medium in which the colloidal particles are dispersed.

Question 7.
Define miscible and immiscible liquids.
Answer:
Miscible liquid :
A liquid is said to be miscible if it dissolve completely in another liquid.
Ex : Alcohol is miscible in water.

Immiscible liquid :
An immiscible liquid is one which doesn’t dissolve but forms a layer over another liquid and can be separated easily.
Ex : Oil is immiscible in water.

Question 8.
Define element and compound. Give examples.
Answer:
Element:
Element can be defined as a basic form of matter that cannot be broken down into simpler substances by chemical reactions.
Ex : lron, gold, silver, sodium, magnesium, etc.

Compound :
Compound can be defined as pure substance that can be separated into two or more components by means of chemical reactions.
Ex : Copper sulphate, ammonium chloride, etc.

Question 9.
How do you appreciate the efforts of scientists in discovering elements?
Answer:

• Generally elements are available in nature in the form of their ores.
• Study of properties of elements lead to the development of civilization.
• The efforts of early alchemists-up to and including Newton, Hennig Brand, Sir Humphry Davy are appreciable for their works in unearthing new elements.
• The definition of element and compound given by Robert Boyle and Lavoisier lead to study the properties of elements and compounds.

Question 10.
Name the technique to separate,
i) butter from curd
ii) salt from sea-water
iii) camphor from salt
Answer:
i) Centrifugation method is used to separate butter from curd .
ii) Crystallisation is used to separate salt from water.
iii) Sublimation technique is used to separate camphor from salt.

Question 11.
What are the factors on which solubility depends on?
Answer:
The factor influence solubility are

1. nature of solute
2. nature of solvent
3. temperature.

Question 12.
What is a concentrated and dilute solution?
Answer:
If the amount of solute is less in a solution, then it is called dilute solution.

If the amount of solute is more in a solution, then the solution is called concentrated solution.

Question 13.
How do you separate following mixture?
a) Iodine from sodium chloride
b) Petrol from water
c) Butter from milk
d) Sugar from water
Answer:
a) Sublimation
b) Separating funnel
c) Centrifugation
d) Evaporation

Question 14.
Identify pure substance and mixture from this figure. Why?

Answer:
a) Fig ‘a’ is a pure substance. Because particles are evenly distributed.
b) Fig ‘b’ is a mixture. Because particles are unevenly distributed.

9th Class Physical Science 3rd Lesson Is Matter Pure? 4 Marks Important Questions and Answers

Question 1.
Define the terms :
a) Solubility
b) Saturated solution
c) Unsaturated solution
d) Concentration
Answer:
a) Solubility :
The amount of solute present in a saturated solution at a certain tem-perature is called its solubility.

b) Saturated solution :
When no more solute can be dissolved in the solution at a certain temperature, it is said to be a saturated solution.

c) Unsaturated solution :
If the amount of solute present in the solution is less than that in the saturated solution is called an unsaturated solution.

d) Concentration of a solution:
The concentration of a solution can be defined as the amount (mass) of solute present in a given amount (mass) of solution or the amount (mass) of solute dissolved in a given volume of the solution.

Question 2.
Compare the properties of suspensions and colloids.
Answer:

Question 3.
Explain the process of preparation of ice cream.
Answer:

1. Ice cream is made by churning a mixture of milk, sugar and flavours.
2. This mixture is slowly chilled to form ice cream.
3. The churning process disperses air bubbles into the mixture by foaming and break up the large ice crystals into tiny particles.
4. The result is a complex substance which contains solids, liquids and gases.
5. This is the ice cream.
6. Ice cream is a colloid.

Question 4.
What is chromatography? What are its uses?
Answer:
Chromatography :
Chromatography is a laboratory technique for the separation of mixtures into its individual components.
Uses:

1. Used to separate components of ink and dyes.
2. Used to separate the coloured pigments in plants.
3. Used to determine the chemical composition of many substances.
4. Used in crime scene investigations.
5. Used in hospitals to detect alcohol levels in a patient’s blood stream.
6. Used by environmental agencies to detect the level of pollutants in water supplies.
7. Used by pharmacists to determine the amount of each chemical found in each product.

Question 5.
Explain the process of separation of components of air briefly.
Answer:

1. If we want oxygen gas from air, we have to separate out all the other gases present in air.
2. The air is compressed by increasing the pressure and then cooled by decreasing the temperature to get liquid air.
3. This liquid air is allowed to warm up slowly in a fractional distillation column where gases get separated at different heights depending upon their boiling points.

Question 6.
Distinguish between mixtures and compounds.
Answer:

Question 7.
Draw a flow chart shows the process of obtaining gases from air.
Answer:

1. We have learnt that air is a homogeneous mixture.
2. It can be separated into its components.
3. The following are the steps involved in separating the components of air.

Question 8.
Draw a flow chart to understand the chemical and physical nature of the matter.
Answer:

Question 9.
Give some daily life experiences where you can observe “Tyndall effect”.
(OR)
What is Tyndall effect? Write any two applications of Tyndall effect.
Answer:
Tyndall effect:
Scattering a beam of light is called the ‘Tyndall effect’.

Daily life experiences:

1. Select a room where the sun light falls directly through a window.
   Close the window in such a way that a slit is left open between the windows.
   We observe a beam of light passing through the slit.
2. While walking on a road having a lot of trees on both sides, when the sun light passes through branches and leaves, we can see the path of dust particles.
3. In cinema halls we can observe the beam of light scatters from projector to the screen.
4. When smoke in the kitchen is exposed to sun light, we can observe the path of smoke particles.

Question 10.
How do you separate the mixture of napthaleine powder and salt powder?
Answer:
Aim :
To separate components in the mixture of napthaleine and salt.

Apparatus:
China dish, cotton plug, funnel, burner.

Method :
Separation mixtures by sublimation.

Procedure :

1. Take 4 table spoons of mixture in a China dish.
2. Take the glass funnel and plug the mouth of the funnel with cotton plug.
3. Invert the funnel over the China dish.
4. Keep the dish on the stand of stove and heat for some time.

Observation :

1. Vapours of napthaleine solidified on the walls of the funnel.
2. Salt remains in the China dish.

AP State Syllabus AP Board 9th Class Physical Science Important Questions Chapter 7 Reflection of Light at Curved Surfaces

AP State Syllabus 9th Class Physical Science Important Questions 7th Lesson Reflection of Light at Curved Surfaces

9th Class Physical Science 7th Lesson Reflection of Light at Curved Surfaces 1 Mark Important Questions and Answers

Question 1.
Which mirror is used as rear-view mirror in the vehicles?
Answer:
Convex mirror is used as rear view mirror in the vehicles.

Question 2.
What is the relation between focal length (f) and radius of curvature (R)?
Answer:
The radius of curvature of a spherical mirror is twice to its focal length.
⇒ R = 2f (or) f = \(\frac{R}{2}\).

Question 3.
Can a virtual image be photographed by a camera?
Answer:
Yes, virtual image can be photographed by a camera.

Question 4.
Complete the diagram and draw the image.

Answer:

Question 5.
Predict and write the reason, why the value of the distance of the object (u) is always negative in the mirror equation.
Answer:
i) Direction of the incident rays is taken as positive (+ve).
ii) Object distance is measured from the pole to the object in the opposite direction of incident rays.

Question 6.
Which property of concave mirror is used in making the solar cooker?
Answer:
Rays coming parallel to the principal axis of a concave mirror is focused at focal point. Based on this property solar cooker is made.

Question 7.
Draw the ray diagram to show the formation of image for the object of height 1 cm. placed at 5 cm. distance, in front of a convex mirror having the radius of curvature R = 5 cm.
Answer:

Question 8.
What is reflection?
Answer:
The light rays falling on a surface are returned into the original medium. This phenomenon is called reflection.

Question 9.
What is the relation between focal length and radius of curvature?
Answer:
Radius of curvature = 2 x focal length
∴ R = 2f (or) f = \(\frac{R}{2}\)

Question 10.
What is the mirror formula for spherical mirrors?
Answer:
The mirror formula is \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}\)
f = focal length of mirror ; u = object distance ; v = image distance

Question 11.
What is a real image? What is a virtual image?
Answer:
Real image :
The image formed due to convergence of light rays. The real image can be caught on the screen.

Virtual image :
The image that we get by extending the rays backwards is called a virtual image. A virtual image cannot be caught on the screen.

Question 12.
What is focal length?
Answer:
The distance between focus and vertex.

Question 13.
What is radius of curvature?
Answer:
The distance between vertex and centre of curvature.

Question 14.
What is magnification?
Answer:
The ratio of size of image to size of object is called magnification.

(OR)

The ratio of image distance to object distance is called magnification.

Question 15.
Why are concave and convex mirrors called spherical mirrors?
Anwer:
The reflecting surface of convex and concave mirror is considered to form a part of the surface of a sphere. So they are called spherical mirrors.

Question 16.
What is a reflecting surface?
Answer:
The surface used for reflection is called reflecting surface.

Question 17.
What is principal axis?
Answer:
The horizontal line which passes through the centre of curvature is called principal axis.

Question 18.
What is meant by converging of light rays?
Answer:
If light rays after reflection meet at a point, then we say the light rays are converging.

Question 19.
When do you say light rays are diverging?
Answer:
If light rays appear as if they are coming from a point after reflection, then we say light rays are diverging.

Question 20.
When does a ray reflect in the same path from a concave mirror?
Answer:
When it passes through centre of curvature.

Question 21.
When a light ray travelling from parallel to principal axis falls on concave mirror, then what is the path of reflected ray?
Answer:
The reflected ray passes through focal point.

Question 22.
Where do you place the vessel in solar cooker?
Answer:
We place the vessel in solar cooker at the focal point.

Question 23.
Name a mirror that can give an erect and enlarged image of an object.
Answer:
Concave mirror can give an erect and enlarged image of an object.

Question 24.
Can a convex mirror burn a paper? If not, why?
Answer:
The rays coming parallel to principal axis after reflection diverge from the mirror. So we cannot burn a paper by using a convex mirror.

Question 25.
Which mirror has wider field of view?
Answer:
A convex mirror has wider field of view, that’s why they are used as rear view mirrors in vehicles.

Question 26.
Why does our image appear thin or bulged?
Answer:
Due to converging or diverging of light rays from the mirror.

Question 27.
Why is angle of incidence equal to angle of reflection when a light ray reflects from a surface?
Answer:
Because light selects the path that takes least time to cover a distance.

Question 28.
Are angle of reflection and angle of incidence also equal for curved surface?
Answer:
Yes, it is equal for curved surfaces like spherical mirrors.

Question 29.
What is a spherical mirror? Give different types of spherical mirrors.
Answer:
If the reflecting surface of mirror is considered to form a part of the surface of sphere, then it is called spherical mirror. Spherical mirrors are of two types :

1. Concave mirror
2. Convex mirror

Question 30.
Write about various distances related to mirrors.
Answer:
The various distances related to mirrors are
1) Focal length (f) :
The distance between vertex and focus is called focal length.

2) Radius of curvature (R) :
The distance between vertex and centre of curvature is called radius of curvature.

Question 31.
We wish to obtain an erect image of an object using a concave mirror of focal length of 15 cm. What should be range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object?
Answer:
The range of distance of object is between 0 and 15 cm.
The image is virtual and erect.
The image is larger than the object.

Question 32.
Name some apparatus which can work on the principle of reflection of light.
Answer:
Plane mirror, spherical mirrors, periscope, kaleidoscope.

Question 33.
If you want to get parallel beam by using concave mirror, then where do you keep the source?
Answer:
The object should be kept at focus because the light rays coming from focus after reflection from mirror travel parallel to principal axis.

Question 34.
If you want to form the image of an object at infinity, then where do you keep the object?
Answer:
The object should be kept at focus; then the image would be formed at infinity.

Question 35.
How do you get a virtual image with a concave mirror?
Answer:
When we place the object between vertex and focus then we will get a virtual image.

Question 36.
Why do dentists use concave mirror?
Answer:
If the object is between mirror and its focus we get enlarged virtual and straight image by using concave mirror. So dentists use this principle to see inner parts of mouth.

Question 37.
A concave mirror produces three times magnified real image of an object placed at 10 cm in front of it. Where is the image located?
Answer:
> 10 cm.

Question 38.
What is your opinion on elevating buildings with mirrors?
Answer:
The mirrors used in elevating buildings are reinforced, tough and laminated glasses. These mirrors provide safety and make the buildings attractive.

Question 39.
Identify the mirror having focal length +15 cm.
Answer:
Convex mirror (since the focal length of convex mirror is taken as positive).

Question 40.
If the focal length of mirror is 10 cm, what is that mirror?
Answer:
The mirror is concave (since the focal length of concave mirror is taken as negative).

Question 41.
Can we focus a sunlight at a point using a mirror instead of magnifying glass?
Answer:
Yes, by using concave mirror we can focus sunlight at a point.

Question 42.
To reduce glaze of surroundings the windows of some department stores, rather than being vertical, slant inward at the bottom. How does this reduce glaze?
Answer:
This slant reflects the sunlight further down towards the ground, then it would happen as if they are vertical.

Question 43.
Why do we prefer a convex mirror as a rear-view mirror in the vehicles?
Answer:
It gives erect and small image and covers large distance.

Question 44.
An object is placed at a distance 8 cm from a concave mirror of radius of curvature 16 cm. What are the characteristics of image?
Answer:
The image is real, inverted, and same size.

Question 45.
What happens when light falls on an opaque object?
Answer:
Some part of light is reflected back and remaining part is absorbed.

Question 46.
What happens when light is reflected from transparent object?
Answer:
Some part of light is reflected and remaining part is partly transmitted or partly absorbed.

Question 47.
Which objects at your home act as spherical mirrors?
Answer:
Objects at home that act as spherical mirrors are :

1. Spoons
2. Spectacles
3. Sink
4. Cooking vessel

Question 48.
Complete the following ray diagram.

Answer:

The light ray passing through centre of curvature falls normal to the concave mirror. So it retraces the same path.

Question 49.
If focal length is 20 cm, then what is radius of curvature of mirror?
Answer:
f = 20 cm
R = 2f = 2 × 20 = 40 cm.

Question 50.
The radius of curvature of a spherical mirror is 20 cm. What is the focal length?
Answer:
Radius of curvature (R) = 20 cm
R 20
Focal length (f) = \(\frac{\mathrm{R}}{2}=\frac{20}{2}\) = 10 cm.

Question 51.
The focal length of convex mirror is 16 cm. What is its radius of curvature?
Answer:
f = 16 cm
R = 2f = 2 × 16 = 32 cm

Question 52.
Write any two uses of concave mirror in our daily life.
Answer:
Uses of concave mirror :

1. Concave mirrors are used by dentists to see enlarged image of tooth.
2. Concave mirrors are used in car head lights.

Question 53.
Write any two uses of convex mirror in our daily life.
Answer:
Uses of convex mirror :

1. Convex mirrors are used as rear view mirrors in vehicles because convex mirrors increase field of view.
2. Convex mirrors are used in street light reflectors as they spread light over greater

Question 54.
Suggest a new use with a spherical mirror.
Answer:
Spherical mirrors are newly adapted in ATMs.

Question 55.
Focal length of a concave mirror is x. Find the sum of focal length and radius of curvature.
Answer:
Focal length = x; Radius of curvature = 2 x focal length = 2x.
The sum of focal length and radius of curvature = x + 2x = 3x.

Question 56.
If the angle between the mirror and incident ray is 40°, then find the angle of reflection.
Answer:
Given that angle between incident ray and mirror = 40°.
Suppose angle of incidence = x.
∴ 40 + x = 90
x = 90 – 40 = 50°.

But we know angle of incidence = angle of reflection
∴ Angle of reflection = 50°.

9th Class Physical Science 7th Lesson Reflection of Light at Curved Surfaces 2 Marks Important Questions and Answers

Question 1.
Your friend has a doubt that whether a concave mirror or a convex mirror is used as a rear view mirror in the vehicles. What questions will you ask to clarify his doubts?
Answer:

• Is the image in a rear-view mirror smaller or larger when compared to real object?
• Which mirror forms smaller image than the object in the given mirrors?

Question 2.
The focal length of a huge concave mirror is 120 cm. A man is standing in front of it at a distance of 40 cm. What are the characteristics of his image in that mirror?
Answer:
i) Image form in the mirror
ii) Virtual image
iii)Erected image
iv) Enlarged image

Question 3.
How can you find out the focal length of concave mirror experimentally when there is no sunlight?
Answer:
Place the object / candle in front of the mirror and adjust the screen to get image on it. Measure the object distance, image distance. Substitute the values (as per sign connection) in mirror formula \(\left(\frac{1}{f}\right)=\left(\frac{1}{u}\right)+\left(\frac{1}{v}\right)\). We get the focal length of mirror.
(OR)
Place the object / candle and the screen at same point in front of the mirror. Adjust this set of material to get sharp image on the screen.

Measure the distance from mirror to object/screen. This distance is the radius of curvature and make it half, it gives focal length of the mirror.

Question 4.
The magnification of the image by the concave mirror is – 1. Mention the four char-acteristics of image from the above information.
Answer:

1. Image will be formed at the centre of curvature (C).
2. Image size is equal to that of the object size. ‘
3. Inverted image.
4. Real image.

Question 5.
Write about different points related to mirrors.
Answer:
The different points related to mirrors are
1) Vertex (P) :
The point where the central axis touches the mirror is called vertex.

2) Focus or focal point (F):
The light rays coming from distinct object appear to meet at point in case of concave mirror and tend to meet at point when drawn backward in case of convex mirror. That point is known as focus or focal point.

3) Centre of curvature (C) :
It is centre of the sphere to which the mirror belongs.

Question 6.
What happens if light rays parallel to principal axis fall on the concave mirror, and draw ray diagram?
Answer:

The light rays that are parallel to the principal axis get reflected such that they pass through the focal point of the mirror. R₁ is such ray in figure.

Question 7.
What happens to a ray which passes through focal point and falls on the concave mirror, and also draw the ray diagram?
Answer:

The light ray which goes through the focal point of the mirror travels parallel to principal axis. R₂ is such ray in figure.

Question 8.
How does an image form due to convex mirror? Draw the ray diagram.
Answer:

• The parallel rays coming from distance object tend to diverge after reflection.
• If we extend the reflected rays backwards they meet at ‘F’, i.e. focal point of the convex mirror.

Question 9.
Which light ray after reflection will travel along the same path in opposite direction? What can be such a ray for a spherical mirror? Draw the ray diagram.
Answer:

1. Any ray that is normal to the surface, on reflection, will travel along the same path but in opposite direction.
2. The line drawn from the centre of curvature of mirror is perpendicular to the tangent at the point, the line meets the curve.
3. So if we draw a ray starting from the tip of the object going through the centre of curvature to meet the mirror, it will get reflected along the same line. This ray is shown as R₃ in the figure.

Question 10.
What happens if an object is placed at centre of curvature of a mirror? Draw the ray diagram.
Answer:

From the ray diagram we conclude that the image of the object will be formed at the same distance as the object and it will be inverted and of the same size. The image is real because it forms on a screen.

Question 11.
Draw the ray diagrams with convex mirror and write rules of ray diagram of convex mirror.
Answer:
Rule -1 :

A ray running parallel to main axis, on meeting the convex mirror will get reflected so as to appear as if it is coming from the focal point.

Rule – 2 :

This is converse of rule 1. A ray going in the direction of focal point after reflection will become parallel to main axis.

Rule – 3 :

A ray going in the direction of the centre of curvature will get reflected back in opposite direction, and looks like that is coming from the centre of curvature.

Question 12.
Why do we use parabolic mirror instead of concave mirror?
Answer:

1. We use parabolic mirror instead of concave mirror because with the concave mirror all the rays coming parallel in it may not be focused at focal point (F).
2. Those rays which are very nearer to principal axis will only be focused at focal point.
3. It is very effective to make the mirror parabolic in order to make all the rays to converge at focus.

Question 13.
The magnification of mirror is given as – 3. What is the inference do you get from this information?
Answer:
Magnification – ve indicates it is an inverted image. So it is a real image.
Magnification 3 indicates the image size is three times the object size. So the image is enlarged. Since it is forming real image the mirror is concave.

Question 14.
Why are we able to see various objects around us?
Answer:
We are able to see various objects around us due to the diffused light reflected from these objects reaches to our eye which gives sense of vision to those objects.

Question 15.
Which type of mirror is used as a reflector in street lamp?
Answer:
The reflectors of the street lamp are made in convex in shape so that reflected rays diverge over the larger area on ground. Therefore convex mirror acts as a reflector in street lamp.

Question 16.
Which type of mirror is used in doctor’s head lamp?
Answer:
Doctors use head lamp to examine nose, throat, teeth, etc. of patients. In this lamp a parallel beam of light is allowed to fall on the concave mirror. The reflected light concentrates on focus on the mirror on a smaller area to be examined. So the concave mirror is used in doctor head lamp.

Question 17.
Would you able to burn a paper using concave mirror?
Answer:

1. Concave mirror focuses the parallel sun rays at focal point of the mirror.
2. So with a small concave mirror we can heat up and burn a paper.

Question 18.
How do you find the focal length of concave mirror?
Answer:

1. Hold a concave mirror such that sunlight falls on it.
2. Take a small paper and slowly move it in front of the mirror until we will get smallest and brightest spot of the Sun.
3. Find the distance between mirror and image of the Sun that will give the focal length of mirror.

Question 19.
See the table and identify the mirrors in each case.

Answer:
Magnification negative indicates that it is inverted image and also real. Magnification -1 means same size. So the mirror which gives real image of same size is concave mirror. So ‘X’ is concave.
The magnification +1 means the image is virtual, erect and same size. So the mirror Y is plane.
The magnification + 0.5 means the image is virtual, erect and diminished. So the mirror Z is convex.

Question 20.
An object is placed at various positions in front of concave mirror of focal length 10 cm. Complete the table by using given information without actually doing the problem.

Answer:

Question 21.
Draw a normal at any point of a concave mirror.
Answer:

Question 22.
Identify the following in a ray diagram showing the reflection of light in a concave mirror.
a) Pole of the mirror
b) Principal axis
c) Centre of curvature
d) Focal point
e) Focal length
Answer:

Question 23.
Complete the ray diagram.

Answer:
First we have to draw normal at point of contact to the concave mirror and then we have to use laws of reflection to draw the reflecting ray.

Question 24.
Figure shows two parallel light rays falling on a convex mirror. Complete the ray diagram.

Answer:

Question 25.
See the belog figure and complete the ray diagram.

Answer:

Question 26.
Assume that an object is kept at a distance of 20 cm in front of a concave mirror. If its focal length is 30 cm, then
a) what is the image distance?
b) what the magnification of mirror in this case?
Answer:
Object distance = u = 20 cm

Question 27.
There is an object in front of convex mirror at a distance of 5 cm. If its focal length is 10 cm, then
a) what is the image distance?
b) what is its magnification?
Answer:

9th Class Physical Science 7th Lesson Reflection of Light at Curved Surfaces 4 Marks Important Questions and Answers

Question 1.
Sudheer wants to find focal length of a concave mirror experimentally.
a) What apparatus does he need?
b) Is the screen required or not? Explain.
c) Draw the table required to tabulate the values found in his experiment.
d) What is the formula used by him to find focal length?
Answer:
a) Apparatus required to Sudheer are

1. Concave mirror,
2. White paper or screen,
3. Scale,
4. V – stand,
5. Candle.

b) Yes, screen is required.
To catch and measure the image distance screen is required.

c) Table for observation and calculation of ‘f’.

d) Focal length \(\Rightarrow \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}} \text { (or) } \mathrm{f}=\frac{\mathrm{uv}}{\mathrm{u}+\mathrm{v}}\)
This is the formula used by him to find a focal length.

Question 2.
Show the formation of image with a ray diagram when an object is placed on the principal axis of a Concave mirror between focus and centre of curvature of the mirror.
Answer:

Question 3.
An object of 6 cm height is placed at a distance of 30 cm in front of a concave mirror of focal length 10 cm. At what distance from the mirror, will the image be formed? What are the characteristics of the image?
Answer:

Question 4.
List the materials required for conducting an experiment to find the focal length of a concave mirror. Explain the experimental process also.
Answer:
a) Material required for conducting an experiment to find the focal length of a concave mirror are

1. concave mirror
   a piece of paper
2. meter scale.

b) Procedure of the experiment:

1. Hold a concave mirror such that sunlight falls on it.
2. Take a small paper and slowly move it in front of the mirror and find out the point where we get the smallest and brightest spot, which will be the image of the sun.
3. Measure the distance of this spot from the pole of the mirror.
4. This distance is the focal length (f) of the mirror.

Another Experiment:
a) Material required :
A candle, paper / screen, concave mirror, V-stand, measuring tape or meter scale.

b) Procedure :
1) Place the concave mirror on V-stand, a candle and meter scale as shown in the figure.

2) Keep the candle at different length from the mirror (10 cm to 80 cm) along the axis and by moving the paper find the position where the sharp image is got on the paper.

3) Measure the image distance (o) and note in the given table.

4) Find the average of focal lengths (f) obtained in the experiment.

5) The average T is the focal length of the given mirror.

Question 5.
An object of height 5 cm is placed at 30 cm distance on the principal axis in front of a concave mirror of focal length 20 cm. Find the image distance and size of the image.
Answer:
Object distance (u) = – 30 cm, Focal length (f) = – 20 cm, Height of object (h_o) = 5 cm, Image distance (v) = ?, Height of image (h_i) = ?

∴ The image is real, inverted with a height of 10 cm.

Question 6.
A student conducted an experiment to observe characteristics of images formed by spherical mirrors and recorded his observations as follows. Observe the table and answer the questions.

i) Above said information belongs to which spherical mirror?
ii) In which situation, magnification is less than 1.
iii) An object of height 8 cm placed at centre of curvature on principal axis, then where do you get the image and what is its height?
iv) “All real images are inverted”. Justify the statement by using above table.
Answer:
i) It is a Concave mirror.
ii) When object is kept beyond ‘C’ then magnification is less than 1.
iii) Image formed at ‘C’. The height of the image is 8 cm.
iv) According to the table if the image is erected image, it is a real image. In all the other cases every real image is virtual image.

Question 7.
In the following cases calculate the magnification values for a concave mirror. Give reason.
a) When the object is at the focal point of the mirror.
b) When the object is between focal point and the pole.
Answer:
In the case of concave mirror
a) When the object is at the focal point of the mirror, then its magnification value is -1.

Reason :
In this case size of the image is large, compared with the object. It is called virtual image. Image is formed behind the mirror so magnification has negative sign.
Nature of the object: It is real, inverted, enlarged and forms at infinity.

b) When the object is between focal point (F) and the pole (P) of the mirror, then its magnification value is +1.

Reason :
In this case image is formed on the same side of the object and it is also virtual image so, the sign of the magnification is positive.

Nature of the object:
It is virtual, erect, enlarged and on the same side of the object.

Question 8.
Write the derivation of mirror formula.
(OR)
Derive \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\).
(OR)
A student wants to find the image distance for a given object distance of a mirror. Then derive a formula for the mirror.
Answer:
Derivation of mirror formula :
In the figure P = pole, C = centre of curvature and F= focus of the concave miror. Object AB is placed beyond C. Image AB’ is formed in between F and C.
From the diagram triangles A’B’C and ABC are similar triangles.
\(\frac{\mathrm{AB}}{\mathrm{A}^{\prime} \mathrm{B}^{\prime}}=\frac{\mathrm{BC}}{\mathrm{B}^{\prime} \mathrm{C}}\) ………………… (1)

Question 9.
Where is the base of the candle going to be in the image when the object is placed on the axis of the mirror beyond ‘C’?
Answer:

1. Any ray starting from a point on the axis and travelling along the axis will reflect on the axis itself.
2. So the base of the image is going to be the axis.
3. If the object is placed vertically on the axis, the image is going to be vertical.
4. Draw perpendicular from point A to axis.
5. The intersection point is the point where the base of the image of the candle is going to be formed

Question 10.
What happens if an object placed at a distance less than the focal length of the concave mirror? Draw the ray diagram.
(OR)
When do you get a virtual image by using a concave mirror and draw the ray diagram?
Answer:

1. The candle object (0) is placed at a distance less than the focal length of the mirror.
2. The first ray (R₁) will start from tip of the object and run parallel to axis to get reflected so as to pass through focal length.
3. The second ray (R₂) is the ray starting from the tip of the object and going through the focal point but it is not possible as such a ray will not meet the mirror.
4. The third ray (R₃), starting from the tip of the object goes to the centre of curvature but that also seem not to be possible.
5. Now consider a ray (R₄) that starts from the tip and goes in such a direction that it would go through the centre of curvature if extended backwards.
6. This ray is normal to surface and so will be reflected along the same line in opposite direction and will go through centre of curvature.
7. The two reflected rays diverge and will not meet.
8. When we extend these rays backward they appear to be coming from one point.
9. As seen from the figure (2) the image will be erect and enlarged and virtual.

Question 11.
A person in a dark room looking through a window can clearly see a person outside in the daylight, whereas the person outside cannot see the person inside. Why?
Answer:

• There is usually some reflection that occurs at an interface between the two materials but most often of light passing through.
• Imagine you are inside in the dark. A person outside in bright sunlight is sending out (reflection) lots of light, most of which would come through the window to you, so you see them clearly.
• Since it is so bright outside, there is also a good amount of light which reflects back towards them.
• This can distract them from little bit of light from you that is going towards them, so they have much harder time seeing you.

Question 12.
What is magnification? Derive an expression for magnification.
Answer:
Magnification :
The ratio of height of image to height of object is called magnification.

Question 13.
What are the rules to be followed while drawing ray diagrams?
Answer:
Various rules to be followed for drawing ray diagrams.
1) A ray parallel to the principal axis, after reflection will pass through the principal focus in case of a concave mirror or appear to diverge from the principal focus in case of convex mirror.

2) A ray passing through the principal focus of a concave mirror or a ray which is directed towards the principal focus of convex mirror, after reflection will emerge parallel to the principal axis.

3) A ray passing through the centre of curvature of a concave mirror or directed in the direction of centre of curvature of a convex mirror, after reflection, is reflected back along the same path.

4) A ray incident obliquely to the principal axis, towards a point P on the concave mirror or a convex mirror, is reflected obliquely. The incident ray and reflected rays follow laws of reflection.

Question 14.
How do you make a parallel beam with an experiment?
Answer:
Aim :
Making a beam of parallel lines.

Material used :
Two pins, thermocol block, candle.

Procedure:

1. Stick two pins on a thermocol block.
2. The pins are exactly parallel to each other.
3. As we can see in the figure, when a source of light is kept very near, we see the shadows diverging (from the base of the pins).
4. As we move the source away from the pins, the divergent angle starts reducing.
5. If we move the source far away, we will get parallel shadows. Thus we get a beam of parallel lines.

Question 15.
Write a table which shows the image formed by a concave mirror for different positions and also give size and nature of image.
Answer:

Question 16.
Ray diagrams of concave mirror.
Answer:
Object is placed ‘Infinitely’ :

Object is placed between ‘P’ and ‘F :

Question 17.
Complete the following ray diagrams and give reasons.

Answer:
a)

Reason :
The light appears to be passing through centre of curvature after reflection from convex mirror retraces the same path.

b)

Reason :
The light ray making certain angle of incidence ‘q’ with principal axis follows laws of reflection and reflects with same angle q.

c)

Reason :
The light ray which travels parallel to principal axis after reflection from convex mirror diverges from mirror and if we extended the ray backwards it passes through principal focus.

d)

Reason :
The light ray which travels parallel to principal axis after reflection from convex mirror diverges from mirror when the light ray drawn backwards passes through focus and second light ray follows laws of reflection (i.e., ∠i = ∠r).

These extended backward light rays meet and form a virtual, diminished and erect image between pole and focus inside the mirror.

Question 18.
Complete the following diagram to obtain image of object AB.

Answer:

1. The light ray which appears to coming from focus after reflection from convex mirror travels parallel to principal axis.
2. The light ray which is incident with certain angle ’x’ at pole ‘P’ reflects with same angle from convex mirror.
3. These two extended light rays meet at B. So AB’ is the image of the object AB.

Question 19.
An object 4 cm in size is placed at 25 cm in front of a concave mirror of focal length 15 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Find the nature and size of image.
Answer:
Given that f = – 15 cm ; u = – 25 cm ; h0 = 4 cm ; v = ?

So the image is enlarged and inverted.

Question 20.
Focal length of a concave mirror is f. The distance from its focal point to the object is P. Find the ratio of heights of image.
Answer:
Concave mirror is a part of spherical mirror.

Question 21.
When do we get a blurred image from a distant object by using concave mirror?
Answer:

1. The diagram shows a few rays starting from the tip of the flame.
2. The reflected rays intersect A. So the reflected image of tip of flame will be at intersection point A.
3. If we hold the paper at any point before or beyond point A (for example at point B), we see that the rays will meet the paper at different points,
4. So the image of the tip of the flame will be formed at different points due to these rays.
5. If we draw more rays emanating from the same tip we will see that point A they will meet but at point ‘B’ they won’t.
6. So the image of the tip of the flame will be sharp if we hold the paper at A and become blurred (due to mixing of multiple images) when we move the paper slightly in any direction (forward or backward).

Question 22.
An object of size 7 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed? Find size and nature of the image.
Answer:
Given, h₀ = 7 cm ; u = – 27 cm ; f = – 18 cm ; v = ?

Question 23.
An object 3 cm high is placed at a distance of 15 cm from a concave mirror, the radius curvature is 20 cm. Find the nature, position and size of the image. (V = -30 cm, m = -2, h₂ = -6 cm)
Answer:
h₀ = 3 cm; u = -15 cm; r = -20 cm; f = \(\frac{r}{2}\) = – 10 cm

Question 24.
Focal length of a concave mirror is 15 cm. An object of length 5 cm is placed in front of this mirror. Draw neat diagrams to find the length and position of image when object is at (I) 5 cm, 0i) 12 cm, (iif) 20 cm, (jv) 35 cm away from the mirror.
Answer:
i) An object is kept at a distance of 5 cm from the mirror.

SCERT AP 9th Class Physics Study Material Pdf Download 9th Lesson తేలియాడే వస్తువులు Textbook Questions and Answers.

AP State Syllabus 9th Class Physical Science 9th Lesson Questions and Answers తేలియాడే వస్తువులు

9th Class Physical Science 9th Lesson తేలియాడే వస్తువులు Textbook Questions and Answers

అభ్యసనాన్ని మెరుగుపరుచుకోండి

ప్రశ్న 1.
1. 2 సెం.మీ వ్యాసార్ధం గల గోళం యొక్క ద్రవ్యరాశి 0.05 కి.గ్రా. అయిన దాని సాపేక్ష సాంద్రత ఎంత? (AS 1)
జవాబు:
గోళం వ్యాసార్థం = 2 సెం.మీ.

ప్రశ్న 2.
ఒక సీసా ఖాళీగానున్నపుడు 20 గ్రాములు. దానిలో నీరు నింపినపుడు 22 గ్రాములు బరువు ఉంది. దానిని నూనెతో నింపినపుడు 21.76 గ్రాములుంటే ఆ నూనె సాంద్రత ఎంత? (AS 1)
జవాబు:
నీటి బరువు = 22 – 20 = 2 గ్రా
నూనె బరువు = 21.76 – 20 = 1.76 గ్రా.

ప్రశ్న 3.
ఒక గ్లాసులోని నీటిలో మంచుగడ్డ తేలుతూ ఉంది (మంచు సాంద్రత 0.9 గ్రా/ఘ. సెం.మీ). ఆ మంచుగడ్డ పూర్తిగా కరిగితే ఆ గ్లాసులోని నీటి మట్టంలో పెరుగుదల ఉంటుందా? (AS 1)
జవాబు:
గ్లాసులోని నీటిమట్టం పెరుగుతుంది.

కారణం :
మంచుగడ్డ సాంద్రత, నీటి సాంద్రతకన్నా తక్కువ ఉండడం వల్ల నీటిపై తేలుతుంది. మంచుగడ్డ కరిగి నీరుగా మారడం వలన గ్లాసులోని నీటిమట్టం పెరుగుతుంది.

ప్రశ్న 4.
నీటిలో కొన్ని వస్తువులు తేలుతాయి. కొన్ని మునుగుతాయి. ఎందుకు? (AS 1)
జవాబు:
నీటిలో వస్తువు మునుగుట, తేలుట అనేది రెండు విషయాలపై ఆధారపడి ఉంటుంది. అవి
1. సాపేక్ష సాంద్రత :
వస్తువు యొక్క సాపేక్ష సాంద్రత 1 కన్నా ఎక్కువ ఉంటే ఆ వస్తువు నీటిలో మునుగుతుంది, లేకుంటే తేలుతుంది.

2. వస్తువుచే తొలగింపబడిన నీటి ద్రవ్యరాశి :
వస్తువు సాపేక్ష సాంద్రత 1 కన్నా ఎక్కువ ఉన్నప్పటికీ, ఆ వస్తువుచే తొలగింపబడిన నీటి ద్రవ్యరాశి ఆ వస్తువు ద్రవ్యరాశికి సమానమైతే ఆ వస్తువు నీటిపై తేలుతుంది.

ప్రశ్న 5.
సాంద్రత, సాపేక్ష సాంద్రతలను వివరించండి. సూత్రాలు రాయండి. (AS 1)
జవాబు:
సాంద్రత : ప్రమాణ ఘనపరిమాణం గల వస్తువు ద్రవ్యరాశిని ఆ వస్తువు యొక్క సాంద్రత అంటారు.

ఘనపరిమాణం సాంద్రత ప్రమాణాలు : గ్రా/సెం.మీ (లేదా) కి. గ్రా/ మీ’.

సాపేక్ష సాంద్రత :
వస్తువు సాంద్రతకు, నీటి సాంద్రతకు గల నిష్పత్తిని ఆ వస్తువు యొక్క సాపేక్ష సాంద్రత అంటారు.

సాపేక్ష సాంద్రతకు ప్రమాణాలు ఉండవు.

ప్రశ్న 6.
నీటి సాంద్రత ఎంత? (AS 1)
జవాబు:
నీటి సాంద్రత = 1 గ్రా/సెం.మీ. (లేదా) 1 కి. గ్రా/ మీ³.

ప్రశ్న 7.
ఉత్థవనం (buoyancy) అనగానేమి? (AS 1)
జవాబు:
ద్రవంలో ఉన్న వస్తువుపై ఊర్ధ్వ దిశలో కలుగజేయబడే బలాన్ని ఉత్పవనం అంటాం. ఈ బలం ఆ వస్తువు వల్ల తొలగించబడిన ద్రవం బరువుకి సమానం.
(లేదా)
వస్తువును ద్రవంలో తేలేటట్లు చేయగల సామర్థ్యమే ఉత్సవనం.

ప్రశ్న 8.
కింద ఇవ్వబడిన పదార్థాలను సాపేక్ష సాంద్రత 1 కన్నా ఎక్కువ గల వస్తువులు, 1కన్నా తక్కువ గల వస్తువులుగా వర్గీకరించండి. (AS 1)
(చెక్క ఇనుము, రబ్బరు, ప్లాస్టిక్, గాజు, రాయి, బెండు, గాలి, బొగ్గు, మంచు, మైనం, కాగితం, పాలు, కిరోసిన, కొబ్బరినూనె, సబ్బు)
జవాబు:

ప్రశ్న 9.
నీరు, పాలలో ఏది అధిక సాంద్రత కలిగినది? (AS 2)
జవాబు:
నీటి సాంద్రత 1 గ్రా./ఘ. సెం.మీ, మరియు పాల సాంద్రత 1.02 గ్రా./ఘ. సెం.మీ. కావున పాల సాంద్రత నీటి సాంద్రతకన్నా కొద్దిగా ఎక్కువ.

ప్రశ్న 10.
నీటిలో ఇనుము మునుగుతుంది. చెక్క తేలుతుంది. ఒకే ఘనపరిమాణం గల ఇనుము, చెక్క ముక్కలను ఒక్కటిగా కట్టి నీటిలో వేస్తే అది మునుగుతుందా? తేలుతుందా? ఊహించండి. ప్రయోగం చేసి మీ ఊహ సరైనదో, కాదో పరీక్షించుకోండి. (AS 2, AS 3)
జవాబు:
ఒకే ఘనపరిమాణం గల ఇనుము, చెక్క ముక్కలను ఒక్కటిగా కట్టి నీటిలో వేస్తే అది మునుగుతుంది.

కారణం :
రెండు వస్తువుల ఫలిత ద్రవ్యరాశి పెరుగుతుంది. తత్ఫలితంగా ఫలిత సాంద్రత కూడా పెరుగుతుంది.

ప్రశ్న 11.
చెక్క యొక్క సాపేక్ష సాంద్రతను కనుగొనండి. కనుగొనే విధానాన్ని వివరించండి. (AS 3)
జవాబు:
ఉద్దేశ్యం : చెక్క యొక్క సాపేక్ష సాంద్రతను కనుగొనుట. (ప్రయోగశాల కృత్యం – 1)

కావలసిన పరికరాలు :
ఓవర్ ఫ్లో పాత్ర, 50 మి.లీ. కొలజాడీ, స్ప్రింగు త్రాసు, చెక్క ముక్క నీరు

విధానం :

1. 50 మి.లీ కొలజాడీ ద్రవ్యరాశిని కొలిచి నమోదు చేయండి.
2. చెక్క ముక్క యొక్క ద్రవ్యరాశిని కనుగొని నమోదు చేయండి.
3. ఓవర్ ఫ్లో పాత్రలో ప్రక్క గొట్టం గుండా నీరు పొర్లిపోయేంత వరకు నీటిని పోయండి.
4. నీరు పొర్లిపోవడం ఆగిపోగానే ఆ గొట్టంకింద 50 మి.లీ.ల కొలజాడీ నుంచండి.
5. ఇప్పుడు చెక్క ముక్కను పాత్రలోని నీటిలో జాగ్రత్తగా జారవిడవండి.
6. చెక్కముక్కను నీటిలో ఉంచగానే పక్కగొట్టంద్వారా కొంతనీరు పొర్లి కొలజాడీలోకి చేరుతుంది.
7. నీరు పొర్లిపోవడం ఆగే వరకు వేచి చూడండి.
8. నీటితో సహా కొలజాడీ ద్రవ్యరాశిని కొలిచి పట్టికలో నమోదు చేయండి.

ప్రశ్న 12.
ఒక ద్రవం యొక్క సాపేక్ష సాంద్రతను ఎలా కనుగొంటారు? (ప్రయోగశాల కృత్యం – 2) (AS 3)
జవాబు:
ఉద్దేశ్యం :
ద్రవం సాపేక్ష సాంద్రతను కనుగొనుట.

కావలసిన పరికరాలు :
50 మి.లీ. ద్రవం పట్టే సీసా, స్ప్రింగ్ త్రాసు, ఏదైనా ద్రవం (దాదాపు 50 మి.లీ.).

విధానం :

1. ముందుగా ఖాళీ సీసా ద్రవ్యరాశి కనుగొనాలి.
2. ఆ ఖాళీ సీసాను నీటితో నింపి మరల ద్రవ్యరాశిని కనుగొనాలి.
3. ఇప్పుడు 50 మి.లీల నీటి ద్రవ్యరాశి = నీటితో నింపిన సీసా ద్రవ్యరాశి – ఖాళీ సీసా ద్రవ్యరాశి
4. సీసా నుండి నీటిని తీసివేసి ఆ సీసాను ఏదైనా ద్రవం (పాలు) తో నింపి దాని ద్రవ్యరాశిని కనుగొనండి.
5. 50 మి. లీ.ల ద్రవం ద్రవ్యరాశి = ద్రవంతో నింపిన సీసా ద్రవ్యరాశి – ఖాళీ సీసా ద్రవ్యరాశి
6. 
7. ఇదే విధంగా ఏ ద్రవం యొక్క సాపేక్ష సాంద్రతనైనా కనుగొనవచ్చును. వివిధ ద్రవాల సాపేక్ష సాంద్రతలను కింది పట్టికలో నమోదు చేయండి.

ప్రశ్న 13.
వివిధ రకాల పండ్లు, కూరగాయల సాపేక్ష సాంద్రతలను కనుగొని జాబితా రాయంది. (AS 3)
జవాబు:

1. ప్రయోగాలు – క్షేత్ర పరిశీలనలు కింద ఉన్న ప్రశ్న (1)లో సూచించిన విధానాన్ని అనుసరించండి.
2. ఈ విధానంలో చెక్క ముక్కకు బదులు వివిధ రకాల పండ్లు, కూరగాయలు వాడండి.
3. వచ్చిన విలువలు కింది పట్టికలో నమోదు చేయండి.

ప్రశ్న 14.
బాల్ పెన్ రీఫిల్ లో లాక్టోమీటర్ తయారుచేయండి. రీఫిల్ నీటిలో నిటారుగా నిలబడడానికి మీరేం చేశారు?(కృత్యం – 2) (AS 5)
జవాబు:

1. ఒక ఖాళీ బాల్ పెన్ రీఫిలను తీసుకోండి. దాని చివర లోహపు ముల్లు ఉండాలి.
2. ఒక లావు పరీక్షనాళికను తీసుకొని, దానిని దాదాపుగా నిండుగా నీటిని తీసుకొని, పటంలో చూపినట్లు రీఫిలను నీటిలో ఉంచండి.
3. రీఫిల్ యొక్క లోహపు ముల్లు కిందికి ఉండేటట్లుగా జాగ్రత్త వహించండి.
4. రీఫిల్ నీటిలో ఎంతవరకు మునిగిందో, అక్కడ పెతో గుర్తు పెట్టండి.
5. బాయిలింగ్ ట్యూబ్ నుండి నీటిని తీసివేసి, పాలను పోయండి.
6. ఆ పాలలో రీఫిలను ఉంచండి.
7. రీఫిల్ పాలలో ఎంతవరకు మునిగిందో, అక్కడ పెతో మరొక గుర్తు పెట్టండి.
8. ఈ రెండు గుర్తులు ఒకే స్థానంలో ఉండవు.
9. ఇదే అభివృద్ధి పరచబడిన లాక్టోమీటరు.
10. రీఫిల్ యొక్క లోహపు ముల్లుకు ఒక బరువును (బెండు లాంటిది) అమర్చినచో రీఫిల్ ఒక పక్కకు వాలకుండా నిటారుగా నీటిలో తేలుతుంది.

ప్రశ్న 15.
పాదరస భారమితి బొమ్మ గీయండి. (AS 5)
జవాబు:

ప్రశ్న 16.
హైద్రాలిక్ బాక్స్ తయారీలో ఉపయోగపడుతున్న పాస్కల్ ఆవిష్కరణను నీవెలా అభినందిస్తావు? (AS 6)
జవాబు:
పాస్కల్ నియమం :
ఏదైనా ప్రవాహి బంధింపబడి ఉన్నప్పుడు దానిపై బాహ్య పీడనం కలుగజేస్తే ఆ ప్రవాహిలో అన్ని వైపులా ఒకే విధంగా పీడనం పెరుగుతుంది.

ఉపయోగము :

1. హైడ్రాలిక్ యంత్రాల తయారీలో ఈ సూత్రము ఉపయోగపడుతుంది.
2. మెకానిక్ షాపులందు వాహనాలను బాగు చేసేటప్పుడు వాడే జాకీలు పాస్కల్ నియమముపై ఆధారపడి పనిచేస్తాయి.
3. ఈ జాకీల వలన మనం కొద్ది బలాన్ని ప్రయోగించి భారీ వాహనాలను కూడా సులభంగా పైకెత్తవచ్చు.

ప్రశంస:

1. కేవలం మెకానిక్ షాపులయందు మాత్రమే కాక ఎక్కడైతే ఎక్కువ బరువులను తక్కువ బలంతో పైకెత్తవలసి ఉంటుందో, ఆ పరిశ్రమలన్నింటిలోను హైడ్రాలిక్ జాక్లను ఉపయోగిస్తారు.
2. శాస్త్రజ్ఞులు కనుగొన్న నియమాలు, సూత్రాలు అనేక నూతన పరికరాల రూపకల్పనకు దోహదపడి మన జీవితాన్ని సుఖమయం చేస్తున్నాయి.
3. దీనివల్ల మనం శాస్త్రజ్ఞుల కృషిని తప్పక అభినందించాలి.

ప్రశ్న 17.
ఉత్సవనం గురించి వివరించిన ఆర్కిమెడీస్ సిద్ధాంతాన్ని నీవెలా అభినందిస్తావు? (AS 6)
జవాబు:
ఆర్కిమెడీస్ సూత్రము :
ఏదైనా ఒక వస్తువును ఒక ప్రవాహిలో పూర్తిగా కాని, పాక్షికంగాగాని ముంచినప్పుడు ఆ వస్తువు తొలగించిన ప్రవాహి బరువుకు సమానమైన ఉత్సవన బలం ఆ వస్తువుపై ఊర్వ దిశలో పనిచేస్తుంది.

ఉపయోగము :
ఈ సూత్రము లోహాల స్వచ్చతను కనుగొనుటకు ఉపయోగపడును.

ప్రశంస:

1. ఆర్కిమెడీస్ స్నానం చేస్తూండగా అకస్మాత్తుగా ఈ సూత్రం కనుగొనుట జరిగినది.
2. ఈ సూత్రం సాయంతో రాజు తనకప్పజెప్పబడిన సమస్యను ఆర్కిమెడిస్ సులభంగా పరిష్కరించగలిగాడు.
3. మన నిత్యజీవితంలో ఎదురయ్యే ఎన్నో సమస్యలకు సమాధానాలుగా అనేక సూత్రాలను, నియమాలను శాస్త్రజ్ఞులు కనుగొనుట జరిగినది.
4. హుస్సేన్‌సాగర్‌లో బుద్ధ విగ్రహం మునిగిపోవుట గురించి వినే ఉంటారు.
5. ఈ విగ్రహాన్ని ఉత్సవన బలం ఆధారంగానే బయటకు తీయగలిగారు.
6. శాస్త్రజ్ఞులు ఆర్కిమెడిసన్ను ఒక మంచి గణిత శాస్త్రవేత్తగా గౌరవించారు.
7. చంద్రునిపై కనుగొన్న ఒక పెద్ద బిలానికి ఆర్కిమెడీస్ పేరు పెట్టడం జరిగినది.
8. కొన్ని శిఖరాలకు కూడా ఆర్కిమెడీస్ శిఖరాలు అని పేరు పెట్టడం జరిగినది.
9. కావున ఆర్కిమెడీస్ కనుగొన్న అనేక విషయాలను మనం అభినందించక తప్పదు.

ప్రశ్న 18.
నీటిలో మునిగే పదార్థాలతో, నీటిలో మునగని పడవలు తయారుచేసే సాంకేతికత నీకు అద్భుతంగా అనిపించిందా? ఎందుకు? (AS 6)
జవాబు:

1. ఇనుము సాపేక్ష సాంద్రత 8.5. ఇది నీటి సాంద్రతకన్నా చాలా ఎక్కువ.
2. కాని అనేక టన్నుల ఇనుముతో తయారుచేయబడిన ఒక ఓడ నీటిలో తేలడం నిజంగా ఒక వింత.
3. ఆర్కిమెడిస్ ఉత్సవన నియమం ప్రకారం ఏ వస్తువైనా ఒక ద్రవంలో ముంచబడినపుడు అది తొలగించే ద్రవం బరువు దాని బరువుకు సమానమైనప్పుడు ఆ వస్తువు ఆ ద్రవంలో తేలుతుంది.
4. కావున ఓడలను, పూర్తిగా నింపబడిన ఓడ బరువు, అది తొలగించే నీటి బరువుకు సమానమయ్యేటట్లు అధిక ఉపరితల వైశాల్యంతో నిర్మిస్తారు.
5. ఈ నిర్మాణంలో కచ్చితమైన కొలతలు, ఎంతో శాస్త్ర విజ్ఞాన నైపుణ్యము ఇమిడి ఉంటాయి.
6. నిజంగా ఈ విధమైన కచ్చితమైన కొలతలు, ఇంతటి విలువైన శాస్త్ర విజ్ఞాన నైపుణ్యాన్ని కలిగియున్న శాస్త్రవేత్తలను, ఈ నియమాలను అందించిన శాస్త్రవేత్తలను అభినందించకుండా ఉండలేము.

ప్రశ్న 19.
మీ దైనందిన జీవితంలో ఆర్కిమెడీస్ సిద్ధాంతాన్ని ఎక్కడెక్కడ పరిశీలిస్తారు? రెండు ఉదాహరణలివ్వండి. (AS 7)
జవాబు:
నిత్యజీవితంలో ఆర్కిమెడీస్ నియమ ఉపయోగం :

1. నిత్య జీవితంలో ఆర్కిమెడీస్ సూత్రం అనేక విధాలుగా ఉపయోగపడుతుంది.
2. నీటిపై తేలే చేపలు, నీటిలో ఈదే మనుషులు, నీటిపై తేలే మంచు పర్వతాలు, ఓడలు మొదలగునవి ఆర్కిమెడీస్ ఉత్సవన నియమాన్ని పాటిస్తాయి.
3. గాలిలో బెలూను ఎగురవేయడం కూడా ఆర్కిమెడీస్ సూత్ర వినియోగమే.
4. అలాగే బావిలో నుండి నీటితో నిండిన బకెట్ ను లాగేటప్పుడు, ఆ బకెట్ నీటి ఉపరితలానికి వచ్చే వరకు బరువును కోల్పోయినట్లనిపిస్తుంది. ఇది కూడా ఉత్తవన బలం యొక్క ఫలితమే.
5. నీటిలో బాతు ఈదడం కూడా ఆర్కిమెడీస్ సూత్రానికి ఉదాహరణ.

ప్రశ్న 20.
మీ దైనందిన జీవితంలో పాస్కల్ నియమాన్ని ఎక్కడెక్కడ పరిశీలిస్తారు? రెండు ఉదాహరణలివ్వండి. (AS 7)
జవాబు:
పాస్కల్ నియమం యొక్క నిత్యజీవిత ఉపయోగాలు :

1. హైడ్రాలిక్ జాక్స్
2. హైడ్రాలిక్ పంపులు
3. హైడ్రాలిక్ లిఫ్టులు
4. హైడ్రాలిక్ క్రేన్లు
5. సైఫన్
6. బావులు
7. డ్యాములు

ప్రశ్న 21.
50 గ్రా. ద్రవ్యరాశి గల ఒక పదార్థ ఘనపరిమాణము 20 ఘ. సెం.మీ. నీటి సాంద్రత 1 గ్రా./ఘ. సెం.మీ. అయితే ఆ పదార్థం నీటిలో మునుగుతుందా? తేలుతుందా? అది తొలగించే నీటి బరువు ఎంత? (AS 1)
జవాబు:

నీటి సాంద్రత : 1 గ్రా/సెం.మీ³
పదార్థ సాంద్రత, నీటి సాంద్రతకన్నా ఎక్కువ. కావున ఆ వస్తువు నీటిలో మునుగుతుంది.
ఆ వస్తువు సాపేక్ష సాంద్రత = 2.5 గ్రా/సెం.మీ³/1 గ్రా/సెం.మీ³ = 2.5
వస్తువు సాపేక్ష సాంద్రత = వస్తువు బరువు / వస్తువు ఘనపరిమాణమునకు సమాన ఘనపరిమాణము గల నీటి బరువు

వస్తువుచే తొలగింపబడిన నీటి ద్రవ్యరాశి = 20 గ్రా.

ప్రశ్న 22.
వాతావరణ పీడనం 100 కిలో పాస్కల్ ఉన్నపుడు నీటిలో 10 మీ. లోతున పీడనం ఎంత ఉంటుంది? (AS 1)
(పాస్కల్ = న్యూటన్/మీ²) (100 కిలో పాస్కల్ = 10⁵ పాస్కల్ = 10⁵ న్యూటన్/మీ² = 1 అట్మాస్పియర్).
జవాబు:
వాతావరణ పీడనం P = 100 కిలో పాస్కల్
నీటి ద్రవ్యరాశి : 1 గ్రా/సెం.మీ³
h లోతులో పీడనం P_h = P₀ + ρ h g
= 100 + 10 × 1 × 9.8
= 100 + 98 – 198 కిలో పాస్కల్

ప్రశ్న 23.
ఇనుమును నీటిలో తేలేటట్లు చేయగలవా? ఎలా? (AS 3)
జవాబు:
ఇనుమును నీటిలో మునిగేటట్లు చేయవచ్చును.

విధానం :

1. ఒక ఇనుప ముక్కను తీసుకొని దానిని ఒక నీరుగల జాడీలో జారవిడవండి.
2. ఇనుప ముక్క నీటిలో మునుగుటను గమనిస్తాము.
3. ఒక సన్నని ఇనుప రేకును తీసుకొని దానిని నాలుగు మడతలు వేసి నీటిలో వేయండి.
4. ఇది కూడా నీటిలో మునుగుట గమనిస్తాము.
5. ఇప్పుడు ఇనుప రేకు యొక్క మడతలు విప్పదీసి, దానిని ఒక గిన్నెలాగా మడిచి ఆ గిన్నెను నీటిలో వేయండి.
6. ఆ గిన్నె నీటిలో తేలుటను గమనిస్తాము.

కారణం :
ఇనుప గిన్నెచే తొలగింపబడిన నీటి ద్రవ్యరాశి, ఆ ఇనుప గిన్నె బరువుకన్నా తక్కువ అవడం చేత ఇనుప గిన్నె నీటిపై తేలింది.

ప్రశ్న 24.
మీరు వివిధ ఘన, ద్రవ పదార్థాల సాపేక్ష సాంద్రతలను కనుగొన్నారు. వాటిని వాటి సాపేక్ష సాంద్రతల ఆరోహణ క్రమంలో రాయండి. (AS 4)
జవాబు:

ప్రశ్న 25.
వాహనాలలో వాడే ఆయిల్ బ్రేకులు బ్రాహప్రెస్ నియమాన్ని (పాస్కల్ నియమాన్ని) పాటిస్తాయి. మరి ఎయిర్ బ్రేకులు .. ఎలా పనిచేస్తాయి? వాహనాలలో ఎయిర్ బ్రేకులు పనిచేసే విధానాన్ని గురించి సమాచారాన్ని సేకరించండి. (AS 4)
జవాబు:

1. ఎయిర్ బ్రేకులు శక్తి నిత్యత్వం అనే నియమంపై ఆధారపడి పనిచేస్తాయి.
2. సాధారణంగా రైలు పరిగెత్తుతున్నపుడు గతిశక్తి పుడుతుంది. ఈ గతిశక్తిని తగ్గిస్తే రైలు ఆగిపోతుంది.
3. గాలినుపయోగించి గతిశక్తిని ఉష్ణశక్తిగా మార్చడం ద్వారా రైలును ఆపగలుగుతున్నారు.
4. ఎయిర్ బ్రేకుల వ్యవస్థను పటంలో చూపడమైనది.
5. ఇందులోని ముఖ్య భాగాలు: కంప్రెసర్, ప్రధాన రిజర్వాయర్, డ్రైవరు వద్దనుండే బ్రేకు వాల్వు, బ్రేకు గొట్టం , ట్రిపుల్ వాల్వు, ఆక్టిలరీ రిజర్వాయర్, బ్రేకు సిలిండర్, బ్రేకు బ్లాకు.

పనిచేయు విధానం:

1. డ్రైవరు బ్రేకు వాల్వును నొక్కగానే బ్రేకు గొట్టంలోని గాలి పీడనం బయటకు నెట్టివేయబడును.
2. ట్రిపుల్ వాల్వు ఈ పీడనం బయటకు నెట్టివేయబడడాన్ని గుర్తిస్తుంది.
3. ఇప్పుడు బ్రేకు సిలిండర్‌కు, ఆక్టిలరీ రిజర్వాయర్‌కు మధ్యగల అనుసంధానం తెరుచుకోబడి, ఆర్డీలరీ రిజర్వాయర్ ద్వారా బ్రేక్ సిలిండర్‌ లోనికి గాలి నెట్టబడుతుంది.
4. ఈ గాలి పీడనం, ముషలకాన్ని ముందుకు నెట్టడం ద్వారా, చక్రాలకు దగ్గరలోనున్న ముషలకాలు ముందుకు నెట్టబడి, చక్రాలను ఆపుతాయి.
5. ఈ విధంగా ఎయిర్ బ్రేకులు పనిచేస్తాయి.

9th Class Physical Science 9th Lesson తేలియాడే వస్తువులు Textbook InText Questions and Answers

9th Class Physical Science Textbook Page No. 144

ప్రశ్న 1.
మీ వద్ద 30 ఘ. సెం.మీ. పరిమాణం గల దిమ్మె ఒకటి, 60 ఘ. సెం.మీ. పరిమాణం గల దిమ్మె ఒకటి ఉన్నాయనుకోండి. అవి ఏయే పదార్థాలతో తయారయ్యా యో నీకు తెలియదు. కాని 60 ఘ. సెం.మీ. పరిమాణం గలది ఎక్కువ బరువుంది. ఈ సమాచారంతో ఆ రెండు దిమ్మెలలో దేని సాంద్రత ఎక్కువో చెప్పగలరా?
జవాబు:
ఒక వస్తువు సాంద్రతను చెప్పాలంటే ఆ వస్తువు ద్రవ్యరాశి మరియు ఘనపరిమాణములు తెలిసియుండాలి. కాని పై సందర్భములో కేవలం ఘనపరిమాణము మాత్రమే తెలుసు. కాని ఆ రెండు వస్తువుల ద్రవ్యరాశులు తెలియవు కావున దేని సాంద్రత ఎక్కువో చెప్పలేము.

9th Class Physical Science Textbook Page No. 155

ప్రశ్న 2.
ఎ) “టారిసెల్లీ” భారమితిని చంద్రునిపై ఉంచితే ఏమి జరుగుతుంది?
జవాబు:
చంద్రునిపై వాతావరణ పీడనం లేదు కావున “టారిసెల్లి” భారమితిని చంద్రునిపై ఉంచితే పాదరస స్థంభం ఎత్తు ‘సున్న’ అవుతుంది.

బి) భారమితిలో పాదరస మట్టానికి కొంచెం దిగువగా గాజు గొట్టానికి ఒక రంధ్రం చేయబడి అందులో ఒక “పిడి” బిగించబడి ఉందనుకుందాం. ఆ రంధ్రం నుండి ఆ పిడిని తొలగిస్తే ఏం జరుగుతుంది?
జవాబు:

1. పాదరస స్థంభం పైన “శూన్య ప్రదేశం” ఉంటుంది. కావున పాదరసం పైన ఎటువంటి పీడనం ఉండదు.
2. అంతేగాక గాజు గొట్టంలోని పాదరస మట్టం యొక్క ‘భారం’ దానిపై వాతావరణ పీడన ఫలితంగా గిన్నెలోని పాదరసంవల్ల కలిగే బలానికి సమానంగా ఉంటుంది.
3. అందువల్ల పాదరస స్థంభం యొక్క ఎత్తులో ఎటువంటి మార్పు రాదు.

సి) భారమితిలో పాదరసానికి బదులుగా మనం నీరు ఎందుకు వాడకూడదు? ఒకవేళ మీరు వాడాలంటే గాజు గొట్టం పొడవు ఎంత ఉండాలి?
జవాబు:
భారమితిలో పాదరసానికి బదులుగా నీరు వాడలేము. ఎందుకంటే
1) నీరు ఉష్ణోగ్రత, పీడనములలోని అతి స్వల్ప మార్పులకు వ్యాకోచ, సంకోచాలు చెందదు.
2) నీరు వాడాలంటే గాజు గొట్టం పొడవు సుమారు 10 మీ. కంటే ఎక్కువ ఉండాలి. ఇది చాలా అసౌకర్యంగా ఉంటుంది. ఒకవేళ నీటిని తీసుకుంటే, పాదరస స్థంభం ఎత్తు

డి) భూమి చుట్టూ ఉన్న మొత్తం వాతావరణ పీడనం బరువు కనుక్కోండి. (భూ వ్యాసార్థం 6400 కి.మీ.)
జవాబు:
భూమి చుట్టూ ఉన్న మొత్తం వాతావరణ పీడనం బరువు = వాతావరణ పీడనం × భూ ఉపరితల వైశాల్యం

9th Class Physical Science Textbook Page No. 159

ప్రశ్న 3.
ఎ) స్వచ్ఛమైన నీటిలో కంటే ఉప్పునీటిలో మీరు సులభంగా తేలుతారు. ఎందుకు?
జవాబు:
ఉప్పునీటి సాంద్రత స్వచ్ఛమైన నీటి సాంద్రత కంటే ఎక్కువ.

బి) ద్రవంలో ముంచబడిన వస్తువుపై పార్వ దిశలో ఉత్సవన బలం ఎందుకుండదు?
జవాబు:
ఉత్సవన బలం ఊర్ధ్వ బలం మాత్రమే. వస్తువు ద్రవంలో ముంచబడినది అంటే దాని బరువు ఉత్సవన బలంకంటె ఎక్కువున్నది అని అర్థం. కావున పార్శ్వ దిశలో ఉత్సవన బలం ఉండదు.

సి) ఒకే పరిమాణం గల ఒక ఇనుప దిమ్మె, ఒక అల్యూమినియం దిమ్మెలను నీటిలో ముంచితే దీనిపై ఉత్సవన బలం అధికంగా ఉంటుంది?
జవాబు:
అల్యూమినియం దిమ్మెపై కన్నా ఇనుప దిమ్మెపై ఉత్సవన బలం అధికంగా ఉంటుంది. ఎందుకనగా ఇనుము సాంద్రత అల్యూమినియం సాంద్రత కన్నా ఎక్కువ.

డి) ఒక చెక్క దిమ్మెపై ఇనుప ముక్కను ఉంచి చెక్కదిమ్మె నీటిలో సాధారణ స్థితికంటే ఎక్కువ మునిగేటట్లు చేశారు. ఒకవేళ ఇనుప ముక్కను చెక్కదిమ్మెకు వేలాడదీస్తే చెక్కదిమ్మె ఎంతవరకు మునుగుతుంది? మొదటకంటే ఎక్కువ లోతుకా? తక్కువ లోతుకా?
జవాబు:
మొదటకంటే ఎక్కువ లోతుకు మునుగుతుంది.

9th Class Physical Science Textbook Page No. 143

ప్రశ్న 4.
‘ఒక సరదా కృత్యం చేద్దాం’ అనే కృత్యాన్ని నిర్వహించారు కదా… ఈ కింది ప్రశ్నలకు సమాధానం రాయండి.
ఎ) కిరోసిన్ నీటిపై తేలుతుందా? లేక నీరు కిరోసిన్ పై తేలుతుందా?
జవాబు:
కిరోసిన్ నీటిపై తేలుతుంది.

బి) ఏయే వస్తువులు కిరోసిన్ పై తేలుతున్నాయి?
జవాబు:
గుండీలు, అగ్గిపుల్లలు, చిన్న చిన్న కాగితం ఉండలు వంటివి కిరోసిన్ పై తేలుతున్నాయి.

సి) ఏయే వస్తువులు కిరోసిన్లో మునిగి నీటిపై తేలుతున్నాయి?
జవాబు:
మైనం కిరోసిన్లో మునుగుతుంది, కాని నీటిపై తేలుతుంది.

డి) ఏయే వస్తువులు నీటిలో మునిగాయి?
జవాబు:
గుండు సూదులు, చిన్న రాళ్ళు, ఇసుక వంటివి నీటిలో మునిగాయి.

ఇ) పరీక్షనాళికలో ఏయే వస్తువులు ఎలా అమరాయో తెలిపే పటాన్ని గీయండి.
జవాబు:

ఎఫ్) ఎందుకు కొన్ని వస్తువులు తేలుతున్నాయి? కొన్ని మునుగుతున్నాయి?
జవాబు:
ఈ విధమైన ప్రవర్తనకు ఆయా వస్తువుల సాంద్రత ప్రధాన కారణం.

ప్రశ్న 5.
గాజు గోళీకన్నా బరువైన చెక్కముక్కలు నీటిలో ఎందుకు తేలుతున్నాయి?
జవాబు:
నీటి సాంద్రతతో పోల్చినపుడు చెక్క యొక్క సాంద్రత తక్కువగాను, గాజు (గోళీ) యొక్క సాంద్రత ఎక్కువగాను ఉంటుంది. అందువల్ల చెక్క నీటిపై తేలుతుంది.

ప్రశ్న 6.
అసలు ‘బరువు’, ‘తేలిక’ అంటే ఏమిటి?
జవాబు:
‘బరువు’, ‘తేలిక’ అనేవి వస్తువు యొక్క సాంద్రత మీద ఆధారపడి నిర్ణయించబడతాయి. ఒకే ఘనపరిమాణం గల రెండు వస్తువులను తీసుకున్నపుడు వాటిలో ఏది ఎక్కువ ద్రవ్యరాశిని కలిగియుంటుందో దానిని ‘బరువైన’ వస్తువుగా చెబుతాము.

9th Class Physical Science Textbook Page No. 147

ప్రశ్న 7.
ప్రయోగశాల కృత్యం 2 ఆధారంగా ఈ కింది ప్రశ్నలకు సమాధానం రాయండి.
ఎ) కొబ్బరినూనెను నీటితో కలిపితే ఏది పైన తేలుతుంది?
జవాబు:
కొబ్బరినూనె పైన తేలుతుంది.

బి) కిరోసిన్లో చెక్కముక్కను పడవేస్తే మునుగుతుందా? తేలుతుందా? కారణం చెప్పండి.
జవాబు:
చెక్కముక్కను కిరోసిన్లో పడవేస్తే వెంటనే తేలుతుంది. కారణం చెక్క యొక్క సాంద్రత కిరోసిన్ సాంద్రతకన్నా తక్కువ ఉంటుంది. కాని కొంత సేపటి తర్వాత, చెక్కముక్క కిరోసినను పీల్చుకొని కిరోసిన్లో మునుగుతుంది.

సి) మైనం ముక్క నీటిలో తేలుతుందని, మరొక ద్రవం ‘X’ లో మునుగుతుందని అంటే ‘X’ ద్రవం యొక్క సాపేక్ష సాంద్రత 1 కన్నా ఎక్కువ ఉంటుందా? తక్కువ ఉంటుందా?
జవాబు:
మరొక ద్రవం ‘X’ యొక్క సాపేక్ష సాంద్రత 1 కన్నా తక్కువ ఉంటుంది. కారణం :

1. నీటి సాపేక్ష సాంద్రత = 1
2. మైనం యొక్క సాపేక్ష సాంద్రత 1 కన్నా తక్కువ.
3. కావున మైనం నీటిపై తేలును.
4. కాని మైనం, మరొక ద్రవం ‘X’ లో మునుగును.
5. కావున ఆ ద్రవం యొక్క సాపేక్ష సాంద్రత మైనం యొక్క సాపేక్ష సాంద్రత కన్నా తక్కువ ఉండాలి.

ప్రశ్న 8.
పాలకు నీరు కలిపితే ఆ మిశ్రమం యొక్క సాంద్రత పాల సాంద్రతకన్నా ఎక్కువ ఉంటుందా? లేక తక్కువ ఉంటుందా?
జవాబు:
పాలకు నీరు కలిపితే ఆ మిశ్రమం యొక్క సాంద్రత పాల సాంద్రతకన్నా తక్కువ ఉంటుంది.

ప్రశ్న 9.
సమాన ఘనపరిమాణం గల రెండు సీసాలలో ఒక దానిలో స్వచ్ఛమైన పాలని, మరొక దానిలో నీళ్ళు కలిపిన పాలని పోస్తే ఏసీసా బరువుగా ఉంటుంది?
జవాబు:
స్వచ్ఛమైన పాలు గల సీసా బరువుగా ఉంటుంది.

9th Class Physical Science Textbook Page No. 152

ప్రశ్న 10.
చిన్న చిన్న ఇనుప ముక్కలు నీటిలో మునుగుతున్నప్పటికీ, ఇనుము మరియు స్టీలు వంటి పదార్థాలతో చేయబడిన పెద్ద పెద్ద నౌకలు నీటిలో ఎలా తేలుతున్నాయో వివరించగలరా?
జవాబు:

1. ఆర్కిమెడీస్ ఉత్సవన నియమం ప్రకారం, ఏదైనా వస్తువు ద్రవంలో ముంచబడినపుడు ఆ వస్తువుచే తొలగించబడిన నీటి బరువు, ఆ వస్తువు బరువుకు సమానమైనపుడు ఆ వస్తువు ఆ ద్రవంలో తేలుతుంది.
2. కావున నౌకలను, వాటి బరువుకు సమానమైన బరువుగల నీటిని తొలగించే విధంగా అధిక ఉపరితల వైశాల్యంతో నిర్మిస్తారు.

ప్రశ్న 11.
ఒక లోహపు ముక్కకన్నా అంతే ద్రవ్యరాశి గల ఆ లోహంతో తయారుచేయబడిన గిన్నె ఎందుకు ఎక్కువ నీటిని పక్కకు తొలగిస్తుంది?
జవాబు:
లోహపు గిన్నె యొక్క ఉపరితల వైశాల్యం, లోహపు ముక్క యొక్క ఉపరితల వైశాల్యం కన్నా ఎక్కువ. అందువల్ల లోహపు గిన్నె ఎక్కువ నీటిని పక్కకు తొలగిస్తుంది.

9th Class Physical Science Textbook Page No. 153

ప్రశ్న 12.
గాజు గొట్టంలో పాదరస మట్టం ఎందుకు 76 సెం.మీ. ఉంటుంది?
జవాబు:
గాజు గొట్టంలోని పాదరస మట్టం యొక్క “భారం” దానిపై వాతావరణ పీడన ఫలితంగా గిన్నెలోని పాదరసం వల్ల కలిగే బలానికి సమానంగా ఉంటుంది. కావున గొట్టంలోని పాదరసం బరువు, గిన్నె పైనున్న వాతావరణ పీడనానికి సరిగ్గా సమానమయ్యేవరకు గొట్టంలోని పాదరసమట్టం మారుతూ ఉంటుంది. ఇది 76 సెం.మీ వద్ద స్థిరంగా ఉంటుంది.

9th Class Physical Science Textbook Page No. 157

ప్రశ్న 13.
రాయి నీటిలో మునిగినపుడు దాని భారాన్ని కోల్పోయినట్లు ఎందుకు అనిపిస్తుంది?
జవాబు:
నీటిలో ముంచబడిన రాయిపై ఊర్ధ్వదిశలో కలుగజేయబడిన ఉత్సవన బలం వలననే దానిపై భూమ్యాకర్షణ బలం, తగ్గినట్లయి ఆ రాయి బరువు కోల్పోయినట్లనిపిస్తుంది.

పరికరాల జాబితా

నీరు, కిరోసిన్, గుండీలు, గుండుసూదులు, అగ్గిపుల్లలు, చిన్న రాళ్లు, చిన్న కాగితం ఉండలు, ఇసుక, మైనం ముక్కలు, గాజు గోళీలు, చెక్క ముక్కలు, పెన్సిల్ రబ్బరు, చెక్కదిమ్మె, గాజు స్లెడులు, ఇనుప సీలలు, ప్లాస్టిక్ ఘనాలు, అల్యూమినియం sheet, రాళ్లు, బెండ్లు, పాలు, కొబ్బరినూనె, ఖాళీ బాల్ పెన్ రీఫిల్, ఖాళీ ప్లాస్టిక్ సీసా, బకెట్, నీరు, గాజు గ్లాస్, బీకరు, దూది, రాయి, పరీక్ష నాళిక, ఓవర్ ఫ్లో పాత్ర, 50 మి.లీ. కొలజాడీ, సాధారణ త్రాసు, బరువులు, స్ప్రింగ్ త్రాసు, సాంద్రత బుడ్డి, లావు పరీక్ష నాళిక, పాస్కల్ నియమాన్ని ప్రదర్శించే నమూనా

9th Class Physical Science 9th Lesson తేలియాడే వస్తువులు Textbook Activities (కృత్యములు)

కృత్యం – 1

సాంద్రతలను పోల్చడం :

ప్రశ్న 1.
సాంద్రత, సాపేక్ష సాంద్రతలను ఒక కృత్యం ద్వారా పోల్చుము.
జవాబు:

1. ఒకే పరిమాణం గల రెండు పరీక్షనాళికలను తీసుకొని ఒకదానిలో నీరు, మరొక దానిలో నూనె నింపండి.
2. వాటి బరువులు కనుగొనండి.
3. నూనెతో నింపిన పరీక్షనాళిక బరువు ఎక్కువ ఉన్నట్లుగా గుర్తిస్తాము.
4. దీనిని బట్టి నూనె సాంద్రత ఎక్కువ ఉన్నట్లుగా తెలుస్తుంది.
5. ఒకే పరిమాణం గల చెక్క, రబ్బరు దిమ్మెలను తీసుకోండి.
6. వాటి బరువులు కనుక్కోండి.
7. చెక్క దిమ్మె, రబ్బరు దిమ్మెకన్నా ఎక్కువ బరువు ఉన్నట్లు గమనిస్తాము.
8. రెండు వస్తువుల సాంద్రతలను పోల్చాలంటే వాటిని సమాన ఘనపరిమాణంలో తీసుకొని వాటి ద్రవ్యరాశులను పోల్చడం ఒక పద్ధతి. అయితే ఇది అన్నిరకాల ఘనపదార్థాలకు వీలుపడకపోవచ్చు.
9. దీనికొరకు ప్రతి వస్తువు సాంద్రతను నీటి సాంద్రతతో పోల్చి చూసే ఒక సులభమైన పద్ధతి ఉంది. దీనినే సాపేక్ష సాంద్రత అంటారు.

కృత్యం – 3

నీటి సాంద్రత కన్నా అధిక సాంద్రత కలిగిన పదార్థంతో తయారైన వస్తువులు నీటిలో తేలుతాయా?

ప్రశ్న 2.
నీటి సాంద్రత కన్నా అధిక సాంద్రత కలిగిన పదార్థంతో తయారైన వస్తువులు నీటిలో తేలుతాయా? ఒక కృత్యం ద్వారా నిరూపించండి.
జవాబు:
1) కింది పట్టికలో సూచించిన విధంగా కొన్ని వస్తువులను సేకరించండి.

2) ప్రతి వస్తువును ఒకదాని తర్వాత మరొకటిగా ఒక గ్లాసులోని నీటిలో వేసి, అవి మునుగుతాయో, తేలుతాయో గమనించండి.

3) మీ పరిశీలనలను కింది పట్టికలో నమోదు చేయండి.

a) ప్రయోగ క్షేత్ర పరిశీలనలు (1) లో సూచించిన విధంగా ప్రతి వస్తువు యొక్క సాపేక్ష సాంద్రతలను కనుక్కోండి.
b) కొన్ని వస్తువులు నీటిలో మునుగుటను, కొన్ని వస్తువులు తేలుటను గమనిస్తాము.
c) జామెట్రీ బాక్సు వంటిది ఇనుముతో చేసినదైనప్పటికీ, నీటిపై తేలుటను గమనిస్తాము.
d) కావున వస్తువు నీటిలో మునుగుట, తేలుట అనేది ఆ వస్తువు యొక్క సాపేక్ష సాంద్రత పైనే కాదు, ఆ వస్తువు ఉపరితల వైశాల్యం పైన కూడా ఆధారపడి ఉంటుందని తెలుస్తుంది.

కృత్యం – 4

వస్తుభారం, తొలగింపబడిన నీటిభారాలు సమానమా?

ప్రశ్న 3.
నీటిలో తేలే వస్తువు విషయంలో, ఆ వస్తువు బరువు దానిచే తొలగింపబడిన నీటి భారానికి సమానంగా ఉంటుందని చూపండి.
జవాబు:

1. ఒక బీకరును తీసుకొని దాని భారాన్ని త్రాసుతో కొలిచి నమోదు చేయండి.
2. ఓవర్ ఫ్లో పాత్రలో నీటిని నింపి, దాని పక్క గొట్టం గుండా నీరు పొర్లిపోవడం ఆగేంతవరకు వేచిచూడండి.
3. త్రాసులో తూచిన బీకరును తీసి ఓవర్ ఫ్లో పాత్ర పక్క గొట్టం కింద ఉంచండి.
4. ఒక చెక్క దిమ్మెను తీసుకొని మొదటగా దానిని నీటిలో తడిపి, తర్వాత దానిని ఓవర్ ఫ్లో పాత్రలోని నీటిలో నెమ్మదిగా జారవిడవండి.
5. చెక్కదిమ్మెను నీటిలో విడవగానే పొర్లిన నీరు’ బీకరులో చేరుతుంది.
6. ఇప్పుడు బీకరు బరువును నీటితో సహా కనుక్కోండి.
7. రెండవసారి కనుగొన్న బీకరు బరువునుండి, మొదటిసారి కనుగొన్న బీకరు బరువును తీసివేస్తే చెక్కదిమ్మెచే తొలగించబడిన నీటి బరువు వస్తుంది.
8. ఇప్పుడు చెక్కదిమ్మెను ఓవర్ ఫ్లో పాత్ర నుండి తీసివేసి, ఆరనిచ్చి, దాని బరువును కనుక్కోండి.
9. చెక్కదిమ్మె బరువు, ఆ చెక్కదిమ్మెచే తొలగింపబడిన నీటి బరువులు సమానమని మనకు తెలుస్తుంది.
10. ఇదే ప్రయోగాన్ని వివిధ రకాల వస్తువులతో చేసి మీ పరిశీలనలను కింది పట్టికలో నమోదు చేయండి.

కృత్యం – 5

అల్యూమినియంను తేలేటట్లు చేద్దాం :

ప్రశ్న 4.
అల్యూమినియంను తేలేటట్లు చేసే విధానాన్ని వివరింపుము.
జవాబు:

1. పలుచటి అల్యూమినియం రేకును కొద్దిగా తీసుకోండి.
2. దానిని 4 – 5 మడతలు మడవండి.
3. దానిని నీటిలో పడవేసి పరిశీలించండి. అది మునుగుటను గమనిస్తాము.
4. తర్వాత అల్యూమినియం రేకును బయటికి తీసి, దానిని తెరిచి ఒక గిన్నెవలె తయారుచేయండి. దానిని నీటిలో ఉంచి పరిశీలించండి.
5. అది తేలుటను గమనిస్తాము.
6. గిన్నె బరువును కనుక్కోండి.
7. ఆ అల్యూమినియం గిన్నెచే తొలగింపబడిన నీటి బరువును కనుక్కోండి.
8. ఈ రెండు బరువులు సమానంగా ఉండడాన్ని గమనించండి.
9. కావున ఒక వస్తువు బరువు, దానిచే తొలగింపబడిన నీటి బరువుకు సమానమయినపుడు ఆ వస్తువు నీటిలో తేలుతుంది.

కృత్యం – 6

ద్రవాలలో ఊర్ధ్వముఖ బలాన్ని పరిశీలిద్దాం :

ప్రశ్న 5.
ద్రవం వస్తువులపై ఊర్ధ్వముఖ పీడనాన్ని కలుగజేస్తుందని ఋజువు చేయండి.
జవాబు:

1. ఒక ఖాళీ ప్లాస్టిక్ సీసాను తీసుకొని దానికి గట్టిగా మూతను బిగించండి.
2. ఆ సీసాను ఒక బకెట్ లోని నీటిలో ఉంచండి.
3. అది నీటిలో తేలుతుంది.
4. ఆ సీసాను పటంలో చూపినట్లు నీటిలోకి అదమండి. పై దిశలో ఒత్తిడి కలుగుతున్నట్లు అనిపిస్తుంది.
5. సీసాను ఇంకా కిందికి అదమండి. పై దిశలో పనిచేసే బలం పెరుగుతున్నట్లుగా అనిపిస్తుంది.
6. ఇప్పుడు సీసాను వదిలేయండి. అది నీటి ఉపరితలంపైకి దూసుకు వస్తుంది.
7. ఊర్ధ్వ దిశలో పనిచేసే నీటి యొక్క ఈ బలం నిజమైనది మరియు పరిశీలించడానికి అనువైనది.
8. ఒక వస్తువు ఉపరితల ప్రమాణ వైశాల్యంపై పనిచేసే బలాన్ని “పీడనం” అంటారు.

కృత్యం – 7

గాలి పీడనాన్ని పరిశీలిద్దాం :

ప్రశ్న 6.
గాలి పీడనాన్ని పరిశీలించడానికి ఒక కృత్యాన్ని వివరించండి.
జవాబు:

1. ఒక గాజుగ్లాసును తీసుకొని దానిలో అడుగుభాగాన కొంత దూదిని అంటించండి.
2. గ్లాసును తలకిందులుగా చేసి పటంలో చూపినట్లు ఒక పాత్రలోని నీటిలో అడుగువరకు ముంచండి.
3. తర్వాత గ్లాసును అలాగే బయటకు తీయండి.
4. గ్లాసులోని దూది తడవకుండా ఉండడాన్ని గమనిస్తాము.
5. గ్లాసులోని గాలి యొక్క ఒత్తిడి నీటి పై పనిచేసి గ్లాసులోనికి నీరు చేరకుండా అడ్డుకుంది.
6. నీటి ఉపరితలంపైన ప్రమాణ వైశాల్యంలో కలుగజేయబడిన ఈ గాలి ఒత్తిడిని గాలి పీడనం అంటారు.

కృత్యం – 8

ఉత్ల్ఫవన బలాన్ని కొలవగలమా? ప్రయత్నిద్దాం !

ప్రశ్న 7.
ఉత్ల్ఫవన బలాన్ని ఎలా కొలుస్తారు?
జవాబు:

1. ఒక రాయిని స్ప్రింగు త్రాసుకు కట్టి దాని బరువును కనుగొనండి.
2. ఒక బీకరులో సగం వరకు నీటిని తీసుకోండి.
3. స్ప్రింగు త్రాసుకు వేలాడదీయబడిన రాయిని నీటిలో ముంచండి.
4. ఇప్పుడు స్ప్రింగు త్రాసు రీడింగు నీటిలో ముంచబడిన రాయి బరువును తెలుపుతుంది.
5. నీటిలో మునిగినప్పుడు రాయి బరువు మొదట ఉన్న బరువుకన్నా తగ్గినట్లుండడం గమనిస్తాము.
6. ఆ రాయి కోల్పోయినట్లనిపించే బరువుని కొలవడం ద్వారా ఆ ద్రవం కలిగించిన ఉత్సవన బలాన్ని కొలవగలుగుతాము.

కృత్యం – 9

రాయి చేత తొలగింపబడిన నీటి బరువును కొలుద్దాం:

ప్రశ్న 8.
ఆర్కిమెడీస్ ఉత్తీవన సూత్రాన్ని పేర్కొని నిరూపించుము.
(లేదా)
ఆర్కిమెడిస్ సూత్రం తెలిపి దానిని ప్రయోగ పూర్వకంగా నీవెలా ఋజువు చేస్తావో రాయండి.
జవాబు:
ఆర్కిమెడీస్ ఉత్తవన సూత్రం:
ఏదైనా ఒక వస్తువును ఒక ప్రవాహిలో పూర్తిగా గాని, పాక్షికంగా గాని ముంచినప్పుడు ఆ వస్తువు తొలగించిన ప్రవాహి బరువుకు సమానమైన ఉత్సవన బలం ఆ వస్తువుపై ఊర్ధ్వ దిశలో పనిచేస్తుంది.

నిరూపణ:

1. ఒక రాయిని తీసుకొని స్ప్రింగ్ త్రాసుతో దాని బరువును తూచండి.
2. ఒక ఓవర్ ఫ్లో పాత్రను తీసుకొని దాని పక్క గొట్టం వరకు నీరు పోయండి.
3. పటంలో చూపినట్లు ఆ పక్క గొట్టం కింద కొలతలు గల బీకరును ఉంచండి.
4. ఇప్పుడు స్ప్రింగు త్రాసుకు వేలాడదీసిన రాయిని ఓవర్ ఫ్లో పాత్రలో పూర్తిగా ముంచండి.
5. స్ప్రింగు త్రాసు రీడింగును, బీకరులోని నీటి కొలతను నమోదు చేయండి.
6. స్ప్రింగు త్రాసు రీడింగు నీటిలో ముంచబడిన రాయి బరువును, బీకరులోని నీటి కొలత రాయి వలన తొలగించబడిన నీటి ఘనపరిమాణాన్ని తెలుపుతుంది.
7. స్ప్రింగు త్రాసు యొక్క రెండు రీడింగులలోని తేడా, ఆ రాయి నీటిలో కోల్పోయినట్లనిపించే బరువుకు సమానం.
8. బీకరులోని నీటి బరువును కనుక్కోండి.
9. తగ్గినట్లనిపించే రాయి బరువు, ఆ రాయిచే తొలగింపబడిన నీటి బరువు సమానంగా ఉంటాయి.
10. ఇది ఆర్కిమెడీస్ సూత్రానికి నిరూపణ.

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.4 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.4

Question 1.
Determine which of the following polynomials has (x + 1) as a factor.
i) x³ – x² – x + 1
Solution:
f(- 1) = (- 1)³ – (- 1)² – (- 1) + 1
= -1 – 1 + 1 + 1 = 0
∴ (x + 1) is a factor.

ii) x⁴ -x³ +x² – x + 1
Solution:
f(- 1) = (- 1)⁴ – (- 1)³ + (- 1)² – (- 1) + 1
= 1 + 1 + 1 + 1 + 1= 5
∴ (x + 1) is not a factor.

iii) x⁴ + 2x³ + 2x² + x + 1
Solution:
f(- 1) = (-1)⁴ + 2 (- 1)³ + 2 (- 1)² + (-1) + 1
= 1 – 2 + 2 – 1 + 1 = 1
∴ (x + 1) is not a factor.

iv) x³ – x² – (3 – √3)x + √3
Solution:
f(- 1) = (- 1)³ – (- 1)² – (3 – √3)(-1) + √3
= – 1 – 1 + 3 – √3 + √3 = 1
∴ (x + 1) is not a factor.

Question 2.
Use the factor theorem to determine whether g(x) is a factor of f(x) in each of the following cases:
i) f(x) = 5x³ + x² – 5x – 1; g(x) = x + 1
[Factor theorem : If f(x) is a polynomial; f(a) = 0 then (x – a) is a factor of f(x); a ∈ R]
Solution:
g(x) = x+ 1 = x- a say
∴ a = – 1
f(a) = f(- 1) = 5 (- 1)³ + (- 1)² – 5 (- 1) – 1
= -5 + 1 + 5 – 1 = 0
∴ x + 1 is a factor of f(x).

ii) f(x) = x³ + 3x² + 3x + 1; g(x) = x + 1
Solution:
g(x) = x + 1 = x – a
∴ a = – 1
f(a) = f(- 1) = (- 1)³ + 3 (- 1)² + 3(-1) + 1
= -1 + 3 – 3 + 1 =0
∴ f(x) is a factor of g(x).

iii) f(x) = x³ – 4x² + x + 6;
g(x) = x – 2
Solution:
g(x) = x- 2 = x- a
∴ a = 2
f(a) = f(2) = 2³ – 4(2)² + 2 + 6
= 8 – 16 + 2 + 6 = 0
∴ g(x) is a factor of f(x).

iv) f(x) = 3x³+ x² – 20x +12; g(x) = 3x – 2
Solution:
g(x) = 3x – 2 = \(x-\frac{2}{3}\) = x – a
∴ a = 2/3

v) f(x) = 4x³+ 20x²+ 33x + 18; g(x) = 2x + 3
Solution:
g(x) = 2x + 3 = x + \(\frac{3}{2}=\) = x – a
∴ a = -3/2

∴ g(x) is a factor of f(x).

Question 3.
Show that (x – 2), (x + 3) and (x – 4) are factors of x³ – 3x² – 10x + 24.
Solution:
Given f(x) = x³ – 3x² – 10x + 24
To check whether (x – 2), (x + 3) and (x – 4) are factors of f(x), let f(2), f(- 3) and f(4)
f(2) = 2³ – 3(2)² – 10(2) + 24
= 8- 12-20 + 24 = 0
∴ (x – 2) is a factor of f(x).

f(- 3) = (- 3)³ – 3(- 3)²– 10(- 3) + 24
= – 27 – 27 + 30 + 24 = 0
∴ (x + 3) is a factor of f(x).

f(4) = (4)³ – 3 (4)² – 10 (4) + 24
= 64 – 48 – 40 + 24
= 88 – 88
= 0
∴ (x – 4) is a factor of f(x).

Question 4.
Show that (x + 4), (x – 3) and (x – 7) are factors of x³ – 6x² – 19x + 84.
Solution:
Let f(x) = x³ – 6x² – 19x + 84
To verify whether (x + 4), (x – 3) and (x – 7) are factors of f(x) we use factor theorem.

Let f(- 4), f(3) and f(7)
f(- 4) = (- 4)³ – 6 (- 4)² – 19 (- 4) + 84
= -64 – 96 + 76 + 84
= 0 .
∴ (x + 4) is a factor of f(x).

f(3) = 3³ – 6(3)² – 19(3) + 84
= 27 – 54 – 57 + 84
= 0
∴ (x – 3) is a factor of f(x).

f(7) = 7³ – 6(7)² – 19(7) + 84
= 343 – 294 – 133 + 84
= 427 – 427
= 0
∴ (x – 7) is a factor of f(x).

Question 5.
If both (x – 2) and \(\left(x-\frac{1}{2}\right)\) of px² + 5x + r, show that p = r.
Solution:
Let f(x) = px²+ 5x + r
As (x – 2) and \(\left(x-\frac{1}{2}\right)\) are factor of f(x), we have f(2) = 0 and f(1/2) = 0
∴ f(2) = p(2)² + 5(2) + r
= 4p + 10 + r = 0
= 4p + r
= – 10 ………………(1)

⇒ p + 10 + 4r = 0
⇒ p + 4r = – 10 ………………. (2)
From (1) and (2);
4p + r = p + 4r
4p – p = 4r – r
3p = 3r
∴ P = r

Question 6.
If (x² – 1) is a factor of ax⁴ + bx³ + cx² + dx + e, show that a + c + e = b + d = 0.
Solution:
Let f(x) = ax⁴ + bx³ + cx² + dx + e
As (x – 1) is a factor of f(x) we have
x² – 1 = (x + 1) (x – 1) hence f(1) = 0 and f(-1) = 0
f(1) = a + b + c + d + e = 0 ……………. (1)
and f(-1) = a- b + c- d + e = 0
⇒ a + c + e = b + d
Substitute this value in equation (1)
a + c + e + b + d=0
b + d + b + d=0
2 (b + d) = 0
⇒ b + d = 0
∴ a + c + e = b + d = 0

Question 7.
Factorise
i) x³ – 2x² – x + 2
Solution:
Let f(x) = x³ – 2x² – x + 2
By trial, we find f(l) = 1³ – 2(1)² – 1 + 2
= 1 – 2 – 1 + 2
= 0 .
∴ (x – 1) is a factor of f(x).
[by factor theorem]
Now dividing f(x) by (x – 1).

f(x) = (x – 1) (x² – x – 2)
= (x – 1) [x² – 2x + x- 2]
= (x – 1) [x (x – 2) + 1 (x – 2)]
= (x – 1) (x – 2) (x + 1)

ii) x³ – 3x² – 9x – 5
Solution:
Let f(x) = x³ – 3x² – 9x – 5By trial,
f(- 1) = (- 1)³ – 3(- 1)² – 9(- 1) – 5
=-1 – 3 + 9 – 5
=0
∴ (x + 1) is a factor of f(x).
[ ∵ by factor theorem]
Now dividing f(x) by (x + 1).

f(x)=(x + 1)(x² – 4x – 5)
But x²– 4x – 5 = x² – 5x + x – 5
= x (x – 5) + 1 (x – 5)
=(x – 5)(x + 1)
∴ f(x)=(x + 1)(x + 1)(x – 5)

iii) x³ + 13x² + 32x + 20
Solution:
Let f(x) = x³ + 13x² + 32x + 20
Let f(- 1)
= (- 1)³ + 13 (- 1)² + 32 (- 1) + 20
= – 1 + 13 – 32 + 20 = 33 – 33 = 0
∴ (x + 1) is a factor of f(x).
[ ∵ by factor theorem] Now dividing f(x) by (x + 1).

iv) y³ + y² – y – 1
Let f(y) = y³ + y² – y – 1
f(1) = 1³+ 1²– 1 – 1 = 0
(y – 1) is a factor of f(y).
Now dividing f(y) by (y – 1).

∴ f(x) = (x + 1)(x² + 12x + 20)
But (x² + 12x + 20) = x²+ 10x + 2x + 20
=x(x + 10)+2(x + 10)
=(x + 10)(x + 2)
∴f(x) = (x + 1)(x + 2)(x + 10)

Question 8.
If ax² + bx + c and bx² + ax + c have a common factor x + 1 then show that c = 0 and a = b.
Solution:
Let f(x) = ax² + bx + c and g(x) = bx² + ax + c given that (x + 1) is a common factor for both f(x) and g(x).
∴ f(-1) = g(- 1)
⇒a(- 1)² + b(- 1) + c
= b(- 1)² + a (- 1) + c
⇒ a – b + c = b – a + c
⇒ a + a = b + b
⇒ 2a = 2b
⇒ a = b
Also f(- 1) = a – b + c = 0
⇒ b – b + c = 0
⇒ c = 0

Question 9.
If x² – x – 6 and x² + 3x – 18 have a common factor x – a then find the value of a.
Solution:
Let f(x) = x² – x – 6 and
g(x) = x² + 3x – 18
Given that (x – a) is a factor of both f(x) and g(x).
f(a) = g(a) = 0
⇒ a² – a – 6 = a² + 3a – 18
⇒ – 4a = – 18 + 6
⇒ – 4a = – 12
∴ a = 3

Question 10.
If (y – 3) is a factor of y³– 2y²– 9y + 18, then find the other two factors.
Solution:
Let f(y) = y³– 2y² – 9y + 18
Given that (y – 3) is a factor of f(y).
Dividing f(y) by (y – 3)

∴ f(y) = (y – 3) (y + y – 6)
But y² + y – 6
= y² + 3y – 2y – 6
= y (y + 3) – 2 (y + 3)
= (y + 3) (y – 2)
∴ f(y) = (y – 2)(y – 3)(y + 3)
The other two factors are (y – 2) and (y + 3).

AP State Syllabus AP Board 9th Class Physical Science Solutions Chapter 5 What is inside the Atom? Textbook Questions and Answers.

AP State Syllabus 9th Class Physical Science Solutions 5th Lesson What is inside the Atom?

9th Class Physical Science 5th Lesson What is inside the Atom? Textbook Questions and Answers

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Question 1.
What are the three subatomic particles? (AS 1)
Answer:
The three sub-atomic particles are electrons, protons, and neutrons.

Question 2.
Compare the subatomic particles electron, proton, and neutron. (AS 1)
Answer:

Question 3.
What are the limitations of J.J. Thomson’s model of the atom? (AS 1)
Answer:
The main limitation of J.J. Thomson’s model of atom was that he is unable to explain, how the positively charged particles are shielded from negatively charged particles without getting neutralized.

Question 4.
What were the three major observations Rutherford made in the gold foil experiment? (AS 1)
Answer:
The three major observations Rutherford made in the gold foil experiment were

1. Most of the space inside the atom is empty.
2. All the positive charge must be concentrated in a very small space within the atom called nucleus.
3. The size of the nucleus is very small as compared to the size of the atom.

Question 5.
Sketch Rutherford’s atomic model. Why is Rutherford’s model of the atom called the planetary model? (AS 5)
Answer:

Rutherford’s model is called planetary model because the motion of the electrons around the nucleus resembles the motion of the planets around the sun.

Question 6.
Put tick (✓) against correct choice and cross (✗) against wrong choice: (AS 1)
i) In Rutherford’s gold foil experiment, majority of alpha particles passed directly through the gold foil. This observation leads to which conclusions?
a) The positively charged region of the atom is very small. (✗)
b) The atom must consist of empty space. (✓)
c) The alpha particles makes a direct hit on the positively charged region of the atom. (✗)
d) The positively charged region of the atom is very dense. (✗)

ii) In Rutherford’s gold foil experiment, occasionally the alpha particles veered from a straight line path. This observation leads to which conclusion?
a) The positively charged region of the atom is very small. (✗)
b) The majority of the space in the atom is empty. (✗)
c) The alpha particle makes a direct hit on the positively charged region of the atom. (✗)
d) The positively charged region of the atom is very dense. (✓)

Question 7.
Which one of the following is the correct electronic configuration of sodium? (AS 1)
a) 2, 8
b) 8, 2, 1
c) 2, 1, 8
d) 2, 8, 1
Answer:
d) 2, 8, 1

Question 8.
Give the main postulates of Bohr’s model of an atom. (AS 1)
Answer:
The main postulates of Bohr’s model of atom are

1. Only certain special, discrete orbits of electrons are allowed inside the atom. These orbits or shells are called energy levels.
2. While revolving in these discrete orbits the electrons do not radiate energy and this helps that the electrons do not crash into the nucleus.
3. These orbits or shells are represented by the letters K, L, M, N, ……… or the numbers n = 1, 2, 3, …………

Question 9.
Compare all the proposed models of an atom given in this chapter. (AS 1)
Answer:
In this chapter, four atomic models were discussed. The main postulates of those models are

1. Dalton’s proposal :
a) Atoms are indivisible.
b) Atoms of an element are all identical to each other and different from the atoms of other elements.

2. Thomson’s proposal:
a) An atom is considered to be a sphere of uniform positive charge and electrons are embedded into it.
b) The total mass of the atom is considered to be uniformly distributed throughout the atom.

3. Rutherford’s proposal:
a) All the positively charged material in an atom formed a small dense centre, called the nucleus of the atom.
b) Negatively charged electrons revolve around the nucleus in well defined orbits.
c) Size of the nucleus is very small as compared to the size of the atom.

4. Neils Bohr’s proposal :
a) Electrons are revolving around the nucleus in special, discrete orbits called en¬ergy levels or shells.
b) While revolving in these discrete orbits the electron do not radiate energy and this helps that the electrons do not crash into the nucleus.
c) These orbits or shells are represented by the letters K, L, M, N,…. or the numbers n = 1, 2, 3,

Question 1o.
Define valency by taking examples of nitrogen and boron. (AS 1)
Answer:
Valency: The number of electrons present in outer most orbit of an atom is called its valency. Valency of Nitrogen :
a) Atomic number of nitrogen is 7.
b) The distribution of electrons is 2, 5.
c) The outer most orbit has 5 electrons.
d) Hence its valency should be 5. But it is easier to nitrogen to gain 3 electrons than to loose 5 electrons for becoming octet.
e) Hence the valency of nitrogen is ‘3’.

Valency of Boron :
a) Atomic number of boron is 5.
b) The distribution of electrons is 2, 3.
c) The outer most orbit has 3 electrons.
d) Hence the valency of boron is 3.

Question 11.
State the valencies of the following elements : magnesium and sodium. (AS 1)
Answer:
Magnesium :
a) Atomic number of magnesium is 12.
b) Distribution of electrons is 2, 8, 2.
c) Hence the valency is 2.

Sodium :
a) Atomic number of sodium is 11.
b) Distribution of electrons is 2, 8, 1.
c) Hence the valency is 1.

Question 12.
If Z = 5, what would be the valency of the element? (AS 2)
Answer:
1) If Z = 5, the distribution of electrons is 2, 3.
2) Hence the valency is ‘3’.

Question 13.
Write the atomic number and the symbol of an element which has mass number 32 and the number of neutrons 16 in the nucleus. (AS 1)
Answer:
Mass number (A) = 32 ; Number of neutrons (N) = 16
Number of protons (Z) = A- N = 32-16 = 16
∴ Atomic number =16
The element is sulphur.
The symbol of sulphur is S’.

Question 14.
Cl- has completely filled K and L shells. Explain. (AS 1)
Answer:
Atomic number of Cl is 17, but Cl^– has one more electron when compared with Cl atom. Distribution of electrons in Cl^– is

K shell can accommodate 2 electrons and L shell cab accommodate 8 electrons accord¬ing to the formula 2n2.
Hence the K and L shells are completely filled.

Question 15.
What is the main difference among the isotopes of the same element? (AS 1)
Answer:
The main difference between isotopes of the same element is
a) The number of neutrons is different.
b) Their physical properties are different but the chemical properties are similar.

Question 16.
For the following statements, write T for True and F for False. (AS 1)
a) J.J. Thomson proposed that the nucleus of an atom contains only nucleons.
b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.
c) The mass of an electron is 1836 times that of proton.
Answer:
a) False
b) False
c) True

Question 17.
Fill in the missing information in the following table. (AS 4)

Answer:

Question 18.
How do you appreciate the efforts made by scientists to explain the structure of atom by developing various atomic models? (AS 6)
Answer:

• Structure of atom, till today it is mysterious and challenging the scientists.
• We have to appreciate the scientists right from Lavoisier, who proposed law of conservation of mass, Proust who proposed law of constant proportions, John Dalton for his first model of atom, Rutherford for giving planetary model of atom and Neils Bohr for his model of atom.
• Till today scientists are trying to know the existence of more and sub-atomic particles besides electrons, protons and neutrons.
• Hence the efforts of scientists are highly appreciable, for making our lives comfortable and leaving many challenges before us to unveil or discover them.

Question 19.
Geeta got a doubt, “Why does atomic nucleus contain proton and neutrons? Why can’t electrons and neutrons be in it”. Can you help to clarify her doubt? Explain. (AS 1)
Answer:
Nucleus contains protons and neutrons inside it but not electrons and neutrons. If it would have happen, then

1. the alpha particles in the Rutherford’s alpha particle scattering experiment would have not been deflected or scattered.
2. the idea of nucleus would have not been evolved because the mass of electron is negligible, it is most unstable.

Question 20.
Collect information about various experiments conducted and theories proposed by scientists starting from John Dalton to Neils Bohr. Prepare a story with a title “History of atom”. (AS 4)
Answer:
History of atom
John Dalton proposed atomic theory based on law of conservation of mass and law of constant proportion as :

1. Atoms were indivisible.
2. Atoms of an element are all identical to each other and different from the atoms of other elements.

Later on various experiments conducted by Thomson, Goldstein, etc. proved that atom is divisible and consists sub-atomic particles like electrons, protons and neutrons. Based on this J.J. Thomson proposed a model of atom in 1898.

According to Thomson,

1. An atom is considered to be a sphere of uniform positive charge and electrons are embedded into it.
2. The total mass of the atom is considered to be uniformly distributed throughout the atom.
3. The negative and the positive charges are supposed to be balance out and the atom as a whole is electrically neutral.

This model is also called as plum pudding model or watermelon model.

Thomson’s student Ernest Rutherford conducted alpha particle scattering experiment and got the results which were not in favour of Thomson’s model. Based on his experiment, Rutherford proposed a model of atom. According to him,

1. All the positively charged material in an atom formed a small dense centre, called the nucleus of the atom. The electrons were not a part of nucleus.
2. Negatively charged electrons revolve around the nucleus in well – defined orbits like planets revolve around the sun.
3. The size of nucleus is very small as compared to the size of the atom.

This model could not account for stability of atom, as revolving electron must lose energy and eventually crash into the nucleus, as a result matter would not exist in the form that we see it now.

In 1913, Neils Bohr proposed another model to overcome Rutherford’s defect. According to Bohr,

1. Only certain special, discrete orbits of electrons are allowed in side the atom. These orbits or shells are called energy levels.
2. While revolving in these discrete orbits the electrons do not radiate energy and this helps that the electrons do not crash into the nucleus.
3. These orbits or shells are represented by K, L, M, N ………… or the numbers 1, 2, 3,

This model could not predict the spectra of atoms.

Hence this journey continues

9th Class Physical Science 5th Lesson What is inside the Atom? InText Questions and Answers

9th Class Physical Science Textbook Page No. 75

Question 1.
Why are the atoms of different elements different?
Answer:
Nature and properties of elements depends on the arrangement of atoms. We know that different elements behave differently. This is due to the difference in their atoms.

Question 2.
Is there anything inside atom that make them to be same or different?
Answer:
The arrangement of sub atomic particles inside the atom is responsible to make them to be same or different.

Question 3.
Are atoms indivisible?
Answer:
No, atom is divisible. There are many sub-atomic particles inside the atom according to the recent experiments.

9th Class Physical Science Textbook Page No. 77

Question 4.
If an atom consists of sub-atomic particles like protons, neutrons and electrons, how are they arranged in the atom ?
Answer:
The arrangement of sub-atomic particles like protons, neutrons and electrons has been explained by many scientists like Rutherford, Neils Bohr, etc. According to them, atom consists a central mass called nucleus. Nucleus consists protons and neutrons. Electrons revolve around the nucleus in fixed shells.

9th Class Physical Science Textbook Page No. 80

Question 5.
Why is atom stable?
Answer:
In an atom, the number of protons in the nucleus is equal to the number of electrons out side the nucleus. Hence the positive and negative charges in an atom are equal. So, atom is electrically neutral. So, atom is stable. But the stability of atom was explained by Neils Bohr in a different way.

Question 6.
Can you suggest any other arrangement of subatomic particles in the atom which prevents the revolving electron to fall into the nucleus?
Answer:
Electrons have to revolve around the nucleus in definite orbits such that the centripetal and centrifugal forces acting on the electron must be equal in magnitude and opposite direction. Then the revolving electron do not fall into the nucleus.

9th Class Physical Science Textbook Page No. 82

Question 7.
How many electrons can be accommodated in each shell of an atom?
Answer:
The number of electrons that can be accommodated in each shell of an atom depends on the shell number. First shell (K) consists 2 electrons, second (L) shell consists 8 electrons, third (M) shell consists 18 electrons, fourth shell (N) consists 32 electrons, and so on.

Question 8.
Can a particular shell have just one electron?
Answer:
No, shell has just one electron.

Question 9.
What is the criteria to decide number of electrons in a shell?
Answer:
The number of electrons in a shell can be decided by using a formula 2n². (Where n is shell number).

9th Class Physical Science Textbook Page No. 83

Question 10.
What is the valency of oxygen that you can calculate by the method discussed above?
Answer:

• Oxygen has 8 electrons in its atom. The distribution of electrons is 2, 6.
• The outer most shell consists 6 electrons, this number is hear to 8.
• Hence the valency of oxygen is 8 – 6 = 2.

9th Class Physical Science Textbook Page No. 85

Question 11.
Should we consider the number of neutrons as a characteristic of an atom?
Answer:
The mass of an atom which is a characteristic of an atom depends on the number of neutrons and protons that its nucleus contains. Hence the number of neutrons can be considered as a characteristic of an atom.

9th Class Physical Science Textbook Page No. 76

Question 12.
An atom is electrically neutral. But the electrons present in it are negatively charged particles. If only negative charges were present, the atom would not be neutral. Then, why are atoms considered to be neutral?
Answer:

• This is the idea before Rutherford’s model.
• According to Rutherford’s model, number of protons inside the nucleus and number of electrons outside the nucleus are equal.
• Hence the net negative charge is equal to net positive charge. So, the atom is electrically neutral.

9th Class Physical Science Textbook Page No. 80

Question 13.
Compare Rutherford and Thomson’s models of the atom on the following basis :
1) Where is the positive charge placed?
2) How are the electrons placed?
3) Are they stationary inside the atom or moving?
Answer:

1. According to Thomson, the positive charge is uniformly distributed throughout the atom. Whereas according to Rutherford, the positively charged protons are inside the nucleus.
2. According to Thomson, electrons are embedded in the positively charged atom, but according to Rutherford, electrons are revolving around the nucleus in welldefined orbits.
3. According to Thomson, electrons are stable inside the atom but according to Rutherford, electrons are moving inside the atom.

9th Class Physical Science Textbook Page No. 83

Question 14.
Phosphorus and sulphur show multiple valency. See table 2. Why do some elements show multiple valency? Discuss with your Mends and teachers.
Answer:

• For sulphur, the number of electrons in outer most orbit is 6.
• Hence the valency should be (8 – 6 =) 2.
• But sulphur exists in so many forms.
• In the excited state, these 6 electrons also tend to participate in the bond formation.
• Hence sometimes it shows the valency 6. Ex : SO₂, SO₃, etc.
• Same situation happens for phosphorus. Ex . PCl₃, PCl₅, etc.

9th Class Physical Science 5th Lesson What is inside the Atom? Activities

Activity – 1

Question 1.
Sketch the structure of atom as you imagine.

We learnt about electron, proton and neutron.
a) Suppose you had to arrange them in an atom, how do you do it?
Answer:
Many arrangements are possible. Think that atom looks like a room, we can arrange the particles in alternating rows.

b) In how many ways can you arrange these sub-atomic particles in a spherical shape?
Answer:
Protons are positively charged, electrons are negatively charged and neutrons are
neutral. Hence neutrons and protons can be kept nearer and electrons can be kept farther or near the edge of the sphere. This is only an assumption. We can arrange in so many ways like this.

AP State Syllabus AP Board 9th Class Physical Science Solutions Chapter 9 Floating Bodies Textbook Questions and Answers.

AP State Syllabus 9th Class Physical Science Solutions 9th Lesson Floating Bodies

9th Class Physical Science 9th Lesson Floating Bodies Textbook Questions and Answers

Improve Your Learning

Question 1.
A solid sphere has a radius of 2 cm and a mass of 0.05 kg. What is the relative density of the sphere? (AS 1)
Answer:
Radius of the sphere = 2 cm

Question 2.
A small bottle weighs 20 g when empty and 22 g when filled with water. When it is filled with oil it weighs 21.76 g. What is the density of oil? (AS 1)
Answer:
Weight of water = 22 – 20 = 2 gm
Weight of oil = 21.76 – 20 = 1.76 gm

Question 3.
An ice cube floats on the surface of a glass of water (density of ice = 0.9 g/cm3). When the ice melts will the water level in the glass rise? (AS 1)
Answer:
Yes, the water level rises.

Reason :
The ice cube floats on water, because its density is less than the density of water. When ice cube melts, it becomes water, so that the water level rises.

Question 4.
The volume of 50 g of a substance is 20 cm³. If the density of water is 1 g/cm³, will the substance sink or float when placed on the surface of water? What will be the mass of water displaced by the substance? (AS 1)
Answer:

∴ Weight of the water displaced by the substance = 20 g

Question 5.
Find the pressure at a depth of 10 m in water if the atmospheric pressure is 100 kPa. [1Pa = 1 N/m²] [100kPa = 10⁵ Pa = 10⁵ N/m² = 1 atm.] (AS 1)
Answer:
Depth ‘h’ = 10 m ; Atmospheric pressure P₀ = 100 kPa
Density of water p = 1 gm/cm³ = 1 kg/m³
Pressure at a depth ‘h’ is P = P₀ + hρg
= 100 + 10 x 1 x 9.8
= 100 + 98
= 198 kPa

Question 6.
Why do some objects float on the water? And some sink? (AS 1)
Answer:

• Floating or sinking of objects on water depends on two factors.
  a) Relative density
  b) Weight of the water displace by the object
• If the relative density of an object is greater than 1, the object sinks otherwise it floats.
• Eventhough the relative density is greater than 1, if the weight of the water displaced by the object equal to the weight of the object itself, the object floats on the water.

Question 7.
Explain density and relative density and write their formulae. (AS 1)
Answer:
Density :
Density is defined as mass per unit volume.
Density = \(\frac{\text { Mass }}{\text { Volume }}\)

Unit of density is gm/cm³ or kg/m³.

Relative density :
Relative density of an object is the ratio of density of the object to the density of water.

Question 8.
What is the value of density of water? (AS 1)
Answer:
Value of density of water = 1 gm/cm³ (or) 1 kg/cm³.

Question 9.
Find the relative density of wood. Explain the process. (Lab Activity 1) (AS 3)
Answer:
Aim :
To find the relative density of wood.

Materials required :
Overflow vessel, 50 ml measuring cylinder, spring balance, wooden block, water.

Procedure:

1. Weigh the 50 ml measuring cylinder and note its weight.
2. Weigh the wooden block and note its weight.
3. Pour water in the overflow vessel until it starts dripping from its beak.
4. When water stops dripping from the beak, place the 50 ml measuring cylinder under it.
5. Slip the wooden block gently into the overflow vessel, ensuring that the water does not splash out.
6. Once the wooden block is in the overflow vessel, water flows out of the beak and collects in the 50 ml cylinder.
7. Wait till the flow of water from beak, stops.
8. Weigh the cylinder with the water that overflowed and record the weight.

Question 10.
Which is denser, water or milk? (AS 2)
Answer:
The density of water is 1 gm/cc and that of milk is 1.02 gm/cc. Hence milk is slightly denser than water.

Question 11.
What is buoyancy? (AS 1)
Answer:
Buoyancy is the upward force that a fluid exerts on an object less denser than itself.
(or)
Buoyancy is the ability of an object to float in a liquid.

Question 12.
Classify the following things into substances having relative density > 1 and relative density < 1. Wood, iron, rubber, plastic, glass, stone, cork, air, coal, ice, wax, paper, milk, kerosene, groundnut oil, soap. (AS 1)
Answer:

Question 13.
How can you appreciate the technology of making ships float, using the material which sink in water? (AS 6)
Answer:

• We know that a piece of iron sinks in water.
• The relative density of iron is 8.5 which is many times more than water.
• A ship that floats on water is made up of tonnes of iron. It is really wonderful.
• According to the Archimedes principle of buoyancy, any object can float when its weight is equal to the weight of water displaced by it.
• Hence ships are made with a larger surface area so that it displaces the water, whose weight is equal to the weight of loaded ship.
• It involves very sharp scientific calculations and a large engineering technology.
• Really such type of technologies are highly appreciable and the scientists who formulates all these are also really great.

Question 14.
Can you make iron float? How? (AS 3)
Answer:
Yes, we can make iron to float on water.
Procedure:

1. Take a piece of iron and drop it in a vessel of water.
2. We observe that the iron piece sinks in water.
3. Take a thin foil of iron and fold it into four folds.
4. Drop it in water. It sinks.
5. Now unfold the foil and bend it in the form of a bowl. [You can use an iron tin also]
6. Now drop the bowl in water.
7. If floats on water.

Reason:
The weight of water displaced by iron bowl (iron tin) is less than the weight of the iron bowl (iron tin).

Question 15.
How can you find the relative density of a liquid? (Lab Activity – 2) (AS 3)
Answer:
Aim :
To find the relative density of a liquid.

Materials required :
Small bottle of 50 ml capacity (the bottle should weigh not less than 10 gm), spring balance, any liquid (milk or oil or kerosene) about 50 ml.

Procedure:

1. Find the weight of empty 50 ml bottle.
2. Fill the bottle with water and weigh it.
3. Find the weight of 50 ml water.
4. Remove water from the bottle and fill it with any liquid (say milk).
5. Weigh the bottle with liquid.
6. Weight of 50 ml liquid = Weight of bottle with liquid – Weight of empty bottle

Question 16.
Find the relative density of different fruits and vegetables and make a list. (AS 3)
Answer:

• Follow the procedure in Q.No (11) of A.S – (3).
• Use different fruits and vegetables in the place of wooden block.
• Write the observed values in the following table.

Question 17.
Make a lactometer with ball point refill. What would you do to make the refill stand vertically straight? (Activity – 2) (AS 5)
Answer:

• Take an empty ball pen refill. It should have a metal point.
• Take a boiling tube and fill it with water.
• Put the refill in with metallic point inside the water.
• Use a pen to mark the point on the refill to show the part which is above the water surface.
• Pour out the water from the boiling tube and fill it with milk.
• Float the refill in the milk.
• Puf the record mark on the refill, at the point showing the part which is above the surface of the milk.
• These two marks are not at the same place. This is the improvised lactometer.
• We have to attach a small weight at the bottom of the refill to make it stand vertically straight.

Question 18.
Draw the diagram of a mercury barometer. (AS 5)
Answer:

Question 19.
Write a note on Pascal’s discovery in helping to make hydraulic jacks. (AS 6)
Answer:
Pascal’s principle:
External pressure applied to an enclosed body of fluid is transmitted equally in all directions throughout the fluid volume and the walls of the containing vessel.

Use :

1. This principle is used in designing and working of hydraulic jacks.
2. Hydraulic jacks are useful to lift heavy objects like cars and other vehicles in automobile work shops when the vehicles are to be required or repaired.
3. Here very less force is used to lift such heavy vehicles.

Appreciation :

1. Hydraulic jacks are used not only in automobile work shops, but also in so many industries where heavy weights are to be lifted with a little force.
2. All the comforts which we are enjoying now are the efforts of scientists who discovered the laws and principles.
3. Hence we have to appreciate the efforts of Pascal for his contribution in designing these principles.

Question 20.
Write a note on Archimedes discovery of force of buoyancy. (AS 6)
Answer:
Archimedes principle:
When a body is immersed in a fluid, it experiences an upward force of buoyancy equal to the weight of fluid displaced by the immersed portion of the body.

Use :
The principle is used to determine the purity of metals.

Appreciation :

1. This principle was discovered by Archimedes eventually when he was taking bath.
2. With this principle he could solve the problem assigned by the king to him.
3. So many problems in our life may be solved by so many scientific principles discovered by various scientists.
4. You might have heard about falling of the statue of Budha in Hussain Sagar.
5. That statue was lifted by using the principle of buoyancy.
6. Archimedes is thought to be so important as a mathematician that scientists honoured him.
   a) A large hole or crator on the moon is named after Archimedes.
   b) Some mountains on the moon are called the monte – Archimedes.
7. Hence the efforts of Archimedes in discovering such type of principle may be highly appreciated.

Question 21.
You found the relative densities of some solids and some liquids by doing some activities. List the solids and liquids in increasing order of relative density. (AS 4)
Answer:

Question 22.
Iron sinks in water, wood floats in water. If we tie an iron piece to wood piece of the same volume, buritlle and drop it in water, would bundle sink or float? Make a guess and find out whether your guess is correct or wrong with an experiment. Give reasons. (AS 2, AS 3)
Answer:
The body sinks in water.

Reason :
The combined mass of the system increases, so the combined density also increases. Hence the body sinks in water.

Question 23.
Air brakes in automobiles work on Pascal’s principle. What about air brakes? Collect the information about the working process of air brakes. (AS 4)
Answer:

• Air brakes works on the principle of conversion of energy. Generally in trains, while it is moving -fit .produces kinetic energy. This kinetic energy has to be reduced to make the train stop. Here air is used to reduce kinetic energy by converting it into heat energy.
• The system of air brakes in trains has been shown in the figure.
• The important parts are compressor, main reservoir, driver’s brake value, brake pipe, triple value, auxiliary reservoir, brake cylinders, and brake block.

Working:

1. When the driver placed the value in application position, the air pressure in the brake pipe escapes.
2. The loss of pressure is detected by the slide value in the triple value.
3. Now a connection between the auxiliary reservoir and the brake cylinder has been opened and the air in the auxiliary reservoir feeds through into the brake cylinder.
4. The air pressure forces the piston to move against the spring pressure and causes the brakes to be applied to the wheels.

Question 24.
Where do you observe Archimedes principle in daily life? Give two examples.
Answer:
Daily life application of Archimedes principle :

1. Archimedes principle of buoyancy is applied in our daily life in many ways.
2. Fish, human swimmers, ice bergs and ships float follow Archimedes principle of buoyancy.
3. Rise of balloon in air also follows Archimedes principle.
4. While dragging water from a well, the bucket filled with water seems to be weightless till it reaches the surface of the water in the well. This is also due to buoyancy.
5. Swimming of duck in water is also an example of Archimedes principle.

Question 25.
Where do you observe Pascal’s principle in daily life? Give a few examples.
(OR)
Write any one application of Pascal’s principle in daily life.
Answer:
Daily life application of Pascal’s principle :
Pascal’s principle is applied in the working of

1. Hydraulic jacks
2. Hydraulic lifts
3. Hydraulic pumps
4. Hydraulic cranes
5. Siphons
6. Artesian wells
7. Water towers and dams

9th Class Physical Science 9th Lesson Floating Bodies InText Questions and Answers

9th Class Physical Science Textbook Page No. 141

Question 1.
a) Did kerosene float above the water or did water float above the kerosene?
Answer:
Kerosene floats above the water.

b) Which objects float in kerosene?
Answer:
Plastic buttons, match stick, tiny paper balls, wax, etc. floats in kerosene.

c) Which objects sink in kerosene but float on water?
Answer:
Wax sink in kerosene but float on water.

d) Which objects sink in water?
Answer:
Pins, small pebbles, sand, etc. sink in water.

e) Draw a diagram of the tube, showing the results of your activity.
Answer:

f) Why did different objects behave differently?
Answer:
The density of objects is the main reason to behave differently.

9th Class Physical Science Textbook Page No. 144

Question 2.
1) What is the relative density of wood?
Answer:
0.8

2) What is the relative density of glass?
Answer:
1.2

3) Which is denser, rubber or plastic?
Answer:
Rubber is denser than plastic.

4) Which is denser, wood or cork?
Answer:
Wood is denser than cork.

5) Do objects that have a relative density less than 1 Sink in water or float on it?
Answer:
Float on water.

6) Do the objects that sink in water have a relative density less than 1 or more than 1?
Answer:
More than 1.

7) Classify the above materials as denser than stone and less denser than the stone.
Answer:
Denser than the stone : iron, nails.
Less denser than the stone :
eraser, wood, glass slides, plastic cube, Aluminium, glass marbles, cork.

8) What relationship do you find between the relative density of objects and floating sinking of the objects?
Answer:
If the relative density is less than 1 the object of less density will float above the other in the water.

9th Class Physical Science Textbook Page No. 145

Question 3.
1) Which liquid will float on top if groundnut oil is poured over water?
Answer:
Groundnut oil float on the water.

2) If we put a wooden block in kerosene, will it float or sink? Give reasons for your answer.
Answer:

• Wooden block will sink in the kerosene.
• Because, the density of the wooden block is more than the kerosene.

3) A piece of wax floats in water but the same piece of wax sinks in a liquid say liquid ‘X’. Will the relative density of liquid . ‘X’ be less than 1 or greater than 1? How can you say?
Answer:

• The relative density of liquid ‘X’ is less than 1.
• The density of wax is less than 1 because it floats on the water.
• The density of liquid ‘X’ is less than the density of the wax this means less than 1. because wax sinks in that liquid ‘X’.

4) If we mix some water in milk, will the relative density of the mixture be less than or more than the relative density of milk?
Answer:

• We know that relative density of milk is more than 1 and density of milk is more ‘ than water.
• If we mix some water in milk, the density of the mixture will decreases and the relative density will also be less than the relative density of milk.

5) If we take two bottles of equal volume and pour pure milk in one and milk mixed with water in the other, which one will be heavier?
Answer:

• Pure milk is heavier than water mix milk.
• This is due to density of milk is more than water.

9th Class Physical Science Textbook Page No. 151

Question 4.
Why is the height of mercury column nearly 76 cm in the tube?
Answer:
Air pressure is the weight of air in the atmosphere above the reservoir (bowl of mercury). So, the level of mercury continues to change until the weight of mercury in the glass tube is exactly equal to the weight of the air above the reservoir, which is 76 cm.

9th Class Physical Science Textbook Page No. 154

Question 5.
What happens if we replace this cylindrical liquid column with another object which is made up of a material whose density is equal to the density of liquid?
Answer:
We know that the pressure difference in the liquid,
P₂ – P₁ = hρg

Since F = PA and W = mg
We get F = W (Values of displaced liquid)

1. Here F’ is the force applied on the object and ‘W’ is the weight of the liquid.
2. So, the”force applied on the object by the liquid is equal to the weight of the displaced liquid.

9th Class Physical Science Textbook Page No. 155

Question 6.
Why does the stone lose weight when it is immersed?
Answer:

• Suspend a stone from a spring balance. Note the reading of the spring balance.
• Take a beaker half-filled with water.
• Now immerse the stone in the water, note the reading of the spring balance.
• We notice that the stone, when immersed appears to lose some weight.
• The immersed stone appears to lose weight because the force of buoyancy.
• Thus the apparent loss of weight must be equal to the force of buoyancy acting on the immersed stone.

9th Class Physical Science Textbook Page No. 142

Question 7.
Let us suppose you have two blocks and you do not know what material they are made of. The volume of one block is 30 cm3 while the other is 60 cm3. The second block is heavier than the first. Based on this information, can you tell which of the two blocks is denser?
Answer:
No, we cannot say which of the two blocks is denser, because any one of the quantity
i. e., either volume or weight must be same.

9th Class Physical Science Textbook Page No. 153

Question 8.
a) What would happen if Toricelli’s experiment is done on moon?
b) A stopper is inserted in the small hole of the glass tube of the mercury barometer below the top level of the mercury in it. What happens when you pull out the . stopper from the glass tube?
c) Why don’t we use water instead of mercury in Toricelli experiment? If we are ready to do this experiment, what length of tube is needed?
d) Find the weight of the atmosphere around the earth (take the radius of earth as 6400km)
Answer:
a) If Toricelli’s experiment had been done on moon, the height of mercury column will be zero. Because there is no atmosphere on the moon.

b) The mercury level does not change, because there is vacuum above the mercury level. Hence no pressure is on the mercury. Also, the weight of mercury column must be equal to the air pressure above the reservoir. Hence there will be no change in the height of the mercury column.

c) We cannot use water instead of mercury in Toricelli experiment because, if we want to use water, we have to take the glass tube of length nearby above 10 m, which is inconvenient.
If water is taken,
P₀ = ρhg
1 01 yin5
1.01 × 10⁵ = 1 × 10³ × h × 9.8 ⇒ h = = 0.1030 × 10² = 10.3 m

d) Weight of atmosphere = Atmospheric pressure x Surface area of the earth

9th Class Physical Science Textbook Page No. 157

Question 9.
a) Why is it easier for you to float in saltwater than in freshwater?
b) Why is there no horizontal buoyant force on a submerged body?
c) Two solid blocks of identical size are submerged in water. One block is iron and the other is aluminium. Upon which is the buoyant force greater?
d) A piece of iron when placed on a block of wood, this makes the wood to float lower in the water. If the iron piece is suspended beneath the wood block, would it float at the same depth? Or lower or higher?
Answer:
a) Salt water is denser than freshwater.
b) Buoyant force is the upward force only. The body is submerged means, its weight is more than the buoyant force. Here there will be no horizontal buoyant force.
c) Buoyant force on iron block is more than that of aluminium block.
d) It floats on higher depth than in the first case.

9th Class Physical Science 9th Lesson Floating Bodies Activities

Activity – 1

Question 1.
Comparing density – relative density.
Take two test tubes of the same size and fill one to the brim with water and the other with oil.
a) Which will weigh more?
Answer:
The test tube with oil will weigh more.

b) Which liquid is denser?
Answer:
Oil is denser than water.
Take two equal sized blocks made of wood and rubber.

c) Which of these two blocks is heavier?
Answer:
Wooden block is heavier.

d) Which one is denser?
Answer:
Wooden block is denser than rubber.

Activity – 3

Question 2.
Do the objects denser than water float in it? Prove it with an activity.
Answer:

• Collect some objects listed in the table below.
• Place them one by one in a glass of water and observe whether they sink or float in water.
• Record your observations in the table.

Observations:

• Find the relative densities of all the objects.
• We observe that some objects floats and some objects like iron nail, glass marble, stone sinks in water.
• Geometry box made of iron floats on water, though its relative density is greater than 1. This is due to its larger surface area.
• Geometry box made of iron floats on water though it is made up of a substance denser than water.
•  We can say that the floating or sinking of an object does not depend not only on its relative density, but also its surface area which displaces the water.

Activity – 4

Question 3.
Show that for a floating object, the weight of the object is equal to the weight of water displaced by it.
Answer:

• Take a beaker and weigh it. Note down its weight.
• Fill water in an overflow jar, wait until the water stops dripping from the outlet of the overflow jar.
• Now place the beaker below the outlet of the overflow jar.
• Take a wooden block, moisten it with water and gently drop it into the overflow jar.
• Water will flow out of the overflow jar and collects in the beaker kept under the overflow jar.
• Measure the weight of beaker with water.
• Subtract the weight of beaker from this. The value gives the weight of water displaced by wooden block. Note it.
• Now remove wooden block from the overflow jar, make it cry and weight.
• We can observe that the weight of wooden block is equal to weight of water displaced by it.
• Do the same experiment with some other substances and record your observations in the table.

Activity – 5

Question 4.
Explain how does metal aluminium floats on the water with an activity.
Answer:

• Take a small sheet of aluminium foil.
• Fold it four or five times, pressing the foil tight after each fold.
• We can find that the aluminium foil will sink in the water.
• Now unfold the aluminium foil and make it as a small bowl.
• We can find that the bowl will float on the water.
• The metal bowl displace larger amount of water than a metal piece.
• Weight of the displaced water is more than the metal sheet.
• So, bowl will float on the water.

Activity – 6

Question 5.
Prove that the water exerts upward force on objects.
Answer:

1. Take an empty plastic bottle.
2. Put the cap on it tightly.
3. Place the bottle in a bucket of water.
4. The bottle will float.
5. Push the bottle into the water by your hand as shown in figure.
6. We feel some upward force.
7. Try to push it further down. We feel increase in the upward force.
8. Now release the bottle.
9. It bounce back to the surface of water.
10. Here the upward force of water a real, observable force.
11. This force acting on unit area of the surface of an object is called static pressure of the water.

Activity – 7

Question 6.
Describe an activity to observe the air pressure.
Answer:

1. Take a glass tumbler.
2. Stick some cotton at the bottom of it.
3. Immerse it inversely in water upto the bottom of the container as shown in figure.
4. Take out the tumbler from water.
5. We observe that the cotton attached at the bottom of the tumbler is not wet.
6. This is due to the force of air which is applied on water by the air present in the tumbler and stops water from entering the tumbler.
7. This force on unit area of water is the pressure of air.

Activity – 8

Question 7.
How can you measure the force of buoyancy and how much?
(OR)
Why does and stone lose weight when it is immersed in water?
Answer:

• Suspend a stone from a spring balance.
• Note that the reading of the spring balance.
• The reading gives the weight of the stone.
• Take a beaker half filled with water.
• Now immerse the stone in the water.
• Note the reading of the spring balance.
• The reading gives the weight of the immersed stone.
• We may notice that the stone, when immersed, appears to lose some weight.
• The immersed stone appears to lose weight because the force of BUOYANCY, exerted on the stone by the water in the upward direction.
• Thus the apparent loss of weight must be equal to the force of buogancy acting on the immersed stone.
• The lose of weight of stone is equal to weight of the water displaced by the stone, this is the force of buoyancy.

Activity – 9

Question 8.
‘State and prove Archimedes principle of buoyancy.
Answer:

Archimedes principle :
Archimedes principle states that when a body is immersed in a fluid, it experiences an upward force of buoyancy equal to the weight of fluid displaced by the immersed portion of the body.

Proof:

1. Suspend a stone from a spring balance.
2. Note the reading on the spring balance. This reading is the weight of the stone.
3. Take an overflow jar with water and place a graduated beaker below the beak as shown in the figure.
4. Now immerse the stone in the water and note the reading on the spring balance.
5. Measure the volume of the water that overflows into the graduated beaker.
6. The reading of the spring balance gives the weight of the immersed stone.
7. The beaker reading gives the volume of water displaced by the stone.
8. The difference in the two readings of spring balance gives the apparent loss of weight of the stone.
9. Now weigh the water in the graduated beaker.
10. We observe that the apparent loss of weight of the stone is equal to the weight of the water displaced by the stone.
11. Hence Archimedes principle is proved.

AP State Syllabus AP Board 9th Class Physical Science Solutions Chapter 7 Reflection of Light at Curved Surfaces Textbook Questions and Answers.

AP State Syllabus 9th Class Physical Science Solutions 7th Lesson Reflection of Light at Curved Surfaces

9th Class Physical Science 7th Lesson Reflection of Light at Curved Surfaces Textbook Questions and Answers

Improve Your Learning

Question 1.
“Laws of reflection of light are not applicable to curved surfaces” Is it correct?
Answer:
No, laws of reflection of light are applicable to curved surfaces also.

Question 2.
How do you find the focal length of a concave mirror?
(OR)
What is the name given to distance between pole and focal point? How would you able to find that distance by using an activity? That distance is called focal length.
Answer:

To find the focal length of a concave mirror –

1. Hold a concave mirror perpendicular to the direction of sunlight.
   Take a small paper and slowly move it in front of the mirror and find out the point where you get smallest and brightest spot, which is image of the sun.
   The rays coming from sun parallel to the concave mirror are converging at a point.
   This point is called Focus or focal point (F) of the concave mirror.
   Measure the distance of the spot from the vertex (point P) of the mirror.
   This distance gives the focal length of the concave mirror.

Question 3.
Where will the image form when we place an object, on the principal axis of a concave mirror at a point between focus and centre of curvature?
Answer:

• When we place an object on the principal axis of a concave mirror at a point between focus and centre of curvature then the image will form at a point beyond the centre of curvature.
• The image is enlarged, inverted and real.

Question 4.
Find the distance of the image, when an object is placed on the principal axis at a distance of 10 cm in front of a concave mirror whose radius of curvature is 8 cm.
(OR)
If the object is placed on the principal axis at a distance of 10 cm in front of a concave mirror with curvature is 8 cm, what is the distance of the image?
Answer:
Object distance u = – 10 cm
Radius of curvature (R) = – 8 cm
∴ Focal length f = \(\frac{R}{2}=\frac{-8}{2}\) = – 4 cm
Image distance = v = ?

∴ The image distance = 6.67 cm.
i.e., Real image is formed at same side of the mirror.

Question 5.
State the differences between convex and concave mirrors.
(OR)
Distinguish between convex and concave mirrors.
(OR)
Ramu observed a mirror which is used by drivers to see the vehicles behind. Srinu observed a mirror which is used by dentist to see teeth inside the mouth. What are those two mirrors and distinguish them?
Answer:
Those mirrors are (1) Convex, (2) Concave

Question 6.
Distinguish between real and virtual images.
(OR)
Varun sees his image in a concave mirror. When he moves away he was unable to see his image. Write the differences between those two images.
Answer:

Question 7.
How do you get a virtual image using a concave mirror?
(OR)
Ramu moved an object towards a concave mirror. When he reached certain point he was able to see his image. Where does he had placed the object inorder to see the object in the mirror?
Answer:

• Place the object at a distance less than the focal length of the mirror.
• Draw ray diagram as shown in the figure.
• When the rays are extended, they seem to be coming out from a point on the other side of the mirror. (G)
• We cannot find the image on the screen, but we can see the image in the mirror. (IG)
• This is the virtual, erect and magnified image.

Question 8.
What do you know about the terms given below related to the spherical mirrors?
a) Pole
b) Centre of curvature
c) Focus
d) Radius of curvature
e) Focal length
f) Principal axis
g) Object distance
h) Image distance
i) Magnification
Answer:
a) Pole :
The centre of the spherical mirror is called pole of the mirror.

b) Centre of curvature :
The normal from a concave mirror converge at a point, that point is called centre of curvature.

c) Focus :
A point on the principal axis where a beam of light parallel to the principal axis either converges into or appears to diverge from after reflection from the mirror is called focus (or) focal point (F).

d) Radius of curvature :
The distance between pole and centre of curvature is called radius of curvature. It is denoted by ‘R’.
R = 2f

e) Focal length :
The distance of the focus from the pole is called the focal length (f) of the mirror.

f) Principal axis :
The horizontal line which passes through the centre of curvature and pole.

g) Object distance :
The distance between object and mirror is called object distance (u).

h) Image distance :
The distance between image and mirror is called image distance (v).

i) Magnification :
The ratio of size of image to the size of object is known as magnification.

Question 9.
Write the rules for sign convention.
(OR)
Vinay wants to solve a problem related to mirrors. What are rules to be followed by him in order to solve the problem?
Answer:

Sign conventions used in mirror equation :

1. All distances should be measured from the pole.
2. All distances measured in the direction of incident ray are to be taken as positive.
3. The distances measured in the direction opposite to incident ray are to be taken as negative.
4. Height of object (h₀) and height of image (h_i) are positive, if measured above principal axis and negative if measured below principal axis.

Question 10.
The magnification produced by a convex mirror is – 1. Do you agree it?
Answer:
The magnification produced by a convex mirror is – 1.1 agree with this statement.

Question 11.
Imagine that spherical mirrors were not known to human beings. Guess the consequences.
(OR)
What are the consequences occur when the spherical mirrors were not known to human beings?
Answer:

• Spherical mirrors are used by human beings in their daily life.
• Convex mirrors are used in hospitals, hotels, stores, apartments, roads, ATMs, computer monitors, rare-view mirrors in vehicles, and also in sun glasses,
• Whereas concave mirrors are used by ENT specialists, dentists to see inner parts of mouth,
• Car head lights are used to get parallel beam of light,
• Solar cookers are also made from concave mirrors, So everywhere we are using these spherical mirrors,
• So without these spherical mirrors the life of human beings is miserable,

Question 12.
By observing steel vessels and different images in them, Surya, a third class student, asked his elder sister Vidya some questions. What may be those questions?
(OR)
By observing the image on steel vessels, Surya asked his sister some questions. What may be those questions?
Answer:

• Why do we see our image in some vessels which bulged outwards?
• Why are we not able to see our image in some vessels which bulged inwards?
• Which type of vessels are useful to reflect light on wall and form our image on wall?
• Can we see the images in all vessels?
• Is these images real or virtual?
• The vessels which are bulged inside behaves like which type of mirrors?

Question 13.
How do you form a diminished image by a concave mirror on a screen?
Answer:
If we place an object beyond centre of curvature of a concave mirror, diminished image will be formed between focus and centre of curvature,

Question 14.
How do you find the focal length of a concave mirror in the lab?
Answer:
Aim :
To find out focal length of a concave mirror,

Material required :
A candle, paper, concave mirror, V-stand, measuring tape, or meter scale.

Procedure:
1) Place the concave mirror on V-stand, a candle, and meter scale as shown in figure.

2) Keep the candle at different distances from the mirror (10 cm to 80 cm) along the axis and by moving the paper (screen) find the position where you get the sharp image on paper.

3) Note the distance of candle from the mirror (u) and distance of the image from the mirror (v) in the given table.
Here u = object distance, v = image distance, f = focal length.

4) Find the average of T. This value of T is the focal length of the given concave mirror.

Question 15.
What do you infer from the experiment which you did with concave mirrors and measured the distance of object and distance of image?
Answer:
I observed the following points from the experiment with concave mirrors, to form images at different places, they are :

1. As the object distance increases, the image distance decreases.
2. As the object distance increases, the size of image decreases.

Question 16.
Draw a ray diagram to show the process of converging of the parallel beam by a concave mirror by taking four parallel incident rays.
Answer:

Question 17.
Draw suitable rays by which we can guess the position of the image formed by a concave mirror.
Answer:
Suitable rays by which we can guess the position of the image formed by a concave mirror.

Rule 1 :
All the rays, that are parallel to the axis, get reflected such that they pass through the focal point of the mirror.

Rule 2 :
A ray that passes through the focal point of the mirror will, travel parallel to the axis after reflection.

Rule 3 :
A ray, passing through the centre of curvature of the mirror and incidents on the mirror, after reflection will travel along the same line, but in opposition direction.

Question 18.
Show the formation of image with a ray diagram, when an object is placed on the •principal axis of a concave mirror away from the centre of curvature.
Answer:

Question 19.
Make a solar heater / cooker and explain the process of making.
(OR)
Which cooker uses solar energy to cook food and explain its making?
(OR)
Explain the construction and working of a solar cooker.
Answer:

Construction:

1. Make a wooden / iron frame in dish shape. It is solar cooker.
2. Cut the acrylic mirror sheets into 8 or 12 pieces in the shape of isosceles triangles with a height equal to the radius of a dish antenna.
3. The bases of 8 or 12 triangles together make the circumference of the dish.
4. Stick the triangle mirrors to the dish as shown in figure.
5. Thus solar heater/cooker is ready.

Working:

1. Face it towards the sun.
2. Find its focal point and place a vessel at that point.
3. It will get heated. Everyone cook rice in that vessel.

Question 20.
To form the image on the object itself, how should we place the object in front of a concave mirror? Explain with a ray diagram.
(OR)
Draw ray diagram to explain the image formation on the concave mirror if the object is in front of it.
Answer:

• The object should be placed on centre of curvature of concave mirror.
• When the object placed at centre of curva¬ture after reflection from the mirror the light rays converge at centre of curvature.
• So a real and inverted image is formed on the object itself. Ray diagram

Question 21.
How do you appreciate the role of spherical mirrors in our daily life?
(OR)
Write the usage of spherical mirror in daily life situations.
Answer:

• Spherical mirrors are useful in our daily life in many ways.
• Convex mirrors are used as rear view mirrors in cars, scooters, buses, etc. This helps us to see the traffic behind the vehicle, which avoids accidents while taking turns.
• Big convex mirrors are used as shop security mirrors.
• Concave mirrors are used by dentists, opthamologists, to see the smaller parts of teeth, eyes, and ears.
• Concave mirrors are also used in solar heating devices.
• Concave mirrors are used as shaving mirrors to see a large image of the chin (or) face.
• Concave mirrors are used as doctor’s head mirrors to focus light coming from a lamp on to the body parts of the patient to be examined by the doctor.
• So, I appreciate the role of spherical mirrors in our daily life.

Question 22.
How do doctors use concave mirrors?
Answer:

• Dentists and ENT specialist doctors use concave mirrors to get a magnified clearer view of the teeth, skin and ear cavities.
• A concave mirror converges the rays of light falling on it in a direction parallel to the principal axis, on to its focus. This is the reason why ENT specialists use it as a head mirror.

Question 23.
Why do we prefer a convex mirror as a rear-view mirror in the vehicles?
Answer:

• Convex mirror always gives an erect and diminished image.
• The image distance will be smaller than the object distance.
• A convex mirror has a wider field of view than a plane mirror.
• Hence we prefer convex mirror as rear-view mirror in the vehicles, as we have to observe the vehicles which are at the back of our vehicle.

Question 24.
Complete the table-1 which is related to experiment done by a concave mirror.

Answer:

From the given data, m = \(\frac{40 \mathrm{~cm}}{40 \mathrm{~cm}}\) cm = 1.
Hence, R = 40 cm, f = 20 cm.

Question 25.
How can you show the diverging and the converging of light by using laser lights?
Answer:
Aim :
To show diverging and converging of light by using laser lights.

Material required :
Concave mirror, convex mirror, laser lights-2, screen, V-stand, Agarbathi.

Procedure:

• Place a concave mirror on a V-stand and place the V-stand on a table.
• Take two laser lights.
• Focus the light rays of laser lights parallel to the axis of the concave mirror.
• The light rays (beams) incident on the concave mirror are reflected back and converge at one point.
• Place the screen to catch that converging point of reflected light rays.
• Light a Agarbathi near the table.

Observation :
We can observe path of the incident and reflected rays clearly in the smoke of Agarbathi.

• Now place a convex mirror on the stand.
• Again pass the laser light rays parallel to the principal axis at the convex mirror.

Observation :
We can observe diverging light rays. We cannot catch any converging point on the screen.

Question 26.
Collect information about the history of spherical mirrors in human civilization. Display it in your classroom.
Answer:

• The idea of mirror came into existence long back when people saw their images in water, on poluted surfaces, etc.
• The history of mirrors starts in the 6000 BC. The earliest man made mirrors were pieces of poluted stone.
• The first glass mirrors were produced in 1A.D. by Romans.
• The first modern silver-glass mirror was created by Justus Von Liebig, a German chemist in 1835.
• The invention of glass blowing method during the 14th century led to the discovery of spherical mirrors, which increased the popularity of glass mirrors.
• By the end of 18^th century, decorative mirrors have widely used.
• New, cheaper techniques of mirror production in the 19th century led to a great proliferation in their use.

Question 3.
Think about the objects which act as a concave or convex mirrors in your surroundings. Make a table and display it in your classroom.
(OR)
Collect some objects from your surroundings that act as concave, convex mirrors and display them your classroom.
Answer:
Object used as convex and concave mirror in our surroundings.

Question 4.
How will our image be in concave and convex mirrors? Collect photographs and display in your classroom.
Answer:

• In concave mirrors our image is thin and enlarged.
• As we move away from mirror the image will be diminished and become pointed at the pole.
• In convex mirror, our image is bulged and size of image is diminished.
• As we move away from the mirror, the image is further diminished.

Question 5.
How do you appreciate the use of reflection of light by a concave mirror in making of TV antenna dishes?
Answer:

• The parabolic shape of dish reflects the signals to the dish’s focal point.
• Mounted on brackets at the dish’s focal point is a device called a feed horn.
• This feed horn is a wave guide that gathers the signals at or near the focal point and conducts them to a LNB (Low-Noise Block down converter).
• The LNB converts these electromagnetic waves into electrical signals and shifts to the receiver i.e. T.V. set.
• This is all possible only with the help of parabolic dish antennas (concave shape).
• We know, the T.V. plays an important role in our daily life. So the inventions of the utilities of concave shapes are more helpful for us.

Question 6.
A convex mirror with a radius of curvature of 3 m is used as rear view in an automobile. If a bus is located at 5 m from this mirror, And the position, nature and size of the image.
Answer:

Position = 1.15 m behind the mirror. Nature = Virtual and erected.
∴ Image is erect and smaller in size by a factor of 0.23.

Question 7.
An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
(OR)
An object 5 cm high is placed at a distance of 10 cm from a convex mirror, the focal length of the mirror is 15 cm. Find the nature, position and size of the image.
(v = +6m ; m = 0.6, h₂ = 3 cm)
Answer:
Given, u = – 10 cm ; f = 15 cm ; v =?

So the Image is virtual and seen in the mirror.
Magnification m = \(-\frac{v}{u}=\frac{-6}{-10}=\frac{6}{10}\) = 0.6
Position : 6 cm behind the mirror.
Nature : Virtual and erect image.
So the image is erect and diminished.

Question 8.
Write answers to the following questions based on the data given in the table – 3.
1) What changes will come gradually in the size of an image, when we move an object away from the concave mirror?
2) In which situation inverted images formed by a concave mirror?
3) If the centre of curvature of a concave mirror is 10 cm, where the object should be placed to get an image at centere of curvature?
Answer:
1) a) If we move an object from a concave mirror to its focus (F), we will get a virtual image.
b) If we place the object at ‘F, the size of the image is infinity.
c) If we move, the object from ‘F to centre of curvature the size of the image is decreases gradually but the size of the image is bigger than that of the size of the object.
d) At ‘C the size of the image is same as the size of the object.
e) From ‘C’ to infinity distance the size of the image is smaller than the siz4 of$he object and decreases gradually.
f) Hence, the size of the image decreases if we move an object away from the focus fo the concave mirror.

2) If we place an object beyond focus (F) inverted image will be formed.

3) At 10 cm distance (or) at u = 10 cm.

9th Class Physical Science 7th Lesson Reflection of Light at Curved Surfaces InText Questions and Answers

9th Class Physical Science Textbook Page No. 111

Question 1.
Did you burn a paper by using a magnifier?
Answer:
Yes.

Question 2.
What causes burning of the paper, while doing it?
Answer:
Converging light rays through magnifier at one point on the paper.

Question 3.
Can you use a plane mirror to burn a paper instead of a magnifier lens?
Answer:
Cannot.

Question 4.
Can’t? Why?
Answer:
Plane mirror cannot converge light rays.

Question 5.
Which type of mirrors may be used to converge light rays?
Answer:
Concave mirrors.

Question 6.
Which type of reflected surface used by Archimedes?
Answer:
Curved reflected surfaces.

9th Class Physical Science Textbook Page No. 112

Question 7.
Can we burn a paper by using light rays of the torch light?
Answer:
Cannot.

Question 8.
Can’t? Why?
Answer:
The light rays coming from the torch light are not parallel rays. Hence, they cannot converge at one point on the paper with high intensity.

Question 9.
Which type of incident rays are required to get high intensity and point sized converging point?
Answer:
Parallel light rays.

Question 10.
Which line is taken as a base to draw a reflected ray to the given incident ray?
Answer:
Normal taken as base.

9th Class Physical Science Textbook Page No. 114

Question 11.
Does a concave mirror from image at focus in all situations?
Answer:
No

9th Class Physical Science Textbook Page No. 114

Question 12.
See the given figure, a set of parallel rays are falling on a convex mirror. What conclusions can you draw from this?

Answer:
The parallel rays are diverging after reflection. If we extend the reflected rays backwards, they meet at F which is focal point of convex mirror. The image formed is virtual, straight and point sized.

Question 13.
Will you get a point image if you place a paper at the focal point?
Answer:
We can get highly enlarged image at infinity when we place paper at focal point.

9th Class Physical Science Textbook Page No. 118

Question 14.
How can you find the focal length of a convex mirror?
Answer:

• The focal length of a convex mirror can be determined by introducing a convex lens between the object and the convex mirror.
• An image can be obtained with the help of a convex lens side by side with object when the convex mirror reflects the rays along the same path.
• Then, the radius of curvature (R) of the mirror is the distance between the screen and the mirror.

• The focal length ‘f of the convex mirror is calculated using the formula R (Radius of curvature)
  

Question 15.
Can we get a magnified image by a convex mirror?
Answer:
No, convex mirror cannot form magnified image.

9th Class Physical Science Textbook Page No. 119

Question 16.
How can one see an image formed on the object itself?
Answer:
When an object placed at centre of curvature the image should form at C in inverted position. So the image formed on the object itself.

Question 17.
Do you get an image when object is placed at F? Draw a ray diagram. Do the experiment.
Answer:

When an object placed at F, we can get the image at an infinite distance.

9th Class Physical Science 7th Lesson Reflection of Light at Curved Surfaces Activities

Activity – 1

1. Observation the light rays on concave and convex surfaces.
Answer:
Procedure :

1. Take a rectangular shaped (3″ x 6″) Acrylic sheet.
2. Hold the Acrylic sheet with your palm without bending it.
3. Ask your friend to focus light with a torch light on the sheet.
4. Now adjust the sheet as reflected rays are caught on a wall.
5. Now bend slowly your palm as the sheet bends inwards in the shape of concave.
6. Again ask your friend to focus the light.
7. Now bend slowly your palm as the sheet bends outwards in the shape of convex.
8. Again ask your friend to focus the light.

Observation :

1. In the first case, when without bending the sheet, reflected light is not converged at one place. Because the sheet was acted as a plane mirror.
2. In the second case, reflected light converged at one place. Because the acrylic sheet is bent inwards and it is acted as a concave mirror.
3. In the third case, reflected light rays are not converged, of course they diverged with low intensity. Because the sheet was acted as a convex mirror.

Conclusion :
Concave surfaces converge light rays and convex surfaces diverge.

Activity – 2

Question 2.
Show that a powerful source far away is needed to get parallel rays of light.
Answer:

1. Stick two pins on a thermocole block.
2. The pins are exactly parallel to each other.
3. When a source of light is kept very near, we see the shadows diverging (from the base of the pins).
4. As we move the source away from the pins, the angle of divergence gets reduced.
5. If we move the source far away we will get parallel shadows. But the light intensity reduces.
6. That means to get a beam of parallel rays all that we need is a powerful source far away.

Activity – 3

Question 3.
Identify the focus or focal point of concave mirror.
(OR)
How do you able to find the focal point of a given concave mirror? Explain it with an activity.
Answer:

1. Hold a concave mirror perpendicular to the direction of sunlight.
2. Take a small paper and slowly move it in front of the mirror and find out the point where we get smallest and brightest spot image of the sun.
3. The rays coming from the sun parallel to the concave mirror are converging at a point.
4. This point is called Focus or focal point (F) of the concave mirror.

Activity – 4

Question 4.
Find the normal to a curved surface.
(OR)
How do you find normal to a curved surface ? What is the name given to wavering point of normals, explain.
Answer:

• Take a small piece of thin foam or rubber (like the sole of a slipper).
• Put some pins in a straight line on the foam.
• All these pins are perpendicular to the plane of foam.
• If the foam is considered as a mirror, each pin would represent the normal at that point.
• Any ray incident at the point where the pin makes contact with the surface will reflect at the same angle as the incident ray made with the pin-normal.
• If we bend the foam piece inwards, we will notice that all the pins tend to converge at a point.
• If we bend the foam piece outwards, the pins seem to move away from each other or they diverge.
• This gives us an idea about spherical mirror.
• A concave mirror will be like the rubber sole bent inwards (fig. b) and the convex mirror will be like the rubber sole bent outwards (fig. c).

• For a concave mirror, like these pins all normals will converge towards a point. This point is called centre of curvature (C) of the mirror.
• For the ray R, the incident angle is the angle it makes with the radius (normal) shown as Zi and the reflected angle is shown as Zr. We know by first law of reflection Zi = Zr.

Lab Activity

Question 5.
Write the experimented method in measure the distances of object and image using concave mirror. And write the table for observations.
(OR)
Write an experimental activity which gives the information about types of images and measuring the object distance and image distance.
Answer:
Aim :
Observing the types of images formed by a concave mirror and measuring the object distance and image distance.

Material required :
A candle, paper, concave mirror, V-stand, measuring tape or meter scale.

Procedure:

1) Place the concave mirror on V-stand, a candle and meter scale as shown in figure.
2) Keep the candle at different distances from the mirror (10 cm to 80 cm) along the axis and by moving the paper find the position where we get sharp image on paper.

3) Note down the observations in table -1

4) Since we know the focal point and centre of curvature, we can re-classify the above observations as shown in the table – 2.

5) Then note down the observations in table – 2

AP State Syllabus AP Board 9th Class Physical Science Solutions Chapter 3 Is Matter Pure? Textbook Questions and Answers.

AP State Syllabus 9th Class Physical Science Solutions 3rd Lesson Is Matter Pure?

9th Class Physical Science 3rd Lesson Is Matter Pure? Textbook Questions and Answers

Improve Your Learning

Question 1.
Which separation techniques will you apply for the separation of the following? (AS 1)
a) Sodium chloridfe from its solution in water.
b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride.
c) Small pieces of metal in the engine oil of a car.
d) Different pigments from an extract of flower petals.
e) Butter from curd.
f) Oil from water.
g) Tea leaves from tea.
h) Iron pins from sand.
i) Wheat grains from husk.
j) Fine mud particles suspended in water.
Answer:

Question 2.
Write the steps you would use for making tea. Use the words given below and write the steps for making tea. (AS 7)

Answer:

• Take a cup of milk (solvent) in a tea kettle.
• Add one table spoon of sugar (solute), and one table spoon of tea powder (insoluble) to the solvent.
• Heat the tea kettle on the stove.
• The sugar (solute) dissolves in the milk (solvent) and the tea powder remains undissolved.
• Now filter the solution so formed.
• The filtrate is the tea (solution).
• The residue remained in the sieve is the insoluble component of tea powder.

Question 3.
Explain the following giving examples. (AS 1)
a) Saturated solution
b) Pure substance
c) Colloid
d) Suspension
Answer:
a) Saturated solution :
When no more solute can be dissolved in the solution at a certain temperature, it is said to be a saturated solution.

In a saturated solution, equilibrium with the undissolved solute at a certain temperature.

Ex :

1. Take 50 ml of water in a cup.
2. Add one spoon of sugar to the cup and stir still it dissolves.
3. Keep on adding sugar to the water in the cup and stir till no more sugar can be dissolved.
4. The solution so formed is a saturated solution.

b) Pure substance :
A substance is pure i.e., homogeneous if the com-position doesn’t change, no matter which part of the substance we take for examination.

Ex :

1. Take a small part of pure gold biscuit as a sample.
2. The composition is found to be same throughout it.

c) Colloid :
Colloids are heterogeneous mixtures in which the particle size is too small to be seen with the naked eye, but is big enough to scatter light.
Ex : Milk, butter, cheese, cream, gel, etc.

d) Suspension :
Suspension is a heterogeneous mixture in which the solute particles didn’t dissolve and the particles are visible to naked eye.
Ex : Syrups, chalk powder mixed with water, etc.

Question 4.
Classify each of the following as a homogeneous or heterogeneous mixture. Give reasons. (AS 1)

Answer:

Question 5.
How would you confirm that a colourless liquid given to you is pure water? (AS 1)
Answer:

• Observe the smell. We should not find any smell.
• Observe with a naked eye we should not find any suspended particles or fumes or air bubbles.
• Pass a beam of light. It should not scatter.
• The temperature should be normal.
  Then the given colourless liquid is pure water.

Question 6.
Which of the following materials fall in the category of a “pure substance”? Give reasons. (AS 1)
a) Ice
b) Milk
c) Iron
d) Hydrochloric acid
e) Calcium oxide
f) Mercury
g) Brick
h) Wood
i) Air
Answer:

• Except brick and wood remaining materials given in the list can be treated as pure substances.
• Take any small part of ice, milk, iron, hydrochloric acid, calcium oxide, mercury and air and test for their components.
• We find that the composition is same throughout them.

Question 7.
Identify the solutions among the following mixtures. (AS 1)
a) Soil
b) Sea water
c) Air
d) Coal
e) Soda water
Answer:
The solutions are : sea water, air and soda water.

Question 8.
Which of the following will show “Tyndall effect”? How can you demonstrate “Tyndall effect” in them? (AS 1, AS 3)
a) Salt solution
b) Milk
c) Copper sulphate solution
d) Starch solution
Answer:
Milk shows Tyndall effect.

Demonstration :

1. Prepare the milk, copper sulphate, salt and starch solutions in different beakers.
2. Allow a beam of light through each of them.
3. The path of the light beam is clearly visible to us through milk.
4. The path of the light beam is not visible through remaining.
5. This experiment will be effective if it is performed in a dark room.

Question 9.
Classify the following into elements, compounds and mixtures. (AS 1)
a) Sodium
b) Soil
c) Sugar solution
d) Silver
e) Calcium carbonate
f) Tin
g) Silicon
h) Coal
i) Air
j) Soap
k) Methane
l) Carbondioxide
m) Blood
Answer:

Question 10.
Classify the following substances in the below given table. (AS 1)

Answer:

Question 11.
Take a solution, a suspension, a colloidal dispersion in different beakers. Test whether each of these mixtures shows the Tyndall effect by focusing a light at the side of the container. (AS 3)
Answer:

• Take sugar solution (solution), starch solution (suspension) and milk solution (colloidal dispersion) in three different beakers.
• Focus a beam of light by torch or a laser beam at the side of each container and observe.
• We can see that the path of beam of light is clearly visible through all the solutions.
• Hence all the three solutions show ‘Tyndall effect”.

Question 12.
Draw the figures of arrangement of appatus for distillation and fractional distillation. What do you find the major difference in these apparatus? (AS 5)
Answer:

The main difference between these two apparatus is that a fractionating column is fitted in between the distillation flask and the condenser.

Question 13.
Determine the mass by mass percentage concentration of a 100 g salt solution which contains 20 g salt. (AS 1)
Answer:
Mass of salt = 20 g; Mass of salt solution = 100g

Question 14.
Calculate the concentration in terms of mass by volume percentage of the solution containing 2.5 g potassium chloride in 50 ml of potassium chloride (KCl) solution. (AS 1)
Answer:
Mass of potassium chloride = 2.5 g
Volume of potassium chloride solution = 50 ml

9th Class Physical Science 3rd Lesson Is Matter Pure? InText Questions and Answers

9th Class Physical Science Textbook Page No. 40

Question 1.
Can you give few more examples of this kind?
Answer:
Some more examples of homogeneous mixtures are sugar solution, lemon squash, fruit juices, syrups and tonics used in medicine, etc.

Question 2.
Can you prove this with an experiment?
Answer:

• Take some thick milk in a test tube.
• Pass a beam of light from torch or a laser light.
• We cannot observe the path of light through the solution.

Question 3.
If the solution is diluted, can the path of light be visible?
Answer:

• Take some thick milk in a test tube.
• Dilute it by adding some water.
• Now pass a beam of light from torch or a laser light.
• We cannot observe the path of light through the solution.

Question 4.
What would happen if you add a little more solute to a solvent?
Answer:
The solution becomes concentrated.

Question 5.
How do you determine the percentage of the solute present in a solution?
Answer:

• Take 100 ml of water in a beaker.
• Take 50 g. of sugar in a plate.
• Add a spoon of sugar to water and stir it tell the sugar dissolve in water.
• Go on adding sugar till you reach a situation that the sugar cannot be dissolved in water.
• Now weigh the sugar remained in the plate.
• Subtract thin weight from 50 g. The weight so obtained is dissolved in water.
• Hence the maximum amount of solute present in 100 ml of solvent is the percentage of solute (solubility).

9th Class Physical Science Textbook Page No. 44

Question 6.
Did you ever observe this phenomenon in the cinema halls?
Answer:
In cinema halls when we observe the projector while the movie is running, we can observe the phenomenon of “Tyndall effect”. We can see the beams of light in which dust particles also observed.

9th Class Physical Science Textbook Page No. 46

Question 7.
Is the mixture heterogeneous? Give reasons.
Answer:
The mixture of ammonium chloride and salt is a heterogeneous mixture. Even though these two are white in colour their particles do not mix.

Question 8.
How do we separate the salt and ammonium chloride?
Answer:
We can separate the salt and ammonium chloride by the method of sublimation.

9th Class Physical Science Textbook Page No. 49

Question 9.
Can you give any examples where we use fractional distillation technique?
Answer:
We use this technique in separating the components of crude oil i.e., petrol, naphthalene, kerosene, greese, etc.

9th Class Physical Science Textbook Page No. 38

Question 10.
How does a laundry dryer squeeze out water from wet clothes?
Answer:

• The laundry dryer contains a cylindrical vessel with holes on its walls.
• When wet clothes are dropped in it, it is rotated with high speed with the help of an electric motor.
• Due to centrifugation, the water from the clothes reaches to the walls of the cylinder and comes out through the holes.
• Hence the clothes are dried up.

9th Class Physical Science Textbook Page No. 40

Question 11.
a) “All the solutions are mixtures, but not all mixtures are solutions”. Discuss about the validity of the statement and give reasons to support your argument.
Answer:

• You take any solution like salt solution, sugar solution, air, etc. all are homogeneous mixtures.
• Consider a mixture of sand and iron fehlings. It is not homogeneous. Hence this is not a solution.

b) Usually we think of a solution as a liquid that contains either a solid, liquid or a gas dissolved in it. But, we can have solid solutions. Can you give some examples?
Answer:
Examples of solid solutions are :

1. Steel used in constructions (a homogeneous mixture of iron and carbon).
2. Brass (a homogeneous mixture of zinc and copper).

9th Class Physical Science Textbook Page No. 43

Question 12.
1) Have you ever observed carefully the syrup that you take for cough? Why do you shake it before consuming?
2) Is it a suspension or colloidal solution?
Answer:

1. The syrup used for cough will be shook before consuming because it consists some undissolved particles settled down.
2. Hence cough syrup is a suspension.

9th Class Physical Science Textbook Page No. 45

Question 13.
Is there any difference between a true solution and colloidal solution? If you find the differences, what are those differences?
Answer:
Differences between true solutions and colloidal solutions :

Question 14.
Why do we use different separation techniques for mixtures like grain and husk as well as ammonium chloride and salt though both of them are heterogeneous mixtures? What is the basis for choosing a separation technique to separate mixtures?
Answer:
The basis for choosing a separation technique to separate mixtures is the property of a component in the mixture i.e., solubility in water, evaporation, appearance, etc.

9th Class Physical Science Textbook Page No. 47

Question 15.
Is it possible to find out adulteration of kerosene in petrol with this technique?
Answer:
The adulteration of kerosene in petrol can be found by using density meter.

9th Class Physical Science Textbook Page No. 50

Question 16.
a) Arrange the gases present in air in increas you observe?
Answer:

b) Which gas forms the liquid first as the air is cooled?
Answer:
Oxygen forms the liquid first as the air is cooled.

9th Class Physical Science 3rd Lesson Is Matter Pure? Activities

Activity – 1

Question 1.
How can we separate cream from milk?
Answer:

• Take some milk in a vessel.
• Spin it with a milk churner for some time.
• After some time you observe, separation of a paste like solid out of the milk.
• The paste like solid is called cream.

Activity – 2

Question 2.
Explain a demonstration to identify homogeneous and heterogeneous mixtures.
Answer:

• Take two test tubes.
• Now add one tea spoon of salt to both the test tubes.
• Fill one test tube with water and another with kerosene and stir them.
• In the first test tube (water), the salt dissolves completely.
• This is a homogeneous mixture.
• In the second test tube (kerosene), the salt is not dissolved.
• This is a heterogeneous mixture.

Activity – 3

Question 3.
Describe an activity to prepare saturated and unsaturated solutions.
Answer:
Preparation of saturated solution :

1. When no more solute can be dissolved in the solution at a certain temperature, it is said to be a saturated solution.
2. Take 50 ml of water in an empty cup.
3. Add one spoon of sugar to the water in the cup.
4. Stir the water until it dissolves.
5. Keep on adding sugar to the cup and stir till no more sugar can be dissolved in it.
6. Thus formed solution is called saturated solution.
7. In a saturated solution, equilibrium with the undissolved solute at a certain temperature.

Preparation of unsaturated solution :

1. If the amount of solute present in a solution is less than that in the saturated solution, is called an unsaturated solution.
2. Now take the solution prepared by you into a beaker.
3. Heat that solution slowly by 5 to 6°C above the room temperature.
4. The undissolved solute dissolves.
5. Add some more sugar to this solution.
6. You notice that more sugar dissolves in it easily when it is heated.
7. Thus we prepared an unsaturated solution.

Activity – 4

Question 4.
What are the factors affecting the rate of dissolving ? How do you prove them?
Answer:
Factors affecting the rate of dissolving are :

1. Temperature of the solvent.
2. Size of the solute particles.
3. Stirring the solution.

Proof:

1. Take three glass beakers and fill each of them with 100 ml of water.
2. Add two spoons of salt to each beaker.
3. Place the first beaker undisturbed.
4. Stir the contents of the second beaker.
5. Heat gently the third beaker.
6. In all the cases, the salt dissolves but the time taken to dissolve is different.
7. When the beaker is heated, the salt dissolved quickly.
8. When we stir the contents, the salt dissolved but slower than heating.
9. When we observe the undisturbed beaker, the salt dissolves but at the slowest rate.
10. This shows that the temperature of the solvent, size of the solute particles, stirring of contents are the factors affecting the rate of dissolving.

Activity – 5

Question 5.
Describe an experiment to identify suspensions and colloids.
Answer:

• Take some chalk powder in a test tube.
• Take a few drops of milk in another test tube.
• Add water to these samples and stir with a glass rod.
• Now do the following steps and write your observations in the table given.

Step 1 :
Direct a beam of light from a torch or a laser beam on the test tubes. Observe the path of the light through the solutions.

Step 2 :
1) Leave the mixture undisturbed for some time.
2) See whether the solute settles down after some time.

Step 3 :
Filter the mixtures and observe any residue found on the filter paper.
Now read your observations :

Observations :

1. In the chalk mixture, the particles of chalk settled at the bottom of the test tube and on filtration, we can observe a residue on the filter paper.
2. Hence the chalk mixture is a suspension.
3. In the milk mixture, the particles of milk are uniformly spread throughout the mixture and no residue is found on the filter paper.
4. Hence milk mixture is a colloidal solution.

Activity – 6

Question 6.
Describe an example for the separation of mixtures by sublimation.
Describe a method of separating ammonium chloride from the mixture of ammonium chloride and common salt.
Answer:
Aim :
To separate ammonium chlo-ride from the mixture of ammonium chloride and common salt.

Materials required :
China dish, funnel, cotton, ammonium chloride, common salt and stove.

Procedure:

1. Take one table spoon of ammo-nium chloride and one table spoon of common salt and mix them.
2. Take the mixture in a China dish.
3. Take a glass funnel.
4. Plug the mouth of the funnel with cotton.
5. Invert the funnel over the dish.
6. Heat the dish on the stove and observe the walls of the funnel.

Observations :
Initially we find vapours of ammonium chloride and then solidified ammonium chloride on the walls of the funnel.

Activity – 7

Question 7.
Describe a method to separate the dye present in ink.
(or)
Describe an example for the separation of a mixture by the process of evaporation.
Answer:
Aim :
To separate the dye present in ink by the process of evaporation.

Materials required :
Beaker, watch glass, water, ink and stove.

Procedure :

1. Take a beaker and fill it to half its volume with water.
2. Keep 3, glass on the mouth of a beaker.
3. Put few drops of ink on the watch glass.
4. Heat the beaker and observe the watch glass.

Observations:

1. We observe some fumes coming from the watch glass.
2. Continue heating till you do not observe any further change on the watch glass.
3. A small residue will be remained on the watch glass.

Inference :

1. We know that ink is a mixture of a dye in water.
2. The residue remained on the watch glass is the dye present in the ink.

Lab Activity

Question 8.
Describe paper chromatography activity to observe the colours present in a marker ink.
(OR)
How can you perform chromatography activity in your laboratory.
Answer:
Aim :
Separating the components of ink using paper chromatography.

Materials required :
Beaker, rectangular shaped filter papers, black marker (non-permanent), water, pencil and cello tape.

Procedure:

1. Draw a thick line just above the bottom of the filter paper using the marker.
2. Pour some water into the beaker.
3. Hang the paper strip with help of a pencil and tape in such a way that it should just touch the surface of water.
4. Make sure that the ink line or mark does not touch the water.
5. Allow the water to move up the paper for 5 minutes and then remove the strip from water.
6. Let it dry.
7. Repeat the process with green marker, a permanent marker, etc.

Observations :

1. When black marker is used, we observe different colours like red, green, violet, black, etc. on the filter paper after drying.
2. When green marker is used, we observe yellow, blue, green colours on the filter paper.
3. When permanent marker is used, we cannot find any change in the mark.

Activity – 8

Question 9.
How do you separate water and kerosene from the mixture of kerosene and water?
Answer:
Aim :
To separate water and kerosene from the mixture of kerosene and water.

Materials required :
Kerosene, water, separating funnel, beakers.

Procedure:

1. Pour the mixture of kerosene and water in a separating funnel.
2. Let it stand undisturbed for some time, so that the layers of oil and water are formed.
3. Open the stopcock of the separating funnel and pour out the lower layer of water carefully.
4. Close the stopcock of the separating funnel as the oil reaches the stop cock.

Principle involved :
The immiscible liquids separate out into layers depending on their densities.

Activity – 9

Question 10.
Explain the method of separation of two miscible liquids by distillation.
Answer:
Aim :
To separate two miscible liquids (water and acetone) by distillation.

Materials required :
Stand, distillation flask, thermometer, condenser, beaker, acetone and water, one holed rubber cork.

Procedure:

1. Take a mixture of acetone and water in a distillation flask.
2. Fit it with a thermometer and clamp it to stand.
3. Attach the condenser of the flask on one side.
4. On the other side of the condenser keep a beaker to collect distillate.
5. Heat the mixture slowly.
6. Keep a close watch on the thermometer.
7. The acetone vapourizes and condenses in the condenser.
8. The acetone can be collected from the condenser outlet.
9. Water remains in the distillation flask.
10. The separation technique used above is called distillation.

Activity – 10

Question 11.
How do you separate copper metal from the mixture of copper sulphate and aluminium?
Answer:

• Take a concentrated solution of copper sulphate into a beaker.
• Drop an aluminium foil in the beaker.
• After some time, we observe a layer of copper deposited on the aluminium foil.
• The solution becomes colourless.
• A chemical reaction takes place among the copper ions present in the solution with aluminium and copper metal is separated.

AP State Syllabus AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules Textbook Questions and Answers.

AP State Syllabus 9th Class Physical Science Solutions 4th Lesson Atoms and Molecules

9th Class Physical Science 4th Lesson Atoms and Molecules Textbook Questions and Answers

Improve Your Learning

Question 1.
Draw the diagram to show the experimental setup for the law of conservation of mass. (AS 5)
(OR)
Draw the experimental arrangement used in verifying law of conservation of mass. Write the law of conservation of mass.
Answer:

Question 2.
Explain the process and precautions in verifying law of conservation of mass. (AS 5)
(OR)
Explain the procedure to prove in a chemical reaction the mass neither destroyed.

Answer:
Aim :
To verify law of conservation of mass.

Material required :
Sodium sulphate,Barium chloride, distilled water, conical flask, spring balance, small test tube, rubber cork, thread, retort stand.

Procedure:

1. Prepare a solution of sodium sulphate by dissolving approximately 2 gm of sodium sulphate in 100 ml distilled water in a 250 ml conical flask.
2. Prepare a Barium chloride solution by dissolving approximately 2 gm of potassium iodide in 100 ml water in another conical flask.
3. Take 100 ml solution of sodium sulphate in 250 ml conical flask.
4. Also take 4 ml solution of Barium chloride in test tube.
5. Hang the test tube in the flask carefully without mixing the solutions. Put a cork on the flask.
6. Weigh the flask with its contents carefully by spring balance.
7. Now tilt and swirl the flask, so that the two solutions mix.
8. Weigh the flask again by the spring balance.

Observations:

1. Weight of flask and contents before mixing = m₁ g
2. Weight of flask and contents after mixing = m₂ g

Conclusion :

1. We have observed that the two weights i.e., mj and m2 are equal.
2. This proves the law of conservation of mass.

Precautions:

1. Care should be taken while handling chemicals.
2. Glass apparatus may slip and break down. Hence make sure that they should not slip from your hands.
3. Contents of the conical, ffhsk should not mix before weighing first time.
4. Tie a thick thread to the conical flask, so that it will not slip while weighing.

Question 3.
15.9g of copper sulphate and 10.6g of sodium carbonate react together to give 14.2g of sodium sulphate and 12.3 g of copper carbonate. Which law of chemical combination is obeyed? How? (AS 1, AS 2)
Answer:
Reactants:
Mass of copper sulphate = 15.9 g ; Mass of sodium carbonate = 10.6 g
Total mass of reactants = 15.9 + 10.6 = 26.5 g

Products:
Mass of sodium sulphate = 14.2 g ; Mass of copper carbonate = 12.3 g
Total mass of products = 14.2 + 12.3 = 26.5 g
∴ Total mass of reactants is equal to total mass of products. This is the “Law of conservation of Mass”.

Question 4.
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. Which law of chemical combination will govern your answer? (AS 1, AS 2)
Answer:

1. Let x g of carbon dioxide is added to 112 g of calcium oxide.
2. The product is 200 g of calcium carbonate.
3. According to law of conservation of mass, ,
   Total mass of reactants = Total mass of products
   x+ 112 = 200 g
   x = 88 g
   ∴ 88 g of carbon dioxide is used.

Question 5.
0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.144 g of oxygen and 0.096 g of boron. Calculate the percentage composition of the compound by weight. (AS 1)
Mass of compound of oxygen and boron = 0.24 g
On analysis,
Mass of oxygen in the compound = 1.44 g
Mass of boron in the compound = 0.096 g

Question 6.
In a class, a teacher asked students to write the molecular formula of oxygen. Shamita wrote the formula as 02 and Priyanka as O. Which one is correct? State the reason. (AS 1, AS 2)
Answer:
Shamitha’s answer is correct.
Reason :

1. Oxygen is diatomic.
2. Two atoms of oxygen combine to form oxygen molecule.
3. Hence the formula of oxygen molecule will be ‘O₂‘.

Question 7.
Imagine what would happen if we do not have standard symbols for elements. (AS 2)
(OR)
Is it necessary to use symbols for elements? Write your opinion.
Answer:

• Chemistry involves a lot of reactions.
• If we do not have symbols, we have to write their names to represent the reactions.
• This is very tedious work.
• To avoid this, we need standard symbols to elements, which are universally accepted.
• In advanced studies, balancing of equations, atoms present in a compound, etc. will not be understood without symbols.
• Simply chemistry will not be developed unless symbols, formulae, etc. are not known.

Question 8.
Mohith said “H₂ differs from 2H.” Justify. (AS 1)
Answer:
H₂ is the hydrogen molecule in which two hydrogen atoms are combined to form one hydrogen molecule.

2H is the hydrogen atom. Here 2 hydrogen atoms are ready to participate in chemical reaction.

Question 9.
Lakshmi gives a statement “CO and Co both represent element”. Is it correct? State reason. (AS 1, AS 2)
Answer:
Lakshmi’s statement is incorrect.

Reason :

1. CO stands for carbon monoxide, a compound, which consists carbon and oxygen atoms.
2. This can be identified with the help of both C and O are capital (upper case) letters.
3. Co stands for cobalt, an element.
4. This can be identified with the help of ‘C’ is capital (upper case) letter and ‘o’ small (lower case) letter.

Question 10.
The formula of water molecule is H₂O. What information do you get from this formula? (AS 1)
Answer:

• Water is a combination of hydrogen and oxygen.
• Two hydrogen atoms and one oxygen atom combine to form one water molecule.
• Molecular weight of water molecule is 18. [Hydrogen 1, Oxygen 16. H₂O ⇒ 2 × 1 + 16=18]
• 18 g of water molecule contains 6.022 × 10²³ particles in it.
• Valency of hydrogen is 1 and oxygen is 2.

Question 11.
How would you write 2 molecules of Oxygen and 5 molecules of Nitrogen? (AS 1)
Answer:
2 molecules of oxygen → 2O₂
Reason :

1. Oxygen is diatomic element.
2. Two oxygen atoms combine to form one oxygen molecule.
3. The formula of oxygen molecule is O₂.

5 molecules of nitrogen → 5N₂
Reason :

1. Nitrogen is also diatomic element.
2. Two nitrogen atoms combine to form one nitrogen molecule.
3. Molecular formula of nitrogen is N₂.

Question 12.
The formula of a metal oxide is MO. Then write the formula of its chloride. (AS 1)
Answer:

• The valency of oxide is 2 i.e., O^-2.
• The formula of a metal oxide is given as MO.
• Hence the valency of the given metal must be 2 i.e., M⁺².
• Valency of chloride is 1 i.e., C^–.
• Therefore according to criss-cross method, the formula of given metal chloride will be MCl₂.

Question 13.
Formula of calcium hydroxide is Ca(OH)₂ and zinc phosphate is Zn₃(PO₄)₂. Then write the formula to calcium phosphate. (AS 1)
(OR)
Formula of calcium hydroxide is Ca(OH)₂ and zinc phosphate is Zn₃(PO₄)₂. Then write the valencies of calcium and phosphate and then write the formula of calcium phosphate.
Answer:

• Formula of calcium hydroxide is Ca(OH)₂.
• From criss-cross method we know that the valency of calcium is 2 i.e., Ca⁺² and hydroxide is 1 i.e., OH^–.
• Formula of zinc phosphate is Zn₃(PO₄)₂.
• Valency of Zn is 2 i.e., Zn^+2, and valency of phosphate is 3 i.e., PO₄^-3.
• Now the formula of calcium phosphate according to criss-cross method is Ca₃(PO₄)₂.

Question 14.
Find out the chemical names and formulae for the following common household substances. (AS 1)
a) Common salt
b) Baking soda
c) Washing soda
d) Vinegar
Answer:

Question 15.
Calculate the mass of the following. (AS 1)
a) 0.5 mole of N₂ gas
b) 0.5 mole of N atoms
c) 3.011 × 10²³ number of N atoms
d) 6.022 × 10²³ number of N₂ molecules
Answer:
a) 0.5 mole of N₂ gas :
Mass of one mole of N₂ gas = 28 g. (∵ Molecular wt. of N₂ = 28)
Mass of 0.5 mole of N₂ gas = 28 × 0.5 = 14 g

b) 0.5 mole of N atoms :
Mass of one mole of N atoms = 14 g (∵ Atomic wt. of N = 14)
Mass of 0.5 mole of N atoms = 14 × 0.5 = 7 g

c) 3.011 × 10²³ number of N atoms :
Mass of 6.022 × 10²³ number of N atoms = 14 g

d) 6.022 × 10²³ number of N₂ molecules :
Mass of 6.022 × 10²³ number of N₂ molecules = 28 g

Question 16.
Calculate the number of particles in each of the following. (AS 1)
a) 46 g of Na
b) 8 g of O₂
c) 0.1 mole of hydrogen
Answer:
a) 46 g of Na :
Atomic weight of Na = 23
Number of particles in 23 g of Na atom = 6.022 × 10²³

b) 8 g of O₂ :
Molecular weight of O₂ is 32.
Number of particles in 32 g of O₂ molecule = 6.022 × 10²³

c) 0.1 mole of hydrogen :
Atomic weight of hydrogen is 1.
Number of particles in 1 mole of hydrogen = 6.022 × 10²³
Number of particles in 0.1 mole of hydrogen= \(\frac{0.1}{1}\) × 6.022 × 10²³ = 6.022 × 10²²

Question 17.
Convert into moles. (AS 1)
a) 12 g of O₂ gas
b) 20 g of water
c) 22 g of carbon dioxide
Answer:
a) 12 g of O₂ gas :
Molecular weight of O₂ is 32.
∴ Number of moles of 32 g of O₂ gas = 1

b) 20 g of water :
Molecular weight of water (H₂O) is 18.
Number of moles of 18 g of water = 1
Number of moles of 20 g of water = \(\frac{20}{18}\) x 1 =1.11

c) 22 g of carbon dioxide :
Molecular weight of carbon dioxide (CO₂) is 44.
∴ Number of moles of 44 g of CO₂ = 1

Question 18.
Write the valencies of Fe in FeCl₂ and FeCl₃. (AS 1)
Answer:

1. In FeCl₂, the valency of Fe is 2.
2. In FeCl₃, the valency of Fe is 3.

Question 19.
Calculate the molar mass of sulphuric acid (H₂SO₄) and glucose (C₆H₁₂O₆). (AS 1)
Answer:
a) Formula of sulphuric acid is H₂SO₄.
Molecular mass of H₂SO₄ = 2 × 1 + 1 × 32 + 4 × 16 = 2 + 32 + 64 = 98 u
∴ Molar mass of H₂SO₄ = 98 g

b) Formula of glucose is C₆H₁₂O₆.
Molecular mass of C₆H₁₂O₆ = (6 × 12) + (12 × 1) + (6 × 16) = 72 + 12 + 96 = 180 u
∴ Molar mass of C₆H₁₂O₆ = 180 g.

Question 20.
Which has more number of atoms – 100 g of sodium or 100 g of iron? Justify your answer. (Atomic mass of sodium = 23 u, atomic mass of iron = 56 u) (AS 1)
Answer:
100g of sodium has more number of atoms than 100g of iron.

Justification :
1) Atomic mass of sodium = 23 u
23 g of sodium contains 6.022 × 10²³ atoms.
100 g of sodium contains = \(\frac{100}{23}\) × 6.022 × 10²³ = 26.1826 × 10²³ atoms of sodium.

2) Atomic mass of iron = 56 u
∴ 56 g of iron contains 6.022 × 10²³ atoms.
100 g of iron contains = \(\frac{100}{56}\) × 6.022 × 10²³ = 10.7535 × 10²³ atoms of iron.

Question 21.
Complete the following table. (AS 1)

Answer:

Question 22.
Fill the following table. (AS 1)

Answer:

Question 23.
Make placards with symbols and valencies of the atoms of the elements separately. Each student should hold two placards, one with the symbol in the right hand and the other with the valency in the left hand. Keeping the symbols in place, students should criss-cross their valencies to form the formula of a compound.
Answer:
Student’s activity.

Question 24.
Take empty blister packs of medicines. Cut them into pieces having

Hence the formula of sodium carbonate will be Na₂ CO₃
Hence the forrnu A. Student’s activity.

9th Class Physical Science 4th Lesson Atoms and Molecules InText Questions and Answers

9th Class Physical Science Textbook Page No. 56

Question 1.
Does the weight of iron rod increase or decrease, on rusting?
Answer:
The weight of iron rod decreases on rusting.

Question 2.
Where does the matter charcoal go?
Answer:
The charcoal, on burning, gives off CO₂ which is mixed in atmosphere. The residue is remained as ash.

Question 3.
Wet clothes dry after some time – where does the water go?
Answer:
Water evaporates and mixed in the atmosphere.

Question 4.
What happens to magnesium on burning it in air?
Answer:
Magnesium on burning in air gives a bright light and ash is remained. The ash is magnesium oxide.

Question 5.
What happens to sulphur on burning it in air?
Answer:
Sulphur on burning, changes its state and colour.

9th Class Physical Science Textbook Page No. 57

Question 6.
Did you observe any precipitate in the reaction?
Answer:
In the flask, a reaction takes place between lead nitrate and potassium iodide.

Question 7.
Do you think that a chemical reaction has taken place in the flask? Give reason.
Answer:
Yes, the contents in the flask are changed as lead iodide and potassium nitrate.

9th Class Physical Science Textbook Page No. 58

Question 8.
Do the weights of the flask and its contents change during the activity?
Answer:
The weights of the flask and its contents do not change before and after reaction.

Question 9.
What are your conclusions?
Answer:
Mass was neither created nor destroyed.

Question 10.
What do you observe from table – 1?
Answer:
The components of a compound are mixed at same proportions in any sample.

9th Class Physical Science Textbook Page No. 59

Question 11.
What difference do you observe in percentage of copper, carbon, and oxygen in two samples?
Answer:
The percentage of copper, carbon and oxygen are same in two samples, i.e., they are mixed at same proportions.

9th Class Physical Science Textbook Page No. 60

Question 12.
Are elements also made of atoms?
Answer:
When the particles of a substance contain only one type of atoms, that substance is called an element. In elements the smallest particle that exist may be atoms or molecules.

9th Class Physical Science Textbook Page No. 62

Question 13.
How do we write the symbols for calcium, chlorine, chromium?
Answer:
We have only 26 alphabets in English, but there are over 100 known elements. We cannot write the same symbol for carbon, calcium, chromium, etc.

9th Class Physical Science Textbook Page No. 63

Question 14.
Would you be able to recognise the elements of the table – 2, have symbols of this category?
Answer:
Yes. They are iron, gold, sodium, and potassium.

9th Class Physical Science Textbook Page No. 64

Question 15.
Observe the atomicity and fill the following table.

Answer:

Question 16.
What is valency?
Answer:
Every element reacts with other element according to its combining capacity, which we call as its valency.

9th Class Physical Science Textbook Page No. 66

Question 17.
Can you write the formula of carbon dioxide and carbon monoxide? Try to write formula for them as we have done in case of water molecule.
Answer:
Carbon dioxide :
The elements present are carbon and oxygen. One atom of carbon and one atom of oxygen are present in a molecule of carbon monoxide. Hence the formula of carbon monoxide is CO.

Carbon dioxide :
The elements present are carbon and oxygen. One atom of carbon and 2 atoms of oxygen are present in a molecule of carbon dioxide. Hence the formula of carbon dioxide is CO₂.

9th Class Physical Science Textbook Page No. 69

Question 18.
How many molecules are there in 18 grams of water?
Answer:
6.022 × 10²³ molecules are there in 18 grams of water.

Question 19.
How many atoms are there in 12 grams of carbon?
Answer:
6.022 × 10²³ atoms are there in 12 grams of carbon.

9th Class Physical Science Textbook Page No. 58

Question 20.
Do you get the same result if the conical flask is not closed?
Answer:

1. No, we cannot get the same result.
2. When the conical flask is not closed, some gases will leave out the flask during chemical reaction.

Question 21.
Recall the burning of the magnesium ribbon in air. Do you think mass is conserved during this reaction?
Answer:

1. Yes, but we cannot observe the conservation of mass.
2. When the experiment is conducted in a closed container where there is no scope for oxygen to escape, we can observe the conservation of mass. But in this condition this experiment is not possible.

9th Class Physical Science Textbook Page No. 59

Question 22.
100 g of mercuric oxide decompose to give 92.6 g of mercury and 7.4 g of oxygen. Let us assume that 10 g of oxygen reacts completely with 125 g of mercury to give mercuric oxide. Do these values agree with the law of constant proportions?
Answer:
Proportion of oxygen = 7.4 : 10
Proportion of mercury = 92.6 : 125
\(\Rightarrow \frac{7.4}{10}=\frac{92.6}{125} \Rightarrow 0.74=0.74\)
∴ They follow law of constant proportions.

Question 23.
Discuss with your friends if the carbon dioxide that you breathe out and the carbon dioxide they breathe out are identical. Is the composition of the carbon dioxide of different sources same? (Page – 73)
Answer:
Same.
This can be justified with the help of law of constant proportions.

9th Class Physical Science Textbook Page No. 60

Question 24.
Which postulate of Dalton’s theory is the result of the law of conservation of mass?
Answer:
First postulate of Dalton’s theory i.e. “Matter consists of indivisible particles called atoms”, is the result of law of conservation of mass.

Question 25.
Which postulate of Dalton’s theory can explain the law of constant proportions?
Answer:
Third postulate of Dalton’s theory i.e. “Atoms of a given element have identical mass and chemical properties. Atoms of different elements have different masses and chemical properties”, is the result of law of constant proportions.

9th Class Physical Science 4th Lesson Atoms and Molecules Activities

Activity – 1

Question 1.
Some elements and their possible symbols are given. Correct them and give reasons for your corrections.
Answer:

Activity – 2

Question 2.
Write the symbols for given elements.

Answer:

AP State Syllabus AP Board 9th Class Physical Science Solutions Chapter 11 Sound Textbook Questions and Answers.

AP State Syllabus 9th Class Physical Science Solutions 11th Lesson Sound

9th Class Physical Science 11th Lesson Sound Textbook Questions and Answers

Improve Your Learning

Pick out the correct answer :

Question 1.
When we say sound travels in a medium (AS 1)
A) the medium travels
B) the particles of the medium travel
C) the source travels
D) the disturbance travels
Answer:
D) the disturbance travels

Question 2.
A sound wave consists of (AS 1)
A) number of compression pulses only
B) number of rarefaction pulses only
C) number of compression and rarefaction pulses one after the other
D) vacuum only
Answer:
C) number of compression and rarefaction pulses one after the other

Question 3.
Hertz stands for oscillations per (AS 1)
A) second
B) minute
C) hour
D) milli second
Answer:
A)second

Question 4.
When we increase the loudness of sound of a TV, the property of sound that changes is (AS 1)
A) amplitude
B) frequency
C) wavelength
D) speed
Answer:
A) amplitude

Question 5.
The characteristic of the sound that describes how the brain interprets the frequency of sound is called (AS 1)
A) pitch
B) loudness
C) quality
D) sound
Answer:
A) pitch

Question 6.
In a stethoscope, sound of heart beats travel through stethoscope tube (AS 1)
A) by bending along the tube
B) in a straight line
C) undergoing multiple reflections
D) all of the above
Answer:
C) undergoing multiple reflections

Question 7.
Explain the following terms : (AS 1)
a) amplitude
b) wavelength
c) frequency
Answer:
a) Amplitude :
The maximum variation in density or pressure from the mean value is called amplitude.
(or)

The maximum disturbance of particles of a medium from their mean position is called amplitude.

b) Wavelength :
The distance between two consecutive compressions or two consecutive rarefactions is called the wavelength of a sound wave, denoted by W. Wavelength is measured in ‘meters’.

c) Frequency :

1. The number of oscillations of the density of the medium at a place per unit time is called the frequency of the sound wave.
2. Frequency is denoted by ‘o’.
3. The S.I. unit of frequency is ‘Hertz’.

Question 8.
Deduce the relation between wavelength, frequency, and speed of sound. (AS 1)
Answer:
1) Speed of sound can be defined as the distance by which a point on the wave, such as a compression or rarefaction, travels in unit time.

2) Let the distance travelled by a wave in T seconds = X metres

3) The distance travelled by a wave in 1 second = \(\frac{\lambda}{\mathrm{T}}\) meters

4) Thus by definition of speed of wave, v = \(\frac{\lambda}{\mathrm{T}}\) ……….. (1)

5) We know that frequency and time period are related as υ = \(\frac{1}{T}\) ……….. (2)

6) From (1) and (2) we get v = λ .υ
∴ Speed of sound = Frequency × Wavelength

Question 9.
How are multiple reflections of sound helpful to doctors and engipeeps? (AS 7)
Answer:
1) Doctors use multiple reflections of sound to hear the sounds produced with in the body using stethoscope.

2) Doctors can see the images of patient’s organs like liver, gall bladder, uterus, etc. to know the abnormalities in their functioning, using ultrasounds.

3) Engineers use the reflections of sound in designing concert halls and cinema halls.

4) Generally the ceilings of concert halls, conference halls, cinema halls are designed such that sound after reflection reaches all corners of the hall as shown in the figure.

Question 10.
Name two quantities that vary periodically at a place in air as a sound wave travels through it. (AS 1)
Answer:
The two quantities that vary periodically at a place in air as a sound wave travels through it are density and pressure of particles.

Question 11.
Which has larger frequency – infrasonic sound or ultrasonic sound? (AS 7, AS 2)
Answer:

• Infrasonics are the sounds of frequency less than 20 Hz.
• Ultrasonics are the sounds of frequency greater than 20 kHz.
• Hence the ultrasonics have larger frequency.

Question 12.
The grandparents and parents of two-year-old girl are playing with her in a room. A sound source produces a 28 kHz sound. Who in the room is most likely to hear the sound? (AS 2, AS 7)
Answer:

• The two-year-old girl is able to hear the sound.
• Children can hear sounds of somewhat higher frequencies up to 30 kHz.

Question 13.
Does the sound follow same laws of reflection as light does? (AS 1)
Answer:

• Reflection of sound follows the same laws as the reflection of light when sound is reflected.
• The directions in which the sound is incident and reflected make equal angles with the normal to the reflecting surface.

Question 14.
Why is soft furnishing avoided in concert halls? (AS 7)
Answer:

• Sound reflects like the reflection of light.
• But unlike to light, sound reflects more on rough surfaces than soft surfaces.
• In concert halls, sound must undergo multiple reflections, so as to reach all corners of the hall.
• Hence for better reflection, soft furnishing is avoided in concert halls.

Question 15.
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? (AS 1)
Answer:
Frequency of source A = 1 kHz; Frequency of source B = 30 kHz

As the speed of wave increases with frequency and both the waves have same amplitude, the sound produced from source B has larger power.

Question 16.
What do you understand by a sound wave? (AS 1)
Answer:

• Sound is produced from a vibrating body.
• It travels through air in the form of a wave.
• Sound waves are longitudinal.

Question 17.
Define the wavelength of a sound wave. How is it related to the frequency and the wave speed? (AS 1)
Answer:
Wavelength :
The distance between two consecutive compressions or rarefactions is called wavelength.

Relation between wavelength, frequency, and wave speed :

1. Speed of sound can be defined as the distance by which a point on the wave, such as a compression or rarefaction, travels in unit time.
2. Let the distance travelled by a wave in T seconds = λ metres
3. The distance travelled by a wave in 1 second = \(\frac{\lambda}{\mathrm{T}}\) meters
4. Thus by definition of speed of wave, v = \(\frac{\lambda}{\mathrm{T}}\) ………… (1)
5. We know that frequency and time period are related as o = \(\frac{1}{T}\) ……….. (2)

Question 18.
Explain how echoes are used by bats to judge the distance of an obstacle in front of them. (AS 1)
Answer:

• Bats search out prey and fly in dark night by emitting and detecting reflections of ultrasonic waves.
• The high pitched ultrasonic squeaks of the bat are reflected from the obstacles or prey and returned to bat’s ear.
• The nature of reflections tells the bat where the obstacle or prey is and what it is like.
• The bats use ultrasound for navigation and location of the food in dark.

Question 19.
With the help of a diagram describe how compression and rarefaction pulses are produced in air near a source of sound. (AS 5)
Answer:

• Consider a vibrating membrane of a musical instrument like a drum or tabla.
• As it moves back and forth, it produces a sound.
• The figure shows the membrane at different instants and the condition of the air near it at those instants.
• As the membrane moves forward, it pushes the particles of air in the layer in front of it.

• So, the particle of air in the layer get closer to each other, hence the density increases.
• This layer of air pushes and compresses the layer next to it and so on.
• We call this disturbance as compression pulse.
• When the membrane moves backward, it drags back the layer of air near it. Hence the density decreases.
• The particles of air in the next layer on the right move into fill this less dense area.
• This is a rarefaction pulse moves to right.
• As the membrane moves back and forth repeatedly, compression and rarefaction pulses are produced, one after the other.
• These two pulses travel one behind the other, carrying the disturbance with it.

Question 20.
How do echoes in a normal room affect the quality of the sounds that we hear? (AS 7)
Answer:

• Echo is a reflected sound, arriving at the position of listener more than 0.1s after the direct sound.
• Quality is the characteristic of a sound which enables us to distinguish between musical notes emitted by different musical instruments.
• In a normal room, if echo is formed, we can hear multiple sounds, at same time.
• Our ear cannot perceive and judge the sound from where it is coming.
• So, quality of sound does not work here.

Question 21.
Explain the working and applications of SONAR. (AS 1)
Answer:

• SONAR stands for Sonographic Navigation And Ranging.
• This is a method for detecting and finding the distance of objects under water by means of reflected ultrasonic waves.

Working of SONAR:

1. SONAR system consists of a transmitter and a detector in the “Observation Centre” on board of a ship.
2. From the observation centre, ultrasonic waves of high frequency are sent in all directions under the water through transmitter.
3. These waves travel in straight lines till they hit an object such as a submarine, a sunken ship, etc.
4. The waves are then reflected and are received back by the receiver at the observation centre.
5. The study of these reflected waves gives information about the direction of the object located.
6. The time between sending ultrasonic wave and receiving its echo, the distance of the object is calculated.
7. Reflections from various angles can be utilized to determine the shape and size of the object.

Mathematical expression :

1. Let’d’ be the distance between SONAR and an underwater object.
2. ‘t’ be the time between sending an ultrasonic wave and receiving its echo.
3. ‘u’ be the speed of sound in water.
4. The total distance covered by the wave from the SONAR to the object and back is 2d.
5. From the equation s = ut ⇒ 2d = ut ⇒ d = \(\frac{ut}{2}\).

Application:
Marine geologists use this method to determine the depth of the sea and to locate underwater hills and valleys.

Question 22.
Find the time period of a source of a sound wave whose frequency is 400 Hz. (AS 1)
Answer:
Frequency υ = 400 Hz
Time period T =?

Question 23.
A sound wave travels at a speed of 340 m/s. If its wavelength is 2 cm, what is the frequency of the wave? Will it be in the audible range? (AS 1)
Answer:
Speed of sound v = 340 m/s.; Wavelength λ = 2 cm = 0.02 m.
Frequency υ =?

The audible range of sound wave is 20 Hz to 20 kHz.
Hence this is in the audible range.

Question 24.
Given that sound travels in air at 340 m/s, find the wavelength of the waves in air produced by a 20 kHz sound source. If the same source is put in a water tank, what would be the wavelength of the sound waves in water? (AS 7)
Speed of sound in water = 1,480 m/s.
Answer:
In air:
Speed of sound wave (v) = 340 m/s ; Frequency of source of sound (o) = 20 kHz
Wavelength of the sound wave λ =?
v = υλ

∴ Wavelength of the sound wave in air = 17 m

Same source is kept in water :
∴ Speed of sound in water (v) = 1480 m/s
Frequency of sound wave (p) = 20 kHz
Wavelength of sound wave λ =?
v = υλ

∴ Wavelength of the sound wave in water = 74 m

Question 25.
A man is lying on the floor of a large, empty hemispherical hall, in such a way that his head is at the centre of the hall. He shouts “Hello!” and hears the echo of his voice after 0.2 s. What is the radius of the hall ? (Speed of sound in air 340 m/s) (AS 7)
Answer:
Let the distance travelled by the sound wave = 2d m

As the head of the man is at the centre of hemispherical room, then ‘d’ is the radius of the hall.
∴ Radius of the hall = 34 m

Question 26.
“We know that sound is a form of energy. So, the large amount of energy produced due the sound pollution in cosmopolitan cities can be used to our day-to-day needs of energy. It also helps us to protect biodiversity in urban areas”. Do you agree with this statement? Explain.
Answer:

1. Sound is a form of mechanical energy.
2. So, the mechanical energy can be converted into electrical energy.
3. Experiments are going on this concept.
4. If this is successful, we have the following benefits.
   a) Sound pollution can be controlled.
   b) Conventional methods of producing electrical energy from coal or water will cause in loss of biodiversity. This can be avoided.
   c) Natural resources like water can be protected.
   d) Increasing needs of energy can be overcome by this method.

Question 27.
How do you appreciate efforts of a musician to produce melodious sound using a musical instrument by simultaneously controlling frequency and amplitude of the sounds produced by it.
Answer:

• The sounds which produce pleasing effect on the ear are called musical sounds.
• Any instrument which produces musical sound is called musical instrument.
• The person who plays a musical instrument to produce melodious sound is called a musician.
• The musician must have control on breathing, concentration on the output of the sound, which is a very hard task.
• For this the musician needs a lot of practise.
• With the musician’s practise and knowledge over musical notes only we can hear melodious sound otherwise it could only be a noise.
• Hence the efforts of a musician are highly appreciable.

Question 28.
You might have observed that sometimes your pet dog starts barking though no one is seen near in its surroundings or no disturbance heard nearby. Does this observation raise any doubts in your mind about the peculiar behaviour of dog after your understanding about ‘range of hearing the sound’. If yes, write them.
Answer:

• Dogs can hear sounds of frequencies up to about 50 kHz, which is ultrasonic.
• After hearing this ultrasonics, a dog will bark panicly, though no one is seen near.
• I understood this after studying about ‘range of hearing the sound’.
• Before the knowledge of ‘range of hearing the sound’, I felt that the dogs are barking by seeing some devils, which is a misconception.
• Scientific knowledge helps us to know reasons for many misconcepts.

Question 29.
Find out the names of animals (and their photographs from internet) which communicate using infra-sonic or ultra-sonic sound and prepare a scrap book.
Answer:
Scrap book

9th Class Physical Science 11th Lesson Sound InText Questions and Answers

9th Class Physical Science Textbook Page No. 184

Question 1.
How does sound reach our ears from the source of its production?
Answer:
Soufid travels in the form of waves. It reaches our ears from the source of its production, in the form of waves.

Question 2.
Does it travel by itself or is there any force bringing it to our ears?
Answer:
Sound does not travel by itself. When a sound is produced, the kinetic energy of the source, vibrates the nearest particles in the medium. These particles transfers energy and finally, it reaches our ears.

Question 3.
What is sound? Is it a force or an energy?
Answer:
Sound is a form of energy.

Question 4.
Why don’t we hear sounds when our ears are closed?
Answer:
When our ears are closed, the energy in the form of waves reaches our ear, but it cannot make the eardrum to vibrate. Hence we cannot hear the sound.

Question 5.
Why is the light ray dancing, after sound is made in the tin?
Answer:
The dancing of light ray, after the sound is made in the tin shows that the sound travels in the form of waves from the source of its production.

Question 6.
What do you infer from this?
Answer:
Sound is a form of energy which can travel in the form of waves through the medium.

Question 7.
Can we say that sound is a form of mechanical energy?
Answer:
Yes, sound is a form of mechanical energy.

9th Class Physical Science Textbook Page No. 185

Question 8.
Do you hear any sound?
Answer:
We cannot hear any sound.

Question 9.
Do you see any vibrations in the tuning fork?
Answer:
Yes, we can sense the vibrations in the tuning fork.

Question 10.
What do you conclude from the above activity?
Answer:
We observe that vibrating tuning fork produces sound.

Question 11.
Can you produce sound without vibration in the body?
Answer:
We cannot produce sound without vibration in the body.

Question 12.
Give some examples of vibrating bodies which produce sound.
Answer:
Drums, tabla, calling bell, school bell, etc.

Question 13.
What part of our body vibrates when we speak?
Answer:
When we speak, vocal cord vibrates in our body.

Question 14.
Do all vibrating bodies necessarily produce sound?
Answer:
All vibrating bodies produce sound, but we cannot hear some of them, due to the limit of audible frequency.

9th Class Physical Science Textbook Page No. 186

Question 15.
If sound travels in the form of a wave then what is the pattern?
Answer:
Sound travels in the form of longitudinal or transverse waves in the air or in the other material.

9th Class Physical Science Textbook Page No. 188

Question 16.
What do you say about sound waves in air by the above activity?
Answer:
From the above activity, we can say that there involves change in the density of medium while sound waves are travelling in air.

Question 17.
Are they longitudinal or transverse?
Answer:
Sound waves in air are longitudinal.

Question 18.
Does sound get reflected at the surface of a solid?
Answer:
Yes, sound gets reflected at the surface of a solid as in the case of reflection of light.

Question 19.
What happens if you lift your tube slightly above the table?
Answer:
If we lift the tube slightly we cannot hear the sound clearly.

Question 20.
Are able to listen to the sound? If not why?
Answer:
We are unable to hear the sound. If we lift one of the pipes then the pipe carrying incident sound, the pipe carrying reflected sound will not be in the same plane. Hence we cannot hear the sound.

9th Class Physical Science Textbook Page No. 195

Question 21.
Do hard surfaces reflect sound better than soft ones?
Answer:
Generally, hard surfaces reflect sound better than soft surfaces. But sound reflects quite well from rough surfaces than polished surfaces.

9th Class Physical Science Textbook Page No. 187

Question 22.
Do compressions and rarefactions in sound wave travel in same directions or in opposite directions? Explain.
Answer:

• Compressions and rarefactions in a sound wave will be in opposite direction.
• In a compression, all the particles come close, so the density and pressure increases.
• In a rarefaction, all particles drag back, so the density and pressure decreases.
• In a microscopic view of particle, the compression and rarefaction travel in opposite directions.

9th Class Physical Science Textbook Page No. 191

Question 23.
Does the frequency of sound waves depend on the medium in which it frawels? How?
Answer:
Yes.

2) As speed of wave differs from medium to fnedium, the frequency also changes, keeping the wavelengths constant.

Question 24.
The frequency of source of sound is 10 Hz. How many times does it vibrate in one minute?
Answer:
Number of vibrations per second = 10
Number of vibrations in one minute = 10 × 60 = 600

Question 25.
Gently strike a hanging bell (temple bell) and try to listen to the sound produced by it with a stethoscope keeping it both at bottom portion and top portion of the bell. Is the pitch and loudness of the sound same at the two portions? Why?
Answer:
No. The bob of the bell strikes at the bottom portion of the bell. Hence the pitch and loudness are high.

Question 26.
During a thunderstorm if you note a 3 second delay between the flash of lightning and sound of thunder. What is the approximate distance of thunderstorm from you ?
Answer:
Time taken to reach the sound = 3 s ; Speed of sound in air = 343.2 m/sec.
Distance of thunderstorm = 343.2 x 3 = 1029.6 m

9th Class Physical Science Textbook Page No. 194

Question 27.
Two girls are playing on identical stringed instruments. The strings of the both instruments are adjusted to give notes of same pitch. Will the quality of two notes be same? Justify your answer.
Answer:
If the two girls are playing with same instruments, then the quality is same. If the girls are playing with different stringed instruments the quality will be different.

Quality is the characteristic which enables us to distinguish between musical notes emitted by different musical instruments.

Question 28.
What change, would you expect in the characteristic of a musical sound when we increase its frequency one instance and amplitude in another instance?
Answer:
When frequency is increased, the pitch of sound increases. When amplitude is increased, its loudness increases.

9th Class Physical Science Textbook Page No. 195

Question 29.
What could be the reason for better reflection of sound by rough surfaces than polished surfaces?
Answer:

• Sound reflects better on rough surfaces than polished surfaces.
• The rough surface reflects sound in all directions, so it can reach in many directions.

9th Class Physical Science Textbook Page No. 196

Question 30.
Why is an echo weaker than the original sound?
Answer:

• Echo is the reflected sound.
• While it travels back, it gradually loses its energy.
• Hence the echo is weaker than original sound.

Question 31.
In a closed box if you say hello, the sound heard will be Hellooooo ………. What does it mean?
Answer:
In a closed box, the multiple reflections of sound has no way to go out side. Hence we will hear the sound as hellooooooo ………

9th Class Physical Science Textbook Page No. 197

Question 32.
What is the advantage of having conical openings in horns, megaphones, etc?
Answer:
The conical openings in horns, megaphones, etc. will guide the reflected sound waves in forward direction and spreads towards the audience.

Question 33.
Why do we put cushions on the chairs, carpet on the floor, straw materials on the walls in cinema halls?
Answer:

• These materials absorb unnecessary reflections of sound, so that we can hear more clearly.
• As these material absorb reflected sound waves, the people outside the cinema halls, cannot hear the sound produced in the cinema hall.

9th Class Physical Science Textbook Page No. 199

Question 34.
What is the benefit of using ultrasound over light waves in the above applications?
Answer:

• Light waves cannot penetrate in the internal organs like liver, kidney, etc.
• Light wave after reflection do not form image.
• Ultrasound waves can penetrate through internal organs like liver, kidney, etc.
• After reflection, ultrasound waves produce image. So, in ultrasonography and surgeries, ultrasound waves are better than light waves.

9th Class Physical Science 11th Lesson Sound Activities

Activity 1

Question 1.
How can .you say that the sound is a form of energy?
Answer:

1. Take a tin can and remove both ends to make a hollow cylinder.
2. Take a balloon and stretch it over the can.
3. Wrap a rubber band around the balloon.
4. Take a small mirror and stick it on the balloon.
5. Take a laser light and let it fall on the mirror.
6. After reflection the light spot is seen on the wall.
7. Now shout directly into the open end of the can and observe the dancing light.

Observations:

1. When sound is made, the energy produced from the sound vibrates the membrane of the balloon, resulting in the dancing of light ray.
2. This shows that sound is a form of energy which travels in air.

Activity – 2

Question 2.
Prove that the sound is produced from a vibrating source.
Answer:

1. Attach a small piece of steel wire to one of the prongs of a tuning fork as shown in the figure.
2. Strike the tuning fork with a rubber hammer.
3. While it is vibrating, try to draw a straight line on a piece of smoked glass as quick as possible with, it.
4. Keep the end of the wire in such a way just it touches the glass.
5. A line is formed in the form of a wave.
6. Repeat the experiment when the tuning fork is not vibrating and observe the difference in the line.

Observations:

1. We have produced vibrations in the tuning fork by striking it with a hammer.
2. Thus the vibrating tuning fork produces sound.
3. Thus the sound is produced by vibrating bodies.

Activity – 3

Question 3.
а) How do you demonstrate the formation of compressions and rarefactions in a slinky?
Answer:

1. Take a slinky.
2. It is a spring-shaped toy which can be extended or compressed very easily.
3. Lay it down on a table or the floor as shown in figure.
4. Ask a friend to hold one end.
5. Pull the other end to stretch the slinky and then move it to and fro along its length.
6. We will see alternate compressions and rarefactions of the coil.
7. This is similar to the pattern of varying density produced in a medium when sound passes through it.

b) Mow do you demonstrate the formation of crests and troughs in a slinky?
Answer:

1. Hang a slinky from a fixed support.
2. Hold it gently at the lower end and quickly move your hand sideways and back.
3. This will cause a hump on the slinky near the lower end.
4. The hump travels upwards on the slinky as shown in the figure.
5. The humps formed alternately are known as crests and troughs.

Activity – 4

Question 4.
Describe an activity to listen the reflected sound.
Answer:

1. Take two long, identical tubes and place them on
2. Ask your friend to speak softly into one tube while you use the other tube to listen.
3. Adjust the tube until you hear the best sound.
4. You will find that you hear your friend’s voice best when the tube makes equal angles with a normal to the wall.
5. This shows that reflection of sound follows the laws of reflection of light.
6. Lift your tube slightly above the table.
7. You will not be able to listen the voice clearly, because the plane carrying the incident wave and reflected wave are changed.
8. Repeat the experiment by placing flat objects of different materials (steel and plastic trays, a card board, a tray wrapped with cloth, etc.) against the wall and observe the changes in the sound.

Observations:

1. Reflection of sound follows the laws of reflection of light.
2. When the plane carrying incident wave and reflecting wave changes, the reflected sound cannot be heard clearly.
3. Hard or rough surfaces reflect the sound better than soft surfaces.

AP State Syllabus AP Board 9th Class Physical Science Solutions Chapter 6 Chemical Reactions and Equations Textbook Questions and Answers.

AP State Syllabus 9th Class Physical Science Solutions 6th Lesson Chemical Reactions and Equations

9th Class Physical Science 6th Lesson Chemical Reactions and Equations Textbook Questions and Answers

Improve Your Learning

Question 1.
What is a balanced chemical equation? Why should chemical equations be balanced?
Answer:
1) A chemical equation in which the number of atoms of different elements on the reactants side (left side) are same as those on product side (right side) is called a balanced chemical equation.
Ex : Zn + 2HCl → ZnCl₂ + H₂ ↑

2) All the chemical equations must balance because atoms are neither created nor destroyed in chemical reactions.

3) The number of atoms of each element before and after reaction must be the same.

4) According to the law of conservation of mass, the total mass of the substances that are taking part in a chemical reaction must be the same before and after the reaction.

Question 2.
Balance the following chemical equations.
a) NaOH + H₂SO₄ → Na₂SO₄ + H₂O
b) Hg (NO₃)₂ + KI → Hgl₂ + KNO₃
c) H₂ + O₂ → H₂O
d) KClO₃ → KCl + O₂
e) C₃H₈ + O₂ → CO₂ + H₂O
Answer:
a) 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
b) Hg (NO₃)₂ + 2 KI → Hgl₂ + 2KNO₃
c) 2H₂ + O₂ → 2H₂O
d) 2KClO₃ → 2KCl + 3O₂
e) C₃Hg + 5O₂ → 3CO₂ + 4H₂O

Question 3.
Write the balanced chemical equations for the following reactions.
a) Zinc + Silver nitrate → Zinc nitrate + Silver
b) Aluminium + Copper chloride → Aluminium chloride + Copper
c) Hydrogen + Chlorine → Hydrogen chloride
d) Ammonium nitrate → Nitrous Oxide + Water
Answer:
a) Zn + 2AgNO₃ → Zn(NO₃)₂ + 2Ag
b) 2Al + 3CuCl₂ → 2AlCl₃ + 3Cu
c) H₂ + Cl₂ → 2HCl
d) NH₄NO₃ → N₂O + 2H₂O

Question 4.
Write the balanced chemical equations for the following and identify the type of reaction in each case.
a) Calcium hydroxide(aq) + Nitric acid(aq) → Water(l) + Calcium nitrate(aq)
b) Magnesium(sJ + Iodine → Magnesium Iodide(s)
c) Magnesium(s) + Hydrochloric acid(aq) → Magnesium chloride(aq) + Hydrogen^
d) Zinc(s) + Calcium chloride(aq) → Zinc Chloride(aq) + Calcium(s)
Answer:
a) Ca(OH₂) + HNO₃ → H₂O + Ca(NO₃)₂
This is double decomposition reaction.

b) Mg + I₂ → Mgl₂
This is chemical combination reaction.

c) Mg + 2HCl → MgCl₂ + H₂ ↑
This is chemical displacement reaction.

d) Zn + CaCl₂ → ZnCl₂ + Ca
This is chemical displacement reaction. This reaction is not possible, because calcium is more reactive than zinc.

Question 5.
Write an equation for decomposition reaction where energy is supplied in the form of heat / light / electricity.
Answer:
Chemical decomposition reaction : A chemical reaction in which a single substance splits into two or more substances is called chemical decomposition.
For decomposition reaction energy is supplied in the form of :

Question 6.
What do you mean by precipitation reaction?
Answer:
A reaction in which insoluble substance in water is formed as product is called precipitation reaction.

Question 7.
How does chemical displacement reactions differ from chemical decomposition reaction? Explain with an example for each.
Answer:

Question 8.
Name the reactions taking place in the presence of sunlight.
Answer:
1) In the presence of sunlight plants prepare their food by taking C02 from the air and H20 from the soil with their chloroplasts of the green leaves. This reaction is called photosynthesis.

Question 9.
Why does respiration considered as an exothermic reaction? Explain.
Answer:

• We need energy to stay alive.
• We get this energy from food we eat.
• During digestion, food is broken down into simpler substances.
• For example, rice and potato contains starch.
• The starch breaks down to form glucose.
• This glucose combines with oxygen in the cells of our body and releases energy, which helps to do the various works.
• During this process, energy is given out. Hence this reaction can be called exothermic reaction.
• The special name of this reaction is respiration.
• So respiration is considered as exothermic reaction.
• C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + Q (Energy)

Question 10.
What is the difference between displacement and double displacement reactions? Write equations for these reactions.
Answer:

Question 11.
MnOz + 4HCl → MnCl₂ + 2H₂O + Cl₂ ↑
In the above equation, name the compound which is oxidized and which is reduced.
Answer:
In the above equation, HCl compound is oxidized and MnO₂ is reduced.

Question 12.
Give two examples for oxidation-reduction reaction.
Oxidation :
Oxidation is a reaction that involves the addition of oxygen or loss of hydrogen or electrons.

Reduction :
The process in which a substance loses oxygen or gains hydrogen or electrons is known as reduction.

Question 13.
In the refining of silver, the recovery of silver from silver nitrate solution involved displacement by copper metal. Write the reaction involved.
Answer:

• Cu_(s) + 2 AgNO_3(aq) > Cu(NO₃)_2(aq) + 2Ag_(s)
• This is redox reaction, copper is the reducing agent and the silver is reduced.
• Electrons from the copper metal are transferred to the silver.
• This reaction can also be called a displacement reaction because copper displaces silver as it is more reactive.

Question 14.
What do you mean by corrosion? How can you prevent it?
Answer:

• When some metals are exposed to moisture, acids, etc. they tarnish due to the formation of respective metal oxide on their surface. This process is called “corrosion”.
• Corrosion can be prevented by painting, oiling, greasing, galvanizing, chrome plating or making alloys.
• Galvanizing is a method of protecting iron from rusting by coating them a thin layer of zinc.

Question 15.
Explain rancidity.
Answer:

• Oxidation reactions in food material that were left for a long period are responsible for spoiling of food. This process is called “rancidity”.
• When these processes occur in food, undesirable odours and flavours can result.
• Rancidity is an oxidation reaction.

Question 16.
Balance the following chemical equations including the physical states,
a) C₆H₁₂O₆ → C₂H₅OH + CO₂
b) Fe + O₂ → Fe₂O₃
c) NH₃ + Cl₂ → N₂H₄ + NH₄Cl
d) Na + H₂O → NaOH + H₂
Answer:
a) C₆H₁₂O_6(s) → C₂H₅OH_(l) + CO_2(g)
b) Fe_(s) + O_2(g) → Fe₂O_3(s)
c) NH_3(aq) + Cl_2(g) → N₂H_4(l) + NH₄Cl_(aq)
d) 2Na_(s) + 2H₂O_(l) → NaOH + H₂

Question 17.
Balance the chemical equation by including the physical states of the substances for the following reactions.
a) Barium chloride and Sodium sulphate aqueous solutions react to give insoluble Barium sulphate and aqueous solution of Sodium chloride.
b) Sodium hydroxide reacts with Hydrochloric acid to produce Sodium chloride and water.
c) Zinc pieces react with dilute Hydrochloric acid to liberate Hydrogen gas and forms Zinc chloride.
Answer:
a) BaCl_2(aq) + Na₂SO_4(aq) → BaSO_4(s)↓ + NaCl_(aq)
b) NaOH_(aq) + HCl_(aq) → NaCl_(aq) + H₂O_(l)
c) Zn_(s) + 2HCl_(aq) → ZnCl_2(aq) + H_2(g)

Question 18.
A shiny brown coloured element ‘X’ on heating in air becomes black in colour. Can you predict the element ‘X’ and the black coloured substance formed? How do you support your predictions?
Answer:
The brown coloured element is copper (Cu). On heating copper reacts with oxygen
present in the atmosphere to form copper oxide which is black in colour.
The reaction is shown below.

If we pass hydrogen gas over hot copper oxide we will notice that black coating on copper turns brown because copper oxide loses oxygen to form copper.
This will support our prediction.

Question 19.
Why do we apply paint on iron articles?
Answer:

• Ferrous reacts with oxygen in the air and form iron oxide.
  2 Fe + O₂ → 2 FeO
• This reaction is called corrosion. It spoils the iron articles by rusting.
• Corrosion of iron articles can be prevented or minimized by shielding the metal surface from oxygen and moisture.
• It can be prevented by applying paint on the articles.

Question 20.
What is the use of keeping food in air tight containers?
Answer:

• Oxidation is defined as the interaction of oxygen molecules with all the different substances from metal to living tissue which may come into contact with it.
• When fats and oils are oxidized they become rancid. Their smell and taste change.
• Keeping food in airtight containers helps to slow down oxidation process.
• So, manufacturers of potato chips usually flush bags of chips with gas such as nitrogen to prevent the chips from getting oxidized.

Fill in the Blanks

1. The decomposition of vegetable into compost is an example of ……………. reaction.
2. The chemical reaction in which energy is absorbed to form a new compound is called ………….
3. The reaction 2N₂O → 2N₂ + O₂ is an example for ………….. reaction.
4. The reaction Ca + 2H₂O → Ca(OH)₂ + H₂ ↑ is an example for ………….. reaction.
5. The substances that are present on left side of a chemical equation are called
6. The arrow mark between the products and reactants of a chemical equation shows of the reaction.
Answer:

1. oxidation
2. endothermic reaction
3. decomposition
4. displacement
5. reactants
6. direction

7. Match the following :

Answer:

Multiple Choice Questions:

1. Fe₂O₃ + 2Al → Al₂O₃ + 2Fe
The above reaction is an example of:
A) Combination reaction
B) Decomposition reaction
C) Displacement reaction
D) Double decomposition reaction
Answer:
C) Displacement reaction

2. What happens when dil. hydrochloric acid is added to iron filings? Choose the correct answer.
A) Hydrogen gas and iron chloride are produced.
B) Chlorine gas and iron hydroxide are produced.
C) No reaction takes place.
D) Iron salt and water are produced.
Answer:
A) Hydrogen gas and iron chloride are produced.

3. 2PbO_(s) + C_(s) → 2Pb_(s) + CO_2(g)
Which of the following statements are correct for the above chemical reaction?
i) Lead is reduced
ii) Carbon dioxide is oxidized
iii) Carbon is oxidized
iv) Lead oxide is reduced
A) (i) and (ii)
B) (i) and (iii)
C) (i), (ii) and (iii)
D) all
Answer:
B) (i) and (iii)

4. The chemical equation
BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl represents following type of chemical reaction.
A) displacement
B) combination
C) decomposition
D) double-displacement
Answer:
D) double-displacement

5. The reaction of formation hydrogen chloride from hydrogen and chlorine repre¬sents following type of chemical reaction i
A) decomposition
B) displacement
C) combination
D) double-displacement
Answer:
C) combination

9th Class Physical Science 6th Lesson Chemical Reactions and Equations InText Questions and Answers

9th Class Physical Science Textbook Page No. 90

Question 1.
What changes do you notice generally?
Answer:
I noticed two types of changes generally. They are :

1. Physical change,
2. Chemical change.

Question 2.
“Coal is burnt”, “crackers are burnt” ………. changes Are they physical changes (or) chemical changes?
Answer:
They are all chemical changes.

Question 3.
Are they (coal, crackers) temporary changes or permanent changes?
Answer:
They are permanent changes.

Question 4.
How do we know a chemical reaction has taken place?
Answer:
While we observe the following things, we can conclude that a chemical reaction has taken place.

1. A change that changes state and colour of substance.
2. A change that releases heat energy.
3. A change which forms an insoluble substance as precipitate.
4. A change that liberates a gas.

9th Class Physical Science Textbook Page No. 92

Question 5.
Do the atoms of each element on left side equal to the atoms of the element on the right side of the equation?
Answer:
Yes. The atoms of each element in left side are equal to the atoms of their corresponding element.

9th Class Physical Science Textbook Page No. 94

Question 6.
Are the atoms of all elements of reactants present in products?
Answer:
Yes. The atoms of all elements of reactants are present in products.

Question 7.
2 C₃H₈ + 10 O₂ → 6 CO₂ + 8 H₂O.
Is it a well balanced equation as per rules ? How do you say?
Answer:
Yes, it is a balanced equation.

9th Class Physical Science Textbook Page No. 105

Question 8.
Did you notice the colour coating on silver and copper articles?
Answer:
Yes. I noticed. To prevent from corrosion they are colour coated like that.

9th Class Physical Science Textbook Page No. 106

Question 9.
How can we prevent the spoiling of food?
Answer:
The spoilage of food can be prevented by adding preservatives like vitamin C and vitamin E.

9th Class Physical Science 6th Lesson Chemical Reactions and Equations Activities

Activity – 1

Question 1.
Write an activity when calcium oxide reacts with water.
What type of reaction is this ? Write balanced chemical equation.
Answer:

1. Take about 1 gm of quick lime (CaO) in a beaker.
2. Add 10 ml of water to this.
3. Touch the beaker with your finger.
4. The beaker is hot.
5. The reason is that the calcium oxide reacts with water and releases heat energy.
6. Calcium oxide dissolves in water producing colourless solution [Ca(OH)2],
7. Dip the red litmus paper in it.
8. Red litmus paper changes into blue colour.
9. We conclude that the solution obtained is a basic solution.
10. It is a chemical combination reaction.
    CaO_(s) + H₂O_(l) → Ca(OH)₂

Activity – 2

Question 2.
What change do you observe by mixing of sodium sulphate solution and barium chloride solution?
(OR)
Which type of reaction is this (mixing of sodium sulphate and barium chloride solutions)?
Answer:

• Take about 100 ml of water in a beaker.
• Dissolve a small quantity of sodium sulphate (Na₂SO₄) in it.
• Take about 100 ml of water in another beaker.
• Dissolve a small quantity of barium chloride (BaCl₂) in it.
• Add these (Na₂SO₄, BaCl₂) two solutions.
• We will get a white precipitate of barium sulphate.
  Na₂SO₄ + BaCl₂ → BaSO₄↓ + 2NaCl
• It is a double displacement reaction.

Activity – 3

Question 3.
Formation of H₂ gas by the action of dil. HCl and Zn pieces.
(OR)
What happens if dilute HCl is added to zinc granules. Explain the process with an experiment. What type of reaction is this? Write the balanced chemical equation for this process.
Answer:

Hydrogen gas experiment:

1. Take a few zinc granules in a conical flask.
2. Add about 5 ml of dilute hydrochloric acid to the conical flask.
3. Observe the changes in the conical flask.
4. Keep a burning matchstick near the mouth of the conical flask.
5. The light of burning matchstick is put off with ‘pop’ sound.
6. This indicates the H₂ gas has released in this reaction.
7. When we touch the bottom of the conical flask with our finger, we feel hot.
8. So, this reaction produces heat.
9. It is a displacement reaction
   Zn_(s) + 2HCl_(aq) → ZnCl_2(aq) + H_2(g)

Activity – 4 Chemical Combination

Question 4.
Write an activity on burning of magnesium ribbon in the presence of air.
(OR)
Write an activity which shows burning of magnesium ribbon is a chemical combination reaction.
Answer:

1) Take about 3 cm long piece of magnesium ribbon.
2) Rub the magnesium ribbon with sand paper.
3) Hold it with a pair of tongs.
4) Burn it with a spirit lamp or burner.
5) Magnesium burns in oxygen by producing dazzling white flame and changes into white powder (magnesium oxide).

7) In this reaction magnesium and oxygen combine to form a new substance magnesium oxide.
8) A reaction in which single product is formed from two or more reactants is known as chemical combination reaction.

Activity – 5 Decomposition Reaction

Question 5.
How can you prove that CO₂ is released on heating CaCO₃?
(OR)
When calcium carbonate is heated it releases certain gas. What is the gas that has been released? How do you identify that gas? Write the activity of this reaction.
Answer:

• Take a pinch of Calcium Carbonate (lime stone) in a boiling tube.
• Heat the boiling tube over the flame of spirit lamp or burner.
• Now take a burning matchstick near the mouth of delivery tube.

Heating of calcium carbonate and testing the gas evolved with burning matchstick

• We can observe that the matchstick kept near the mouth of the boiling tube will be put off.
• In the above activity, on heating CaC03, it decomposes to Calcium oxide (CaO) and Carbon dioxide (CO₂).
  
• When a decomposition reaction is carried out by heating, it is called thermal decomposition reaction.

Activity – 6

Question 6.
Write an activity to show that nitrogen dioxide (NO^ is released by heating of lead nitrate.
(OR)
Write an activity which shows thermal decomposition reaction.
Answer:

1. Take about 0.5 g of lead nitrate powder in a boiling test tube.
2. Hold the boiling tube with a test tube holder.
3. Heat the boiling tube over a flame.
4. We observe that on heating of lead nitrate brown fumes of nitrogen dioxide are released.
5. On heating lead nitrate, it decomposes and gives lead oxide, oxygen and nitrogen dioxide.

6) Reaction :

7) This is also an example for thermal decomposition reaction.

Activity – 7

Question 7.
Write an activity to show dissociate of water into hydrogen and oxygen.
(OR)
Write an activity showing decomposition of a compound in the presence of electricity.
(OR)
Draw a neat diagram representing electrolysis of water.
How do you prove that water contains hydrogen and oxygen in the ratio of 2 : 1?
(OR)
Explain electrolysis of water.
(OR)
What are the materials required for the experiment to show the chemical decomposition of water ? Write the procedure of the experiment. Name the products which we get in this reaction.
(OR)
Draw a neat diagram showing the electrolytic decomposition reaction of water. Write the balanced chemical equation of the above reaction.
Answer:

1. Take a plastic mug.
2. Drill two holes at its base.
3. Fit two ‘one holed rubber stoppers’ in these holes.
4. Insert two graphite electrodes in these rubber stoppers.
5. Connect the electrodes to 9V battery.
6. Fill the mug with water, so that the electrodes are immersed.
7. Add few drops of dilute sulphuric acid to water.
8. Take two test tubes filled with water and invert them over the two graphite electrodes.
9. Switch on the current and leave the apparatus undisturbed for sometime.
10. We will notice that the liberation of gas bubbles at both the electrodes.
11. These bubbles displace the water in the test tubes.
12. We also observe that the volume of gas collected in the both test tubes is different.
13. Once the test tubes are filled with gases take them out carefully.
14. Test both the gases separately by bringing a burning candle near the mouth of each test tube.
15. The gas which occupies high volume makes to burn with blue flame and put off candle flame with ‘puf sound is hydrogen gas.
16. The gas which occupies low volume and makes to burn candle brightly is oxygen gas.
17. In the above activity on passing the electricity, water dissociates to hydrogen and oxygen gases. This is called electrolytic composition reaction.
    

Activity – 8

Question 8.
Write an activity to observe silver bromide decomposes in the presence of sunlight.
(OR)
Write an activity for photochemical reaction.
Answer:

• Take some quantity of silver bromide on a watch glass.
• When we observe the colour of silver bromide it is light yellow in colour.
• Place the watch glass in sunlight for sometime.
• Silver bromide decomposes to silver and bromine in sunlight.
• After sometime we observe that the colour of silver bromide changes into grey colour.

• Light yellow coloured AgBr changes into Ag and Br.
• This decomposition reaction occurs in the presence of sunlight and such reactions are called photochemical reactions.
• The above decomposition reactions require some energy to convert reactants to products.
• This type of reactions are called endothermic reactions.
• 

Activity – 9

Question 9.
Write an activity to show that zinc displaces hydrogen from dil. hydrochloric acid.
Answer:

1. Take a small quantity of zinc dust in a conical flask.
2. Add dilute hydrochloric acid slowly.
3. Now take a balloon and tie it to the mouth of the conical flask.
4. We can observe that the gas bubbles coming out from the solution and the balloon bulges out.
5. Zinc pieces react with dilute hydrochloric acid and liberate hydrogen gas.
6. The element zinc has displaced hydrogen from dilute hydrochloric acid. This is one of the examples for displacement reaction.
7. Equation : Zn_(s) + 2 HCl_(aq) → ZnCl_2(aq) + H_2(g) ↑

Activity – 10

Question 10.
Write an activity to show that iron displaces copper from copper sulphate.
(OR)
Write an activity about how you conduct an experiment to show that more reactive metals replace less reactive metals from their compounds. (OR)
Why is an iron nail kept in a solution brownish? Explain the activity.
Answer:

1. Take two iron nails and clean them by rubbing with sand paper.
2. Take two test tubes and mark them as A and B.
3. Take about 10 ml of copper sulphate solution in each test tube.
4. Dip one iron nail in copper sulphate solution of test tube A.
5. Keep the test tube with iron nail undisturbed for 20 minutes.
6. Keep the other iron nail and test tube aside. ;
7. Compare the colours of the solutions in the test tubes.
8. Now take out the iron nail from copper sulphate solution.
9. Keep the iron nail and test tubes A and B side by side.
10. Compare with the other iron nail that has been kept aside.
11. We will observe that the iron nail dipped in copper sulphate solution becomes brownish.
12. Blue colour of copper sulphate solution in test tube ‘A’ fades.
13. Iron is more reactive than copper, so it displaces copper from copper sulphate.
14. This is one of the examples for displacement reaction.
15. Equation : Fe_(s) + CuSO_4(aq) → FeSO_4(aq) + Cu_(s)

Activity – 11 Double displacement reaction

Question 11.
Write an activity for the formation of lead iodide and potassium nitrate.
(OR)
Your friend has a doubt about chemical double displacement reaction. How can you clarify his/her doubt by showing an experiment? Explain.
(OR)
What happens if lead nitrate solution is added to potassium iodide solution? Explain the activity.
Answer:

It forms yellow precipitate of lead iodide.

1. Take a pinch of lead nitrate and dissolve in 5.0 ml of distilled water in a test tube.
2. Take a pinch of potassium iodide in another test tube and dissolve in distilled water.
3. Mix lead nitrate solution with potassium iodide solution.
4. We observe that a yellow coloured substance which is insoluble in water, is formed.
5. This insoluble substance is known as precipitate.
6. The precipitate is Lead Iodide.
7. Equation : Fb(NO₃)_2(aq) + 2 KI_(aq) → Pbl_2(s) + 2KNO_3(aq)
8. In the above reaction, lead ion and potassium ion exchange their places each other.
9. Lead ion combines with iodide ion and forms Pbl2 as precipitate and KN03 remains in the solution.
10. Such reaction is double displacement reaction.

Activity – 12

Question 12.
Write an activity to show that the oxidation of copper to copper oxide.
(OR)
Write an activity for example of oxidation and reduction.
Answer:

• Take about 1.0 gram of copper powder in a China dish.
• Keep the China dish on a tripod stand containing wire gauze.
• Heat it with a bunsen burner or with a spirit lamp.
• We can find that the surface layer of copper becomes black.
• On heating copper reacts with oxygen present in the atmosphere to form copper oxide.

•  Here copper combines with oxygen to form copper oxide.
• Here oxygen is gained and the process is called oxidation.
• Now pass hydrogen gas over hot copper oxide.

• We can observe that the black coating on copper turns brown because copper oxide loSes oxygen to form copper.
• In this process oxygen is lost and the process is called ”Reduction”.

AP State Syllabus AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion Textbook Questions and Answers.

AP State Syllabus 9th Class Physical Science Solutions 2nd Lesson Laws of Motion

9th Class Physical Science 2nd Lesson Laws of Motion Textbook Questions and Answers

Improve Your Learning

Question 1.
Explain the reasons for the following. (AS 1)
a) When a carpet is beaten with a stick, dust comes out of it.
Answer:

1. The dust particles in the carpet are at rest.
2. When the carpet is beaten with a stick, the state of rest of the dust particles is disturbed.
3. Due to inertia, the dust particles comes out.

b) Luggage kept on the roof of a bus is tied with a rope.
Answer:

1. Luggage kept on the roof of a bus is in the state of rest.
2. As the bus moves, the luggage also moves with a velocity equal to the velocity of the bus.
3. If the bus suddenly stops, the luggage resists to change its state of motion.
4. Hence due to inertia it will fall down.
5. To avoid this, the luggage is tied with a rope.

c) A pace bowler in cricket runs in from a long distance before he bowls.
Answer:

1. When he runs in from a long distance, he gains momentum of inertia.
2. Due to this larger inertia, larger force is applied in a short interval of time.
3. Hence the momentum will be more.

Question 2.
Two objects have masses 8 kg and 25 kg. Which one has more inertia? Why? (AS 1)
Answer:

1. The object with mass 25 kg has more inertia.
2. The resistance to change the state of object will be more for a body of larger mass.

Question 3.
Keep a small rectangular shaped piece of paper on the edge of a table and place an old five rupee coin on its surface vertically as shown in the figure below. Now give a quick push to the paper with your finger. How do you explain inertia with this experiment?

Answer:

1. The coin and the paper are in inertia of rest.
2. When we give a quick push to the paper, paper comes to inertia of motion and the coin remains in its original state i.e., inertia of rest.
3. As a result, the paper will come out and the coin remains on the table without changing its position.

Question 4.
If a car is travelling westwards with a.constant speed of 20 m/s, what is the resultant force acting on it? (AS 1, AS 7)
Answer:

1. A car is moving with a constant speed.
2. Hence the net force on the car is zero both in horizontal and vertical directions.

Question 5.
What is the momentum of a 6.0 kg bowling ball with a velocity of 2.2 m/s? (AS 1)
Answer:
Mass of the ball (m) = 6.0 kg
Velocity of the ball (v) = 2.2 m/s
Momentum (p) = nv = 6.0 kg × 2.2 m/s = 13.2 kg m/s (or) 13.2 N-s

Question 6.
Two people push a car for 3 sec, with a combined net force of 200 N. (AS 1)
a) Calculate the impulse provided to the car.
Answer:
F_net = 200 N ; ∆t =3 sec
Impulse ∆p = F_net. ∆t = 200 × 3 = 600 N – sec.

b) If the car has a mass of 1200 kg, what will be its change in velocity?
Answer:
Mass of the car (m) = 1200 kg.; Net force (F_net) = 200 N
Time ∆t = 3 sec. ; Change in velocity ∆v =?

Question 7.
What force is required to produce an acceleration of 3 m/sec2 in an object of mass 0.7 kg? (AS 1)
Answer:
Mass of the object (m) = 0.7 kg.; Acceleration (a) = 3 m/sec²
Force required (F) = ?
F = ma = 0.7 x 3 = 2.1 N

Question 8.
A force acts for 0.2 sec on an object having mass 1.4 kg initially at rest. The force stops to act but the object moves through 4 m in the next 2 seconds, find the magnitude of the force. (AS 1)
Answer:

Velocity after 0.2 sec v = u + at = 0 + 0.2 a = 0.2a ……….. (1)
After 0.2 s, the body moves with uniform velocity, acceleration is zero, because force is removed.
∴ Velocity v = \(\frac{s}{t}=\frac{4}{2}\) =2 m/s. ………. (2)
From (1) & (2)
v = 0.2a ⇒ 2 = 0.2a
⇒ a = \(\frac{2}{0.2}\) = 10 m/s²
∴ Force applied F = ma = 1.4 kg × 10 m/s² = 14 N.

Question 9.
An object of mass 5 kg is moving with a velocity of 10 ms^-1. A force is applied so that in 15 s, it attains a velocity of 25 ms^-1. What is the force applied on the object? (AS 1)
Answer:
Mass (m) = 5 kg ; Initial velocity (u) = 10 m/s.; Time (t) = 15 s
Final velocity (v) = 25 m/s.
Acceleration a = \(=\frac{v-u}{t}=\frac{25-10}{15}=\frac{15}{15}=1 \mathrm{~m} / \mathrm{s}^{2}\).
Force applied on the object F = ma = 5 × 1=5 N.

Question 10.
Find the acceleration of body of mass 2 kg from the figures shown. (AS 1)

Answer:
1) Force 30 N is acting downwards on weight of (2 × 10) = 20 kg.
The acceleration a = \(\frac{30-20}{2}=\frac{10}{2}\) = 5 m/s²

2) m₁ = 2 kg, m₂ = 3 kg.
m₂ pulls the body mt with a weight 3 × 10 = 30 N.
∴ Acceleration of m₁ = \(\frac{30-20}{3+2}=\frac{10}{5}\)
= 2 m/s².

Question 11.
Take some identical marbles. Make a path or a track keeping your notebooks on either side so as to make a path in which marbles can move. Now use one marble to hit the other marbles. Take two, three marbles and make them to hit the other marbles. What can you explain from your observations? (AS 5)
Answer:

1. When one marble is hit by another marble, both the marbles move with some velocity.
2. When the marble is hit by two, three marbles, all marbles move with a velocity which is more than in the previous case.
3. As we are hitting with more marbles, the mass increases. So that the net momentum also increases.

Question 12.
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? (AS 1, AS 7)
Answer:
Mass m = 30 kg. ; Force F = 450 N
Acceleration a = ?
F = ma
∴ a = \(\frac{\mathrm{F}}{\mathrm{m}}=\frac{450}{30}\) =15 m/sec²
∴ The required acceleration =15 m/sec²

Question 13.
An 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/sec²? (AS 1, AS 7)
Answer:
Mass of the vehicle, m = 1500 kg.; Acceleration (-a) = 1.7 m/sec²
Force, F =?
F = m (-a) = 1500 × (-1.7) = ( – ) 2550 N
∴ The force between the vehicle and road is 2550 N, in the direction opposite to that of the vehicle.

Question 14.
If a fly collides with the windshield of a fast moving bus, is the impact force experienced, same for the fly and the bus? Why? (AS 1, AS 2)
Answer:
The impact force experienced by the fly will be more, because the mass of fly is negligible when compared to the mass of the bus.

Question 15.
A truck is moving under a hopper with a constant speed of 20 m/sec. Sand falls on the truck at a rate 20 kg/s. What is the force acting on the truck due to falling of sand? (AS 1, AS 7)
Answer:
Mass of the sand falling on the truck in 1 sec = 20 kg
Constant speed of the truck = 20 m/s
Acceleration in 1 sec, a = \(\frac{\Delta v}{\Delta t}=\frac{20}{1}\) = 20 m/sec²
Force applied on the truck, F = ma = 20 kg x 20 m/sec² = 400 N

Question 16.
Two rubber bands stretched to the standard length cause an object to accelerate at 2 m/sec². Suppose another object with twice the mass is pulled by four rubber bands stretched to the standard length. What is the acceleration of the second object? (AS 1)
Answer:
First object:
Let the force applied by two rubber bands = F₁ Newton
Mass of the object = m₁ kg; Acceleration a₁ = 2 m/sec²
We know F = ma
F₁ = m₁ × 2
⇒ F₁ = 2m₁ …….(1)

Second object:
The force applied by 4 rubber bands = 2F₁ Newton
Mass of the object = 2m₁ kg ; Acceleration a₂ =?
We know F = ma
2F₁ = 2m₁. a₂

∴ Acceleration of the second object = 2 m/sec²

Question 17.
Illustrate an example of each of the three laws of motion. (AS 1)
Answer:
First law of motion :
A body continues its state of rest or of uniform motion unless a net force acts on it.
Ex:

1. When the bus which is at rest begins to move suddenly, the person standing in the bus falls backward.
2. When you are travelling in bus, the sudden stop of the bus makes you fall forward.

Second law of motion :
The rate of change of momentum of a body is directly proportional to the net force acting on it and it takes place in the direction of net force.
Ex : Place a ball on the veranda and push it gently. Then the ball accelerates from rest. Thus, we can say that force is an action which produces acceleration.

Third law of motion :
If one object exerts a force on the other object, the second object exerts a force on the first one with equal magnitude but in opposite direction.

Ex :

1. When birds fly, they push the air downwards with their wings, and the air pushes back the bird in opposite upward direction.
2. When a fish swims in water, the fish pushes the water back and the water pushes the fish with equal force but in opposite direction.
3. A rocket accelerates by expelling gas at high velocity. The reaction force of the gas on the rocket accelerates the rocket in a direction opposite to the expelled gases.

Question 18.
Two ice-skaters initially at rest, push of each other. If one skater whose mass is 60 kg has a velocity of 2 m/s. What is the velocity of other skater whose mass is 40 kg? (AS 1, AS 7)
Answer:
Mass of first skater m₁ = 60 kg.;
Velocity of first skater v₁ = 2 m/s.
Mass of second skater m₂ = 40 kg.; Velocity of second skater v₂ =?
As the two skaters push each other, the resultant momentum will become zero.
The resultant momentum m₁v₁ + m₂v₂ = 0

∴ Velocity of second skater is 3 m/s, but in the direction opposite to the first skater.

Question 19.
A passenger in moving train tosses a coin which falls behind him. It means that the motion of the train is (AS 7)
a) Accelerated
b) Uniform
c) Retarded
d) Circular motion
Answer:
a) Accelerated

Question 20.
A horse continues to apply a force in order to move a cart with a constant speed. Explain. (AS 1)
Answer:

1. The cart moves when the force (in the form of pulling) is applied by the horse.
2. As the horse and cart are moving, the net momentum will be zero at any instance of time.
3. Hence when the horse comes to rest, the cart also comes to rest.
4. To avoid this and to move the cart with a constant speed, the horse must apply force continuously.

Question 21.
A force of 5 N produces an acceleration of 8 m/sec² on a mass m, and an acceleration of 24 m/sec² on a mass m₂. What acceleration would the same force provide If both the masses are tied together? (AS 1)
Answer:
For the first mass (m₁)
Force F = 5 N
Acceleration a = 8 m/s²
We know, F = ma
5 = m, . 8
m₁= \(\frac{5}{8}\) kg

For the second mass (m₂)
Force F = 5 N
Acceleration a = 24 m/sec²
We know, F = ma
5 = m₂.24
m₂ = \(\frac{5}{24}\) kg
When both the masses are tied together and the same force is applied, then

Question 22.
A hammer of mass 400 g, moving at 30 m/s, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer? (AS 1)
Answer:
Mass of the hammer (m) = 400 g = 0.4 kg. ; Velocity of the hammer (v) = 30 m/s.
Momentum (∆p) = 30 × 0.4 N – s.
The nail stops the hammer with in a time 0.01 s.
∴ ∆t = 0.01 s.
The stopping force of the nail on the hammer.

Question 23.
System is shown in figure. Assume there is no friction. Find the acceleration of the liloc-ks-and tension in the string. Take g = 10 m/s² (AS 1)

Answer:
m₁ = 3 kg, m₂ = 3 kg.
1) Acceleration ’a’ and tension T on m₁ are shown in figure.
2) Acceleration a and tension T will be as shown in figure.

Normal force = weight 3s – T = 3a ………… (2)
∴ only tension applied
Tension T = 3 kg × a = 3a
From (1) & (2)
3g – 3a = 3a ⇒ 3g = 6a ⇒ a = \(\frac{3 g}{6}=\frac{3 \times 10}{6}\) = 5 m/s².
Tension T = 3a = 3 × 5 = 15 N.

Question 24.
Three identical blocks, each of mass 10 kg, are pulled as shown on the hoii ;ontal frictionless surface. If the tension (F) in the rope is 30 N, what is the acceleration oi each block? And what are the tensions in the other ropes? (Neglect the masses of the ropes) (AS 1)

Answer:
Three blocks, each of mass 10 kg are pulled by a rope.
∴ Total mass = 10 + 10 + 10 = 30 kg
Force applied by the rope, F = 30 N
∴ Acceleration of each block, a = \(\frac{F}{m}=\frac{30}{30}\) = 1 m/sec².

Tension in first rope (T₁)
First rope pulls only one block whose mass is 10 kg with an acceleration 1 m/sec².
∴ F = ma = 10 kg × 1 m/sec² = 10 N

Tension in second rope (T₂)
Second rope pulls two blocks, each of mass 10 kg.
∴ Total mass = 10 + 10 = 20 kg.
Acceleration, a = 1 m/sec²
Force, F = ma = 20 × 1 = 20 N

Question 25.
A ball of mass’m’ moves perpendicularly to a wall with a speed v, strikes it and rebounds with the same speed in the opposite direction. What is the direction and magnitude of the average force acting on the ball due to the wall? (AS 7)
Answer:
According to the Newton s third law of motion,
Force exerted by ball on the wall = – (Force exerts by the wall on the ball)
∴ F_B.W = – F_W.B
Force exerted by ball :
Mass of ball = m, speed = v, F_BW = ma = \(\frac{\mathrm{m} \cdot \mathrm{v}}{\mathrm{t}}\)
As the wall is at rest and exerts some force on the ball of mass m, then it moves in the other direction with the same speed.

Question 26.
Divya observed a horse pulling a cart. She thought that cart also pulls the horse with same force in opposite direction. As per third law of motion, the cart should not move forward. But her observation of moving cart raised some questions in her mind. Can you guess what questions are raised in her mind? (AS 2)
Answer:

• According to Newton’s third law, when horse pulls a cart, the cart also pulls the horse with same force but in opposite direction. So the cart has to stop. Why is the cart moving?
• What is the effect of friction of ground on the cart and horse?
• Does the cart continue to move even if the horse stops pulling it?
• What makes the cart to move continuously?
• Does it become an isolated system?
• What is the action and reaction in this system?

Question 27.
How do you appreciate Galileo’s thought of “any moving body continues in the stale only until some external force acts on it”, which is contradiction to the Aristotle’s belief of “any moving body naturally comes to rest”? (AS 6)
Answer:

• Science is dynamic.
• All theories can change time to time so that the science and technology will be developed.
• Aristotle’s belief proved to be wrong only by the experiments conducted by Galileo.
• So anybody can challenge the existing theories with proper experimentation.
• Aristotle’s and Galileo’s contradictory thought lead Newton to propose most popular laws of motion.
• Newton’s third law of motion is the basic principle in rocket launching.
• Nowadays we are enjoying the results of satellites launched by rockets.
• Hence comfortable life is the effort on experiments, theories, and calculations made by scientists with a zeal to invent new.

9th Class Physical Science 2nd Lesson Laws of Motion InText Questions and Answers

9th Class Physical Science Textbook Page No. 24

Question 1.
Do all the bodies have the same inertia?
Answer:
The inertia of all bodies is not same. It depends on the mass of the object.

Question 2.
What factors can decide the inertia of a body?
Answer:
Mass is the factor, which decides the inertia of a body.

9th Class Physical Science Textbook Page No. 26

Question 3.
Is the acceleration increased when net force is increased?
Answer:
Yes, the acceleration increased, as we increased the net force without changing mass of the object.

9th Class Physical Science Textbook Page No. 29

Question 4.
What do you notice from the readings in the spring balances?
Answer:
The two spring balances stretch up to a certain limit equally.

Question 5.
Are the readings of two spring balances the same?
Answer:
Yes, the readings of two spring balances are equal.

Question 6.
Are we able to make the spring balances to show different readings by pulling them simultaneously in opposite directions? Why not?
Answer:
When same force is applied in both the directions, we are unable to make the spring balances to show different readings because the action and reaction are same in magnitude and opposite in direction. When we use two forces with different magnitudes, then the spring balances can show different readings.

9th Class Physical Science Textbook Page No. 30

Question 7.
Does the rocket exert a force on the gas expelled from it?
Answer:
The rocket also exerts a force on the gas expelled from it.

9th Class Physical Science Textbook Page No. 32

Question 8.
Why does a pole vault jumper land on thick mats of foam?
Answer:
A thick mat of foam reduces the force of impact of the jumper, so that he doesn’t have any damage to his body.

Question 9.
Is it safe to jump on sand rather than a cement floor? Why?
Answer:

• It is safe to jump on sand rather than a cement floor.
• A soft and more cushioned landing surface provides a greater stopping distance because of the longer time taken to stop.

9th Class Physical Science Textbook Page No. 24

Question 10.
You may have seen the trick where a tablecloth is jerked from a table, leaving the dishes that were on the cloth nearly in their original positions.
a) What do you need to perform this successfully?
Answer:
We need a table, a cloth and some massive objects to perform this activity. The performer drag the cloth from the table very skillfully.

b) Which cloth should we use? Is it cloth made of thick cotton or thin silk?
Answer:
We have to use a thin silk cloth to perform this activity.

c) Should the dishes possess large mass or small mass?
Answer:
The dishes must possess large mass. We should not use lighter objects like plastic cups, etc.

d) Is it better to pull the cloth with a large force or pull it with a gentle and steady force?
Answer:
The cloth must be pulled with a gentle force, but with a sudden jerk.

Question 11.
What is the velocity of a small object that has separated from a rocket moving in free space with velocity 10 km/s?
Answer:
When a small object is separated from another object which is moving with a certain velocity. The small object also moves with a velocity equal to that of the object from which it is separated. Hence, the speed of the small object is 10 km/s.

9th Class Physical Science Textbook Page No. 27

Question 12.
Observe the following diagram.

What is the upper limit of weight that a strong man of mass 80 kg can lift as shown in figure?
Answer:
Total force mg = N + T (N = normal force, T = tension)
As the person is standing on the floor, the normal force N = 0.
80 = 0 + T
∴ T = 80
∴ The upper limit of the weight that the person in the figure can lift is 80 kg.

Question 13.
What is the momentum of a ceiling fan when it is rotating?
Answer:
Ceiling fan when it is rotating, possesses angular momentum.
Angular momentum L = mvr or mr²ω.

Question 14.
Is it possible to move in a curved path in the absence of a net force?
Answer:
A body comes into curved path, when centripetal force real force acts on it. Immediately after coming into curved path, an imaginary force which acts away from the centre i.e., centrifugal force comes into existence. These two forces are equal in magnitude and opposite in direction. Hence the net force is zero. So it is possible to move in a curved path in the absence of a net force.

Question 15.
Prove that the tension throughout the string is uniform when the mass of string is considered to be zero.
Answer:

Let a body of mass m is suspended through a string. The weight of the object mg acts downwards. Now tension in the string T = mg + m_sg when m_s is the mass of the string.
Here m_s is considered as zero.
Hence T_A= mg + 0 = mg; T_B = mg ; T_C = mg ; T_D = mg
∴ The tension throughout the string is uniform when the mass of string is considered to be zero.

9th Class Physical Science Textbook Page No. 31

Question 16.
The force exerted by the earth on the ball is 8 N. What is the force on the earth by the ball?
Answer:

• The force exerted by the earth on the ball is 8 N.
• The force exerted by the ball on the earth is – 8 N.

“According to Newton’s third law, if a body A exerted a force p on another body B, the B exerts a force -p on A, the two forces acting along the same line”.

Question 17.
A block is placed on the horizontal surface. There are two forces acting on the block. One, the downward pull of gravity and other a normal force acting on it. Are these forces equal and opposite? Do they form action – reaction pair? Discuss with your friends.
Answer:
These two forces form action-reaction pair.

Question 18.
Why is it difficult for a fire fighter to hold a hose that ejects large amount of water at high speed?
Answer:
A large amount of water with high speed ejects from the hose of a fire engine, produces a large force in forward direction. According to action-reaction, the hose moves back with the same force. But the fire fighter has to resist that reaction force. Hence it becomes very difficult for him.

9th Class Physical Science Textbook Page No. 33

Question 19.
A meteorite burns in the atmosphere before it reaches the earth’s surface. What happens to its momentum?
Answer:
The momentum of the meteorite becomes zero. It doesn’t touch the ground as it burns in the atmosphere. So no mass of the meteorite hits the ground.

Question 20.
As you throw a heavy ball upward, is there any change in the normal force on your feet?
Answer:
The normal force on the feet changes its direction and acts in upward direction. As a result we raise our foot while throwing the ball.

Question 21.
When a coconut falls from a tree and strikes the ground without bouncing. What happens to its momentum?
Answer:
Its momentum doesn’t change but its impact force will be very less because it is not bouncing.

Question 22.
Air bags are used in’the cars for safety. Why?
Answer:
When a car hits another vehicle, the air bags immediately comes in between the persons in the car and the wind shield of the car, to prevent damage to life of passengers.

9th Class Physical Science 2nd Lesson Laws of Motion Activities

Activity – 1

Question 1.
Explain the motion of a pen cap kept on a thick paper ring.
Answer:

1. Make a circular strip from a thick paper.
2. Balance the hoop on the centre of the mouth of the bottle.
3. Now balance a pen cap on the paper hoop aligning it on the centre of the bottle’s mouth.
4. Give the paper hoop a sharp push with your finger as fast as you can.
5. We observe that the pen cap suddenly falls into the bottle.
6. As we push paper hoop, we applied force on the paper hoop. So it changed its state from rest to motion.
7. Pen cap cannot change its state of rest.
8. Due to gravitational force, the pen cap falls into the bottle.

Activity – 2

Question 2.
Explain the motion of the carrom coins hit by a striker.
Answer:

1. Make a stack of carrom coins on the carrom board.
2. Give a sharp hit at the bottom of the stack with striker.
3. We can find that the bottom coin will be removed from the stack.
4. The other coins in the stack will slide down.
5. When we apply force on the bottom coin, the coin will move, due to change in the state of rest.
6. The stack of remaining coins does not fall vertically due to inertia.

Activity – 3

Question 3.
Show that the object with larger mass has greater inertia.
Answer:

1. Take two rectangular wooden blocks with different masses.
2. Place them on a straight line drawn on a floor.
3. Give the same push at the same time to both the blocks with the help of a wooden scale.
4. We observe that the block with small mass will accelerate more and goes farther.
5. The block with large mass accelerates less and moves shorter, due to high inertia.
6. This shows that the bodies of higher mass have high inertia.

Activity – 4

Question 4.
Show that the larger the net force greater the acceleration.
Answer:

1. Gently push a block of ice on a smooth surface and observe how the object speeds up, in other words how it accelerates.
2. Now increase the net force and observe change in its speed.

Observation : The acceleration increases.

Conclusion : If the net force is larger, then the accelerations greater.

Activity – 5

Question 5.
Show that the larger the mass smaller the acceleration.
Answer:

1. Apply a force on an ice block.
2. It undergoes some acceleration.
3. Now take a block of ice with greater mass.
4. Then apply almost the same force on the ice block which has greater mass.

Observations :

1. In both cases the object accelerates.
2. But we can observe in the second case, it will not speed up as quickly as before.

Conclusion : If the mass is larger, then the acceleration is smaller.

Activity – 6

Question 6.
Pulling two spring balances.
Answer:
Let’s take two spring balances of equal calibrations. Connect the two spring balances as shown in figure. Pull the spring balances in opposite directions as shown in figure.

Observation :
There is no change in the reading of spring balances. We are not able to make the spring balances to show different readings by pulling them simultaneously in opposite directions.

Conclusion :
According to third law of motion, when an object exerts a force on the other object, the second object also exerts a force on the first one which is equal in magnitude but opposite in direction.

The two opposing forces are known as action and reaction pair. Newton’s third law explains what happens when one object exerts a force on another object.

Activity – 7

Question 7.
Describe the preparation of a balloon rocket. What is the principle involved in it?
Answer:

Preparation of a balloon rocket :

1. Inflate a balloon and press its neck with fingers to prevent air escaping from it.
2. Pass a thread through a straw and tape the balloon on the straw.
3. Hold one end of the thread and ask your friend to hold the other end of the thread.
4. Now release air from balloon by removing fingers from the neck of the balloon.
5. The balloon moves like a rocket towards the other end

Principle involved in it:

1. Newton’s third law of motion is the principle.
2. As the air in the balloon moves backward, the balloon moves forward.

Lab Activity

Question 8.
Describe an activity to study the action and reaction forces acting on two different objects.
Answer:
Aim : To show the action and reaction forces acting on two different objects.

Material required : Test tube, rubber cork cap, Bunsen burner, laboratory stand and thread.

Procedure :

1. Take a test tube and put a small amount of water in it.
2. Place a rubber cork cap at its mouth to close it.
3. Now suspend the test tube horizontally to a stand with the help of two strings.
4. Heat the test tube with a bunsen burner until water vapourize and the rubber cork cap blows out.

Observations :

1. Observe the movement of test tube when cork cap blows out.
2. As the cork cap blows out in forward direction, the test tube recoils back.
3. We can observe the velocities of cork cap and recoil of test tube to be same.

Activity – 8

Question 9.
Show that the impulse will be less on a soft and cushioned surface.
Answer:

• Take two eggs.
• Drop them from a certain height, so that one egg falls on a concrete floor and the other on a cushioned pillow.
• We observe that the egg that falls on a concrete floor will break.

• The reason is large force acts on the egg for short interval of time.
  ∆p = F_net∆t
• The egg which falls on a cushioned pillow doesn’t break, because a smaller force acts on the egg for a longer time.
  ∆p= F_net ∆t
• This shows that the impulse (∆P) will be less on a soft and cushioned surface.

Note : Even if the ∆p is the same in both cases, the magnitude of the net force (F_{net</sub) acting on the egg determines whether the egg will break or not.}