Category: Class 9

Std 9 Science Chapter 11 Reflection of Light Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 11 Reflection of Light Notes, Textbook Exercise Important Questions and Answers.

Class 9 Science Chapter 11 Reflection of Light Question Answer Maharashtra Board

Class 9 Science Chapter 11 Reflection of Light Textbook Questions and Answers

1. Answer the following questions.

a. Explain the difference between a plane mirror, a concave mirror and a convex mirror with respect to the type and size of the images produced.
Answer:

b. Describe the positions of the source of light with respect to a concave mirror in
1. Torch light
2. Projector lamp
3. Floodlight
Answer:
(a) Torch light: The source of light is placed at the focus.
(b) Projector lamp : The source of light is placed at the centre of curvature.
(c) Flood light : The source of light is placed just beyond the centre of curvature.

c. Why are concave mirrors used in solar devices?
Answer:

• Solar devices like solar cooker or solar water heater use solar energy to cook food or heat water.
• When sun rays fall on the concave mirror, they converge and come together in the focal plane.
• Due to convergence, the intensity of sun rays increases and the food or water is heated faster. Hence, concave mirrors are used in solar- devices.

d. Why are the mirrors fitted on the outside of cars convex?
Answer:

• A convex mirror is used as rear view mirror because they form erect, virtual, and diminished images.
• This, allows the driver to view a large area in a small mirror.

e. Why does obtaining the image of the sun on a paper with the help of a concave mirror burn the paper?
Answer:

• When sunrays fall on the concave mirror, they converge and come together in the focal plane.
• Due to convergence, the intensity of sunrays increases.
• Hence, image of the sun on a paper with the help of concave mirror bums the paper.

f. If a spherical mirror breaks, what type of mirrors are the individual pieces?
Answer:

• When a spherical mirror breaks into smaller pieces, the radius of curvature and focal length does not change.
• Hence, it will continue to behave like a spherical mirror only.

2. What sign conventions are used for reflection from a spherical mirror?
Answer:
According to the Cartesian sign convention, the pole of the mirror is taken as the origin. The principal axis is taken as the X-axis of the frame of reference. The sign conventions are as follows.

1. The object is always kept on the left of the mirror. All distances parallel to the principal axis are measured from the pole of the mirror.
2. All distances measured towards the right of the pole are taken to be positive, while those measured towards the left are taken to be negative.
3. The distance measured vertically upwards from the principal axis are taken to be positive.
4. The distance measured vertically downwards from the principal axis are taken to be negative.
5. The focal length of a concave mirror is negative while that of a convex mirror is positive.

3. Draw ray diagrams for the cases of images obtained in concave mirrors as described in the table on page 122.

Answer:
(a) A ray diagram for object at infinity for a concave mirror.

(b) A ray diagram for object beyond centre of curvature for a concave mirror.

An object beyond centre of curvature for a concave mirror

(c) A ray diagram for object at the centre of curvature for a concave mirror.

Object at centre of Curva fu re be a concave mirror.

(d) A ray diagram for object between F and C for a concave mirror.

Object between F & C for a concave mirror

(e) A ray diagram for obj ect at focus for a concave mirror.

Object at focus for a concave mirror.

(f) A ray diagram for object between pole and focus for a concave mirror.

4. Which type of mirrors are used in the following?
Periscope, floodlights, shaving mirror, kaleidoscope, street lights, headlamps of a car.

Answer:

5. Solve the following examples

a. An object of height 7 cm is kept at a distance of 25 cm in front of a concave mirror. The focal length of the mirror is 15 cm. At what distance from the mirror should a screen be kept so as to get a clear image? What will be the size and nature of the image?
Solution:
Given: Object size (h₁) = 7 cm
Object distance (u) = -25 cm
Focal length (f) = -15cm
To find: Image distance (u) = ?
Image size (h₂) = ?

The screen should be kept 373 cm in front of the mirror. The image is real.

The height of the image is 10.5 cm, it is an inverted and enlarged image.

b. A convex mirror has a focal length of 18 cm. The image of an object kept in front of the mirror is half the height of the object. What is the distance of the object from the mirror?
Solution:
Given: Image size (h₂) = ¹/₂ h₁
Focal length (f) = 18 cm
To find: Object distance (u) ?

The object is placed in front of the convex mirror at a distance of 18 cm.

c. A 10 cm long stick is kept in front of a concave mirror having focal length of 10 cm in such a way that the end of the stick closest to the pole is at a distance of 20 cm. What will be the length of the image?
Solution:
Given: Object size (h₁) = 10 cm
Object distance (u) = -20 cm
Focal length (f) = -10 cm

The height of the image is 10 cm and it is a real and inverted image.

6. Three mirrors are created from a single sphere. Which of the following:
pole, centre of curvature, radius of curvature, principal axis – will be common to them and which will not be common?
Answer:

• Centre of curvature and Radius of curvature will be common for all three pieces.
• Pole and Principal axis will not be common.

Class 9 Science Chapter 11 Reflection of Light Intext Questions and Answers

Class 9 Science Chapter 11 Reflection Of Light Question 1.
What is light
Answer:
Light is a form of electromagnetic radiation that produces the sensation of vision.

9th Class Science Chapter 11 Reflection Of Light Exercise Question 2.
What is a mirror?
Answer:
A mirror is a reflecting surface which reflects light and creates clear images.

9th Class Science Chapter 11 Reflection Of Light Exercise Answer Question 3.
Principal Focus of Concave and Convex Mirror.
Answer:

9th Class Science Chapter 11 Reflection Of Light Notes Question 4.
If we hold a page of a book in front of a mirror, we see laterally inverted letters in the mirror. Why does it happen?
Answer:

• When we hold a page of a book in front of the mirror, the image of the words appear laterally inverted.
• The image of every point of the word is formed behind the mirror at the same distance from the mirror
• Because of this the left and right side of the image is interchanged.
• Hence, if we hold a page of a book in front of a mirror, we see laterally inverted letters in the mirror.

11 Reflection Of Light Exercise Question 5.
Which letters of the English alphabet form images that look the same as the original letters?
Answer:
A, H, I, M, O, T, U, V, W, X, Y

9th Science Chapter 11 Reflection Of Light Exercise Question 6.
When a person stands in front of a plane mirror, how is the image formed? What is the nature of the image?
Answer:

• The image of a person is formed from every point of the source, thereby forming an extended image of the whole source.
• The image formed would be virtual, upright and left-right reversed.

Answer the following questions:

Reflection Of Light Class 9 Questions And Answers Question 1.
Place two plane mirrors at an angle of 90a to each other. Place a small object between them. Images will be formed in both mirrors. How many images do you see? Now change the angle between the mirrors as given in the following table and count the number of images each time. How is this number related to the measure of the angle?
Answer:
The Relation between the angle between the mirrors and the number of images formed is given by
\(n=\frac{360^{\circ}}{\mathrm{A}}-1\)
n = number of images,
A = angle between the mirrors

Class 9 Science Chapter 11 Reflection of Light Additional Important Questions and Answers

Can you recall?

9th Class Science Chapter 11 Reflection Of Light Question 1.
What is meant by reflection of light and what are the types of reflection?
Answer:
The bouncing back of light when it hits an opaque surface is called reflection of light. The two types of reflection are regular and irregular reflection.

Reflection Of Light Class 9 Exercise Answers Question 2.
What are the laws of reflection.
Answer:

• The incident ray, reflected ray and normal all lie in the same plane at the point of incidence.
• The angle of incidence is equal to the angle of reflection.
• The incident ray and the reflected ray lie on opposite sides of the normal.

Choose and the correct option:

Class 9th Science Chapter 11 Reflection Of Light Question Answer Question 1.
If the reflected rays do not actually meet, such an image is called as image.
(a) real
(b) virtual
(c) magnified
(d) inverted
Answer:
(b) virtual

Class 9 Science Chapter 11 Reflection Of Light Exercise Solutions Question 2.
In a plane mirror, the perpendicular distance of the image from the mirror is equal to
(a) the perpendicular distance of the source from the object.
(b) the perpendicular distance of the source from the mirror.
(c) the parallel distance of the source from the object.
(d) the parallel distance of the source from the mirror.
Answer:
(b) the perpendicular distance of the source from the mirror

Reflection Of Light Class 9 Notes Pdf Maharashtra Board Question 3.
The image formed in a convex mirror is always
(a) virtual, smaller and behind the mirror
(b) virtual, smaller and in front of the mirror
(c) real, smaller and behind the mirror
(d) real, smaller and in front of the mirror
Answer:
(a) virtual, smaller and behind the mirror

Reflection Of Light Class 9 Solutions Question 4.
images can be displayed on a screen.
(a) Virtual
(b) Real
(c) Virtual and erect
(d) Virtual and inverted
Answer:
(b) Real

9th Class Science Chapter 11 Reflection Of Light Exercise Pdf Question 5.
A concave mirror is also called as a mirror.
(a) converging
(b) diverging
(c) plane
(d) outward curved
Answer:
(a) converging

9th Science Chapter 11 Reflection Of Light Question 6.
The centre of the mirror surface is called its
(a) pole
(b) centre of curvature
(c) principal axis
(d) focus
Answer:
(a) pole

Class 9 Science Chapter 11 Question Answer Reflection Of Light Question 7.
According to the new sign convention, the of the mirror is taken as origin.
(a) focus
(b) pole
(c) optical centre
(d) centre of curvature
Answer:
(b) pole

Class 9 Science Chapter 11 Reflection Of Light Exercise Question 8.
A convex mirror is also called as a mirror.
(a) converging
(b) plane
(c) diverging
(d) inward curved
Answer:
(c) diverging

Reflection Of Light Class 9 Maharashtra Board Question 9.
In order to see the full image of a person standing in front of a mirror, the minimum height of the mirror must be
(a) same height as that of the person
(b) double the height of the person
(c) half the height of the person
(d) quarter the height of the person
Answer:
(c) half the height of the person

Reflection Of Light Exercise 9th Class Question 10.
If the inner surface of the spherical mirror is reflecting, then it is a mirror, and if the outer surface is reflecting then it is mirror.
(a) convex, concave
(b) convex, plane
(c) concave, plane
(d) concave, convex
Answer:
(d) concave, convex

9th Std Science Chapter 11 Reflection Of Light Question 11.
The image formed by a concave mirror is
(a) always virtual and erect
(b) always virtual and inverted
(c) virtual if the object is placed between the pole and the focus
(d) virtual if the object is beyond the focus
Answer:
(c) virtual if the object is placed between the pole and the focus

Question 12.
No matter how far you stand from a spherical mirror, your image appears erect. The mirror may be
(a) plane
(b) concave
(c) convex
(d) either plane or convex
Answer:
(d) either plane or convex

Question 13.
In case of a concave mirror, an erect image is
(a) real and enlarged
(b) real and diminished
(c) virtual and diminished
(d) virtual and enlarged
Answer:
(d) virtual and enlarged

Question 14.
A rear view mirror of a car is
(a) plane mirror
(b) concave mirror
(c) convex mirror
(d) cylindrical mirror
Answer:
(c) convex mirror

Question 15.
An image of an object placed at infinite distance from a concave mirror is formed at
(a) the focus of the mirror
(b) behind the mirror
(c) centre of curvature
(d) infinity
Answer:
(a) the focus of the mirror

Question 16.
A ray of light parallel to principal axis after reflection from concave mirror passes through
(a) centre of curvature
(b) focus
(c) pole
(d) optical centre
Answer:
(b) focus

Question 17.
The image made by a plane mirror is a image.
(a) real
(b) virtual
(c) inverted
(d) diminished
Answer:
(b) virtual

Question 18.
The size of the image of an object placed at the focus of a concave mirror is
(a) erect
(b) very large
(c) same size
(d) diminished
Answer:
(b) very large

Question 19.
For virtual images, the height is while for real images, it is
(a) positive, positive
(b) negative, positive
(c) negative, negative
(d) positive, negative
Answer:
(d) positive, negative

Find the odd man out:

Question 1.
Torches, flood lights, head lamps of vehicles, rear view mirror.
Answer:
Rear view mirror – In rear view mirrors, convex . mirror is used. Concave mirrors are used in the rest.

Question 2.
Side mirrors of cars, parking mirrors, flood lights, mirror fitted in shops.
Answer:
Flood lights – In flood lights concave mirror is used. Convex mirrors are used in the rest.

Question 3.
Virtual and enlarged, virtual and diminished, real and inverted, real and magnified
Answer:
Virtual and diminished type of image is not formed by a concave mirror. All the other types of images are formed by a concave mirror.

Question 4.
Image is laterally inverted, image is of same size, image is at same distance, image is diminished.
Answer:
Image is diminished is not a characteristic of image formed in a plane mirror. Rest of them are characteristics of plane mirror.

Answer the following in one sentence:

Question 1.
What kind of mirror will a doctor use to concentrate on teeth, eyes, ears etc.?
Answer:
The doctor will use a concave mirror to concentrate on teeth, eyes, ears etc.

Question 2.
What do the nature, position and size of the image depend on?
Answer:
The nature, position and size of the image depend upon the distance of the object from the reflecting surface.

Question 3.
Give the expression for mirror formula.
Answer:
\(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Question 4.
State any four uses of concave mirror.
Answer:
Concave mirrors are used in torches, headlights, shaving mirrors, dentists’ mirrors, solar devices etc.

Question 5.
What are the two types of spherical mirror?
Answer:
Convex mirror and concave mirror are the two types of spherical mirror.

Match the columns:

Question 1.

Answer:
(1 – c),
(2 – d),
(3 – a),
(4 – b)

Question 2.

Answer:
(1 – b),
(2 – c),
(3 – a)

State whether the following statements are true or false. Correct the false statements:

(1) If the mirrors are kept at right angle to each other, then the number of images formed will be 4.
(2) A convex mirror is used in flood lights.
(3) A concave mirror always forms a magnified image.
(4) Images formed by convex mirrors are always virtual.
(5) The distance between the focus and the pole is called the radius of curvature.
(6) Reflection from a spherical mirror obeys laws of reflection.
(7) The reflecting surface of a concave mirror is curved.
(8) Distances measured in the direction of the incident light are taken as positive.
(9) If the image is erect, the height of the image is negative.
(10) A real image can be displayed on a screen.
(11) A concave mirror always forms a real and inverted image.
(12) Doctors use diverging beam of light to study teeth, ears and eyes.
Answer:
(1) False. if the mirrors are kept at right angle to each other then the number of images formed will be 3.
(2) False. a concave mirror is used in flood lights.
(3) False. a concave mirror can sometimes form a diminished image as well.
(4) True
(5) False. the distance between the focus and the pole is called the focal length.
(6) True
(7) True
(8) True
(9) False. if the image is erect, the height of the image is positive.
(10) True
(11) False. a concave mirror can also form a virtual and erect image.
(12) False. doctors use a converging beam of light to study teeth, ears and eyes.

Give scientific reasons:

Question 1.
A concave mirror is called a converging mirror.
Answer:

• When rays of light parallel to the principal axis are incident on concave mirror, they converge.
• After convergence, they meet at one point on the principal axis, hence concave mirror is called converging mirror.

Question 2.
Concave mirrors are used in torches and in car headlights.
Answer:

1. Concave mirrors are used in torches and car headlights because when a source of light is placed at the focus of a concave mirror, a parallel beam of light rays is obtained.
2. This helps us to see things upto a considerable distance in the darkness.

Question 3.
A dentist uses a concave mirror while examining teeth.
Answer:

• A concave mirror produces an erect, virtual and magnified image of an object placed between its pole and focus.
• A dentist uses this principle to get a clear and distinct image of teeth, hence, a dentist uses a concave mirror.

Solve the following numerlcals.

Tips for solving numerical:

• Object distance (u) is always -ve
• If Image distance (u) is +ve then image is behind the mirror and virtual. if u is -ve then image is in front of the mirror and real.
• Object height (h₁) is always +ve since it is erect.
• Image height (h₂) can be +ve for virtual and -ve for real.

Type – A

Question 1.
A bird is sitting in front of two plane mirrors inclined at an angle of 600 to each other. How many images does the bird see in the mirror?
Solution:
Given : Angle between mirror A = 60⁰
To find: Number of images formed n = ?

The brrd sees 5 images in the mirror.

Question 2.
A coin is kept in front of two plane mirrors inclined to each other. If 3 images of the coin are seen then what is the angle A between the mirrors?
Solution:
Given: no. of images formed n =3
To find: Angle between mirror A =?


The mirrors are inclined atan angle of 900 to each other.

Question 3.
An image is formed 5 cm behind a convex mirror of focal length 10 cm. At what distance is the object placed from the mirror?
Solution:
Given: Image distance (u) = 5 cm
Focal length (f) = 10 an
To find: Object distance (u) = ?

The object is placed at a distance of 10 cm in front of the mirror.

Question 4.
An object placed 20 cm in front of a convex mirror is found to have an image 15cm behind the mirror. Find the focal length of the mirror.
Solution:
Given: Object distance (u) = -20 cm
Image distance (u) = 15 cm
To find: focal length (f) = ?


The focal length of the convex mirror is 60 cm.

Numerical For Practice

Question 5.
An object is placed at a distance of 36 cm from a concave mirror of focal length 12 cm. Find the image distance.
Answer:
-18 cm

Question 6.
An arrow is placed at a distance of 25 cm from a diverging mirror of focal length 20 cm. Find the image distance.
Answer:
11.1 cm

Type – B

Question 1.
An object 4cm in height is placed at a distance of 36 cm from a concave mirror. The image is formed 18 cm in the front of the mirror. Find the height of the image.
Solution:
Given: Object height (h₁) = 4 cm
Image distance (u) = -18 cm
Object distance (u) = -36 cm
To find: Height of image (h₂) = ?

The height of the image is 2 cm and it is inverted.

Question 2.
An object 2 cm high is placed at a distance of 16cm from a concave mirror which produces a real image 3 cm high. Find the image distance.
Solution:
Given: Object height (h₁) = 2 cm
Object distance (u) = -16 cm
Image height (h₂) = -3 cm
To find: Image distance (u) = ?

The image is formed at a distance of 24 cm in front of the mirror.

Numericals For Practice

Question 3.
An object 10cm in height is placed at a distance of 36 cm from a concave mirror. 1f the image is formed at a distance of 18 cm in front of the mirror, find the height of the image.
Answer:
-5cm

Question 4.
A converging mirror forms a real image of height 4 cm of an object of height 1 cm placed 20 cm away from the mirror. Find the image distance.
Answer:
-80cm

Type – C

Question 1.
Rajashree wants to get an inverted image of height 5 cm of an object kept at a distance of 30 cm from a concave mirror. The focal length of the mirror is 10 cm. At what distance from the mirror should she place the screen? What will be the type of the image, and what is the height of the object?
Solution:
Given:
Focal length = f = -10 cm,
Object distance = u = -30 cm
Height of the image = h₂ = 7 cm
To find: Height of the object = h₁ = ?
Image distance = u =?

Rajashree has to place the screen 15 cm to the left of the mirror.
Magnification formula

The height of the object is 10 cm. Thus, the image will be real and diminished.

Question 2.
A 10 cm long stick is kept horizontally in front of the concave mirror having focal length of 10 cm in such a way that the end of the stick closest to the pole is at a distance of 20 cm. What will be the length of the image?
Solution:
The stick is kept parallel to the Principal axis. Distance between A and P is 20 cm. Say u₁ = 20 cm.
Hence, the other end of the stick is at distance, u₂ = (u₁ + 10) = 30 cm from pole of the mirror.

Using mirror formula for concave mirror,
Solution:

Here, negative signs indicate that images are formed on the left of the mirror.

The length of the image formed ?s given by, u = u₂ – u₁ = 15 – (-20) = 5cm.
The length of the image is 5 cm.

Numerical For Practice

Question 3.
An object 2 cm in height is placed at a distance of 16 cm from a concave mirror. If the focal length of the mirror is 9.6 cm., find the image distance, nature and size of the image.
Answer:
u = -24 an, h₂ = -3 cm; real, inverted and enlarged.

Question 4.
An arrow of 2.5cm height is placed at a distance of 25 cm from a diverging mirror of focal length 20 cm. Find the nature, position and size of the image formed.
Answer:
v = 11.1cm, h₁ = 1.1cm; virtual and in dimirrished form.

Define the following:

Question 1.
Centre of curvature of mirror (C)
Ans.
The centre of the sphere of which the mirror is a parte is called the centre of curvature of the mirror.

Question 2.
The radius of curvature (R)
Answer:
The radius of the sphere of which the mirror is a part, is called the radius of curvature of the mirror.

Question 3.
Pole (P)
Answer:
The centre of the mirror surface is called its pole.

Question 4.
The principal axis of a mirror
Answer:
The straight line passing through the pole and centre of curvature of the mirror is called its principal axis.

Question 5.
The focus of a concave mirror (F)
Answer:
Incident rays which are parallel to the principal axis of a concave mirror, after reflection from the mirror, meet at a particular point in front of the mirror on the principal axis. This point (F) is called the principal focus of the concave mirror.

Question 6.
Focus of a convex mirror (F)
Answer:
Incident rays parallel to the principal axis, after reflection, appear to come from a particular point behind the mirror lying along the principal axis. This point is called the principal focus of the convex mirror.

Question 7.
Focal length of a mirror (f)
Answer:
The distance (f) between the pole and the principal focus of the mirror is called the focal length. This distance is half of the radius of curvature of the mirror. \(f=\frac{R}{2}\)

Answer the following in short:

Question 1.
What are the rules for drawing ray diagrams for the formation of image by spherical mirror?
Answer:
The rules are as follows :

• If an incident ray is parallel to the principal axis, then the reflected ray passes through the principal focus.
• If an incident ray passes through the principal focus of the mirror, the reflected ray is parallel to the principal axis.
• If an incident ray passes through the centre of curvature of the mirror, the reflected ray traces the same path back.

Distinguish between:

Question 1.
Convex mirror and Concave mirror
Answer:

Question 2.
Real image and Virtual image
Answer:

Answer the following questions:

Question 1.
If we keep the mirrors parallel to each other, how many images will we see ?
Answer:
When two mirrors are kept parallel to each other infinite images are formed, this is because light gets reflected infinite times.

Answer in detail:

Question 1.
What sign conventions are used for reflection from a spherical mirror?
Answer:

According to the Cartesian sign convention, the pole of the mirror is taken as the origin. The principal axis is taken as the X-axis of the frame of reference. The sign conventions are as follows.

• The object is always kept on the left of the mirror. All distances parallel to the principal axis are measured from the pole of the mirror.
• All distances measured towards the right of the pole are taken to be positive, while those measured towards the left are taken to be negative.
• Distance measured vertically upwards from the principal axis are taken to be positive.
• Distance measured vertically downwards from the principal axis are taken to be negative.
• The focal length of a concave mirror is negative while that of a convex mirror is positive.

Question 2.
Draw ray diagrams for the image obtained in convex mirrors.
Answer:

Question 3.
In order to see the full image of a person standing in front of a mirror, the minimum height of the mirror must be half the height of the person. Explain.
Answer:

Proof:

1. In the figure, the point at the top of the head, the eyes and a point at the feet of a person are indicated by H, E and F respectively.
2. R and S are midpoints of HE and EF respectively.
3. The mirror PQ is at a height of NQ from the ground and is perpendicular to it. PQ is the minimum height of the mirror in order to obtain the full image of the person.

For this, RP and QS must be perpendicular to the mirror.

Minimum height of the mirror
PQ = RS
= RE + ES
\(=\frac{\mathrm{HE}}{2}+\frac{\mathrm{EF}}{2}=\frac{\mathrm{HF}}{2}\)
= Half of the person’s height.

Question 4.
Determine the sign of magnification in each of the 6 cases in the table and verify that they are same using formulae
\(\mathbf{M}=\frac{h_{2}}{h_{1}}\) and \(\mathbf{M}=\frac{-v}{u}\)
Answer:

Question 5.
Explain the images formed by concave mirrors with respect to position of the image and object and also the Nature and size of image
Answer:

Balbharati Maharashtra State Board 9th Std Science Textbook Solutions 

• Information Communication Technology (ICT) Class 9 Science Textbook Solutions
• Reflection of Light Class 9 Science Textbook Solutions
• Study of Sound Class 9 Science Textbook Solutions
• Carbon: An Important Element Class 9 Science Textbook Solutions
• Substances in Common Use Class 9 Science Textbook Solutions
• Life Processes in Living Organisms Class 9 Science Textbook Solutions
• Heredity and Variation Class 9 Science Textbook Solutions
• Introduction to Biotechnology Class 9 Science Textbook Solutions
• Observing Space: Telescopes Class 9 Science Textbook Solutions

Std 9 Science Chapter 12 Study of Sound Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 12 Study of Sound Notes, Textbook Exercise Important Questions and Answers.

Class 9 Science Chapter 12 Study of Sound Question Answer Maharashtra Board

Class 9 Science Chapter 12 Study of Sound Textbook Questions and Answers

1. Fill in the blanks and explain.

a. Sound does not travel through ……………………….……….. .
b The velocity of sound in steel is ……………………….………… than the velocity of sand in water.
c. The incidence of ……………………….………… in daily life shows that the velocity of sound is less than the velocity of light.
d. To discover a sunken ship or objects deep inside the sea, ……………………….………… technology is used.

2. Explain giving scientific reasons.

a. The roof of a movie theatre and a conference hall is curved.
Answer:

• Sound waves get reflected from the walls and roof of a room multiple times. This causes a single sound to be heard not once but continuously. This is called reverberation.
• Due to reverberation, some auditoriums or some particular seats in an auditorium have inferior sound reception. This can be compensated with curtains.
• Ceilings of these halls are made curved so that sound after reflecting from the ceiling, reaches all parts of the hall and the quality of sound improves.

b. The intensity of reverberation is higher in a closed and empty house.
Answer:

• Reverberation occurs due to multiple reflections of sound.
• The furniture in the house acts as a sound-absorbing material.
• So if the house is closed and empty, a reflection of sound will be maximum and hence, intensity of reverberation is higher.

c. We cannot hear the echo produced in a classroom.
Answer:

• For distinct echoes, the minimum distance of the reflecting surface from the source of sound must be 17.2 m.
• Benches in the classroom are sound absorbing materials which prevent echo of sound.
• Because of these two reasons echo is not heard in a classroom.

3. Answer the following questions in your own words.

a. What is an echo? What factors are important to get a distinct echo?
Answer:

• An echo is the repetition of the original sound because of reflection by some surface.
• At 22°C, the velocity of sound in air is 344 m/s.
• Our brain retains a sound for 0.1 seconds Thus, for us to be able to hear a distinct echo, the sound should take more than 0.1 seconds after starting from the source to get reflected and. come back to us.
• We know that,
  Distance = speed x time
  = 344 m/s x 0.1 s
  = 34.4 m
• Thus, to be able to hear a distinct echo, the reflecting surface should be at a minimum distance of half of the above, i.e. 17.2 m.
• As the velocity of sound depends on the temperature of air, this distance depends on the temperature.

b. Study the construction of the Golghumat at Vijapur and discuss the reasons for the multiple echoes produced there.
Answer:

• Goighumat with a height of 51 metres and diameter of 37 metres with 3 metres thick walls is spread over approximately 1700 square metres.
• This meets the conditions for echo i.e. : 17.2 metres minimum.
• The dome of the golghumat is curved and hence, sound reflects multiple times before reaching the observer.
• This is the reason for multiple echoes being produced.

c. What should be the dimensions and the shape of classrooms so that no echo can be produced there?
Answer:

1. Dimensions: The distance between opposite walls in a classroom must be less than 17.2 m so that the reflected sound returns to the observer within 0.1 s.
2. Shape: The classrooms should have curved ceilings and walls so that the reflected sound is directed towards the observer instantly within 0.1 s

4. Where and why are sound-absorbing materials used?
Answer:
The sound absorbing materials are used in :

• School, cinema hall, concert hall, houses or places where quality of sound is important.
• In the absence of sound absorbing material the sound will undergo multiple reflection causing reverberation of sound.

5. Solve the following examples.

a. The speed of sound in air at O °C is 332 m/s. If it increases at the rate of 0.6 m/s per degree, what will be the temperature when the velocity has increased to 344 m/s?
Answer:
Given:
Initial speed of sound at 0°C 332 m/s.
Final speed of sound -344 m/s.
Rate of increase per degree rise in temp. = 0.6m/s
To find:
Temperature when speed is 344m/s
Formulae:
Increase in temperature

Temperature when the speed of sound is 344 m/sis 20°C

b. Nita heard the sound of lightning after 4 seconds of seeing it. What was the distance of the lightning from her? (The velocity of sound in air is 340 m/s?)
Answer:
Given : Speed of sound (v) = 340 m/s
Time taken (f) = 4 sec
To find : Distance (s) = ?

The lightning has struck at a distance of 1360 m from the observer.

c. Sunil is standing between two walls. The wall closest to him is at a distance of 360 m. If he shouts, he hears the first echo after 4 s and another after another 2 seconds.
1. What is the velocity of sound in air?
2. What is the distance between the two walls? (Ans: 330 m/s; 1650 m)
Answer:
Given:
Distance of the closer wall (S₁) = 660 m
Time of echo from closer wall = 4 sec
∴ Time taken (t₁) = 4/2 sec = 2 sec
Time of echo from distant wall = 6 sec
∴ Time taken (t₂) = 6/2 sec = 3 sec
To find :
Velocity of sound in air (y) =?
Distance between two walls (S₁ + S₂) = ?

The velocity of sound in air is 330 mIs and the distance between two walls is 1650 m.

d. Hydrogen gas is filled in two identical bottles, A and B, at the same temperature. The mass of hydrogen in the two bottles is 12 gm and 48 gm respectively. In which bottle will sound travel faster? How may times as fast as the other? (Ans: In A; Twice)
Answer:
In A; Thrice

e. Helium gas is filled in two identical bottles A and B. The mass of the gas in the two bottles is 10 gm and 40 gm respectively. If the speed of sound is the same in both bottles, what conclusions will you draw? (Ans: Temperature of B is 4 times the temperature of A.)
Given:
Mass of Helium in bottle A = (mA) = 10gm
Mass of Helium in bottle B = (mB) = 40gm


The temperature of B is 4 times the temperature of A

Class 9 Science Chapter 12 Study of Sound Intext Questions and Answers

Study Of Sound Class 9 Notes Maharashtra Board Question 1.
How does the velocity of sound depend on its frequency?
Answer:
The velocity of sound is directly proportional to its frequency
ν = υ λ
when ν = velocity
υ = frequency
λ = wavelength

9th Class Science Chapter 12 Study Of Sound Exercise Question 2.
The molecular weight of oxygen gas (O₂) is 32 while that of hydrogen gas (H₂) is 2. Prove that under the same physical conditions, the velocity of sound in hydrogen is four times that in oxygen.
Answer:
Given:
Molecular wt of Oxygen (M_o) =32
Molecular wt of hydrogen (M_H) = 2
To Find:
V_H = 4 v_o


Hence, proved that velocity of sound in hydrogen is four times that in oxygen.

Answer the following:

Study Of Sound Class 9 Maharashtra Board  Question 1.
How will you reduce reverberation in public halls or buildings?
Answer:
(i) Reverberation in public halls or buildings will be reduced by using sound absorbing materials like curtains on wall, carpets on the floor.
(ii) By keeping the windows open, as sound will not get reflected.

12 Study Of Sound 9th Class Exercise  Question 2.
How is ultrasound used in medical science?
Answer:

• Sonography: Sonography technology uses ultrasonic sound waves to generate images of internal organs of the human body.
• Echocardiography: Echocardiography is a test that uses ultrasonic sound waves to produce live images of your heart.

9th Science Chapter 12 Study Of Sound Exercise Question 3.
To hear the echo distinctly, will the distance from the source of sound to the reflecting surface be same at all temperatures? Explain your answer.
Answer:

• No,the distance from the source of sound to the reflecting surface will not be the same at all temperatures.
• Velocity of sound is directly proportional to the square root of temperature.
• One of the conditions of echo is that the time interval between the original and reflected sound should be more than 0.1 sec.
• So if the temperature increases, the velocity of sound increases and the reflected sound reaches in less than 0.1 sec.
• So for echo to be heard the distance between the observer and the reflecting surface has to increase.

9th Science Chapter 12 Study Of Sound Question 4.
When is the reflection of sound harmful?
Answer:

• Reflected sound of high intensity called as noise is disturbing and harmful to the ears.
• When sound reverberates i.e it undergoes multiple reflections, poor quality of sound is produced.

9th Class Science Chapter 12 Study Of Sound Question Answer Question 5.
What kind of waves are created when a stone is dropped in water ?
Answer:

• When a stone is dropped in water, the particles of water oscillate up and down.
• These oscillations are perpendicular to the direction of propagation of the wave, such waves are called transverse waves.

Answer the following question:

12 Study Of Sound 9th Class Question 1.
Observe the graph/ diagram and discuss your observation.
Answer:

1. Fig. A shows changes in density. The region where particles are crowded is called compression and where they are far apart are rarefaction.
2. Fig. B show change in pressure. The lines represent layers of air. The regions when lines are crowded are high pressure regions while when they are far apart are of low pressure.
3. Fig. C shows changes in density or pressure. The crest represents high pressure region while trough represents low pressure region.

Answer the following question:

Study Of Sound Class 9 Question Answer Question 1.
How are the frequencies of notes sa, re, ga, ma, pa, dha, ni related to each other?
Answer:
The frequencies of notes sa, re, ga, ma, pa, dha, ni are related in the ratio.

i.e if first Sa is 240Hz then the next Sa will be 240 x 2 = 480Hz

Class 9 Science Chapter 12 Study Of Sound Exercise Question 2.
What is the main difference between the frequencies of the voice of a man and that of a woman?
Answer:

• Voice of a woman is high pitch i.e shorter wavelength and higher frequency
• Voice of man is low pitch i.e larger wavelength and smaller frequency.

Question 3.
Try this;

(a) In the above activity, what will happen if you lift one of the tubes to some height?
Answer:
If one of the tubes is lifted, angle of incidence will not be equal to angle of reflection, hence, the sound will not be clearly audible.

(b) Measure the angle of incidence 01 and the angle of reflection 02. Try to see if they are related in any way.
Answer:
Angle of incidence is same as the angle of reflection.

Class 9 Science Chapter 12 Study of Sound Additional Important Questions and Answers

Can you recall?

12.Study Of Sound Question 1.
How is the direction of the oscillation of the particles of the medium related to the direction of propagation if the sound wave?
Answer:

• Sound travels as a longitudinal wave.
• In a longitudinal wave, the particle of the medium oscillate parallel to the direction of propagation of the wave.

Choose and write the correct option:

Class 9 Science Chapter 12 Study Of Sound Question 1.
The unit of frequency is ……………………………… .
(a) Hertz
(b) m/s²
(c) Decibels
(d) m/s
Answer:
(a) Hertz

Study Of Sound Class 9 Exercise Question 2.
The normal hearing range for humans is ……………………………… .
(a) 0 Hz to 20 Hz
(b) greater than 20,000 Hz
(c) 20 Hz to 20,000 Hz
(d) none of these
Answer:
(c) 20 Hz to 20,000 Hz

Class 9th Science Chapter 12 Study Of Sound Question Answer  Question 3.
Sound will not travel through ……………………………… .
(a) Vacuum
(b) Liquid
(c) Solid
(d) Gases
Answer:
(a) vacuum

Class 9 Science Chapter 12 Study Of Sound Question Answer Question 4.
SI unit of ………………………………. is Hertz (Hz).
(a) Wavelength
(b) Frequency
(c) Speed of wave
(d) Velocity
Answer:
(b) frequency

Reflection Of Sound Class 9 Question 5.
The velocity of sound is inversely proportional to the ……………………………… .
(a) Pressure
(b) Square root of temperature
(c) Square root of density
(d) Humidity
Answer:
(c) square root of density

Question 6.
Sound waves with frequency greater than 20 kHz are called ……………………………… .
(a) Infrasound
(b) Ultrasound
(c) Sonic
(d) Damped sound
Answer:
(b) ultrasound

Question 7.
The loudness of a sound depends upon ……………………………… .
(a) Amplitude
(b) Speed
(c) Density
(d) Wavelength
Answer:
(a) Amplitude

Question 8.
……………………………… are used in sonography.
(a) High frequency ultrasound
(b) Stationary waves
(c) High frequency infrasound
(d) High frequency micro waves
Answer:
(a) High frequency ultrasound

Question 9.
The ……………………………… receives the vibrations coming from the membrane and converts them into electrical signals which are sent to the brain through the nerve.
(a) Cochlea
(b) Tympanic cavity
(c) Stapes
(d) Pinna
Answer:
(a) Cochlea

Find the odd one out:

Question 1.
Bats, rats, cats, dolphins
Answer:
Cats: cannot produce ultrasonic sound.

Question 2.
Clothes, paper, curtains, mirror
Answer:
Mirror: is a good reflector of sound, while others are poor reflectors.

Question 3.
Submarines, icebergs, internal organ, sunken ships.
Answer:
Internal organ: sonography is used , while for others sonar system is used.

Question 4.
Temperature, density, molecular weight, pressure
Answer:
Pressure: for a fixed temperature, the speed of sound does not depend on the pressure of the gas, all other factors affect speed of sound.

Answer in one sentence:

Question 1.
How can one produce sound?
Answer:
Vibration set up in an object produces sound (or) sound is produced when an object is disturbed and starts vibrating.

Question 2.
What is velocity of sound wave ?
Answer:
The distance covered by a point on the wave in unit time is the velocity of the sound wave.

Question 3.
What is the minimum distance of the reflecting surface to hear an echo ?
Answer:
To be able to hear a distinct echo, the reflecting surface should be at a minimum distance of 17.2 m.

Match the columns:

Question 1.

Answer:
(1-b- iii),
(2– a- i),
(3 – d – v),
(4 – e – ii),
(5 -c- iv)

Question 2.

Answer:
(1 -b – ii),
(2 -c – i),
(3-d – iv),
(4 – a – iii)

Name the following:

Question 1.
A form of energy which produces sensation of hearing in our ears.
Answer:
Sound energy

Question 2.
Repetitions of sound due to reflection .
Answer:
Echo

Question 3.
The audible range of sound for human being.
Answer:
20 Hz to 20,000 Hz

Question 4.
A method to obtain images of internal organs of the human body.
Answer:
Sonography

Question 5.
The matter or substance through which sound gets transmitted.
Answer:
Solid, liquid, gases

Question 6.
Three major parts of the ear.
Answer:
External ear, the middle ear and the inner ear.

Question 7.
Any two examples in which infrasound is produced.
Answer:
Pendulum, earthquake.

Question 8.
Name the living beings that can produce ultrasound.
Answer:
Bats, dolphins, mice.

Give scientific reasons:

Question 1.
Bats can navigate in dark.
Answer:

• The ultrasonic sound produced by bats, gets reflected on hitting an obstacle.
• This reflected sound is received by their ears and they can locate the obstacle and estimate its distance even in the dark.
• Hence, bats can navigate in dark.

Question 2.
A SONAR system is installed in a ship.
Answer:

• A SONAR system determines the depth of the sea.
• It locates underwater hills, valleys, icebergs, submarines and sunken ships. It also locates the positions of other ships or submarines.
• Hence a SONAR system is installed in a ship.

Question 3.
Sound travels faster in iron than in air.
Answer:

• Sound requires a material medium for its propagation and travels in the form of a longitudinal wave.
• The denser the medium, faster is the propagation of sound.
• Hence, sound travels faster in iron than in air.

Solve the following:

Type – A

Formula:
\(\text { (i) Velocity }=\frac{\text { distance }}{\text { time }}\)

Question 1.
Ultrasonic waves are transmitted downwards into the sea with the help of a SONAR. The reflected sound is received after 4 s. What is the depth of the sea at that place? (Velocity of sound in seawater = 1550 m/s)
Answer:
Given:
Time to hear echo = 4 sec
Time taken by sound waves to reach the bottom 4 of sea (t) = 4/2 sec = 2 sec
Velocity of sound in sea water (v) = 1550 m/s
To find:
Depth of sea(s) = ?

The depth of the sea at that place is 3100 m.

Question 2.
A person standing near a hill fires a gun and hears the echo after 1 second. If speed of sound in air is 340 m/s. Find the distance between the hill and the person.
Answer:
Given:
Time to hear echo = 1 sec 1
Time taken (t) = 1/2 sec
Velocity of sound (v) = 340 m/s
To find:
Distance (s) = ?

Distance between the person and hill is 170 m.

Numerical For Practice

Question 3.
If you hear the thunder 20 seconds after you see the flash of lightning, how far from you has the lightning occurred? (Speed of sound in air = 340 m/s)
Answer:
6800m

Question 4.
Aboy observes smoke from a cannon 3 seconds before he hears the bang. If the cannon is 1020 m from the observer, find the velocity of sound.
Answer:
340 rn/s

Question 5.
A soldier standing between the two buildings fires a gun. He heard the echo of the sounds from the first building after 2 seconds and echo from the second building after 3 seconds. Find the distance between two buildings. (Speed of sound in air = 340 m/s)
Answer:
850m

Type – B

Question 1.
Sound waves of wavelength 1 cm have a velocity of 340 mIs in air. What is their frequency? Can this sound be heard by the human ear?
Answer:
Given:
wave length (λ) = 1cm = 1/100
Velocity of sound (v) = 340 m/s
To fInd :
frequency (u) = ?
Formulae:
ν = υ λ
Solution:

The frequency of the sound waves is 34000 Hz. The frequency is higher than 20000 Hz and therefore, this sound cannot be heard by the human ear.

Question 2.
How long will it take for a sound wave of 25 cm wavelength and 1.5 kHz frequency, to travel a distance of 1.5 km?
Answer:
Given:
frequency (u) = 1.5 kHz = 1500 Hz

\(\begin{array}{l}
=\frac{1500}{375} \\
=4 \mathrm{sec}
\end{array}\)
The sound wave takes 4 sec to travel the distance of 1.5 km.

Question 3.
Calculate distance travelled by a sound wave having frequency 1000 Hz and wavelength 0.25 m, if it travels for 5 seconds in a certain medium.
Answer:
Given:
frequency (u) = 1000 Hz
wavelength (λ) = 0.25 m
time (t) = 5 seconds
To find :
Distance (d) =?
Formulae:
ν = υ λ
Solution:

The distance travelled by the sound wave is 1250 m.

Question 4.
The audible range of sound is 20 Hz to 20000 Hz. At 22°C in air speed of sound is 344 mIs. Express the range of sound in terms of wavelength by calculating the respective values.
Answer:
Given:
frequency (u₁) 20 Hz
frequency ( u₂) = 20,000 Hz
velocity (v) = 344 rn/s
To find :
Wavelengths λ₁ and λ₂ = ?
Formulae:
ν = υ λ
Solution:


Audible range of wavelength of sound is from 17.2 x 10^-3 m to 17.2 m.

Numerical For Practice

Question 5.
A sound wave has frequency 320 Hz and wavelength 0.25 m. How much distance will it travel in 10 second?
Answer:
The distance travelled is 800 m.

Type – C

Question 1.
Hydrogen gas is filled in two identical bottles, A and B, at the same temperature. The mass of hydrogen in the two bottles is 12 gm and 48 gin respectively. In which bottle will sound travel faster? How many timés as fast as the other?
Answer:
Given:
Mass of hydrogen in bottle A (m_A) = 12gm
Mass of hydrogen in bottle B(m_B) = 48gm
To find:
In which bottle sound travels faster.
Formulae:


Since both bottles are identical hence, the volume is the same, i.e. v
Dividing (j) and (ii),

(i) Vivacity of sound will be more in bottle A.
(ii) Velocity of sound in bottle A (V_A) is twice of that in bottle B (v_B)

Numerical For Practice

Question 2.
Argon gas is filled in two identical bottles X and Y. The mass of the gas in the two bottles is 5 gm and 25gm respectively. If the speed of sound is the same in both bottles, what conclusions will you draw?
Answer:
(Temperature of Y is 5 times the temperature of X.)

Type – D

Numerical For Practice

Question 1.
Velocity of sound in air at 0°C is 332nVs. It increases by 0.6ni/s for each °Celsius rise in temperature. At what temperature of ait the velocity will be 359m1s?
Answer:
45°C

Question 2.
Velocity of sound In air at 0°C is 332m/s It increases by 0.6mIs for each degree Celsius rise In temperature. What will be the velocity of sound at 60°C?
Answer:
368 rn/s

Define the following:

Question 1.
Wave length (λ)
Answer:
The distance between two consecutive compressions (or crests) or two consecutive rarefactions (or troughs) is called the wavelength.

Question 2.
Amplitude (A)
Answer:
The maximum value of pressure or density is called amplitude.

Question 3.
Frequency (υ)
Answer:
The frequency of a sound wave is defined as the number of complete oscillations of density (or pressure of the medium) per second.

Question 4.
Time Period (T)
Answer:
The time taken for one complete oscillation of pressure or density at a point in the medium is called the time period.

Question 5.
Echo
Answer:
An echo is the repetition of the original sound because of reflection by some surface.

Question 6.
Transverse waves
Answer:
Oscillations of the particles of the medium vibrate at right angles to the direction of propagation of the wave are called transverse waves.

Question 7.
longitudinal waves
Answer:
The particles of the medium oscillate about their central or mean position in a direction parallel to the propagation of wave is called as longitudinal waves.

Question 8.
Velocity of wave
Answer:
The distance covered by a point on the wave (for example the point of highest density or lowest density) in unit time is the velocity of the sound wave.’

Distinguish between:

Question 1.
Infrasound and Ultrasound
Answer:

Question 2.
Transverse waves and Longitudinal waves
Answer:

Question 3.
Consider two cases
(A) whistle of train (B) roar of a lion

(I) In which case the sound is high pitch?
Answer:
Whistle of a train is high pitch as compared to roar of a lion, as the frequency is higher.

(II) What is the real cause of sound production? Explain with examples.
Answer:

• Vibrations in the object are responsible to produce a sound.
• Vibration is a rapid to and fro motion of an object.
• Sometimes the vibrations may be strong enough to be seen by eyes, e.g. string vibrations in string instruments, vibration on mobile phone, blowing air in the cap of your pen by holding it near the lips.

(III) Three sounds 5 Hz, 500 Hz and 50,000 Hz are produced by different sources.
(a) Which sound will be heard by humans?
(b) Which sounds may be produced by bats?
(c) Which sounds may be produced by elephants?
Answer:
(a) 500 Hz – Humans can hear sounds in the range of 20 Hz-20,000 Hz
(b) 50,000 Hz – Bats produce ultrasonic sounds above 20,000 Hz
(c) 5 Hz – Elephants can produce infrasonic sounds below 20 Hz

Question 6.
Suppose you and your friend are on the moon. Will you be able to hear any sound
Answer:
Sound waves need a material medium for their propagation. Since there is no atmosphere on the moon, we cannot hear any sound on the moon.

Answer in detail:

Question 1.
What are the factors on which velocity of sound in gaseous medium depend?
Answer:
The velocity of sound in a gaseous medium depends on the physical conditions i.e. the temperature, density of the gas and its molecular weight.

1. Temperature (T): The velocity of sound is directly proportional to the square root of the temperature of the medium. This means that increasing the temperature four times doubles the velocity.
   \(\text { v } \alpha \sqrt{\mathrm{T}}\)
2. Density(p): The velocity of sound is inversely proportional to the square root of density. Thus, increasing the density four times, reduces the velocity to half its value.
   \(\mathrm{v} \alpha \frac{1}{\sqrt{\rho}}\)
3. Molecular weight (M): The velocity sound is inversely proportional to the square root of molecular weight of the gas. Thus, increasing the molecular weight four times, reduces the velocity to haff its value.
   \(\mathrm{v} \alpha \frac{1}{\sqrt{\mathrm{M}}}\)

Question 2.
What are the uses of ultrasonic sound?
Answer:
Uses of ultrasonic sound are as follows:

• For communication between ships at sea.
• To join plastic surfaces together.
• To sterilize liquids like milk by killing the bacteria in it so that the milk keeps for a longer duration.
• Echocardiography which studies heartbeats, is based on ultrasonic waves (Sonography technology).
• To obtain images of internal organs in a human body.
•  In industry to clean intricate parts of machines where hands cannot reach.
• To locate the cracks and faults in metal blocks.

Question 3.
Explain with the help of a neat labelled diagram the working of human ear.
Answer:

• The ear is an important organ of the human body.
• When sound waves fall on the eardrum, it vibrates and these vibrations are converted into electrical signals which travel to the brain through nerves.
• The ear can be divided into three parts:
  (a) Outer ear
  (b) Middle ear
  (c) Inner ear.

(a) Outer ear or Pinna
The outer ear collects the sound waves and passes them through a tube to a cavity in the middle ear. Its peculiar funnel like shape helps to collect and pass sounds into the middle ear.

(b) Middle ear
There’ is a thin membrane in the cavity of the middle ear called the eardrum. When a compression in a sound wave reaches the eardrum, the pressure outside it increases and it gets pushed inwards. The opposite happens when a rarefaction reaches there. The pressure outside decreases and the membrane gets pulled outwards. Thus, sound waves cause vibrations of the membrane.

(c) Inner ear
The auditory nerve connects the inner ear to the brain. The inner ear has a structure resembling the shell of a snail. It is called the cochlea. The cochlea receives the vibrations coming from the membrane and converts them into electrical signals which are sent to the brain through the nerve. The brain analyses these signals.

Question 4.
Write a short note on SONAR
Answer:

(i) SONAR is the short form for Sound Navigation and Ranging. It is used to determine the direction, distance and speed of an underwater object with the help of ultrasonic sound waves. SONAR has a transmitter and a receiver, which are fitted on ships or boats.

(ii) The transmitter produces and transmits ultrasonic sound waves. These waves travel through water, strike underwater objects and get reflected by them. The reflected waves are received by the receiver on the ship.

(iii) The receiver converts the ultrasonic sound into electrical signals and these signals are properly interpreted. The time difference between transmission and reception is noted. This time and the velocity of sound in water give the distance from the ship, of the object which reflects the waves.

(iv) SONAR is used to determine the depth of the sea. SONAR is also used to search underwater hills, valleys, submarines, icebergs, sunken ships etc.

Question 5.
Write a short note on Sonography. How is it misused?
Answer:

• Sonography technology uses ultrasonic sound waves to generate images of internal organs of the human body.
• This is useful in finding out the cause of swelling, infection, pain, condition of the heart, the state of the heart after a heart attack as well as the growth of foetus inside the womb of a pregnant woman.
• This technique makes use of a probe and a gel.
• The gel is used to make proper contact between the skin and the probe so that the full capacity of the ultrasound can be utilized.
• High-frequency ultrasound is transmitted inside the body with the help of the probe.
• The sound reflected from the internal organ is again collected by the probe and fed to a computer which generates the images of the internal organ.
• As this method is painless, it is increasingly used in medical practice for correct diagnosis.
• This technique is used by many people to find out gender of an unborn baby and this often leads to the incidence of female foeticide.

Balbharati Maharashtra State Board 9th Std Science Textbook Solutions 

• Information Communication Technology (ICT) Class 9 Science Textbook Solutions
• Reflection of Light Class 9 Science Textbook Solutions
• Study of Sound Class 9 Science Textbook Solutions
• Carbon: An Important Element Class 9 Science Textbook Solutions
• Substances in Common Use Class 9 Science Textbook Solutions
• Life Processes in Living Organisms Class 9 Science Textbook Solutions
• Heredity and Variation Class 9 Science Textbook Solutions
• Introduction to Biotechnology Class 9 Science Textbook Solutions
• Observing Space: Telescopes Class 9 Science Textbook Solutions

Std 9 Science Chapter 10 Information Communication Technology (ICT) Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT) Notes, Textbook Exercise Important Questions and Answers. BALRAMCHIN Pivot Point Calculator

Class 9 Science Chapter 10 Information Communication Technology (ICT) Question Answer Maharashtra Board

Class 9 Science Chapter 10 Information Communication Technology (ICT) Textbook Questions and Answers

1. Fill in the blanks to complete the statements. Justify the statements.

a. While working with a computer we can read the information stored in its memory and perform other actions in ………………………. memory.
b. While presenting pictures and videos about the works of scientists, we can use ……………………… .
c. To draw graphs based on the quantitative information obtained in an experiment, one uses ……………………… .
d. The first generation computers used to shut down because of ……………………… .
e. A computer will not work unless ………………………. is supplied to it.

2. Answer the following questions.

a. Explain the role and importance of information communication in science and technology.
Answer:

• ICT plays a key role in creating, displaying, collecting, processing and communicating information in the field of science and technology.
• Following is the importance of ICT in science and technology:
  (a) Access to wide range of information
  (b) Storing of Data
  (c) Processing of Data
  (d) Securing work files
  (e) Proper representation of data

b. Which application software in the computer system did you find useful while studying science, and how?
Answer:

• Microsoft word: To write down the information collected and making a document for further evaluation.
• Microsoft excel : To draw graph based on the obtained numerical information from the experiment.
• Internet explorer: To search for information in finding out the solution or solving the queries by reading the available information.

c. How does a computer work?
Answer:

d. What precautions should be taken while using various types of software on the computer?
Answer:

• Antivirus must be installed.
• Software should be legal and from a trusted place.
• Application should be scanned before using.
• Pirated Software should not be used
• Provide all necessary data to obtain the best possible results.

e. Which are the various devices used in information communication? How are they used in the context of science?
Answer:

• Various devices used in information communication are: Computers, Laptops, Mobiles, Radios, Television, etc.
• Computers, Laptops and Mobiles: Help in accessing, collecting, processing, communicating, sharing and storing of information. It helps in determining the appropriate conclusions in all fields, including the field of science.
• Television: Help in getting information about the new and innovative technology.

3. Using a spreadsheet, draw graphs between distance and time, using the information about the movements of Amar, Akbar and Anthony given in the table on page 4, in the lesson on Laws of Motion. What precautions will you take while drawing the graph?

Answer:
Precautions to be taken while drawing a graph:

• The data should be kept in tabular form.
• Whenever there is ‘drag and fill’ option used, ‘smart tag’ option should be used after ‘drag data’ to fill data as required.
• Entered data should be formatted in the manner required.
• Various types of graphs can be created by using the same data, so appropriate graph should be selected.
• Chart titles and axes titles should be updated as per the data.

4. Explain the differences between the different generations of computers. How did science contribute to these developments?
Answer:
Generation: 1st
Time Period: 1946 – 1956
Development: Vacuum Tubes Characteristics:

• Huge in size
• Expensive
• Lot of electricity consumption
• Heat generation

Generation: 2nd
Time Period: 1956 – 1963
Development: Transistors
Characteristics:

• Frequent shutdowns
• Superior to 1st Generation
• Small in size and fast
• Cheaper as compared to 1st Generation
• Less consumption of electricity

Generation: 3rd
Time Period: 1963 -1971
Development: IC
Characteristics:

• Keyboards and monitors
• OS
• Smaller and still cheaper

Generation: 4th
Time Period: 1971 – 2010
Development: Microprocessor Characteristics:

• Use of Internet
• GUI
• Introduction of portable devices like mobiles, laptops, etc.

Generation: 5th
Time Period: 2010 – Till Date
Development: Artificial Intelligence (AI) Characteristics:

• Voice recognition
• Sensors
• Nano technology

1st Generation computers occupied the entire room, but due to advancement in science and technology, today’s computer fits into our pockets.

Initially computers needed a specific language to interact but today we use voice recognition for the same.

In these ways, science has contributed in making the computers faster, smaller, cheaper and much more useful.

5. What devices will you use to share with others the knowledge that you have?
Answer:
Devices like radios, televisions, pendrives, computers, laptops, mobiles, landlines, hard drives, CDs, memory cards help us in sharing our knowledge with others.

6. Using information communication technology, prepare powerpoint presentations on at least three topics in your textbook. Make a flowchart of the steps you used while making these presentations.
Answer:
Steps for preparation of PowerPoint presentations:

7. Which technical difficulties did you face while using the computer? What did you do to overcome them?
Answer:

• Lagging: Due to lot of applications running at the same time, the computer starts lagging and becomes slow. Closing a few applications helped solve the problem of lagging.
• Viruses and Bugs: Cybercrimes are rising daily, even from single mail the computer can be attacked by viruses. Installing a valid antivirus helps solve the problem of viruses and bugs.
• Breach of Privacy: Confidential information being accessed by anyone is the breach of privacy. Putting privacy setting in place helps solve the problem.
• Physical Damage: Hardware over a period of time might get physically damaged. Taking precautions while using will help to solve the problem.

Class 9 Science Chapter 10 Information Communication Technology (ICT) Intext Questions and Answers

Question 1.
Make a list of various hardware and software items of a computer,
Answer:
Hardware: Mouse, Keyboard, Pendrive, Monitor and other parts of computer.
Software: Operating Systems, Application Programs, Antivirus, etc.

Answer the following questions:

Question 2.
Which devices do we directly or indirectly use for collecting, sharing, processing and communicating information?
Answer:

• Computers
• Laptops
• Mobiles
• Memory Cards
• Pendrives
• Landlines
• Hard disks etc.

Question 3.
How is information communication technology important for dealing with explosion of information?
Answer:

• Information explosion means a situation whère information is available in abundance, in other words, too much information.
• Devices like computers, laptops help us in easier accessment of information that we need from all the data.

Class 9 Science Chapter 10 Information Communication Technology (ICT) Additional Important Questions and Answers

Choose and write the correct option:

Question 1.
…………………….. includes communication devices and the use of those devices as well as the services provided with their help.
(a) Operating System
(b) Office
(c) Computers
(d) Information Communication Technology
Answer:
(d) Information Communication Technology

Question 2.
Computers have gone through …………………….. generations.
(a) 5
(b) 7
(c) 10
(d) 8
Answer:
(a) 5

Question 3.
First generation of Computers were considered to be present in the period of ……………………. .
(a) 2000 – 2001
(b) 1901 – 2001
(c) 1946 -1959,
(d) 1996 – 2001
Answer:
(c) 1946-1959

Question 4.
Full form of RAM is ……………………. .
(a) Roaming Application Memory
(b) Random Accessible Media
(c) Random Access Memory
(d) None of the above
Answer:
(c) Random Access Memory

Question 5.
Full form of ROM is ……………………. .
(a) Roaming Only Memory
(b) Random Output Media
(c) Read Only Memory
(d) None of the above
Answer:
(c) Read Only Memory

Question 6.
…………………….. is raw information.
(a) Program
(b) Memory
(c) Data
(d) Operating System
Answer:
(c) Data

Question 7.
…………………….. are used for sharing information.
(a) Telephones
(b) Hard disks
(c) RAM
(d) ROM
Answer:
(a) Telephones

Question 8.
Hard disks are used for …………………….. information.
(a) storing
(b) communicating
(c) sharing
(d) all of the above
Answer:
(a) storing

Question 9.
Computers are used for …………………….. information.
(a) storing
(b) managing
(c) sharing
(d) all of the above
Answer:
(d) All of the above

Question 10.
RAM and ROM are 2 types of …………………….. memory.
(a) external
(b) internal
(c) physical
(d) garbage
Answer:
(b) internal

Question 11.
The information stored in ROM is only …………………….., changes cannot be made.
(a) external memory
(b) readable
(c) accessible
(d) physical
Answer:
(b) readable

Question 12.
…………………….. is a group of commands to be given to the computer.
(a) Program
(b) Memory
(c) Data
(d) Operating System
Answer:
(a) Program

Question 13.
…………………….. communicates between the computer and the person working on it.
(a) Program
(b) Memory
(c) Data
(d) Operating System
Answer:
(d) Operating System

Match the columns:

Answer:
(1 – b),
(2 – d),
(3 – e),
(4 – a),
(5 – c)

State whether the following statements are true or false and if false, correct the false statements:

(1) RAM and ROM are the types of external memory.
(2) ICT includes communication devices and the use of those devices as well as services provided with their help.
(3) A computer cannot be used without its operating system.
(4) Microsoft Excel is used to make PowerPoint.
(5) Software refers to the set of commands given to the computer.
Answer:
(1) False. RAM and ROM are the types of internal memory.
(2) True
(3) True
(4) False. Microsoft Excel is used to make spreadsheets.
(5) True

Answer the following in one sentence:

Question 1.
Name the computer which was made between 1946 -1959.
Answer:
The ENIAC computer was made in the period of 1946-1959.

Question 2.
Give one example of Input Unit.
Answer:
Keyboard.

Question 3.
Name the 3 major parts of the processing units.
Answer:

• Memory unit
• Control unit
• ALU unit

Question 4.
What precautions need to the taken care while entering formula into the excel?
Answer:
While using a formula, the ‘=’ sign should be typed first. Similarly, no space should be inserted while typing any formula.

Question 5.
What is Internet Explorer?
Answer:
This is a kind of Search Engine. It helps to find the information we want from all the information available on the internet.

Question 6.
What is a PDF?
Answer:
A PDF or Portable Document Format file can be used to view the file to print it or to handle files.

Question 7.
What is C-DAC?
Answer:
C-D AC, is a well-known Centre for Development of Advanced Computing, situated in Pune.

Write the Full forms of the following abbreviation:

Question 1.
ICT
Answer:
Information Communication Technology

Question 2.
OS
Answer:
Operating System

Question 3.
RAM
Answer:
Random Access Memory

Question 4.
ROM
Answer:
Read Only Memory

Question 5.
CPU
Answer:
Central Processing Unit

Question 6.
DOS
Answer:
Disk Operating System

Question 7.
PDF
Answer:
Portable Document Format

Question 8.
ALU
Answer:
Arithmetic Logical Unit

Question 9.
GUI
Answer:
Graphical User Interface

Question 10.
C-DAC
Answer:
Centre for Development of Advanced Computing

Question 11.
ISCII
Answer:
Indian Script Code for Information Interchange

Define the following:

Question 1.
Memory
Answer:
Memory is the place for storing data obtained from the input and also the generated solution or answer by the computer.

Question 2.
RAM
Answer:
RAM is created from electronic components and can function only as long as it is supplied with electricity.

Question 3.
ROM
Answer:
Information stored in ROM can only be read and changes cannot be made to the information originally stored here.

Question 4.
Operating System
Answer:
It is a program which provides a means of communication between the computer and the person working on it. It is called the DOS (Disk Operating System).

Question 5.
Program
Answer:
A program is a group of commands to be given to a computer.

Question 6.
Data and Information
Answer:
Data is information in its raw (unprocessed) form.

Question 7.
Hardware
Answer:
Hardware consists of all the electronic and mechanical parts used in computers.

Question 8.
Software:
Answer:
Software refers to the commands given to the computer, information supplied to it (input) and the results obtained from the computer after analysis (output).

Give scientific reasons:

Question 1.
Computer cannot function without its operating system.
Answer:

• Operating system is like a link between the computer and the person working on it.
• Operating system manages all the activities performed by the computer.
• Without the operating system, the user won’t be able to input any data or run any program. Thus, a computer cannot run without an operating system.

Question 2.
ROM is a Read Only Memory.
Answer:

• ROM also known as Read Only Memory is a part of internal memory of a computer where the information stored can only be read.
• ROM helps store data permanently for a long period of time and the information stored cannot be deleted.
• Thus, data in a ROM can only be read and cannot be altered or modified and hence, it is called as Read Only Memory.

Complete the table:
Answer:

Answer the following questions:

Question 1.
What precautions will you take when entering data?
Answer:

1. As far as possible, the data should be kept in tabular form. Different types of data should be entered in different cells. Data should be entered neatly and in one ‘flow’. Unnecessary space and special characters should not be used.
2. Many times we ‘drag and fill’ data. At such times, the ‘smart tag’ can be used after ‘drag data’ to fill any data in any manner as required.
3. Once the data has been entered, it can be formatted in different ways. Similarly, we can perform different types of calculations, using different formulae.
4. While using a formula, the ‘=’ sign should be typed first. Similarly, no space should be inserted while typing any formula.

Answer in detail:

Question 1.
Write in short about the opportunities in the field of ICT.
Answer:
(i) Software Field: This is an important field. Having accepted the challenge of creating software, many companies have entered this field. The opportunities in the software field can be classified as follows – application program development, software package development, operating systems and utility development, special purpose scientific applications.

(ii) Hardware Field: Today, there are several companies in our country too, which make computers. They sell computers that they have themselves made. Others sell computers brought from outside as well as repair them and take maintenance contracts to keep computers in big companies working efficiently without a break. Plenty of jobs are available here. There are job opportunities in hardware designing, hardware production, hardware assembly and testing, hardware maintenance, servicing and repairs, etc.

(iii) Marketing: There are many establishments which make and sell computers and related accessories. They need good sales personnel who are experienced in the working of computers as well as skilled in marketing.

(iv) Training: The training of new entrants for various jobs is a vast field. It is very important to have dedicated teachers who are competent in the field of computers.

Question 2.
Write in short about the industries conducting research in the field of computers.
Answer:

• C-DAC, the well-known Centre for Development of Advanced Computing, situated in Pune, is the leading institute in India that conducts research in the field of computers.
• The first Indian supercomputer was made with help from this institute. Valuable guidance for making this computer (the Param computer) was received from the senior scientist Vijay Bhatkar. Param means the supreme.
• This computer can perform one billion calculations per second. It is used in many fields like space research, movements in the interior of the earth, research in oil deposits, medicine, meteorology, engineering, military etc.
• C-DAC has also participated in developing the ISCII code for writing different language scripts. (Indian Script Code for Information Interchange).

Question 3.
Use Microsoft Word to create a document and write equations.
Answer:

• Click on the Microsoft word 2010 icon on the desktop.
• Select the ‘New option in the ‘File’ tab, and then select the ‘Blank document’ option.
• Type your material on the blank page on the screen using the keyboard. Use the language, font size, bold, etc. options in the Home tab to make the typed material attractive.
• To type equations in the text, select the ‘Equation’ option in the ‘Insert’ tab.
  
• Select the proper equation and type it using mathematical symbols.
  

Balbharati Maharashtra State Board 9th Std Science Textbook Solutions 

• Information Communication Technology (ICT) Class 9 Science Textbook Solutions
• Reflection of Light Class 9 Science Textbook Solutions
• Study of Sound Class 9 Science Textbook Solutions
• Carbon: An Important Element Class 9 Science Textbook Solutions
• Substances in Common Use Class 9 Science Textbook Solutions
• Life Processes in Living Organisms Class 9 Science Textbook Solutions
• Heredity and Variation Class 9 Science Textbook Solutions
• Introduction to Biotechnology Class 9 Science Textbook Solutions
• Observing Space: Telescopes Class 9 Science Textbook Solutions

9th Standard Maths 2 Practice Set 9.1 Chapter 9 Surface Area and Volume Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 9.1 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 9 Surface Area and Volume.

Class 9 Maths Part 2 Practice Set 9.1 Chapter 9 Surface Area and Volume Questions With Answers Maharashtra Board

Question 1.
Length, breadth and height of a cuboid shape box of medicine is 20 cm, 12 cm and 10 cm respectively. Find the surface area of vertical faces and total surface area of this box.
Given: For cuboid shape box of medicine,
length (l) = 20 cm, breadth (b) = 12 cm and height (h) = 10 cm.
To find: Surface area of vertical faces and total surface area of the box
Solution:
i. Surface area of vertical faces of the box
= 2(l + b) x h
= 2(20+ 12) x 10
= 2 x 32 x 10
= 640 sq.cm.

ii. Total surface area of the box
= 2 (lb + bh + lh)
= 2(20 x 12+ 12 x 10 + 20 x 10)
= 2(240 + 120 + 200)
= 2 x 560
= 1120 sq.cm.
∴ The surface area of vertical faces and total surface area of the box are 640 sq.cm, and 1120 sq.cm, respectively.

Question 2.
Total surface area of a box of cuboid shape is 500 sq.unit. Its breadth and height is 6 unit and 5 unit respectively. What is the length of that box?
Given: For cuboid shape box,
breadth (b) = 6 unit, height (h) = 5 unit Total surface area = 500 sq. unit.
To find: Length of the box (l)
Solution:
Total surface area of the box = 2 (lb + bh + lh)
∴ 500 = 2 (6l + 6 x 5 + 5l)
∴ \(\frac { 500 }{ 2 }\) = (11l + 30)
∴ 250= 11l + 30
∴ 250 – 30= 11l
∴ 220 = 11l
∴ 220 = l
∴ \(\frac { 220 }{ 11 }\) = l
∴ l = 20 units
∴ The length of the box is 20 units.

Question 3.
Side of a cube is 4.5 cm. Find the surface area of all vertical faces and total surface area of the cube.
Given: Side of cube (l) = 4.5 cm
To find: Surface area of all vertical faces and the total surface area of the cube
Solution:
i. Area of vertical faces of cube = 4l²
= 4 (4.5)² = 4 x 20.25 = 81 sq.cm.
ii. Total surface area of the cube = 6l²
= 6 (4.5)²
= 6 x 20.25
= 121.5 sq.cm.
∴ The surface area of all vertical faces and the total surface area of the cube are 81 sq.cm, and 121.5 sq.cm, respectively.

Question 4.
Total surface area of a cube is 5400 sq. cm. Find the surface area of all vertical faces of the cube.
Given: Total surface area of cube = 5400 sq.cm.
To find: Surface area of all vertical faces of the cube
Solution:
i. Total surface area of cube = 6l²
∴ 5400 = 6l²
∴ \(\frac { 5400 }{ 6 }\) = l²
∴ l² = 900
ii. Area of vertical faces of cube = 4l²
= 4 x 900 = 3600 sq.cm.
∴ The surface area of all vertical faces of the cube is 3600 sq.cm.

Question 5.
Volume of a cuboid is 34.50 cubic metre. Breadth and height of the cuboid is 1.5 m and 1.15 m respectively. Find its length.
Given: Breadth (b) = 1.5 m, height (h) = 1.15 m
Volume of cuboid = 34.50 cubic metre
To find: Length of the cuboid (l)
Solution:
Volume of cuboid = l x b x h
∴ 34.50 = l x b x h
∴ 34.50 = l x 1.5 x 1.15

= 20
∴ The length of the cuboid is 20 m.

Question 6.
What will be the volume of a cube having length of edge 7.5 cm ?
Given: Length of edge of cube (l) = 7.5 cm
To find: Volume of a cube
Solution:
Volume of a cube = l²
= (7.5)³
= 421.875 ≈ 421.88 cubic cm
∴The volume of the cube is 421.88 cubic cm.

Question 7.
Radius of base of a cylinder is 20 cm and its height is 13 cm, find its curved surface area and total surface area, (π = 3.14)
Given: Radius (r) = 20 cm, height (h) = 13 cm
To find: Curved surface area and
the total surface area of the cylinder
Solution:
i. Curved surface area of cylinder = 2πrh
= 2 x 3.14 x 20 x 13
= 1632.8 sq.cm

ii. Total surface area of cylinder = 2πr(r + h)
= 2 x 3.14 x 20(20 + 13)
= 2 x 3.14 x 20 x 33 = 4144.8 sq.cm
∴ The curved surface area and the total surface area of the cylinder are 1632.8 sq.cm and 4144.8 sq.cm respectively.

Question 8.
Curved surface area of a cylinder is 1980 cm² and radius of its base is 15 cm. Find the height of the cylinder. (π = \(\frac { 22 }{ 7 }\))
Given: Curved surface area of cylinder = 1980 sq.cm., radius (r) = 15 cm
To find: Height of the cylinder (h)
Solution:
Curved surface area of cylinder = 2πrh
∴ 1980 = 2 x \(\frac { 22 }{ 7 }\) x 15 x h
∴ \(h=\frac{1980 \times 7}{2 \times 22 \times 15}\)
∴ h = 21 cm
∴ The height of the cylinder is 21 cm.

Maharashtra Board Class 9 Maths Solutions

• Surface Area and Volume Practice Set 9.1 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9.2 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9.3 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9 Class 9 Maths Solutions

9th Standard Maths 2 Practice Set 9.2 Chapter 9 Surface Area and Volume Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 9.2 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 9 Surface Area and Volume.

Class 9 Maths Part 2 Practice Set 9.2 Chapter 9 Surface Area and Volume Questions With Answers Maharashtra Board

Question 1.
Perpendicular height of a cone is 12 cm and its slant height is 13 cm. Find the radius of the base of the cone.
Given: Height (h) = 12 cm, length (l) = 13 cm
To find: Radius of the base of the cone (r)
Solution:
l² = r² + h²
∴ 13² = r² + 12²
∴ 169 = r² + 144
∴169 – 144 = r²
∴ r² = 25
∴ r = √25 … [Taking square root on both sides]
= 5 cm
∴ The radius of base of the cone is 5 cm.

Question 2.
Find the volume of a cone, if its total surface area is 7128 sq.cm and radius of base is 28 cm. ( π = \(\frac { 22 }{ 7 }\))
Given: Radius (r) = 28 cm,
Total surface area of cone = 7128 sq.cm
To find: Volume of the cone
Solution:
i. Total surface area of cone = πr (l + r)
∴ 7128= y x 28 x (l + 28)
∴ 7128 = 22 x 4 x(l +28)
∴ l + 28 = \(\frac { 7128 }{ 22\times 4 }\)
∴ l + 28 = 81
∴ l = 81 – 28
∴ l = 53cm

ii. Now, l² = r² + h²
∴ 53² = 28²+ h²
∴ 2809 = 784 + h²
∴ 2809 – 784 = h²
∴ h² = 2025
∴ h = \(\sqrt { 2025 }\) …… [Taking square root on both sides]
= 45 cm

= 22 x 4 x 28 x 15
= 36960 cubic.cm
∴ The volume of the cone is 36960 cubic.cm.

Question 3.
Curved surface area of a cone is 251.2 cm² and radius of its base is 8 cm. Find its slant height and perpendicular height, (π = 3.14)
Given: Radius (r) = 8 cm, curved surface area
of cone = 251.2 cm²
To find: Slant height (l) and the perpendicular height (h) of the cone
Solution:
i. Curved surface area of cone = πrl
∴ 251.2 = 3.14 x 8 x l

∴ l= 10 cm

ii. Now, l² = r² + h²
∴ 10² = 8² + h²
∴ 100 = 64 + h²
∴ 100 – 64 = h²
∴ h² = 36
∴ h = √36 … [Taking square root on both sides]
= 6 cm
∴ The slant height and the perpendicular height of the cone are 10 cm and 6 cm respectively.

Question 4.
What will be the cost of making a closed cone of tin sheet having radius of base 6 m and slant height 8 m if the rate of making is ₹ 10 per sq.m?
Given: Radius (r) = 6 m, length (l) = 8 m
To find: Total cost of making the cone
Solution:
i. To find the total cost of making the cone of tin sheet, first we need to find the total surface area of the cone.
Total surface area of the cone = πr (l + r)
= \(\frac { 22 }{ 7 }\) x 6 x (8 + 6)
= \(\frac { 22 }{ 7 }\) x 6 x 14
= 22 x 6 x 2 = 264 sq.m

ii. Rate of making the cone = ₹ 10 per sq.m
∴ Total cost = Total surface area x Rate of making the cone
= 264 x 10
= ₹ 2640
∴ A The total cost of making the cone of tin sheet is ₹ 2640.

Question 5.
Volume of a cone is 6280 cubic cm and base radius of the cone is 20 cm. Find its perpendicular height, (π = 3.14)
Given: Radius (r) = 20 cm,
Volume of cone = 6280 cubic cm
To find: Perpendicular height (h) of the cone
Solution:

∴ The perpendicular height of the cone is 15 cm.

Question 6.
Surface area of a cone is 188.4 sq.cm and its slant height is 10 cm. Find its perpendicular height (π = 3.14).
Given: Length (l) =10 cm, curved surface area of the cone = 188.4 sq.cm
To find: Perpendicular height (h) of the cone
Solution:
i. Curved surface area of the cone = πrl
∴ 188.4 = 3.14 x r x 10

ii. Now, l² = r² + h²
∴ 10² = 6² + h²
∴ 100 = 36 + h²
∴ 100 – 36 = h²
∴ h² = 64
∴ h = \(\sqrt { 64 }\) … [Taking square root on both sides]
= 8 cm
∴ The perpendicular height of the cone is 8 cm.

Question 7.
Volume of a cone is 1232 cm³ and its height is 24 cm. Find the surface area of the cone. (π = \(\frac { 22 }{ 7 }\))
Given: Height (h) = 24 cm,
Volume of cone = 1232 cm³
To find: Surface area of the cone
Solution:

∴ r² = 49
∴ r = \(\sqrt { 49 }\) … [Taking square root on both sides]
= 7 cm

ii. Now, l² = r² + h²
∴ l² = 7² + 24²
= 49 + 576 = 625
∴ l = \(\sqrt { 625 }\) … [Taking square root on both sides]
= 25

iii. Curved surface area of cone = πrl
= \(\frac { 22 }{ 7 }\) x 7 x 25
= 22 x 25
= 550 sq.cm
∴The surface area of the cone is 550 sq.cm.

Question 8.
The curved surface area of a cone is 2200 sq.cm and its slant height is 50 cm. Find the total surface area of cone. (π = \(\frac { 22 }{ 7 }\))
Given: Length (l) = 50 cm, curved surface area of cone = 2200 sq.cm
To find: Total surface area of the cone
Solution:
i. Curved surface area of cone = πrl

ii. Total surface area of cone = πr (l + r)
= \(\frac { 22 }{ 7 }\) x 14 x (50 + 14)
= \(\frac { 22 }{ 7 }\) x 14 x 64
= 22 x 2 x 64
= 2816 sq.cm
∴ The total surface area of the cone is 2816 sq.cm.

Question 9.
There are 25 persons in a tent which is conical in shape. Every person needs an area of 4 sq.m, of the ground inside the tent. If height of the tent is 18 m, find the volume of the tent.
Given: For the tent,
height (h) = 18m,
number of people in the tent = 25,
area required for each person = 4 sq.m
To find: Volume of the tent
Solution:
i. Every person needs an area of 4 sq.m, of the ground inside the tent.
Surface area of the base of the tent = number of people in the tent × area required for each person
= 25 × 4
= 100 sq.m

ii. Surface area of the base of the tent = πr²
∴ 100 = πr²
∴ πr² = 100

iii. Volume of the tent= \(\frac { 1 }{ 3 }\) πr²h
= \(\frac { 1 }{ 3 }\) x 100 x 18 …….[∵ πr² = 100]
= 100 x 6
= 600 cubic metre
∴ The volume of the tent is 600 cubic metre.

Question 10.
In a field, dry fodder for the cattle is heaped in a conical shape. The height of the cone is 2.1 m and diameter of base is 7.2 m. Find the volume of the heap of the fodder. If it is to be covered by polythene in rainy se&son then how much minimum polythene
sheet is needed? (π = \(\frac { 22 }{ 7 }\) and \(\sqrt { 17.37 }\) = 4.17 ]
Given: Height of the heap (h) = 2.1 m.
diameter of the base (d) = 7.2 m
∴Radius of the base (r) = \(\frac { d }{ 2 }\) = \(\frac { 7.2 }{ 2 }\) = 3.6 m
To find: Volume of the heap of the fodder and polythene sheet required
Solution:
i. Volume of the heap of fodder = \(\frac { 1 }{ 3 }\)πr²h
= \(\frac { 1 }{ 3 }\) x \(\frac { 22 }{ 7 }\) x (3.6)² x 2.1
= \(\frac { 1 }{ 3 }\) x \(\frac { 22 }{ 7 }\) x 3.6 x 3.6 x 2.1
= 1 x 22 x 1.2 x 3.6 x 0.3
= 28.51 cubic metre

ii. Now, l² = r² + h²
= (3.6)² + (2.1)²
= 12.96 + 4.41
∴ l² =17.37
∴ l² = \(\sqrt { 17.37 }\) .. .[Taking square root on both sides]
= 4.17 m

iii. Area of the polythene sheet needed to cover the heap of the fodder = Curved surface area of the conical heap
= πrl
= \(\frac { 22 }{ 7 }\) x 3.6 x 4.17
= 47.18 sq.m
∴ The volume of the heap of the fodder is 28.51 cubic metre and a polythene sheet of 47.18 sq.m will be required to cover it.

Maharashtra Board Class 9 Maths Solutions

Maharashtra Board Class 9 Maths Solutions

• Surface Area and Volume Practice Set 9.1 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9.2 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9.3 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9 Class 9 Maths Solutions

9th Standard Maths 2 Practice Set 9.3 Chapter 9 Surface Area and Volume Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 9.3 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 9 Surface Area and Volume.

Class 9 Maths Part 2 Practice Set 9.3 Chapter 9 Surface Area and Volume Questions With Answers Maharashtra Board

Question 1.
Find the surface areas and volumes of spheres of the following radii
i. 4 cm
ii. 9 cm
iii. 3.5 cm (π = 3.14)
i. Given: Radius (r) = 4 cm
To find: Surface area and volume of sphere
Solution:
Surface area of sphere = 4πr²
= 4 x 3.14 x 4²
∴ Surface area of sphere = 200.96 sq.cm
Volume of sphere = \(\frac { 4 }{ 3 }\)πr³
= \(\frac { 4 }{ 3 }\) x 3.14 x 4²
∴ Volume of sphere = 267.95 cubic cm

ii. Given: Radius (r) = 9 cm
To find: Surface area and volume of sphere
Solution:
Surface area of sphere = 4πr²
= 4 x 3.14 x 9²
∴ Surface area of sphere = 1017.36 sq.cm

∴ Volume of sphere = 3052.08 cubic cm

iii. Given: Radius (r) = 3.5 cm
To find: Surface area and volume of sphere
Solution:
Surface area of sphere = 4πr²
= 4 x 3.14 x (3.5)²
∴ Surface area of sphere = 153.86 sq.cm
Volume of sphere = \(\frac { 4 }{ 3 }\) πr³
= \(\frac { 4 }{ 3 }\) x 3.14 x (3.5)³
∴ Volume of sphere = 179.50 cubic cm

Question 2.
If the radius of a solid hemisphere is 5 cm, then find its curved surface area and total surface area, (π = 3.14)
Given: Radius (r) = 5 cm
To find: Curved surface area and total surface area of hemisphere
Solution:
i. Curved surface area of hemisphere = 2πr²
= 2 x 3.14 x 5²
= 2 x 3.14 x 25
= 50 x 3.14
= 157 sq.cm.

ii. Total surface area of hemisphere = 3πr²
= 3 x 3.14 x 5²
= 235.5 sq.cm.
∴ The curved surface area and totai surface area of hemisphere are 157 sq.cm, and 235.5 sq.cm, respectively.

Question 3.
If the surface area of a sphere is 2826 cm² then find its volume. (π = 3.14)
Given: Surface area of sphere = 2826 sq.cm.
To find: Volume of sphere
Solution:
i. Surface area of sphere = 4πr²
∴ 2826 = 4 x 3.14 x r²
2826 = 282600 = 900
∴ \( r^{2}=\frac{2826}{4 \times 3.14}=\frac{282600}{4 \times 314}=\frac{900}{4}\)
∴ r² = 225
∴ r = \(\sqrt { 225 }\) … [Taking square root on both sides]
= 15 cm

ii. Volume of sphere = \(\frac { 4 }{ 3 }\) πr³
= \(\frac { 4 }{ 3 }\) x 3.14 x 15³
= \(\frac { 4 }{ 3 }\) x 3.14 x 15 x 15 x 15
= 4 x 3.14 x 5 x 15 x 15
= 14130 cubic cm.
∴ The volume of the sphere is 14130 cubic cm.

Question 4.
Find the surface area of a sphere, if its volume is 38808 cubic cm. (π = \(\frac { 22 }{ 7 }\))
Given: Volume of sphere = 38808 cubic cm.
To find: Surface area of sphere
Solution:

∴ r³ = 441 x 21 = 21 x 21 x 21
∴ r = 21 cm … [Taking cube root on both sides]
ii. Surface area of sphere = 4πr²
= 4 x \(\frac { 22 }{ 7 }\) x 21
= 4 x \(\frac { 22 }{ 7 }\) x 21 x 21
= 4 x 22 x 3 x 21
= 5544 sq.cm.
∴ The surface area of sphere is 5544 sq.cm.

Question 5.
Volume of a hemisphere is 18000π cubic cm. Find its diameter.
Given: Volume of hemisphere = 1 8000π cubic cm.
To find: Diameter of the hemisphere
Solution:
i. Volume of hemisphere = \(\frac { 2 }{ 3 }\) πr³

= 9000 x 3
∴ r³ = 27000
∴ r = \(\sqrt [ 3 ]{ 27000 }\) … [Taking cube root on both sides]
= 30 cm

ii. Diameter = 2r
= 2 x 30 = 60 cm
∴ The diameter of the hemisphere is 60 cm.

Maharashtra Board Class 9 Maths Solutions

Maharashtra Board Class 9 Maths Solutions

• Surface Area and Volume Practice Set 9.1 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9.2 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9.3 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9 Class 9 Maths Solutions

9th Standard Maths 2 Practice Set 9 Chapter 9 Surface Area and Volume Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 9 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 9 Surface Area and Volume.

Class 9 Maths Part 2 Practice Set 9 Chapter 9 Surface Area and Volume Questions With Answers Maharashtra Board

Question 1.
If diameter of a road roller is 0.9 m and its length is 1.4 m, how much area of a field will be pressed in its 500 rotations? ( π = \(\frac { 22 }{ 7 }\))
Given: For road roller,
diameter (d) = 0.9 m, length (h) = 1.4 m
To find: Area of a field pressed in 500 rotations
Solution:
i. Since, area of field pressed in 1 rotation of road roller = curved surface area of road roller
∴ Curved surface area of the road roller = 2πrh
= πdh ,..[∵ d = 2r]
= \(\frac { 22 }{ 7 }\) x 0.9 x 1.4 7
= 22 x 0.9 x 0.2
= 3.96 sq.m.

ii. Area of land pressed in 1 rotation = 3.96 sq.m.
∴Area of land pressed in 500 rotations = 500 x 3.96
= 1980 sq.m.
∴ 1980 sq.m, land will be pressed in 500 rotations of the road roller.

Question 2.
To make an open fish tank, a glass sheet of 2 mm gauge is used. The outer length, breadth and height of the tank are 60.4 cm, 40.4 cm and 40.2 cm respectively. How much maximum volume of water will be contained in it ?
Given: Thickness of the glass = 2 mm,
outer length of the tank = 60.4 cm,
outer breadth of the tank = 40.4 cm,
outer height of the tank = 40.2 cm
To find: Volume of water fish tank contains
Solution:

i. Thickness oldie glass = 2 mm.
= \(\frac { 2 }{ 10 }\) cm
= 0.2 cm
Outerlengthofthetank = 60.4 cm
∴ Inner length oldie tank (l) = Outer length – thickness oldie glass on both sides
= 60.4 – 0.2 – 0.2
= 60cm
Outer breadth oldie tank = 40.4 cm
∴ Inner breadth of the tank (b) = 40.4 – 0.2 – 0.2
= 40 cm
Outer height of the tank = 40.2 cm
∴Inner height of the tank (h) = 40.2 – 0.2
= 40 cm

ii. Maximum volume of water that can be contained in the tank = volume of the tank
= l x b x h
= 60 x 40 x 40
= 96000 cubic cm.
∴ The fishtank can contain maximum of 96000 cubic cm. water in it.

Question 3.
If the ratio of radius of base and height of a cone is 5 : 12 and its volume is 314 cubic metre. Find its perpendicular height and slant height (π = 3.14).
Given: Ratio of radius of base and height of a cone = 5 : 12,
Volume = 314 cubic metre
To find: Perpendicular height (h) and slant height (l)
Solution:
i. The ratio of radius and height of cone is 5 : 12
Let the common multiple be x.
∴ Radius of base (r) = 5x
Perpendicular height (h) = 12x

∴ x³ = 1
∴ x = 1 … [Taking cube root on both sides]
∴ r = 5x = 5(1) = 5m
h = 12x = 12(1) = 12 m

ii. Now, l² = r² + h²
= 5² + 12²
= 25 + 144
∴l² = 169
∴ l = \(\sqrt { 169 }\) … [Taking square root on both sides]
= 13 m
The perpendicular height and slant height of the cone are 12 m and 13 m respectively.

Question 4.
Find the radius of a sphere if its volume is 904.32 cubic cm. (π = 3.14)
Given: Volume of sphere = 904.32 cubic cm.
To find: Radius of a sphere
Solution:
Volume of sphere = \(\frac { 4 }{ 3 }\) πr³
∴ 904.32 = \(\frac { 4 }{ 3 }\) x 3.14 x r³

= 216
∴ r = \(\sqrt [ 3 ]{ 216 }\) … [Taking cube root on both sides]
= 6 cm
∴ The radius of the sphere is 6 cm.

Question 5.
Total surface area of a cube is 864 sq.cm. Find its volume.
Given: Total surface area of cube = 864 sq. cm
To find: Volume of cube
Solution:
i. Total surface area of cube = 6l²
∴ 864 = 6l²
∴ l²= \(\sqrt [ 864 ]{ 6 }\)
∴ l² = 144
∴ l = \(\sqrt { 144 }\) … [Taking square root on both sides]
= 12 cm

ii. Volume of cube = l²
= 12³
= 1728 cubic cm.
∴ The volume of cube is 1728 cubic cm.

Question 6.
Find the volume of a sphere, if its surface area is 154 sq.cm.
Given: Surface area of sphere = 154 sq. cm.
To find: Volume of sphere
Solution:
i. Surface area of sphere = 4πr²


∴ The volume of sphere is 179.67 cubic cm.

Question 7.
Total surface area of a cone is 616 sq.cm. If the slant ‘height of the cone Is three times the radius of its base, find its slant height.
Given: Total surface area of a cone = 616 sq.cm., slant height of the cone is three times the radius of its base
To find: Slant height (l)
Solution:
i. Let the radius of base be r cm.
∴ Slant height (l) = 3r cm
Total surface area of cone = πr (l + r)
∴ 616 = πr(l + r)
∴ 616 = \(\sqrt [ 22 ]{ 7 }\) x r x (3r + r)
∴ 616 = \(\sqrt [ 22 ]{ 7 }\) x 4r²

∴ r² = 49
∴ r = \(\sqrt { 49 }\) … [Taking square root on both sides]
= 7

ii. Slant height (l) = 3r = 3 x 7 = 21 cm
∴ The slant height of the cone is 21 cm.

Question 8.
The inner diameter of a well is 4.20 metre and its depth is 10 metre. Find the inner surface area of the well. Find the cost of plastering it from inside at the rate ₹ 52 per sq.m.
Given: Inner diameter (d) = 4.2 m,
To find: depth (h) = 10 m,
rate of plastering = ₹ 52 per sq.m.
Inner surface area and total cost of plastering
Solution:
i. Inner curved surface area of the well = 2πrh
= πdh …[∵ d = 2r]
= \(\sqrt [ 22 ]{ 7 }\) x 4.2 x 10
= \(\sqrt [ 22 ]{ 7 }\) x 42
= 22 x 6
= 132 sq.m.

ii. Rate of plastering = ₹52 per sq.m.
∴ Total cost = Curved surface area x Rate of plastering
= 132 x 52 = ₹6864
∴ The cost of plastering the well from inside is ₹6864.

Question 9.
The length of a road roller is 2.1 m and its diameter is 1.4 m. For levelling a ground 500 rotations of the road roller were required. How much area of ground was levelled by the road roller? Find the cost of levelling at the rate of ₹ 7 per sq.m.
Given: For road roller,
diameter (d) = 1.4 m,
length (h) = 2.1 m
number of rotations required for levelling the ground = 500,
rate of levelling = ₹ 7 per sq. m.
To find: Area of ground leveled by the road roller and cost of levelling
Solution:
i. Since, area of ground levelled in 1 rotation of road roller = curved surface area of road roller
∴Curved surface area of the road roller = 2πrh
= πdh …[∵ d = 2r]
= \(\frac { 22 }{ 7 }\) x 1.4 x 2.1
= 22 x 0.2 x 2.1
= 9.24 sq.m.

ii. Area of ground levelled in 1 rotation = 9.24 sq.m.
∴Area of ground levelled in 500 rotations = 9.24 x 500
= 4620 sq.m.

iii. Rate of levelling ₹ 7 per sq.m.
∴Total cost = Area of ground levelled x Rate of levelling
= 4620 x 7
= ₹32340
∴ The road roller levels 4620 sq.m. land in 500 rotation, and the cost of levelling is ₹32340.

Maharashtra Board Class 9 Maths Chapter 9 Surface Area and Volume Practice Set 9 Intext Questions and Activities

Question 1.
Curved surface area of cone. (Textbook pg. no. 116)

Circumference of base of the cone = 2πr
As shown in the figure (c), make pieces of the net as small as possible. Join them as shown in the figure (d),. By joining the small pieces of net of the cone, we get a rectangle ABCD approximately.
Total length of AB and CD is 2πr.
∴ length of side AB of rectangle ABCD is πr and length of side CD is also πr.
Length of side BC of rectangle = slant height of cone = l.
Curved surface area of cone is equal to the area of the rectangle.
∴ curved surface area of cone = Area of rectangle = AB x BC = πr x l = πrl

Question 2.
Prepare a cylinder of a card sheet, keeping one of its faces open. Prepare an open cone of card sheet which will have the same base-radius and the same height as that of the cylinder. Pour fine sand in the cone till it just fills up the cone. Empty the cone in the cylinder. Repeat the procedure till the cylinder is just filled up with sand. Note how many coneful of sand is required to fill up the cylinder. (Textbook pg, no 117)

Answer:
To fill the cylinder, three coneful of sand is required.

Question 3.
Finding total surface area of sphere. (Textbook pg, no 120)

i. Take a sweet lime (Mosambe), Cut it into two equal parts.

ii. Take one of the parts. Place its circular face on a paper. Draw its circular border. Copy three more such circles. Again, cut each half of the sweet lime into two equal parts.

iii. Now you get 4 quarters of sweet lime. Separate the peel of a quarter part. Cut it into pieces as small as possible. Try to cover one o’f the circles drawn, by the small pieces. Observe that the circle gets nearly covered.
The activity suggests that,
Curved surface area of a sphere = 4πr²

∴ Curved surface area of a sphere = 4 x Area of a circle

Question 4.
Make a cone and a hemisphere of cardsheet such that radii of cone and hemisphere are equal and height of cone is equal to radius of the hemisphere.
Fill the cone with fine sand. Pour the sand in the hemisphere. How many cones are required to fill the hemisphere completely ? (Textbook pg. no. 121)

Answer:
To fill the hemisphere, two coneful of sand is required.

Maharashtra Board Class 9 Maths Solutions

• Surface Area and Volume Practice Set 9.1 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9.2 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9.3 Class 9 Maths Solutions
• Surface Area and Volume Practice Set 9 Class 9 Maths Solutions

9th Standard Maths 2 Problem Set 8 Chapter 8 Trigonometry Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 8 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 8 Trigonometry.

Class 9 Maths Part 2 Problem Set 8 Chapter 8 Trigonometry Questions With Answers Maharashtra Board

Question 1.
Choose the correct alternative answer for the following multiple choice questions.

i. Which of the following statements is true?
(A) sin θ = cos (90 – θ)
(B) cos θ = tan (90 – θ)
(C) sin θ = tan (90 – θ)
(D) tan θ = tan (90 – θ)
Answer:
(A) sin θ = cos (90 – θ)

ii. Which of the following is the value of sin 90°?
(A) \( \frac { \sqrt { 3 } }{ 2 }\)
(B) 0
(C) \(\frac { 1 }{ 2 }\)
(D) 1
Answer:
(D) 1

iii. 2 tan 45° + cos 45° – sin 45° = ?
(A) 0
(B) 1
(C) 2
(D) 3
Answer:
2 tan 45° + cos 45° – sin
\( =2(1)+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=2\)
(C) 2

iv. \( \frac{\cos 28^{\circ}}{\sin 62^{\circ}}\) =?
(A) 2
(B) -1
(C) 0
(D) 1
Answer:
\( \frac{\cos 28^{\circ}}{\sin 62^{\circ}}\)

(D) 1

Question 2.
In right angled ∆TSU, TS = 5, ∠S = 90°, SU = 12, then find sin T, cos T, tan T. Similarly find sin U, cos U, tan U.

Solution:
i. TS = 5, SU = 12 …[Given]
In ∆TSU, ∠S = 90° … [Given]
∴ TU² = TS² + SU² …[Pythagoras theorem]
= 5² + 12² = 25 + 144 = 169
∴ TU = \(\sqrt { 169 }\) .. .[Taking square root of both sides]
= 13

Question 3.
In right angled ∆YXZ, ∠X = 90°, XZ = 8 cm, YZ = 17 cm, find sin Y, cos Y, tan Y, sin Z, cos Z, tan Z.

Solution:
i. XZ = 8 cm, YZ = 17 cm …[Given]
In ∆YXZ, ∠X = 90° … [Given]
∴ YZ² = XY² + XZ² .. .[Pythagoras theorem]
∴ 17² = XY² + 8²
∴ 289 = XY² + 64
∴ XY² = 289 – 64
= 225
∴ x = \(\sqrt { 225 }\) .. .[Taking square root of both sides]
= 15

Question 4.
In right angled ∆LMN, if ∠N = θ, ∠M = 90°, cos θ = \(\frac { 24 }{ 25 }\), find sin θ and tan θ. Similarly, find (sin²θ) and (cos²θ).

Solution:
i. cos θ = \(\frac { 24 }{ 25 }\)
In ∆LMN, ∠M = 90°, ∠N = θ

Let the common multiple be k.
∴ MN = 24k and LN = 25k
Now, LN²= LM² + MN² … [Pythagoras theorem]
∴ (25k)² = LM² + (24k)²
∴ 625 k² = LM² + 576k²
∴ LM² = 625k² – 576k²
∴ LM² = 49k²
∴ LM = \(\sqrt { 49{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= 7k

Question 5.
Fill in the blanks.
i. sin 20° = cos
ii. tan 30° x tan  = 1
iii. cos 40° = sin
Solution:
i. sin 20° = cos (90° – 20°) …..[∵ sin θ = cos (90 – θ)]
= cos 70°

ii. tan θ x tan (90 – θ) = 1
Substituting θ = 30°,
tan 30° x tan (90 – 30)° = 1
∴ tan 30° x tan 60° = 1

iii. cos 40° = sin (90° – 40°) …[∵ COS θ = sin (90 – θ)]
= sin 50°

Maharashtra Board Class 9 Maths Solutions Chapter 8 Trigonometry Problem Set 8

Question 1.
Measuring height of a tree using trigonometric ratios. (Textbook pg. no. 101)

This experiment can be conducted on a clear sunny day. Look at the figure given above. Height of the tree is QR, height of the stick is BC.
Thrust a stick in the ground as shown in the figure. Measure its height and length of its shadow. Also measure the length of the shadow of the tree. Using these values, how will you determine the height of the tree?
Solution:
Rays of sunlight are parallel.
So, ∆PQR and ∆ABC are equiangular i.e., similar triangles.
Sides of similar triangles are proportional.
∴ \(\frac { QR }{BC }\) = \(\frac { PR }{ AC }\)
∴ Height of the tree (QR) = \(\frac { BC }{ AC }\) x PR
Substituting the values of PR, BC and AC in the above equation, we can get length of QR i.e., the height of the tree.

Question 2.
It is convenient to do the above experiment between 11:30 am and 1:30 pm instead of doing it in the morning at 8’O clock. Can you tell why? (Textbook pg. no. 101)
Solution:
At 8’O clock in the morning, the sunlight is not very bright. At the same time, the sun is on the horizon and the shadow would by very long. It would be extremely difficult to measure shadow in that case.
Between 11:30 am and 1:30 pm, the sun is overhead and it would be easier to measure the length of shadow.

Question 3.
Conduct the above discussed activity and find the height of a tall tree in your surrounding. If there is no tree in the premises, then find the height of a pole. (Textbook pg. no. 101)

Maharashtra Board Class 9 Maths Solutions

• Trigonometry Practice Set 8.1 Class 9 Maths Solutions
• Trigonometry Practice Set 8.2 Class 9 Maths Solutions
• Trigonometry Problem Set 8 Class 9 Maths Solutions

9th Standard Maths 2 Practice Set 8.2 Chapter 8 Trigonometry Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 8.2 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 8 Trigonometry.

Class 9 Maths Part 2 Practice Set 8.2 Chapter 8 Trigonometry Questions With Answers Maharashtra Board

Question 1.
In the following table, a ratio is given in each column. Find the remaining two ratios in the column and complete the table.

Solution:
i. cos θ = \(\frac { 35 }{ 37 }\) …(i) )[Given]
In right angled ∆ABC,

∠C = θ.

Let the common multiple be k.
∴ BC = 35k and AC = 37k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ (37k)² = AB²+ (35k)²
1369k² = AB² + 1225k²
AB2 = 1369k² – 1225k²
= 144k²
AB = 144k²
AB = \(\sqrt { 2ghK }\)² … [Taking square root of both sides]
= 12k

ii. sin θ = \(\frac { 11 }{ 61 }\) …..(i) [Given]
In right angled ∆ABC, ∠C = θ.


Let the common multiple be k.
AB = 11k and AC = 61k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ (61k)² = (11k)² + BC²
∴ 3721k² = 121k² + BC²
∴ BC² = 3721k² – 121k² = 3600k²
BC = \(\sqrt { 3600{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= 60k

iii. tan θ = 1 = \(\frac { 1 }{ 1 }\) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 1k and BC = 1k
Now, AC² = AB² + BC² …[Pythagoras theorem]
= K² + K²
= 2K²
∴ AC = \(\sqrt { 2{ k }\)

iv. sin θ = \(\frac { 1 }{ 2 }\) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 1k and BC = 2k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ 2K² = K² + BC²
∴ 4K² = K² + BC²
∴ BC² = 4K² – K² = 3K²
∴ BC = \(\sqrt { 3{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= \(\sqrt { 3{ k }\)

v. cos θ = \(\frac { 1 }{ \sqrt { 3 } } \) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 1k and BC = √3k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ (√3K)² = AB² + K²
∴ 3K² = 3K² – K² = 2K²
∴ AB = \(\sqrt { 2{ k }^{ 2 } }\) .. .[Taking square root of both sides]
AB = √2K

vi. cos θ = \(\frac { 21 }{ \sqrt { 20 } } \) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 21k and BC = 20k
Now, AC² = AB² + BC² …[Pythagoras theorem]
= (21)K² + (20K)²
= 441K² – 400²
= 841K²
∴ AB = \(\sqrt { 841{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= 29K

vii. tan θ = \(\frac { 8 }{ 15 } \) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 8k and BC = 15k
Now, AC² = AB² + BC² …[Pythagoras theorem]
= (8)K² + (15K)²
= 64K² – 225²
= 289K²
∴ AC = \(\sqrt { 289{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= 17K

viii. sin θ = \(\frac { 3 }{ 5 } \) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.



Let the common multiple be k.
∴ AB = 3k and AC = 5k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ (5)K²= (3)K² + BC²
∴ 25K² = 9K² – 225²
∴ BC² = 25K² – 9K²
∴ BC = \(\sqrt { 16{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= 4K

ix. tan θ = \(\frac { 1 }{ 2\sqrt { 2 } }\) ..(i) [Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 1k and AC = 2√2 k
Now, AC² = AB² + BC² …[Pythagoras theorem]
= K² + (2√2 k )²
= K² – 225²
= 25K² + 8K²
= 9K²
∴ AC = \(\sqrt { 9{ k }^{ 2 } }\) .. .[Taking square root of both sides]
= 3K

Question 2.
Find the values of:
i. 5 sin 30° + 3 tan 45°
ii. \(\frac { 4 }{ 5 }\)tan² 60° + 3 sin² 60°
iii. 2 sin 30° + cos 0° + 3 sin 90°
iv. \(\frac{\tan 60^{\circ}}{\sin 60^{\circ}+\cos 60^{\circ}}\)
v. cos² 45° + sin² 30°
vi. cos 60° x cos 30° + sin 60° x sin 30°
Solution:
i. sin 30° = \(\frac { 1 }{ 2 }\) and tan 45° = 1

ii. \(\frac { 4 }{ 5 }\)tan² 60° + 3 sin² 60°

iii. 2 sin 30° + cos 0° + 3 sin 90°
2 sin 30° + cos0° + 3 sin 90° = 2 (\(\frac { 1 }{ 2 }\)) + 1 + 3(1)
= 1 + 1 + 3
∴ 2 sin 30° + cos 0° + 3 sin 90° = 5

iv. \(\frac{\tan 60^{\circ}}{\sin 60^{\circ}+\cos 60^{\circ}}\)

v. cos² 45° + sin² 30°

vi. cos 60° x cos 30° + sin 60° x sin 30°

Question 3.
If sin θ = \(\frac { 4 }{ 5 }\) , then find cos θ.
Solution:
sin θ = \(\frac { 4 }{ 5 }\) .. .(i)[Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ AB = 4k and AC = 5k
Now, AC² = AB² + BC² … [Pythagoras theorem]
∴ (5 k)² = (4k)² + BC²
∴ 25k² = 16k² + BC²
∴ BC² = 25k² – 16k² = 9k²
∴ BC = \(\sqrt { 9{ k }^{ 2 } }\) . .[Taking square root of both sides]
= 3k

Question 4.
If cos θ = \(\frac { 15 }{ 17 }\) , then find sin θ.
Solution:
cos θ = \(\frac { 15 }{ 17 }\) .. .(i)[Given]
In right angled ∆ABC,
∠C = θ.


Let the common multiple be k.
∴ BC = 15k and AC = 17k
Now, AC² = AB² + BC² … [Pythagoras theorem]
∴ (17 k)² = AB² + (15K)²
∴ 289k² = AB² + 225²
∴ AB² = 289k² – 225k²
= 64k²
∴ AB = \(\sqrt { 64{ k }^{ 2 } }\) . .[Taking square root of both sides]
= 8k

Maharashtra Board Class 9 Maths Chapter 8 Trigonometry Practice Set 8.2 Intext Questions and Activities

Question 1.
In right angled ∆PQR, ∠Q = 900. Therefore ∠P and ∠R are complementary angles of each other. Verify the following ratios.
i. sin θ = cos (90 – θ)
ii. cos θ = sin (90 – θ)
iii. sin 30° = cos (90° – 30°) = cos 60°
iv. cos 30° = sin (90° – 30°) = sin 60° (Textbook pg. no. 107)
Solution:
In ∆PQR, ∠Q = 90°, ∠P = θ
∴ ∠R = 90 – θ

i. sin θ = cos (90 – θ)

ii. cos θ = sin (90 – θ)

iii. Let ∠P = θ = 30°
∴ ∠R = 90° – 30°

sin 30° = cos (90° – 30°) … [From (i) and (ii)]
sin 30° = cos 60°

iv. cos 30° = sin (90° – 30°) = sin 60°

∴ cos 30° = sin (90° – 30°) .,.[From (i) and (ii)]
∴ cos 30° = sin 60°

Question 2.
In right angled ∆PQR, ∠Q = 90°, ∠R = θ and if sin θ = \(\frac { 5 }{ 13 }\), then find cos θ and tan θ. (Textbook pg. no. 110)
Solution:
i. Take the given trigonometric ratio as 13k equation (i).
sin θ = \(\frac { 5 }{ 13 }\) .. .(i)[Given]
By using the definition write the trigonometric ratio of sin O and take it as equation (ii).
In right angled ∆PQR, ∠R = θ


Let the common multiple be k.
∴ PQ = 5k and PR = 13k
Find QR by using Pythagoras theorem.
PR² = PQ² + QR² … [Pythagoras theorem]
∴ (13k)² = (5k)² + QR²
∴ 169k² = 25k² + QR²
∴ QR² = 169k² – 25k²
= 144k²
∴ QR = \(\sqrt { 144{ k }^{ 2 } }\) . . . [Taking square root of both sides]
= 12k

Question 3.
While solving the above Illustrative example, why the lengths of PQ and PR are taken 5k and 13k? (Textbook pg. no. 111)
Solution:
\(\frac { PQ }{ PR }\) = \(\frac { 5 }{ 13 }\) … [Given]
Here, the ratio of the lengths of sides PQ and PR is 5 : 13.
The actual lengths of the sides can be any multiple of the ratio. Hence, we consider the multiple k while solving.

Question 4.
While solving the above illustrative example, can we take the lengths of PQ and PR as 5 and 13? If so, then what changes are needed In the writing of the solution. (Tcxtbook pg. no. 111)
Solution:
Yes, we can take lengths of PQ and PR as 5 and 13.
In that case, we will have to take k = 1 and solve the problem accordingly.

Question 5.
Verify that the equation ‘sin² θ + cos² θ = 1’ is true when θ = 0° or θ = 90°.
(Textbook pg. no. 112)
Solution:
sin² θ + cos² θ = 1
i. lf θ = 0°,
LH.S. = sin² θ + cos² θ
= sin² 0° + cos² 0°
= 0 + 1 …[∵ sin 0° = 0, cos 0° = 1]
= R.H.S.
∴ sin² θ + cos² θ = 1

ii. If θ = 90°,
L.H.S.= sin² θ +cos² θ
= sin² 90° + cos² 90°
= 1 + 0 … [ ∵ sin 90° = 1, cos 90° = 0]
= 1
= R.H.S.
∴ sin² θ + cos² θ = 1

Maharashtra Board Class 9 Maths Solutions

• Trigonometry Practice Set 8.1 Class 9 Maths Solutions
• Trigonometry Practice Set 8.2 Class 9 Maths Solutions
• Trigonometry Problem Set 8 Class 9 Maths Solutions

9th Standard Maths 2 Practice Set 8.1 Chapter 8 Trigonometry Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 8.1 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 8 Trigonometry.

Class 9 Maths Part 2 Practice Set 8.1 Chapter 8 Trigonometry Questions With Answers Maharashtra Board

Question 1.
In the given figure, ∠R is the right angle of ∆PQR. Write the following ratios.

i. sin P
ii. cos Q
iii. tan P
iv. tan Q
Solution:

Question 2.
In the right angled ∆XYZ, ∠XYZ = 90° and a, b, c are the lengths of the sides as shown in the figure. Write the following ratios.

i. sin x
ii. tan z
iii. cos x
iv. tan x.
Solution:

Question 3.
In right angled ∆LMN, ∠LMN = 90°, ∠L = 50° and ∠N = 40°. Write the following ratios.

i. sin 50°
ii. cos 50°
iii. tan 40°
iv. cos 40°
Solution:

Question 4.
In the given figure, ∠PQR = 90°, ∠PQS = 90°, ∠PRQ = α and ∠QPS = θ. Write the following trigonometric ratios.
i. sin α, cos α , tan α
ii. sin θ, cos θ, tan θ

Solution:
i. In ∆PQR,

ii. In ∆PQS,

Maharashtra Board Class 9 Maths Chapter 8 Trigonometry Practice Set 8.1 Intext Questions and Activities

Question 1.
In the figure gIven below, ∆PQR is a right angled triangle. Write the names of sides opposite and adjacent to ∠P and ∠R. (Textbook pg no. 102)

Solution:
In right angled ∆PQR,
i. side opposite to ∠P = QR
ii. side opposite to ∠R = PQ
iii. side adjacent to ∠P = PQ
iv. side adjacent to ∠R = QR

Maharashtra Board Class 9 Maths Solutions

• Trigonometry Practice Set 8.1 Class 9 Maths Solutions
• Trigonometry Practice Set 8.2 Class 9 Maths Solutions
• Trigonometry Problem Set 8 Class 9 Maths Solutions