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You can Download Reflection of Light in Spherical Mirrors Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Basic Science Solutions Chapter 18 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 18 Reflection of Light in Spherical Mirrors

The phenomena of light is always a surprising one. From long time ago man has tried to study about light. This chapter explains to draw the figure of the images formed by mirrors, their uses, magnification etc.

Reflection Of Light In Spherical Mirrors Class 8 Notes Spherical mirrors

Images are formed not only in-plane mirrors but also in smooth curved surfaces. Spherical mirrors are mirrors in which the reflecting surface is a part of the sphere. The center of a sphere of which the mirror is a part is the center of curvature. Any I line drawn from the center of curvature to the mirror is normal to the mirror. Radius of curvature of a mirror is the radius of the sphere of which is a part. The reflecting surface of the mirror is called the aperture of a mirror. The midpoint of the reflecting surface is called the pole. The straight line connecting the pole and center of curvature of a mirror is the principal axis of the mirror.

The angle of incident and angle of reflection are equal in spherical mirrors. Rays of light incident on a concave mirror, parallel to the principal axis, passes through a fixed point on the principal axis after reflection. This point is the principal focus of the concave mirror. Rays of light incident on a convex mirror parallel to the principal axis appear to come from a fixed point on the other side of the mirror. The point is the principal axis of the convex mirror.

Focal length of a mirror is the distance from pole of the mirror and the principal’s focus of the mirror. Rays of light coming from infinity get focused on a plane perpendicular to the principal axis. This plane is the focus plane. The focus plane passes through the focus plane.

Reflection Of Light In Spherical Mirrors Class 8 Images formed by spherical mirrors

The image of an object placed different positions in front of a mirror is.formed in different positions. Rays of light incident on a concave mirror, parallel to the principal axis, passes through a fixed point on the principal axis after reflection Rays incident on an a concave mirror through the principal focus are reflected parallel to the principal axis.

Rays incident through the center of curvature reflected through the same path. Rays incident on the pole makes an angle with the principal axis. The ray diagrams of images formed by spherical mirrors are drawn according to the above rules. An object placed between F and C in front of a concave mirror, the image will be real and inverted. If the object is at F the image is formed at infinity. The paths of reflected rays are parallel to each other. If the object is in between F and P the image is formed behind the mirror. The image is erect and virtual.

Reflection Of Light In Spherical Mirrors 8th Magnification

Magnification is the ratio of the height of the image to the height of the object.

Reflection Of Light In Spherical Mirrors Class 8 Notes Pdf Uses of spherical mirrors

Spherical mirrors are used in lighthouses and reflectors.

Reflection of Light in Spherical Mirrors Textbook Questions and Answers

Class 8 Physics Notes Kerala Syllabus Questions 1.
Classify the following statements as to those related to concave mirrors and convex mirrors and tabulate them accordingly.
a. to view the face
b. as makeup mirror
c. as rearview mirrors in vehicles
d. in solar concentrators
e. in periscopes
f. as shaving mirror
Answer:
Concave mirror:

• In solar concentrators
• Makeup mirror
• Shaving mirror

Convex mirror:

• In rearview mirrors of vehicles
• In Searchlights

Plane mirror:

• In periscopes
• To see face

8th Class Physics Notes Pdf Question 2.
Calculate the radius of curvature of a convex mirror of focal length 12 cm.
Answer:
focal length of the mirror = 12 cm
\(\mathrm{f}=\frac{\mathrm{R}}{2} \quad 12=\frac{\mathrm{R}}{2}\)
R = 2 × 12 = 24 cm

Reflection Of Light At Curved Surfaces Questions And Answers 8th Question 3.
A ray of light is made to fall on the pole of a concave mirror making an angle 30° with the principal axis.
a. What is the angle of reflection?
b. Justify your answer.
c. Draw the ray diagram.
Answer:
a. Angle of reflection is 30°
b. angle of incidence and angle of reflection are equal
c. the figure showing angle of incidence and angle of reflection is 210°

Chemistry Class 8 Kerala Syllabus Question 4.
Which type of mirror always gives an erect and diminished image?
Answer:
Convex mirror

Kerala Syllabus 8th Standard Chemistry Notes Question 5.
A ray of light incident on a spherical mirror gets reflected along the same path. If so, show the light incident on the mirror.
Answer:

8th Class Biology Notes Pdf Question 6.
OA is a ray of light incident on a concave mirror.
a. Draw the path of the reflected ray

b. On what basis did you mark the reflected ray?
Answer:

b. In mirrors angle of incidence and angle of reflection are same. The normal to the point of incidence is passed through the center of curva¬ture. The angle between ray of reflection and the normal is the same as angle of incidence.

8th Class Biology Notes Pdf Malayalam Medium Question 7.
Write down the type of mirrors that should be used for getting the following type of images.
a. real and magnified
b. virtual and magnified
c. virtual and diminished
d. real and diminished
Answer:
a. concave mirror
b. concave mirror
c. convex mirror
d. concave mirror

Kerala Syllabus 8th Standard Chemistry Notes Malayalam Medium Question 8.
The height of an object kept 12 cm away from a concave mirror is 1 cm. Calculate the magnifica¬tion if an image of height 2.5 cm is formed in front of the mirror.
Answer:

8th Standard Chemistry Textbook Question 9.
Which type of mirror always gives a virtual and erect-image, b. Is this image magnified or diminished?
Answer:
a. convex mirror
b. Diminished

Reflection of Light in Spherical Mirrors Additional Questions and Answers

8th Class Biology Notes Pdf Kerala Syllabus Question 1.
Complete the table

Answer:
a. 30°
b. 40°
c. 50°
d. 60°

Kerala Syllabus 8th Standard Physics Notes Question 2.
If the radius of curvature of a concave mirror is 24 cm what is its focal length?
Answer:
R = 24 cm

8th Standard Chemistry Textbook Kerala Syllabus Question 3.
Find the radius of curvature of a convex mirror of focal length 0.6 m
Answer:

Basic Science Class 8 Solutions Question 4.
Write the characteristics of the image formed by an object plac¬ed at the center of curvature of a concave mirror?
Answer:
Position: At the center of curvature at the same side
Size : Same as the size of the object Nature: Real, Inverted

Reflection Of Light By Spherical Mirrors 8th Question 5.
Complete the figure

Answer:

Hsslive Guru Physics 8th Standard Question 6.
Complete the table

Answer:
a. Passes through principal focus
b. Seem to come from the principal focus
c, d. Returns parallel to the principal axis
e, f. Returns through the same path
g, h. Reflects in the same angle of incident ray.

Question 7.
Write three differences of real image and virtual image which is made by spherical mirrors
Answer:
Real image:

1. Inverted
2. can be shown on the screen
3. can measure the length of the image and distance to the image

Virtual image:

1. Cannot show on the screen
2. Cannot measure
3. Erect

Question 8.
Write uses of concave mirrors
Answer:

•  As shaving mirror
• As makeup mirror
• As head mirrors used by doctors
• In film projectors

Question 9.
Examine the position off the object given in the figure and table the following peculiarities

a. position of the image
b. size of the image
c. nature of the image
Answer:
a. Behind the mirror.
b. Larger than the object.
c. Erect and virtual

Question 10.
When an object of height 4 cm is placed in front of a concave mirror an image of height 8cm is formed. Find magnification.
Answer:
h_i = 4cm h_o= -8 cm
Magnification = \(\frac{h_{i}}{h_{0}}\) = \(\frac{h_{-8}}{h_{4}}\) = -2

Question 11.

a. Examine the figure and find the magnification
b. What is the height of the object if height of the image is 4cm when the object is placed on the same position in front of the mirror.
Answer:

Question 12.
Write the uses of convex mirror
Answer:

• As reflector in the street light
• As rearview mirror
• In searchlights

You can Download Statistics Questions and Answers, Activity, Notes, Kerala Syllabus 9th Standard Maths Solutions Chapter 13 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Maths Solutions Chapter 13 Statistics

Statistics Textual Questions and Answers

Textbook Page No. 193

Statistics Class 9 Kerala Syllabus Question 1.
The weight of 6 players in a volleyball team are all different and the average weight is 60 kilograms.
i. Prove that the team has at least one player weighing more than 60 kilograms.
ii. Prove that the team has at least one player weighing less than 60 kilograms.
Answer:
i. Total weight of 6 players = 60 × 6 = 360 kg
The team contains players having weights 60, less than 60 or greater than 60. If the weights of all the players are less than 60, the average will also be less than 60. This is not possible. Therefore there will be at least one player having weight greater than 60.

ii. If the weight of all the players are greater than 60, the average will also be greater than 60. Therefore there will be at least one player having weight less than 60.

Hss Live Guru 9th Maths Kerala Syllabus Question 2.
Find two sets of 6 numbers with average 60, satisfying each of the conditions below:
i. 4 of the numbers are less than 60 and 2 of them greater than 60.
ii. 4 of the numbers are greater than 60 and 2 of them less than 60.
Answer:
Total sum = 60 × 6 = 360
i. 20, 30, 40, 50, 100, 120

ii. 5, 15, 70, 80, 90, 100
Other ways are also possible.

Kerala Syllabus 9th Standard Maths Notes Question 3.
The table shows the children in a class, sorted according to the marks they got for a math test.

Calculate the average marks of the class.
Answer:
Total number of children is 45. Repeated addition can be written as multiplication.

Average mark = \(\frac { Total }{ Number }\) = \(\frac { 297 }{ 45 }\) = 6.6

Class 9 Maths Solutions Kerala Syllabus Question 4.
The table below shows the days in a month sorted according to the amount of rainfall in a locality

What is the average rainfall per day during this month?
Answer:

The average of the rain fall per day during that month = \(\frac { Total rain fall }{ Number of days }\)
= \(\frac { 1545 }{ 30 }\) = 51.5mm

Hsslive Guru 9th Maths Kerala Syllabus Question 5.
The details of rubber sheets a farmer got during a month are given below.

i. How many kilograms of rubber did he get a day on average in this month?
ii. The price of rubber is 120 rupees per kilogram. What is his average income per day this month from selling rubber?
Answer:


i. Average Quantity of rubber per day = \(\frac { 381 }{ 30 }\) = 12.77kg
ii. . If the price is Rs. 120 per kg, then average incomeperday = 12.7 × 120 = Rs.1524

Textbook Page No. 197

Kerala Syllabus 9th Standard Maths Solutions Question 1.
Find different sets of 6 different numbers between 10 and 30 with each number given below as mean:
i. 20
ii. 15
iii. 25
Answer:
i. The mean of 6 numbers is 20.
ie; sum = 6 x 20= 120 (Write 3 pairs with sum 40)
i.e., 15, 25, 18, 22, 19, 21

ii. The mean of 6 numbers is 15
i.e; sum =6 x 15=90
(Write 30 pairs with sum 3)
12, 18, 13, 17, 14, 16

iii. Mean is 25
sum = 25 x 6 = 150
(Write 50 pairs with sum 3)
22, 28, 23, 27, 24, 26

Hss Live Guru Class 9 Maths Kerala Syllabus Question 2.
The table below shows the children in a class, sorted according to their heights.

What is the mean height of a child in this class?
Answer:

Mean height = \(\frac { Total height }{ No of children }\)
= \(\frac { 7892 }{ 50 }\) = 157.84 cm

Hss Live Class 9 Maths Kerala Syllabus Question 3.
The teachers in a university are sorted according to their ages, as shown below.

What is the mean age of a teacher in this university?
Answer:

Mean age = \(\frac { Total age }{ No of persons }\)
= \(\frac { 2775 }{ 70 }\) = 39.64

Kerala Syllabus 9th Standard Maths Question 4.
The table below shows children in a class sorted according to their weights.

The mean weight is calculated as 26 kilograms. How many children have weights between 23 and 25 kilograms?
Answer:
Let’s prepare the table for finding the mean by considering the number of children in the group 23 to 25, as ‘x’.

Mean weight = 26 kg

i.e; the number of children having weight between 23 to 25 is 7.

Statistics Exam oriented Questions and Answers

Kerala Syllabus Std 9 Maths Solutions Question 1.
The details of rubber sheets got for a month by a farmer are given in the table.

During this month he got an average of 10 sheets per day. If so in how many days did he get 11kg per day ?
Answer:

Let ‘x’ be the number of days in which he got 11 kg rubber sheet.

∴ He got 11 kgs of sheets for 5 days.

Kerala Syllabus 9th Standard Maths Notes Malayalam Medium Question 2.
In a factory, there are workers belonging to four categories. The average income and the number of workers in each category are given. What is the mean income when all the workers in the four categories are combined?

Answer:

Mean income = \(\frac { 320000 }{ 40 }\)
= Rs. 8000

Kerala Syllabus 9th Standard Maths Solution Question 3.
The daily wages of 10 workers in a factory are given below.
400, 350, 450, 500, 400, 500, 350, 500, 350, 450
If one more person is joined, the mean becomes Rs. 450. What is the daily wage of the new person?
Answer:
Total wages of 10 workers = 4250 Total wages of 11 workers= 11 × 450 = Rs. 4950
Wage of the 11th person = 4950 – 4250

Kerala State Class 9 Maths Solutions Question 4.
Find 10 different numbers between 10 and 30 whose mean is 20.
Answer:
Given mean is 20
Sum 20 × 10 = 200.
We have to find 10 different numbers whose sum is 200 (for this find 5 pairs of sum 40)
(15, 25) (16, 24) (17, 23) (18, 22) (19, 21)
The numbers are 15, 16, 17, 18, 19, 21, 22, 23, 24, 25

9th Maths Notes Kerala Syllabus Question 5.
A table categorizing the workers in an office on the basis of their salary is given below.

Find the mean of salary.
Answer:

Mean income = \(\frac { 349500 }{ 15 }\)
= R.s 23300

Kerala Syllabus 9th Standard Notes Maths Question 6.
i. Find the mean of natural numbers from 1 to 100.
ii. What is the mean of even numbers from 1 to 100? What is the mean of odd numbers?
iii. What is the difference between the means of the first 100 even numbers and odd numbers?
iv. What is the difference between the means of the first 200 even numbers and 200 odd numbers?
Answer:
Sum of the natrural numbers from 1 to n = \(\frac n{ n + 1 }{ 2 }\)


Sum of the first 100 odd numbers
= 1002 = 100 × 100

In general, the difference between the means of n even numbers and n odd numbers is always 1.

Question 7.
A table tabulating the players in a cricket team on the basis of their age is given below.

Calculate the mean age of the players?
Answer:

Mean age of players = 306/12 = 25.5

You can Download अकाल में सारस Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Hindi Solutions Unit 4 Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Hindi Solutions Unit 4 Chapter 1 अकाल में सारस (कविता)

अकाल में सारस Textual Questions and Answers

अकाल में सारस विश्लेषणात्मक प्रश्न

अकाल में सारस कविता का आशय Kerala Syllabus 9th प्रश्ना 1.
धान की सूखी पत्तियों की गंध’ –से किसका आभास होता है?

उत्तर:
धान के पौधे गर्मियों में सूखते हैं। धान की सूखी पत्तियों की गंध से गर्मियों का आभास मिलता है।

Akal Mem Saras Hindi Poem Kerala Syllabus 9th प्रश्ना 2.
सारसों ने जल भर कटोरे को क्यों न देखा होगा?

उत्तर:
सारसों के लिए कटोरे भर का जल पर्याप्त नहीं है। वे ताल-तलैयों की तलाश कर रहे थे। उनके लिए प्राकृतिक जल स्रोतों की ज़रूरत थी। इसलिए वे जल भर कटोरे को न देखा होगा।

अकाल में सारस Summary Kerala Syllabus 9th प्रश्ना 3.
सारसों ने जाते-जाते शहर की ओर क्यों मुड़कर देखा होगा?

उत्तर:
शायद सारसों को वह जगह पसंदीदा थी। लेकिन उनकी आशा के विरुद्ध वहाँ कोई जलस्रोत न देखा। वे निराश लौटने लगे। वहाँ के लोगों के प्रति उनके मन में दया या घृणा की भावना थी। दया इसलिए कि ये लोग कितने अभागे हैं। घृणा शायद इसलिए कि इन्होंने प्राकृतिक जलस्रोतों का ह्रास किया है।

अकाल में सारस Additional Questions and Answers

अकाल में सारस आशयग्रहण के प्रश्न

Akaal Mein Saras Summary In Hindi Kerala Syllabus 9th प्रश्ना 1.
वे देर तक करते रहे शहर की परिक्रमा’ -सारस किसकी तलाश में शहर की परिक्रमा करते होंगे?

उत्तर:
सारस देश-देशांतर में जानेवाले पक्षी हैं। वे ऋतुओं के बदलने के अनुसार एक देश से दूसरे में चले जाते हैं। यहाँ सारस पानी की तलाश कर रहे हैं।

अकाल में सारस कविता की व्याख्या Kerala Syllabus 9th प्रश्ना 2.
बुढ़िया ने क्या सोचकर पानी लाकर रखा होगा?

उत्तर:
सारसों को देखकर बुढ़िया ने सोचा कि वे पानी की तलाश में हैं। इसलिए बुढ़िया ने कटोरे में पानी लाकर रखा।

अकाल में सारस Grammar

अकाल में सारस व्याकरण के प्रश्न

अकाल में सारस Summary In Hindi Kerala Syllabus 9th प्रश्ना 1.
तुलना करें :

i. तीन बजे दिन में / आ गए वे

ii. वे दिन में तीन बजे आ गए।

iii. कवितांश की संरचना और वाक्य की संरचना का अंतर पहचानें।

iv. कविता को गद्य में बदलने पर शब्दों के क्रम में क्या परिवर्तन आया है?

उत्तर:
वाक्य की संरचना कर्ता, कर्म और क्रिया का पालन किया है। लेकिन कविता की संरचना में यह क्रम नहीं है। कविता की सुंदरता को बढ़ाने के लिए कभी-कुभी शब्दों के क्रम में आवश्यक परिवर्तन करते हैं।

Hss Live Guru 9th Hindi Kerala Syllabus प्रश्ना 2.
इन पंक्तियों को गद्य की संरचना में बदलकर लिखें।

उत्तर:

9th Class Hindi Notes Kerala Syllabus प्रश्ना 3.
लेख लिखें :

1. प्रकृति के जलस्रोतों का नाश होता जा रहा है।
2. साथ ही पर्यावरण का संतुलन भी बिगड़ रहा है।

उत्तर:
अकाल में सारस’ कविता मनुष्यों के बुरे व्यवहार की संकेत करनेवाली कविता है। प्राकृतिक संसाधन केवल मनुष्यों के लिए नहीं है। पशु-पक्षी, मछलियाँ और दूसरे प्राणी भी इनके हकदार है। दरअसल ऋतु बदलने पर सारस पक्षी पानी आदि की तलाश में एक देश से दूसरे देश में चले जाते हैं। लेकिन यहाँ कहीं पानी नहीं है। सारसों को तो पता तक नहीं था कि लोग उन्हें सारस कहते हैं। यहाँ तो पानी और धान की पत्तियाँ सूख गई हैं। उनकी गंध हवा में हैं। यही गंध सारसों के डैनों से नीचे झर रही है। एक बुढ़िया अपने आँगन में जलभरा कटोरा रख देती है।

लेकिन सारसों ने न तो बुढ़िया को देखा। क्योंकि वे जलस्रोतों की तलाश में आए थे। उनकी निगाहों में दया थी या घृणा, हम समझ नहीं पाते। दया शायद इसलिए कि सारस तो उड़कर कहीं और चले जाएँगे, जहाँ पानी होगा। किंतु ये दयनीय लोग जो अपना पानी तक नहीं बचा पाए, अपने खेत-खलिहान और घर-बार छोड़कर, कहाँ और कैसे जाएँगे? और घृणा इसलिए कि अपना पानी तो नष्ट किया ही हमारा भी नष्ट कर दिया। क्योंकि पर्यावरण नष्ट करने के ज़िम्मेदार तो मनुष्य ही हैं।

अकाल में सारस Summary in Malayalam and Translation

Akal प्रश्ना 1.
बरसों बीते
बादलों को इधर
बरसे नहीं।
ये पंक्तियाँ किस हालत की ओर ।
संकेत करती है?

उत्तर:
पानी की कमी एक बड़ी समस्था हो.
पर्यावरण में हुए बदलाव से बारिश
की मात्रा कम होती जा रही है। इस
हालत की ओर ये पंक्तियाँ संकेत देती है 

अकाल में सारस शब्दार्थ

You can Download Unravelling Genetic Mysteries Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Biology Solutions Chapter 6 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Biology Solutions Chapter 6 Unravelling Genetic Mysteries

Unraveling Genetic Mysteries Text Book Questions and Answers

Emergence Of Genetics

Genetics is the branch of science emerged at the beginning of the 20th century. It influences almost all areas of life like diagnosis, therapeutic and food production. Gregor Johann Mendel’s hybridization experiments in pea plants have led to the foundation of genetics. So he is considered as the father of Genetics.

Unravelling Genetic Mysteries Kerala Syllabus 10th Chapter 6 Question 1.
What are the traits that were experimented by Mendel?
Answer:
The traits that were experimented by Mendel

• Height of the plant (Tall – Dwarf)
• Position of the flower (Terminal – Axial)
• Shape of the seed (Round – Wrinkled)
• Colour of the seed (Green-Yellow)
• Colour of the flower (Purple, White)
• Shape of the pod (Inflated – Constricted)
• Colour of the fruit (Yellow-Green)

Experiments Of Mendel

Sslc Biology Chapter 6 Kerala Syllabus Chapter 6 Question 2.
Which trait of the pea plant was considered in these experiments?
Answer:
Height of the pea plant was considered as the trait in this experiment.

Biology Chapter 6 Class 10 Kerala Syllabus Chapter 6 Question 3.
What variant forms of the trait are considered here?
Answer:
Tall and dwarf are the variant forms of the trait considered here.

Unravelling Genetic Mysteries Notes Kerala Syllabus 10th Chapter 6 Question 4.
Which forms of the trait was expressed in the first generation?
Answer:
Tall is the trait was expressed in the first generation.

Sslc Biology Chapter 6 Notes Kerala Syllabus Question 5.
The following illustration showing hybridization experiment in pea plants using symbols for the factors that control traits.
Answer:

Peculiarities of offsprings in the second generation

Mendel self-pollinated the plants obtained in the first generation and produced the second generation. Among the 1064 plants obtained in the second generation, 787 plants were tall and 277 plants were dwarf. The ratio of the result obtained is about 3:1

Statistics In Mendel’S Experiment

10th Class Biology 6th Chapter Kerala Syllabus Question 6.
Complete the table

Answer:
a) Axial
b)3:1
c) Round
d) 3:1

Inferences of Mendel:

• A trait is controlled by the combination of two traits.
• One character is expressed (dominant character) and the other character remains hidden (recessive character) in the offspring first generation.
• The character which remains hidden in the first generation appears in the second generation.
• The ratio of the dominant character and recessive character in the second generation is 3:1

Gene – Allele:
The gene present in the chromosome of the nucleus determines the character. A gene that controls a trait has different forms. The different forms of a gene are called alleles. Generally, a gene has two alleles. When.we illustrate hybridization experiment, the allele that controls the dominant character that is expressed in the first generation is indicated by a capital letter and the allele that controls recessive character is indicated by a small letter.

Sslc Biology Chapter 6 Malayalam Medium Question 7.
Which are the alleles of a tall plant?
Answer:
‘TT are the alleles of a tall plant.

Hsslive Guru 10th Biology Kerala Syllabus Chapter 6 Question 8.
Which are the alleles of the dwarf planets?
Answer:
‘tt’ are the alleles of the dwarf planet.

Sslc Biology Focus Area Questions And Answers Kerala Syllabus Chapter 6 Question 9.
How do the allele combination of the first generation differ from parental plants?
Answer:
The allele combination ‘Tt’ is responsible for the character height in first-generation instead of TT. One character is expressed and the other character remains hidden in the offsprings.

Kerala Genetics Question 10.
How can you differentiate alleles that control the dominant character and recessive character?
Answer:
The allele that controls dominant character generally indicated by capital letter and the allele that controls recessive character is indicated by a small letter.

Biology Class 10 Kerala Syllabus Kerala Syllabus Chapter 6 Question 11.
Observe the illustration showing the hybridization experiment conducted by Mendel on two traits namely height and color of flowers. Complete the illustration and table suitably based on the indicators, analyze illustration and write down inferences.

Answer:

Tall, red flower TtRr
Self-pollination first-generation TtRr x TtRr

Hss Live Guru 10th Biology Kerala Syllabus Chapter 6 Question 12.
What are the characters expressed in the offsprings of the first generation? Which are the recessive ones?
Answer:
Expressed characters – Tall and red flower
Recessive characters – Dwarf and white flower.

Kerala Syllabus 10th Standard Biology Notes Pdf Chapter 6 Question 13.
Are there new combination of characters different from parents appeared in the second generation? Which are they?
Answer:
Yes. New combination of characters different form parents appeared in the second generation. The appearance of new combination of characters in offsprings is due to the independent assortment of each character.
New combination of characters appeared in second generation

• Dwarf, red flower plants
• Tall, white flower plant

DNA (Deoxyribonucleic Acid)

Sslc Biology Chapter Wise Questions And Answers Kerala Syllabus Chapter 6  Question 14.
What are the peculiarities of double-helical model of DNA presented by Watson and Crick?
Answer:
James Watson and Francis Crick presented the double-helical model of DNA in 1953. As per the double-helical model, DNA contains two strands. Two strands of DNA are made of phosphate and pentose sugar and steps with nitrogen bases. The nitrogen bases are adenine, thymine, guanine, and cytosine.

Structure of DNA molecule

DNA molecule contains two long strands with deoxyribose sugar, phosphate, and steps with nitrogen bases. Nitrogen bases are molecules that contain nitrogen and are alkaline in nature. The nitrogen bases are of four types. In DNA the bases adenine pairs with thymine and guanine pairs with cytosine. One deoxyribose sugar molecule one phosphate molecule and one nitrogenous base join together to form a nucleotide. Nucleotides are the basic unit of DNA. Since DNA has four kinds of nitrogen bases, DNA has four kinds of nucleotides.

Kerala Syllabus 10th Standard Biology Pdf Chapter 6 Question 15.
Observe the illustration and complete its second strand.

Answer:

Terminal Axial Question 16.
Compare the structure of DNA and RNA and complete the table.

Answer:

How Do Genes Act?

Question 17.
Observe the following illustration and write down the inferences in the science diary.

Answer:
DNA does not participate directly in protein synthesis. RNA is the molecule that carries information form DNA to ribosomes and controls protein synthesis. This RNA is messenger of DNA, it is called messenger RNA or mRNA. Besides mRNA, there are tRNA (Transfer RNA) that carry amino acids to the ribosomes and rRNA (Ribosomal RNA) that are seen associated with ribosomes. Protein molecule is synthesized by the combined activities of all these molecules.

Question 18.
Prepare a flow chart of Protein synthesize.
Answer:

There are 46 chromosomes in human beings. Of these 44 are somatic chromosomes and two are sex chromosomes. A somatic chromosome pair contains two identical chromosomes. Thus in human beings, there are 22 pairs of somatic chromosomes. Sex chromosomes are two types. They are called x chromosomes and y chromosomes. Females have 44 + xx and genetics of variation that of male 44 + XY.

Genetics Of Variation

Crossing over in chromosomes: A source of variation:

During the initial phase of meiosis, chromosomes pair and exchange their parts. This process is called crossing over. As a result of this, part of a DNA crosses over to become the part of another DNA. This causes a difference in the distribution of genes. When these chromosomes are transferred to the next generation, it causes the expression of new characters in offsprings.

Combination of Allele during fertilization:
When gametes undergo fusion, the combination of allele changes. This causes the expression of characteristics in offsprings that are different from parents. Thus fertilization causes variations int the next generation.

Mutation

Question 19.
What is mutation?
Answer:
Mutation is a sudden heritable change in the genetic constitution of an organism.

Question 20.
What are the causes of mutations?
Answer:
The defects in the duplication of DNA, certain chemicals and radiations.

Question 21.
What is the importance of mutations?
Answer:
Certain mutations are harmful and some are helpful for survival of the organisms. Mutations lead to variations in characters. Mutation has great relevance in evolution.

Is The Child – Male Or Female

Observe illustration and write down the inferences in the science diary.

Question 22.
Is there any difference in the number of chromosomes in male and female.
Answer:
No. In male and female, there is not any difference in the number of chromosomes. In men and women 46 chromosomes are present.

Question 23.
Which chromosome is different in male and female?
Answer:
Sex chromosome is different. The sex chromosome of female is are XX and in male are XY.

Question 24.
What is the possibility for the birth of a male or a female child? Discuss.
Answer:
The possibility for the birth of a male or a female child is more or less equal. If the X chromosomes in male unite with X chromosome in female the offspring will be female and if with the Y chromosome In female the offspring will be male.

Question 25.
Is it fair to criticize mothers who deliver only female- children. Substantiate your opinion scientifically.
Answer:
The XY chromosomes of the father determine whether the child is male or female. Child with XX sex chromosomes is female arid with XY sex chromosomes is made. So male sex chromosomes have greater importance than female sex chromosomes in sex determination. So the view of her husband and relatives is wrong.

Difference In Colour

Melanin, a pigment-protein imparts color to the skin. The difference in gene function is the reason for the color difference of skin. This is simply an adaption to live under sun.

Let Us Assess

Question 1.
The nitrogen base absent in RNA.
a) Admin
b) Thymine
c) Uracil
d) Cytosine
Answer:
b) Thymine

Question 2.
Arrange the stages of protein synthesis in the form of a flow chart.
1. Combines amino acid.
2. mRNA reaches ribosomes
3. mRNA is formed
4. Amino acids are carried to the ribosomes.
Answer:
mRNA is formed → mRNA reaches ribosomes → amino acids are carried to the ribosomes → combines amino acids.

Question 3.
Observe the hybridization experiment given below

a) Prepare an illustration of this hybridization experiment using symbols
b) Prepare an illustration for the second generation.
Answer:

b) Parental plants (Self-pollination of first-generation) F2 → Gg × Gg

Extended Activities

Question 1.
‘ Prepare an excerpt including information on scientist who made contributions in the progress of genetics. (Hints – Gregor Johann Mendel, Walter, S. Sutton, Boveri, Friedrich Meischer, Johannsen, Avery, James Watson, Francis Crick, Marshall, Nirenberg, Har Gobind Khorana)
Answer:
Milestones in the History of Genetics

Question 2.
Prepare models of DNA and RNA using locally available materials and present them in a science exhibition.

Unraveling Genetic Mysteries More Questions And Answers

Question 1.
The father of genetics?
Answer:
Gregor Johann Mendel

Question 2.
Mendel conducted the process of hybridization using one pair of contrasting characters. In all his experiments only one character was expressed. So which was the method he adopted to find out recessive character?
Answer:
Mendel self-pollinated the plants obtained as the first generation.

Question 3.
When Mendel conducted experiment using one pair of contrasting characters, the plants obtained in the F2 generation is was in ………. ratio.
Answer:
3:1

Question 4.
Find out the dominant characters in plants having TTRR, TTRr, TtRR, TtRr traits.
Answer:
All plants are tall with red flowers.

Question 5.
Hereditary factors which Mendel had described are now known as …………….
Answer:
Genes

Question 6.
What are genes?
Answer:
Genes are the specific units of DNA that control metabolic activities and responsible for the specific characters of any organism.

Question 7.
“Offsprings of the same parents show differences among themselves” Give reasons for this?
Answer:
During fertilization alleles from the chromosomes of gametes segregate and causes change in the allele combination. This change causes variations in the offsprings. So offsprings of the same parents also show differences.

Question 8.
When a woman gave birth to girl children in her consecutive deliveries, her husband and relatives blamed her. Evaluate this social situation and write your opinion.
Answer:
The XY chromosomes of the father determine whether the child is male or female. Child with XX sex chromosomes is female and with XY sex chromosomes is male. So male sex hormones have greater importance than female sex chromosomes in sex determination. So the view of her husband and relatives is wrong.

Question 9.
What is the possible ratio for the birth of a male or a female child? What is the reason for this?
Answer:
The possibility for the birth of a male or a female child is more or less equal. If the X chromosomes in male unite with X chromosome in female the offspring will be female and if with the Y chromosome in female the offspring will be male.

Question 10.
Which are the two types of nucleic acids?
Answer:
DNA, RNA

Question 11.
What is the figure shows?
Answer:
a) a DNA molecule
b) a RNA molecule
c) a nucleotide
d) a chromosome
Answer:
a) nucleotide

Question 12.
Identify the picture? From where is it seen?

Answer:
DNA molecule. DNA molecules are seen in chromosome.

Question 13.
Identify the nucleotides presented in DNA and RNA. Which factor help you to identify this?
Answer:

B and C are nucleotides seen in DNA because the adenine and thymine are seen in DNA. A and C are nucleotide seen in RNA because Adenine and Uracil are seen in RNA.

Question 14.
Give examples for different kinds of RNA seen in the cell?
Answer:
mRNA(messenger RNA), tRNA (transfer RNA), rRNA (ribosomal RNA).

Question 15.
What is the following illustration represents?
Answer:

Answer:
Protein synthesis of DNA.

Question 16.
Explain the role of mRNA in protein synthesis.
Answer:
DNA does not participate directly in protein synthesis. It unwinds and mRNA is synthesized which carries the information from DNA to ribosomes. Based on the information in mRNA protein is synthesized by proper adding amino acids.

Question 17.
Observe the diagram and find out the names of the labeled parts a. b and c.
Answer:

Answer:
a) mRNA
b) Ribosomes
c) Protein molecule

Question 18.
Observe the figures and identify the process given below.

Answer:
Crossing over

Question 19.
‘ While some mutations are harmful, some of them are helpful.” Analyze the statement.
Answer:
This statement is correct. Certain mutations are harmful and some are helpful for survival of the organism. Mutations also lead to evolution.

Question 20.
‘‘Some specific processes during Meiosis helps to create variation in characters among organisms Analyse this statement and explain the process.
Answer:
Homologous chromosomes from father and mother pair exchange chromosomal material during meiosis. Tbi3 is “Crossing Over” in meiosis.

Question 21.
Diagrammatically represent with symbols the First generation of progenies of Tall and Dwarf pea plants when cross-pollinated as in Mendel’s first stage of Experiment (March 2015)
Answer:

Question 22.
Observe the flow chart given below. (Model 2015)
Answer:
i) RNA is synthesized
↓
ii) DNA unwinds
↓
iii) RNA combines with ribosome
↓
iv) RNA comes out through nuclear membrane.
↓
v) Protein molecule is formed.
↓
vi) Different amino acids are formed.
↓
a) Arrange the flow orderly.
b) Identify the process.
Answer:
a) DNA unwinds
↓
RNA Is synthesized
↓
RNA comes out through nuclear membrane.
↓
RNA combines with ribosome
↓
Different amino acids are formed.
↓
Protein molecule is formed
b) Protein synthesis or gene action

Question 23.

(Model 2014)
Observe the diagram Identify the process which occurs in Chromosomes during meiosis.
Answer:
Crossing over

Question 24.
A pea plant with genetic constitution TtRr (Tall plant which produces red flower) was subjected to self-pollination. The genetic constitution of some of the progenies obtained are given below.
Write the expressed character of the given progenies based on their genetic constitution. (Model 2014)
(a) TTRr
(b) ttrr
(c) ttRr
(d) Ttrr
Answer
a) TTRr – tall plant with red flower
b) ttrr – dwarf plant with white flower
c) ttRr – dwarf plant with red flower
d) Ttrr – tall plant with white flower

Question 25.
Mr. Rajan decided to divorce his wife by arguing that she is incapable to give birth to a boy child.
a) Can you agree with Mr. Rajan?
b) Give scientific explanation, to this problem with the help of an illustration showing the role of sex chromosomes in determining sex in human beings.
Answer:
a) I cannot agree with Rajan

Y chromosome from male decide the sex of child

Question 26.
Find the odd one. Write down the common feature of the others. (March 2014)
a) Bypass surgery, ECG, EEG, Pacemaker
b) Adenine, Cytosine, Thymine, Uracil
Answer:
a) EEG: Related with the treatment of heart diseases
b) Uracil: Nitrogen bases in DNA/Thymine: Nitrogen bases in RNA.

Question 27.
Gopalettan is trying to develop new varieties of pea plants in his garden. Given below is the illustration of the experiment which he conducted. Observe this and answer the questions that follow. (Model 2013)

a) Which law of hereditary can be used to explain the reason of violet flowered plants in the F1 generation?
b) Among these, which are the dominant and recessive characters?
c) What will be the characters appearing in the F2 generation if the plants in the F1 generation are self-pollinated? In what ratio?
Answer:
a) Law of dominance. (When a pair of contrasting characters combines, only one character is expressed, while the other remains hidden.)
b) Dominant is violet and recessive is white.
c) Yellow and green seeds in 3:1 ratio.

Question 28.
Fill up the blanks in the table showing gene action in protein synthesis. (Model 2013)

Answer:
a) RNA is being synthesized
b) RNA comes out through nuclear membrane
c) RNA1 combines with ribosomes

Question 29.
Rearrange B and C according to the data given in A. (Model 2013)

Answer:

Question 30.
A portion of DNA molecule is shown below. Find out the missing nitrogen base pair from those given below. (Model 2012)

Answer:
(b) C – G

Question 31.
The fusing of gametes during self-pollination of F1 in pea plants with two separate characters combined together were tabulated; (Model 2012)

A) Fill up the blanks.
B) How many white-flowered plants will be formed if 16 pea plants in F₂ generation were produced?
Answer;
A) (a) TtRR
(b) Ttrr
(c) ttRR
(d) ttrr
B) 4

Question 32.
Given below is the illustration showing how sex determination is taking place in man. (March 2012)

a) Observe the illustration and examine the 4 types of possible offsprings. Specify their sex chromosomes.
b) What are the inferences you arrive at from this illustration?
c) What is the probable ratio of formation of male child and female children in man ? Illustrate your answer.
d) Which are the sex-determining chromosomes in man?
Answer:
a) (i) and (iii) are female children having XX chromosomes
(ii) and (iv) are male children having XY chromosomes
b) The sex chromosomes of male determine the sex of a child. The probability of formation of male and female children is almost equal (1:1)
c) 1: 1
d) Male XY and female XX.

Question 33.
Using given indicators, construct RNA nucleotide and anyone DNA nucleotide. (March 2012)

Answer:

Unraveling Genetic Mysteries Questions & Answers

Question 1.
Find the word pair relationship and fill in the blanks appropriately. (Question Pool – 2017)
a) DNA: Thymine
RNA:……………..
b) Adenine: Thymine
Guanine:…………………
c) The character which is expressed: dominant
The character which remains hidden:……………
Answer:
a) Uracil
b) Cytosine
c) Recessive character

Question 2.
Given below are certain indicators exhibited by Anu in her slide presentation while conducting seminar on the topic ” Emergence of Genetics”. What explanations would you give for these indicators?
a) Heredity
b) Variation
c) Genetics
d) The father of Genetics
Answer:
a) Transmission of features of parent to offspring.
b) Features seen in offspring that are different from their parents.
c) The branch of science that deals with heredity and variation.
d) Gregor John Mendel

Question 3.
The note prepared by Shahana on Mendel’s inferences during the classroom analysis of Mendel’s hybridization experiment in pea plants, based on a single trait is given below. Analyze the statements in the note and correct those that are wrong ones.
a) A trait is controlled by a specific factor.
b) A character is expressed and the other remains hidden in the first generation.
c) The character that remains hidden, in the first generation does not appear in the second generation.
d) The ratio of characters in the second generation is 3: 1
Answer:
a) One trait is controlled by the combination of two factors.
b) The characters that remain hidden in the first generation appears in the second generation.

Question 4.
Fill in the blanks in the illustration given below. (Question Pool -2017)

Answer:
A) tt
B) t
C) Tt
D) dwarf

Question 5.
Complete the flowchart illustrating the location of gene by using the information given in the box: (Question Pool – 2017)
nucleus, gene, DNA, cell, chromosome

Answer:
A – Cell
B – Nucleus
C – Chromosome
D – DNA

Question 6.
Analyze the article and answer the questions. The experiments performed by Gregor John Mendel in pea plants led to the emergence of a new branch of science that has today grown and expanded to a great extent. This branch of science has untravelled several mysteries regarding the similarities and variations found in the characters of organisms. (Question Pool – 2017)
a) Which branch of science does the article refer to?
b) List out any 4 traits selected by Mendel for performing hybridization in pea plants.
Answer:
a) Genetics
b) Height, colour of the seed, colour of the flower, shape of the seed, color of the fruit, shape of the pod

Question 7.
Observe the illustration given below and answer the questions. (Question Pool-2017)

a) Identify the dominant character
b) How does the parental plant with green colored seed and the plant in the first generation differ in their alleles.
c) Describe alleles.
Answer:
a) Green
b) Alleles in parental plant – G, G Allele in the first generation – G, g
c) Different forms of a gene

Question 8.
Fill in the blanks in the illustration related to chromosomes in man. (Question Pool-2017)

Answer:
A) Autosomes
B) 2
C) XY

Question 9.
Identify the word pair relationship and fill in the blanks: (Question Pool -2017)
Female : 44 + XX
Male: …………….
Answer:
44 + XY

Question 10.
Fill in the blanks in the illustration. (Question Pool-2017)

Answer:
A. ggww
B. GW
C. Green colored round seed

Question 11.
The indicators given below are about the plant in the first generation formed as a result of the hybridization between a tall plant with red flowers and a dwarf plant with white flowers. (Question Pool-2017)

a) Identify the alleles in the plant related to the trait ’tallness’.
b) Identify the gametes formed from this plant.
Answer:
a) T, t

Question 12.
Complete the illustration of the second generation obtained from the hybridization in which two traits of a plant are considered. (Question Pool-2017)
Indicators:
Dominant character – Tallness, red color of flower Recessive character – Dwarfness, white color of flower

Answer:
A. TTRr
B. TtRr
C.TTRr
D. TtRr
E. TtRr
F. ttRR
G. ttRr
H. TtRr

Question 13.

Given below are some of the offspring obtained by self-pollinating the above plant. Analyze the offspring and answer the questions.
a) ttRr
b) ttRR
c) TTrr
d) ttrr
i) Identify the dominant characters in each of the offspring?
ii) What explanation would you give for the expression of characters in the offspring which were hidden in the parental plant?
Answer:
i. a) dwarf, red flower
b) dwarf, red flower
c) tall, white flower
d) dwarf, white flower
ii. The expression of characters in the offsprings which were hidden in the parental plant is due to the independent assortment of each character.

Question 14.
Analyze the illustration of a nucleotide molecule and answer the questions. (Question Pool-2017)

a) Identify A and B in the illustration.
b) ‘Nucleotides are found in DNA alone’. What is your opinion regarding these statements? Substantiate.
Answer:
a) A – Phosphate
B – Sugar
b) 1. does not agree
2. Like DNA, RNA is also made up of nucleotides

Question 15.
Genes which are the specific units of DNA control the metabolic activities and are also responsible for specific characters. They control the process of protein synthesis. Binu has a doubt on the above note. (Question Pool – 2017)
‘Does the RNA have no role in protein synthesis?’ What explanation would you give to Binu’s doubt? Substantiate.
Answer:

• RNA has role
• DNA is not directly involved in protein synthesis
• mRNA is formed from DNA.
• mRNA that carries information from DNA, controls the protein synthesis
• tRNA carries amino acids to ribosomes
• rRNA associated with ribosomes also have a role in protein synthesis.

Question 16.
The components and features of nucleic acid are given below. Analyze them and complete the table. (Question Pool-2017)
a) ribose sugar
b) double helical shape
c) Uracil
d) one strand
e) deoxyribose sugar
f) thymine

Answer:

Question 17.
Observe the nucleotide strands given below and answer the questions. (Question Pool-2017)

a) Identify the strand that is found in DNA only.
b) Identify the strand that can be found in both DNA and RNA.
c) What is a nucleotide?
Answer:
a) B
b) A
c) A unit of sugar, phosphate and nitrogen base / Component of nucleic acid

Question 18.
The stages in the process of protein synthesis are given below. Prepare a flowchart using the stages. (Question Pool-2017)
a) tRNA carries different kinds of amino acids to the ribosome.
b) mRNA reaches outside the nucleus.
c) mRNA forms from DNA
d) Amino acids are added based on the information in mRNA
e) mRNA reaches ribosome.
f) Proteins are synthesized.
Answer:
c → b → e → a → d → f

Question 19.
A part of the article. Variations in ourself is given below: The features seen in offspring that are different form their parents are called variations. Certain processes taking place in the initial phase of meiosis are responsible for such variations. (Question Pool-2017)
a) Which process, as mentioned in the article, is responsible for variations?
b) How does this process bring about variations?
Answer:
a) Crossing over of chromosomes
b) 1. Part of a DNA crosses over to become the part of another DNA.
2. This causes difference in the distributions of genes
3. When these chromosomes are transferred to the next generation, new characters are expressed.

Question 20.
The process of crossing over of chromosomes that takes place in the initial phase of meiosis is illustrated below. Analyse-it and answer the questions. (Question Pool – 2017)

a) Arrange the stages appropriately.
b) This process brings about variations in offspring. How?
Answer:
a) C, A, B
b) 1. Part of a DNA crosses Over to become the part of another DNA.
2. This causes difference in the distributions of genes.
3. When these chromosomes are transferred to the next generation, new characters are expressed.

Question 21.
Given below is an illustration regarding sex determination Observe the illustration and answer the questions. (Question Pool-2017)

a) Complete A, B, C in the illustration.
b) What is the possibility of the formation of a male child or a female child? Explain.
Answer:
a) A) 44 + XY
B) 22 + X
C) 22+ Y
b) Equal chance. The number of male gametes with X chromosome and those with Y chromosome are equal.
Egg with the X chromosome has equal chance to combine with sperm having Y chromosome and those having X chromosome.

Question 22.
The practice of blaming those mothers who give birth to girl children exists even today. (Question Pool – 2017)
a) As a science student, how will you respond to this situation? Substantiate.
Answer:

• No, it is wrong to blame the mother
• The possibility of the birth of male or female child is equal.
• The gender of the child is determined by the XY chromosome of the father.

Question 23.
The chromosomes from the father determine whether the child is male or female. Evaluate this statement on a scientific basis. (Question Pool-2017)
Answer:
Males have 2 types of sex chromosomes. (X, Y) Females have only one type of sex chromosome (X, X) Sex determination is based on the type of male gamete that fuses with the egg. If the male gamete with Y chromosome fuses with the egg, then male child is born, if the male gamete with X chromosome fuses with the egg, then female child is born.

Question 24.
Given below is a placard exhibited in a school rally organized ‘Against Racism’. It is not racial difference that makes the skin color different;
This is an adaptation to live under the sun.
a) How ; will you. explain the difference in skin color of people living in different parts of the world?
b) What attitude should be adopted by a scientifically enlightened society towards the idea in the placard? Substantiate.
Answer:
a) 1. Melanin, a pigment-protein imparts color to skin.
2. It is due to the difference in gene function,
b) 1. Skin color is an adaptation to live under the sun.
2. Races among making are only cultural.
3. Scientifically, all men are of the same race.
4. Consider all men as equal, without any racial difference.

Question 25.
Vipin wrote the following as situations that create variations in organisms. Choose the right ones. (Question Pool – 2017)
a) Mutation
b) Formation of mRNA
c) Crossing over of chromosomes
d) Action of rRNA
Answer:
a) Mutation
c) Crossing over of chromosomes

Question 26.
The components of nucleic acids are given below. Answer the questions through illustrations using these components: (Question Pool – 2017)

a) Illustrate the nucleotide which is found only in RNA.
b) Illustrate the nucleotide which is found only in DNA.
Answer:

Question 27.
Analyze the nitrogen bases given below and write the nitrogen base pairs found in DNA. (Question Pool-2017)

Answer:
Thymine – Adenine.
Guanine – Cytosine

Question 28.
‘Gene itself is allele; allele itself is gene. (Question Pool – 2017)
Answer:
Statement is party correct.
Each character is controlled by pair factors called genes.
Different forms of a gene are called alleles. Generally, a gene has two alleles. ,
Alleles can be of the same type (TT) or different types (Tt).
If the alleles are of different types, only one trait represented by anyone of the alleles get expressed.

Question 29.
Offsprings may vary in characters from their parents.
a) What are reasons of this variation in the light of genetics?
b) How does the changes take place during meiosis cause variations in next generation?
c) How does the chemical substances and the radiations cause variation in characters?
Answer:
a) Crossing over and mutations.
b) When a part of a particular DNA become the part of another DNA, the sequencing of nucleotides in the DNA become differs and hence variation may occur in the offsprings.
c) Mutation may occur due to chemicals and radiations.

Question 30.
Observe the illustration related to Mendel’s experiment based on two contrasting characters.
a) Complete the illustration appropriately

b) What are the characters observed in the second generation?
Answer:

b) Tall with red flowers = 9
Tall with white flower = 3
Dwarf with red flowers = 3
Dwarf with white flowers = 1

Question 31.
The different stages of protein synthesis given below. Rearrange them appropriately. (Orukkam – 2017)
a) tRNA carries different types of amino acids.
b) mRNA come out from the nucleus.
c) mRNA is formed from DNA.
d) Amino acids are joined together based on the messages in mRNA.
e) mRNA reaches in ribosome.
f) Protein is synthesized.
Answer:
c → b → e → a → d → f

Question 32.
Observe the figure and answer the questions given below. (Orukkam – 2017)

a) Name the process shown in the figure.
b) Write the importance of this process.
Answer:
a) Crossing over
b) Causes variation

Question 33.
Write the role of mRNA and tRNA in protein synthesis. (Orukkam – 2017)
Answer:
mRNA carries messages for protein synthesis from DNA to the ribosomes.
tRNA carries amino acids to ribosomes according to the message in the mRNA.

Question 34.
What is mutation? Write the reasons? (Orukkam – 2017)
Answer:
Mutation is a sudden heritable change occur in the genetic material (chromosomes)
Reasons: – Radiations, chemicals, changes in the replication of DNA.

You can Download The World of Carbon Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Chemistry Solutions Part 2 Chapter 7 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Chemistry Solutions Chapter 7 The World of Carbon

The World of Carbon Textual Questions and Answers

Activities In The Text

Kerala Syllabus 9th Standard Chemistry Notes Question 1.
Find the position of carbon in the periodic table and the complete the table.

Answer:

Hss Live Guru 9th Chemistry Kerala Syllabus Question 2.
What are allotropes?
Answer:
Different forms of the same element having different physical properties but with same chemical properties are known as Allotropes and this phenomenon is called Allotropy.

9th Class Chemistry Chapter 7 Notes Kerala Syllabus Question 3.
What are the characteristics of diamond.
Answer:

• Very hard
• Transparent
• Not a conductor of electricity
• High thermal conductivity
• High refractive index

9th Chemistry Notes Kerala Syllabus  Question 4.
Write the uses of diamond
Answer:

• It is used to make ornaments.
• It is used for cutting glass.

9th Standard Chemistry Kerala Syllabus Question 5.
Explain the structure of diamond with Figure?
Answer:
In diamond each carbon atom is linked by covalent bond with four other carbon atom surrounding it. This strong bonding is responsible for the hardness of diamond. Due to the absence of free electron in this crystal structure, it does not conduct electricity

Chemistry 9th Class Notes Kerala Syllabus Question 6.
Write the important properties of Graphite?
Answer:

• It is Gray in color
• It is good conductor of electricity
• It is a smooth solid
• It does not vapourize
• It is Lustrous
• It is nonvolatile

Hss Live Guru Chemistry 9th Kerala Syllabus Question 7.
Give reason Graphite is used as a lubricant to reduce friction in machine parts?
Answer:
It is a smooth solid. It does not vapourize. It also has high melting point

Hsslive Guru Chemistry Class 9 Kerala Syllabus  Question 8.
Explain the structure of Graphite?
Answer:

In Graphite each carbon atoms is united with three surrounding carbon atom through covalent bond and forms a sheet-like structure. These sheets or layers are stacked one above the other to form three dimensional structure.

Each layer is made up of hexagons there is no covalent bonding between the layers. These layers are held together by weak Vander Waal’s physical forces. Hence these layers can slide over one another.

Hsslive Guru 9 Chemistry Kerala Syllabus Question 9.
What are amorphous carbon?
Answer:
Cock, coal, charcoal, bone charcoal, etc. are non-crystalline allotropes of carbon. These are commonly called amorphous carbon.

Kerala Syllabus 9th Chemistry Notes Question 10.
Which carbon compound is present in the atmosphere?
Answer:
Carbon dioxide

Hss Live 9th Chemistry Kerala Syllabus Question 11.
Which carbon compound is produced by the combustion of fuels?
Answer:
Carbon dioxide

9th Class Chemistry Notes Kerala Syllabus Question 12.
How carbon dioxide is prepared of in the laboratory?
Answer:
Carbon dioxide is prepared in the laboratory by the action of marble piece or calcium carbonate with dil.
HCl CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Hsslive Guru Std 9 Chemistry Kerala Syllabus Question 13.
Which are the reactants are used to prepare carbon dioxide (CO₂) in the laboratory?
Answer:
CaCO₃ and HCl

Hsslive Guru 9th Chemistry Kerala Syllabus Question 14.
Complete the equation of the reaction
Answer:
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Chemistry Solutions Class 9 Kerala Syllabus Question 15.
How can we identify that the gas formed here is C02?
Answer:

1. Show a burning splinter into the gas. It gets extinguished.
2. lt turns lime water in milky

Hsslive Guru Chemistry 9 Kerala Syllabus Question 16.
Which properties of CO₂ are familiar to you?
Answer:
Extinguishes fire

Hsslive 9th Chemistry Kerala Syllabus Question 17.
What are the properties of carbon dioxide known to you? Tick the correct ones given below.
Answer:

• Coloured / Colourless✓
• Supports combustion / Does not support combustion✓
• Denser than air / Density higher than air✓
• Has a characteristic odor/ Odourless ✓

Kerala Syllabus 9th Standard Chemistry Question 18.
Whether the aqueous solution of CO₂ is acidic or alkaline?
Answer:
Acidic (carbonic acid, H₂CO₃)

Hsslive Class 9 Chemistry Kerala Syllabus Question 19.
Write the chemical formulae and uses of some carbonates.
Answer:
Na₂CO₃ – Soda ash
CaCO₂ – limestone, marble.

9th Std Chemistry Notes Kerala Syllabus Question 20.
How carbonates can be identified.
Answer:
Add some dilute HC; to the given salt. If a colorless gas that turns lime-water milky is formed, that salt will be a carbonate. The gas formed is CO₂.

Question 21.
The variety of carbon compounds is essential for the existence of life on the earth. Figure shows the ways in which carbon dioxide is exchanged over the earth. This is known as carbon cycle.

Name the process by which carbon dioxide is utilized by plants?
Answer:
Photosynthesis

Question 22.
What are the activities that increase the amount of carbon dioxide in air?
Answer:
Combustion, Respiration, Decomposition

Question 23.
Is the tremendous increase in the amount of CO₂ in atmosphere advantages?
Answer:
No, It causes greenhouse effect. Greenhouse effect is the reason for global warming.

Question 24.
What is green house effect?
Answer:
The process of increasing atmospheric temperature due to the increase in the amount of carbon dioxide in the atmosphere is called green house effect.

Question 25.
What is global warming?
Answer:
As a result of the green house effect, the average temperature of the earth and atmosphere increases. This is known as global warming.

Question 26.
Discuss the consequences of global warming in the following.
1. In ice layers
2. In ocean islands
3. In the field of agriculture
4. In the climate
Answer:

1. Ice layers melt and rivers will be flooded
2. The ocean islands will be submerged.
3. Fields of agriculture will be submerged
4. Causes rise in atmospheric temperature.

Question 27.
Suggest some measures to resist global warming effectively.
Answer:

1. Limit the use of fossil fuels
2. Plant more trees

Question 28.
Give the uses of carbon dioxide?
Answer:

• It is used in fire extinguisher
• It is used for the preparation of soda water/soft drink
• It is used for the preparation of washing soda, baking soda, etc.
• It is used for the preparation of fertilizers like urea
• Used for the preparation of carbogen which is used for artificial breathing. Carbogen is 95% 02 and 5% C02
• Dry ice is solid carbon dioxide, which is used in stage shows for creating special effects resembling clouds.

Question 29.
How carbon monoxide is formed?
Answer:
Carbon dioxide is the gas formed when carbon reacts with oxygen.
However, if the relative amount of carbon increases or that of oxygen decreases the reaction takes place as given below.
2C + O₂ → 2CO

• The gas thus formed is carbon monoxide. It is a poisonous gas.
• It is formed by the incomplete combustion of carbon in a limited supply of air.

Question 30.
How carbon monoxide becomes fatal?
Answer:
When carbon monoxide is inhaled, it reacts with the hemoglobin in the blood and forms carboxy hemoglobin. As a result, the oxygen-carrying capacity of blood decreases leading even to death.

Question 31.
What measures can be taken to avoid situations that produce carbon monoxide?
Answer:

• Avoid incomplete combustion
• Ensure the supply of oxygen.
• Service motor vehicles regularly

Question 32.
Explain the use of carbon monoxide?
Answer:

• Used as a gaseous fuel
• Industrially important gases like water gas (A mixture of CO and H₂) producer gas (A mixture of CO and N₂)
• Used as reducing agent in metallurgy

Question 33.
Write the chemical formulae of washing soda baking soda and marble.
Answer;
Washing Soda (Na₂CO₃.10H₂O)
Baking soda (NaHCO₃)
Marble (CaCO₃)

Question 34.
What is organic compounds?
Answer:
Organic compounds are carbon compounds except the inorganic compounds like CO,CO₂, carbonates, bicarbonates etc.

Question 35.
How many electrons are there in the outermost shell of carbon?
Answer:
Four

Question 36.
What is the valency of carbon?
Answer:
4

Question 37.
Complete the table given below.

Answer:

Question 38.
What are hydrocarbon?
Answer:
Hydrocarbons are compounds containing only car¬bon and hydrogen.

Question 39.
What is catenation?
Answer:
Catenation is the ability of the atoms of an element to combine among themselves. In comparison to other elements, the ability for catenation is very high for carbon.

Question 40.
What are characteristics responsible for the increase in number of carbon compounds?
Answer:

• Valency of carbon is 4
• Ability of catenation is high
• Single, double and triple bonds are possible be-tween carbon atoms.
• Carbon atoms combine together to form many straight-chain, ring or branched-chain compounds.

Let’S Assess

Question 1.
The names of some allotropes of carbon, their properties and uses are given in the table, but not in the correct order Match them suitably.

Answer:
Diamond:

• Manufacture of ornaments
• Transparent
• High refractive index

Graphite:

• Electric conductor
• Smooth
• Lubricant

Question 2.
Some statements related to carbon monoxide and carbon dioxide are given. Classify them correctly.
a) formed as a result of the incomplete combustion of carbon compounds’
b) aqueous solution shows acidic nature.
c) poisonous gas
d) used in fire extinguishers
e) an be used as a fuel
f) formed as a result of the complete combustion of carbon compounds.
g) can be prepared from carbonates and bicarbonates.
h) is a component of producer gas and water gas.
Answer:
a) carbon monoxide
b) Carbon dioxide
c) Carbon monoxide
d) Carbon dioxide
e) Carbon monoxide
f) Carbon dioxide
g) Carbon dioxide
h) Carbon monoxide

Question 3.
a) Write the chemical formula of calcium carbonate.
b) Which gas is formed when calcium carbonate reacts with acids?
c) What is the name of an aqueous solution of this gas?
Answer:
a) CaCO₃
b) Carbon dioxide
c) Soda water (carbonic acid)

Question 4.
Graphite, which is an allotrope of carbon, is a con-ductor of electricity. But diamond, another allotrope is not a conductor of electricity. Why?
Answer:
diamond each carbon atom is linked by covalent bond with four other carbon atom surrounding it. This strong bonding is responsible for the hardness of diamonds. Due to the absence of free electron in this crystal structure, it does not conduct electricity

Question 5.
Write the structure of a straight-chain and a a ring hydrocarbon having four carbon atoms.
Answer:

Extended Activities

Question 1.
Arrange the objects as shown in the figure and conduct the experiment. Based on your observations, what is the conclusion that you reach at?

Answer:
The bulb glows because pencil led or graphite is a good conductor of Electricity.

Question 2.
Lighted candles of different lengths are arranged in a trough as shown in the figure. Pour a saturated solution of sodium bicarbonate (baking soda) into the trough. Add a little vinegar to the solution. What do you observe? Give reasons for the observation.

Answer:
Candle with smallest height is extinguished first and the candle with highest height is extinguished last. The reason is the density of carbon dioxide is greater than that of air. Carbon dioxide is produced when baking soda react with vinegar.

Question 3.
Let’s make a fire extinguisher Arrange the apparatus as shown in the figure (a). Add the vinegar contained in a test tube to the sodium bicarbonate (baking soda) solution (figure b) by tilting the wash bottle. Introduce the resultant gas to a candle flame. Record your observation. What is your inference?

Answer:
The candle flame gets extinguished. The reason is when Baking soda reacts with vinegar to produce carbon dioxide.

You can Download The Biology of Movement Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Biology Solutions Part 2 Chapter 6 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Biology Solutions Chapter 6 The Biology of Movement

The Biology of Movement Textual Questions and Answers

The Biology Of Movement 9th Chapter 6 Question 1.
Are exercise and games necessary?
Answer:
On physical strength increases as we involve in interesting exercises such as games. Exercise reduces mental stress and helps us to work energetically.

9th Class Biology Notes Kerala Syllabus Chapter 6 Question 2.
Prepare a note on how exercise is beneficial to the body?
Biology Answer:
Exercise helps us in many ways.
It increases blood circulation all over the body. Cardiac muscles become strong. More capillaries are formed in muscles. Increases the efficiency of muscles. Stored fat is broken down thereby reduces obesity. Sweats more and so more waste is eliminated through sweat. Exchange of respiratory gases becomes more effective. Vital capacity increases.

Kerala Syllabus 9th Standard Biology Notes Chapter 6 Question 3.
‘Exercise helps our respiratory system more healthy.’ Do you agree with this statement? Substantiate your answer?
Answer:
Exercise increases our vital capacity and exchange of respiratory gases becomes more effective,

Involuntary Movements

Kerala Syllabus 9th Class Biology Notes Chapter 6 Question 4.
Prepare a table about voluntary movements and involuntary movements?

Answer:

Hss Live Guru 9th Biology Chapter 6 Question 5.
What do you mean by voluntary movements?
Answer:
The movements which occur according to our will is called voluntary movements.

9th Biology Notes Kerala Syllabus Chapter 6 Question 6.
Define involuntary movements?
Answer:
The movements which are not controlled by our will is called involuntary movements.

Types Of Muscles

9th Class Biology Chapter 6 Notes  Question 7.
Which muscle make voluntary movements possible?
Answer:
Skeletal muscle

Hss Live Guru Biology 9 Chapter 6 Question 8.
Striated muscle have cells
Answer:
Cylindrical

9th Standard Biology Notes Chapter 6 Question 9.
Where do you find smooth muscles in human body?
Answer:
Smooth muscles are seen in internal organs like the stomach, small intestine and in blood vessels.

Class 9 Biology Notes Kerala Syllabus Chapter 6 Question 10.
Smooth muscles are also known as
Answer:
Nonstriated muscles

Kerala Syllabus 9th Standard Biology Notes Pdf Chapter 6 Question 11.
Shape of smooth muscle is
Answer:
Spindle

Biology Notes For Class 9 Kerala Syllabus Chapter 6 Question 12.
Cardiac muscles are seen on the
Answer:
Walls of the heart

Kerala Syllabus 9th Standard Biology Solutions Chapter 6 Question 13.
…….. & ……… makes involuntary movements possible.
Answer:
Smooth muscle and cardiac muscle

Kerala Syllabus 9th Standard Biology Notes Malayalam Medium Question 14.
Skeletal muscle: Cylindrical shape
…………………..: Spindle shape
Answer:
Smooth muscle

Kerala Syllabus 9th Std Biology Solutions Chapter 6 Question 15.
Skeletal muscle: striated muscle
………………….: nonstriated muscle
Answer:
Smooth muscle

9th Standard Biology Question 16.
Different types of muscles and their characteristics. Prepare a table.
Answer:

Muscles Fatigue

Question 17.
What do you mean by muscle fatigue?
Answer:
When we are engaged in continuous and strenuous exercises, lactic acid accumulates in the muscles due to anaerobic respiration. This increases acidity in muscles and slows down the action of many enzymes associated with muscle contraction. As a result, muscles get exhausted and temporarily lose their power of contraction. This condition is called muscle fatigue.

Bones and Movement

Question 18.
The human skeleton system consists of bones.
Answer:
206

Question 19.
Based on the position, the human skeleton can be divided into ………. & …………
Answer:
Axial skeleton and appendicular skeleton.

Question 20.
Number of bones in the human skull is
Answer:
29

Question 21.
How many bones are there in human ribs?
Answer:
12 × 2 = 24

Question 22.
Hind limbs: 60 bones
………………: 26 bones
Answer:
Vertebral column

Question 23.
There are bones in the pelvic girdle of human beings.
Answer:
1 × 2 = 2

Question 24.
Muscles which contracts on folding the forelimb?
Answer:
Flexor muscle

Question 25.
Complete the illustration given below

Answer:

Question 26.
Muscle which contracts on extending the forelimb?
Answer:
Extensor muscle

Question 27.
Muscle which relaxes on folding the forelimb?
Answer:
Extensor muscle

Question 28.
Muscle which relaxes on extending the forelimb?
Answer:
Flexor muscle

Question 29.
What is antagonistic muscle?
Answer;
A movement is effective and complete when muscles work in unison with bones. In forelimb, when one muscle contracts the other muscle relaxes. These types of muscles which are opposite in action are called antagonistic muscles.

Question 30.
The basis of almost all the movements of the body is the proper functioning’ of ………….
Answer:
Antagonistic muscles

Joints And Movements

Question 31.
Complete the table of skeletal joints. Which shows its position and peculiarities.

Answer:

Structure Of Joint

Question 32.
…………. are the meeting place of two bones
Answer:
Joints

Question 33.
Explain the function of joints
Answer:
joints help in the movement of bones. Joints give more flexibility to bones to move. The nature of movements varies with the nature of joints.

Question 34.
…………. secretes synovial fluid
Answer:
Synovial membrane

Question 35.
…………. covers and protects the joints
Answer:
Capsule

Question 36.
…………. reduces friction between the bones
Answer:
Cartilage

Question 37.
What is the function of synovial fluid?
Answer:
Synovial fluid functions as a lubricant between the bones.

Question 38.
What is the function of ligaments?
Answer:
Ligaments ensure that bones are not displaced and holds them in position.

Question 39.
What are the functions of the skeletal system?
Answer:
Skeletal system facilitating movements, maintains posture, helps in hearing, protects our internal organs from damage, produces blood cells and maintains the mineral homeostasis.

Skeletal And Muscular Disorders

Rheumatic Arthritis:

• Caused by infection in joints, injuries, degenerative changes due to old age.
• Damage to cartilage
• Severe pain, incapable of moving joints

Dislocation:

• Displacement of bones in joints
• Damage to ligaments
• Severe pain oedema and difficulty in movements

Sprain:

• The stretching or breaking of ligaments
• Severe pain and oedema

Osteoporosis:

• A condition in which bones become brittle and cause fracture
• This may be due to the deficiency of calcium, defects in metabolic activities and deficiency of Vitamin D

Muscular dystrophy:

• A condition that leads to degeneration of muscles due to various reasons.
• Muscles become weak
• Generally, affect boys.

Skeleton Outside the Muscles

Locomotion Without Skeleton

Question 40.
Different types of movement in organisms which move without skeleton.
Answer:

Locomotion And Movement

Movement is the displacement occurring in any part of the body. Displacement of the entire body is called locomotion

The diversity of locomotion in animal world:

Do plants move?

Question 41.
Plants exhibit movements in response to various
Answer:
Stimuli

Question 42.
What are the various stimuli which cause movements in plants?
Answer:
Light, gravity, water, touch, chemicals, etc. are the various stimuli which cause movements in plants.

Question 43.
Complete the table relating to the plant movement

Answer:

Question 44.
Identify the type of movement, roots grow towards water
Answer:
Hydrotropism

Question 45.
What do you mean by tropic movement?
Answer:
If the direction of plant movements is in accordance with the direction of stimulus it is called tropic movements.

Question 46.
Is there any relation between stimulus and the direction of movement in mimosa?
Answer:
No. Hence it is nastic movement.

Question 47.
What do you mean by nastic movement?
Answer:
If the direction of plant movement is not in accordance with the stimulus, it is called nastic movement.

Question 48.
Write some examples for nasty plant movements from your surroundings.
Answer:
Movements of Mimosa pudica, prayer plant, venus flytrap, etc.

Let Us Assess

Question 1.
What is the reason for muscle fatigue?
a) Lack of glucose in muscle cells
b) Lack of oxygen in muscle cells
c) Increase in the level of carbon dioxide in muscle cells
d) cellular respiration ceases
Answer:
b) Lack of oxygen in muscle cells

Question 2.
Observe the figure and answer the following questions. What changes do you observe in the growth of root and stem in a plant, if it is kept stationary as shown in the figure for a few days? Why?

Answer:
Roots grow towards gravity and the stem grows against gravity.

Question 3.
Identify the odd one giving reason
a) Coconut trees near a river bend towards the river
b) Root of trees near a well grows towards the well
c) Leaves of touch-me-not fold when we touch it
d) Roots of plants grows towards gravity
Answer:
Leaves of touch-me-not fold when we touch it because it is a nastic movement.

You can Download Prisms Questions and Answers, Activity, Notes, Kerala Syllabus 9th Standard Maths Solutions Chapter 11 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Maths Solutions Chapter 11 Prisms

Prisms Textual Questions and Answers

Textbook Page No. 171

Kerala 9th Maths Solutions Chapter 11 Question 1.
The base of a prism is an equilateral triangle of perimeter 15 centimetres and its height is 5 centimetres. Calculate its volume.
Answer:
Base perimeter of an equilateral triangle = 15 cm.
Base edge = 15/3 = 5 cm
Height = 5 cm
Volume = Base area x Height

Kerala Syllabus 9th Standard Maths Notes Chapter 11 Question 2.
A hexagonal hole of each side 2 metres is dug in the school ground to collect rainwater. It is 3 metres deep. It now has water one metre deep. How much litres of water is in it?
Answer:
One side of the hexagon = 2 m
It is given that the depth of the pit is 3 metre but the water level is only in 1 metre. So we take the height as 1 metre.
Volume of water = Volume of the hexagonal prism = Area of the hexagon × height

Area of the regular hexagon is equal to six times the area of equilateral triangle.

Kerala Syllabus 9th Standard Maths Solutions Chapter 11 Question 3.
A hollow prism of base a square of side 16 centimetres contains water 10 centimetres high. If a solid cube of side 8 centimetres is immersed in it, by how much would the water level rise?
Answer:
When a solid cube is immersed into water, the volume of water is raised.
Sum of the volume of the water at first time and volume of the solid cube immersed equally to the product of base area and height of water level now.
Volume of the water at first = Base area × height
= 16 × 16 × 10 = 2560 cm3
Volume of the solid cube when edges are 8 cm
= 8 × 8 × 8 = 512 cm3
Height at first = 10 cm
Water level raised = 12 -10 = 2 cm

Textbook Page No. 174

Class 9 Maths Solution Kerala Syllabus Chapter 11 Question 1.
The base of a prism is an equilateral triangle of perimeter 12 centimetres and its height is 5 centimetres. What is its total surface area?
Answer:
Base perimeter of equilateral triangular prism = 12 cm
Base side = 12/3 = 4 cm
Lateral surface area of equilateral trian-gular prism = Base perimeter × Height
=12 × 5 = 60 cm²
Base area of equilateral triangular prism = \(\frac { √3 }{ 4 }\) × 4² = 4√3 = 6.92 cm²
Total surface area of equilateral triangular prism =60 + 2 × 6.92 = 73.84 cm²

Kerala Syllabus 9th Standard Notes Maths Chapter 11 Question 2.
Two identical prisms with right tri-angles as base are joined to form a rectangular prism as shown below:

What is total surface area?
Answer:
The sides of the rectangular prism are
Length = 12 cm
Breadth = 5 cm and
Height = 15 cm.
Base area = 2 × perimeter of rectangle
2 × 12 × 5 = 120 cm²
Lateral surface area = base perimeter × height
= 2 (length + breadth) × height = 2(12 + 5) × 15
= 2 × 17 × 15 = 510 cm²
Total surface area = 120 + 510 = 630 cm²

Class 9 Maths Notes Kerala Syllabus Chapter 11 Question 3.
A water trough in the shape of a prism has trapezoidal faces. The dimensions of a base are shown in this picture:

The length of the trough is 80 centimetres. It is to be painted inside and outside. How much would be the cost at 100 rupees per square metre?
Answer:
Now we add the area of two edge faces and three lateral faces ( 1 on bottom and 2 on sides).
In figure, AB = 50 cm, CD = 70 cm

EC = \(\frac { 70 – 50 }{ 2 }\) = 10 cm
BE = 24 cm
BC² = BE² + EC2 = 24² + 10²
= 576 + 100 = 676
BC = √676 = 26 cm.
Area of the trapezoidal faces of the water

= 12 × 120 = 1440 cm² Area of two trapezoidal faces
= 2 × 1440 = 2880 cm² Area of two rectangular faces on sides
= 2 × 80 × 26=4160 cm²
Area of rectangular face at bottom
=50 × 80 = 4000 cm²
Total Area= 2880 + 4160 + 4000 = 11040 cm² = 1.1 m²
Total area to be painted
= 2 × 1.1 =2.2 m²
Total cost = 2.2 × 100 = 220
= Rs. 220

Textbook Page No. 176

9th Standard Maths Notes Kerala Syllabus Chapter 11 Question 1.
The base radius of an iron cylinder is 15 centimetres and its height is 32 centimetres. It is melted and re-cast into a cylinder of base radius 20 centimetres. What is the height of this cylinder?
Answer:
Volume of the first cylinder
= π R2 H = π (15)² x 32 = 7200 π
Volume of the melted and recast cylinder = π r2h = π (20)² x h = 400 π h
Volume remains constant when melted.
400 π h=7200π
h = \(\frac { 7200π }{ 400π }\) =18
Height of the second cylinder = 18 cm

9th Class Maths Notes Malayalam Medium Chapter 11 Question 2.
The base radii of two cylinders of the same height are in the ratio 3:4. What is the ratio of their volumes?
Answer:
Let r₁ r₂ be the radii of two cylinders,
then r₁ : r₂ = 3 : 4,

Kerala Syllabus 9 Standard Maths Chapter 11 Question 3.
The base radii of two cylinders are in the ratio 2:3 and their heights are in the ratio 5:4.
i. What is the ratio of their volu-mes?
ii. The volume of the first cylinder is 720 cubic centimetres. What is the volume of the second?
Answer:
i. If the base radius of the first cylinder is 2r then base radius of the second one is 3r.
Height of the first cylinder is 5h and second cylinder is 4h.
Volume of the first cylinder
= Base area × height .
= π × 2r × 2r × 5h = 20 π r² h.
Volume of the second cylinder
= π × 3r × 3r × 4h=36 πr²h.
The ratio between the volumes
= 20πr²h : 36πr²h = 20 : 36.= 5 : 9

ii. The volume of the first cylinder is 720 cm3 is given. Let consider the volume of the second cylinder be x, then the ratio between the volumes is 5: 9
5 : 9 = 720 : x
5 × x = 9 × 720
x = 1296
Volume of the second cylinder = 1296 cm³

Textbook Page No. 178

Question 1.
The inner diameter of a well is 2.5 metres and it is 8 metres deep. What would be the cost of cementing its inside at 350 rupees per square metre?
Answer:
Area of the part cemented = Base perimeter × height

= 20π = 20 × 3.14 = 62.8cm
Total cost of cementing = 62.8 × 350 = Rs. 21980

Question 2.
The diameter of a road roller is 80 centimetres and it is 1.20 metres long:

What is the area of levelled surface, when it rolls once?
Answer:
Radius of roller = 40 cm
Length of roller (height) = 1.20 m = 120 cm
Area of leveled surface, when it rolls once
= Curved surface area of the roller
= 2 × π × Radius × Height
= 2 × 3.14 × 40 × 120 = 30144 cm² = 3.0144 m²

Question 3.
The base area and the curved surface area of a cylinder are equal. What is the ratio of the base radius and height?
Answer:
Curved surface area of the cylinder = Base perimeter x height = 2 πrh
Base perimeter = πr²
If it is equal, 2 π rh = πr²
2rh = r × r, 2h = r
i.e., The radius is twice the height.

Prisms Exam oriented Questions and Answers

Question 1.
A cylinder has base radius 4 cm and height 10 cm. Then find its lateral surface area.
Answer:
Perimeter of a circle with radius 4 cm = 2 × π × 4 = 8π cm.
Curved surface area of the cylinder = Base perimeter × height = 8 π × 10= 80 π cm²

Question 2.
The volume and base area of a square prism are 3600 cubic centimetre and 144 cm² respectively. What is its total surface area ?
Answer:
Volume of the prism = 3600 cm³
Base area × height = 3600 cm³
144 × height = 3600
height = 3600/144 = 25 cm
Surface area = = Base area + Lateral surface area
Lateral surface area = Base perimeter × height
= 12 × 4 × 25 = 1200 cm²
(Base area = 144, One side = √144 =12 cm)
Total surface area = 144 + 1200 = 1344 cm²

Question 3.
A cylinder has height 20 cm and base radius 4 cm. Then find its vol-ume.
Answer:
When the square of the radius is multiplied by π we get the area of the circle.
Base area of the cylinder
= π × 4² = 16 π cm²
The height of the cylinder is 20cm Volume = 16 π × 20 = 320 π cm³

Question 4.
The base edge of a square prism is 15 cm. The total surface area is 1950 m².
i. What is its height ?
ii. Calculate the volume.
Answer:
i .Base edge = 15 cm
Total surface area = Base area + Lateral surface area
= 2a² + 4ah (base egde is ‘a’)
1950 = 2 × 15² + 4 × 15 × h
1950 = 450 + 60h
60h = 1950 – 450 = 1500
h = 1500/60 = 25 cm

ii. Volume = a² × h = 15 × 15 × 25
= 5625 cm³

Question 5.
A square-shaped plot has length 24 m. A pond of 4 m length, 3 m breadth and 1.5 m height is dug here. If the sand dug out is levelled equally in the remaining area of the plot. Find the height of the levelled sand.
Answer:
Area of the land
= 32 × 24
= 768 m²

Area of the pond = 4 × 3 = 12m²
Area of the remaining places = 768 – 12 = 756 m².
Volume of the sand dug out = 4 × 3 × 1.5
Height of soil = \(\frac { Volume }{ Base area }\)
= \(\frac { 4 × 3 × 1.5 }{ 756 }\) = 0.0238m = 2.38 cm

Question 6.
Two Aluminium sheets of length 10 cm and breadth 6 cm are folded to make two cylindrical vessels. One is made by folding lengthwise and the other breadthwise, which will have maximum volume?
Answer:

Question 7.
A cylinder made of metal has radius 18 cm and height 40 cm. When this melts how many cylinders can be made of radius 2 cm and height 5 cm?
Answer:
Volume of the first c ylinder = π r²H = π (18)² × 40 = 12960 π cm³
Volume of the newly made cylinder = π (2)² × 5 = 20π cm³
Number of the newfy made cylinders
= \(\frac { 12360π }{ 20π }\) = 648 cylindres

Question 8.
If a wooden piec e is in the shape of square prism has base 12 cm and height 70 cm. What is the maximum volume of the cylinder that can be carved out of it?
Answer:
Diameter of the largest cylinder =12cm Radius = 6 cm, Height = 70 cm
Volume of the cylinder
= base area × height
= πr²h = π × 6 × 6 × 70 = 7912.8 cm³

Question 9.
A tin with length 40 cm width 20 cm and height 20 cm, which is in the shape of a quadrangular prism has sugar-filled in it. If the sugar is measured using a cylindrical vessel with radius 4 cm and height 15 cm, then how many times can the sugar be measured using the vessel?
Answer:
Volume of the quadrangular prism = 50 × 40 × 20
Volume of the cylinder = πr²h.
= 3.14 × 4 × 4 × 15
Number of times the sugar can be measured = \(\frac { 50 × 40 × 20 }{ 3.14 × 4 × 4 × 15 }\) = 53 times

Question 10.
A prism is made by cutting cardboard as shown in the figure.

a. What is the name of the prism?
b. What will be the area of the required cardboard for making the rectangle form?
Answer:
a.Triangular prism
b. Lateral area = base perimeter × height = (15 + 13 + 14) × 20 = 42 × 20
= 840 cm²

Question 11.
Two tins in the shape of cylinder has radii 15 cm and 10 cm respectively. The heights are 25 cm and 18 cm respectively.In the both cylinders, the ghee are filled. When it transferred into another cylindrical-shaped tin. If there is ghee in the bigger tin with height 30 cm. Then find its radius.
Answer:
‘Volume of the ghee in the first tin
= πr²h = π × 15 × 15 × 25 = 5625π cm³
Volume of the ghee in the second tin
= πr²h = π × 10 × 10 × 18 = 1800 π cm³
Total volume = 5625 π + 1800 π
=7425 π cm3 Volume of the ghee in the bigger tin
= πr²h = 7425 π
h =30
πr²h = πr² × 30 = 7425 π
r² = \(\frac { 7425 π }{ 30π }\) = 247.5
\(r=\sqrt{247.5}=15.73 \mathrm{cm}\)

Question 12.
The diameter of a water tank in the shape of a cylinder is 3 m and height 4 m. How many litres of water will the tank hold?
Answer:
Radius = 1.5 m, Height = 4m
Volume of the cylinder
= Base area × height = πr²h = 3.14 × 1.5 × 1.5 × 4 = 28.26 m³
= 28.26 × 100 × 100 × 100 cm³
= 28260000 cm³ = 28260000/1000 = 28260 liter

Question 13.
If a box in the shape of a square prism has length 25 cm, 20 cm width and 7-litre volume. What will be its height?
Answer:
Volume = 7 litre = 7000 cm³
lbh = 7000
25 × 20 × h = 7000
h = \(\frac { 7000 }{ 25 × 20 }\) = 14 cm

Question 14.
If the base length, width, height of a quadrangular prism are 37.5 cm, 18 cm, 40 cm respectively. Find the area of cube which has same volume as that of this prism.
Answer:
Volume of quadrangular prism = base area × height = 37.5 × 18× 40 = 27000 cm³
Volume of quadrangular prism = Volume of cube
∴ Volume of cube = 27000 cm³
Let one side of a cube be x, then a³ = 27000 ; a = 30cm
30 × 30 × 30 = 27000, Hence
Surface area of the cube = 6a² = 6 × 30 × 30 = 5400 cm²

Question 15.
The diameters of two-cylinder are in the ratio 2 : 3 and their heights in the ratio 5: 4. If the volume of the first cylinder is 400 cm³, then find the volume of the second cylinder.
Answer:
The diameters of two-cylinder are in the ratio 2 : 3 so the radius of the cylinders are also 2:3.
Assume that radius of first cylinder is 2r and second cylinder be 3r.
Assume that height of first cylinder is 5h and second cylider be 4h.
Volume of the first cylinder = πr2h
= π × 2r × 2r × 5h = 20 πr²
Volume of the second cylinder = πr2h
= π × 3r × 3r × 4h=36 πr²h
Ratio between volumes
= 20πr²h : 36πr²h = 20 : 36 = 5 : 9
Volume of the first cylinder is 400.
If the volume of second cylinder be x, then
5 : 9 = 400: x; 5x = 400 × 9
x = \(\frac { 400 × 9 }{ 5 }\) = 80 × 9 = 720 cm³

You can Download Diversity for Sustenance Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Basic Science Solutions Chapter 13 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 13 Diversity for Sustenance

Diversity for Sustenance Textbook Questions and Answers

Diversity For Sustenance Kerala Syllabus 8th Chapter 13 Biosphere

Biosphere is the part of earth where life exists. Living world contains plants, animals, microorganisms, etc. Abiotic factors are also essential for the existence of living world. Sun is the ultimate source of energy in living world. Green plants convert light energy to chemical energy by photosynthesis.

Illustration (Text Book Page No:182)

Diversity For Sustenance Class 8 Kerala Syllabus Chapter 13 Question 1.
Discuss and complete the illustration given below suitably.

Diversity For Sustenance Pdf Kerala Syllabus 8th Chapter 13 Ecology

Ecology is the study of interrelationship of organisms among themselves and with their environment.

Basic Science Class 8 Chapter 13 Kerala Syllabus Chapter 13 Producers

Plants that perform photosynthesis are the producers.

Hss Live Guru Biology 8 Kerala Syllabus Chapter 13 Consumers

Organisms that directly or indirectly depend on green plants for energy are called consumers. Animals that directly depend plants are called primary consumers. Those dependent on primary consumers are the secondary consumers. The organisms depend on secondary consumers are called tertiary consumers.

Indicators (Text Book Page No: 183)

Hss Live Guru 8th Biology Kerala Syllabus Chapter 13 Question 1.
How do food chain and food web differ from each other?
Answer:
The chain of animals that eat and being eaten constitute food chain. But in nature different food chains are interrelated and this network is called food web.
Eg. Food Chain

8th Class Biology Notes Pdf Kerala Syllabus Chapter 13 Question 2.
Is a single organism involved in more than one food chain?
Answer:

• Same organism belongs to different food chains.
• Beneficial. No species increase or decrease beyond a level.

8th Class Physics Notes Kerala Syllabus Chapter 13 Question 3.
Is the possibility of an organism becoming food to more Is the possibility of an organism becoming food to more than one organism helpful to the existence of the food chain?
Why?
Answer:
If a particular species increase its number, the animals that forms its food get destroyed. It cause food scarcity and thus they themselves destroyed.

Diversity For Sustenance Notes Kerala Syllabus 8th Chapter 13 Question 4.
How does the variation in the number of a particular organism in the food chain affect the existence of other organisms?
Answer:
The number of an organisms decrease it adversely effects the existences of another group that depend them for their food. The increase and decrease in the number of organisms adversely affect the equilibrium of environments.

Hss Live Guru 8 Biology Kerala Syllabus Chapter 13 Trophic Level

Trophic level indicates the position of an organism in a food chain. Green plants belong to first trophic level. All food chains start from green plants. Herbivores are in II trophic level and Carnivores are included in III trophic level.

The Illustration (Text Book Page No: 184)

Basic Science For Class 8 Chapter 13 Question 5.
Did you read the note on trophic level?
Complete the illustration by including the organisms of the food web at various trophic levels
Answer:
Tertiary Consumers – Eagle, Mongoose
Secondary Consumers – Frog, Snake
Primary Consumers – Grasshopper, Rat
Producers – Grass, Paddy

Hsslive Guru Biology 8th Kerala Syllabus Chapter 13 Question 6.
Does the same organism occupy more than one trophic level?
Answer:
The same organism is included in different trophic levels as the complexity of food web increases.

Kerala Syllabus 8th Standard Chemistry Notes Chapter 13 Question 7.
Is there any possibility of a fifth trophic level?
Answer:
The number of trophic levels is an ecosystem is not constant. Even though in nature the food chains are not too long. This is to reduce the loss of energy during transmission.

Kerala Syllabus 8th Standard Physics Notes Chapter 13 Question 8.
How does the elimination of organisms from the higher trophic levels affect the ecosystem?
Answer:
The loss of organisms in higher levels cause tremendous increase in the number of organisms in the lower levels. It disrupt the equilibrium of environment.

Interactions in the Ecosystem

Many relations exist in nature. It maintains the equilibrium and stability of the ecosystem.
Food relations between organisms are good examples for these interactions.
Predation: One is benefitted. Other is harmed,
eg. Tiger and Deer.
Parasitism: One is benefitted. Other is harmed,
eg: Mango tree and Loranthus.
Competition: Both are harmed first. Later the winner is benefitted.
eg: Paddy and Weeds.
Mutualism: Both are benefitted
eg: Sea anemone and Hermit crab.
Commensalism: One is benefitted and the other is neither benefitted nor harmed.
eg : Mango tree and Vanda

Biodiversity

Biodiversity is the sum total of all the diverse species of organisms and their ecosystems (habitats)
Biodiversity has 3 different levels such as ecosystem diversity species diversity and genetic diversity.
Walter G. Rosen is the scientist who used the term ‘biodiversity’ for the first time.

Indicators (Text Book Page No: 186)

Hsslive Guru 8th Class Kerala Syllabus Chapter 13 Question 9.
Are all ecosystems alike in biodiversity?
Answer:
No.

Question 10.
Are all organisms seen in an ecosystem also seen in another ecosystem?
Answer:
Organisms adapted to the conditions of particular ecosystems. The physical and chemical structure of each ecosystem is different. So organisms seen in one ecosystem may not be present in another ecosystem.

Question 11.
What is the need for protecting natural ecosystems?
Answer:
Natural ecosystem are to be conserved for the existence and conservation of organisms. Ecosystems are the treasure houses of biodiversity. They provide innumerable services. Essential services, Ecological services, supporting services, and cultural services.

Indicators (Text Book Page No: 188)

Question 12.
Large scale destruction of ecosystems
Answer:
Birds are primary the victims of changes occurring in the ecosystem

Question 13.
Overexploitation of the natural resources
Answer:
The unwise interference of human beings destroys our ecosystem with rich biodiversity. If adversely affects the bird diversity in our locality. Many species of birds disappeared due to habitat loss. The pesticides like DDT, endosulfan used in agricultural field kills or drives away the birds that come
in search of food.

Conservation of Biodiversity

Conservation of organisms within their natural habitat is termed as in-situ conversation.
eg: Wildlife sanctuaries, National Parks, Community Reserves, etc.
Conservation of organisms outside their natural habitats is termed ex-situ conservation.
eg : Zoological garden, botanical garden, gene bank

Indicators (Text Book Page No: 183)

Question 14.
What is the scope of ex-situ conservation?
Answer:
It is possible to conserve the endangered animals by keeping them in specialized environment and by providing suitable conditions for reproduction. Rare species of plants can be conserved. Seeds, gametes, etc. can be collected and make we when necessary.

Question 15.
What is the significance of gene banks?
Answer:
Gene banks are research centers. Here special arrangements are these to collect seeds, gametes, etc. and to preserve them for long periods. Animals can be recreated when necessary.

Let US assess (Text Book Page No: 195) 

Question 16.
Phytoplankton – zooplankton – fish – seal – shark
a) In which trophic level is the secondary consumer of this food chain included?
b) Rewrite the food chain in such a way that the organism in the third trophic level figures in the second trophic level.
Answer:
a. In 3^rd trophic level.
b. algae → fish → duck

Question 17.
Find the odd one out from the following. Justify your answer.
a) Quagga, Malabar civet cat, Nilgiri Tahr, Lion-tailed macaque.
b) Eravikulam, Mathikettan shola, Periyar, Silent Valley
Answer:
a. Quaaga, Extinct
b. Periyar – Wildlife Sanctuary Others are national parks.

Question 18.
Examine the statements given below and rewrite if there are errors.
a) Extinct species are included in the Red Data Book.
b) WWF is an organisation working with the objective of protection of biodiversity.
c) Gene banks are included in in-situ conservation.
Answer:
In red data book endangered organisms are included
b. Right / True
c. Seed bank, sperm bank etc. are ex-situ conservation methods.

Diversity for Sustenance Additional Questions & Answers

Question 19.
Classify the following as producers and consumers.
Lizard, Planktons, Paddy, Calotes, Carrot, Grasshopper, Tortoise, Algae, Snake
Answer:

Question 20.
Find out suitable example for the animal relations mentioned.
i. Parasitism
ii. Mutualism
iii. Commensalism
Crops × Weeds
Mango tree × Vanda
Mango tree × Loranthus
Fish × Heron
Hermit crab × sea anemone
Answer:
i. Mango tree and Loranthus
ii. Hermit Crab – Sea anemone
iii. Mango tree and Vanda.

Question 21.
Complete the illustration Suitably

Answer:
a. Environmental / Ecological services
b. Cultural services
c. Food, Medicine
d. Nutrient Cycle, Pollination

Question 22.
Which are the different types of conservation of biodiversity?
Answer:
These are mainly two types.

1. In-situ conservation in which organisms are conserved within their natural environment
2. Ex-situ conservation in which animals are protected out their natural environment.

Question 23.
Classify the following into Ex-situ and In-situ.
(Zoological Gardens, Sacred Groves, Gene banks, Biosphere Reserves, Botanical Gardens, National parks)
Answer:

Question 24.
Expand the following terms.
Answer:
JNTBGRI – Jawaharlal Nehru Tropical Botanic Garden and Research Institute.
MBG – Malabar Botanical Garden RGCB – Rajiv Gandhi Centre for Biotechnology.

Question 25.
Why do all the food chains start from green plants?
Answer:
The basis of all food chains is the plants. They are the producers. These are eaten by the herbivores which in turn are eaten by the carnivores.

Question 26.
The members at the successively higher levels are lesser in number and larger in size in a food chain. What about their body weight? What would be the reason for that?
Answer:
In the successively higher levels in the food chain the number of consumers decreases and the size of their body increases. The number of producers will be very large. The number of animals which feed on them is less in number. But their body size increases.

Question 27.
Hay → Horse
Paddy → Fowl → Fox
Phytoplankton → Tadpole → Fish → Man
Grass → GrasshopperFrog → Snakes Vulture
Examine the food chain given above and classify them as primary consumers, secondary and tertiary consumers.
Answer:
Producers: Hay, Paddy, Phytoplankton, Grass
Primary consumer: Horse, Fowl, Tadpole, Grasshopper
Secondary consumers: Fox, Fish, Frog, Snake
Tertiary consumers: Vulture, Man

Question 28.
What will happen if the number of herbivores increases?
Answer:
If the number of herbivores increase they will eat away all the grass and shrubs and they will have to face shortage of food. The destruction of grass and shrubs will cause soil erosion and the top fertile soil will be washed away.

Question 29.
Animals which are facing extinction.
Answer:

• Wild goat
• Musk deer
• Indian wild Ass
• Lion-tailed Monkey
• Lion
• Rhinoceros
• The large Indian Bustard
• Tiger
• Kashmir deer
• Himalayan Tig
• Silver owl
• Panda

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Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 12 Why Classification?

Why Classification? Textbook Questions and Answers

Classification

Classification is the grouping of organisms on the basis of similarities and differences. Classification using suitable criteria makes the study of organisms more easy. Many criteria are used in classification.
Eg: size, beauty, speed, type of teeth, claws, etc.

Taxonomic Keys

Scientific indicators used to recognize and classify plants and animals are called Taxonomic keys. Dichotomous keys is the most popular among these. Each indicator contains 2 options for selection. By selecting the characteristic feature of the organism to be identified it can be recognized and classified.

Indicators (Text Book Page No: 170)

Why Classification Class 8 Kerala Syllabus Chapter 12 Question 1.
Peculiarity of dichotomous keys
Answer:
Dichotomous key is the most popular among these. Each indicator contains 2 options for selection. By selecting the characteristic feature of the organism to be identified it can be recognized and classified.

Why Classification Kerala Syllabus Chapter 12 Taxonomy

Taxonomy is the branch of science that deals with the identification and classification of organisms according to similarities and differences. Organisms are given scientific names. Carl Linnaeus laid foundation stone for classification. In all organism that including human beings are placed in different levels of classification. It was scientist named carl Linnaeus who fixed taxonomic hierarchy and provided a scientific base for classification. Hence he is known as the father of taxonomy.

Scientists and their Contribution to Taxonomy

Aristotle:

Father of Biology Classified animals as red-blooded and non-red blooded

Theophrastus:

Father of Botany Grouped Plants as animals, biennials, and perennials

Charaka:

Father of Ayurveda Author of ‘Charaka Samhita’

John Ray:

– Used the term ‘Species’ for the first time
– Recorded more than 18000 plants in his book ‘Historia Generalis Plantarum’

Carls Linnaeus:

– Father of Modern Taxonomy.
– Suggested different levels of Classification
– Introduced Binomial Nomenclature

Why Classification Notes Kerala Syllabus Chapter 12 Indicators (Text Book Page No: 173)

Hss Live Guru Biology 8 Kerala Syllabus Chapter 12 Question 2.
Which are the organisms included in kingdom Animalia?
Answer:
Cockroach, Butterfly, Bird, Rabbit, Cat, Tiger, Lion, bear

Why Classification Class 8 Notes Kerala Syllabus Chapter 12 Question 3.
Which organisms are excluded at each consecutive level? Why?
Answer:

• Eliminated from Phylum Chordates — Butterfly, Cockroach
  Reason — Animals with vertebral columns alone are included in the phylum chordates.
• Eliminated from Class Mammalia
  — Bird (Pigeon) Reason — Animals that give birth to young ones alone included in this group.
• Eliminated from the order Carnivora — Rabbit
  Reason — Carnivores alone included in this order.
• Eliminated from the family Felidae — Bear
  Reason — It does not have retractile claws.
• Eliminated from the Genus Felis — Lion, tiger
  Reason — Animals having small body and without roaring sound are included.
• Eliminated from the Species domestic — Wild Cat
  Reason — It has the basic features of cat

Basic Science Class 8 Chapter 12 Solution Kerala Syllabus Chapter 12 Question 4.
At what levels of this illustration can humans be included?
Answer:
Man can be included in Class Mam-malia and Phylum Chordata.

Binomial Nomenclature

Binomial nomenclature is the scientific naming of organisms. Scientific name consists of two words. First word indicates genus and second word indicates species. Scientific name of man – Homo sapiens. Earlier two-kingdom classification was in practice. Accordingly, the organisms were broadly classified into planate and anemia. Later Rober. H. Whittaker classified organisms into 5 Kingdoms.
Eg : Monera, Protista, Fungi, Plantae, Animalia.
Another Scientist Carl Vaus added ‘domain’ above kingdom and expanded it into 6 kingdom classification.

Indicators (Text Book Page No: 176)

Class 8 Basic Science Chapter 12 Kerala Syllabus Chapter 12 Question 5.
Limitations of two-kingdom classification.
Answer:
Bacteria, Fungus, etc were not inclu­ded in two kingdom classification.

Hss Live Guru 8th Physics Kerala Syllabus Chapter 12 Question 6.
Possibilities of five kingdom classification.
Answer:
In five kingdom classification bacteri­a, amoeba, fungus, plants, and animals were included in separate kingdoms according to their characteristics.

8th Std Physics Notes Kerala Syllabus Chapter 12 Question 7.
Circumstances that led to the formulation of six kingdom classification.
Answer:
In ancient times knowledge regarding the characteristics of micro organ­isms was limited. It was found out that the cell structu­re and physiology of archaebacteria belongs to kingdom monera are quite different from other bacteria. Hence kingdom monera was divided into two kingdoms – Archae and Bacteria. Be­sides another level namely ‘domain’ was added above kingdom. Thus 6 king­dom classification came into existence

8th Standard Biology Kerala Syllabus Chapter 12 Question 4.
Completing the table (Text Book Page No: 177)

Answer:

Indicators (Text Book Page No: 178)

Basic Science Class 8 Chapter 12 Kerala Syllabus Question 8.
What are the peculiarities of virus?
Answer:
Viruses have no specific cell structure. Genetic material and a protein sheath alone are present. It is difficult to destroy them. They live only in living cells. They are dead or inactive outside the cell. It multiplies inside the host cell and destroys it.

Hss Live Guru 8 Biology Kerala Syllabus Chapter 12 Question 9.
Can virus be included in any of the classification methods we have discussed earlier? Why?
Answer:
As viruses have no cellular structure, it is not possible to include in any of the classifications mentioned.

Let US assess (Text Book Page No: 179) 

Basic Science For Class 8 Chapter 12 Kerala Syllabus Question 10.
Identify the word pair relation and fill in the blanks
a. Five kingdom classification: Robert H.Whittaker
Six kingdom classification :
b. Charaka: Charaka Samhita
John Ray
Answer:
a. Carl Vaus
b. Historia Generalis Plantarum

Kerala Syllabus 8th Standard Biology Notes Chapter 12 Question 11.
Hints about some organisms are given below. Name the
kingdom to which these organisms belong:
a. Multicellular heterotrophic organisms with a nucleus and capacity for locomotion.
b. Multicellular, heterotrophic, non-motile organisms with a nucleus.
c. Unicellular organisms with a nucleus.
d. Multicellular, autotrophic, non-motile organisms with a nucleus.
Answer:
a. Animals
b. Fungi
c. Amoeba
d.Plants

Kerala Syllabus 8th Standard Chemistry Notes Chapter 12 Question 12.
Write from the table the name of the organism which has more resemblances with tiger. Give explanations for your answer.

Answer:
Lion having the scientific name Pantheraleo.
Lion and Tiger belongs to same Genus ‘Panthera’

Why Classification? Additional Questions and Answers

Basic Science Class 8 Ch 12 Kerala Syllabus  Question 13.
Identify the word pair relation and fill the blanks.
a. 2 Kingdom Classification: Carls Linnaeus:: 5 kingdom Classification: …………….
b.Mushrooms: Fungi:: Bacteria: ……………..
c. Aristotle – Father of Biology Carls Linnaeus – …………….
d. Golden shower: Cassia fistula::…………..: Corvus splendens
e. Charaka: Father of Ayurveda::……………: Father of Botany
Answer:
a. R H Whittaker
b. Monera
c. Father of Taxonomy
d. Crow
e. Theophrastus

Hss Live Guru 8th Biology Kerala Syllabus Chapter 12 Question 14.
2. Identify the odd one and write the characteristic features of others.
a. Lion, Tiger, Rabbit, Cat.
b. Genus, Order, Carl Linnaeus, Phylum.
Answer:
a. Rabbit – Others including order Carnivora
b. Carl Linnaeus – He is the Father of Modern Taxonomy. Others are different levels of classification.

Question 15.
Which organism is most suit¬able for the following indicators (amoeba, bacteria, virus, Fungus)
1. Lives only in living cell
2. Pathogen
3. Only genetic material and a protein covering.
Answer:
Virus

Question 16.
Find out the scientists suitable to the statements given.
i. Author of Charaka Samhitha
ii. Author of Historia Generalis Plantarum
iii. Father of Modern Taxonomy
iv.Father of Biology
Answer:
i. Charaka
ii. John ray
iii. Carl Linnaeus
iv. Aristotle

Question 5.
Complete the table

Answer:

• Elephas Maximus / Elephas Indicus
• Pavo Cristatus
• Canis Familiaris
• Hibiscus Rosasinensis
• Azadiracuta Indica
• Oryza sativa

Question 17.
What are limitations in the system of classification of Carl Linnaeus?
Answer:
Some lower organisms share the characters of both animals and plants. So it is difficult to recognize them as plants or animals. Linnaeus considered only plants and animals for his classification. Microscopic organisms like bacteria, fungus, protozoa, etc were not included either in plant kingdom or in animal kingdom. Certain animals which were included in the classification of Linnaeus show characters of both the animal and plant kingdoms. Eg: Euglena protozoan shows locomotory movements like animals, but it contains chlorophyll like plants.

Question 18.
7. Hints about some organisms are given below. Name the kingdom to which these organisms belong.
a. Multicellular heterotrophic organisms with a nucleus and capacity for locomotion.
b. Multicellular heterotropic, non-motile organisms with a nucleus.
c. Unicellular organisms with a nucleus.
d. Multicellular, autotroph, non-motile organisms with a nucleus
Answer:
a. Animalia
b. Fungi
c. Protista
d. Plantae

Question 19.
Write the scientific name of following animals and plants. Coconut, paddy, wheat, crow, mango, grapes
Answer:
Coconut – Cocos nucifera
Paddy – Oryza sativa
Wheat – Triticum aestivum
Crow – Corvus splendens
Mango – Mangifera indica
Grapes – Vitis vinifera

Question 20.
What are two important characters of species?
Answer:
1. Species is a group of organisms that can freely interbreed to produce fertile offsprings.
2. A group of organisms that closely resemble each other in structure biochemical makeup and external characteristics but which are genetically different. In one species there may be subspecies.

Question 21.
The method of classification adopted by Whittaker is much better than the method adopted by Carl Linnaeus the father of the science of classification. What is your response to this statement?
Answer:
The classification of Linnaeus had only two kingdoms, plants, and animals. He did not consider bacteria, protozoa, fungus, etc. Certain characters of animals considered by Linnaeus for classification are found in both the kingdoms, eg. photosynthesis found in plants is seen in some animals like Euglena. Certain characters of animals can be seen in plants also, eg certain types of algae. But Whittaker adopted the method of having five kingdoms including protozoa and bacteria.

• Monera,
• Protista
• fungi
• Plantae
• animalia

Question 22.
Bacteria does not have a well defined Nucleus. Viruses are also like Bacteria. Why is it not possible to include viruses under Monera.
Answer:
Viruses exhibit living nature only when they enter the host cells. On other occasions, they do not exhibit living nature. But organisms in Monera are not like that.

Question 23.
How do the levels of classification of plants made by Carl Linnaeus differ from the levels of classification of Animals?
Answer:
Not much differences are there. In the place of ‘Order’ in animal classification, plant classification has series. And in the place of ‘Phylum’ in animal classification, plant classification has ‘Division’.

Question 24.
Find the level of classification and complete the given table.

Answer:
1. Kingdom
2.Phylum Chordata
3. Mammalia
4. Mammalia
5. Species

Question 25.
Cell is the smallest unit of life. But there are certain organisms that live without cell too. Analyze this statement.
Answer:
Life is not possible without a cell. Viruses do not have cells. As it is so, it does not have life when it outside a living cell. As it enters a living cell, it will show the features of life forms. It makes use of the components of the host cell and continues to live. Though it does not have a cell, it can continue its life only after entering a host cell and by making use of its components.

Question 26.
Observe the given statement, and write correct answer if you find false statements.
a. Aristotle is the Father of Biology.
b. John Ray is the Father of Botany.
c. Carl Linnaeus used the term ‘species’ for the first time.
d. Charaka proposed binomial nomenclature.
Answer:
a. True
b. False, Theophrastus
c. False, John Ray
d. False, Carl Linnaeus

Question 27.
What is the relationship between taxonomic keys and dichotomous key?
Answer:
Taxonomic keys are scientific indicators used to identify and classify plants and animals. Dichotomous keys is one of the popular taxonomic keys.