Prasanna

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Kerala State Syllabus 8th Standard Hindi Solutions Unit 2 Chapter 3 मेरे बच्चे को सिखाएँ (पत्र)

मेरे बच्चे को सिखाएँ पाठ्यपुस्तक के प्रश्न और उत्तर

Mere Bache Ko Sikhaye Notes 8th Kerala Syllabus प्रश्ना 1.
“मेहनत से कमाया एक पैसा भी, हराम में मिली नोटों की गड्डी से कहीं अधिक मूल्यवान होता है।” अपना दृष्टिकोण प्रकट करें।

उत्तर:
यह कथन बिलकुल ठीक है। हराम की चीजें हमारे हक का नहीं है। मेहनत से कमाया पैसा ही मूल्यवान है। मानव को ईमानदारी के साथ जीना है। मेहनत ईमानदारी में चार चाँद लगाता है।

Kerala Syllabus 8th Standard Hindi Notes प्रश्ना 2.
‘बदमाशों को आसानी से काबू में किया जा सकता है। ऐसा क्यों कहा होगा?

उत्तर:
बदमाशों के अंदर भी कुछ सच्चाई होती है। उपदेश और सत्संग के द्वारा उन सच्चाइयों को बाहर ला सकते हैं। इसलिए ऐसा कहा गया है।

Hindi State Syllabus 8th Standard प्रश्ना 3.
‘नकल करके पास होने से फेल होना बेहतर है’ इस प्रस्ताव से क्या आप सहमत है? क्यों?

उत्तर:
मैं इससे शतप्रतिशत सहमत हूँ। नकल से मिली जीत में ज्ञान की गहराई नहीं होती। यह तत्काल लाभ दे सकता है। लेकिन भविष्य में इससे कोई मुनाफ़ा नहीं होता। इसलिए फेल से सीख लेना ही बेहतर है।

8th Standard Hindi Guide Kerala Syllabus प्रश्ना 4.
‘भीड़ से अलग होकर अपना रास्ता बनाना’ का मतलब क्या है?

उत्तर:
भीड़ एक ही मानसिकता के आधार पर चलती है। भीड़ की मानसिकता से अलग होकर सोचने से ही नई दृष्टि और नए विचार मिलते हैं। इस नए दृष्टिकोण से ही सामाजिक प्रगति संभव होती है। संसार के सभी महत् व्यक्ति इस प्रकार सोचनेवाले थे। इसलिए उन्हें संसार में बदलाव ला सका।

मेरे बच्चे को सिखाएँ Textbook Activities

8th Standard Hindi Notes State Syllabus प्रश्ना 1.
लघु-लेख लिखें।
‘सफल जीवन’ विषय पर लघु-लेख लिखें।

उत्तर:
सफल जीवन
जीवन को सफल बनाने के लिए मनुष्य को आत्मविश्वास, दृढ़संकल्प, अदम्य उत्साह और लगन चाहिए। केवल पढ़ने से ही नहीं, अच्छे चरित्र के निर्माण में भी मानव को ध्यान देना चाहिए। उसे परिश्रमी होना चाहिए। उसे यह समझना चाहिए कि नुशासन जीवन को सफल और उज्ज्वल बनाने के लिए आवश्यक है। व्यक्ति को सादा जीवन और उच्च विचार का आदर्श ग्रहण करना चाहिए। उसे बुरी आदतों और बुरे सहवास से बचकर रहना भी होगा। अपने पाठों को लगन से पढ़ना, बड़ों के सदुपदेशों का पालन करना, बड़ों से आदर और छोटों से प्यार करना आदि की आवश्यकता है। उसे पथभ्रष्ट करनेवाली बातों से बचकर रहना भी चाहिए। जो व्यक्ति इस प्रकार का जीवन बिताएगा, वह अपने जीवन में सफल बनेगा।

मेरे बच्चे को सिखाएँ मेरी रचना में

उचित चौकार में ✓ लगाएं।


विषय का विश्लेषण किया है।

प्रस्तुतीकरण में क्रमबद्धता है।

उचित भाषा का प्रयोग किया है।

अपना दृष्टिकोण प्रस्तुत किया है।

उचित शीर्षक दिया है।

मेरे बच्चे को सिखाएँ Summary in Malayalam and Translation

मेरे बच्चे को सिखाएँ शब्दार्थ Word meanings

Students can Download Maths Chapter 12 Proportion Questions and Answers, Summary, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 9th Standard Maths Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Maths Solutions Chapter 12 Proportion in Malayalam Medium

Proportion Questions and Answers in Malayalam

You can Download पिता का प्रायश्चित Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Hindi Solutions Unit 2 Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Hindi Solutions Unit 2 Chapter 2 पिता का प्रायश्चित (संस्मरण)

पिता का प्रायश्चित पाठ्यपुस्तक के प्रश्न और उत्तर

Pitha Ka Prayaschit Questions And Answers Kerala Syllabus 8th प्रश्ना 1.
वह झूठ बोला, “कार तैयार नहीं थी, इसलिए देर हो गई।” इस तरह झूठ बोलना क्या सही है? क्यों?

उत्तर:
झूठ बोलना कभी भी सही नहीं है। क्योंकि यह एक बुरी आदत है। झूठ बोलने से तत्काल फ़ायदा हो सकता है। लेकिन इससे भविष्य में नुकसान ही होगा।

Pitha Ka Prayaschit Notes Kerala Syllabus 8th प्रश्ना 2.
“घर तक की अठारह मील की दूरी पैदल चलकर ही तय करूँगा।” मनीलाल गाँधी के इस निर्णय से आप सहमत हैं? क्यों?

उत्तर:
मनीलाल गाँधी का यह निर्णय बिलकुल सही है। क्योंकि बेटे की गलती का कारण वे अपने को मानते हैं। इसके द्वारा उन्होंने अपने बेटे को अपनी गलती पर सोचविचार करने मौका दिया।

पिता का प्रायश्चित Textbook Activities

Pitha Ka Prayaschit In Malayalam Kerala Syllabus 8th प्रश्ना 1.
सही मिलान करें।


उत्तर:
वर्ष — बरस
सुदूर — दूरदराज
प्रदेश — इलाका
अवसर — मौका
प्रतीक्षा — इंतज़ार
ढूँढ़ — तलाश

Prayashchit Hindi Lesson Question Answer Kerala Syllabus 8th प्रश्ना 2.
अर्थभेद समझें।

डरबन से 18 मील दूर एक आश्रम में रहता है।

डरबन से 18 मील दूर एक आश्रम में रहता था।

मैं और मेरी दो बहिनें हमेशा शहर जाने की इंतज़ार में रहते हैं।

मैं और मेरी दो बहिनें हमेशा शहर जाने की इंतज़ार में रहते थे।

वहाँ दूर तक गन्ने के खेत हैं।

वहाँ दूर तक गन्ने के खेत थे।

Prayaschit Questions And Answers Kerala Syllabus 8th प्रश्ना 3.
पत्र लिखें।

संस्मरण कैसा लगा? पुत्र की गलती पर पिता ने अपने आप को सज़ा दी। इसी दर्द के एहसास से अरुण गाँधी ने यह निर्णय लिया- मैं कभी झूठ नहीं बोलूँगा। अपना दर्द वह दोस्त से बाँटे बिना नहीं रह सका। उसने मित्र को पत्र लिखा। वह पत्र कल्पना करके लिखें।

उत्तर:

डरबन
20 अगस्त 1950

प्रिय मित्र,
नमस्कार।
तुम कैसे हो? सोचता हूँ कुशल से हो। हम यहाँ डरबन में खुशी से जी रहे हैं। अपने जीवन के एक विशिष्ट बात बताने के लिए मैं यह चिट्ठी लिख रहा हूँ। कल पिताजी को मेरी गलती पर प्रायश्चित करना पड़ा। हुआ यह कि पिताजी को शहर में कल एक मीटिंग थी। उन्हें मैंने कार से शहर छोड़ा। शाम पाँच बजे उन्हें लेने जाना था। लेकिन बेन जॉन का सिनेमा देखकर मैं समय भूल गया। देरी के कारण पूछने पर झूठ बोला कि कार ठीक करके गैरेज से नहीं मिला। लेकिन पिताजी बात पहले ही समझ गए थे।

पिताजी ने मेरे झूठ को अपनी गलती माना। वे प्रायश्चित करते हुए घर तक का रास्ता पैदल चले। यह देखकर मुझे बहुत दुख हुआ। मैं यह निश्चय किया हूँ कि आइंदा झूठ नहीं बोलूंगा। अगर पिताजी मुझे कोई सज़ा दी होती तो मैं ऐसा कोई निर्णय नहीं लेता। मैं यह घटना कभी नहीं भूलूँगा। उसकी याद ज़िंदगी में मुझे सही रास्ते पर ज़रूर ले जाएगी।

अपना दोस्त
अरुण गाँधी।

सेवामें

अरविंद
वर्धा आश्रम
पोरबंदर
गुजरात
भारत

पत्र लिखते समय ध्यान दें…
स्थान और तारीख है।
उचित संबोधन है।
विषय का सही संप्रेषण है।
स्वनिर्देश है।
पता है।

पिता का प्रायश्चित मेरी रचना में

उचित चौकोर में ✓ लगाएँ।


स्थान और तारीख हैं।

उचित संबोधन है।

विषय का सही संप्रेषण है।

स्वनिर्देश है।

पता है।

पिता का प्रायश्चित Summary in Malayalam and Translation

पिता का प्रायश्चित शब्दार्थ Word meanings

You can Download Chemical Bonding Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Chemistry Solutions Part 1 Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Chemistry Solutions Chapter 2 Chemical Bonding

Chemical Bonding Textual Questions and Answers

Kerala Syllabus 9th Standard Chemistry Notes Chapter 2 Question 1.
What peculiarity do you see in the electronic configuration of noble elements except Helium? Except Helium all other elements have 8 electrons in the outermost shell, hence shall be considered to be chemically stable.

The arrangement of eight electrons in the outermost shell of atom is called octet electron configuration. In Helium atom there is only one shell. The maximum number of electrons in the first shell is 2. Hence two-electron pattern system of Helium also stable.

Kerala Syllabus 9th Standard Chemistry Chapter 2 Question 2.
The electronic configuration of some elements are given below.

Is the number of electrons in the outermost shell of these elements the same as that of the elements in Table (2.1).
Answer:
No

(i) You are familiar with the compounds of these elements. Write the names of some compounds?
Answer:
Magnesium chloride, Sodium oxide, Sodium chloride.

(ii) How are atoms in these compounds held together?
Answer:
Strong attractive force

(iii) What is meant by Chemical Bonding?
Answer:
The attractive force that holds the atoms together in the formation of a molecule is called chemical bonding.

Ionic Bonding

Class 9 Chemistry Chapter 2 Notes Kerala Syllabus Question 3.
In the formation of sodium chloride which atoms are combing.
Answer:
Sodium, chlorine

Chemical Bonding Notes Class 9 Kerala Syllabus Question 4.
How many electrons are there in the outermost shell of sodium atom?
Answer:
1

Chemical Bonding Questions And Answers Class 9 Kerala Syllabus Question 5.
How many electrons are there in outermost shell of chlorine?
Answer:
7

Chemical Bonding Questions And Answers Pdf Class 9 Kerala Syllabus Question 6.
How do chlorine and sodium attain stability?
Answer:
Sodium donates one electron to chlorine to become sodium ion [Na+] and chlorine become chloride ion [Cl]

Chemical Bonding Class 9 Notes Pdf Kerala Syllabus Question 7.
Analyze the electron transfer in each atom during the formation of sodium chloride.
Answer:

Chemical Bonding Class 9 Pdf Kerala Syllabus Question 8.
Draw the electron dot diagram of the transference of electron of sodium atom and chlorine atom. The diagram represents only electrons in the outermost shell because they are the only electrons participating- in chemical bonding.
Answer:

Questions On Chemical Bonding Class 9 Kerala Syllabus Question 9.
Complete Table 2.3 by examining the arrangement of electrons before and after the chemical reaction during the formation of sodium chloride.

a) Which atom donates electron? How many electrons?
b) Which atom accepts electron? How many electrons?
Answer:
a) Sodium, one electron
b) Chlorine, one electron

Kerala Syllabus 9th Standard Chemistry Notes English Medium Question 10.
Electron transfer during the formation of sodium chloride can be written in the form of an equation
Na → Na⁺+1e^–
Cl + 1e^– → Cl^–
Answer:
During the formation of sodium chloride sodium atom donates electron and gets converted to sodium ion (Na⁺) chlorine accepts an electron to form chloride ion (Cl^– ). Through this sodium and chlorine atoms complete an octet in their outermost shell to attain stability.

The oppositely charged ions thus formed are held together by electrostatic force of attraction. This attractive force is called Ionic Bond. Sodium chloride contains ionic bond.

Kerala Syllabus 9th Standard Chemistry Chapter 1 Question 11.
Define Ionic Bond?
Answer:
Ionic bond is a chemical bond formed by electron transfer in an ionic bond, the ions are held together by the electrostatic force of attraction between the oppositely charged ions.

Kerala Syllabus 9th Standard Chemistry Solutions Question 12.
Explain the formation of magnesium oxide from magnesium and oxygen?
Analyze the electron dot diagram and complete the table.
Answer:

To attain stability magnesium donates 2 electrons to become magnesium ion (Mg₂⁺) and – oxygen become [O₂^– ] ion. This type of bonding is ionic bonding.

Chemical Bonding Notes Class 9 Pdf Kerala Syllabus Question 13.
How the ionic bond formation of sodium oxide is represented?
[Hint: Atomic No. of sodium 11, oxygen 8]
Answer:

Question 14.
Draw the electron dot diagram of following compounds. [Hint: Atomic No. Na=11, F=9, Mg=12]
Answer:
1. Sodium Flouride [NaF]

Question 15.
Define ionic compounds?
Answer:
Compounds formed by ionic bonding are called ionic compound.

Covalent Bonding

Fluorine [F₂], Chlorine [Cl₂] oxygen [O₂] Nitrogen [N₂] etc. are diatomic molecules. Let us examine the formation of these molecules.
The Bohr atom model of fluorine is given in figure.

Question 16.
Write the atomic number of fluorine?
Answer:
9

Question 17.
The electronic configuration of Fluorine
Answer:
2, -7 ‘

Question 18.
How many electrons are required for one fluorine atom to attain the octet?
Answer:
1

Question 19.
Is there a possibility of transferring electrons from one fluorine atom to another fluorine atom?
Answer:
No.

Question 20.
How can the two fluorine atoms attain an octet arrangement?
Answer:
By sharing of electrons

Question 21.
The manner in which the two fluorine atoms in a fluorine molecule undergo chemical bonding is illustrated in fig. 2.6.
Answer:

Question 22.
What happens during the formation of fluorine molecule electron transfer or electron sharing?
Answer:
Electron sharing

Question 23.
How many pairs of electrons are shared?
Answer:
One pair

Question 24.
How covalent bonds are formed?
Answer:
The chemical bond formed as a result of the sharing of electrons between the combining atoms is called a covalent bond.

Question 25.
How single bonds are formed?
Answer:
Single bonds are formed by sharing one pair of electrons. It is represented by a small line between the symbols of the combining element, eg. Fluorine molecules can be represented as F – F. The atomic number of chlorine is 17.

Question 26.
Write down the electronic configuration?
Answer:
2, 8, 7

Question 27.
Draw the electron dot diagram of the formation of chlorine molecule by combining two chlorine atoms?
Answer:

Here one pair of electrons – sharing hence single bond is formed.

Question 28.
Examine the diagram illustrating the chemical bonding in the molecule of oxygen and nitrogen.
Answer:

In oxygen molecules, two pairs of electrons are shared hence this type of covalent bond is double bond.
In nitrogen molecule, three pairs of electrons are shared hence this type of covalent bond is triple bond.

Question 29.
Complete table 2.5 given below related to covalent bonding.

Answer:

Question 30.
Draw the chemical bond formation of hydrogen chloride [HCI]

a) How many electron pairs are shared?
b) Represent chemical bond by using symbols?
Answer:
a) One pair of electrons
b) H – Cl

Question 31.
Examples of some covalent compounds are given draw the chemical bonds of the compound by using electron dot diagram.
a)CH₄
b) HF
c)H₂O
Answer:
a)

Electronegativity

Question 32.
Define electronegativity?
Answer:
In a covalent bond the relative ability of each atom to attract the bonded pair of electrons towards itself is called electronegativity.

Question 33.
Who proposed the electronegativity scale?
Answer:
Linus Pauling

Question 34.
Some compounds and their nature are shown in table (2.6) complete the table by finding out the electron negativity difference between the constituent elements.

Answer:

Generally, the electronegativity difference of the component elements in a compound is 1.7 or more it shows ionic character. If it is less than 1.7 it shows covalent character.

Polar Nature

Question 35.
Consider the case of hydrogen molecule [HCl]
a) What is the electronegativity of hydrogen?
b) What is the electronegativity of chlorine?
c) The atomic nucleus of which of these elements has a greater tendency to attract the shared pair of electrons?
d) The chlorine atom with a higher electronegativity attracts the shared pair of electrons towards its nucleus. As a result, the chlorine atom in hydrogen chloride develops a partial negative charge g: (delta negative) and hydrogen atom develops a partial positive charge δ+ (delta positive) it can be represented below
Answer:
a) 2.2
b) 3.16
c) Chlorine
d) \(\begin{array}{l}{\delta^{+} \quad \quad \delta^{-}} \\ {H-C^{\prime}}\end{array}\) Compounds having partial electron charge separation with the molecule are called polar compound. HF, HBr, H20 are example of polar compounds.

Question 36.
Explain the properties of Ionic compounds and covalent compounds.
Answer:

Valency:

Question 37.
What is meant by valency?
Answer:
Valency is the combining capacity of the atoms of an element. It can be treated as the number of electrons lost gained or shared by an atom during chemical combination.

Question 38.
In the formation of sodium chloride- sodium donates one electron, chlorine accepts one electron write the valencies of each element?
Answer:
1

Question 39.
In the formation of magnesium oxide- How many electrons are donated by magnesium?
Answer:
2

Question 40.
How many electrons are accepted by oxygen?
Answer:
2

Question 41.
How is valency and electron transfer related in this case?
Answer:
Same

Question 42.
In the formation of hydrogen chloride, how many electron pairs are shared?
Answer:
One pair

Question 43.
What will be the valency of each atom?
Answer:
1

Question 44.
Complete the table given below analyze the change in the electronic arrangement of elements during the formation of each compound. Find how they are related to valency.

Answer:

From Valency to Chemical Formula

The chemical formula of some compounds are given
Sodium chloride – NaCl
Magnesium chloride – MgCl₂
Aluminium chloride – AICl₃
Carbon tetrachloride – CCl₄

Question 45.
The symbol of some elements and there valencies are given. Write the chemical formula of the compounds formed by them.

Answer:

Question 46.
Why does the number of chlorine atoms differ in these compounds? Try to find out by analyzing the valency of the elements Na, Mg, Al, Cl and C. Analyse Table 2.9
Answer:

Examine the above table and identify how to write the chemical formula from valency. Compare your findings with the following.

• First write the element with lower electronegativity.
• Exchange the valency of each element and write as suffix.
• Divide the suffix with the common factor.
• If the suffix is 1, it need not be written.

Let Us Assess

Question 1.
Complete the table given below and answer the following questions (symbols used are not true)

a) Which element in the table is the most stable one? Justify your answer.
b) Which element donates electrons in chemical reaction?
c) Write the chemical formula of the compound formed by combining element S with P.
Answer:

a) R – it contains an octet configuration in the outermost shell.
b) S
c) The valency of S = 2 and P is 1, hence the chemical formula

[The element in which the electron lose can be written first]

Question 2.
Electronegativity values of some elements are given. Using these values, find whether the following compounds are ionic or covalent.
(Electronegativity of Ca = 1, O = 3.5, C = 2.5, S = 2.58, H = 2.2, F = 3.98)
i) Sulphur dioxide (S02)
ii) Water (H₂O)
iii) Calcium fluoride (CaF₂)
iv) Carbon dioxide (CO₂)
Answer:
i) SO₂
Electronegativity difference = 3.5 – 2.58 = 0.92
If the electronegativity difference is less than 1.7, it shows covalent character.
ii) Water (H₂O)
Electronegativity difference = 3.5 – 2.2 = 1.3
If the electronegativity difference is less than 1.7 it forms covalent compounds.
iii) CaF₂
Electronegativity difference = 3.98 -1.0 = 2.98
If the electronegativity difference is greater than 1.7 if forms ionic compounds.
iv) CO₂
Electronegativity difference = 3.5 – 2. 5 = 1.0
If the electronegativity difference is less than 1.7 it shows covalent compounds.

Question 3.
Some elements and their valencies are given

a) Write the chemical formula of barium chloride
b) Write the chemical formula of zinc oxide
c) The chemical formula of calcium oxide is CaO. What is the valency of calcium?
Answer:

c) The valency of calcium is 2.

Question 4.
Examine the following chemical equations and answer the questions.
(Hint: Atomic Number Mg = 12 Cl = 17)
Mg + Cl₂ → MgCl₂
Mg → Mg²⁺ + ……..
Cl + 1e^– → ………..
a) Complete the chemical equations.
b) Which is the cation? Which is the anion?
c) Which type of chemical bond is present in MgCl₂?
Answer:
a) Mg → Mg^2- + 2e
Cl + 1e^– → Cl^–
b) The cation or positively charged ion is Mg²⁺ and the anion or negatively charged ion is Cl^–
c) Ionic Bonding

Extended Activities

Question 1.
Draw the electron dot diagram of chemical bonds in methane (CH₄) and ethane (C₂H₆).
Answer:

Question 2.
P, Q, R, S are four elements. Their atomic numbers are 8, 17, 12 and 16 respectively. Find the type of chemical bond in these compounds formed by combining the following pairs of elements. Construct and exhibit the type of bonds using different. substances (eg. pearls) (Electronegativity values:
P = 3.44, Q = 3.16, R = 1.31, S = 2.5)
1) P, R
2) P, S
3) Q, R
Answer:
1. PR
Electronegativity difference = 3.44 – 1.31 = 2.13
The electronegativity difference is greater 1.7 it shows ionic compound.
2. PS
Electronegativity difference = 3.44 – 2.58 = 0.86
The electronegativity difference is less than 1.7 it shows covalent compound.
3. QR
The electronegativity difference = 3.16 – 1.31 = 1.85
The electronegativity difference is greater than 1.7 it shows ionic compound.

Question 3.
Perform the experiments arranging the apparatus as shown in figure.

Record your observations and identify what type of compounds sodium chloride and sugar are
Answer:
When electricity is passed through sodium chloride solution in which carbon rod is immersed hydrogen and chlorine gas are produced. It is an ionic compound. In the second when electricity is passed through sugar solution there is no charge. Hence it belongs to covalent compound.

You can Download Construction of Quadrilaterals Questions and Answers, Activity, Notes, Kerala Syllabus 8th Standard Maths Solutions Chapter 6 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Maths Solutions Chapter 6 Construction of Quadrilaterals

Construction of Quadrilaterals Text Book Questions and Answers

Textbook Page No. 108

Construction Of Quadrilaterals Class 8 State Syllabus Question 1.
Can you draw these patterns of squares in your notebook?


Solution:
1. Draw a square of side 6 cm. Draw lines horizontally and vertically 2cm apart. Rub off unwanted parts. We get the required pattern.

2. Draw a square of side 7 cm. In it draw horizontal and vertical lines at intervals of 4 cm, 2 cm and 1 cm. Erase the unwanted part. We get the required pattern.

3. Draw a square of diameter 6 cm. (Draw a circle of diameter 6 cm and draw two perpendicular diameters. Join their ends.) Mark the points on diagonal 2 cm a part. Join the points as in the following figure. Rub off the unwanted part. We get the required pattern.

Text Book Page No. 111

Class 8 Mathematics Construction Of Quadrilaterals Kerala Scert Solutions Question 2.
Draw the figures below in your notebook.


Solution:
1. Draw a rectangle of diagonal 9 cm and angle between the diagonal and one side is 30°. Mark a point on the diagonal at a distance 6 cm from one end of the diagonal. Draw lines perpendicular to the sides of the rectangle through this point. Erase unwanted parts, we get the required pattern.

2. Construct an equilateral triangle with side 3 cm.

The other two sides of a triangle are made by equal diagonals of the rectangle. Then the diagonal of the rectangle is 6 cm. Draw a rectangle in a horizontal position with diagonals 6 cm and the angle between the diagonals 60°. Draw another rectangle of the same measure in the vertical position at the middle of the first rectangle. Draw the wanted part in bold lines. Erase unwanted part.

Construction Of Quadrilaterals Class 8 Kerala Syllabus Question 3.
Draw a rectangle CBD, which di agonal (AB) is 6 cm and AC making an angle 30° with AB. The arc with centre at B and BC as radius and the arc with centre at A and AC as radi us meet at E. Complete ∆ AEB. Draw an arc with centre A and EB as radi us. Also draw arc with B as centre and AE as radius meet at F. Complete the rectangle AEBF.

Text Book Page No. 114

8th Class Maths Construction Of Quadrilaterals Kerala Syllabus Question 3.
Draw a rhombus of diagonals 5.5 cm and 3 cm in your notebook.
Solution:
Draw a line of length 5.5 cm, and find its midpoint by drawing the perpendicular bisector. Mark the points on the upper and lower part of the bisector.
Mark the points on the upper and lower part of the bisector line at a distance 1.5 cm from the intersecting point of the first line and perpendicular bisector. Join these points to the end of the first line.

Construction Of Quadrilaterals 8th Class Kerala Syllabus Question 4.
Draw also a rhombus of diagonals 5.5 cm and 3.5 cm.
Solution.
It is difficult to measure 1.75 cm (half of 3.5 cm) using scale, so draw a rectangle of 5.5 cm and breadth of 3.5 cm. By drawing the perpendicular bisectors find the midpoints of the sides. By joining the midpoints of the sides, we get a rhombus.

Textbook Page No. 117

Class 8 Construction Of Quadrilaterals Kerala Syllabus Question 5.
Draw these figures
Solution:
1. Two equal rhombuses :


Solution:
1. From the figure two sides of A BCD are equal, angles opposite these sides are also equal. We can calculate them as 50° each. In the same way find other angles in the figure.

Draw a line BD vertically, 3 cm long. At D draw angles of 50° on both sides. At B also draw angles of 50° on both sides. Then we get a rhombus ABCD. Extend BC to G such that BC = CG and extend DC to E such that DC = CE. Draw GE. Draw angle of 50° at G and E to find F.

2. Draw a circle of radius 2 cm. Divide the centre of the circle into angles of 60° each. These lines meet the circle at the points A, B, C, D, E and F.

Draw arcs of 2 cm from A and B to get G. Similarly find H and I. Draw the required parts and rub off unwanted parts.

Construction of Quadrilaterals Class 8 Question 3. Draw a circle of radius 2 cm. Mark the points A, B, C, D, E, F, G, and H on the circle by making 45° angles at the centre. Draw arc of 2 cm A and B to get I. Similarly find J, K and L. We get the required figure.

4. Draw AC, 4 cm long and mark its midpoint B. Since all are rhombuses, ABI is a equilateral triangle. Its angle are 60° each.

In the rhombus BCDJ, ∠ JBC = ∠JDC = 60°, ∠BCD = ∠BJD = 120°
In the rhombus BJFI, ∠IBJ =∠IFJ = 60°, ∠BJF = ∠FIB = 120°. Draw each rhombus and complete the pattern.

5. Draw a line AB, 4 cm long and mark its midpoint C. CJFI is a square, all its angle are 90° each. Also calculate ∠ICA = 45° and ∠CAH = 135°. Taking measures of the sides and angles draw each rhombus.

6. Draw a square of side 3 cm. And draw two parallelograms with sides 3 cm, 2 cm and angle between them 45°, on the sides of the square.

Text Book Page No. 124

Construction Of Quadrilaterals Class 8 Solutions Kerala Syllabus  Question 6.
Draw the figures below :
1. Three equal isosceles trapeziums:



Solution:
1. Draw DE, 5 cm long and mark G on DE such that DG = 2 cm (length of AB) ∆ ABC and ∆ DEF are equilateral triangles. Their each angle is 60°. Find B by drawing ∠EGB = ∠BEG = 30°. Draw ∠ADG = 30°and BA = 2 cm to get A. Now we got one trapezium. Draw the other two trapezium in the same way.

2. Draw two circle of radius 1 cm and 3 cm with the same centre. Divide the circumferance of the both circle into 6 equal parts. And join them to obtain six equal isosceles trapeziums.

3. Draw two parallel sides AB = 8 cm, CD = 4 cm. AE= FB = 2 cm, EF = 4 cm. Also GD = HC = 2 cm. ∠B + ∠B CD = 180°. The angles at C are equal. They are equal to B. So ∠B = \(\frac{180}{3}\) = 60°. ∠A = 60°. Now draw the pattern.

4. Draw a rectangle with length 8 cm and breadth 4 cm. Divide this rectangle into two squares with side 4 cm. Consider one square and half of second. Mark the midpoint of the sides. Join as in the figure.

5. Draw a square of side 8 cm. And draw lines horizontally and vertically 2cm apart. Complete the figure.

Text Book Page No. 128

8th Class Maths Notes Kerala Syllabus Question 7.
Draw the quadrilaterals shown below.

Solution:
1.Draw AB = 5 cm. Draw AD such that ∠A = 80° and AD =3 cm. Mark C such that ∠D = 120° and DC = 4 cm. Join BC.

2. Draw AB = 5 cm. Draw AD such that ∠A = 60° and AD = 3 cm. Mark point C such that ∠B = 80° and ∠D = 100°. Join DC and BC and complete the quadrilateral ABCD.

3. Draw a ABD with AB = 7 cm, BD = 8 cm and AD = 4 cm. Mark the point C at a distance 6 cm from A and 5 cm from D. Join BC and CD.

Construction of Quadrilaterals Additional Questions and Answers

Class 8 Maths Construction Of Quadrilaterals Kerala Syllabus Question 1.
In the figure, ABCD is a square whose diagonals intersect at O. If AD = 10 cm, find the length of BD and CD ?

Solution:
Tn a square diagonal are equal and perpendicular bisectors of each other.
OD = 5 cm and OC = 5 cm.
CD² = OD² + OC² = 5² +5² = 50 cm
CD = \(\sqrt{50}\) = BD = \(5\sqrt{2}\) cm.

Kerala Syllabus 8th Standard Maths Notes Question 2.
Draw quadrilateral PQRS, PQ = 7 cm, QR = 5 cm, RS= 4cm, ∠Q = 60° and ∠R = 140°.
Solution:
Draw PQ = 7 cm and draw QR of length 5 cm which makes an angle 60° with PQ. Draw RS such that RS = 4 cm and ∠R = 140°.

8th Standard Maths Notes State Syllabus Question 3.
(a) Write any two peculiarities of the diagonals of a square.
(b) The length of a diagonal of a square is 7 cm. Draw the square.
Solution:
(a) The diagonals of a square are equal.
The diagonals bisect each other perpendicularly.
(b)

Question 4.
In rhombus PQRS , PR = 7 cm and Question = 5 cm. Construct rhombus PQRS.
Solution:

Question 5.
In the figure, ABCD is a parallelogram. ∠D = 80°. Find all other angles?

Solution:
ABCD is a parallelogram
Opposite angles are equal ∠B = 80°.
Sum of the angles on the same side is 180°.
∠A + ∠B = 180°.
∠A = 180° – 80° = 100°
And ∠C = 100° (opposite angles are equal).

Question 6.
In the figure, ABCD is a parallelogram. Find x, y, z.

Solution:
∠Y = 112° (opposite angles are equal)
In ADC, ∠x + ∠y + 40 = 180° (sum of angles in a triangle)
∠x + 112° + 40° = 180°
∠x = 180° – 152° = 28°
∠z = 28°(transversal alternate interior angles are equal).

Question 7.
Construct a quadrilateral ABCD, AB = 6 cm, BC = 3 cm, CD = 2 cm, AD = 4 cm and AC = 5 cm.
Solution:
Draw AB = 6 cm. Then find C by drawing arcs of radius 5 cm and 3 cm from A and B. Find D by drawing arcs of radius 4 cm and 2 cm from A and C. Join BC, CD and AD to get quadrilateral ABCD.

Question 8.
Construct a quadrilateral ABCD, AB = 8 cm, BC = 6 cm, CD = 5.5 cm, DA = 3 cm and ∠B = 50°.
Solution:
Draw AB = 6 cm. Draw BC such that BC = 6 cm and ∠B = 50°. Draw an arc of radius 5.5 cm with C as centre and another arc of radius 3 cm with A as centre. Mark the point of intersection as D. Join CD and AD.

Question 9.
In parallelogram ABCD, the diagonals AC and BD intersect at O. AC = 6.5 cm, BD = 7 cm and ∠AOB = 100°. Construct the parallelogram.
Solution:

Question 10.
The diagonals of a rhombus are of lengths 16 cm and 12 cm. What is its perimeter?
Solution:

In right angled AOB,
AO = 8 cm, BO = 6 cm, ∠AOB = 90°
AB² = AO² + BO² = 8² + 6²
= 64 + 36 = 100
Side, AB = \(\sqrt{100}\) = 10 cm
Perimeter = 4 × 10 = 40 cm.

Question 11.
Draw the following patterns, a. 6 equal rhombuses :
(a) 6 equal rhombuses:

(b) 3 equal rhombuses

Solution:
(a) Draw a circle of radius 2 cm with centre O. Divide the centre of circle into angles of 60° each. These lines meet the circle at the points A, B, C, D, E and F. Draw arc of 2 cm from A and B to get G. In the same way find H, I, J, K, L. Draw needed part.

(b) Draw square of side 4 cm. Draw rhombuses with side 4 cm and an gle 30° on top and bottom side of the square. Complete the figure.

(c) Angles around the point at which three rhombuses joined together is 120° each. Since one angle of the rhombus is 120°, another angle 60°. Draw three rhombuses with side 4 cm and angle 60°. Complete the figure.

(d) Draw a semicircle of radius 2 cm with centre O. Divide the centre of circle into angles of 45° each .These lines meet the circle at the point A, B, C, D and E. Draw arc of 2 cm from A and B to get F. In the same way find G, H and I. Complete the figure.

(e) Draw a rectangle of length 6 cm and breadth 3 cm. Draw two rhombuses on both side of the rectangle which makes angle 45° and 135° with length and breadth respectively.

Question 12.
In a parallelogram ABCD, find x, y, z from the adjoining figure.

Solution:
ABCD is a parallelogram,
∠C = 45° (Opposite angles are equal)
∠C + Z = 180° (linear pair)
Z = 180° – 45° = 135°
45° + Y = 180° (sum of the angles on the same side is 180°)
Y = 180° – 45° = 135°
sinceY = 135°
X = 135°(Opposite angles are equal)

You can Download Chemical Messages for Homeostasis Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Biology Solutions Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Biology Solutions Chapter 3 Chemical Messages for Homeostasis

Chemical Messages for Homeostasis Text Book Questions and Answers

Chemical Messages For Homeostasis Kerala Syllabus 10th Question 1.
Which are the hormones you know? List them?
Answer:

• Insulin
• Thyroxine
• Oestrogen

Hormones:
The endocrine glands play a vital role in coordinating and controlling life activities. Secretions of endocrine glands are called hormones. There secretions are chemical substances that belong to different categories such as proteins, peptides, steroids, fatty acids, etc. Endocrine glands do not have particular ducts carry hormones to various tissues. Hence they are known as ductless glands. Hormones are transported through blood. As these substances regulate cellular activities, they can be called chemical messages to cell.

Chemical Messages for Homeostasis Hormones In Target Cells

The cell which are acted upon by hormones are called target cells. Only cells having specific receptors can receive a particular hormone. A hormone-receptor complex is formed by the combination of each hormone molecule and its receptor. Following this, enzymes are activated within the cell. As a result, certain changes occur in cellular activities.

After Digestion

Pancreas helps in the digestive process. It functions as an endocrine gland too. It secretes two hormones namely insulin and glucagon.
The beta cells in the Islets of Langerhans glucagon.

Action of insulin and glucagon:

Sslc Biology Chapter 3 Questions And Answers Kerala Syllabus Question 2.
Complete the illustration by including the production of hormones that regulate the level of glucose.
Answer:

Sslc Biology Chapter 3 Notes Kerala Syllabus Question 3.
How is the level of glucose in the blood maintained while fasting? Discuss
Answer:
When the level of glucose increases in blood the cells in the Islets of Langerhans produce insulin which converts the excess glucose into glycogen, protection and lipids.

Diabetes Mellitus

Diabetes is clinically referred to as a condition when the level of glucose before breakfast is above 126 mg/100ml of blood. It is caused either by decreased production of insulin or it$ malfunctioning. Symptoms: Increased appetite and thirst, Frequent urination, Traces glucose in urine Diabetes can be controlled through medicine, diet control and insulin injections.

Biology Class 10 Chapter 3 Notes Kerala Syllabus Question 4.
The increase of glucose in blood is said to be diabetes. Shouldn’t one be more energetic if the glucose level in his/her blood rises? What is your opinion? Write them down in your science diary.
Answer:
No. One be more should not energetic if the glucose level in his or her blood. Persons with diabetes experience loss of body weight, weakening of muscles and tiredness.

Regulation Of Metabolism

The anabolic and catabolic processes taking place in the body are metabolism. The thyroid gland is the endocrine gland contrails the metabolic process.

Functions of thyroxine:

10th Class Biology Chapter 3 Kerala Syllabus Question 5.
How would be body activities be affected if sufficient amount of thyroxine is not produced?
Answer:

• Low energy production
• Bloating of body
• Slowing down of heartbeat
• Loss of appetite, lethargy
• Dry skin

Undersecretion of thyroxine – Hypothyroidism:
The deficiency of thyroxine during the fetal stage or infancy leads to mental retardation and stunted growth. This condition is cretinism. Lack of thyroxine in adults leads to myxoedema.

Symptoms of Hypothyroidism:

• Low metabolic rate
• Sluggishness
• Sleeplessness
• Increase in body weight
• Hypertension
• Oedema

Oversecreation of thyroxine – Hyperthyroidism:
The condition in which all life activities controlled by thyroxine are accelerated due to the excessive production of thyroxine is referred to hyperthyroidism.

Symptoms of Hyperthyroidism:

• High metabolic rate
• Rise in body temperature
• Excessive sweating
• Increased heartbeat
• Sleeplessness
• Weight loss
• Emotional imbalance

Goitre:
Iodine is essential for the production of thyroxine. The production of thyroxine is obstructed in the absence of iodine. In an attempt to produce more thyroxine; the thyroid gland enlarges. This condition is called goitre.

Chemical Messages For Homeostasis Notes Pdf Kerala Syllabus Question 6.
What is the importance of thyroxine in controlling life activities?
Answer:
Thyroxine is a hormone that influences metabolism in our body to a great extent.

Biology 3rd Chapter 10th Class Kerala Syllabus Question 7.
What are the problems caused by excessive production of thyroxine?
Answer:

• Energy production increases and body weight decreases
• Increased heartbeat
• Increased appetite
• Shivering of hands and profuse sweating
• Persistent hyperthyroidism may lead to Graves disease, characterized by exophthalmic goiter.

Sslc Biology Chapter 3 Notes Pdf Kerala Syllabus Question 8.
What are the problems due to thyroxine deficiency?
Answer:
The deficiency of thyroxine retards mental and physical growth of children. This condition is called cretinism. In adults the deficiency of thyroxine results in a disease called myxoedema.

Sslc Biology Chapter 3 Solutions Kerala Syllabus Question 9.
How is iodine related to thyroid gland?
Answer:
Iodine is essential for the production of thyroxine. The production of thyroxine is obstructed in the absence of iodine.

Calcitonin:
It helps in maintaining the level of calcium in blood by depositing excess calcium in bones and by preventing the mixing of calcium with blood, from the bones.

Parathyroid Gland

The parathyroid gland is situated behind the thyroid gland. This gland secretes a hormone called parathormone. The function of this hormone is to raise the level of calcium in blood

Sslc Biology Chapter Wise Questions And Answers Kerala Syllabus Question 10.
Complete the illustration showing the maintenance of the level of calcium in blood by the action of calcitonin and parathormones
Answer:

The hormone only upto youth:
Thymus gland:
The thymus gland is situated just below the sternum. The major function of thymus gland is to control, the activities and maturation of lymphocytes which help to impart immunity. This gland secretes thymosin, which is active during infancy. Hence it is known the ‘youth hormone’.

During Emergencies

These glands are situated above the kidneys. The outer part of the adrenal gland is known as the cortex and inner part is medulla. The adrenal cortex secretes aldosterone, cortisol and sex hormones. Adrenal medulla secretes epinephrine and norepinephrine.

Class 10 Biology Notes Chapter 3 Kerala Syllabus Question 11.
The structure of the adrenal glands and the hormones produced by them are illustrated below. On the basis of the indicators given, discuss and write down the notes in the science diary.

i) Hormones secreted by the adrenal cortex
ii) The function of cortisol
iii) Maintenance of salt-water balance in the body
iv) The function of epinephrine and norepinephrine during emergencies.
Answer:
(i) Aldosterone and sex hormones are the hormones secreted by the adrenal cortex.
ii) The synthesis of glucose from protein and fat controls inflammation and allergy, slows down the action of defense cells.
iii) Aldosterone is the hormone that helps to maintains salt-water balance in the body by restating the lose of Na⁺ ions and by promoting the elimination of K⁺ ions through sweat, urine, etc.
iv) Epinephrine acts along with the sympathetic nervous system during emergencies. Thus we can resist or withdraw ourselves from such situations. Norepinephrine acts along with epinephrine.

Biological Clock

Pineal gland is seen in centre of the brain. It secretes the hormone, melatonin which helps in maintaining the rhythm of our daily activities. The production of melatonin is high at night and low during the day. When the level of melatonin increases we feel sleepy and when it decreases we wake up. This hormone also controls reproductive activities of organisms.

Behind Growth

Pituitary gland is a bibbed gland situated just below the hypothalamus in the brain. The anterior lobe produces tropic hormones which regulate the functions of other glands. The posterior lobe stores the hormones which are produced in the hypothalamus.

Chemical Messages For Homeostasis Questions And Answers Question 12.
The hormones produced by the anterior lobe is listed in table. Analyse-it and complete the following worksheet in the science diary.


Answer:
A – Stimulates the activity of thyroid giand
B – AdrenoCortico Tropic Hormone(ACTH)
C – Gonado Tropic Hormone(GTH)
D – Stimulates the activity of ovaries
E – Production of milk
F – Growth Hormone (GH) or Somatotropic Hormone (STH)

Hss Live Guru 10th Biology Kerala Syllabus Question 13.
How the variation in the production of somatotropin affects growth.
Answer:
Somatotropin promotes growth of the body during its growth phase. If the production of this hormone increases during the growth phase, it leads to the excessive growth of the body. This condition is gigantism. It causes another stage called dwarfism when its production decreases during the growth phase. Acromegaly is the condition caused by the excessive production of somatotropin after the growth phase. It is characterized by the growth of the bones on face, jaws and fingers.
The Posterior Lobe of pituitary gland – A storage center

The hormones oxytocin and vasopressin, which are secreted from the posterior lobe of the pituitary are actually produced in the neuro-secretory cells of the hypothalamus. The posterior lobe stores these two hormones and Releases them into blood when required.

Hss Live Guru 10 Biology Kerala Syllabus Question 14.
Observe the table and write down your inferences in the science diary.

Answer:
Oxytocin and vasopressin are secreted from the hypothalamus and stored in the posterior lobe of pituitary gland. Through the connecting nerve fibers they are transported to pituitary gland. Oxytocin facilitates childbirth by stimulating the contraction of smooth muscles in the uterine wall and facilitates lactation. Vasopressin helps in the reabsorption of water in the kidneys.

Biology Chapter 3 Class 10 Kerala Syllabus Question 15.
Observe illustration given below which shows the action of vasopressin in kidneys. Based on the indicators given discuss and write a note in science diary.

i) The functions of vasopressin in kidneys
ii) The reason for excessive production of urine during the rainy season
iii) The role of vasopressin in preventing loss of water from the body.
iv) Diabetes insipidus
Answer:
i) The hormone vasopressin regulator reabsorption of water and minerals from the glomerular filtrate.. When there is a reduction in the amount of water in blood, the production of vasopression increased. It accelerates the rate of reabsorption of water. When the quantity of water in blood increases the production of vasopressin decreases. As a result, the rate of reabsorption is also reduced.

ii) During rainy season, sweat production reduces. So the water loss through sweat is decreased. It wingcase the quantity of water in our body Comparatively high. Such situations demand elimination of excess water through urine.

iii) The production of vasopressin increases when there is a need to reduce loss of water through urine. As a result of this, more water is reabsorbed to the blood from kidneys, Thereby the loss of water through urine is reduced and regulate water loss in our body.

iv) The rate of resorption of water in the kidney is decreased when there is no sufficient amount of vasopressin. Hence excess amount of urine is excreted. This condition is called diabetes insipidus Symptoms include frequent urination.

Behind Sexual Characteristics

Testes and ovary, the male and female sex organs respectively, secrete different types of hormones.

Prepare a table by including hormone, centre of production and function:

Prime Controller

Oxytocin and vasopressin are secreted by the hypothalamus. In addition to this hypothalamus controls the pituitary gland by secreting a variety of releasing hormone are inhibitory hormones.

Biology Chapter 3 Class 10 Notes Kerala Syllabus  Question 16.
Observe the illustration given below on the functions of releasing hormones and inhibitory hormones. On the basis of indicatiors discuss and writes it down in the science diary.

i) Action of releasing hormone
ii) Influence of tropic hormones in different glands.
iii) Action of inhibitory hormones
Answer:
i) Stimulates the anterior lobe of the pituitary and secretes tropic hormones.
ii) Tropic hormones stimulate the production of hormones of certain other important glands.
iii) Inhibits the production of tropic hormones in the anterior lobe of the pituitary gland.

Chemical Messages For Communication

Pheromones: Pheromones are chemical substances that are secreted in trace amounts to the surrounding in order to facilitate communication among organisms. Pheromones help in attracting mates, to inform the availability of food, to determine the path of travel and to inform about dangers.
Eg:- Musk in the nusk deer, civet on in civet cat, Bombycol in female silkworm.

Plant Hormones

There are certain chemical substances in plant cells to control and co-ordinate life activities. These are also called plant growth regulators.

Question 17.
Observe the illustration, which show plant hormones and their functions and complete the following table suitably.


Answer:
a) Cell growth, cell elongation, fruit formation
b) Controls the dormancy of embryo in the seeds, dropping to leaves and fruits wilting of leaves, flowering, etc.
c) Gibberellins
d) Promotes cell division cell growth
e) Ethylene.

Artifical Plant Hormones

Auxins: NaphtheleneAceticAcid (NAA), Indol Butyric Acid (IBA), etc., are used for sprouting and the prevention of dropping of premature fruits. 2.4-D (2, 4-Dichloro phenoxy acetic acid) is used as a weedicide.

Gibberellins: It is used for increasing fruit size in grapes and apple and also for preventing ripening of fruits to assist in marketing.

Abscisic acid: As it accelerates the dropping of fruit, it is used for harvesting fruits at the same time.

Ethylene: Ethylene is used for the flowering of pineapple plants at a time and for the ripening of tomato, lemon, orange, etc. Ethyphon, a chemical which is available in liquid form gets transformed into ethylene when used in rubber trees, and it increases the production of latex.

Chemical Messages for Homeostasis Let Us Assess

Question 1.
Identifying the word-pair relationship and fill in the blank.
Thyroxine: Thyroid gland
Epinephrine:…………..
Answer:
Adrenal gland

Question 2.
Analyze the information given in the box and answer the questions.
X- The production of this hormone is more in night and less in day time.
Y – Hormones from the adrenal gland work along with the sympathetic system.
(a) Identify and name the hormone ‘X’ and its gland.
(b) Identify the hormones indicated as ‘Y’.
Answer:
a) Melatonin pineal gland
b) Epinephrine (Adrenaline)
c) Norepinephrine (Noradrenline)

Question 3.
Analyse the illustration and complete the table appropriately.


Answer:

Question 4.
The hormone that helps in the reabsorption of water in the kidneys.
a) TSH
b) ACTH
c) ADH
d) GTH
Answer:
ADH

Chemical Messages for Homeostasis Extended Activities

1. Conduct a seminar on the topic – The role of the Endocrine system in maintaining homeostasis’.
Main points:-

• Situations which lead to change in homeostasis
• How is homeostasis reinstated
• Harmonious co-existence

2. Conduct a debate on ‘Use of artificial plant hormones – problems and possibilities’.

3. Collect information about novel laboratory tests related to diagnosis of diabetes and conduct an exhibition on World Diabetes Day.

Chemical Messages for Homeostasis More Questions and Answers

Question 1.
Correct the sentence if it is wrong
1. Endocrine glands are ductless glands
2. The alpha cells in the Islets of Langerhans secrete insulin.
3. Aldosterone slows down the action of defense cells.
4. Anti Diuretic hormone helps in the reabsorption of water in the kidneys.
5. Hypothalamus secretes inhibitory hormones which stimulate the anterior lobe of the pituitary gland.
Answer:
1. Endocrine glands are ductless glands
2. The beta cells in the islets of Langerhans secrete insulin.
3. Cortisol slows down the action of defense cells.
4. Anti Diuretic hormone helps in the reabsorption of water in the kidneys.
5. Hypothalamus secretes releasing hormones that stimulate the anterior lobe of the pituitary gland.

Question 2.
Endocrine glands are called ductless glands. Why?
Answer:
Endocrine glands do not have particular ducts to carry hormones to various tissues. Hence they are called ductless glands.

Question 3.
Name the hormone-producing centers situated in the brain?
Answer:
Hypothalamus, pituitary, Pineal

Question 4.
The gland which is active only during infancy?
Answer:
Thymus

Question 5.
Though hormones reach every part of the body through the blood, all hormones do not act upon all cell. Explain the reason.
Answer:|
The cell which are acted upon by hormones are called target cells. Only cells having specific receptors can receive a particular hormone. A hormone-receptor complex is formed by the combination of each hormone molecule and its receptor. Following this, enzymes are activated withfn the cell. As a result, certain changes occur in cellular activities.

Question 6.
Name the digestive gland which is also functioning as an endocrine gland?
Answer:
Pancreas

Question 7.
What is the normal level of glucose in blood? How is this level maintained?
Answer:
The normal level of glucose is 70-110 mg/100 ml blood. The level of glucose in blood is maintained by the combined action of insulin and glucagon of the Islets of Langerhans tissues of the pancreas. Insulin, released from the beta cells of Islets of Langerhans, helps to reduce blood sugar by accelerating the process of cellular uptake of glucose and Conversion of glucose in to glycogen. When blood glucose level falls, glucagon, released from the alpha cells of Islets of Langerhans, converts glycogen to glucose and synthesizes glucose from amino acids.

Question 8.
Suppose a person is fasting in a day and takes heavy food on the very next day. How is the level of glucose in his body is maintained in these two days?
Answer:
While fasting glucagon converts glycogen or amino acids into glucose. When taking heavy food insulin enhances cellular uptakes of glucose and converts glucose into glycogen.

Question 9.
Diabetic patients frequently take insulin injections. Give reason?
Answer:
Insulin is helpful to reduce the excess glucose in the blood and to maintain its normal level

Question 10.
If the level of glucose increases one feels hunger, thirsty and fatigue instead of becoming energetic. Give reason?
Answer:
Increasing the level of glucose in blood adversely affects the normal functioning of the cells.

Question 11.
A doctor advised one of his patients to use iodized salt and to include more leafy vegetables and marine items in his diet. What should be reason for this recommendation?
Answer:
To prevent goitre. Deficiency of iodine may cause Goitre, a disorder affects on thyroid gland.

Question 12.
Under secretion of thyroxine: Hypothyroidism
Over secretion of thyroxine: ……………..?
Answer:
Hyperthyroidism

Question 13.
Overproduction, as well as underproduction of the hormone thyroxine, may lead to disorders’. Substantiate.
Answer:
Deficiency of thyroxine (Hypothyroidism) leads to cretinism in infants and myxoedema in adults. Excess production of thyroxine (Hyperthyroidism) leads to a condition, known as the Graves disease.

Question 14.
Persistent hyperthyroidism may leads to ……………. disease characterized by bulging of the eye balls.
Answer:
Graves disease

Question 15.
What is the normal level of calcium in the blood? How is this level maintained?
Answer:
9-11 mg/100 ml blood.
When the level of calcium in blood increases, thyroid gland secretes a hormone named calcitonin. It lowers the level of calcium in blood by depositing excess calcium in bones and by preventing the mixing of calcium with blood form the bones. When the level of calcium in blood decreases, parathyroid gland secretes parathormone. It increases blood calcium by reabsorbing it from the kidneys and also preventing the deposition of calcium in bones.

Question 16.
Complete the flow chart

Answer:
A. Medulla
B. Aldosterone
C. Cortisol
D. Norepinephrine

Question 17.
Overproduction of parathormone can weaken the bones. Why?
Answer:
The hormone, parathormone prevents the deposition of calcium in bones resulting its weakening.

Question 18.
The hormone which can be used to prevent allergy and inflammation? Can this hormone be given to diabetic patients? Why?
Answer:
Cortisol of adrenal gland. It cannot be given to diabetic patients as it increases the level of glucose in blood.

Question 19.
The pineal gland is known as the ‘biological clock’ in the body. Why?
Answer:
Melatonin, the secretion of the pineal gland helps to . maintain rhythm of our daily activities. Therefore pineal gland is called as the biological clock.

Qn. 20
What are the hormones of hypothalamus stored in the posterior lobe of pituitary gland? Mention its functions.
Answer:
Oxytocin – Facilitates childbirth by stimulating the contraction of smooth muscles in the uterine wall and also facilitates lactation Vasopressin (Anti Diuretic Hormone) – Helps in the reabsorption of water in the kidneys.

Question 21.
Give reasons.
Some times certain pregnant women need to take oxytocin injection.
Answer:
Oxytocin facilitates childbirth by stimulating the contraction of smooth muscles in the uterine wall. It also facilitates lactation.

Question 22.
Point out the functions of releasing hormones and inhibitory hormones.
Answer:
Releasing Hormones: Stimulate the anterior lobe of the pituitary to secretes tropic hormones and other hormones.
Inhibitory Hormones: Inhibit the production of tropic hormones and other hormones from the anterior lobe of the pituitary gland.

Question 23.
What is the reason behind the difference in the quantity of urine during summer and rainy season?
Answer:
The production of vasopressin is high during summer season where water loss is excessive through sweat. But its production is less during winter and rainy seasons and there is difference in the quantity of urine during summer and rainy seasons.

Question 24.
Why do Vasopressin is known as antidiuretic hormone (ADH)?
Answer:
Because vasopressin retains the quantity of water by inducing the kidneys to reabsorb it.

Question 25.
Complete the following table related with the hormonal functions of our sex organs.

Answer:

Question 26.
Identify the hormone defects concern with the following hints.
a) Insulin injection
b) Treatment using thyroxine
c) Food and medicine containing calcium
d) Seafood, vegetable and iodized salt.
Answer:
a) Diabetes
b) Myxoedema
c) Osteoporosis
d) Goiter

Question 27.
How is homeostasis of the body maintained?
Answer:
Homeostasis of the body maintained by the combined action of the quick nervous system and the slow endocrine system.

Question 28.
How are pheromones useful to animals?
Answer:
Pheromones help in attracting mates, to inform the availability of food, to determine the path of travel and to inform about dangers.

Question 29.
A farmer says pest control is made possible using pheromones. Can you say how?
Answer:
Artificial pheromones are used for pest control in agricultural field.

Question 30.
Identify the plant hormone that performs the following functions.
a) flowering and growth of leaves
b) ripening of fruits
c) dropping of leaves and fruits
d) growth of terminal bud.
Answer:
a) Gibberellin
b) Ethylene
c) Abscisic acid/ ethylene in excess amount.
d) Auxin

Question 31.
Site examples of situations where artificial plant hormones are applied widely.
Answer:
Ethylene is used for the flowering of pineapple plants at a time and for the ripening of tomato, lemon, orange, etc.
Ethyphon, a chemical which is available in liquid form gets transformed into ethylene when used in rubber trees, and it increases the production of latex.

Auxins: Naphthalene Acetic Acid (NAA), Indol Butyric Acid (IBA) etc. are used for sprouting and the prevention of dropping of premature fruits. 2,4- D (2, 4-Dichloro phenoxy acetic acid) is used as a weedicide.

Gibberellins: Used for increasing fruit size in grapes and apple and also for preventing ripening of fruits to assist in marketing.

Abscisic acid: As it accelerates the dropping of fruit, it is used for harvesting fruits at the same time.

Question 32.
The following figure shows the relationship of hypothalamus with an endocrine gland. (Model 2016)

a) Write down the name of endocrine gland marked as X
b) Write down the name of hormone produced the A and B.
c) Mention the functions of hormones produced the B.
Answer:
Answer:
a) Pituitary gland
b) A – Tropic hormone
B – Oxytocin and vasopressin
c) Oxytocin helps to contraction of smooth muscles and vasopressin helps in the reabsorption of water in the kidneys.

Question 33.
Artificial hormones should be handled with care. What is your opinion?
Answer:
This statement is correct. Though artificial hormones are useful they should be handled with care as they are chemicals, which may cause health and environmental issues.

Question 34.
…………….. is used for increasing fruit size in grapes and apple.
Answer:
Gibberellins

Question 35.
………….. is a plant hormone, used for harvesting fruits in a field at the same time.
Answer:
Abscisic acid

Question 36.
The quantity of urine excreted by a person in different seasons is given below. Analyse-it and answer the following questions. (Model 2016)

a) Write down the climate B and C
b) Analyse the difference shown in B and C and write down its reasons.
c) Which hormone is responsible for the excretion of excess water through urine.
Answer:
a) B – Rainy season or winter season, C – Summer season
b) In rainy season production of vasopressin is less it decreases the reabsorption of water in the kidneys. So raises the quantity of urine. In summer season production of vasopressin increases. It increases the reabsorption of water in kidneys and lowers the quantity of urine.
c) ADH or vasopressin

Question 37.
Given below is the blood test result of a person. Analyze the result and answer the following questions? (Model 2016)
Glucose – 200mg/100ml
Calcium -11 mg/100ml
a) Name the disease of the man mentioned in the test report.
b) Write down the name of hormone which related to this disease.
c) What is the cause of this disease?
Answer:
a) Diabetes mellitus
b) Insulin
c) It is caused either by the decreased production of insulin or its malfunctioning

Question 38.

a) Complete the table based on the hormone somatotropin (Model 2014)
b) This hormone is not a tropic hormone. Why?
Answer:
a) (i) X-dwarfism 1) become dwarfs due to stunted growth of bones
ii) Y – gigantism 2) Growing tall with a heavy body
iii) Z-acromegaly 3) enlargement of internal organs and thickening of bones, especially in hands feet and face.

b) Somatotropin does not induce any other endocrine gland to release its hormone

Question 39.
The quantity of urine excreted by a person in different seasons is given below. Analyse-it and answer the following questions. (Model 2014)

a) Which is the coldest season?
b) Which hormone is responsible for the variation in quantity of urine?
c) How this hormone regulates water level of the body.
Answer:
a) Season 3
b) ADH/Vasopressin
c) This hormone promotes reabsorption of water from renal tubules when normal level of water in blood decreases. The rate of reabsorption of water in the kidney is decreased when there is no sufficient amount of vasopressin.

Question 40.
Some hormones are given below. Make them into 4 pairs. Give reasons for pairing. (Model 2014)


basis pairing:- Products of same gland

Question 41.
Observe the chart (March 2013)

Write down the climate A and B
Answer:
A-Summer season
B – Rainy season or winter

Question 42.
Changes in the number of hormones produced will affect our bodily activities. Write down the changes occur in our body by the increase and decrease of the hormones given below.
a) Parathormone
b) Vasopressin
Answer:
a) Increase of parathormone – Bones fragile stones in urinary tract, high blood calcium
Decrease of parathormone – Blood calcium level decrease and it leads to tetany

b) Vasopressin
Production of vasopressin increases It accelerate the rate of reabsorption of water from kidney. So the loss of water through urine is reduced.
Production of vasopressin decreased The rate of reabsorption is reduced and more water discharged out through urine.

Question 43.
Rearrange B, C and D according to the data given A

Answer:

Question 44.
“It is now that I understand why the cock crows early in the morning every day”. Anu said this during a classroom discussion on the rhythm of physiological activities.
a) Which is the hormone that regulates such activities?
b) Which gland secretes this hormone.
c) Write down more examples for such activities
Answer:
a) Melatonin
b) Pineal gland
c) It regulates the rhythm of life, reproductive activities of organisms with definite reproductive periods.

Question 45.
Fill up the blanks (Model 2012)

Answer:
a) Thyroxine
b) Thyroid
c) Insulin
d) Diabetes
e) Pituitory gland
f) Dwarfism

Question 46.
Fill up the blanks

Answer:
a) – Gibberellin
b) – Helps in the ripening of fruits
c) – Abscisic acid

Chemical Messages for Homeostasis SCERT Questions and Answers

Question 1.
Observe the illustration given below and explain how hormones act in target cells.

Answer:
The cell which are acted upon by hormones are called target cells. Only cells having specific receptors can receive a particular hormone. A hormone-receptor complex is formed by the . combination of each hormone molecule and its receptor. Following this, enzymes are activated within the cell. As a result, certain changes occur in cellular activities.

Question 2.
Some statements relate to endocrine system are given below. (Question Pool 2017)
A. Hormones are the secretions of endocrine glands.
B. Hormones are transported through lymph.
C. Hormones are transported through blood.
D. All the harmonies produced by the endocrine glands are proteins.
a) Choose the correct statement.
b) Imagine that particular hormone is not entering a particular cell. What may be the reason? Formulate two hypotheses.
Answer:
a) A, C
c) Receptors of that hormone in not in the cell

Question 3.
Examine the graph indicating the blood glucose level of different individuals before breakfast. (Question Pool 2017)

a) Which individual is affected by diabetes mellitus?
b) Write two actions of insulin to prevent the rise in the level of glucose in blood.
c) Why do people having diabetes mellitus experience extreme fatigue?
Answer:
a) (B)
b) 1. Enhances the entry glucose into the cell.
2. Converts glucose to glycogen in liver and muscles.
c) Sufficient quantity of glucose i not reaching the cell. Energy production decreases. Excess amount of glucose is eliminated through urine.

Question 4.
Case sheets of two patients are given below. Analyze them and answer the questions. (Question Pool 2017)

a) Which are the diseases whose symptoms are indicated above?’
b) Write the reasons for the diseases.
Answer:
a) Case -1 cretinism;
Case – 2 graves disease

b) Case -1 reasons
Deficiency of thyroxine during foetal stage and infancy.
Case-2 reasons
1. Persistent hyperthyroidism
2. Excessive production of thyroxine.

Question 5.
Analyse the table given below. Rearrange column Band C according to the indicators in Column A. (Question Pool 2017)

Answer:
1 – (b) – (r)
2 – (c) – (p)
3 – (a) – (q)

Question 6.
Honey bees and termites live in colonies. (Question Pool 2017)
a) Name the chemical substance which helps them to live together.
b) Mention two uses of these chemical substances.
Answer:
a) Pheromones
b) 1. attracting mates
2. informing availability of food
3. determining the path of travel
4. informing the dangers

Question 7.
Observe the diagram and answer the questions. (Question Pool 2017)

a) Which endocrine gland does ‘X’ indicate?
b) Which are the two hormones produced by the gland to control the physical activities with the sympathetic system?
Answer:
a) Adrenal gland
b) Epinephrine, Norepinephrine

Question 8.
Maintenance of the level of calcium in the blood is illustrated below. Analyse-it and answer the following questions. (Question Pool 2017)

a) Name the hormone indicated as X’.
b) Which gland produces the hormone ‘Y’?
c) Write another activity performed by ‘X’ to raise the level of calcium in blood.
Answer:
a) Parathormone
b) Thyroid gland
c) Helps in the reabsorption of calcium from kidneys.

Question 9. (
Observe the diagram of the endocrine gland given below and answer the question. (Question Pool 2017)

a) Name the part indicated as A and B.
b) Name the hormones synthesized by A. Explain their action.
Answer:
a) A Medulla
B Cortex
b) Epinephrine, Norepinephrine
Epinephrine – Helps to tide over emergency situations
Norepinephrine – acts along with epinephrine

Question 10.
An individual loses large quantities of water through urine (Question Pool 2017)
a) Which could be the disease?
b) Analyze the conditions that lead to this disease.
Answer:
a) Diabetes insipidus
b) ADH is synthesized by hypothalamus.
ADH increases the reabsorption of water into the kidney.
Synthesis of ADH decreases.

Question 11.

a) Identify X and Y. (Question Pool 2017)
b) What is the function of ‘Y’?
Answer:
a) Portal vein – X
Posterior lobe of pituitary – Y
b) Stores the hormones vasopressin and oxytocin synthesized by hypothalamus and releases them into blood when required.

Question 12.
Given in the table below is to growth hormone. Complete the table suitably. (Question Pool 2017)

Answer:
a) dwarfism
b) Excessive production of growth hormone during the growth phase.
c) Excessive production of somatotropin after the growth phase.
d) Growth of the bones on face, jaws and fingers.

Question 13.
Given below is a doctor’s comment at a seminar conducted as part of Diabetic day.
“In diabetic patients, the blood glucose level before breakfast is above 126mg/100ml.
Analyse the statement and enlist the reasons.
Answer:

• Decreased production of insulin
• Malfunctioning of insuline
• Destruction of Beta Cells
• Inactive insulin

Question 14.
Given below are a few statements related to hormones. Pick out the correct ones. (Question Pool 2017)
a) Estrogen helps to maintain embryo in the uterus.
b)Progesterone facilitates childbirth.
c) Prolactin helps in the production of milk.
d) Oxytocin n faci itates I a citation.
Answer:
c, d

Question 15.
Analyze the statements given below and write the reason. (Question Pool 2017)
a) Oxytocin is injected in pregnant women during childbirth, (delivery)
b) Feels sleepy during night, wakeup when day breaks.
Answer:
a) Facilitates childbirth by stimulating the contraction of smooth muscles in the uterine wall.
b) When the level of melatonin increases at night, we feel sleepy,
We wake up when the level of melatonin decreases during the day.

Question 16.
Analyse the table and identify the correct pair. (Question Pool 2017)

Answer:
a) Somatotropin decreases during growth phase – dwarfism

Question 17.
Observe the table, re-arrange column Band C according to column A.

Answer:
1 – (b) – (S)
2 – (d) – (P)
3 – (a) – (Q)

Question 18.
A farmer named Balan cultivated oranges in his orchard. Now the trees are full of oranges. The price of oranges is Rs. 80/kg. (Question Pool 2017)
A) This farmer wants to harvest all fruits together.
B) Ripen them together.
a) Suggest two artificial plant hormones to satisfy the A, B needs of the farmer.
b) Uncontrolled use of plant hormones must be controlled. Evaluate this statement.
Answer:
A) a) A – Abscisic acid
B – Ethylene
b) Though artificial hormones are useful they should be handled with care as they are chemicals. Uncontrolled use of it may cause health and environmental issues.

Question 19.
Analyse the indicators and answer the question given below. (Question Pool 2017)
Indicators
Accelerates the growth and development of the brain in the foetal stage and infancy.
a) Which hormone are the indicators about?
b) Construct a flow chart relating the action of hypothalamus and pituitary in the synthesis of this hormone.
Answer:
a) Thyroxine
b)

Question 20.
Artificial plant hormones are used extensively in the agricultural sector. (Question Pool 2017)
Write the name and function of two artificial plants. hormones belonging to the category, auxin.
Answer:
NAA- Sprouting, prevention of premature fall of fruits.
IBA – -do-
2, 4 – D – Weedicide

Question 21.
Artificial plant hormones are used extensively in the agricultural sector. Write a short note on the advantages and disadvantages of these (Question Pool 2017)
Answer:
Advantages:

• Sprouting
• Prevents premature fall of fruits
• Medicinal action
• Increases size of fruits
• Ripening of fruits
• Increases production of latex in rubber trees
• Harvesting fruits at the same time.
• Prevents early ripening of fruits

Disadvantages:

• Environmental issues
• Health issues

Question 22.
Choose the correct statement related to pheromones from those given below. (Question Pool 2017)
a) Pheromones are chemical substances secreted inside the body for communication.
b) This is the message to attract mates, determining the path of travel, etc.
c) Musk in the civet cat is a pheromone.
d) Bombycol is the pheromone secreted by the female silkworm.
Answer:
b, d

Question 23.
Analyze the box given below and complete the table suitably. (Question Pool 2017)

Answer:

Question 24.
Indicators related to the endocrine glands are given below. Analyze them and answer the questions. (Question Pool 2017)
1. Situated just below the sternum.
2. Active during infancy.
But constricts at puberty.
a) Name this endocrine gland?
b) Which is the hormone synthesized by this gland?
c) Write the function of this hormone.
Answer:
a) Thymus gland
b) Thymosin
c) Controls the activities and maturation of lymphocytes which help to impart immunity.

Question 25.
Given below is the illustration showing the hormones synthesized by the anterior lobe of the pituitary gland. Complete it Suitably. (Question Pool 2017)

Answer:
a) Stimulates thyroid gland
b) ACTH
c) Production of milk
d) Enhances growth

Question 26.
Teacher: The TSH hormone synthesized by the pituitary gland acts on the thyroid gland. It is transported to the thyroid gland through blood. All hormones are transported like this through blood. (Question Pool 2017)
Amirtu: Can all the hormones synthesized by the pituitary gland reach the thyroid gland and act there? What is your answer for Ammu’s doubt?
Answer:
Receptors to receive other hormones synthesized by the pituitary gland are absent in the thyroid gland.

Question 27.
Plant hormones and their functions are given in two boxes below. Pair them suitably (Question Pool 2017)

Answer:
(a) – (iv)
(b) – (ii)
(c) – (i)
(d) – (iii)

Question 28.
The problems faced by two farmers are below. Suggest two artificial plant hormones to overcome this. (Question Pool 2017)
Satheesh: Excessive growth of weeds in the agricultural field.
Saneesh: Premature fall of fruit in the mango orchard.
Answer:
Satheesh: 2, 4- D
Saneesh: NAA /IBA

Question 29.
Observe the illustration given below and answer the questions. (Question Pool 2017)

a) Write the names of the hormones ‘X’ and Y\
b) Mention two actions that take place in A and B.
c) Name the gland which synthesises X and Y.
Answer:
a) X-Insulin; Y-Glucagon
b) A-Converts glycogen to glucose
B – Converts glucose to glycogen
c) Pancreas

Question 30.
Identify the word pair relationship and fill in the blanks. (Question Pool 2017)
a) Civet cat:………………
Silkworm: Bombycol
b) Breaks up stored food: Gibberellins
helps in fruit ripening: ……………….
c) Vasopressin: Diabetes insipidus
Insulin: ………………
d) Dwarfism: somatotropin
Myxoedema: ………………..
Answer:
a) Civetone
b) Ethylene
c) Diabetes mellitus
d) Thyroxine

Question 31.
Pick the odd one out. Write the common features of the others. (Question Pool 2017)
a) Increases metabolic rate, increases energy production regulates growth in children, promotes production of milk.
b) Goitre, Acromegaly, Hypothyroidism, Hyperthyroidism.
c) Cortisol, Vasopressin, Epinephrine, Norepinephrine.
d) Ethylene, Cytokinin, Auxin, Pheromones.
Answer:
a) Increases the production of milk: all others are the activities of thyroxine.
b) Acromegaly: All others are disorders/diseases, related to thyroid gland
c) Vasopressin: All others are hormones of adrenal gland
d) Pheromones: All others plant hormones

Question 32.
Choose the correct statement. (Question Pool 2017)
a) Synthesis of vasopressin increases if the level of water in the blood increases.
b) Thyroid-stimulating hormone stimulates the activity of the thyroid gland.
c) Synthesis of insulin increases if the blood glucose level rises.
d) Deficiency of thyroxine causes cretinism in adults.
Answer:
b, c

Question 33.
Maintenance of the level of calcium in blood is illustrated below. Analyse-it and answer the questions. (Question Pool 2017)

a) Which are the hormones indicated as ‘X’, ‘Y’?
b) Write the actions performed by ‘X’ in the bone and ‘Y’ in the kidney.
c) How does the deficiency of ‘Y’ affect the process of blood clotting?
Answer:
a) X – Calcitonin: Y – Parathormone
b) Action of X : Deposits excess calcium in bones.
Action of Y : Reabsorbs calcium into the blood in the kidney.
c) Deficiency of Y decreases the level of calcium in blood.
As calcium is required for blood clotting, the clotting process becomes slow.

Question 34.
Make suitable word pairs from the words given below. (Orukkam – 2017)

Answer:
Vasopressin- Diabetes insipidus
Dwarfism – Somatotropin
Cretinism – Thyroxin

Question 35.
Complete the illustration using the words given in the box. (Orukkam – 2017)

Answer:

Question 36.
Observe the illustration and answer the following questions? (Orukkam – 2017)

a) Identify the parts marked as A , B and C ?
b) Name the hormones indicated as 1, 2, 3, 4 and 5?
c) What are the functions of the hormones Oxytocin and Prolactin?
d) What are the abnormalities caused by the difference in the production rate of the hormone marked as 1?
Answer:
a) A-Anterior lobe of pituitary
B – posterior lobe of pituitary
C – Hypothalamus

b) 1 – Somatotropin /growth hormone
2 – Vasopressin /ADH
3 – Tropic hormones
4, 5 – TSH/ACTH

c) Oxytocin facilitates childbirth by the contrac¬tion of smooth muscles in the uterine wall and also facilitates lactation. Vasopressin helps in the reabsorption of water in the kidney to prevent water loss through urine.

d) Dwarfism, Gigantism and Acromegaly

Question 37.
Identify the word pair relationship and complete the following. (Orukkam – 2017)
a) Alpha cells: Glucagon
Beta Cells: …………….
b) Prolactin: Production of milk
……………….: Facilitate lactation
c) Parathyroid: Parathormone
Thyroid
Answer:
a) Insulin
b) Oxytocin
c) Calcitonin

Question 38.
All hormones are being transported through the blood and reach all cells of the body, but all hormones are not functioning in all cells. Why? (Orukkam – 2017)
Answer:
Each hormone act only its target tissue, where spe¬cific receptors present to accept the same hormone.

Question 39.
The increased or decreased level of thyroxin may disrupt the homeostasis of the body. Explain? (Orukkam – 2017)
Answer;
Due to hypothyroidism (eg. cretinism) low metabolic rate, sluggishness, sleeplessness, increase in body weight, hypertension, oedema, etc.
Due to hyperthyroidism (eg. Graves disease) high metabolic rate, increased heartbeat, rise in body tem¬perature, sweating, sleeplessness, loss of weight, emotional imbalance.

Question 40.
Bees and termites are maintaining the colony life by using some chemical substances as chemical messages. (Orukkam – 2017)
a) What are these chemical substances?
b) Write the other uses of these chemical substances?
c) Give other examples for these chemical substances?
Answer:
a) Pheromones
b) To attract mates, to inform about food or dangers, to live in colonies, to follow one afterthe other.
c) Civetone in civet cat, Bombycol in female silkworm moth.

You can Download Money Maths Questions and Answers, Activity, Notes, Kerala Syllabus 8th Standard Maths Solutions Chapter 5 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Maths Solutions Chapter 5 Money Maths

Money Maths Text Book Questions and Answers

Textbook Page No. 90

Money Maths Class 8 Kerala Syllabus Chapter 5 Question 1.
Sandeep deposited Rs 25000 in a bank which pays 8% interest compounded annually. How much would he get back after two years?
Solution:
Interest in the first year
= 25000 × \(\frac{8}{100}\)
=Rs 2000
Principal in the second year
= 25000 + 2000
= 27000
Interest in the second year
= 27000 × \(\frac{8}{100}\)
= 2160
Total amount gets back at the end of 2 year
= 27000 + 2160
= 29160

Money Math Class 8 Kerala Syllabus Chapter 5 Question 2.
Thomas took out loan of 15000 rupees from a bank which charges 12% interest, compounded annually. After 2 years, he paid back 10000 rupees. To settle the loan, howmuch should he pay at end of three years?
Solution:
Amount borrowed = 15000
Interest for the first year
= 15000 × \(\frac{12}{100}\) = 1800
Loan for the second year
= 15000 + 1800 = Rs.16800
Interest for the second year
= 16800 × \(\frac{12}{100}\) = 2016
Loan for the end of second year
= 16800 + 2016 = 18816
Amount payback at the end of second year = 10000
Loan for the third year
= 18816 – 10000 = 8816
Interest for the 3rd year
= 8816 × \(\frac{12}{100}\) = 1057.92
Amount to be repair at the end of third year
= 8816 + 1057.92 = Rs.9873.92

Money Maths Class 8 Solutions Kerala Syllabus Chapter 5 Question 3.
Rs 200 was got as interest from a bank for 2 years at the rate 5%. Compute the compound interest for the same principal for 2 years at the same rate of interest.
Solution:
Interest for 2 years = Rs 200
Interest for one year = Rs 100
Excess amount in the 2nd year as the compound interest
= 100 × \(\frac{5}{100}\) = 5
Compound interest for two years = Rs. 205

Textbook Page No. 92

Class 8 Money Maths Kerala Syllabus Chapter 5 Question 1.
Anas deposited Rs 20000 in a bank where compound interest in computed 6% annually. How much amount he will get at the end of 3rd year?
Solution:
Amount Anas gets at the end of third year

Money Maths Class 8 Questions And Answers Kerala Syllabus Question 2.
Diya Deposited Rs. 8000 in a bank where compound interest is computed annually at 10%. Rs. 5000 was withdrawn by her at the end of 2 years. How much amount will he there in the account of Diya after 1 more year?
Solution:
Amount deposited = 8000
Rate of interest = 10%
Amount in the account of Diya at the end of 2 years
= 8000 × (1 + \(\frac{10}{100}\))²
= 8000 × \(\frac{110}{100}\) × \(\frac{110}{100}\) = Rs.9680
Amount withdrawn =Rs. 5000
Principal for the 3rd year
= 9680 – 5000 = Rs.4680
Amount she get at the end of 3 rd year
= 4680 (1 + \(\frac{10}{100}\))
= 4680 × \(\frac{110}{100}\)
= Rs. 5148

Money Maths Class 8 Pdf Kerala Syllabus Chapter 5 Question 3.
Varun borrowed rupees 25000 from a bank where compound interest in computed at 11% annually. He repaid Rs 10000 at the end of 2 years. How much amount he has to repay after one more year?
Solution:
Amount borrowed = 25000
Rate of interest = 11%
Total liability after 2 year

Amount he repaired = Rs.10000
Principal for the third year
= 30802.50 – 10000
= Rs.20802.50
Amount he has to repay at the end of third year
= 20802.50 × (1 + \(\frac{11}{100}\))
= 20802.50 × \(\frac{111}{100}\)
= Rs. 23090.78

Textbook Page No. 93

Hss Live Guru 8 Maths Kerala Syllabus Chapter 5 Question 1.
Arun deposited Rs 5000 in a bank where compound of interest is computed half yearly. Mohan deposited RS 5000 in a bank where compound interest is computed quarterly she rate of interest at both the banks is 6%. Money was withdrawn by both of them after an year. How much amount Mohan got more than Arun.
Solution:
Amount deposited by Arun = 5000.
Rate of interest = 6%
Amount Arun gets after an year
= 5000 × (1 + \(\frac{8}{100}\))²
= 5000 × \(\frac{103}{100}\) × \(\frac{103}{100}\)
= 5304.50
Amount deposited by Mohan
= Rs 5000
Rate of interest = 6%
Amount Mohan gets after 1 year

= Rs 5306.82
Amount Mohan gets more than
Arun = 5306.82 – 5304.50
= Rs 2.32

Kerala Syllabus 8th Standard Maths Notes Pdf Chapter 5 Question 2.
Rs. 16000 was borrowed by a man from a bank where compound interest is computed quarterly. Annual interest rate is 10%. How much he has to pay after 9 months to clear his liabilities?
Solution:
Amount borrowed = Rs 16000
Interest rate = 10 %
9 Months = 3 quarterly years
Amount to be repaired after 9 months
= 16000 × (1 + \(\frac{2.5}{100}\))³
= 16000 × (\(\frac{102.5}{100}\))³
= 16000 × \(\frac{102.5}{100}\) × \(\frac{102.5}{100}\) × \(\frac{102.5}{100}\)
= 17230.25

Hss Live Guru 8th Maths Kerala Syllabus Chapter 5 Question 3.
Manu deposited Rs 15000 in a financial institution. Interest is calculated in every 3 months and added to the amount. Rate of interest in 8 %. How much he gets after 1 year.
Solution:
Amount deposited = Rs 15000
Rate of interest = 8%
Amount he gets after 1 year

Kerala Syllabus 8th Standard Notes Maths Chapter 5 Question 4.
John deposited Rs 2500 on 1st January in a co-operative bank. Bank computes compound interest half yearly. Annual interest rate is 6%. Again he deposited Rs 2500 on 1st July? How much amount he will have in his account at the end of the year?
Solution:
Amount John deposits on 1st January = Rs.2500
Principal after the half year up to July
1 st = 2500 (1 + \(\frac{3}{100}\))
= 2500 × \(\frac{103}{100}\)
= Rs.2575
Amount deposited in July 1st
= Rs 2500
Principal after July 1st= 2575 + 2500 = 5075
Amount he gets at the end of the year 3
= 5075 × (1 + \(\frac{3}{100}\))
= 5075 × \(\frac{103}{100}\)
= Rs 5227.25

Hss Live Guru Class 8 Maths Kerala Syllabus Chapter 5 Question 5.
Ramlath deposited Rs 30,000 in a financial institution where compound interest is computed in every four months. The ann¬ual interest rate in 9%. How much amount Ramlath gets after 1 year?
Solution:
Amount deposited =Rs 30,000
Rate of interest = 9 %
Amount gets after 1 year
= 30000 (1 × \(\frac{3}{100}\))³
= 30000 × \(\frac{103}{100}\) × \(\frac{103}{100}\) × \(\frac{103}{100}\)
= Rs.32781.81

Textbook Page No. 95

Hss Live Guru 8 Kerala Syllabus Chapter 5 Question 1.
The e-waste increases by 15 % every year, according to the study report. There was around 9 crore tonnes of e-waste in 2014. Then how many tonnes of e-waste will be there in 2020.
Solution:
E-waste in 2014 – 9 crore tonnes
Rate of increase = 15 %
The E-waste in 2020 = 9 × (1 + \(\frac{15}{100}\))⁶
= 9 × (\(\frac{115}{100}\))⁶
= 20. 82 crore tonnes

Hss Live 8 Maths Kerala Syllabus Chapter 5 Question 2.
A T.V manufacturer reduces the price of a particular model by 5% every year. The current price of this model is Rs. 8000. What would be the price after 3 years?
Solution:
The present price of the T.V = Rs. 8000
rate of reduction = 5% ,
Price after 2 years = 8000 (1 – \(\frac{5}{100}\))²
= 8000 × (\(\frac{95}{100}\))²
= 8000 × \(\frac{95}{100}\) × \(\frac{95}{100}\)
= Rs.7220

Class 8 Maths Hsslive Kerala Syllabus Chapter 5 Question 3.
Tiger is our national animal. Ac-cording to a statistics the number of tigers is reduced annually by 3%. There are 1700 tigers in 2011 according to the senses or tiger protection authority. Then how many tigers will be there in 2016?
Solution:
The number of tigers in 2011 = 1700
Reduction rate = 3%
No. of tigers in 2016 after 5 years of the senses

= 1459.84
= 1460

Money Maths Additional Questions and Answers

Kerala Syllabus 8th Standard Maths Notes Chapter 5 Question 1.
Nirmala deposited Rs. 20,000 in a bank where compound interest of rate 7% is computed annually Rs. 5000 was with drawn after one year. How much amount she will get after 2 years.
Solution:
Amount deposited in the bank = Rs. 2000
Rate of interest = 7%
principal after one year
= 20000 + [20000 × \(\frac{7}{100}\)] = 20000 + 1400 = 21400
Amount withdrawn after one year = Rs. 5000 .
Principal for the 2-nd year
= 21400 – 5000 = Rs. 16400
Amount she gets after 2 years
= 16400 + [16400 × \(\frac{7}{100}\)] = 16400 + 1148
= Rs. 17548

Question 2.
Aswathy deposited Rs. 10,000 in a bank where compound interest is calculated at the rate of 10%. Rs.5000 was deposited in the beginning of the 2-nd year and Rs.5000 in the beginning of the 3-rd year. How much amount she gets after 3 years?
Solution:
Amount deposited in the first year = Rs. 10,000
rate of interest = 10%
Amount at the end of first year
= 10000 + [10000 × \(\frac{10}{100}\)] = 10000 + 1000 = Rs. 11000
Principal’for the 2-nd year
= 11000 + 5000 = Rs. 16000
Amount at the end of the 2-nd year
= 16000 + [16000 × \(\frac{10}{100}\)] = 16000 + 1600 = Rs. 17600
Principal for the 3-rd year
= 17600 + 5000 = Rs.22600
Amount at the end of 3-rd year
= 22600 + [22600 × \(\frac{10}{100}\)] = 22600 + 2260 = Rs. 24860
Amount Aswathy will get at the end of 3-rd year = Rs. 24860

Question 3.
Raj an borrowed Rs. 15000 from a co-operative bank for business purpose. The bank computes 9% interest. How much money she has to repay after 5 months.
Solution:
Amount borrowed = Rs. 15000
Rate of interest = 9%
Time duration = 5 month months
Interest .
= 15000 × \(\frac{9}{100}\) × \(\frac{5}{12}\) = 562.50
Amount he has to repay
= 15000 + 562.50
= Rs 15562.50

Question 4.
Rasiya deposited Rs. 20,000 in a bank where compound interested is computed half yearly. If the rate of interest is 11%, How much amount she will get after an year?
Solution:
Capital for first year = Rs. 20,000
Rate of interest = 11 %
Interest of first half year
= 20000 × \(\frac{11}{100}\) × \(\frac{1}{2}\) = Rs. 1100
Principal of 2-nd half year
= 20000 + 1100
= 21100
Interest of the 2-nd half year
= 21100 × \(\frac{11}{100}\) × \(\frac{1}{2}\) = Rs. 1160.5
Amount she gets after one year
= 21000 + 1160.5
= Rs. 22160.5

Question 5.
The price of a car is rupees 5 lakh and it depreciates by 6% every year. What would be the price after 2 year ?
Solution:
Here the price every year is 6% less than the previous years price.
First year’s price = Rs 500000
First year’s depreciation
= 500000 × \(\frac{6}{100}\) = Rs. 300000
Second year’s price = Rs 4,70000
Second year’s depreciation
= 470000 × \(\frac{6}{100}\) = 28200
The price of the car after 2 years
= Rs 4,70,000 – Rs 28200
= Rs 441800

Question 6.
The simple interest of an amount is Rs 50 for two years and the compound interest is Rs 55. The rate of interest is same in both the cases. Find the rate? Find the amount?
Solution
The simple interest of the amount for two years = Rs 50
The simple interest of the amount for 1 year = Rs 25
The compound interest of the amount for 2 years = Rs 55
The interest in the 2 nd year
= 55 – 25 = Rs 30
Interest for Rs 25 = Rs 5

Question 7.
A company which manufactures computers increases its production by 10% every year. In 2009 the company produced 80,000 computers. How many computers would it produced in 2011?
Solution:
Here the number of computers produced every year is 10% more than the number produced the year before. So starting from 80,000. We have to find the number of computers produced every year after that for two years.
Number of computers produced in 2009 = 80,000
Number of computers produced in 2010
= 80000 + 80000 × \(\frac{10}{100}\)
= 80000 + 8000 = 88000
Number of computers produced in 2011
= 88000 + 88000 × \(\frac{10}{100}\)
= 88000 + 8800 = 96800

Question 8.
Ramu borrowed Rs 50,000 from a bank at the interest rate of 10% for agricultural purpose. The interest rate will be reduced by 5% if he repays the amount properly within 2 years. If he fails to repay in time there will be fine as 1%. Ramu could not repay the amount in time. How much amount Ramu repair?
Solution:
Amount borrowed =Rs 50,000
Rate of interest = 10%
Interest rate including fine = 10 + 1=11%
Amount to be repaid
= 50000 + [50000 × \(\frac{11}{100}\) × 2]
= 50000 + 11000 = 61000
= Rs 61000

Question 9.
Raju has Rs 800 with him. He spent 25 % of it. How much amount left with him?
Solution:
Total amount = Rs 800
Amount spent = 800 × \(\frac{25}{100}\) = 200
Amount left with Raju = 800 – 200
= Rs 600

Question 10.
Balu and Ramu decided to borrow Rs 15000 each for a joint business. Balu borrowed Rs 15000 from a financier Who imputes Rs 5 per month for Rs 100 as interest. Ramu borrowed Rs 15000 from a bank where compound interest of 12 % is computed. How much money both of them have to repay after 2 years?
Solution:
Interest Balu has to pay after 2 years
= 15000 × \(\frac{60}{100}\) × 2 = Rs. 18000
(Interest for Rs 100 per month in Rs 5. Interest for Rs 100 in an year = 5 × 12
= Rs. 60.
The rate of interest = 60%)
Amount Balu has to repay after 2 years
= 15000 + 18000
= Rs 33000
In the case of Ramu interest is computed as compound interest.
Principal for first year = Rs 15000
Interest for 1 st year
= 15000 × \(\frac{12}{100}\) × 1 = Rs 1800
Principal for 2 nd year
= 15000 + 1800
= Rs 16800
Interest for the second year = 16800 × \(\frac{12}{100}\) × 1 = Rs. 2018
Amount Ramu has to repay
= 16800 + 2018
= Rs 18816

Question 11.
The population of Kerala increases by 3% every year. The current population is 5 crore. What would he the population after 2 years.
Solution:
Current population = 50000000
Percentages increase in every year = 3%
The population in Kerala after 3 year
= 50000000 [1 + \(\frac{3}{10}\)]²
= 50000000 × \(\frac{(103)^{2}}{10000}\)
= 530,45000

Question 12.
A financial company claims that it charges only 20% interest on loans. But if a person takes out a loan of Rs 100 he would get only Rs 80, after subtracting the annual interest of rupees 20 at the outset. And he has to pay back Its 100 after 1 year. How much is their real interest?
Solution:
A borrower gets only Rs 80 when he borrows Rs 100
He has to repay Rs 100.
ie. Rs.20 as additional.
Real interest
\(\frac{20}{80}\) × 100 = 25%

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

Kerala State Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures

Circle Measures Textual Questions and Answers

Textbook Page No. 131

Circle Measures Class 9 Kerala Syllabus Question 1.
Prove that the circumcentre of an equilateral triangle is the same as its centroid.
i. Calculate the length of a side of an equilateral triangle with vertices on a circle of diameter 1 centimetre.
ii. Calculate the perimeter of such a triangle.
Answer:
Perpendicular bisector of sides of a triangle that can meet at a point is its circumcenter

Since the triangle is equilateral the perpendicular bisectors of the sides are also median. Since the triangle is equilateral the perpendicular bisectors of the sides are also centroid. That is circumcentre of an equilateral triangle is the same as its centroid.
i. Length of one side of an equilateral triangle is.

Circle Measures Class 9 Scert Chapter 9 Kerala Syllabus Question 2.
Calculate the perimeter of a square with vertices on a circle of diameter 1 centimetre.
Answer:
In the square ABCD
AO = 1/2 cm
In A AEO, ∠EAO = 45°
∠AEO = 90°
Since AO = 1/2 cm
AE = \(\frac{1}{2 \sqrt{2}} \mathrm{cm}\)
(Since Δ AEO is a isosceles right triangle, hypotenuse is √2 times of a perpendicular side.)

Perimeter = \(=4 \times \frac{\sqrt{2}}{2}=2 \sqrt{2} \mathrm{cm}\)

Circle Measures Class 9 Chapter 9 Kerala Syllabus Question 3.
Calculate the perimeter of a regu¬lar hexagon with vertices on a circle of diameter 1 centimetre.
Answer:
If we draw diagonals through centre of circle inside the regular hexagon it divides the regular hexagon into 6 equilateral triangles.
Let diameter be 1 cm
OA = 1/2 cm
OA = OB = AB.
therefore
One side = 1/2 cm
Perimeter of regular hexagon = 6 × 1/2 = 3 cm

Textbook Page No. 134

Hss Live Guru 9th Maths Chapter 9 Kerala Syllabus Question 1.
The perimeter of a regular hexagon with vertices on a circle is 24 centimetres.
i. What is the perimeter of a square with vertices on this circle?
ii What is the perimeter of a square with vertices on a circle of double the diameter?
iii. What is the perimeter of an equilateral triangle with vertices on a circle of diameter half that of the first circle?
Answer:
Perimeter of a regular hexagon = 24 cm
Length of one side of a regular hexagon = 24/6 = 4 cm
Length of one side of a regular hexagon is equal to the radius of the circle.
i. Diagonal of a square = 8 cm Diagonal of a square is equal to the diameter of the circle.
Let a be the side of the square
a² + a² = 8²
2a² = 64
a² = 64/2 = 32
a= 4√2
∴ One side of a square = 4√2 cm
Perimeter of a square = 4 × 4√2 cm = 16√2 cm

ii. The perimeter of a square with verti¬ces on a circle of double the diameter 2 × 16√2 = 32 √2 cm
(The perimeters of circles are scaled by the same factor as their diameters.)

iii. One side of an equilateral triangle with vertices on a circle of half the diameter of the first circle
\(=2 \sqrt{2^{2}-1^{2}}=2 \sqrt{3} \mathrm{cm}\)
Perimeter = 3 × 2√3 = 6√3 cm

Circles Class 9 State Syllabus Chapter 9 Kerala Syllabus Question 2.
A wire was bent into a circle of diameter 4 centimetres. What would be the diameter of a circle made by bending a wire of half the length?
Answer:
The ratio between the perimeters are equal to the ratio between their diameters. The perimeter of the first circle is twice the perimeter of the second circle.
Therefore diameter of the second circle is half of the diameter of the first circle. Diameter of the second circle.
= 4/2 = 2 cm

Kerala Syllabus 9th Standard Maths Solutions  Question 3.
The perimeter of a circle of diameter 2 metres was measured and found to be about 6.28 metres. How do we compute the perimeter of a circle of diameter 3 metres, without measuring?
Answer:
If diameter is 2 meters, perimeter is 6.28 meter.
If the diameter is 1 metre, perimeter is 6.28/2 meter.
If the diameter is 3 metres, perimeter = \(\frac{6.28}{2} \times 3=9.42 m\)

Textbook Page No. 137

Hss Live Guru Maths 9 Chapter 9 Kerala Syllabus Question 1.
In the pictures below, a regular hexagon, square and a rectangle are drawn with their vertices on a circle. Calculate the perimeter of each circle.

Answer:
a. AB = 2 cm
In the figure triangle are
equilateral triangles,
therefore
radius OA = 2 cm
Perimeter of circle = 2 πr
= 2 × π × 2 = 4π cm

b. ABCD is a square
AB = BC = 2 cm, ∠5 = 90°
AC = \(\sqrt{2^{2}+2^{2}}=\sqrt{8}=2 \sqrt{2}\)

Radius of circle = 1/2 × 2√2
= √2 cm
Perimeter of circle = 2π × √2 cm
= 2√2 π cm

c. PR = \(\sqrt{2^{2}+(1.5)^{2}}\)
= \(\sqrt{6.25}=2.5 \mathrm{cm}\)
Radius of circle = 1/2 × 2.5 = 1.25 cm
Perimeter of circle = 2 × π × 1.25
= 2.5 π cm

Kerala Syllabus 9th Standard Maths Notes Question 2.
An isosceles triangle with its vertices on a circle is shown in this picture.

What is the perimeter of the circle
Answer:
Consider the centre of circle as O and triangle as ABC
OC = Radius of circle = r
OD = 4 – r
AD = 2 cm
AO = r

Therefore, in triangle AOD
(AO)² = (AD)² + (OD)²
r²=2² + (4 – r)²
r²= 4 +16 – 8r + r²
8r=20;
r = 20/8 = 5/2 = 2.5 cm
∴ Perimeter = 2π × r = 2π × 2.5 cm
= 5π cm

Circles Class 9 Kerala Syllabus Chapter 9 Question 3.
In all the pictures below, the centres of the circles are on the same line. In the first two pictures, the small circles are of the same diameter.

Prove that in all pictures, the perimeters of the large circle is the sum of the perimeters of the small circles.
Answer:
a. Smaller circles have same diameters. Consider the diameter as d, perimeter of smaller circle
= π × diameter = π d
Perimeter of two small circles = 2πd

Diameter of the large circle = d + d = 2d
Perimeter of the large circle
= π × 2d = 2πd
Therefore perimeters of the large circle is the sum of the perimeters of the small circles.

b. Let d be the diameter of one small circle, then perimeter = π d
Sum of perimeter of three small circles =3 π d
Diameter of the large circle = d + d + d = 3d
Perimeter of the large circle = π × 3d = 3 π d
Therefore perimeters of the large circle is the sum of the perimeters of the small circles.

c. In figure diameter of three circless are different, let consider the diameters of small circles are p, q and r.

Perimeter of first small circle = πp
Perimeter of second small circle = πq
Perimeter of third small circle = πr
Sum of perimeters of three small circles
= πp + πq + πr = π (p + q + r)
Diameter of large circle = p + q + r
Perimeter of large circle = π (p + q + r)
Therefore also here the perimeters of the large circle is the sum of the perimeters of the small circles.

Hss Live Class 9 Maths Chapter 9 Kerala Syllabus Question 4.
In this picture, the circles have the same centre and the line drawn is a diameter of the large circle. How much more is the perimeter of the large circle than the perimeter of the small circle

Answer:
If ‘r’ is the radius of the small circle,
radius of the large circle = r + 1
Perimeter of the small circle = 2 π r
Perimeter of the large circle = 2 π (r + 1) = 2 π r + 2 π
i.e., perimeter of the large circle is 2 π units more than the perimeter of the small circle.

Textbook Page No. 141

Hss Live Guru 9 Maths Chapter 9 Kerala Syllabus Question 1.
In the pictures below, find the difference between the areas of the circle and the polygon, up to two decimal places.

Answer:
i. Radius of the small circle = 2cm
Area = π × 2² = 3.14 × 4
= 12.56 cm²
Daigonal of the square = 4 cm
One side of the square = 4/√2 cm
Area of the square = \(\frac{4}{\sqrt{2}} \times \frac{4}{\sqrt{2}}=\frac{16}{2}=8 \mathrm{cm}\)
Differences between the areas = 12.56 – 8 = 4.56 cm

ii. Radius of the circle = 2 cm
Area= π × 2² = 3.14 × 4 = 12.56 cm²
One side of the regular hexagon=2 cm
Area of the regular hexagon = \(6 \times \frac{\sqrt{3} \times 2^{2}}{4}=6 \sqrt{3}\)
= 6 × 1.73 = 10.38 cm²
Differences between the areas = 12.56 – 10.38 = 2.18 cm²

9th Std Kerala Syllabus Maths Solutions Question 2.
The pictures below show circles through the vertices of a square and a rectangle

Calculate the areas of the circles
Answer:
i. One side of a square is 3 cm, therefore its diagonal is 3√2 cm
Diameter of the circle = 3√2 cm

ii. Rectangle inside the circle having length 4 cm and breadth 2 cm.
Diagonal = \(\sqrt{4^{2}+2^{2}}=\sqrt{16+4}\)

Diameter of the circle = √20 cm
Radius = \(\frac{\sqrt{20}}{2} \mathrm{cm}\)
Area = \(\pi \times\left(\frac{\sqrt{20}}{2}\right)^{2}=\pi \times \frac{20}{4}\)
= 5 π cm²

Hsslive Guru 9th Maths Chapter 9 Kerala Syllabus Question 3.
Draw a square and draw circles cen¬tered on the corners, of radius half the side of the square. Draw another square formed by four of the first square and a circle just fitting into it.

Prove that the area of the large circle is equal to the sum of the areas of the four small circles
Answer:
In the figure length of one side of the square is 2r
Radius of one small circle = r
Perimeter of one small circle = π r²

Radius of four small circles = 4 π r²
One side of a square in the second figure = 2r + 2r = 4r
Radius of the circle in the figure
= 4r/2 = 2r
perimeter of the circles in the figure = π × (2r)²
= π × 2r × 2r = 4πr²
The area of the large circle is equal to the sum of the areas of the four small circles

Hsslive 9th Maths Chapter 9 Kerala Syllabus Question 4.
In the two pictures below, the squares are of the same size.

Prove that the green regions are of the same area.
Answer:
Let ‘a’ be the side of the square in the picture.
Area of the square = a²
If the four sectors in the vertices are joined together, a circle is formed because the radius of each sector is a/2.
Area of the shaded part = Area of the square – area of the circle.

ii. In the second picture diameter of the
circle = a Radius = a/2
Area of the shaded part = Area of the square – area of the circle.

i.e., Areas of the shaded portions are equal.

Hsslive Guru 9 Maths Chapter 9 Kerala Syllabus Question 5.
Parts of circles are drawn inside a square as shown in the picture below. Prove that the area of the blue region is half the area of the square.

Answer:
Let ‘a’ be the side of the square
Area of the blue part in the half portion of the square is equal to half of the area of the circle having diameter ‘a’.
Area of the blue part in the half portion of the square \(=\frac{1}{2} \times \pi\left(\frac{a}{2}\right)^{2}=\frac{\pi a^{2}}{8}\)
We must subtract area of two circles
having diameter a/2 from the half the
area of the square to get the area of remaining blue shaded part.
Area of remaining blue shaded part.
= \(\frac{a^{2}}{2}-2 \times \frac{1}{4} \times \pi\left(\frac{a}{2}\right)^{2}\)
Area of blue shaded part.
= \(\frac{\pi a^{2}}{8}+\frac{a^{2}}{2}-\frac{\pi a^{2}}{8}=\frac{a^{2}}{2}\)
i.e., area of the blue part is equal to half the area of the square.

Hss Guru 9 Maths Chapter 9 Kerala Syllabus Question 6.
In the figure, semicircles are drawn with the sides of a right triangle as diameter.

Prove that the area of the largest semicircle is the sum of the areas of the smaller ones.
Answer:

In ΔABC, ∠5 = 90°
According to Pythagoras principle,
AB² + BC² = AC² ………. (1)
Radius of the semicircle with diameter
AB = AB/2
Area of the semicircle with diameter AB

= \(\frac{1}{2} \times \pi \times \frac{A C^{2}}{4}=\frac{\pi}{8} A C^{2}\)
Sum of the areas of the smaller semicircles = \(\frac{\pi}{8} A B^{2}+\frac{\pi}{8} B C^{2}=\frac{\pi}{8}\left(A B^{2}+B C^{2}\right)\)
= π/8 AC² = Area of the largest semicircle

Textbook Page No. 148

Hsslive Maths Class 9 Chapter 9 Kerala Syllabus Question 1.
In a circle, the length of an arc of central angle 40° is 3 π centimetres. What is the perimeter of the circle? What is its radius?
Answer:

Question 2.
In a circle, the length of an arc of central angle 25° is 4 centimetres.
i. In the same circle, what is the length of an arc of central angle 75°?
ii. In a circle of radius one and a half times the radius of this circle, what is the length of an arc of central angle 75°?
Answer:
i. Length of the arc having central angle 25° = 4 cm
Three times of 25 is 75.
Length of the arc having central angle 75° = 4 x 3 = 12 cm

ii. Length of the arc having central angle 75° and radius r = 12 cm
Length of the arc having central angle 75° and radius 1 1/2 r
= \(12 \times 1 \frac{1}{2}=18 \mathrm{cm}\)

Question 3.
From a bangle of radius 3 centimetres, a piece is to be cut out to make a ring of radius ^ centimetres.
i. What should be the central angle of the piece to be cut out?
ii. The remaining part of the bangle was bent to make a smaller bangle. What is its radius?
Answer:
Perimeter of the bangle having radius 3 cm = 6 π cm.
Perimeter of the bangle having radius 1/2 cm = π cm.
π is the 1/6 part of 6 π.
Therefore the central angle of the piece to be cut out = 360 × 1/6 = 60°
ii. Length of the remaining part of the bangle = 6π – π = 5π cm
Radius of the other small bangle = 5 π / 2π = 2.5 cm

Question 4.
The picture shows the parts of a circle centred at each vertex of an equilateral triangle and passing through the other two vertices.

What is the perimeter of this figure?
Answer:
Since the triangle is equilateral, each angle is 60°. There are in each side is in a circle of radius 4 cm and the central angle is 60°.

Question 5.
Parts of circles are drawn, centred at each vertex of a regular octagon and a figure is cut out as show below:

Calculate the perimeter of the fig¬ure.
Answer:
Sum of the angles in an octagon
= (n-2) × 18o° = 6x 180°= 1080° One angle of the regular octagon
= 1080 – 8 = 135°
One side of the regular octagon = 2 cm. Radius if the sectors having centre is each vertices of the circle= 1 cm The second picture shows the cut-down form of 8 sectors having centre angle 135° and radius 1 cm.
The perimeter is found by calculating the length of 8 arc having radius 1 cm and central angle 135°.

Textbook Page No. 151

Question 1.
What is the area of a sector of central angle 120° in a circle of radius 3 centimetres? What is the area of a sector of the same central angle in a circle of radius 6 centimetres?
Answer:
Area of the sector in the circle having radius 3 cm and angle of center 120°
π × 3² × \(\frac { 120 }{ 360 }\) = 3π cm²
Area of the sector in the circle having radius 6 cm and angle of center 120°
= π × 6² × \(\frac { 120 }{ 360 }\) = 12π cm²

Question 2.
Calculate the area of the green coloured part of this picture.

Answer:
In the picture area of the shaded part is the difference between the area of the two sectors.
Area of the large sector

Area of the shaded part = 9.42 – 4.19 = 5.23 cm²

Question 3.
Centred at each corner of a regular hexagon, a part of a circle is drawn and a figure is cut out as shown below:

What is the area of this figure?
Answer:
The area of the cut-down portion = Area of the regular hexagon – Area of 6 sectors Area of the regular hexagon

Area of the sector = 120°/360° part of area of the circle
(One angle of a regular hexagon is 120°)

Area of the 6 sectors = \(\frac{6 \times \pi}{3}=2 \pi \mathrm{cm}^{2}\)
Area of cut down portion = 6 √3 – 2 π cm²

Question 4.
The picture below shows two circles, each passing through the centre of the other:

Calculate the area of the region common to both.
Answer:
Consider the picture given below, we can divide the circle into two part by using the line AB, the area of the two portions are same.
That is we get the total area by multi¬plying area of the sector by 2.

We can find out the area of part above the line AB .
Given AB = 2cm
Circles having equal
radius. So,
AC = BC = 2 cm.
AABC is an equilateral triangle so angles are 60° each.
Add the area of sectors having centre A and B .
The area of ΔABC include twice, so we will subtract it once.
Area of sectors having centre A

Area of sectors having centre B

Area of ΔABC

Area of the part above the line AB = 2.09 + 2.09 – 1.7 = 2.45 cm²
The area of the region common to both = 2 × 2.45 = 4.90 cm²

Question 5.
The figure shows three circles drawn with their centres on each vertex of an equilateral triangle and passing through the other two ver¬tices;

Find the area of the region common to all three.
Answer:
The area of the common part = Area of three sectors – 2 × area of an equilateral triangle having side 2 cm

= 2 × 3.14 – 2 × 1.73
= 6.28 – 3.46
= 2.82 cm²
The area of the region common to all three = 2.82 cm²

Circle Measures Exam Oriented Questions And Answers

Question 1.
In the picture PQRS is a square of side 10 cm. A, B, C and D are midpoints of the sides of the square

Semi circles are drawn inside the square.
a. Compute the area of the square.
b. Compute the area of the semi-circle.
c. Compute the area of the shaded part.
Answer:
a. Since one side of the square is 10 cm,
Area = side x side = 10 × 10= 100 cm²

b. The diameter of one semicircle = half of the side of the square
Diameter = 5 cm
∴ Radius = 5/2 cm
Since the four semicircles are equal. Area of the 4 semicircles
= Area of 2 circles = 2 × πr²

c. Area of the shaded part = Area of the square – Area of four semicircles.

Question 2.
If a circular shaped dining table has an area of 31400 cm², find its radius. What will be its perimeter ?
Answer:
Area of the table = 31400 cm²

Question 3.
In the given picture shown two semicircular shaped iron bar can be cut down from a rectangular shaped iron bar. Calculate the area of the remaining shaded portion.

Answer:
Area of the rectangle = 24 × 14 = 336 cm²
Area of two semicircle = Area of a complete circle
Diameter of the circle = 14 cm
Radius = 7 cm
Area of the circle = π r²
= π × 72 =49
π = 153.86 cm2 Area of the remaining portion
= 336 – 153.86= 182.14 cm²

Question 4.
In the figure A, B, C and Dare the points on the square which touches the circle. If the radius of the circle is 6.
a. What is the length of one side of the square?
b. Find the area of the circle.
c. What will be the area of the shaded portion?

Answer:
a.Diameter of the circle = 6 × 2 = 12 cm
Side of the square = 12 cm
b. Area of the circle = n r²
= n × 6 × 6 = 36n =36 × 3.14 =113.04 cm²
c. Area of the square = 12 × 12 = 144 cm²
Area of the shaded portion = 144 – 113.04 = 30.96 cm²

Question 5.
In the ACB is the arc drawn by taking O as the centre and OA as the radius. Then find the area of the shaded region.

Answer:
The area of sector which has 7 cm radius and 90° central angle =

Area of the right angled triangle AOB
\(=\frac{7 \times 7}{2}=24.5\)
Area of the shaded part = 38.455 – 24.5 = 13.965 cm²

Question 6.
In the pictures given below, find the area of the shaded part?

Answer:
’The radius of the sector in the picture (1) is 5cm and its central angle is 40°.
Area of the shaded part \(=\pi \times 5^{2} \times \frac{40}{360}=\frac{25}{9} \pi \mathrm{cm}^{2}\)
The radius of the sector in the picture (2) is 6cm and its central angle is 300° (360 – 60).
Area = \(\pi \times 6^{2} \times \frac{300}{360}=36 \pi \times \frac{5}{6}\)
= 30 π cm²

Question 7.
What amount of reeper is needed to enclose a circular shaped dining table of area 6.28 cm²?
Answer:

Question 8.
A wheel which has 20 cm radius is rotating forward. After 10 rotations what distance will the wheel travel forward?
Answer:
Radius of the wheel = 20 cm
When the wheel rotates once it will travel the distance same as its area
Perimeter of the wheel = 2 × π × radius = 2 × 3.14 × 20 = 125.64cm
The distance travelled forward when the wheel rotates once = 125.64 cm
The distance travelled forward when the wheel rotates 10 times
= 125.64 × 10 = 1256.4 cm

Question 9.
In the picture the central angles of both the sectors are equal. Sum of the radii of the sectors is 18 cm.
Area of the shaded part is 18 π cm². Find the radii of the sector.

Answer:
‘Let ‘r’ be the radius of the small sector and ‘R’ be the radius of the large sector.
R + r = 18 …………. (1)
Area of the shaded portion = Area of the large sector – area of the small sector

Question 10.
In the figure O is the radius of the circle and OABC is a rectangle. OA = 8 cm, OC = 15 cm. Hence find the Area of the shaded portion

Answer:

Area of the shaded portion

Question 11.
The wheel of a vehicle has a diameter 60 cm. For this vehicle travels a distance of 200 m, how many times must this wheel rotates.
Answer:
‘The distance travelled when the wheel is rotated once =2 π r = 2 × π × 30
= 188.4 cm = 1.884 m.
The time required for the wheel to travel a distance of 200m = 200/1.884 = 106.16 = 106

Question 12.
In the below-given figures, there are two circles with the same centre. Then find the area of the second region.

Answer:
a. Area of the shaded portion =Area of the outer circle – Area of the inner circle
= π R² – π F= π × 10² – π × 8²
= 100π – 64π = 36π
=36 × 3.14 = 113.04 cm²

b. Outer radius = 7 + 2 = 9 cm Inner radius = 7 cm
Area of the shaded portion = π × 92- π × 72 = 81π – 49π = 32π = 32 × 3.14 =100.48 cm²

c. Outerradius = 10.5
Inner radius = 10.5 – 1 = 9.5
Area of the shaded portion = π × (10.5)² – π × (9.5)²
= π × (10.5² – 9.5)²
= π × (10.5² – 9.5²)
= π × (10.5 + 9.5) (10.5 – 9.5) = π × 20 × 1 = 62.8 cm²

d. Outer radius = \(\frac { 16π }{ 2π }\) = 8
Inner radius = \(\frac { 14π }{ 2π }\) = 7
Area of the shaded portion

Question 13.
Two semicircular pieces are cut out from a rectangular sheet. Find the area of the remaining portion.

Answer:
Area, of the rectangle = 50 × 20 = 1000 cm2 If the two semicircles cut out are joined it becomes a circle
Its radius = 20/2 = 10 cm
Area of the portion cut out
= πr² = π × 10 × 10 = 3.14 × 10 × 10 = 314 cm²
Area of the remaining portion
= 1000 – 314 = 686 cm²

Question 14.
If a square, equilateral triangle, regular hexagon and circle have the same perimeter. Which of these has the largest area ?
Answer:
If a square, equilateral triangle, regular hexagon and circle has a perimeter of 12 cm.
Equilateral triangle
One side = 12/3 = 4


Circle has the largest area.

Question 15.
In the figure, if ACB is the arc of circle having O as the centre and OA as the O, radius. Then find the area of the shaded portion

Answer:
Radius of the sector = 8cm Central angle = 90°

OBA is a right angled triangle
∴ Area of ΔOBA = 1/2 × 8 × 8 = 32 cm²
Area of shaded portion = 50.24 – 32 = 18.32 cm²

Question 16.
The area of an equilateral triangle is 17300 cm². Draw circles with radius half the length of one side of the triangle and vertices as the centre of the circle. Calculate the area of the shaded portion.
Answer:
Area of the equilateral triangle

One side = 200 cm,
Radius of the circle = 100 cm
Central angle of each sector = 60°
Area of one sector = \(\frac{\pi r^{2} \times 60^{\circ}}{360}\)
Area of the three sectors

Question 17.
In the figure, a side of the regular hexagon has length 20 cm. Find the area of the shaded portion.

Answer:
We have learnt that length of a side of the regular hexagon drawn inside a circle will be equal to the radius of the circle.
So the radius of the circle will be 20 cm.
Area of the circle = πr² = π × 20 × 20 = 400π = 1256 cm²
Area of the regular hexagon \(=\frac{6 \times \sqrt{3} \times a^{2}}{4}\)

Area of shaded portion = 1256 – 1039.2 = 216.8 cm²

Question 18.
If radius of a sector is 7 cm and its perimeter is 2.5 cm. Then find its area.
Answer:
Area of sector
= 2 × radius + length of the arc = 25
Length of the arc = 25 – 2 × radius = 25 – 14 = 11 cm
Let us check the ratio between the length of the arc and its area
The length of the arc: Area of the sector

11 : area of the sector = 2:7
Area of two sectors =7 × 11
Area of the. sector = \(\frac{7 \times 11}{2}\) = 38.5 cm²

Question 19.
In the figure if the length of one side of a square is 12 cm, then find the area of the shaded portion.

Answer:
Area of the square = 12² = 144
Diameter of the circle = 12; r = 6 Area of the circle
= πr² = π × 6 × 6 = 3.14 × 6 × 6 = 113.04 cm²
Area of shaded portion = 144 – 113.04 = 30.96 cm²

Kerala State Syllabus 10th Standard Social Science Notes Chapter 2 World in the Twentieth Century

Twentieth century is the period that influenced world history greatly. Imperialism was the developed form of capitalism which emerged in Europe after the Industrial Revolution. When capitalism developed into imperialism, it faced many crises. When the imperialist powers entered into mutual competitions in order to conquer the world, conflict became widespread. The international problems which surfaced during this period caused mutual mistrust and enmity. The conflicts among the imperialist powers ultimately led the entire world to a war. The growth of Fascism and Nazism, Second World War, international efforts for peace, cold war, non-alignment and globalisation are the other topics discussed in this unit.

→  Industrial Revolution: The changes which took place in England from the 18th century onwards came to be known as the Industrial Revolution. The basic feature of this was that human labour was substituted by machines.

→  Capitalism : The economic system in which production and distribution are controlled by capitalists with the aim to increase profit.

→  Imperialism : The practice of extending a nation’s political, economic and cultural dominance on another nation is imperialism.

→  Triple Alliance and Triple Entente : Military alliances that fought in the First World War.

→  Nationalism : A nation is defined as a people settling in a definite territory, speaking a common language, having a common culture and historical tradition. The ideology and programme of action based on this concept is called nationalism.

→  Aggressive nationalism : The policy of invading neighbouring countries, considering one’s nation as supreme and justifying whatever be the actions of the nation.

→  Pan-Slav Movement : The movement started under the leadership of Russia to unite the Slavic People of Serbia, Bulgaria, Greece, etc. in Eastern Europe.

→  Pan-German Movement : The movement started under the leadership of Germany to establish her dominance in Central Europe and Balkan provinces and to unite the Teutonic people.

→  Revenge Movement : The movement started under the leadership of France to regain her territories of Alsace-Lorraine which were captured by Germany in the Franco-Prussian war of 1871.

→  Treaty of Versailles : The treaty signed with Germany by the victorious powers after the First World War.

→  World Economic Depression : The economic crisis that started in 1929 and affected the whole world.

→  League of Nations : The international organisation formed after the first world war to maintain peace in the world.

→  Fascism, Nazism : The political ideology that supported dictatorship, racial superiority, aggressive nationalism and single party rule.

→  Munich Pact : The agreement that approved the claim of Germany over Sudetanland, a part of Czechoslovakia.

→  Policy of appeasement: Capitalist countries like Britain and France considered Soviet Union, being a socialist country, as their chief enemy and did not prevent fascist attacks. This policy which encouraged fascist attacks is known as policy of appeasement.

→  Non-Aggression Pact : The agreement signed between Germany and Soviet Union in 1939, by which they agreed not to attack each other and to partition Poland.

→  Teutonic People : The Germanic people are also called Teutonic peoples. Originally they belonged to Northern Europe. They spoke languages of the Germanic branch of the lndo-European language family.

→  Pearl Harbour Attack : The attack of Japan in 1941 on Pearl Harbour, the American naval base in the islands of Hawaii.

→  Hiroshima, Nagasaki : Japanese cities where atom bombs were dropped in 1945 by USA.

→  Hibakusha: The surviving victims of the atomic bombings of Hiroshima and Nagasaki.

→  United Nations Organisation : The international organisation formed after the Second World War to prevent war and maintain peace in the world.

→  Decolonization : The process of the colonies of Asia and Africa securing freedom from imperialist control.

→  Cold War : The enmity based on ideological conflict and diplomatic confrontations between US bloc and Soviet bloc.

→  Bipolar politics : USA led the Capitalist bloc and Soviet Union led the Socialist bloc after the Second World War. This ideological division between the power blocs is called bipolar politics.

→  Military Pacts : Military agreements formed among capitalist bloc and socialist bloc after the Second World War.

→  Non-alignment : The policy adopted by the newly independent countries of Asia and Africa not to join the power blocs and to follow an independent foreign policy.

→  Zionism: The movement that started to establ ish a homeland for the Jews.

→  PLO : Palestinian Liberation Organisation was a movement with the objective of establishing a nation for the Palestinians.

→  Oslo Pact: The pact signed between Israel and Palestine under the mediation of USA. By this, Israel agreed to recognise Palestine as a free nation.

→  Glasnost : The administrative reform started under Mikhail Gorbachev in Soviet Union to implement openness in political processes.

World In The Twentieth Century Notes

→  Perestroika: The administrative reform started under Mikhail Gorbachev to restructure the economic system of Soviet Union.

→  Unipolar world : USA emerged as a global power and centre of world politics following the disintegration of Soviet Union. The world order dominated by the USA is called unipolar world.

→  Neo imperialism: The multinational companies began to interfere in the economic, social and cultural sectors of the newly independent countries of Asia and Africa and Latin America for serving the interests of capitalist countries. This is known as neo imperialism.

→  Globalisation: The policy of transfer of products, ( services, raw materials, capital, latest technology and human resources across the borders of countries without any restriction.

→  Liberalisation: The policy of adoption of liberal regulations and taxation systems to facilitate the import of multinational products to domestic markets.

→  Privatisation : The policy of privatisation of public sector undertakings to promote private sector. (The process of reducing the role of public sector in the economy and increasing the role of private sector is known as privatisation).

World in the Twentieth Century – Famous Persons

→  Francis Ferdinand : The heir to the throne of Austria who was assassinated in June 1914 at Sarajevo. This was the immediate cause for the First World War.
→  Benito Mussolini : Leader of fascist reign in Italy.
→  Adolf Hitler: Leader ofNazi reign in Germany.
→  Matteotti: Eminent socialist thinker of Italy who opposed fascism. ‘
→  Woodrow Wilson : The US President who gave leadership to the formation of League of Nations.
→  Nelson Mandela ; Leader of anti – imperialist struggle in South Africa.
→  Quami Nkrumah : Leader of anti – imperialist struggle in Ghana.
→  Jomo Kenyatta : Leader of the freedom movement in Kenya.
→  Bernard Baruch : The American economist who first used the word ‘cold war’.
→  Architects of Non : Aligned movement:

• Jawaharlal Nehru – India
• Gamal Abdul Nasser – Egypt
• Marshal Tito-Yugoslavia .
• Ahmed Sukarno – Indonesia

→ Yasser Arafat: Founder President of Palestinian Liberation Organisation.

→  Mikhail Gorbachev : The last President of Soviet Union.

World in the Twentieth Century – Important Years and Events

• 1871 – Franco Prussian War
• 1904 – Moroccan Crisis
• 1912 – Balkan Crisis
• 1914 – Assassination of Francis Ferdinand
• 1914-18 – First World War
• 1919 – Paris Peace Treaty: Treaty of Versailles
• 1924 – Mussolini in power
• 1929 – World Economic Depression
• 1933 – Hitlar as Chancellor of Germany
• 1938 – Munich Pact
• 1939-45 – Second World War
• 1941 – Pearl Harbour Attack
• 1945 – Atom bombs dropped at Hiroshima and Nagasaki in Japan by USA
• 1945 – Formation ofUNO (October24,1945)
• 1948 – Formation of Israel
• 1955 – Emergence of Non-Aligned Movement
• 1991 – Disintegration of Soviet Union
• 1993 – Oslo Pact

Kerala Syllabus 10th Standard Social Science Notes

You can Download Soldiers of Defense Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Biology Solutions Chapter 5 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard BiologySolutions Chapter 5 Soldiers of Defense

Soldiers of Defense Text Book Questions and Answers

Sslc Biology Chapter 5 Kerala Syllabus Question 1.
Our surrounding are full of microorganisms. Most of them are pathogens too. Though we live in the midst of germs are we susceptible to diseases? What may be the reason?
Answer:
Numerous germs are present in our surroundings, that have the capacity to cause diseases. We are often in contact with them. There are several mechanisms in the human body which prevent the entry of germs. So we don’t get infected always.

Biology Chapter 5 Class 10 Kerala Syllabus Question 2.
What are the mechanisms in the body which prevent the entry of pathogens?
Answer:

• A protein called keratin in skin, sebum, and acids.
• Mucus in the trachea
• Cilia in the bronchus
• Hydrochloric acid in the stomach.
• Cough and sneezing.
• The wax in the ear.
• The enzyme lysozyme in tears and saliva.
• Blood, Lymph.

Sslc Biology Chapter 5 Notes Kerala Syllabus Body Coverings And Secretions

Biology Class 10 Chapter 5 Kerala Syllabus Question 3.
Skin is referred to as a “fort of resistance”, why?
Answer:
Keratin makes the skin a thick fort which prevents germs from entering it. So the skin is referred to as a fort of resistance.

10th Class Biology 5th Chapter Kerala Syllabus Question 4.
What is the function of cilia and mucus in the respiratory tract?
Answer:
Mucus in the trachea prevents the entry of germs into the lungs. The cilia in the bronchus wipe out dust that enters it.

Kerala Syllabus 10th Standard Biology Chapter 5 Question 5.
What are the methods in ears, eyes, and saliva to prevent germs?
Answer:
The enzyme lysozyme present in the tears and saliva are fight against germs. The wax in the ear prevents pathogens.

Class 10th Biology Chapter 5 Notes Kerala Syllabus Question 6.
What is the role of hydrochloric acid in the stomach to prevent germs that enter the body through food?
Answer:
Since hydrochloric acid is present stomach, the germs that enter through food and water are destroyed.

Class 10 Biology Chapter 5 Notes Kerala Syllabus Question 7.
Which are the secretions that help to defend pathogens? Analyze illustration and complete the table.


Answer:

Body Fluids And Defense

• Body fluids like blood and lymph play an important role in defense mechanisms.
• Controlling the entry of germs into the body.
• Neutralizing germs and the toxic substances they produce, preventing their multiplication.

White blood cells and Defense actions

Inflammatory Response

Lysozyme Antibodies Question 8.
Based on the indicators, analyze the following illustration. Write your inference in the science diary.

Answer:
The cells that get damaged by a wound or an infection produce certain chemical substances. These substances dilate the blood vessel thereby increasing the blood flow. Blood plasma and white blood cells reach the wound site. This is the reason for the swelling of the wound site. This defense mechanism is known as inflammatory response.

Question 9.
What is the advantage of the dilation of blood vessels at the wound site?
Answer:
The cells that get damaged by a wound or an infection produce certain chemical substances. These substances dilate the blood vessels thereby increasing the blood flow.

Label the Specializations of the Plasma Membrane Question 10.
Is inflammatory response a defense activity? Why?
Answer:
Inflammatory response is a defense activity. Inflammation formed in the body due to the changes in the wall of blood capillaries in a part of the body that affected a wound. When germs enter through – the wound, changes occur in the capillary wall of that part. It leads to inflammation. Flow of blood through these capillaries increases and as a result more leucocytes come out from the capillaries and destroy the germs by engulfing them. The affected parts swell and become red-colored due to the arrival of more blood at the affected part of the capillaries.

Question 11.
Prepare the flowchart which showing the stages of inflammatory response.
Answer:
Germs enter through wound → Produces chemical messages → Blood vessels dilate → White blood cells from the blood vessel reach the wound site → White blood cells destroy the germs.

Phagocytosis

Phagocytosis is the process of engulfing and destroying germs. The cells engaged in this process are called phagocytes. (Phago – to engulf, cyte – cell) Monocytes and neutrophils are phagocytes.

Question 12.
Stages of phagocytosis
Answer:

Question 13.
Complete the flow chart by analyzing illustration showing the stages of phagocytosis.

Answer:

Blood Clotting

Blood clotting is a defense mechanism to prevent the loss of blood through wounds. In this process fibrin, the plasma protein forms a fibrous network. Blood cells get entangled in the network to form a blood clot.

Question 14.
Analyze the following illustration that details the stages of blood clotting

Answer:
When a person gets a cut or wound the blood that flows out from the wound changes from a liquid to a gel, forming a clot which plays the wound and prevents further bleeding.

When the platelets come into contact with the atmospheric air, they burst and liberate thromboplastin. It converts prothrombin in the plasma to thrombin. This thrombin converts soluble fibrinogen molecules into insoluble fibrin. This fibrin filament form-fine network over the wound and trap blood corpuscles and platelets to form a clot. The clot seals the wounds and stops bleeding.

Healing Of Wounds

Healing of the wound is a stage after inflammatory response and blood clotting. When wound occurs new tissues are formed in place of the tissues damaged by the wound. In such situations, the wound scar does not remain. In cases when new tissues cannot be formed, the connective tissue heals the wound. In such situations, the wound scar remains.

Question 15.
In some situations, the wound scar remains. Why?
Answer:
When wound occurs, new tissues cannot be formed, then connective tissues heal the wound. In such situations, the wound scar remains.

Fever, A Defense Mechanism

Question 16.
The normal body temperature is 37°c (98.6°F). Body temperature rises during fever. Is it a disease or a symptom. Analyze the flowchart given and write your inferences in the science diary.

Answer:
Fever is not a disease. But it is a type of resistance activity. Though the body can control the multiplication of germs through mechanism like raising body temperature. The chemical substances produced by the white blood cells raises the body temperature. If the rise in body temperature persists for a long time, it may badly affect the internal organs including the brain. Hence it is necessary to seek medical assistance immediately.

Question 17.
Fever is the rise in the body temperature. Is it beneficial to the body?
Answer:
Our normal body temperature is 36.9°c. This temperature is suitable for the multiplication of germs. When infection occurs body rises the temperature through fever to reduce the capacity of multiplication of them.

Lymphocytes – The Warrior

Specific defense is the system which identifies and destroys pathogens. White blood cells known as lymphocytes are capable of destroying the pathogens in this way. Lymphocytes are of two types namely B lymphocytes and T lymphocytes. B lymphocytes mature in them bone marrow. T lymphocytes mature in the thymus gland.

B – Lymphocytes

B lymphocytes produce certain chemical substances to act against antigens. The chemical substances which act against antigens are called antibodies.

Antibodies destroy the pathogens in three different ways

1. Destroy the bacteria by disintegrating their cell membrane.
2. Neutralize the toxin of the antigens.
3. Destroy the pathogens by stimulating other white blood cells.

T – Lymphocytes

T lymphocytes stimulate other defense cells of the body. Moreover, these cells are capable of destroying cancer cells and cells affected by virus.

Question 18.
Complete illustration showing the defense mechanisms of blood.

Answer:

• Inflammatory response
• Phagocytosis
• Blood clotting

Lymph And Defense

The lymph formed from the blood and reabsorbed into blood has a prominent role in defense mechanisms, lymph contains plenty of lymphocytes. They destroy the disease-causing bacteria in lymph nodes and spleen.

Immunization

Defense mechanisms become slow when germs enter the body. This causes the spread and multiplication of germs. Immunization is the artificial method to make the defense cells alert against the attack of pathogens.

Question 19.
What are vaccines?
The substances used for synthesizing antibodies are called vaccines.

Question 20.
Which components of vaccine act as antigens?
Answer:
The components from alive or dead or neutralized germs neutralized toxins or cellular parts of the pathogens will be the component of each vaccine.

Question 21.
How do vaccines induce immunity?

Answer:
In induced immunity antibodies which can act against pathogens or toxins produced by them are synthesized in the body itself. The body prepares antibodies to act against these foreign bodies.

Treatment – Final Defense

Question 22.
Which are the different methods of treatment that we depend on?
Answer:

• Ayurveda
• Sidda
• Unani
• Naturopathy
• Homeopathy
• Allopathy (Modern medicine)

Question 23.
Observe the pictures given below and write the name and use of this equipment for diagnosis.

Answer:
A. Stethoscope – to measure heartbeat
B. Thermometer – To measure body temperature
C. Sphygmomanometer – To record blood pressure.

Question 24.
Given below is the table including a few other modern equipments and their uses.
Answer:

Laboratory Tests

The report of a test showing the quantity of different factors in blood

Question 25.
Identify the specializations in medicine and the related areas and complete the table

Answer:

Antibodies

The scientist Alexander Fleming, who first synthesized antibiotics in 1928. Antibiotics are used to resist bacterial diseases.

Question 26.
Antibiotics are very helpful but use of it should be with great care. Why?
Answer:

• Regular use develops immunity in pathogens against antibiotics
• Destroys useful bacteria in the body.
• Reduces the quantity of some vitamins in the body.
• Some of them cause allergy, problems to stomach, bones, and kidneys.

Question 27.
Is it proper to use antibiotics without recommendation by a doctor? Why? Discuss. Write your inferences in the science diary.
Answer:
No. Though antibiotics are effective medicines, their regular use brings many side effects. Therefore use medicines only by the instruction of the doctor. Doctors prescribe medicine by considering the dose, method of use, period of use, age of the patient, etc. indiscriminate use of them causes health problems. Regular use develops immunity in pathogens against antibiotics, destroys useful bacteria in the body and reduces the quantity of some vitamins. So self-treatment without the instruction of the doctor is not good.

First Aid

Question 28.
Observe figures A, B and C and identify the instance in which the following type of first aid is given.

Answer:

• A – Breathing has stopped but heartbeat has not as in drowning electric shock, choking gas, suffocation, etc.
• B – Bone and ligament injuries; and fractures
• C – Choking occurs when an object swallowing

Blood Transfusion

The transfer of blood from one person to another is called blood transfusion. Certain instances such as blood is lost excessively in accidents, affected with diseases like blood cancer and surgical operations require blood transfusion.

Different Types of Blood Group

A, B, AB, O are the main blood groups. Carl Landsteiner proposed blood grouping on the basis of the presence or absence of A, B antigens seen on the surface of the red blood cells. The blood group in which Rh factor is present are positive blood groups and those without Rh factor are negative blood groups.

Question 29.
Can a patient receive blood from any person?
Answer:
No. Blood of certain persons cannot be received by others. The antigen present in the received blood and antibody in the recipient’s blood will react each other to form blood clot.

Question 30.
Observe the table and identify the various types of blood group, antigens, and antibodies present in them
Answer:

Question 31.
Prepare posters on the greatness of donating blood
Answer:

Defense Mechanisms In Plants

Question 32.
Complete the illustration by including different defense mechanisms in plants.

Answer:

Let Us Assess

Question 1.
Which among the following is not included in non-specific body defense?
a) production of sebum
b) action of hydrochloric acid in the stomach
c) action of B lymphocytes
d) action of lysozyme in saliva
Answer:
c) action of B lymphocytes

Question 2.
Write the functions of the two types of lymphocytes in the defense mechanism of the body.
Answer:
B lymphocytes produce antibodies and it destroys the antigens, T- lymphocytes stimulate white blood cells and also destroys cancer cells.

Question 3.
What is the basis of grouping blood Into different types? Everybody cannot receive blood of all groups. Why?
Answer:
The basis of blood grouping is the presence of antigen seen on the surface of the red blood cells. When an antigen reaches one s blood, it stimulates defense activity to produce antibody. The antigen and antibody react each other and form a blood clot. Hence everyone cannot receive blood from aH blood groups.

Soldiers of Defense More Questions And Answers

Question 1.
Identify the diagram and mention the defense process taking place in A

Answer:
Skin.
A is the outermost Keratin layer. Keratin is a protein, it blocks the entry of germs.

Question 2.
Respiratory track is always free from germs. Why?
Answer:
Mucus in the trachea prevents the entry of germs into the lungs. The cilia in the bronchus wipe out dust that enters it. Cough and sneezing help to expel foreign bodies from the respiratory tract. So respiratory tract is always free from germs.

Question 3.
Complete the table suitably.

Answer:
a) Ear
b) Hydrochloric acid
c) eye / mouth
d) The outermost layer blocks the entry of germs/ Keratin / Sebum / Acids.

Question 4.
Is swelling of the wound site helpful or not? why?
Answer:
Yes, It is helpful. The wounds and cuts occur in the skin, that area swells and blood vessels dilate It increases the blood flow and more white blood cells can come out through the enlarged pores and destroy the germs.

Question 5.
Generally, bacteria are useful but some of them are pathogenic. How?
Answer:
After entering the body they multiply by binary fission and produce certain toxic substances, which either disrupt the cellular activities or destroy the cell itself.

Question 6.
Following are certain steps of a defense process identify the process.
1. Phagocytes reach near the pathogens.
2. Engulf pathogens in the membrane sac
3. Membrane sacs combine with lysosome.
4. The enzyme in the Iysosome destroys the pathogens.
5. Expels the remnants from phagocyte.
Answer:
Phagocytosis

Question 7.
Observe the illustration and identify the process.

Answer:
Phagocytosis

Question 8.
What are the factors needed for blood clotting?
Answer:
Prothrombin, Fibrinogen in plasma, Calcium ions, Vitamin K, Red blood cells, Platelets.

Question 9.
Blood clotting is a defense mechanism to prevent the loss of blood through wounds. Mention the different stages of this process.
Answer:

• Tissues of the wounded part degenerate to form the enzyme thromboplastin.
• Thromboplastin converts prothrombin in the plasma to thrombin.
• Thrombin converts the fibrinogen in the plasma to fibrin.
• Blood clot is formed by the entangling of platelets and red blood cells in the fibrin network.

Question 10.
Complete the flowchart showing the blood clotting.

Answer:
A) Thrombin
B) Fibrin

Question 11.
“Fever is not a disease, it is a defense mechanism.” Analyze the statement.
Answer:
Yes. The presence of toxin produced by the pathogens stimulate the white blood cells and hence the white blood cells produce chemical substance that raises the body temperature. The rise in body temperature reduces the rate of multiplication of pathogens and increases the rate of phagocytosis.

Question 12.
How is antibody destroy germs?
Answer:
Antibody destroys the bacteria by disintegrating their cell membrane and neutralize the toxin of the antigens by stimulating other white blood cells.

Question 13.
Define the following
1. Antigen
2. Antibody
3. Antibiotic
Answer:
Any foreign body that stimulates the defense mechanism is called an antigen. The chemical substance produced by the lymphocytes act against antigen is an antibody. Antibiotics are medicines used to resist bacterial diseases.

Question 14.
The graph shown below represents the difference in the number of two bacteria when taken a particular antibiotic by a patient.

Answer:
a) This antibiotic is effective and reduces the number of pathogens. But its antibiotics. It also destroys useful bacteria in the body.
b) Further use of these antibiotics is not effective because the harmful bacteria got resistance against it.

Question 15.
Complete the boxes according to the given hint.

Answer:
A – Regular use develops immunity in pathogens against antibiotics
B – Destroys useful bacteria in the body.
C – Reduces the quantity of some vitamins in the body.

Question 16.
Can a person with ‘A’ group blood receive blood from ‘B’ group person? Or it take place vice versa? Give reason for this.
Answer:
A person with ‘A’ group blood cannot receive or donate blood with ‘B’ group person. Because the antigen present in the received blood and antibody in the recipient’s blood will react each other and form a blood clot (Coagulation).

Question 17.
Give more examples of vaccine
Answer:

Question 18.
“Germs, both alive and dead are used to get immunity”. Substantiate the statement with vaccines used for rabies and tuberculosis. (March 2015)
Answer:
Germs, both alive and dead are used as vaccines. ‘ Dead germs are utilized in rabies vaccine which acts against rabies. Live, but inactivated vaccines are used in BCG vaccine against tuberculosis.

Question 19.
Observe the following figure and answer the given questions.

a) Label A and B.
b) How did they protect our body? (March 2015)
Answer:
a) A – small hair, B – Sebaceous gland
b) Since sebum is oily, water does not stick on to the skin. Covering of hair protects the body from cold and heat and also prevents the entry of foreign bodies.

Question 20.
Whichever be the type of germs infected, the initial symptom appear in human body will be the fever. Give reason. (March 2014)
Answer:
Bacterial infection produces toxic substances in body, body temperature is suitable for bacterial growth, in order to control the growth of bacteria body rises the temperature, fever is not a disease.

Question 21.
Constant use of antibiotics is not good for health. This is the opinion of Rahim.
a) Do you agree with his opinion? Why?
b) Give two specific examples for justifying your answer. (Model 2013)
Answer:
a) I agree with this. Constant use of antibiotics results a few side effects.
b) Constant use of antibiotics may destroy useful bacteria in the body, develop resistance in bacteria against antibiotics or reduces the level of certain vitamins in the body.

Question 22.
How does the influence of the following action blocking germs.
a) Rise in body temperature.
b) Low oil content on skin.
c) Swelling occurs near wound.
d) Lymphocytes produce Antibodies. (March 2013)
Answer:
a) To resist the strengthening or increasing of causative organisms.
b) Waterproof and oily, germs cannot grow.
c) Flow of blood through the capillaries increases and more leucocytes comes out from the capillaries and destroy the germs by engulfing them.
d) Lymphocytes produce antibodies to destroy germs.

Question 23.
Whichever be the type of germs infected, the initial symptom appear in human body will be the fever Give reason. (March 2013)
Answer:
Bacterial infection produces toxic substances in body body temperature is suitable for bacterial growth, in order to control the growth of bacteria body rises the temperature, fever is not a disease.

Question 24.
The graph representing the difference in the number of two bacteria, while applying a particular antibiotic on a patient is shown below

A. Analyze the graph and record the findings.
B. Is this antibiotic effective against the bacteria? Why? (Model 2012)
Answer:
A. The harmful bacteria decrease in number in the first few weeks. Later they increase in number. Number of useful bacteria are decreasing gradually.
B. The antibiotic is not effective because the harmful bacteria got resistance against it. Moreover, number of useful bacteria decreases.

Soldiers of Defense Questions and Answers

Question 1.
Which among the following is the odd one? Why? Lymphocyte, Monocyte, Neutrophil, Basophil, Eosinophil (Question Pool-2017)
Answer:
Lymphocyte – Involved in specific defense

Question 2.
Skin is the largest sense organ of the body. It helps! us to sense heat, cold, touch, pressure, etc and it j acts as a soldier of defense of the body,
a) Does the skin have significance in defense as mentioned above? Justify. (Question Pool – 2017)
Answer:
Yes. The outermost Keratin, the protein layer blocks the entry of germs; sebum and some acids in the: skin-are disinfectants,

Question 3.
A table indicating primary level defense is given below. Arrange column B based on column A. (Question Pool – 2017)

Answer:
i) – c
ii) – d
iii) – a
iv) -b

Question 4.
Which among the following is the odd one and why? (Question Pool – 2017)
a) The Mucus of trachea destroys the pathogens.
b) The wax in the ear destroys pathogens.
c) Neutrophil destroys pathogens by engulfing them.
d) Lysozyme present in Saliva destroys pathogens
Answer:
C, Secondary defense

Question 5.
Nimisha’s hand got injured in an accident. After some time the wound area got swollen.
a) What is this type of activity known for?
b) Is it a defense mechanism? Why?
Answer:
a) Inflammatory response
b) Yes
Secondary level defense
Process to destroy pathogens in the body

Question 6.
Using the following statements,-prepare a flow chart of inflammatory response. (Question Pool-2017)
a) Production of chemical messages.
b) White blood cells destroy pathogens.
c) Blood vessels dilate.
d) Pathogens enter into the wound.
e) White blood cells come out from blood vessels.
f) Blood flow increases
Answer:
d
a
c
f
e
b

Question 7.
The given illustration includes white blood cells which act as a part of nonspecific defense. Fill up the blanks and complete the word web. (Question Pool -2017)

Answer:
A – Neutrophil /Monocyte
B – Stimulates other white blood cells / dilates blood vessels
C – Eosinophil
D – Engulfs and destroys germs

Question 8.
When there is an injury or wound, the blood vessel of that part dilates. (Question Pool – 2017)
a) What is its benefit?
b) Which white blood cell dilates the blood vessel?
Answer:
a) The cells that get damaged by a wound or an infection produce certain chemical substances. These substances dilate the blood vessels thereby increasing the blood flow. Blood plasma and white blood cells reach the wound site and it destroys the germs,
b) Basophil

Question 9.
Observe the given illustration and answer the following questions. (Question Pool-2017)

Answer:
a) Which is the process indicated in the illustration?
b) Which are the.white blood cells involved in the process?
c) Is it a specific defense mechanism? Justify
Answer:
a) Phagocytosis
b) Neutrophil, Monocyte
c) No
does not identify and destroy pathogens that enter to the body.

Question 10.
The flow chart given below indicates a type of defense mechanism occurring in the body. (Question Pool – 2017)

a) Complete the flow chart
b) Which process is it related to?
Answer:
a) i) Engulfs pathogen in the membrane sac
ii) The enzyme in the lysosome destroys the pathogens
iii) Expels the remnants
b) Phagocytosis

Question 11.
Blood clotting is a defense mechanism. Analyze the statement. (Question Pool – 2017)
Answer:

• Prevents the entry of germs through wound
• Prevents bleeding through wounds

Question 12.
Prepare the flow chart of the clotting of blood using the following statements. (Question Pool-2017)
a) Thromboplastin converts prothrombin to thrombin.
b) Blood flows from the wound.
c) Blood clot is formed.
d) Thrombin converts fibrinogen to fibrin.
e) Tissues degenerate to form the enzyme called thromboplastin.
f) The red blood cells and platelets entangle in the fibrin network.
Answer:
d
a
c
f
e
b

Question 13.

a) Identify A
b) B is a vitamin and C is an enzyme. Name them.
c) How does the lack of B or C affect the consequent chemical process? (Question Pool – 2017)
Answer:
A – Prothrombin
B – Vitamin K, C – Thromboplastin
C – Thrombin not formed
fibrinogen not converted to fibrin

Question 14.
Blood clot is formed by the entangling of red blood cells and platelets in the fibrin network.
White blood cells are not involved in this process. What explanation will you give for this? (Question Pool-2017)
Answer:

• White blood cells do not have a definite shape
• They come out through the fibrin network

Question 15.
One of the scars of the wound obtained by Binu while playing football remained even after 10 years. What explanation will you give for the scar remaining as such? (Question Pool-2017)
Answer:
When wound occurs, new tissues cannot be formed, then connective tissues heal the wound. In such situations, the wound scar remains

Question 16.
Fever is a defense mechanism. Is the statement correct? Justify your answer. (Question Pool-2017)
Answer:
Yes. The presence of toxin produced by the pathogens stimulate the white blood cell and hence the white blood cells produce chemical substance that raises the body temperature. The rise in body temperature reduces the rate of multiplication of pathogens and increases the rate of phagocytosis.

Question 17.
Complete the illustration (Question Pool-2017)

Answer:
a) B – lymphocyte
b) T – Lymphocyte
i) Stimulates white blood cell and destroys pathogens
ii) Destroys the bacteria by disintegrating their cell membrane
iii) Stimulates defense cells
iv) Destroys the cell which is affected by virus.

Question 18.
After attending a class on immunity, Arun raised a question to his teacher. (Question Pool-2017)
“In spite of so many defense mechanisms in the body, why are we still affected by diseases?
a) What explanation will you give for Arun’s doubt?
Answer:

• Bad habits
• Unhealthy food habits
• Unhygienic
• Excess pathogens

Question 19.
The use of some modern equipment are given below. Identify the equipment. (Question Pool-2017)
a) To record electric waves in the brain.
b) To record electric waves in the heart muscle
c) To understand the structure of internal organs using ultrasonic sound waves.
Answer:
a) EEG
b) ECG
c) Ultrasound scanner

Question 20.
The doctor prescribed antibiotics to Sunil who is affected with cholera, but not to Anil who is affected with chickenpox. What is the reason? (Question Poo1 -2017)
Answer:
Antibiotics are used to prevent bacterial diseases. Chickenpox is a viral disease. Cholera is a bacterial disease. So the doctor prescribed antibiotics to Sunni

Question 21.
Enlist the demerits of antibiotics for Jose who is preparing for a seminar on the topic “The merits and demerits of Antibiotics. (Question Pool -2017)
Answer:

• The frequent use of antibiotics produces disease defense in pathogens.
• Destroys useful bacteria in the body.
• Reduces the level of some vitamins in the body.

Question 22.
Ashiq who met with an accident was in need of blood. Antigen A and D and Antibody b was identified in his blood. (Question Pool -2017)
a) Name his blood group?
b) Whose blood, among the following can be accepted by ashiq?
(i) Venu = A⁺
(ii) Amal- AB⁺
(iii) Suhara – AB^–
(iv) Anoop – A^–
Answer:
a ) A⁺
b) (i)VenuA⁺
(ii) Anoop A^–

Question 23.
The table given below indicates blood groups.

Answer:
i) A
ii) a
iii) AB
iv) No
v) O
vi) No

Question 24.
Box A includes the major components of vaccines and box B includes the diseases against which they are used. Match them appropriately. (Question Pool -2017)

Answer:
i) -d
ii) -c
iii) -b
iv) -a

Question 25.
Ravi prepared an illustration showing defense mechanisms in plants. Complete it. (Question Pool – 2017)

Answer:
a) Prevents the entry of germs which have crossed the cell wall, through cell membrane.
b) Bark
c) Cuticle in leaves
d) Cell wall

Question 26.
“This mode of treatment is a lifestyle in tune with nature rather than a mere method of treatment” This is a statement regarding a well-known mode of treatment.
a) Name the treatment.
b) Apart from this, name any two well-known modes of treatment. (Question Pool – 2017)
Answer:
a) Ayurveda
b) 1. Allopathy
2. Homeopathy/ etc.

Question 27.
Prepare two suitable placards to conduct an awareness rally in association with World Blood Donation day. (Question Pool-2017)
Answer:
2 placards contain appropriate concepts
Example: Donate blood Donate Life
Blood donation – Nothing to loose profits – Life

Question 28.
Match the following pairs (Question Pool – 2017)
a) T-lymphocyte: Thymus gland
B – lymphocyte:………………
b) EEG: to record electric waves in brain
………… to record electric waves in heart muscles
c) First Antibiotic: Alexander Flemming
First vaccine:……………..
d) Heartbeat: Stethoscope
Blood pressure:……………..
e) Antigen: Red blood cells
Antibody:…………………
Answer:
a) Bone marrow
b) ECG
c) Edward Jenner
d) Sphygmomanometer
e) Plasma

Question 29.
Given below is an equipment used for disease diagnosis. (Question Pool-2017)

a) Identify the equipment
b) What is its use?
c) Name another equipment that works on the same principle.
Answer:
a) ECG
b) To record electric waves in heart muscles
c) EEG

Question 30.
“it is possible to build up a healthy society with hospitals, doctors, and medicines” This is Bashir’s opinion. Evaluate it (Question Pool -2017)
Answer:

• The opinion of Bashir is wrong
• Nutritious food
• Healthy lifestyle
• Hygiene, These are the factors which build up a healthy society.

Question 31.
“It is not necessary to detect blood groups if we can accept blood from anyone”. This was an argument put forward by Sivaprasad in a discussion on blood transfusion.
a) What is the base of blood group determination?
b) Can a person receive any blood from anyone? Why? (Question Pool – 2017)
Answer:
a) The presence of antigens A and B on the surface of Red blood cells,
b) 1. Not possible
2. When a foreign antigen reaches one’s blood, it stimulates the defense activity
3. The antigen present in the received blood and the antibody in the recipients, blood will react each other to forms a blood clot (coagulation)

Question 32.
Given below is the picture of white blood cells which are parts of specific defense? 32 (Question Pool – 2017)

a) Identify A and B
b) What is the role of A in specific defense?
c) Give anyone difference between A and B
Answer:
a) A – T- lymphocyte
B – B lymphocyte
b) 1. Stimulates other defense cells
2. Destroys cancer cells and virus affected cells.
c) B lymphocytes matured at bone marrow T lymphocytes matured at thymus

Question 33.
Analyze the following statement and answer the following questions. 33 (Question Pool-2017)
When there is a wound, the body temperature rises.
a) What is the significance of white blood cells in this activity?
b) How does immunity become possible through a rise in temperature?
Answer:
a) The chemical substances produced by white blood cells rise body temperature
b) The presence of toxin produced by the pathogens stimulate the white blood cell and hence the white blood cells produce chemical substance that raises the body temperature. The rise in body tempera¬ture reduces the rate of multiplication of pathogens and increases the rate of phagocytosis.

Question 34.
Statements related to nonspecific defense and specific defense are given below. Identify the type of the defense and mark them using the letters N and S respectively.
(Orukkam – 2017)
a) The cilia in bronchus wipe out dust that enters it.
b) Destroy the bacteria by disintegrating their cell membrane.
c) The blood vessels near the wound diabetes.
d) The rise in body temperature reduces the rate of multiplication of pathogens.
e) B lymphocytes produce certain chemical substances against antigens.
f) Eosinophil produces chemical substances needed for inflammatory responses.
g) T lymphocytes destroys cancer cells.
h) The enzyme lysozyme present in tears destroys germs.
i) T lymphocytes destroy cancer cells.
j) Phagocytes engulf and destroy germs.
Answer:
a) N
b) S
c) N
d) N
e) S
f) N
g) S
h) N
i) S
j) N

Question 35.
Our body has the capacity to destroy germs those enter the body by breaking the first level defense Write your comment on this statement? (Orukkam – 2017)
( Hints: inflammation, different types WBCs and their functions, phagocytosis)
Answer:
The statement is true. When germs enter to the body the blood vessels diates (inflammation). This helps the white blood cells to act at the site A few white blood cells destroy germs by phagocytosis.

Question 36.
The basis of blood grouping is the presence of antigens in red blood cells. Complete the table given below based on this statement. (Orukkam-2017)
Blood groups

Answer:

a) B lymphocytes produce antibodies against antigens.
b) We can use antigens as vaccines for the formation of antibodies in advance.
c) Neutralized toxins – Diphtheria
Alive but neutralized germs – Measles
Cellular parts of pathogens – Hepatitis B Killed germs – Cholera

Question 37.
Complete the illustration suitability related to antibiotics (Orukkam – 2017)

Answer:
A) Prevent bacterial diseases
B) Prolonged use may develop immunity in germs
C) Destruction of useful bacteria
D) Deficiency of certain vitamins in the body

Question 38.
The wound scar does not remain always. Write reason? (Orukkam – 2017)
Answer:
If the wound is filled with same tissue, no wound scar occurs there.

Question 39.
Fill in the blanks by observing the relationship in the first pair. (Orukkam – 2017)
a) EEG: To record electric waves in the brain
ECG:……………………………..
b) Rabies: Killed germs
Typhoid:………………………..
Answer:
a – Records electric waves in the heart muscles
b – Alive but neutralized germs.

Question 40.
Name the first vaccine? Who developed this? Write the situation which leads to the development of vaccine? (Orukkam – 2017)
Answer:
Smallpox vaccine, developed by Edward Jenner. He observed no smallpox disease in people who had affected cowpox earlier.

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Kerala SSLC Maths Model Question Paper 3 Malayalam Medium