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Students can Download Social Science Part 2 Chapter 8 For a Safer Future Questions and Answers, Summary, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 9th Standard Social Science Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Social Science Solutions Chapter 8 Population, Migration, Settlements in Malayalam Medium

Population, Migration, Settlements Questions and Answers in Malayalam

You can Download मेरे बच्चे को सिखाएँ Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Hindi Solutions Unit 2 Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

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 Ocean and Man Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Social Science Solutions Part 2 Chapter 5 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Social Science Solutions Part 2 Chapter 5 Ocean and Man

Ocean and Man Textual Questions and Answers

Ocean And Man Class 9 Kerala Syllabus Question 1.
Identify the location of each ocean from the world map. List the straits, bays and the seas of each ocean with the help of an atlas.
Answer:

Kerala Syllabus 9th Standard Social Science Notes Question 2.
The following table contains the names of some major islands and peninsulas in the world. With the help of an atlas find out the names of the oceans to which they below.
Answer:

1. Indian Ocean
2. Indian Ocean
3. Pacific Ocean
4. Atlantic Ocean
5. Atlantic Ocean

Kerala Syllabus 9th Standard Social Science Notes Pdf Question 3.
Complete the table using fig. 5.7 in textbook


Answer:

9th Standard Social Science Notes Pdf Kerala Syllabus Question 4.
The equatorial regions; record a high amount of salinity as compared to the polar regions why?
Answer:
The temperature is high in equatorial regions compared to polar regions. Density increases as temperature rises. High density is associated with high salinity. There is possibility of high evaporation in equatorial regions. That is why equatorial regions record high amount of salinity.

9th Standard Social Science Notes Kerala Syllabus  Question 5.
Why is salinity is less at river mouths?
Answer:
Salinity is less at river mouths because of huge amount of freshwater added from hundred of rivers.

9th Std Social Science Notes Kerala Syllabus Question 6.
Which are the warm and cold currents of the Atlantic Ocean? Identify the continents near which they flow.

Answer:

Warm currents	Continents near which they flow
1. North Atlantic current	1. Europe
2. Gulf Stream	2. North America
3. Florida current	3. North America
4. North equatorial currents	4. South America & Africa
5. Equatorial counter currents	5. South America & Africa
6. South Equatorial currents	6. South America & Africa
7. Brazilian current	7. South America

Cold current	Continent near which they flow
1. Labrador current	1. North America
2. West wind drift	2. South America
3. Benguela current	3. Africa
4. Canaries current	4. Europe of Africa

Kerala Syllabus 9th Standard Social Science Notes English Medium Question 7.
Complete the following table based on the currents of the Indian Ocean.


Answer:

Currents	Warm/Cold	Direction
1. South equatorial current	1. Warm	1. From East to West
5. South West Monsoon current	5. Warm	2. From West to East
6. Agulhas current	6. Warm	3. From North to West
7. West Australian current	7. Cold	4. From South to North
8. West wind Drift	8. Cold	5. From West to East

Kerala Syllabus 9th Standard Social Science Notes Malayalam Medium Question 8.
You have learnt the uses of Oceans. Conduct a seminar on the topic ‘influence of Oceans in human life’.
Answer:
“Influence of Oceans in Human Life”
Oceans play significant role in the life of human beings. We cannot neglect oceans because oceans are useful in

• Influencing our climate.
• Providing mineral deposits.
• Helping power generation.
• Providing source of food.

We shall now explain the influence of Ocean in detail.
Climate:
Oceans have a decisive role in controlling the climate along the coastal regions. The sea breeze during the day and the land breeze in the night regulate the temperature over the coasts. Oceans play a part in the formation of weather phenomena like rain, wind, and cyclones. Generally, the coastal regions have moderate climate, whereas severe summer and winter prevail in regions away from the sea.
Oceans as a source of food:
Fish is an important item of food. Fishing is major
activity in Japan, Peru, China, Norway, and the United States of America. Marine organisms are the source of many medicines. They are used for the production of antibiotics, steroids, and vitamins.

Ocean and Man Model Questions and Answers

9th Standard Social Science Notes Pdf In English Kerala Syllabus Question 9.
Nearly ………… % of earth’s surface area is covered with water.
Answer:
71%

9th Class Social Science Textbook Kerala Syllabus Question 10.
Name the Oceans
Answer:
a) The Pacific Ocean
b) The Atlantic Ocean
c) The Indian Ocean
d) The Artie Ocean
e) The Antarctic Ocean

9th Standard Social Science Textbook Kerala Syllabus Question 11.
Match the following

A	B
Pacific ocean	Southern ocean
Atlantic ocean	Warton trench
Indian ocean	Puerto Rico
Antarctic ocean	Challenger Deep

Answer:

A	B
Pacific ocean	Challenger Deep
Atlantic ocean	Puerto Rico
Indian ocean	Warton trench
Antarctic ocean	Southern ocean

Kerala Syllabus 9th Standard Social Science English Medium Question 12.
Distinguish between bay and strait.
Answer:
The portion of the sea surrounded by land on three sides is called a bay. On the other hand, the narrow stretch of sea between two landmasses is known as strait.

Question 13.
Complete the table

Islands	Peninsula
Sri Lanka
Maldives
Green land
Sumatra

Answer:

Islands	Peninsula
Sri Lanka	Indian Peninsula
Maldives	Arabian Peninsula
Green land	Alaska Peninsula
Sumatra	Labrador Peninsula

Question 14.
Define Peninsula
Answer:
The landmasses surrounded by sea on three sides are called peninsula.

Question 15.
Identify the important features of seawater.
Answer:

• Temperature
• Salinity
• Density

Question 16.
Where do you find the highest Ocean temperature?
Answer:
Between 10° latitudes on either side of the equator

Question 17.
As we more away from the equator, temperature considerably.
a) Increases
b) Decreases
c) First increases and then decreases
d) Remains constant
Answer:
a) Increases

Question 18.
What is the reason for the variation in temperature over different latitudinal zones?
Answer:
Variation in the amount of insolation received on the earth is the major reason, The Ocean currents and winds also influence the temperature of seawater.

Question 19.
Define Salinity?
Answer:
The concentration of salt content in seawater is known as salinity.

Question 20.
The average amount of saltiness of seawater is
a) 2.5%
b) 3.5%
c) 4.5%
d) 5.5%
Answer:
b) 3.5%

Question 21.
Name the major contents of seawater
Answer:

• Sodium chloride
• Magnesium chloride
• Magnesium Sulphate

Question 22.
Point out the conditions leading to variations in salinity.
Answer:

• Salinity will be more in landlocked seas.
• Salinity increases in areas of high evaporation.
• Salinity decreases in areas where snow meltwater reaches in large quantity.
• Salinity decreases at river mouths.
• Heavy rainfall leads to reduction in salinity.

Question 23.
What causes movements in seawater.
Answer:
The density of seawater is not uniform everywhere. This is due to the variations in salinity and temperature of sea water. Density decreases as temperature increases, and it increases as salinity increases. You have understood that the temperature, salinity and the density of seawater are not uniform everywhere. These variations lead to movements in seawater.

Question 24.
Which are the movements of seawater.
Answer:

• Waves
• Tides
• Ocean currents

Question 25.
What are sea waves?
Answer:
The up and down motion of the water along the surface of the sea is called sea waves.

Question 26.
Mark the missing portion.

Answer:

1. WaveLength
2. Crest
3. Wave height
4. Trough

Question 27.
Name the following:
(i) The summit of the wave
(ii) Bottom part of the wave
(iii) Distance between 2 adjacent crests
(iv) Vertical distance between the crest and the trough
Answer:
(i) Wave crest
(ii) Wave trough
(iii) Wavelength
(iv) Wave height

Question 28.
………. is the reason for waves.
Answer:
The friction exerted by winds on the ocean surface

Question 29.
As the speed of the wind increases, the strength of the wave
a) Decreases
b) Increases
c) Remains constant
Answer:
b) Increases

Question 30.
Name two damages caused by strong waves
Answer:

• Cyclones
• Shelving of shores

Question 31.
Some measures are taken to prevent damage and to protect the lives of people living in the coastal areas. Identify the measures.
Answer:

• Depositing boulders along the seashore.
• Construction of interlocking concrete structures (Pulimuttu)
• Planting of mangroves.

Question 32.
Name the sea waves generated by earthquakes and volcanos
Answer:
Seismic sea waves or tsunami waves.

Question 33.
Prepare a note on mud bank
Answer:
Mudbank is a phenomenon that develops in the Arabian Sea during the onset or the end of the monsoon season. Planktons grow luxuriantly in the turbulent muddy water along the seashore during the monsoon rains. Schools of fish such as shrimp, sardine, and mackerel arrive to feed on the planktons and the mud, giving fishermen a good catch. This phenomenon is known as mud bank.

Question 34.
What do you mean by tides?
Answer:
Tides are he periodic rise and fall of water level in the ocean.

Question 35.
Distinguish between high tide and low tide
Answer:
The rise in the level of ocean water is the high tide and the lowering of the water level is known as the low tide.

Question 36.
What are the reasons for tides?
Answer:
Tides are formed as a result of the gravitational pull exerted by the moon and the sun along with the centrifugal force due to the earth’s rotation.

Question 37.
‘ Illustrate with the help of a figure the occurrence of high tide and low tide.

Answer:
The water level on the part of the earth facing the moon rises. The rise in water level due to the gravitational pull exerted by the moon leads to high tide. You might have noticed that the water level at the opposite side also has risen. The centrifugal force due to the earth’s rotation is the reason for the rise in water level here. It can be seen that the water level goes down at places located 90° away from the places of tidal influence. This is due to the draining of water towards the tidal regions. The phenomenon of fall of water level is known as low tide.

Question 38.
Point out the important effects of tides.
Answer:
High tides and low tides have many effects. Let’s have a look at them.

• The debris dumped along the seashores and ports are washed off to the deep sea.
• The formation of deltas is disrupted due to strong tides.
• Brackish water can be collected in salt pans during high tides.
• The fishermen make use of the tides for going and returning from the sea in catamarans.
• Tidal energy can be used for power generation.
• Ships can be brought to shallow harbors during high tides.

Question 39.
Prepare a note on ocean currents
Answer:
It is a type of seawater movement. Ocean currents are the continuous flow of seawater from one direction to another. They can be classified as warm currents and cold currents. Warm currents are the currents that flow from the tropical or subtropical regions towards the polar or subpolar regions. The water that flows in will be warmer than the water at the destination.

Similarly, cold currents are the currents that flow in from the polar or the subpolar regions towards the tropical or subtropical regions.In this case, the water that flow in will be colder than the water at the destination. The temperature and salinity of seawater varies from ocean to ocean. This difference leads to density difference in seawater. The difference in density is one of the factors that cause ocean currents.

Question 40.
Complete the chart.

Answer:

1. Warm currents
2. Cold currents

Question 41.
What are the effects of ocean currents?
Answer;

• Influence the climate of coastal regions.
• Fog develops in the regions where warm and cold currents meet.
• The regions where the warm and cold currents meet provide favorable conditions for the growth of fish.

Question 42.
Which are the mineral deposits in ocean.
Answer;

• Iron ore
• Coal
• Petroleum
• Natural gas

Question 43
The oil field started in 1974 in Mumbai is known as
Answer:
Mumbai High

Question 44.
Name some countries where fishing is an important activity.
Answer:

• Japan
• Peru
• China
• Norway
• USA

Question 45.
What are the medicinal user of marine organisms?
Answer:
Marine organisms help in production of :-

• antibiotics
• steroids
• vitamines

Question 46.
Prepare a flow chart showing method of purifying seawater.
Answer:

Let Us Assess

Question 47.
Which among the following statements is not related to the Indian Ocean?
a. The southern part of this ocean extends up to the Antarctic Ocean.
b. The average depth is more than that of the Atlan tic Ocean.
c. The Puerto Rico trench is situated in this ocean.
d. It ranks third in area.
Answer:
B & C are not related

Question 48.
which among the following places record the least salinity? why?
1. Land-locked sea,
2. Areas of heavy rainfall
3. Areas of high evaporation
Answer;
Landlocked sea. Because salinity increases in landlocked seas.

Question 49.
Is there any relation between the intensity of waves and the wavelength? substantiate
Answer:
The Intensity of waves is defined as the power delivered per unit area. The unit of intensity will be W.m2. The wave energy comes from simple harmonic motion of its particles. The total energy will equal the maximum kinetic energy. As intensity of waves increases the wavelength also increases.

Question 50.
High tide occurs twice a day. Explain this statement.
Answer:
The rise in water level due to the gravitational pull exerted by the moon leads to high tide. Tides can occur as two high waters and two low water each day. The tides are occurred by the gravitational. The moon’s gravity pulls the ocean surface closer to it and the moon makes two trips around the earth each day. Hence there in high tide twice a day.

Question 51.
Explain spring tides and neap tides with the help of diagrams.

The sun, moon, and earth come in a straight line on full moon and new moon days. The tidal force will be intense due to the combined influence of sun and moon. As a result, the tides formed on these days will be stronger. These are known as spring tides. The moon and the sun will be at an angular distance of 90° from the earth after seven days from the full moon and new moon days. As the sun and the moon attract the earth from an angular distance of 90° the tides casued are weak. Such weak tides are known as neap tides. Note the positions of the earth, moon, and sun in the given diagram.

Question 52.
Oceans play an important role in human life and environment. Justify.
Answer:
Influence of Oceans in human life
Oceans play significant role in the life of human beings. We cannot neglect oceans because oceans are useful in

• Influencing our climate.
• Providing mineral deposits.
• helping power generation.
• Providing source of food.

We shall now explain the influence of ocean in detail.

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.

Element	Atomic mass	Electronic configuration
Magnesium	12	2, 8, 2
Oxygen	8	2, 6
Sodium	11	2, 8, 1
Chlorine	17	2, 8, 7

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:

Element molecule	Shared electron pairs	Chemical bond
F2	One pair	Single bond
Cl2	One pair	Single bond
O2	Two pair	Double bond
N2	Three pairs	Triple bond

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:

Compounds	Electronegativity difference of the elements	Character of compound
Carbon Tetra Chloride [CCl4]	3.44 – 2.55 = 0.89	Covalent bond
Sodium chloride [NaCl]	3.16 – 0.98 = 2.23	Ionic bond
Methane [CH4]	2.55 – 2.20 = 0.35	Covalent bond
Magnesium chloride [MgCl2]	3.16 – 1.31 = 1.85	Ionic bond
Sodium oxide[Na2O]	3.44 – 0.93 = 2.51	Ionic bond

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:

Properties	Ionic compound	Covalent compounds
State	Solid	Found in the three states solids, liquids and gases
Solubility in water	soluble in water	Insoluble in water. But soluble in organic solvent like kerosene, benzene etc.
Electrical Conductivity	conduct electri­city in fused or solution state	Do not conduct Electricity
Melting point Boiling point	High	Generally Low

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:

Compound	Component elements	Atomic number	Electron configur­ation	No. of elect­rons donated accepted/ shared	Valency
NaCl	Na	11	2, 8, 1	1	1
	Cl	17	2, 8 ,7	1	1
MgO	Mg	12	2, 8, 2	2	2
	0	8	2, 6	2	2
HF	H	1	1	1	1
	F	9	2, 7	1	1
CCl4	C	6	2, 4	4	4
	Cl	17	2, 8, 7	1	1

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.

Element	Valency
Cl	1
Li	1
0	2
Zn	2
Ca	2

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)

Element	Atomic number	Electronic configuration
P	9	2, 7
Q	17	…………………………..
R	10	…………………………..
S	12	……………………………

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:

Element	Atomic number	Electronic configuration
P	9	2, 7
Q	17	2, 8, 7
R	10	2, 8
S	12	2, 8, 2

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

Element	Valency
Ba	2
Cl	1
Zn	2
O	2

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 Reactivity Series and Electrochemistry Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Chemsitry Solutions Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Chemistry Solution Chapter 3 Reactivity Series and Electrochemistry

Reactivity Series and Electrochemistry Text Book Questions and Answers

Reactivity Series And Electrochemistry Kerala Syllabus Text Book Page No: 48

→ Which metal reacts vigorously?
Answer:
Sodium.

→ Which gas is formed as a result of this reaction?
Answer:
Hydrogen.

→ Write down its chemical equation.
Answer:
2Na + 2H₂O → 2NaOH + H₂

→ Complete the table (3.1) given below.

Answer:

→ Based on your observation, arrange these metals in the decreasing order of their reactivity.
Answer:
Sodium > Magnesium > Copper
→ 2Mg + O₂ →
Answer:
2Mg + O₂ → 2MgO

Sslc Chemistry Chapter 3 Kerala Syllabus Text Book Page No: 49

→ Which metal among magnesium, copper, gold, sodium and aluminium, loses its lustre at a faster rate?
Answer:
Sodium

→ List the above metals in the decreasing order of their reactivity with air and thereby losing lustre
Answer:
Sodium > Magnesium > Aluminium > Copper > Gold.

Text Book Page No: 50

→ What happened to the Zn rod?
Answer:
Before the experiment the Zn rod was colourless. After the experiment Zn rod became blue due to the deposition of copper.

→ What is the reason for this?
Answer:
When the Zn rod is dipped in CuSO₄ solution, the Cu²⁺ ions in the solution get deposited at the Zn rod as Cu atoms.

→ What is the reason for the change in intensity of the colour of CuSO₄ solution?
Answer:
The blue colour of CuSO₄ solution is due to the presence of Cu²⁺ ions. The change in intensity of the colour of CuSO₄ solution because when the Zn rod is dipped in CuSO₄ solution, the Cu²⁺ ions in the solution get, deposited at the Zn rod as Cu atoms.

→ Which is the metal that gets displaced here?
Answer:
Copper

→ Which is more reactive Zn or Cu?
Answer:
Zn

→ On the basis of the position of Zn and Cu in the reactivity series, can you explain why Cu had been displaced?
Answer:
Zn is placed above Cu in the reactivity series because Zn has a higher reactivity than Cu.

→ Isn’t it due to the higher reactivity of zinc (Zn) when compared to copper (Cu)?
Answer:
Yes.

Chemistry Class 10 Chapter 3 Kerala Syllabus Text Book Page No: 51

→ Is this reaction oxidation or reduction? Why?
Answer:
Oxidation. Because the losing of electrons is called oxidation.

→ The change that happened to Cu²⁺
Answer:
Cu^–2+ + 2e^– → Cu

→ What is the name of this reaction? Why?
Answer:
Reduction. The gaining of electrons is called reduction.
→

Complete this chemical equation by assigning oxidation numbers.
Answer:
2 Ag⁺¹ NO₃^1–+ Cu°→ Cu²⁺ (NO₃)^–1₂ + 2Ag⁰

→ Which metal was oxidised in this case? Which metal was reduced?
Answer:
Metal which was oxidised: Cu
Metal ion which was reduced: Ag⁺

→ Write equations showing oxidation and reduction.
Answer:
Oxidation : Cu⁰ → Cu²⁺ + 2e^–
Reduction : Ag⁺+ le^– → Ag⁰

Reactivity Series And Electrochemistry Sslc NotesText Book Page No: 52

→ Complete the table 3.3.

Answer:

Sslc Chemistry Chapter 3 Notes Pdf Kerala Syllabus Text Book Page No: 53

→ Which electrode has the ability to donate electrons in a cell constructed using these metals?
Answer:
Zn

→ Which one can gain electrons?
Answer:
Cu

→ Identify the chemical reaction that takes place at the Zn electrode. Tick ✓ the right one.
Answer:
Zn(s) → Zn²⁺ (aq) + 2e^–  (✓)
Zn²⁺(aq) + 2e^– → Zn(s)   (✘)

→ What is the reaction taking place here?
Answer:
Oxidation.

→ Write the chemical equation for the reaction taking place at the Cu electrode.
Answer:
Cu²⁺ (aq) + 2e^– → Cu(s)

→ Sketch the cell constructed.
Answer:

→ Note down the reaction of the Galvanic cell.
Answer:
Cu(s) + 2Ag⁺(aq) → Cu²⁺ (aq) + 2Ag(s)
Cu(s) → Cu²⁺(aq) + 2e^– (Anode)
Ag⁺(aq) + le^– → Ag(s) (Cathode)

→ Direction of flow of electrons From Cu to Ag
Answer:
Mark the direction of electron how in the cell illustrated.

→ write the reactions taking place at cathode and anode.
Answer:
At cathode : Ag⁺ + le^– → Ag
At anode : Cu → Cu²⁺ + 2e^–

Hss Live Guru 10th Chemistry Kerala Syllabus Text Book Page No: 55

→ You have used three metals Zn, Cu and Ag. How many cells can be produced using these?
Answer:
Three.

→ Complete the Table 3.4 by writing anode and cathode in each.

Answer:

→ What are the substances obtained when electricity is passed through acidified water?
Answer:
Hydrogen, Oxygen.

→ Do such type of chemical changes happen when electricity is passed through metals?
Answer:
Yes.

Sslc Chemistry Chapter 3 Questions And Answers Kerala Syllabus Text Book Page No: 56

→ To which electrodes are the positive ions attracted during electrolysis?
Answer:
Towards negative electrodes(Cathode)

→ To which electrodes are the negative ions attracted?
Answer:
Towards positive electrodes(Anode),

→ What changes happen to the ions which N are attracted to cathode?
Answer:
Reduction

→ What about the changes happening to the ions attracted to anode?
Answer:
Oxidation

→ Which ion is attracted to the positive electrode (anode)?
Answer:
Chloride ion (Cl^–)

→ What is the chemical reaction taking place there?
Answer:
2Cl^– (aq) → Cl₂(g) + 2e^–

Text Book Page No: 57

→ Which is the gas liberated at the anode?
Answer:
Chlorine(Cl₂)

→ Which is the ion attracted to the negative electrode (cathode)? Write the change happening to it?
Answer:
Na⁺ ions. These ions accept one electron and changes to sodium atom. That is sodium ions are reduced.

→ Which is the metal deposited at the cathode?
Answer:
Sodium (Na)

→ Which are the ions attracted to the positive electrode?
Ans.
Cl^–,OH^–

→ Which are the ions attracted to the negative electrode?
Answer:
Na⁺,
H₃O⁺,
H₂O.

Text Book Page No: 59

→ Which metal is connected to the negative terminal of the battery?
Answer:
Iron.

→ Which metal is connected to the positive terminal of the battery?
Answer:
Copper.

→ Which solution is used as the electrolyte?
Answer:
Copper sulphate solution.

→ What happens to Cu²⁺ ions at the cathode? Complete the equation.
Answer:
Cu²⁺ + 2e → Cu

→ What happened to the copper ions? Oxidation/Reduction?
Answer:
Reduction.

→ Complete the equation given below.
Answer:
Cu → Cu²⁺ + 2e^–

Text Book Page No: 60

→ Find out more examples and extend the list.
Answer:

• Chromium plating is used in motor car etc.
• To make metal coating easily corroding metals to prevent corrosion.
• In ICs (Integrated Circuits) coating of gold /silver is made by electroplating.

Reactivity Series and Electrochemistry Let Us Assess

Kerala Genetics Question 1.
The solutions of ZnSO₄, FeSO₄, CuSO₄ and AgNO₃ are taken in four different test tubes. Suppose, an iron nail is kept immersed in each one
In which test tube the iron nail undergoes a colour change?
What is the reaction taking place here?
Justify your answer. (Refer reactivity series of metals).
Answer:
Iron nail immersed in solution of CuSO₄ and AgNO₃ undergoes a colour change.
i. Fe(s) + CuSO₄(aq) →
FeSO₄ (aq) + Cu (s).
ii. Fe(s) + 2AgNO₃ (aq) →
Fe(NO₃)₂(aq) + 2Ag(s).
Iron displaces Cu from CuSO₄ and Ag from AgNO₃ because Fe has higher reactivity than Cu and Ag.

Genetic Engineering Syllabus Question 2.
Compare the electrolysis of molten potassium chloride and solution of potassium chloride. What are the processes taking place at the cathode and the anode?
Answer:
Molten KCl
KCl (s) → K⁺ + Cl^–
At the negative electrode:
K⁺ + le^– → K (reduction – cathode)
At the positive electrode:
2Cl^– → Cl₂ + 2e^– (oxidation- anode)
Solution of potasium chloride.
At the negative electrode:
2H₂O + 2e^– → H₂ + 2OH^– (cathode).
At the positive electrode:
2Cl^– → Cl₂ + 2e^– (anode).

Future Diary 10th Question 3.
You are given a solution of AgNO₃, a solution of MgSO₄, a Ag rod and a Mg ribbon. How can you arrange a Galvanic cell using these? Write down the reactions taking place at the cathode and the anode.
Answer:
At anode,
Mg (s) → Mg²⁺ (aq) + 2e^–
At cathode,
Ag (aq) + le^– → Ag (s)

Reactivity Series and Electrochemistry Extended Activities

The Reactivity Question 1.
1. Keep two carbon rods immersed in copper sulphate solution. Then pass electricity through the solution.
i. At which electrode does colour change occur anode or cathode?
ii. Is there any change in the blue colour of the copper sulphate solution?
iii. Write down chemical equations for the changes occurring here.
Answer:
i. At cathode
ii. Colour fades
iii.At cathode: Cu²⁺ (aq) + 2e^– → Cu (s)
At anode: 2H₂O → O₂(g) + 4H⁺ (aq) + 2e^–

SSLC Chemistry Chapter 4 Question 2.
When acidified copper sulphate solution is electrolysed oxygen is obtained at the anode. What arrangements are to be made for this? Find the element deposited at the cathode.
Answer:

Element deposited at the cathode: Copper.

Question 3.
a. When Galvanic cells are made using the metals like Mg, Cu, Zn and Ag, what will be the nature of reactions in each cell?
(Reactivity: Mg > Zn > Cu >Ag)
b. How many Galvanic cells can be made by using the metals Ag, Cu, Zn and Mg?
Chapter 7 Biology test Answers:
a. i. Cu-Ag cell
Anode: Cu (s) → Cu²⁺ (aq) + 2e^–
Cathode: Ag⁺ (aq) + le^– → Ag(s)
ii.Zn-Ag cell
Anode: Zn (s) → Zn²⁺ (aq) + 2e^–
Cathode: Ag⁺ (aq) + le^– → Ag (s)
(iii) Mg-Ag cell
Anode: Mg (S) → Mg²⁺ (aq) + 2e^–
Cathode: Ag⁺ (aq) + le^– → Ag (s)
(iv) Zn – Cu cell
Anode: Zn (s) → Zn²⁺ (aq) + 2e^–
Cathode: Cu²⁺ (aq)+ 2e^– → Cu(s)
(v) Mg-Cu cell
Anode: Mg (s) → Mg²⁺ (aq) + 2e^–
Cathode: Cu²⁺ (aq) + 2e^– → Cu (s)
(vi) Mg-Zn cell
Anode: Mg (s) → Mg²⁺ (aq) + 2e^–
Cathode: Zn²⁺(aq) + 2e^– → Zn(s)
b. 6 cells

Reactivity Series and Electrochemistry Orukkam Questions and Answers

Scope of Genetic Engineering Question 1.
1. Take cold water and Hot water in two test tubes, Add one or two drops of phenolphtha lein in it. Drop equally sized Mg ribbon in it.
a. In which test tube pink colour occured sharply?
b.Why did pink colour appear in that test tube so early?
c. Which gas evolved out from both test tubes?
d. Write balanced equation for the above mentioned reaction.
Answer:
a. Pink colour occurred sharply in the test tube with hot water.
b. Temperature is a factor that affects the rate of a reaction. When heated the kinetic energy of molecules increases and hence the rate of chemical reaction also increases which causes the pink colour to appear early.
c. Hydrogen
d. Mg(s) + 2H₂O(l) → Mg(OH)₂(aq) + H₂(g)

SCERT Question Pool 2017 Question 2.
Cut a small sodium metal piece into two, watch it.
a. What change occurred on the surface of sodium metal?
b.Write one word for the process of this type of decomposition.
c. Write down the equations for this.
Answer:
a. After some time the cut piece of sodium will turn dull.
b.Corrosion.
c. 4Na(s) + O₂(g) → 2Na₂O
2Na(s) + 2H₂O(l) → 2NaOH(s) + H₂(g)
2NaOH (s) + CO₂(g) → Na₂CO₃(s) + H₂O(l)

Question 3.
Take equal quantities of dil HCl in five test tubes. Drop Mg, Zn, Fe, Cu in each test tube. Watch carefully.
a. Arrange metals in decreasing order of reactivity.
b.Write balanced equations for each reaction.
SCERT Question Pool Answer:
a. Mg > Zn > Fe > Cu
b. Zn + 2HCl → ZnCl₂ + H₂
Fe + 2HCl → FeCl₂ + H₂
Cu + HCl → No reaction
Mg + 2HCl → Mg Cl₂ + H₂

SCERT Question Pool Question 4.
Some metals and metallic compounds are given in the table. If the metal substitute the metal in the compound put a tick mark in the corresponding column and otherwise a cross mark in the column. Write down correct answer based on the table given below.

Metalsolution	Mg	Cu	Zn	Ag	Fe
CuSO4		×		×
ZnSO4		×	×	×	×
AgNO4				×
MgSO4				×

a. Correct the table if necessary.
b. Is it possible to substitute lower positioned metals by top positioned metals in the reactivity series?
c. What type of reaction is this?
d. Write down balanced equations for all the true sign given in the table.
Answer:
a.

Metalsolution	Mg	Cu	Zn	Ag	Fe
CuSO4	✓	×	✓	×	✓
ZnSO4	✓	×	×	×	×
AgNO4	✓	✓	✓	×	✓
MgSO4	✓	✓	✓	×	✓

b.Yes. It is possible.
c. Substitution reactions.
d.CuSO₄ + Mg → Cu + MgSO₄
CuSO₄ + Zn → ZnSO₄+ Cu
CuSO₄ + Fe → FeSO₄ + Cu
ZnSO₄ + Mg → MgSO₄ + Zn
AgNO₃ + Mg → Ag + MgNO₃
AgNO₃ + Cu → CuNO₃ + Ag
AgNO₃ + Zn → ZnNO₃ + Ag
AgNO₃ + Fe → FeNO₃ + Ag

Human Insulin Gene Question 5.
Draw maximum number of Galvanic cell using substances given in the table.
Salt bridge, Zinc rod, Copper rod, Voltmeter, Aluminium chloride, Copper sulphate, Zinc sulphate, Silver nitrate, Silver rod, Calcium chloride
a. Complete the table based on the figures drawn.

Galvanic Cell	Electrode which Gives Electron	Electrode which Gain Electron

B. Write down the general names used for an electrode which gives electrons,
c. Metals in that electrode in the reactivity series is (in the Top, Bottom)
d. General name of the Electrode which accepts electron.
e. Process of giving electron is
f. Process of Accepting electron is
g.Direction of the flow of Electron
h.Write down the balanced equation taking place in both electrodes.

Galvanic Cell	Electrode which Gives Electron	Electrode which accepts Electron

Answer:

Zn-Ag cell is also possible.
a.

b. Anode.
c. In the top.
d. Cathode.
e. Oxidation.
f. Reduction.
g. From Anode to Cathode.

Chemistry Reactivity Series Question 6.
Take Cupric chloride (CuCl₂) solution in a beaker. Dip two graphic rod in it. Pass 5V electricity through it.
a. Why electricity passes through cupric chloride solution?
b.Which gas evolved out through positive electrode? How did you identify that gas?
c. Which product formed in negative electrode?
d. In which electrode oxidation and reduction take place?
e. Write one word for the process of chemical change happening in a Electrolyte while passing Electricity?
Answer:
a. Electrolytes are substances which conduct electricity in molten states or in aqueous solutions. In molten state ions of CuCl₂ can more freely. These ions are responsible for the Conduction of electricity by electrolytes.
b.Chlorine.
c. Copper.
d.Oxidation takes place at the positive electrode and Reduction takes place at the negative electrode.
e. Electrolysis.

Reactivity Series Class 10 Question 7.
Take 25 ml water in a beaker and the pass electricity through it. Then add little sulphuric acid in it.
a. Why electricity didn’t pass through pure water?
b. What happens when dil H₂SO₄ is added?
c. Which type of ion formed more when sulphuric acid is added in water?
d. Complete the equation of the Ionization 0f H₂SO₄
H₂SO₄ → 2H⁺ +
Based on the equation given below write down the correct answers.
2H⁺ +  + 2H₂O → 2H₃O +  + SO₄^2–
e. Complete the equation.
f. Write down the name of H3O⁺ ion?
g.Which ion is moving towards negative ion?
h.Complete the reaction taking place in the negative electrode.
2H₃O⁺(aq) + 2e^– →  +
i. Whicfrion is having highest oxidation potential?
j. Complete the reaction taking place in positive electrode.
2H₂O →  + 4H⁺
k. Ions remain in the beaker after the electrolysis are
I. What product form when these two combined together
Answer:
a. Since the number of ions is less, pure water does not allow the passage of electricity.
b. When dil H₂SO₄ is added the water becomes a good electrolyte. Hence electricity passes through it. When a little dilute sulphuric acid is mixed with water large quantity of hydronium ions are formed.
c. Hydronium ion (H₃O⁺)
d. H₂SO₄ → 2H⁺ + SO₄^2–
e. 2H⁺ + S0₄^2– + 2H₂O → 2H₃O⁺ + SO₄^2–
f. Hydronium ion
g. H₃O⁺
h. 2H₃O⁺(aq) + 2e^– → H_2(g) + 2H₂O
i. H₂O has the highest oxidation potential when compared to SO₄^2–
j. 2H₂O → O₂(g) + 4H⁺ + 4e^–
k. 4H⁺, SO₄^2–ions.
l. H₂SO₄ is formed when these 2 combine together.

Reactivity Series Chemistry Question 8.
a. Complete the table based on the Electrolysis of molten sodium chloride.

Electrodes	Reaction taking place	product
Anode
Cathode

b. Write down the reaction taking place in each electrodes and products formed in the electrolysis of sodium chloride solution.

Electrodes	Reaction taking place	product
Anode
Cathode

c. Why is hydrogen formed in the cathode instead of sodium?
d. Write one word for a solution undergoes chemical change when electricity passes through it.
e. Write the name of the above process.
f. Write down the uses of above type of reaction.
Answer:

c. When Na+ ion and water are compared reduction occurs to water. Hence H₂ is liberated at cathode.
H⁺+ e → H,
H + H → H2.
d. Electrolyte.
e. Electrolysis.
f. Electroplating, Production of chemicals, Purification of metals.

Reactivity Series and Electrochemistry Evaluation Questions

Take little water in a test tube add two drops of phenolphthalein in it Same quantity of Kerosene is added to the mixture and a small piece of sodium is dipped in it.

Insulin Gene Question 1.
What kind of colour formed in the test tube? Why?
Answer:
Water become pink in colour. Because when phenolphthalein is added to water it becomes alkaline.

Question 2.
Which gas is bubbled on the surface of sodium metal?
Answer:
Hydrogen

Question 3.
Write balanced equation of the reaction between sodium and water.
Answer:
2Na(s) + 2H₂O(l) → 2NaOH (aq) + H₂(g).

Question 4.
What products occurs when Iron reacts with water vapour?
Answer:
Fe₃O₄ (Iron Oxide) and Hydrogen gas.

Question 5.
Lustre of magnesium disappeared fast when it placed in open space why?
Answer:
When Magnesium is kept in open space it reacts with atmosphere air and a light coat of magnesium oxide is formed. This is the reason for Magnesium to lose its lustre.

Question 6.
Verdigris formed on copper utensils disappeared after some days why?
Answer:
The copper in copper utensils react with atmospheric air and forms copper oxide. Due to this verdigris are formed on Copper utensils.

Question 7.
Lustre of aluminium utensils disappear after some days. Why?
Answer:
Aluminium reacts with atmospheric air and Aluminium oxide is formed. This process takes place slowly. Hence Aluminium utensils loose their lustre.

Question 8.
Write down the equation for the reaction between CuSO₄ and iron nail? What type of reaction is this?
Answer:
CuSO₄ + Fe → FeSO₄ + Cu.
Redox reaction.

Reactivity Series and Electrochemistry SCERT Questions and Answers

Question 1.
5ml water is taken in 3 test tubes. Copper, sodium and magnesium of equal mass are dropped in different test tubes. Test tubes having copper and magnesium are heated.
a. Write the observations in the heated test tubes.
b. Write the equation for the reaction in the test tube in which sodium is dropped,
c. Arrange these metals in the decreasing order of their reactivity.
Answer:
a. Mg reacts with hot water liberating hydrogen, Copper does not react with hot water.
b. 2Na + 2H₂O → 2NaOH + H₂.
c. Na > Mg > Cu.

Question 2.
a. Which metal among copper, aluminium and gold loose its metallic lustre at a faster rate? Write the equation of the reaction.
b. Sodium is kept in Kerosene. Why?
Answer:
a. Aluminium, Al + 3O₂ → 2Al₂O₃
b. Na reacts with air (oxygen) and water.

Question 3.
An experimental setup is made to compare the reactions of Mg, Zn and Cu with dilute hydrochloric acid.
a. Write the procedure and observation of the reaction.
b. Which is the gas evolved when zinc reacts with dilute hydrochloric acid?
Answer:
a. Take Mg, Zn and Cu in different test tubes and add dilute HCl to each.
Observation: Magnesium and Zinc reacts with dilute hydrochloric acid copper does not react with the acid.
b.Hydrogen.

Question 4.

a. What are the changes that can be observed with the iron rod and the colour of copper sulphate solution?
b. Write the equations of the oxidation and reduction reactions.
c. What will be the change if silver rod is used instead of iron rod? What is the reason?
Answer:
a. Copper is deposited on iron and the blue.colour of copper sulphate solution decreases.
b. Cathode – Cu²⁺(aq) + 2e^– → Cu (s) (Reduction)
Anode – Fe (s) → Fe²⁺(aq) + 2e^– (Oxidation)
c. No change occurs, Reactivity of silver is less than that of copper. In the reactivity series, the position of silver is below copper.

Question 5.
Sodium reacts with water.
a. Identify the gas evolved in the reaction h If two drops of phenol[ihthaloin is added to the water, what will be colour change of the resultant solution? Explain the reason?
Answer:
a. Hydrogen.
b. Colour changes to pink. Due to the presence of sodium hydroxide (alkaline nature).

Question 6.
Three Galvanic cells are given.

a. Find out the most reactive metal and least reactive metal among them.
b. In cell, which electrode undergoes oxidation why?
c. Write the equation of the redox reaction occurring in cell 3 (Valency of A, B are 2.)
Answer:
a. Most reactive metal A, Least reactive metal B.
b. A Reactivity of A is higher than B.
c. A+ 2e^– → A²⁺,
C²⁺ + 2e^– → C ,
A + C²⁺ → A²⁺ C .

Question 7.
Some metals and salt solutions are given (Cu, Zn, Ag, ZnSO₄, AgNO₃, MgCl₂)
a. Draw the diagram of a Galvanic cell that can be made using these substances.
b. Find out the anode and cathode of this cell and write the chemical equation for the reaction at cathode.
Answer:

b. Anode – Zn,
Cathode – Ag
2Ag⁺+ 2e^–→ 2Ag

Question 8.
Give reasons for the following.
a. Iron vessels are not used a boilers that are used to boil water.
b. Blue vitriol solution is not kept in iron vessels.
Answer:
a. Iron reacted with steam-heated to high temperature

Question 9.
Examine the given electrolytic cell.

a. Which gas is evolved at the positive electrode?
b. Write the oxidation and reduction reactions of this cell.
c. What is the difference in the energy transformation of a Galvanic cell and an electrolytic cell?
Answer:
a. Chlorine / Cl₂
b. 2Cl^– → Cl₂ + 2e^–
Cu²⁺ + 2e^– → Cu
c. Galvanic cell – Chemical energy is converted to electrical energy.
Electrical Cell – Electrical energy is changed to chemical energy.

Question 10.
The solutions in the given table electrolyzed.

List any two areas in which electrolysis is made use of?
Answer:
a. i. Hydrogen.
ii. Chlorine.
iii.Chlorine.
iv. Hydrogen.
b.

• Purification of metals.
• Electroplating.
• Production of chemicals.

Question 11.
The position of iron is below that of zinc in reactivity series. The cell formed by them is given. Correct the mistakes and redraw.

Answer:

Question 12.
Sodium chloride solution is electrolysed using platinum electrodes.
a. Write the chemical equation of the reaction at cathode.
b. What happens when phenolphthalein is added to the solution? State the reason?
Answer:
a. 2H₂O + 2e^– → H₂ + 2OH
b. Colour turns pink. Presence of sodium hydroxide in the solution.

Question 13.
The anode and cathode of two Galvanic cells are given.

Galvanic Cell	Anode	Cathode
Cell 1	Mg	Zn
Cell 1	Zn	Ag

A. Mg → Mg²⁺ + 2e^–
B. Zn²⁺+2e^– → Zn
C. Ag⁺ +le^– → Ag
D. Zn → Zn²⁺ + 2e^–
E. Ag → Ag⁺ + le^–
F. Mg²⁺ +2e^– → Mg
a. Find out the reactions at the anode and cathode for each cell from the above.
b. Which metal can act only as cathode? Why?
Answer:
a. Cell 1: Anode Mg → Mg²⁺ + 2e^–
Cathode Zn²⁺ + 2e^– → Zn
Cell 2: Anode Zn → Zn²⁺ + 2e^–
Cathode Ag⁺ + le^– → Ag
b. Ag. Lesser tendency to give up electron that is it is a less reactive metal.

Question 14.
The chemical reactions of various Galvanic cells are given as incomplete in the table. Complete them.

Answer:
a. Zn → Zn²⁺+ 2e^–
b. Zn+Cu²⁺ → Zn²⁺ + Cu
c. Fe → Fe²⁺ + 2e^–
d. 2Ag⁺ + 2e- → 2Ag
e. Pb²⁺+ 2e^– → Pb
f. Mg + Pb²⁺ → Mg²⁺ + Pb

Question 15.
The picture of a Galvanic Cell is given below.

a. Identify A and B.
b. Give the direction of electron flow?
c. Write the chemical equation at the anode and cathode.
Answer:
a. A – Zn,
B – FeSO₄Solution
b. From Zn to Fe
c. Anode Zn → Zn²⁺ + 2e^–
Cathode Fe²⁺ + 2e^– → Fe

Question 16.
An incomplete table about the electrolysis of different electrolytes are given below. Complete it.

Answer:
a. Cu²⁺,
Cl^–,
H₂O.
b. 2H₂O → O₂ + 4H⁺ + 4e^–
c. Na⁺,
Cl^–
d. Na⁺ + le^– → Na
e. 2Cl → Cl₂ + 2e^–
f. 2H₂O + 2e^– → H₂ + 2OH^–

Question 17.
5mI AgNO₃ is taken in a test tube and a copper rod is dipped into
a. Identify the changes occurring with the copper rod and the solution?
b. Complete the equation of the reaction.
Cu + 2AgNO₃ →  +
c. Write the equations of the oxidation and reduction reactions.
Answer:
a. Silver is deposited in copper rod. Colour of solution changes to blue.
b. Cu + 2 AgNO₃ → Cu(NO₃)₂ + 2Ag
c. 2Ag⁺ + 2e^– → 2Ag (Reduction)
Cu → Cu²⁺ + 2e^– (Oxidation)

Reactivity Series and Electrochemistry Exam Oriented Questions and Answers

Very Short Answer Type Questions (Score 1)

Question 1.
The symbols of certain metals are given below. Arrange them as they are given in the reactivity series.
Mg, Pb, Ag, Cu, Zn, Fe, Au, Sn.
Answer:
Mg, Zn, Fe, Sn, Pb, Cu, Ag, Au.

Question 2.
Analyse the table given below and answer the questions.

a. Find out the metals which are likely to be A, B and C from the box given below.

b. Write down the chemical equation between metal ‘B’ and water.
Answer:
a. A – Mg,
B – Cu,
C – Ca
b. Cu (s) + 2H₂O (l) → Cu OH₂ (aq) + H₂ (g)

Very Short Answer Type Questions (Score 2)

Question 3.
Certain metals are given below:
Ag, Zn, Pb, Sn, Fe
a. When a galvanic cell is constructed using these metals, which one acts only as anode? Give the reason.
b. Draw Zn-Fe cell. Mark the direction of electron flow and write the chemical equation anode.
Answer:
a. Zn. Because Zn has a higher reactivity than the other 4 metals.

Direction of electron flow from Zn to Fe Chemical reaction at anode:
Zn(s) → Zn²⁺(aq) + 2e^–

Question 4.
Zn (s) + 2AgNO₃ (aq) →
Zn(NO₃)₂ (aq) + 2Ag
a. Write the oxidation state of each element in this displacement reaction.
b. Write the chemical equation for oxidation and reduction.
Answer:
a. Zn° (s) + 2Ag¹⁺N⁵⁺ O₃^2– (aq)
→ Zn²⁺(N⁵⁺O^2–₃)₂ (aq) + 2Ag°

b.Oxidation:
Zn° (s) → Zn²⁺ (aq) + 2e^–
(Oxidation means losing of electrons)

Reduction:
Ag¹⁺ (aq) + le^– → Ag (s)
(Reduction means gaining of electrons)

Question 5.
Based on the reactions given below, answer the following questions.
i. Aqueous solution of CuCl₂ undergoes electrolysis using graphite rods.
ii. Molten KCl undergoes electrolysis.
iii. Aqueous solution of NaCI undergoes electrolysis.
a. In which all reactions Cl₂ gas is formed? At which electrode is Cl₂ gas formed?
b. In which reaction H₂ gas is formed? Write the chemical equation of this reaction.
Answer:
a. Cl₂ gas is formed in all the three reactions. Chlorine gas is formed at anode.

b. When aqueous solution of NaCI undergoes electrolysis, hydrogen gas is formed at cathode.
Equation: 2H₂O + 2e^– → H₂ + 2OH^–

Very Short Answer Type Questions (Score 3)

Question 6.
Take water in 4 different beakers and add a small piece of sodium, lead, iron and copper in each.
a. In which all solutions gas bubbles will be formed? Which gas is formed?
b. Which solution will turn pink on adding phenolphthalein? Why?
c. Write the chemical equation between this metal and water.
Answer:
a. Gas bubbles are formed in the beaker containing sodium. Hydrogen is tlie gas formed.

b. Beaker containing sodium turns pink because sodium reacts with water and forms an alkali, sodium hydroxide. Alkalies turn phenolphthalein pink.

c. 2Na (s) + 2H₂O (l) → 2NaOH (aq) + H₂ (g)

Question 7.
Analyse the picture given below.

a. Identify ‘A’.
b. Write the chemical equation at ‘B’.
c. Add few drops of phenolphthalein to the remaining solution after electrolysis. What change will take place? Why?
Answer:
a. A – cathode.

b. Chemical equation at ‘B’:
2Cl^– → Cl₂ + 2e^–

c. Solution turns pink. Because, after electrolysis K⁺ and OH^– ions are present in the remaining solution. That means KOH, an alkali is formed.

Question 8.
The flow of electron in certain galvanic cells are given below:
i. Cu → Ag
ii. Ag → Zn
iii.Na → Mg
iv.Fe → K
a. Choose the incorrect ones.
b. Explain your answer.
Answer:
a. ii. Ag → Zn,
iv. Fe → K are the incorrect ones.

b. Zn has a higher reactivity than Ag. Hence Zn undergoes oxidation (loses electrons). So the direction of electron flow is from Zn to Ag. Similarly, K has higher reactivity than Fe. Hence direction of electron flow is from K → Fe.

Question 9.
Which of the chemical reactions given below are wrong? Explain the reason.
a. Cu (s) + 2HCl (aq) →
CuCl₂(aq) + H₂ (g)
b. Mg(s) + Pb(NO₃)₂ (aq) →
Mg(N0₃)₂ (aq) + Mg (s)
c. 3Fe (s) + 4H₂O (l) →
Fe₃O₄(s) + 4H₂(g)
Answer:
Equations (a) and (c) are wrong.
a. Copper cannot displace hydrogen from acids because it is placed below hydrogen in the reactivity series.

c. Fe reacts only with superheated steam. Fe does not react with water in liquid state.

Question 10.
Write the chemical equation of the electrolysis of water to which little sulphuric acid (H₂SO₄) is added.
Answer:
H₂SO₄ (l) + 2H₂O (l) → 2H₃O (aq) + SO₂^2–(aq)
at cathode:
2H₃O+ (aq) + 2e^– → H₂(g) + 2H₂O (l) (reduction)
at anode:
2H₂O (l) → O₂(g) + 4H^– (aq) + 4e^– (oxidation)

Long Answer Type Questions (Score 4)

Question 11.
Analyse the reactions and answer the following questions:

a. Which among the following test tubes will undergo a chemical reaction?
b. What are these chemical reactions called?
c. Explain the oxidation and reduction reactions taking place here including the chemical equation?
Answer:
a. Chemical reaction takes place in test tube (ii) only.
b. These chemical reactions are called displacement reactions. .
c. Chemical equation: (including oxidation states)
Mg° (s) + Fe²⁺S⁶⁺O4^2– (aq) →
Mg²⁺S⁶⁺O^2–₄ (aq) + Fe° (s)
At Mg : Mg°(s) → Mg²⁺(aq) + 2e^– (oxidation)
At Fe²⁺, Fe²⁺ (aq) + 2e^– →  Fe°(s) (reductipn)

Question 12.
A students observation is given below:
i. When Zn is put in salt solution, Na gets deposited over Zn.
ii. Au reacts with water vapour and hydrogen gas is formed.
iii. Al reacts with acid and forms hydrogen gas.
iv. Mg reacts with hot water and forms hydrogen gas.
a. Which statements are incorrect?
b. Give reason for your answer.
Answer:
a. Statements (i) and (ii) are wrong.

b. i. Zn cannot displace sodium. Because sodium has a higher reactivity than Zn.
ii. Au does not react with water vapour.

Question 13.
“Sodium cannot be kept open in atmospheric air, and cannot be stored in water. So it is stored in kerosene.” Give explanation for the above statement with its chemical equation.
Answer:
Sodium which is placed above in the reactivity series reacts vigorously with water and oxygen.
4Na (s) + O₂ (g) → 2Na₂O (s)
2Na (s) + 2H₂O (l) →
2NaOH (aq) + H₂(g)
This NaOH reacts with CO₂ in the atmospheric air and forms sodium carbonate.
2NaOH (s) + CO₂ (g) → Na₂CO₃ + H₂O (l)
To avoid the reaction with O₂, H₂O and CO₂, sodium is stored in kerosene.

Question 14.
The direction of electron flow in certain galvanic cells are given below. (Symbols are not real)
i. B → A,
ii. E → C,
iii.D → E,
iv.A → D.
a. Arrange the metals A, B, C, D and E in the decreasing order of their reactivity.
b. Choose the reaction taking place at ‘C’ in cell (ii) E → C. Give the reason.
i. C⁺ (aq) + le^– → C(s)
ii. C (s) → C⁺ (aq) + le^–
iii.C (s) → C⁺ (aq) + le^–
Answer:
a. B,
A,
D,
E,
C.
b.C⁺ (aq) + le^– → C (s) is the correct equation. Because ‘E’ has higher reactivity than. So ‘C’undergoes reduction.

Question 15.
Give reason for the following.
a. CuSO₄ solution is not stored in iron vessels.
b. Buttermilk is not stored in aluminium vessels.
Ans.
a. Fe displaces copper from copper sulphate solution and forms FeSO₄.
b. Aluminium reacts with acid in the buttermilk.

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 Mathematics of Chance Questions and Answers, Activity, Notes, Kerala Syllabus 10th Standard Maths Solutions Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Maths Solutions Chapter 3 Mathematics of Chance

Mathematics of Chance Text Book Questions and Answers

Textbook Page No. 71

Mathematics Of Chance Questions And Answers Kerala Syllabus 10th Standard Question 1.
A box contains 6 black and 4 white balls. If a ball is taken from it. What is the probability of it being black? And the probability of it being white?
Answer:
Total no. of balls = 10
No. of black balls = 6
No. of white balls = 4
Probability of it to be black = \(\frac { 6 }{ 10 }\) = \(\frac { 3 }{ 5 }\)
Probablitiy of to be white = \(\frac { 4 }{ 10 }\) = \(\frac { 2 }{ 5 }\)

Mathematics Of Chance Class 10 Kerala Syllabus Kerala Syllabus Question 2.
There are 3 red balls and 7 green balls in a bag, 8 red and 7 green balls in another.
i. What is the probability of getting a red ball from the first bag?
ii. From the second bag?
iii. If all the balls are put in a single bag, What is the probability of getting a red ball from it?
Answer:
i. Total no. of balls in the first bag = 10
No. of red balls = 3
P(red ball) = 3/10

ii. Total no. of balls in the second bag= 15 No. of red balls = 8
P(red ball) = 8/15

iii. Total no. of balls in both bags = 25
Total no. of red balls = 11
P(red ball) = 11/25

Mathematics Of Chance Questions And Answers Pdf Kerala Syllabus 10th Standard Question 3.
One is asked to say a two-digit number. What is the probability of it being a perfect square?
Answer:
Total no. of two-digit numbers = 90
Perfect squares from 10 to 99 are 16, 25, 36, 49, 64, 81.
There are 6 favourable numbers Probability = \(\frac { 6 }{ 90 }\) = \(\frac { 1 }{ 15 }\)

Mathematics Of Chance Extra Questions And Answers Kerala Syllabus 10th Standard Question 4.
Numbers from 1 to 50 are written on slips of paper and put in a box. A slip is to be drawn from it; but before doing so, one must make a guess about the number, either prime number or a multiple of five. Which is the better guess? Why?
Answer:
Probablity of getting prime numbers from 1 to 50 = \(\frac { 15 }{ 50 }\) = \(\frac { 3 }{ 10 }\)
Probability of getting a multiple of 5 from 1 to 50 = \(\frac { 10 }{ 50 }\) = \(\frac { 1 }{ 5 }\)
∴ Better guess would be prime numbers as probability of that is more.

Sslc Maths Mathematics Of Chance Kerala Syllabus 10th Standard Question 5.
A bag contains 3 red beads and 7 green beads. Another contains one red and one green more. The probability of getting a red from which bag is more?
Answer:
Total no. of beads in first bag = 10
No. of red beads in first bag = 3
P(red ball) in first bag = \(\frac { 3 }{ 10 }\)
Total no. of beads in second bag = 12
No. of red beads in second bag = 4
P(red ball) in second bag = \(\frac { 4 }{ 12 }\) = \(\frac { 1 }{ 3 }\)
The probability of getting a red bead is more from the second bag since, \(\frac{1}{3}>\frac{3}{10}\left(\frac{10}{30}>\frac{9}{30}\right)\)

Textbook Page No. 72

In each picture below, the explanation of the green part is given. Calculate in each, the probability of a dot put without looking to be within the green part.

Mathematics Of Chance Pdf Kerala Syllabus 10th Standard Question 1.
A square got by joining the mid points of a bigger square.

Answer:
∴ Probability = \(\frac { 4 }{ 8 }\) = \(\frac { 1 }{ 2 }\)

Class 10 Maths Chapter 3 Mathematics Of Chance Kerala Syllabus Question 2.
A square with all vertices on a circle

Answer:
Side of square = a = 2cm Radius

ie., The area of the square is 2/π the area of the circle.

Probability = \(\frac{a^{2}}{\frac{\pi}{2} a^{2}}=\frac{4}{2 \pi}=\frac{2}{\pi}\)
Probablity ofthe dot falling on the square is \(\frac { 2 }{ π }\) = 0.64

Sslc Maths Chapter 3 Questions And Answers Kerala Syllabus Question 3.
Circle exactly fitting inside a square.

Answer:
If radius of circle is r
Side of the square = 2r
Area of the square = 4r²
Area of the circle= πr²

Mathematics Of Chance Extra Questions Kerala Syllabus 10th Standard Question 4.
A triangle got by joining alternate vertices of a regular hexagon.

Answer:
The area of regular hexagon having side a = \(\frac{3 \sqrt{3}}{2} a^{2}\)
Side of the triangle got by joining alternate vertices = √3a
Area of triangle = \(\frac{\sqrt{3}}{4}(\sqrt{3} a)^{2}=\frac{3 \sqrt{3}}{4} a^{2}\)
∴ Probability of the dot falling on the tri-angle = \(\frac{3 \sqrt{3}}{4} a^{2} \times \frac{2}{3 \sqrt{3} a^{2}}=\frac{2}{4}=\frac{1}{2}\)

Sslc Maths Chapter 3 Solutions Kerala Syllabus 10th Standard Question 5.
A regular hexagon formed by two overlapping equilateral triangles.

Answer:
Area of two equilateral triangles

Area of regular hexagon = \(\frac{3 \sqrt{3}}{2} a^{2}\)
Area of regular hexagon is half of the area of equilateral triangle.
∴ Probability of dot falling on the hexagon

Textbook Page No. 75

Class 10 Maths Chapter 3 Kerala Syllabus Question 1.
Raj ani has three necklaces and free pairs of earrings, of green, blue and red stones. In what all different ways can she wear them? What is the probability of her wearing the necklace and earrings of the same color? Of different colors?
Answer:
Possible ways of wearing:

Necklace	Earring
Green	Green
Green	Blue
Green	Red
Blue	Green
Blue	Blue
Blue	Red
Red	Green
Red	Blue
Red	Red

Raj ani can wear the ornaments in 9 different ways.
Probability of wearing necklace and earring of same colour = \(\frac { 3 }{ 9 }\) = \(\frac { 1 }{ 3 }\)
Probability of wearing necklace and earring of different colour = \(\frac { 6 }{ 9 }\) = \(\frac { 2 }{ 3 }\)

Sslc Maths Chapter 3 Notes Question 2.
A box contains four slips numbered 1, 2, 3, 4 and another box contains two slips numbered 1,2. If one slip is taken from each, what is the probability of the sum of numbers being odd? What is the probability of the sum being even?
Answer:

First box	Second box	Sum
1	1	2
1	2	3
2	1	3
2	2	4
3	1	4
3	2	5
4	1	5
4	2	6

Probability of sum being odd = \(\frac { 4 }{ 8 }\) = \(\frac { 1 }{ 2 }\)
Probability of sum being even = \(\frac { 4 }{ 8 }\) = \(\frac { 1 }{ 2 }\)

Kerala Syllabus 10th Standard Maths Chapter 3 Question 3.
A box contains four slips numbers 1, 2, 3, 4 and another contains three slips numbered 1, 2, 3. If one slip is taken from each, what is the probability of the product being odd? The probability of the product being even?
Answer:

First box	Second box	Product
1	1	1
1	2	2
1	3	3
2	1	2
2	2	4
2	3	6
3	1	3
3      .	2	6
3	3	9
4	1	4
4	2	8
4	3	12

Probability of sum being odd = \(\frac { 4 }{ 12 }\) = \(\frac { 1 }{ 3 }\)
Probability of sum being even = \(\frac { 8 }{ 12 }\) = \(\frac { 2 }{ 3 }\)

10 Th Maths Text Book Questions And Answers Question 4.
From all two-digit numbers with either digit 1, 2 or 3 one number is chosen.
i. What is the probability of both digits being the same?
ii. What is the probability of the sum of the digits being 4?
Answer:
Two digit numbers
11, 12, 13, 21, 22, 23, 31, 32, 33.
i. P (both digits being same) = \(\frac { 3 }{ 9 }\) = \(\frac { 1 }{ 3 }\)
ii. P(sum of digits being 4) = \(\frac { 3 }{ 9 }\) = \(\frac { 1 }{ 3 }\)

10th Class Maths Textbook Answers Question 5.
A game for two players. First, each has to decide whether he wants odd number or even number. Then both raise some fingers of one hand. If the sum is odd, the one who chose odd at the beginning wins; if it is even, the one who chose even wins. In this game, which is the better choice at the beginning, odd or even?
Answer:
The results in the order when the first player raises one finger and second player raises one finger and so on.

Textbook Page No. 78

Maths Chapter 3 Class 10 Kerala Syllabus 10th Standard Question 1.
In class 10A, there are 30 boys and 20 girls. In 10B, there are 15 boys and 25 girls. One student is to be selected from each class.
i. What is the probability of both being girls?
ii. What is the probability of both being boys?
iii. What is the probability of one boy and one girl?
iv. What is the probability of at least one boy?
Answer:
i. Total no. of possible pairs = 50 × 40 = 2000
No. of pairs in which both are girls = 20 × 25 = 500
Probability of both being girls = \(\frac { 500 }{ 2000 }\) = \(\frac { 1 }{ 4 }\)

ii No. of pairs in which both are boys = 30 × 15 = 450
Probability of both being boys = \(\frac { 450 }{ 2000 }\) = \(\frac { 9 }{ 40 }\)

iii. No. of pairs in which one is boy and one is girl = 2000 – (500 + 450) = 2000 – 950 = 1050
Probability of one being boy and one girl = \(\frac { 1050 }{ 2000 }\) = \(\frac { 21 }{ 40 }\)

iv. No. of pairs in which atleast one is boy = 1050 + 450 = 1500
Probability in which atleast one is boy = \(\frac { 1050 }{ 2000 }\) = \(\frac { 3 }{ 4 }\)

Question 2.
One is asked to say a two-digit number.
i. What is the probability of both digits being the same?
ii. What is the probability of the first digit being larger?
iii. What is the probability of the first digit being smaller?
Answer:
i. Probability of two digits being same = \(\frac { 9 }{ 90 }\) = \(\frac { 1 }{ 10 }\)
(11, 22, 33, 44, 55, 66, 77, 88, 99)

ii. Numbers in which first digit is greater than second digit are 10, 20, 21, 30, 31, 32, 40, 41, 42, 43, 50, 51, 52, 53, 54, 60, 61, 62, 63, 64, 65, 70, 71, 72, 73, 74, 75, 76, 80, 81, 82, 83, 84, 85, 86, 87, 90, 91, 92, 93, 94, 95, 96, 97, 98,
45 outcomes, Required Probability = \(\frac { 45 }{ 90 }\) = \(\frac { 1 }{ 2 }\)

iii. Numbers in which first digit is smaller than second digit are
12, 13, 14, 15, 16, 17, 18, 19, 23, 24, 25, 26, 27, 28, 29, 34, 35, 36, 37, 38, 39, 45, 46, 47, 48, 49, 56, 57, 58, 59, 67, 68, 69, 78, 79, 89,
36 total no.of favourable outcomes = 36
Probability =

Question 3.
Each two-digit number is written on a paper slip and these are all put in a box. What is the probability that the product of the digits of a number drawn is a prime number? What if three-digit numbers are used instead?
Answer:
Total two-digit numbers = 90
Product of the digits of a number drawn is a prime number 12, 13, 15, 17, 21, 31, 51, 71.
(1 is not a prime number)
Total number whose product of the digits drawn is a prime number = 8
Probability of product of the digits drawn is a prime number = \(\frac { 8 }{ 90 }\) = \(\frac { 4 }{ 45 }\)
Total three-digit numbers = 900
Product of the digits of a number drawn is a prime number 112, 113, 115, 117, 121, 131, 151, 171.
Total number whose product of the digits drawn is a prime number = 8
Probability of product of the digits drawn is a prime number = \(\frac { 8 }{ 900 }\) = \(\frac { 2 }{ 225 }\)

Question 4.
Two dice with faces numbered from 1 to 6 are rolled together. What are the possible sums? Which of these sums has the maximum probability?
Answer:
Each dice has the following numbers:

Mathematics of Chance Orukkam Questions & Answers

Worksheet 1

Question 1.
How many odd numbers are there below 25
Answer:
12 odd numbers

Question 2.
How many prime numbers are there below 30?
Answer:
10 prime numbers

Question 3.
Find the number of two-digit even numbers?
Answer:
90 two-digit even numbers

Question 4.
How many two digits perfect squares are there?
Answer:
6

Question 5.
Write all three-digit numbers that can be written using the digits 3, 6, 8 without repeating the digits.
Answer:
368, 386, 683, 638, 836, 863

Question 6.
How many multiples of 7 are there in between 100 and 300?
Answer:
First multiple of 7 = 105
Last multiple of 7 = 294
Number of multiples
=

Question 7.
There are 50 children in a class. Thirty of them are girls. There are 40 children in another class. 25 of them are boys. One student is taken from each class at random. What is the number of outcomes? How many outcomes contain both boys. How many outcomes contain both girls. How many outcomes have one boy and one girl?
Answer:
Total pairs = 50 × 40 = 2000
No. of pairs in which both are boys = 20 × 25 = 500
No. of pairs in which both are girls = 30 × 15 = 450
No. of pairs in which one is a boy and the other a girl = 30 × 25 + 20 × 15 = 750 + 300 = 1050

Worksheet 2

Question 8.
A fine dot is placed into the picture with-out looking into it. What is the probability of falling the dot in the small semicircle? What is the probability of falling the dot outside the small semicircle but inside the big semicircle?
Answer:
Let the radius of circle be r, then the radius of

Question 9.
P, Q, R are the midpoints of the sides of triangle ABC. Another triangle is drawn by joining these points. A fine dot is placed into the figure without looking into the picture. What is the probability of falling the dot in triangle PQR?
What is the probability of falling the dot outside the triangle?
Answer:
Area of each small triangle is 1/4^th of area of large triangle ABC.

Probability of the dot falling on triangle PQR is = 1/4
Probability of the dot falling on small triangle is = 1/4
Probability of the dot falling inside the triangle PQR is \(\frac{3}{4}-\frac{1}{4}=\frac{2}{4}=\frac{1}{2}\) less than that of outside the triangle.

Question 10.
What is the probability of occurring 53 Sundays in a leap year
Answer:
Leap year have 366 days.
That is 52 weeks and 2 days.
There two days are
Sunday – Monday, Monday – Tuesday, Tuesday – Wednesday, Wednesday – Thursday, Thursday – Friday, Friday – Saturday, Saturday – Sunday.
∴ Probability for to occur 53 Sunday is 2/7.

Question 11.
You can see a triangle inside a square. ABCD is a square. P, Q are the midpoints of C D and C B. A fine dot is placed into the figure without looking into the figure. What is the probability of falling the dot in triangle APQ?
Answer:

Question 12.
The value of 21, 22, 23… 250 are written in small papers and put it in the box. A paper is taken at random. What is the probability of getting a number having 4 in ones place? What is the probability of falling 8 in the one’s place?
Answer:
The 13 numbers having 4 in one’s place are 22, 26, 210 …, 250.
∴ Probability = \(\frac { 13 }{ 50 }\)
The 12 numbers having 8 in one’s place are 23, 27, 211 …, 247.
∴ Probability = \(\frac { 12 }{ 50 }\)

Worksheet 3

Question 13.
Numbers from 1 to 10 are written in small papers and placed in a box. One number is taken from the box at random. What is the probability of getting a prime number?
Answer:
Probability of getting a prime number = \(\frac { 4 }{ 10 }\) = \(\frac { 2 }{ 5 }\)

Question 14.
Two boxes contain tokens on which numbers 1, 2, 3, 4 are written One token is taken from each box. What is the probability of getting sum of the face numbers a prime number
Answer:
Pairs of numbers in tokens are
(1.1) , (1,2), (1,3), (1,4)
(2.1), (2, 2), (2, 3), (2, 4)
(3, 1), (3,2), (3; 3), (3,4)
(4, 1), (4, 2), (4, 3), (4,4).
Pairs getting sum as prime numbers
(1.1) , (1,2), (1,4), (2, 1), (2, 3), (3, 2), (3, 4), (4, 1), (4, 3)
Probability of getting sum of the face numbers a prime number = \(\frac { 9 }{ 16 }\)

Question 15.
One box contains 8 black balls and 12 white balls. Another box contains 9 black and 6 white balls. One ball is taken from each box at random. What is probability of getting both black? What is the probability of getting both white? What is the probability of getting one black and one white?
Answer:
Total pairs = 20 x 15 = 300
Number of pairs getting both black= 8 x 9 = 72
Probability of getting both black = \(\frac { 72 }{ 300 }\) = \(\frac { 6 }{ 25 }\)
Probability of getting one black and one white

Question 16.
In the figure, a triangle is drawn by joining the alternate vertices of a regular hexagon. A fine dot is placed into the figure at random. What is the probability of falling the dot in the triangle?

Answer:

Probability of the dot falling on triangle

Question 17.
What is the probability of occurring four Wednesdays in 23 consecutive days in a month?
Answer:
23 days = 3 weeks + 2 days
Wednesday comes on Tuesday + Wednesday, Wednesday + Thursday when two days are taken.
Total probabilities (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)
∴ Total probabilities = 7
∴ Probabilities = 2/7

Mathematics of Chance SCERT Question Pool Questions & Answers

Question 18.
One is asked to say a two digit number. What is the probability of being the number not a perfect square? [Score : 3, Time : 3 Minutes]
Answer:
Total number of two digit number : 90 (1)
Total number of two digit perfect squares: 6 (1)
Number of two digit numbers which are not perfeet squares : 90 – 6 = 84,
Probability = \(\frac { 84 }{ 90 }\) = \(\frac { 42 }{ 45 }\) (1)

Question 19.
A bag contains 10 blue balls and 12 yellow balls. Another contains 15 blue balls and 7 yellow balls.
a. What is the probability of getting a yellow ball from the first bag?
b. What is the probability of getting a yellow ball from the second bag?
C. If all the balls are put in a single bag, what is the probability of getting a yellow ball from it? [Score : 4, Time : 4 Minutes]
Answer:
a. Total number of balls in the first bag = 10 + 12 =22, Number of yellow balls = 12
Probability of getting a yellow ball = \(\frac { 12 }{ 22 }\) = \(\frac { 6 }{ 11 }\) (1)

b. Total number of balls in the second bag = 15 + 7 = 22, Number of a yellow ball = 7
Probability of getting a yellow ball = \(\frac { 7 }{ 22 }\) (1)

c. Total number of balls = 22 + 22 = 44
Number of yellow balls = 12 + 7 = 19 (1)
Probability of getting a yellow ball = \(\frac { 19 }{ 44 }\) (1)

Question 20.
A regular hexagon is drawn with its vertices on a circle. Without looking into the I picture, if one put dot in that picture, what is the probability of being the dot not in the regular hexagon? [Score: 4, Time: 3 Minutes] .

Answer:
Area of circle = πr²

Question 21.
In the figure, all the four shaded semicircles have same area. If we put a dot in the figure without looking into it, what is the probability of being the dot in the shaded semicircles? [Score : 3, Time: 5 Minutes]

Answer:
If the radius of the shaded semicircle is r,

Question 22.
What is the probability of getting 5 Sundays in December in a calendar year? [Score : 3, Time : 5 Minutes]
Answer:
There are 31 days in December. That means 4 full weeks and 3 days.
The probable three days are as shown below.

Question 23.

Two semicircles are drawn in a square as shown. If we put a dot in the figure, without looking into it, what is the probability of being the dot in the shaded region? [Score: 3, Time: 5 Minutes]
Answer:
If a is the side of a square,

Mathematics of Chance Exam Oriented Questions and Answers

Short Answer Type Questions (Score 2)

Question 24.
A bag contains 6 red balls, 8 green balls, and 8 white balls. One ball is drawn at random from the bag, find the probability of getting
i. A white or green ball
ii. Neither green ball nor a red ball.
Answer:
Red balls = 6
Green balls = 8
White balls = 8
Total number of balls = 6 + 8 + 8 = 22
a. The probability of getting a white or green ball = 16/20
b. The probability of getting neither green balls nor a red ball = 8/20

Question 25.
20 cards numbered 1, 2, 3, 4, ….19, 20 are put in a box. One boy draws a card from the box. Find the probability that the number on the card is:
i. Prime
ii. Divisible by 3
Answer:
Total number of outcomes = 20
a. Prime numbers from 1 to 17 are 2, 3, 5, 7, 11,13, 17, 19.
Number of outcomes = 8
The probability that the card drawn is prime number = 8/20

b. Numbers are divisible by 3 are 3, 6, 9, 12, 15, 18.
Number of outcomes = 6
The probability that the card drawn is divisible by 3 = 6/20

Question 26.
In a bag, there were 3 white balls and 5 black balls. From this one ball is taken, then
a. What is the probability of being blackball?
b. What is the probability of being white ball?
Answer:
Total no. of balls is 8 and in that 5 of then are black balls.
a. Probability of getting black balls = 5/8
b. 3 of the balls were white, so the probability of getting white balls = 3/8

Question 27.
a. How many two-digit natural numbers are there in all?
b. If we choose one number from the two-digit numbers, what is the probability that the sum of digits of that number will be 10?
Answer:
a. Number of two-digit natural numbers = 90

b. Numbers whose sum of digits will be 10 is 19, 28, 37, 46, 55, 64, 73, 82, 91
Probability that the sum of digits of that number will be 10 = \(\frac { 9 }{ 90 }\) = \(\frac { 1 }{ 10 }\)

Short Answer Type Questions (Score 3)

Question 28.
In selecting a two-digit number up to 50.
a. What is the probability of the digit in the ten’s place to be larger than the digit in the one’s place?
b. What is the probability of the digit in the tens place to be smaller than the digit in the one’s place?
Answer:
Total numbers of two-digit numbers up to 50 = 41
a. Number of numbers with the digit in the tens place to be larger than the digit in the units place = 11
Probability that the number with tens place digit is larger than the digit in the unit place = 11/41

b. Probability that the number with the tenth place digit is smaller than the digit in the unit place = 26/41

Question 29.
A black dice and a white dice are thrown at the same time,
a. Write all the possible outcomes.
b. What is the probability that the sum of the two numbers is to be 8.
c. What is the probability that being the same number to be on both dice?
Answer:
a. Total outcomes
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6) ‘
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6) .
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4.1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5.1) , (5,2), (5,3), (5,4), (5,5), (5,6)
(6.1), (6,2), (6,3), (6,4), (6,5), (6,6)
Total = 36

b. Sum of two numbers is to be 8 = 5
(2,6), (3,5), (4,4), (5,3), (6,2),
Probability = 5/36

c. Probability of being same number =(1,1),(2,2), (3,3), (4,4), (5,5), (6,6),
Total = 6
Probability = \(\frac { 6 }{ 36 }\) = \(\frac { 1 }{ 6 }\)

Question 30.
A box contains 400 electronic toy cars. Among them, 12 are defective. One toy is taken out at random. What is the probability that
a. It is a defective toy.
b. It is a non-defective toy.
Answer:
Total number of cars = 400
Number of defective toys = 12
Number of non-defective toys = 400 – 12 = 388
a. Probability of getting a defective toy = \(\frac { 12 }{ 400 }\) = \(\frac { 3 }{ 100 }\)

b. Probability of getting a non-defective toy = \(\frac { 388 }{ 400 }\) = \(\frac { 97 }{ 100 }\)

Long Answer Type Questions (Score 4)

Question 31.
Natural numbers from 1 to 30 are written on paper slips and kept in a box. If one slip is taken from the box,
a. What is the probability of this number to be even?
b What is the probability of this number to be a multiple of 3?
c. What is the probability of this number to be a multiple of 3 and 5?
d. What is the probability that this number be a natural number?
Answer:

Question 32.
There is one spot at one side of the cube, two on another side, three on the third side and so on. There are spots on all the six faces; in this order. Another cube which is marked in the same way is taken.
a. If both the cubes are thrown, what is the probability that the total number of spots on the upper faces is 6?
b. What is the probability that the sum of the spots on the upper faces is 9?
c. What is the probability that the sum of the spots be one?
d. What is the probability that the sum of the spots be a prime number?
Answer:

Question 33.
In a pack of 52 cards, half are red and the rest are black. There are 4 Suits of 13 cards each and having the signs ‘hearts’, ‘spade’, ‘clubs’ and ‘diamond’. If a card is picked from this pack.
a. What is the probability of it being black?
b. Whatistheprobabilityofitbeingaspade?
c. What is the probability of it being a spade or a diamond?
Answer:
a. Total number of cards = 52
Number of black cards = 26
Probability of the picked card being black = \(\frac { 26 }{ 52 }\) = \(\frac { 1 }{ 2 }\)

b. No. of spade cards = 13
Probability of a picked card being spade = \(\frac { 12 }{ 52 }\) = \(\frac { 1 }{ 4 }\)

c. No. of cards spade or diamond =13 + 3 = 26
Probability of a picked card being spade or diamond = \(\frac { 26 }{ 52 }\) = \(\frac { 1 }{ 2 }\)

Long Answer Type Questions (Score 5)

Question 34.
A man is asked to say a 3 digit number,
a. What is the probability that the first and last digits be equal?
b. What is the probability that the last two digits be ‘O’?
c. What is the probability that the last digits being greater than the first?
Answer:
Total 3 digit numbers = 900
a. Numbers with first and last digits are equal
101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212, 222, 232, 242, 252, 262, 272, 282, 292, 303, 313, 323, 333, 343, 353, 363, 373, 383, 393
There are 90 such numbers.
∴ Probability = \(\frac { 90 }{ 900 }\) = \(\frac { 1 }{ 10 }\)

b. The numbers with last two digits zero are 100, 200, 300,……… 900 total numbers 9.
∴ Probability = \(\frac { 9 }{ 900 }\) = \(\frac { 1 }{ 100 }\)

c. The numbers with the last digit greater than the first digit is 36
∴ probability = \(\frac { 36 }{ 90 }\) = \(\frac { 2 }{ 5 }\)

Question 35.
There are 10 black pearls and 5 white pearls in the box A. There are 8 black pearls and 7 white pearls in box B.
a. Which box has more probability of be ing the pearls black, when a pearl from each of the boxes is taken?
b. What is the probability to get a white pearl from the box A?
c. What is the probability to get a black pearl from box B?
d. If all the pearls in the box B is dropped in the box A, then what is the probability to get a black pearl from it?
Answer:
a. Box A because it has more black pearl

Question 36.
Cards marked with the numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn from this box. Find the probability that the number of the card is
a. an even number
b. a number less than 16
c. a number which is a perfect square
d. a prime number less than 25
Answer:

Mathematics of Chance Memory Map

When probabilities are explained in terms of numbers it is the ratio of number of favorable outcomes to the total number of outcomes.

The least probability will be 0 and the highest will be 1. The probability will be a number between 0 and 1.

If an event can be completed in ‘m’ ways and another event can be completed in ‘n’ ways then both the events can be completed one after the other in m x n ways.

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