Kerala Padavali Malayalam Standard 9 Solutions Unit 3 Chapter 3 Vishwam Deepamayam

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Kerala Padavali Malayalam Standard 9 Guide Unit 3 Chapter 3 Vishwam Deepamayam

Vishwam Deepamayam Questions and Answers, Summary, Notes

Kerala Padavali Malayalam Standard 9 Solutions Unit 3 Chapter 3 Vishwam Deepamayam 1

Kerala Padavali Malayalam Standard 9 Solutions Unit 3 Chapter 3 Vishwam Deepamayam 2
Kerala Padavali Malayalam Standard 9 Solutions Unit 3 Chapter 3 Vishwam Deepamayam 3

Kerala Padavali Malayalam Standard 9 Solutions Unit 3 Chapter 3 Vishwam Deepamayam 4
Kerala Padavali Malayalam Standard 9 Solutions Unit 3 Chapter 3 Vishwam Deepamayam 5
Kerala Padavali Malayalam Standard 9 Solutions Unit 3 Chapter 3 Vishwam Deepamayam 6

Kerala Padavali Malayalam Standard 9 Solutions Unit 3 Chapter 3 Vishwam Deepamayam 7
Kerala Padavali Malayalam Standard 9 Solutions Unit 3 Chapter 3 Vishwam Deepamayam 8
Kerala Padavali Malayalam Standard 9 Solutions Unit 3 Chapter 3 Vishwam Deepamayam 9

Kerala Padavali Malayalam Standard 9 Solutions Unit 3 Chapter 3 Vishwam Deepamayam 10
Kerala Padavali Malayalam Standard 9 Solutions Unit 3 Chapter 3 Vishwam Deepamayam 11
Kerala Padavali Malayalam Standard 9 Solutions Unit 3 Chapter 3 Vishwam Deepamayam 12

Kerala Padavali Malayalam Standard 9 Solutions Unit 3 Chapter 3 Vishwam Deepamayam 13
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Kerala Padavali Malayalam Standard 9 Solutions Unit 3 Chapter 3 Vishwam Deepamayam 15

Kerala Syllabus 9th Standard Social Science Solutions Part 2 Chapter 8 Population, Migration, Settlements

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Kerala State Syllabus 9th Standard Social Science Solutions Part 2 Chapter 8 Population, Migration, Settlements

Population, Migration, Settlements Textual Questions and Answers

Question 1.
List the areas that require analysis of population-related information.
Answer:

  • For planning the food grain production.
  • To generate employment opportunities.
  • To formulate welfare schemes

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Question 2.
What is meant by density of population?
Answer:
The average population of every square kilometre is called as density of population.

Question 3.
Complete the flow chart
Kerala Syllabus 9th Standard Social Science Solutions Part 2 Chapter 8 Population, Migration, Settlements 1
Answer:
Kerala Syllabus 9th Standard Social Science Solutions Part 2 Chapter 8 Population, Migration, Settlements 2

Question 4.
Name the most densely populated state in India.
Answer:
Bihar

Question 5.
Examine the factors influencing the density of population.
Answer:
The high density of population in certain places is mainly due to factors like level topography, moderate climate, fertile soil favoring agriculture are availability of fresh water, etc. Other than these, the increasing employment opportunities in the mineral-rich and industrial regions and also the attractive infrastructure and services provided by urban areas also cause high density of population in such regions. Now you might have understood the cause for imbalance in population density and also the significant influence of geographical factors on the same.

Question 6.
Define population growth!
Answer:
Population growth in the change in population of any particular place over a particular period.

Question 7.
The decadal growth rate of population in India during 2001 – 2011 is ……….
a) 16.7%
b) 17.7%
c) 18.3%
d) 20.6%
Answer:
17 – 7%

Question 8.
Point out the factors causing change in population.
Answer:

  • Birth rate
  • Death rate
  • Migration

Question 9.
Define Migration.
Answer:
Permanent or temporary shifting of residence of people from one place to another is called migration.

Question 10.
Prepare a flow chart showing different levels of migration.
Kerala Syllabus 9th Standard Social Science Solutions Part 2 Chapter 8 Population, Migration, Settlements 3
Answer:
Kerala Syllabus 9th Standard Social Science Solutions Part 2 Chapter 8 Population, Migration, Settlements 4

Question 11.
Distinguish between immigration and emigration.
Answer:
Migration across international boundaries is called international migration. The inward movement of people to a country is called immigration and the outward migration of people from one country to another is called emigration.

Question 12.
Find out the pull factors of migration
Answer:

  • Employment opportunities
  • Higher education facilities
  • Better living standards

Question 13.
Find out the other push factors causing migrations.
Answer:

  • Resource scarcity
  • Unemployment
  • Political unrest
  • Natural calamities
  • Internal conflicts
  • Policy changes of governments
  • War and similar unrest

Question 14.
Name 2 types of settlements
Answer:

  1. Nucleated settlements
  2. Dispersed settlements

Question 15.
Arrange the table

A B
Town More than 10 lakh
City Above 50 lakh
Metropolis Between 1 lakh and 10 lakh
Megacity Less than 1 lakh

Answer:

A B
Town Less than 1 lakh
City Between 1 lakh and 10 lakh
Metropolis More than 10 lakh
Megacity Above 50 lakh

Question16.
What are the major problems faced by urban centers?
Answer:

  • Slums
  • Traffic problems
  • Pollution

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Question 17.
Based on the data collected from the 2011 census, answer the following questions.
i) What is the population of India?
ii) What is the density of population?
iii) Which Indian state has the highest density of population?
iv) What is the decadal growth rate of population?
v) What is the male-female ratio?
vi) Which state has least populated in the country?
vii) What is the anticipated population in 2028?
Answer:
i) 121.06 crores
ii) 382/sq.km
iii) Uttar Pradesh
iv) 17.7%
v) 943
vi) Sikkim (6.07 lakhs)
vii) By 2028 India will become the most popular country in the world.

Question 18.
What is meant by population?
Answer:
The total number of people living in a definite area is called population.

Question 19.
The people of the nation is the real wealth of a nation.’ Write an explanation of this.
Answer:
A country is known outside through the people of that country. It is the people who decide the policy of the country and citizens the natural resources. Hence human resources is the real wealth of a nation.

Question 20.
Is uncontrolled population growth is good for the development of the nation?
Answer:
A nation is known through its people because the resources are properly utilized and the policies are decided by them. Hence human resources is the real wealth of a nation. So we can say that increase in population is favorable for the nation.

Question 21.
Observe the map showing the distribution of population and answer the question.
i) Which are the states that have high population?
ii) Which are the states that have least population?
Answer:
i) Uttar Pradesh, Bihar, Maharashtra
ii) Sikkim, Manipur, Meghalaya,Arunachal Pradesh, Nagaland, Mizoram, Tripura, Himachal Pradesh and Uttaranchal.

Question 22.
Observe the map showing the distribution of population and answer the following questions.
i) Density of population is very high in the northern plains. Explain.
ii) In Peninsular plateau the density of population is moderate. Explain.
ii) How was the density of population in the mountain states? Explain.
Answer:
i) Along the northern plains, the density of population is very high because of the fertility of the soil, road and rail network, etc.
ii) Lack of fertility of the soil and the difficulty in reaching the places are the reasons. At the same time presence of mineral resources have helped the concentration of population in some areas.
iii) In the mountain states, the population is very less due to difficult terrains and the soil is not suitable for cultivation.

Question 23.
How is density of population calculated?
Answer:
density of popuation = \(\frac{\text { Total Population }}{\text { Land Area }}\)

Question 24.
Even though Ghina is the most populated country, the density of population is less than India. Give the reasons for this.
Answer:
Density of population is calculated based on the land area. Hence land area is more in China, the density of population is less.

Question 25.
Complete the following table analyzing the density of population.

Density of population Classification States
Less than 100 Very low population density
Between 101 & 250 Less density
Between 251 & 500 Moderate density
Between 501 & 1000 High density
Above 1000 Very high

Answer:
Kerala Syllabus 9th Standard Social Science Solutions Part 2 Chapter 8 Population, Migration, Settlements 5

Question 26.
Analyze the factors influencing the density of population of a region and prepare a note.
Answer:
Level of topography, moderate climate, fertile soil favoring agriculture and availability of freshwater, etc. are the reasons for the concentration of population in certain regions. Other than there increasing employment opportunities in the mineral-rich and industrial regions and also the attractive infrastructure and services provided by urban areas also cause high density of population in some regions.

Question 27.
What is meant by population growth? What is the decaded population growth rate in India?
Answer:
Increase in the number of people living in a particular area during a definite period of time is called population growth. It is generally calculated in percentages. The decaded growth rate of population in India is 17.7%.

Question 28.
Explain positive and negative population growth.
Answer:
When there is an increase in population, it is called positive growth of population. There are situations where the population of a place declines. This is termed as negative growth of population.

Question 29.
What are the factors that influence the change in population? How do these factors bring change to population?
Answer:
Birth rate, death rate and migration. As birth rate increases, the population increases and when the death rate increases the population decreases. Migration lead to decrease in population at one place and increases in another place.

Question 30.
International migration has two aspects. Explain them.
Answer:
Migration from one country to another country is called international migration. The inward movement of the people is called immigration and the outward movement is called emigration.

Question 31.
There are three international airports in Kerala. What might be the reason for so many airports in this small state?
Answer:
There is a large-scale international migration from Kerala to the Gulf countries and to Europe. This is the reason why there is an increased number of international airports in Kerala.

Question 32.
Prepare a note analyzing the international migration in India.
Answer:
Migration within the country are called international migrations and are done mainly for employment opportunities. People tend to migrate to places within the country for better employment and wage.

Question 33.
What may be the cause for the large-scale migration of people to Kerala?
Answer:
Better employment opportunities and wages are the reasons for the migration of north Indian laborers to Kerala.

Question 34.
Prepare a note elucidating internal migration.
Answer:
Migration within the country from one state to another state is called interstate migration. When people migrate from within the state is called intrastate migration. Migration from one district to another district is called interdistrict migration. This type of migration are of four types.

  1. From village to another village
  2. Fromtpwnto another town
  3. From village to town
  4. From town to village

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Question 35.
Employment opportunity is an important reason for migration. Prepare a note substantiating this statement.
Answer:
Thousands of people have migrated from our country to another country in search of employment opportunities. Employment opportunities in the. developed nations is an important reason. For example, there was a surge in migration to the west Asian countries with petroleum mining.

Question 36.
Other than employment opportunities what are the other factors of migration?
Answer:
Better educational facilities, better living conditions, availability of resources, business opportunities, opportunities in the tourism sector and favorable climate.

Question 37.
What are the factors behind forced migration?
Answer:
Scarcity of resources, unemployment, political anarchy, natural calamities, slavery, war, poverty, and hostile climate.

Question 38.
Prepare a comparative note about voluntary migration and forced migration.
Answer;
The migration of people due to some attractive factors are called voluntary migration. Migration due to adverse circumstances are called forced migration.

Question 39.
Migration may cause crucial changes in the social, cultural and economic sectors of both source regions and destination of the migrants.
Answer:

  • Helps in the sharing of human resources
  • Helps in the flow of foreign currency to the parent country.
  • Leads to overpopulation in certain regions.
  • Causes scarcity of resources.
  • Facilitates exchange of technology
  • Creates more employment opportunities.
  • Weakens social ties among the people.
  • Causes the formation of the slums.
  • Causes spread of communicable diseases.
  • Gets opportunities for higher education.
  • Causes imbalance in the sex ratio
  • Country loses the service of the educated and the youth
  • Results in the exploitation of resources.
  • Increases the intensity of environmental pollution

Question 40.
A few major migrations are mentioned in the table. Put a tick (✓) mark in the appropriate column by identifying the types of migration you have familiarized.
Kerala Syllabus 9th Standard Social Science Solutions Part 2 Chapter 8 Population, Migration, Settlements 6
Answer:
Kerala Syllabus 9th Standard Social Science Solutions Part 2 Chapter 8 Population, Migration, Settlements 7

Question 41.
What is known as settlements?
Answer:
The clusters of permanent or temporary human habitats of different sizes are termed as settlements.

Question 42.
List down the factors influencing the settlements.
Answer:
Favorable climate, availability of water, transport & communication facilities and job opportunities.

Question 43.
Differentiate nucleated settlements and dispersed settlements.
Answer:
ln places where houses are seen in close vicinity are called nucleated settlements. The settlements where houses are located farther apart are called dispersed settlements.

Question 44.
Based on the favorable factors such as accessibility availability of water etc. nucleated settlements take different shapes. Elucidate.
Answer:
Linear Pattern: Settlement pattern that develops parallel to features such as roads, rivers, coastlines, etc.
Circular Pattern: Settlement pattern that develops around features such as water bodies, pastures, places of worship, etc.
Star Pattern: Settlement pattern that develops at places where different road coverage.

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Question 45.
What are urban settlements?
Answer;
The settlements that generally have a high population which is mostly dependent on non-agricultural sectors are called urban settlements.

Question 46.
What is urbanization?
Answer;
The transition of population from rural agrarian economy to urban industrial and service sector economy is termed as urbanization.

Question 47.
Based on what entries is a place classified as urban in India?
Answer:

  • Population above 5,000
  • Density of population above 400 per sq.km
  • 75% or more of the population should be engaged in non-agricultures activities.

Question 48.
List down some problems due to migration from rural areas to urban areas.
Answer:
Slums, traffic problems and pollution

Question 49.
Write any two solutions for problems created due to urbanization.
Answer:
Town planning, rehabilitation, waste management, and awareness programmes.,

Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms

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Kerala State Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms

Kerala Syllabus 9th Standard Maths Decimal Forms Text Book Questions and Answers

Textbook Page No. 26

Question 1.
Write the fractions below in decimal form:

  1. \(\frac{3}{20}\)
  2. \(\frac{3}{40}\)
  3. \(\frac{13}{40}\)
  4. \(\frac{7}{80}\)
  5. \(\frac{5}{16}\)

Answer:
1. \(\frac{3}{20}\)
= \(\frac{3×5}{20×5}\) = \(\frac{15}{100}\) = 0.15

2. 40 = 2 x 2 x (2 x 5)
Multiplied by 5 x 5
(2 x 5) x (2 x 5) x (2 x 5) = 1000
it is a multiple of 10.
\(\frac{3}{10}\) = \(\frac{3x5x5}{40x4x5x}\) = \(\frac{75}{1000}\) = 0.75

3. 40 = 2 x 2 x (2 x 5)
Multiplied by 5 x 5
(2 x 5) x (2 x 5) x (2 x 5) = 1000
It is a multiple of 10.
\(\frac{13}{40}\) = \(\frac{13x5x5}{20}\) = \(\frac{325}{1000}\) = 0.325

4. 80 = 2 x 2 x 2 x (2 x 5)
Multiplied by 5 x 5 x 5
It is a multiple of 10.
(2 x 5) x (2 x 5) x (2 x 5) x (2 x 5) = 1000
\(\frac{7}{80}\) = \(\frac{7x5x5x5}{80x5x5x5}\) = \(\frac{875}{10000}\) = 0.0875

5. 16 = 2 x 2 x 2 x 2
Multiplied by 5 x 5 x 5 x 5 x 5
(2 x 5) x (2 x 5) x (2 x 5) x (2 x 5) = 1000
It is a multiple of 10.
\(\frac{5}{16}\) = \(\frac{5x5x5x5x5}{16x5x5x5x5}\) = \(\frac{3125}{10000}\) = 0.3125

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Question 2.
Find the decimal form of the sums below:

  1. \(\frac{1}{80}\) + \(\frac{1}{25}\) + \(\frac{1}{125}\)
  2. \(\frac{1}{5}\) + \(\frac { 1 }{ { 5 }^{ 2 } } \) + \(\frac { 1 }{ { 5 }^{ 3 } } \) + \(\frac { 1 }{ { 5 }^{ 4 } } \)
  3. \(\frac{1}{2}\) + \(\frac { 1 }{ { 2 }^{ 2 } } \) + \(\frac { 1 }{ { 2 }^{ 3 } } \)

Answer:
1. \(\frac{1}{80}\) + \(\frac{1}{25}\) + \(\frac{1}{125}\)
Denominators of all the fractions should be 125
\(\frac{1×25}{5×25}\) + \(\frac{1×5}{25×5}\) + \(\frac{1×1}{125×1}\)
= \(\frac{25}{125}\) + \(\frac{5}{125}\) + \(\frac{1}{125}\) = \(\frac{31}{125}\)
125 = 5 x 5 x 5
Multiplied by 2 x 2 x 2
(5 x 2) x (5 x 2) x (5 x 2) = 1000
It is a power of 10
\(\frac{31}{125}\) = \(\frac{31x2x2x2}{125x2x2x2}\) = \(\frac{248}{1000}\) = 0.248

2. \(\frac{1}{5}\) + \(\frac { 1 }{ { 5 }^{ 2 } } \) + \(\frac { 1 }{ { 5 }^{ 3 } } \) + \(\frac { 1 }{ { 5 }^{ 4 } } \)
Denominators of all the fractions should be 54
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms img-1

3. \(\frac{1}{2}\) + \(\frac { 1 }{ { 2 }^{ 2 } } \) + \(\frac { 1 }{ { 2 }^{ 3 } } \)
Denominators of all the fractions should be 23
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms img-2

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Question 3.
A two – digit number divided by an other two-digit number gives 5.875. What are the numbers?
Answer:
5875 = \(\frac{5875}{1000}\)
5875 = 5 x 5 x 5 x 47
1000 = 5 x 5 x 5 x 8
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms img-3
5.875 = \(\frac{5875}{1000}\) = \(\frac{5x5x5x47}{5x5x5x8}\) = \(\frac{47}{8}\)
But 8 is not a two digit number. So we multiplied both numerator and denominator by 2, we get
\(\frac{47}{8}\) = \(\frac{94}{16}\)
5.875 = \(\frac{94}{16}\)

Textbook Page No. 30

Question 1.
For each of the fractions below, find fractions with denominators powers of 10 gefting chaser and closer to it and hence write its decimal form:

  1. \(\frac{2}{3}\)
  2. \(\frac{5}{6}\)
  3. \(\frac{1}{9}\)

Answer:
1. \(\frac{2}{3}\) = \(\frac{2×10}{3×10}\) = \(\frac{2}{10}\) x \(\frac{10}{3}\) = \(\frac{2}{10}\) x (3 + \(\frac{1}{3}\))
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms img-4
The fraction \(\frac{6}{10}\), \(\frac{66}{100}\), \(\frac{666}{1000}\), ……… and so on get closer and closer to \(\frac{2}{3}\).
i.e., 0.6, 0.66, 0.666, …………..
\(\frac{1}{2}\) = 0.666

2. \(\frac{5}{6}\) = \(\frac{5×10}{6×10}\) = \(\frac{5}{10}\) x \(\frac{10}{6}\) = \(\frac{5}{10}\) x (1 + \(\frac{4}{6}\)
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms img-5
The fractions \(\frac{80}{100}\), \(\frac{830}{1000}\), \(\frac{8330}{5}\), ………. and so on get closer and closer to \(\frac{5}{6}\).
∴ \(\frac{5}{6}\) = 0.8333………

3. \(\frac{1}{9}\) = \(\frac{1×10}{9×10}\) = \(\frac{1}{10}\) x \(\frac{10}{9}\)
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms img-6
The fractions \(\frac{1}{10}\), \(\frac{11}{100}\), \(\frac{111}{1000}\), ………… and so on get closer and closer to \(\frac{1}{9}\).
∴ \(\frac{1}{9}\) = 0.1111 ……

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Question 2.

  1. Using algebra, explain why \(\frac{1}{10}\), \(\frac{11}{100}\), \(\frac{1}{1000}\), ………… of any number get closer and closer to \(\frac{1}{9}\) of that number.
  2. Use the general principle got above to single-digit numbers to find the decimal forms of \(\frac{2}{9}\), \(\frac{4}{9}\), \(\frac{5}{9}\), \(\frac{7}{9}\), \(\frac{8}{9}\).
  3. What can we say in general about decimal forms with a single digit repeating?

Answer:
1. Let x be the number
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms img-7
The numbers \(\frac{x}{10}\), \(\frac{11x}{100}\), \(\frac{111x}{1000}\), ……… comes closer and closer to \(\frac{x}{9}\).
So the fraction \(\frac{1}{10}\), \(\frac{11}{100}\), \(\frac{111}{1000}\), ……. comes closer and closer to \(\frac{1}{9}\).

2. a.
\(\frac{1}{10}\) part of a number comes closer to its \(\frac{1}{9}\)
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms img-8

b. \(\frac{1}{10}\) part of a number comes closer to its \(\frac{1}{9}\)
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms img-9

c. \(\frac{1}{10}\) part of a number comes closer to its \(\frac{1}{9}\)
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms img-10

d. Since \(\frac{1}{9}\) is closes to \(\frac{1}{10}\)
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms img-11

e. Since \(\frac{1}{9}\) is closes to \(\frac{1}{10}\)
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms img-12

3. The denominator of a fraction is 9 or multiple of 9 then its decimal froms is always a single digit repetition of their own numerator.

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Question 3.

  1. Find the decimal form of \(\frac{1}{11}\).
  2. Find the decimal forms of \(\frac{2}{11}\), \(\frac{3}{11}\)
  3. What is the decimal form of \(\frac{10}{11}\)?

Answer:
1. \(\frac{1}{11}\) = \(\frac{1×10}{11×10}\) = \(\frac{1}{10}\) x \(\frac{10}{11}\)
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms img-13
Continuing like this,
= 0.090909

2. a. \(\frac{2}{11}\) = \(\frac{2×10}{11×10}\) + \(\frac{2}{10}\) x \(\frac{10}{11}\)
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms img-14
Continuing like this,
\(\frac{2}{11}\) = 0.18181818 …….

b. \(\frac{3}{11}\) = \(\frac{3×10}{11×10}\) = \(\frac{3}{10}\) x \(\frac{10}{11}\)
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms img-15
Continuing like this
\(\frac{3}{11}\) = 0.272727…..

3. \(\frac{10}{11}\) = \(\frac{10×10}{11×10}\) = \(\frac{10}{10}\) x \(\frac{10}{11}\)
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms img-16
Continuing like this
\(\frac{10}{11}\) = 0.90909……….

Question 4.
Write the results of the operations below as decimals:

  1. 0.111…. + 0.222 …….
  2. 0.333…. + 0.777 …..
  3. 0.333…. x 0.666…
  4. (0.333….)2
  5. \(\sqrt { 0.444…… } \)

Answer:
1. 0.111 = \(\frac{1}{9}\)
0.222…. = \(\frac{2}{9}\)
0.111 …. + 0.2222 …….. = \(\frac{1}{9}\) + \(\frac{2}{9}\) = \(\frac{3}{9}\)
= 0.3333

2. 0.3333 = \(\frac{3}{9}\)
0.7777…. = \(\frac{7}{9}\)
0.3333 ……. + 0.7777…. = \(\frac{1}{9}\) = \(\frac{2}{9}\) = \(\frac{3}{9}\)
= 1.1111……….

3. 0.3333 = \(\frac{3}{9}\)
0.6666…. = \(\frac{6}{9}\)
0.3333….. x 0.6666…… = \(\frac{3}{9}\) x \(\frac{6}{9}\)
= \(\frac{18}{81}\) = \(\frac{2}{9}\) = 0.2222……..

4. 0.3333 …….. = \(\frac{3}{9}\)
(0.3333)2 ….. = (\(\frac{3}{9}\))2
= \(\frac{3×3}{9×9}\) = \(\frac{1}{9}\) = 0.1111…………

5. 0.4444 ….. = \(\frac{4}{9}\)
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms img-17

Kerala Syllabus 9th Standard Maths Decimal Forms Exam Oriented Question and Answers

Question 1.
Write the deciamal form of \(\frac{1}{6}\)
Answer:
\(\frac{1}{6}\) = 0.1666…….

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Question 2.
Write in deciamals

  1. \(\frac{1}{9}\)
  2. \(\frac{2}{9}\)
  3. \(\frac{1}{7}\)
  4. \(\frac{1}{11}\)
  5. \(\frac{2}{11}\)
  6. \(\frac{1}{12}\)

Answer:

  1. 0.111……
  2. 0.222………
  3. 0.14285…..
  4. 0.090909…..
  5. 0.181818….
  6. 0.08333……

Question 3.
Find the fraction of denominator is a power of 10 equal to each of the fractions below, and then write their decimal forms:
i. \(\frac{1}{50}\)
ii. \(\frac{3}{40}\)
iii.\(\frac{5}{16}\)
iv. \(\frac{12}{625}\)
Answer:
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms img-18

Question 4.
Find the fraction of denominator is a power of 10 getting closer and closer to each of the fractions be low, and then write their decimal forms.

  1. \(\frac{5}{6}\)
  2. \(\frac{3}{11}\)
  3. \(\frac{23}{11}\)
  4. \(\frac{1}{13}\)

Answer:
1. \(\frac{5}{6}\) = \(\frac{5}{10}\) x \(\frac{10}{6}\) = \(\frac{1}{10}\)(\(\frac{50}{6}\))
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms img-19

2. \(\frac{3}{11}\) = \(\frac{3}{100}\) x \(\frac{100}{11}\) = \(\frac{3}{100}\)(9 + \(\frac{1}{11}\))
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms img-20

3. \(\frac{23}{11}\) = \(\frac{23}{100}\) x \(\frac{100}{11}\)
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms img-21

4. \(\frac{1}{13}\) = \(\frac{1}{100}\) x \(\frac{100}{13}\)
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms img-22

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Question 5.

  1. Explain using algebra, that the fractions \(\frac{1}{10}\), \(\frac{11}{100}\), \(\frac{111}{1000}\)… gets closer and closer to \(\frac{1}{9}\)
  2. Using the general principle above on single digit numbers, find the decimal forms of \(\frac{2}{9}\), \(\frac{4}{9}\), \(\frac{5}{9}\), \(\frac{7}{9}\), \(\frac{8}{9}\) (Why \(\frac{3}{9}\) and \(\frac{6}{9}\) left out in this?)
  3. What can we say in general about those decimal forms In which a single digit repeats?

Answer:
1. Let x be the number
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms img-23

2. \(\frac{2}{9}\) – \(\frac{2}{10}\) = \(\frac{2}{90}\)
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms img-24
(\(\frac{3}{9}\), \(\frac{6}{9}\) These fractions having common factor in the numerator and the denominator)

3. Repeated deciamals.

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Question 6.

  1. Find the decimal form of \(\frac{1}{4}\).
  2. Write the decimal form of \(\frac{7}{10}\) + \(\frac{3}{100}\) + \(\frac{4}{1000}\).

Answer:
1. \(\frac{1}{4}\) = 0.25

2. \(\frac{7}{10}\) = 0.7
\(\frac{3}{100}\) = 0.03
\(\frac{4}{1000}\) = 0.004
\(\frac{7}{10}\) + \(\frac{3}{100}\) + \(\frac{4}{1000}\)
= 0.7 + 0.03 + 0.004
= 0.734

Question 7.

  1. Write the decimal forms of \(\frac{1}{3}\) and \(\frac{1}{9}\).
  2. What is the decimal form of (0.3333…..)2?

Answer:
1. \(\frac{1}{3}\) = 0.3333…..
\(\frac{1}{9}\) = 0.1111……

2. (0.3333…..)2 = (\(\frac{1}{3}\))2 = \(\frac{1}{9}\) = 0.1111….

Question 8.
Write the decimal forms of \(\frac{3}{25}\) and \(\frac{1}{8}\).
Answer:
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms img-25

Question 9.
a. Write the decimal form of the fractions and \(\frac{1}{2}\) and \(\frac{2}{5}\).
b. 1f f is a fraction between and \(\frac{2}{5}\) What is k?
c. Write the decimal form of the fraction \(\frac{4}{k}\).
Answer:
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms img-26
If \(\frac{4}{k}\) is between \(\frac{1}{2}\) and \(\frac{2}{5}\), then \(\frac{4}{k}\) is between \(\frac{4}{8}\) and \(\frac{4}{10}\). Then the number is \(\frac{4}{9}\).
∴ k = 9

c. \(\frac{4}{9}\) = 0.4444…….

Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity

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

Kerala State State Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity

Current Electricity Textual Questions and Answers

Activity -1

A positively charged electroscope is connected to the earth through a switch using a conductor.
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 1

Question 1.
What kind of charge is present in this electroscope?
Answer:
Static

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Question 2.
What happens to this charge when the switch is turned on?
Answer:
Charge neutralizes

Question 3.
Will the flow of charge sustain in this arrangement?
Answer:
The charge will not sustain in this arrangement

Activity – 2

A circuit with a cell, a bulb, and a switch is given in the figure.
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 2

Question 4.
Will the flow of electric current sustain in the circuit if it is switched on?
Answer:
Yes, the flow of electric charge will sustain.

Question 5.
What difference is there in the flow of current in both circuits?
Answer:
In the first circuit, there is a flow of charge for a short interval of time. There is a continuous flow of charge in the second.

Question 6.
Complete the table based on different situations as shown in figure.
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 3
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 4
Answer:

Situation Direction of flow/motion
Ball falling down Downwards from a higher level to a lower level
Flow of air From a region of high pressure to a region of low pressure
Flow of water From a higher level to lower level
Flow of heat From a point having higher temperature to that having lower temperature

There should be a difference in energy levels between two points if any type of flow is to occur.

Activity-3

observe the figures
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 5
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 6

Question 7.
If the value is opened, in which one will there be a flow of water and rotation of the wheel?
Answer:
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 7

Question 8.
Why?
Answer:
It is due to the gravitational potential difference that there is a flow of water and consequent rotation of the wheel.

Activity-4

Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 8
A bulb is connected to a switch, using a conductor.

Question 9.
Will the bulb glow if switched on? Why?
Answer:
The bulb will not glow
There is no potential difference between P and Q. Hence there is no flow of current and the bulb does not flow.

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Potential Difference and Current:
There should be a potential difference between two points of a conductor if there is to be flow of current between them. Current flows from a point of high electric potential to a point of low electric potential.

The unit of potential difference is volt (V). Voltameter is the device to measure this. If 1 joule of work is done to move one-coulomb charge from one point to another, then the potential difference between the points is 1 volt.

Activity – 5

Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 9
The pump has been used in such a way that the some quantity of water that flows from A to B per second is returned to Afrom B in the same period of time.

Question 10.
Why is there a continuous flow of water when the value is opened?
Answer:
Here, it is due to the working of the pump, which is an external source of power, that the potential difference was maintained and the flow of water was made possible continuously.

Source of emf

An external source is needed to maintain a potential difference between the ends of a conductor and to maintain the flow of electric current through the conductor. That external source is called source of emf.
Eg: Generator, cell, battery, solar cell ………

Question 11.
Write down the energy change in each.
Answer:
Generator: Mechanical energy → electrical energy
Cell (While discharging): Chemical energy → electrical energy
Battery (While discharging): Chemical energy → electrical energy
Solar cell: Solar energy → electrical energy

Question 12.
Complete the table
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 10
Answer:

Water circuit Pump Water wheel flow of water Value
Electric circuit Cell bulb flow of electric charge Switch

A source of emf is essential to maintain a potential difference between the ends of a conductor and to maintain the flow of current through the conductor.

Activity – 6

Make a circuit given in figure using a voltmeter a 12V, 3W bulb, a cell, and a switch operate it
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 11

Question 13.
How should-you connect a voltmeter in a circuit?
Answer:
The voltmeter should be connected across the points (parallel) where the potential difference is to be mea¬sured.

Question 14.
In what mode are the cells connected within the remote control of a TV?
Answer:
In series mode

Question 15.
If 4 cells of 1.5 V each are connected in series what is the total voltage?
Answer:
4 × 1.5 = 6V

Question 16.
How can you connect four cells of 1.5V each to get 3V? What is the advantage of doing so?
Answer:
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 12
If connected in this mode, we get electric current for a long time without variation in voltage.
Combination of cells:
A battery is a combination of two or more cells connected in a suitable manner. Cells can be connected in two ways.
1. Series connection
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 13
This is the method of connecting cells one after the other in such a way that the positive of one cell is connected to the negative of another cell.
Salient features:

  • The total emf is the sum of the emf of all the cells.
  • The current passing through each cell is the same.
  • The internal resistance developed in the circuit by the battery increases.
  • The current in the external circuit increases under high voltage.

2. Parallel connection

Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 14
This is the method of connecting similar potes together.
Salient features:

  • If all the cells have equal emf then the emf of the circuit is the same as that of a single cell.
  • The total current flowing in the circuit splits up and flows through each cell.
  • The internal resistance of the circuit is very low.
  • More current can be made available for a longer time under low voltage.

Electric Current:
Electric current is the flow of electric charges. Current is the quantity of charge that flows through a conductor in a circuit in one second.

Question 17.
If 10 coulomb charge flows in a circuit in 5s, how much is the charge flowing in the circuit in one second?
Answer:
Charge, Q = 10C
Time,t = 5s
Charge flowing in one second = \(\frac { 10 }{ 5 }\) = 2 C/s

Question 18.
If a charge of Q coulomb flows in a time t second, then how much is the quality of charge that flows in one second?
Answer:
Current (I) = \(\frac{\text { Quantity of charge }}{\text { Time taken }}\)
i.e I = Q/t

Question 19.
What is the unit of current?
Answer:
Unit of current = \(\frac{\text { Unit of charge }}{\text { Unit of time }}\)
= C/s OR A

Activity -1

Make a circuit containing an ammeter, switch, cell and a bulb connected in series.
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 15
Repeat the experiment by increasing the number of cells in series.

Question 20.
What change occurred in the ammeter reading when the number of cells was increased?
Answer:
Ammeter reading increases

Question 21.
What about the intensity of light from the bulb?
Answer:
Intensity of light increases

Question 22.
How are the current and the intensity of light related to each other?
Answer:
As current in circuit increases, intensity of light increases.

Question 23.
What is the current in a conductor if 2 C charge flows in 10s?
Answer:
\(\mathrm{I}=\frac{\mathrm{Q}}{\mathrm{t}}=\frac{2 \mathrm{c}}{10 \mathrm{s}}=0.2 \mathrm{C} / \mathrm{s}\)
= 0.2 A

Ammeter

Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 16
Ammeter is a device used to measure the current in a circuit. The positive terminal of it must be con¬nected directly to the positive of the cell and the negative terminal, to the negative of the cell. Ammeter should be connected in series in the circuit The needle of the device moves in accordance with the current in the circuit We can measure the current by checking the position of the needle. Unit of current is ampere (A), it is also written as C/s. mA(milliampere) and µA(microampere) are smaller units of current The symbol of ammeter is
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 17

Ohm’S Law

Make a circuit by including a nichrome wire (30cm), cell, switch, ammeter, and voltmeter.
Measure current (I) and potential difference (V). Repeat the activity by increasing the number of cells in series.
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 18
Analyze the table and record the findings

Question 24.
What change occurred in the circuit when there was a change in voltage?
Answer:
As voltage increases, current increases.

Question 25.
Do you see any peculiarity in the value of V/l? v
Answer:
V/I will be a constant
V ∝ I
V = a constant × I
V/I = a constant
This constant is the resistance of the conductor. This is indicated by the letter R.
∴R = V/I
Ohm’s law:
When temperature remains constant, the current through a conductor is directly proportional to the potential difference between its ends. In other words, the ratio of potential difference to the current is a constant.
Resistors are conductors used to include a particular resistance in a circuit its symbol is
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 19

Question 26.
On the basis of the information gathered from Table 6.5, draw a V-l graph, Mark I in the X – axis and V in the Y – axis.
Answer:
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 20

Question 27.
Is the graph a straight line?
Answer:
Yes, the graph is a

Question 28.
What is the unit of resistance?
Answer:
Unit of resistance = \(=\frac{\text { Unit of voltage }}{\text { Unit of Current }}\)
\(\frac{\text { Volt }}{\text { Ampere }}\) or ohm (Ω )

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Question 29.
1 Ω = 1V/1A From this what do you mean by 1 ohm?
Answer:
If the potential difference between the ends of a conductor is 1V when a current of 1A flows through it, then the resistance of the conductor is 1Ω.

When the potential difference between the ends of conductor is 1 Vand if a current of 1A flows through it, then the resistance of the conductor is 1Ω.
Using the given figure, from equation representing Ohm’s Law.
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 21

Question 30.
Complete the following table based on Ohm’s Law

Voltage (Volt V) Current (1)  ampere (A) Resistance (R)  ohm(Q)
12 …………………. 4
………………….. 2 3
6 3  …………………..

Answer:

Voltage (Volt V) Current (I) ampere (A) Resistance (R) ohm(Q)
12 …… 3 …… 4
…… 6 ….. 2 3
6 3 ……. 2 ……..

Resistors

Arrange a circuit as shown in the figure
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 22
Among the conductors fixed on the wooden plank, PA is iron, PB is aluminum, PC, PD and PE are nichrome. Their lengths are the same. PE has double the length. The thickness of PD is double that of the others. Touch the free end J at A, B, C, D and record in the table the ammeter reading at each distance.

Question 31.
Is the intensity of light from the bulb the same in each situation?
Answer:
No

Question 32.
Is the ammeter reading the same when different conductors of the same length and thickness were included?
Answer:
No. The ammeter reading was not the same

Question 33.
What change has occurred in the ammeter reading when the area of cross-section of the same conductor is altered?
Answer:
When the area of cross-section increases ammeter reading also increases.

Question 34.
Is there a change in the ammeter reading when the length of the same conductor is altered? Record,
Answer:
Ammeter reading decreases with increase in length.

Question 35.
Is the applied potential difference the same in all cases?
Answer:
Yes. The potential difference applied is the same.

Question 36.
According to Ohm’s Law, V/I must be a constant (resistance, R). If so, what is the reason for the changes in the ammeter readings?
Answer:
The reason for the change in the ammeter reading is the variation of resistance of the conductors included in the circuit.

Activity -1

Connect a 6V bulb to a 6V source. Using a multimeter, measure the resistance of the bulb when the circuit is switched off. Switch on the bulb for a short time, then switch it off and immediately measure its resistance.

Question 37.
Is the resistance the same in both the situations?
Answer:
No

Question 38.
When the circuit was switched on, was the temperature of the bulb low or high?
Answer:
High

Question 39.
Did the resistance increase or decrease when the temperature was increased?
Answer:
Resistance increased when the temperature was increased.

Question 40.
List the factors affecting the resistance of a conductor?
Answer:

  • Area of cross-section
  • Nature of the material
  • Length
  • Temperature

Activity-2

In the activity conducted above, touch the free end J at E and slowly slide it from E to P

Question 41.
What change occurred in the intensity of light from the bulb?
Answer:
Intensity of light increases gradually.

Question 42.
What may be the reason behind the change?
Answer:
As the length of the conductor decreases, resistance and current increases.

Question 43.
What is the working principle of a rheostat?
Answer:
If the potential difference is constant, then the current is inversely proportional to the resistance. For a conductor of uniform area of cross-section, the length of a conductor and the resistance are directly proportional.

Rheostat is a device used to regulate the current in a circuit by changing the resistance

Question 44.
What is the symbol of a rheostat?
Answer:
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 23

Question 45.
Given below is a table related to the resistance of a conductor. Complete the table suitably.
Answer:

Question 46.
Analysis the completed table and write down the inferences.
Answer:
The resistance of a conductor increases with the increases in the length of the conductor.
R α l
The resistance of a conductor decreases with increases in the area of cross-section.
R α I/A that is R α I/A
R = a constant × l/A
R = \(\rho \frac{1}{\mathrm{A}}\)
\(\rho\) = RA/l
P is the resistivity of the material the conductor is made of. The length of a conductor of resistance R Q is 1m and its area of cross-section is 1m2. Calculate the resistivity of the material the conductor is made of. length, l= 1m
Area of cross-section, A = 1m2
Resistivity \(\rho=\frac{R A}{1}=\frac{R \times 1}{1} \quad \rho=R\)
Resistivity of a substance is the resistance of the conductor of unit length and unit area of cross-section. The resistivity of a substance is a constant at fixed temperature. But it will be different for different materials.
Unit of resistivity =
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 24
The Unit of resistivity is Qm
Let’s get acquainted with some of the tools related to electric current:
We use many electric devices in our everyday life. Different tools are needed to connect these devices with the electric line and to perform maintenance. They are enlisted here.

Screwdriver
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 25
it helps in fixing and removing the screws Screwdrivers are available in different sizes It is used to combine a wide range of screws with +,* shaped edges.

Electric tester
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 26
It is used to check whether current is coming into the sockets or other devices in the houses Some of these can be used as screwdriver. The bulb in¬side the tester will glow if there is presence of current.

Wire stripper
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 27
It is used to remove insulation of wires while combining insulated electric wires or when they are to be connected to the devices

Pliers
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 28
It is used to join wires by twisting them together or for cutting or removing wires. Pliers are available in different shapes and sizes

Gloves
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 29
While doing the work related to electric power, gloves are worn in the hand as a protection from electric shock.

Multimeter
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 30
It is used to measure current, voltage, and resistance in a circuit and to understand whether the circuit is open, closed or any connection is left. Besides, it also helps to check whether the various elements in an electronic circuit are functioning properly.

Clamp ammeter
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 31
It helps to measure the current in a circuit without connecting wires or devices in the circuit.

Insulation tape
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 32
When connecting the wires or connecting it with a device, this is used to provide insulation in those parts where it has been damaged.

Spanner
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 33
It is used for fixing nut and bolt. They are available in different sizes.

Soldering iron
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 34
It is used to solder electronic components in a circuit

Hammer
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 35
It is used for fixing and removing nails.

Drill machine
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 36
It is used to drill holes on hard surfaces. It can be used to fix and remove screws as well.

Let Us Assess

Question 1.
complete the table properly
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 37
Answer:
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 38

Question 2.
Given below are the diagrams showing the connection of ammeter and voltmeter in a circuit. Of these, which are correct?
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 39
Answer:
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 40

Question 3.
Complete the table. The conductor is made of the same material.
Answer:

Length of conductor Area of cross-section of conductor Resistance of the conductor
1 cm 2 cm2 10 Q
2 cm 2cm2 20 Q
1 cm 4cm2 5Q

Question 4.
In an electrical circuit if 100 J work is done to move 10 C electric charge from point A to the point B, find out the potential difference between the points A& B.
Answer:
100/10 = 10V

Question 5.
6 electric cells are connected in series in an elec¬tronic device which works at 9 V potential difference. Find out emf of one cell.
Answer:
9/6 = 1.5V

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Question 6.
An ammeter that connects to an electronic circuit shows a reading of 2A. Find how many charges flows through the ammeter in 10 s.
Answer:
Q = I x t = 2 x 10 = 20 coulomb

Question 7.
When a conductor is stretched, its length becomes double. Find out how many times the resistance changes.
Answer:
4 times.

Question 8.
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 41
In the given graphs which is the graph depicting Ohm’s Law? Justify your answer.
Answer:
(a) v ∝ I, As V increases, I also increase.

Question 9.
A conductor of 5 Q resistance has length 2m and area of cross-section 2 m2. If so, find out the resistivity of the material of the conductor.
Answer:
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 42

Question 10.
Draw a circuit diagram describing how 6 torch cells should be connected to a bulb and a switch to obtain effective voltage of 9 V.
Answer:
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 43

Current Electricity More Questions

Question 1.
The resistance of a 10cm long wire is 120. If this is folded into two parts of equal length and included in a circuit, how much will be the resistance produced?
Answer:
When folded into two parts, length is halved and area of cross-section is doubled. Due to the decrease in length, resistance is halved. Also due to the decrease in area of cross-section, resistance is again halved.

So effective resistance R = \(12 \times \frac{1}{2} \times \frac{1}{2}=3 \Omega\)

Question 2.
Of the following, which one correctly indicates Ohm’s Law?
Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity 44
Answer:
The second one

Question 3.
A potential difference of 6V is applied across a conductor having 12Ω resistance. How much current will pass through it?
How many times will the current increase if length of the resistor is halved and potential difference is doubled?
Answer:
I = \(\frac { V }{ R }\) = \(\frac { 6 }{ 12 }\) = 0.5A
If length is halved R = 12 × 1/2 = 6
potential difference V = 2 × 6 = 10V
I = \(\frac { V }{ R }\) = \(\frac { 12 }{ 6 }\) = 2A
That is I is increased by 4 times.

Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion

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

Kerala State Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion

Wave Motion Textual Questions and Answers

Wave Motion

Fill half of a trough with water. Place some crumpled paper balls in it. Make ripples on the surface of the water using finger.
Observation:
We can see disturbance spreading from its origin to other place. Water particles move up and down about their mean position without displacement in the direction of propagation of wave. Energy is transferred from particle to particle and spreads everywhere due to wave motion. Wave motion is the propagation of disturbances, produced on one part of a medium by the vibration of its particles, to all its other parts.

Question 1.
Write down examples of wave motion that you see around.
Answer:

  • waves on water
  • waves formed in a slinky
  • waves formed when a rope is moved up and down after tying one of its ends.
  • waves formed on stretched strings.

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Mechanical waves & electromagnetic waves:
Waves can be classified mainly into two types

  1. Mechanical waves: The presence of a medium is required for the transmission of these waves, eg. The waves formed on the surface of water, sound waves etc.
  2. Electromagnetic waves: Electromagnetic wave is a combined form of an electric field and a magnetic field which vary continuously. A medium is not essential for its propagation
    eg. radio waves, lightwaves.

Question 2.
How many types of mechanical waves are there? What are they?
Answer:
There are two types of mechanical waves:
They are

  1. Transverse waves
  2. Longitudinal waves

Transverse Wave
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 1
Tie one end of a rope to a window, Wind a ribbon or paper on the rope ins such a way that you can see it clearly. Hold the other end of the rope and move it up and down. Observe the wave motion on the rope.

Question 3.
How does the ribbon/paper move?
Answer:
Ribbon/ paper moves up and down

Question 4.
In which direction does the wave move?
Answer:
The wave moves forward or horizontally.

Inferences

1. The ribbon moves up and down
2. The ribbon’s position on the rope does not change.
3. The ribbon is vibrating in a direction perpendicular to the direction of propagation of the wave.
Each particle of the wave vibrates in a direction perpendicular to the direction of propagation of the wave.
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 2
4. In this case is the motion of particles parallel or perpendicular to the direction of propagation of the wave? The direction of motion of particles is perpendicular to the direction of propagation of the wave. A transverse wave is a wave in which the particles of the medium vibrate in a direction perpendicular to the direction of propagation of the wave.

Can’t you explain why the disturbances on water could not make the paper boat move away from the shore? The waves formed on the surface of water is transverse wave. The water particles move only up and down. The water particles do not move in the horizontal direction. So the paper boats could not move away from the shore.

Observe the graphic representation of a transverse wave at a particular instant.
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 3

Question 5.
What are crests and troughs?
Answer:
The elevated portions are called crests. The depressed portions are called troughs

Question 6.
In the figure, which are the points of the highest displacement (amplitude)?
Answer:
A, C, E, G, I, K, M

Question 7.
How many crests and troughs are there in the figure?
Answer:
4 crests, and 3 troughs.

Question 8.
Whether all the particles are in the same phase of vibration at a particular time?
Answer:
No

Question 9.
Which are the particles in the same phase of vibration as that of A?
Answer:
E, I, M

Question 10.
What about C?
Answer:
G, K

Question 11.
What is the wavelength of the wave shown in the figure?
Answer:
Wavelength = 4 m

Characteristics of Waves:

  1. Amplitude : Amplitude is the maximum displacement of a particle from its mean position. This is denoted by the letter a.
  2. Wavelength: Wavelength is the distance between two consecutive particles which are in the same phase of vibration. This is equivalent to the distance advanced by the wave by the time a particle has completed one vibration. The Greek letter X (lambda) is used to denote the wavelength. The unit is metre (m).
  3. Frequency: Frequency is the number of vibrations in one second.
    Frequency = \(\frac{\text { number of vibrations }}{\text { Time taken }}\)\(\mathrm{f}=\frac{\mathrm{n}}{\mathrm{t}}\)
    The unit of frequency is hertz (Hz).

Question 12.
What is the frequency of the wave if the particles A makes 100 vibrations is 5 s?
Answer:
\(\begin{array}{l}{\text { Frequency of the wave }=\frac{\text { number of vibrations }}{\text { time }}} \\ {f=n / t=\frac{100}{5}=20 \mathrm{Hz}}\end{array}\)
The equation connecting velocity, wavelength, and frequency of a wave is v=f λ ;
v – Velocity (distance travelled by the wave in one second);
f – frequency (number of vibrations in one second);
x – wavelength (distance between two consecutive particles which are in the same phase of vibration).

The graphical representation of two waves of the same amplitude, generated at specific intervals of time, is given below.
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 4

Question 13.
What is the wavelength of the first wave? What about the second one?
Answer:

  • wavelength of the first wave = 4m
  • wavelength of the second one = 2m

Question 14.
Which wave has a higher wavelength?
Answer:
First wave has a higher wavelength.

Question 15.
Calculate the frequency of each wave if they have traveled this distance (12 m) in 0.25s.
Answer:
Frequency of the first wave
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 5

Question 16.
What change takes place in the wavelength when the frequency increases?
Answer:
As frequency increases, wavelength decreases. Wavelength of a wave with a constant speed decreases with increase in frequency, ie. frequency is inversely proportional to the wavelength.
f ∝ 1/λ

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Question 17.
Observe the graphic representation of a wave motion given below.
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 6
a) What is the amplitude of the wave?
b) What is the wavelength?
c) Calculate the frequency of the wave if it took 0.2s to reach A.
d) Calculate the speed of the wave
Answer:
a) 2 cm
b) 8 m
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 7

Longitudinal Wave

Fix one end of a slinky to a wall. Hang some pieces of paper on the coils at equal distances. Press a few coils on the free end held in the hand and then release
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 8
Observation: The air particles vibrate to and fro when the waves pass through air.
High pressure is experienced in places where the air particles are close. Such a region is the compression (C). Regions of low pressure are the rarefactions (R).

A longitudinal wave is a wave in which the particles of the medium vibrate in a direction parallel to the direction of propagation of the wave. This creates compressions and rarefactions alternately in the medium.

Question 18.
How do we hear a sound from a source?
Answer:
Listen to the sound from an excited tuning fork. The vibrations of the tuning fork make the air particles around it to vibrate.
Sound waves are longitudinal waves
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 9

Question 19.
How many compressions are their in the longitudinal wave shown in the figure?
Answer:
4 compressions

Question 20.
Find out the differences between transverse and longitudinal waves and complete the table.
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 10
Answer:
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 11

Sound

Sound is produced by the vibration of objects. Sound is a longitudinal wave. Sound needs a material medium to travel.
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 12

Question 21.
What do C and R in the figure indicate?
Answer;
C for compressions and R for rarefactions.
Wavelength of longitudinal wave The distance between corresponding points of two consecutive compressions or two consecutive rarefactions is the wavelength of the longitudinal wave.

Question 22.
Find out the wavelength in the figure and write it down.
Answer:
Wavelength = 1 m

Question 23.
What is the speed of the wave if its frequency is 92 Hz?
Answer:
Velocity V = f λ = 92 × 1 = 92 m/s

Speed Of Sound

  • Sound travels through all media.
  • The speed of sound differs from one medium to another.
  • Sound waves involve the vibrations of the particle of the medium. So sound waves can not travel through vacuum.
  • Sound travels faster in solids because the mol¬ecules are closely packed.
  • Speed of sound in liquids is comparatively slow.
  • Speed of sound in gases is much slower because molecules are loosely packed.
  • One of the reasons for the difference in the speed of sound is the difference in the density of the media.
Medium Velocity(m/s) (At 20°C)
Solid Aluminum 6420
Steel 5941
Liquid Pure water 1482
Seawater 1522
Gas Air 343
Helium 965

Question 24.
What are the factors that influence the speed of sound through air?
Answer:

  • Humidity [As humidity increases, speed increases)
  • Density [As density increases, speed decreases]
  • Wind
  • Temperature [As temperature increases, speed increases] ‘
  • Nature of the medium.

Humidity and speed of sound:
The amount of water vapor in the air is humidity. It is less during winter and high in summer. The speed of sound increases with increase in humidity. This is because the density of air decreases with the increase in humidity.

Reflection Of Sound
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 13

Arrange two PVC pipes, a glass plate, and a stop clock as shown in the figure.
We can hear sound through the pipe B.
It is due to the reflection of sound from the glass plate.
Sound can also reflect like light The law of reflection, ∠i = ∠r is true in the case of sound also.

Multiple Reflection Of Sound

Sound getting reflected repeatedly from different objects is multiple reflection.

Situation making use of multiple reflection:

  • Devices like megaphone, horns, musical instruments like shehnai and trumpets, are made in such a way that the sound produced form them travels only in a certain direction without spreading to other directions. In such devices there is a conical shaped open end which enables the reflected sounds to travel in a particular direction alone, thus enabling us to hear it louder.
  • Stethoscope: Helps us to detect beats in the body especially heartbeats.
  • The ceilings of halls are given a curvature: As a result sound undergoes multiple reflection and spreads everywhere in the hall.
  • Soundboards: The curved soundboards placed behind the screen makes the sound undergo multiple reflections and spreads everywhere in the hall. The boards in musical instruments like guitar, violin, etc., also act as soundboards.

Reverberation

Reverberation is the persistence of sound as a result of multiple reflection

Question 25.
If we felt a boom of sound in empty rooms.
a) Which are the regions where sound waves in a room get reflected?
b) Do these repeatedly reflected sound waves reach the ear of a listener simultaneously?
c) Will you be able to hear all these sounds clearly due to the persistence of audibility? Won’t’ you be hearing only a boom of all the sound?
Answer:
a) The sound waves get reflected form the walls, ceilings, floor etc.
b) No
c) We will not be able to hear all these sounds clearly, due to the persistence of audibility. We will be able to hear only a boom of all the sounds. This is as a result of reverberation.
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 14

Persistence of Audibility:
The sensation of hearing produced by a sound is 1
retained for a period of 1/10 s = 0.1 s. This characteristics of the ear is the persistence of audibility. If another sound reaches the ear within 0.1 s, simultaneous hearing of both the sounds is experienced.

Echo

The phenomenon of hearing a sound by reflection from a surface or obstacle, after hearing the original sound is the echo.

Question 26.
What should be the minimum distance to the reflecting surface if the velocity of sound in air is 340 m/s?
Answer:
Due to persistence of audibility, we can hear the first sound and reflected sound separately, only if there is a time gap second between them.
Speed = \(\frac { Distance travelled }{ time }\)
Speed of sound in air = 340 m/s
time = 1/10 s
Distance = speed × time
= 340 × 1/10
= 34m
So the minimum distance to the reflecting surface is half of 34 m.
i.e. = 17 m

HSSLive.Guru

Question 27.
A person who bursts a cracker hears its echo after 1 s. What is the distance to the reflecting surface if the speed of sound in air is 340 m/s?
Answer:
2d = v × t = 340 × 1 = 340
∴ d = 340/2 = 170m

Question 28.
What should be the minimum distance between the source and the reflecting surface in water to identify the echo within water? (speed of sound in water is 1482 m/s)
Answer:
Velocity sound = 1482 m/s
Distance = velocity × time
= 1482 × 1/10
= 148.2 m
So the minimum distance between the source and the reflecting surface in water to identify the echo within water
= 148.2/2
= 74.1 m

Question 29.
Write down the situation in which echo is heard.
Answer:
When we clap our hands from a field or valley having width greater the 17m, we can hear echo. When we talk loudly standing in an auditorium having length more than 17m, we can hear echo.

Acoustics Of Buildings

Acoustics of building is the branch of science that deals with the conditions to be fulfilled in the construction of a building for clear audibility.

Question 30.
Why are the walls made rough in big halls like the cinema theatres?
Answer:
The walls are made rough to avoid the regular reflection of sound. Rough surfaces can absorb sound.

Question 31.
With respect to reflection of sound, what are the problems if the distance between the walls in a room is more than 17 m?
Answer:
If the distance between the walls in a room is more than 17m, the sound waves are reflected repeatedly from the walls, ceiling and floor of the hall, and produce many echos So the sound becomes blurred, distorted and confusing due to overlapping of different sounds.

Question 32.
What are the methods to minimize the problems that occur due to reflection of sound?
Answer:
The methods to minimize the problems that occur due to reflection of sound are

  • Use curtains having many folds.
  • Provide large number of ventilators and windows.
  • Make the walls and roof rough.
  • Make sure that the ratio between the height and width of the hall 2:3.
  • Make the floor rough using carpets.
  • Increase the number of audiences.

Whispering galley:
The Whispering Gallery at St. Paul’s Cathedral in London is the best example for the reflection of, sound. Event if you are only whispering near the circular wall below the dome the sound will be heard loudly anywhere within the gallery. This is due to the multiple reflection of sound from the circular walls. The Gol Gumbas in Bijapur of Karnataka is another example.

Ultrasonic Sound

Sound with a frequency greater than 20000 Hz, ie, above the higher limit of audibility is called ultrasonic sound.

Uses of Ultrasonic waves:
1. Ultrasonic waves are used to clean spiral tubes, machine parts without a definite shape and electronic components. Objects to be cleaned are dipped in a cleaning solution. Ultrasonic waves are passed through this solution. Due to the high frequency of vibration of ultrasonic waves, dust and grease like substances get detached and are removed from the object.
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 15

2. Ultrasonic waves are used to detect cracks and flaws in large metal blocks
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 16
Ultrasonic waves passed through a metal block, are allowed to reach the detectors. If there is any crack or flaw, ultrasonic waves will reflect back from that part. So they do not reach the detectors. Audible sound waves of longer wavelength cannot be used for this purpose as they bend around the corners of the defective location to reach the detectors.
3. Echocardiography: Ultrasonic waves are used for taking images of heart. This is known as Echocardiography (ECG).
4. Ultrasonography: Ultrasonic waves are used for getting images of internal organs such as kidney, liver, gall bladder and uterus. Ultrasonic waves travel through the tissues of the body and get reflected from the region where there is a change in tissue density. These waves are then converted into electrical signals and are used to form images of the organs. This technology is called ultrasonography Ultrasonic waves can crush small stones formed in the kidney into fine grains.
5. Sonar (Sound Navigation and Ranging): Sonar is a device that uses ultrasonic waves to measure the distance, direction, and speed of objects underwater.
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 17
In the figure, ultrasonic waves are sent out from a SONAR which is installed in a ship and they get reflected back after striking an object at the bottom of the sea.

Question 33.
What happens to the ultrasonic waves after striking the object on the seabed?
Answer:
Ultrasonic waves that are reflected back after striking the object reach the detector. The detector converts the ultrasonic waves into electrical signals.

Question 34.
The distance traveled by a wave can be calculated by knowing the speed of ultrasonic sound in seawater and the time taken for the wave to return.
Answer:
Distance = speed × time

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Question 35.
Ultrasonic waves from a ship hits a rock at the bottom of the sea and comes back after 0.5 s. Calculate the distance to the rock from the ship. Consider speed of sound through seawater as 1522 m/s.
Answer:
Distance = speed × time .
= 1522 × \(\frac { 0.5 }{ 2 }\) = 380.5 m

Question 36.
Bats make use of ultrasonic sounds for catching prey. How do bats catch prey? Observe figure and write down in your science diary.
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 18
Answer:
Bats produce ultrasonic sounds and the sound gets reflected after striking the prey. It can receive the waves.

Seismic Waves And Tsunami

Originate from the epicenter of the earthquake. Seismology is the branch of science that deals with the study of seismic waves. Scientists dealing with the study of seismic waves are called seismologists. The intensity of earthquake is measured in Richter scale.
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 19
Seismic waves formed as a result of earthquakes are classified into three: Primary waves (p waves), Secondary waves (S waves) and surface waves. Among these, primary waves travel fastest.
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 20
Secondary waves are slower than the primary waves. In a seismograph, the difference in the arrival time of the primary and secondary waves can be used to determine the approximate distance to the epicenter. The amplitude of the waves, obtained using seismo-graph determines the intensity of an earthquake. Two surface waves, Rayleigh waves that travel only through the Earth’s surface and Love waves, are also formed as a result of earthquake. Surface waves are the reason for major damages caused by earthquake, though the speed-of surface waves is less than that of secondary waves.

Tsunami

Tsunami is a series of gigantic waves in a water body caused by the displacement of large volume of water in the deep regions near the sea bed. Tsunamis are caused by underwater earthquakes, volcanic eruptions, meteorite impacts, and other such disturbances. The term

Tsunami is coined by combining two Japanese words Tsu’ which means ‘harbor’ and ‘nami’ .which means ‘long wave’. In the bay region, the speed of Tsunami ranges from 600 to 800 km/h and their wavelength from 10 to 1000 km. The amplitude is less in the deep sea. Hence Tsunami is not felt by passengers . in ships. As waves approach the seashore, the trough of the waves rubs against the land. As a result the speed and wavelength of the waves drop down suddenly, amplitude increases and the coastal re¬gion gets submerged.

Tsunami height depends on the geographical nature of the coast and the depth of the seabed. As Tsu¬nami approaches the shore from the deep sea, the energy lost is not significant. Hence the magnitude of destruction will be very high. If it is the crest of the wave that first reaches the shore, the waves rise high and if it is the trough that reaches first, the sea will be in a state of retreat. The system that gives advance warning about Tsunami is known as DART (Deep Ocean Assessment and Reporting of Tsunami).

Question 37.
What are the methods to be adopted to escape from Tsunami? Discuss.
Answer:

  • Move to a higher plain taking the unusual receding of the sea from the sea shore as a warning of the approaching Tsunami.
  • Don’t assume for yourself that the danger is over, instead, wait for the official announcement.
  • Try to save your life and not your belongings, as life is precious.
  • If caught in Tsunami, try to save yourself by latching onto some floating objects.

Let Us Assess

Question 1.
Observe the graph
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 21
1. Find out the amplitude of the wave
2. What is the speed of the wave if it travels 800 m in 2 s?
3. What is the frequency of the wave?
Answer:
1. The amplitude = 1. 5 m
2. Speed =  \(\frac { Distance travellaed }{ time }\)
3. V = 800/2 = 400 m/s
λ = 4m
v= 400 m/s
f = V/ λ = 400/4 = 100 Hz

Question 2.
What do you mean by acoustics of buildings? Suggest four steps that can be taken, while constructing buildings, to avoid problems that may occur due to multiple reflection of sound.
Answer:
Acoustics of buildings is the branch of science that deals with the conditions to be fulfilled in the construction of a building for clear audibility.
To avoid problems that may occur due to multiple reflection of sound are

  • Make the floor rough using carpets.
  • Provide large number of ventilators and windows
  • Make the walls and roof rough.
  • Use curtains having many folds.

Question 3.
A sound signal from a ship floating on water hits a rock at the bottom of the sea and comes back to the ship after 4s. Calculate the distance of the rock from the surface of water. The speed of sound in water is estimated to be 1500 m/s
Answer:
V = s/t
Velocity of sound = V = 1500 m / s
time ,t = 4s
Distance travelled, S = v x t =1500 x 4 = 6000 m
= 6000 m
Distance of the rock from the surface of water = 6000/2 m
= 3000 m

Question 4.
Wavelength of a wave that travels with a speed 339 ms is 1.5 km. What will be its frequency?
Answer:
v = 339 m/s
λ = 1.5 km
= 1500 m
V = f λ
f = v/λ = 339/1500
= 0.226 Hz

Question 5.
Wavelength of a sound wave having frequency 2 kHz is 35 cm. How much time will it take to travel a distance 1500 m?
Answer:
f = 2 kHz = 2000 Hz
λ = 35 cm = 0.35 m
v = f λ = 2000 × 0.35
= 700 m/s
t = \(\frac{\text { distance }}{\text { time }}\)
= \(\frac { 1500 }{ 700 }\) = 2.14 s

Question 6.
For a person with normal hearing, the limit of audibility is 20 Hz to 20000 Hz. If so, what will be the limit of wavelength of sound waves that are audible to human beings? Assume that the speed of sound is 340 m/s.
Answer:
f = 20 Hz,
v = f λ ,
λ = v/f = 17m
f = 20000Hz,
v = f λ,
λ = 0.017 m
so limit of wavelength = 0.017 to 17
= 0.017 to 17 m

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Wave Motion More Questions and Answers

Question 1.
Classify the following statements as transverse waves and longitudinal waves.
a) Particles of the medium vibrate perpendicular to the direction of propagation of the wave.
b) Creates pressure difference in the medium
c) Forms in solids, liquids, and gases
d) Lightwaves
e) Seismic waves
f) Forms crest and trough
Answer:
Transverse wave: a, d, f
Longitudinal wave: b, c, e

Question 2.
Observe the figure
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 22
a) What kind of wave motion is shown in the figure? Illustrate your answer
b) Calculate the frequency of this wave if its velocity is 6420 m/s and wavelength is 6 m.
Answer:
a) Longitudinal wave
Here pressure difference is plotted against the y-axis. Pressure difference forms only in longitudinal waves
b) V = 6420 m/s
λ = 6 m
frequency, f = V/λ
= 6420/6 = 1070 Hz

Question 3.
If a person clap his hands stands at a distance of 99m from a wall hear the echo after 0.6s, what is the velocity of sound?
Answer:
Distance travelled by the sound
= 2 x 99 = 198 m
time = 0.65
Velocity = \(\frac{\text { Distance }}{\text { time }}\)
V = 198/0.6
= 330 m/s

Question 4.
A sound signal of 50 kHz is sent to the bottom of a sea. It returned back after 4S. The velocity of sound in seawater is 1500 m/s. Then,
a) Calculate the depth of the sea
b) What is the wavelength of the wave?
Answer:
a) Suppose the depth of the sea =d
then distance travelled by the wave = 2d
distance – velocity × time
= 1500 x 4 = 6000m
∴ depth = 6000/2
= 3000 m
b) V = f λ
V = 1500 m/s
f = 50 KHz
= 50000Hz
λ = \(\frac{v}{f}=\frac{1500}{50000}=0.03 \mathrm{m}\)

Question 5.
Given below are graphs of sound waves from differ¬ent sources that travels through the same medium. Among these, which one has higher frequency? What is the basis of your conclusion?
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 23
Answer:
Wave with higher frequency is shown in graph B.
As both travel through the same medium, velocity remains the same. But wavelength is inversely proportional to the frequency. ‘B’ has lower wavelength, so B itself has higher frequency.

Question 6.
A sound wave enters water from air. What happens to its wavelength? Why?
Answer:
Wavelength increases. Velocity of sound in water is 1482 m/s and velocity in air is 343 m/s. But the frequency of the sound wave will not change as the medium differs. We know V = f λ. So wavelength should increase.

Question 7.
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 24
A person standing at A claps his hands,
a) Is there any change for occurring echo?
Answer:
a) Yes, can hear echo.

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Question 8.
Figure shows the distance – displacement graph. The wave is formed in 2s.
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 25
a) What is its amplitude?
b) What is its wavelength?
c) What is its frequency?
d) What is its velocity?
Answer:
a) 0.2 m
b) 4 m
c) f = \(\frac { n }{ t }\) = \(\frac { 4 }{ 2 }\) = 2Hz
d) V = f λ = 2 × 4 = 8 m/s

Question 9.
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 26
Answer:
A: Stethoscope
B: Reverberation C: Echo
D: 1) Provide more ventilation and windows
2) Use carpets on the floor

Question 10.
Figure shows the distance and displacement of a wave formed in 0.2s.
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 27
Answer:
a) What is its wavelength?
b) What is the frequency?
c) What is its velocity?
Answer:
a) 5 m
b) f = \(\frac { n }{ t }\) = \(\frac { 3 }{ 0.2 }\) = 15Hz
c) V = f λ = 15 × 5 = 75 m/s

Question 11.
Identify what kind of wave the following are
a) sound wave
b) ripples formed on the surface of water.
c) wave formed due to vibration of tuning fork.
Answer:
a) longitudinal wave
b) transverse wave
c) longitudinal wave

Question 12.
Observe the figure
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 28
a) Which particle is in the same phase of vibration as that of B?
b) What is the distance between these particles called?
c) If the distance between C and E is 25 m, what is the wavelength?
Answer:
a) F
b) wavelength
c) 50 m

Question 13.
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 29
a) How many trough are there?
b) Find out the wavelength?
c) Calculate velocity of the wave if it is traveled within 0.02 S.
Answer:
a) 2
b) 4 m
c) f = \(\frac { n }{ t }\) = \(\frac { 3 }{ 0.02 }\) = \(\frac { 300 }{ 2 }\) = 150 Hz
V = f λ
= 150 × 4 = 600 m/s

Question 14.
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 30
a) What is its amplitude?
b) What is the frequency?
c) Draw the graph of another wave with no change in the frequency and with half of its amplitude.
d) If this wave travels with velocity 300m/s in 1s, calculate the wavelength?
Answer:
a) 0.1 m
b) f = \(\frac { n }{ t }\) = \(\frac { 2 }{ 4 }\) = 0.5 Hz
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 31
d) V = 300 m/s
V = f λ
f = 0.5 Hz
= 300 = 0.5 × λ
λ = 300/0.5 = 600 m

Question 15.
If a sound wave travels 1700 m through a medium in 5 s. Identify the medium. (March 2016)
Answer:
Air
V = \(\frac { s }{ t }\)
= \(\frac { 1700 }{ 5 }\) = 340 m/s

Question 16.
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 32
Given in the graph the points A, B, C, D represents state of vibration of a sound wave. From the below-mentioned options which represent the wavelength.
a) Distance between A and C
b) Distance between A and D
c) Distance between A and B
d) Distance between B and C
Answer:
Distance between A and C

HSSLive.Guru

Question 17.
Select the instrument that works on the principle of multiple reflection of sound,
a) Watthour meter
b) Sonar
c) Stethoscope
d) Decibel meter (March 2015)
Answer:
c) Stethoscope

Question 18.
State what happens to the loudness of sound in the following cases.
a) Density of the medium increases.
b) Distance between the source and the receiver increases.
a) Loudness increases (March 2015)
Answer:
b) Loudness decreases

Question 19.
A sound wave generated in 4 seconds is shown in the graph.
Kerala Syllabus 9th Standard Physics Solutions Chapter 7 Wave Motion 33
If wavelength of the wave is 15m, calculate
a) Frequency of the wave
b) Distance traveled by the wave in 4 seconds.
c) Find out the amplitude of the wave. (March 2015)
Answer:
a) 1 Hz.
b) 60m
c) 1m

Question 20.
A sound was produced. Three seconds later, its echo was heard.
a) What is the distance between the reflecting surface and source? (Speed of sound in air is 340 m/s)
b) Write TWO methods, by which you can reduce the harmful effects of reflection of sound in big halls. (March 2015)
Answer:
a) Distance = Velocity × time
= 340 × 3 = 1020m
So distance between the source and reflecting surface = 1020/2 = 510 m
b) Make the walls rough, provide a large number of ventilators. .

Question 21.
State what change happens to the loudness of sound in the following situations:
a) Amplitude of vibration decreases.
b) The distance between source and receiver decreases (Model 2014)
Answer:
a) Loudness decreases
b) Loudness increases

Question 22.
The wavelength of sound traveling in air with the velocity 330 m/s was found to be 66 m. If so.
a) Find the frequency of sound.
b) By what name are sounds of such frequency known as? (Model 2014)
Answer:
a) V = f λ
330 = f × 66
f = 330/60 = 5.5 Hz
b) Infrasonic ( ∠ 20Hz)

Question 23.
Velocity of sound in air is 340 m/s. Sound waves of wavelength 0.01 m. from a vibrating body reach your ear through air. Will you be able to hear the sound? Justify your answer. (March 2012)
Answer:
V = 340 m/s
λ = 0.01 m
v = u λ
f = V/λ
f = 340/0.01
= 34000 Hz > 20,000 Hz
Human beings cannot hear these sounds. Audible frequency range is 20 Hz to 20000 Hz

Kerala Syllabus 9th Standard Social Science Solutions Part 1 Chapter 7 Kerala: From Eighth to Eighteenth Century

You can Download Kerala: From Eighth to Eighteenth-Century Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Social Science Solutions Part 1 Chapter 7 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Social Science Solutions Part 1 Chapter 7 Kerala: From Eighth to Eighteenth Century

Kerala: From Eighth to Eighteenth-Century Textual Questions and Answers

Question 1.
The chief of Kudi was …………..
Answer:
Kudipathi

Question 2.
Prepare a note on Perumal rule and its characteristics.
Answer:
The Nadus were under the Perumalswho ruled Kerala with their capital at Mahodayapuram (present Kodungaloor). All the 14 nadus from Kolathunadu in the north to the Venad in the south accepted the rule of the Perumal’s. It was during this period that a centralized rule came into being in Kerala for the first time. Rulers from Rajasekharan to Ramakulasekharan ruled during 800 -1122 CE with Mahodayapuram as their capital. Let us examine the characteristics of the rule of the Perumal’s.

  • Perumal’s had representatives called Koyiladhikarikal.
  • In the matters of administration, the Perumals were assisted by Naluthali, the council of Brahmins.
  • Perumal’s had a militia called Ayiram (Thousand).
  • Perumal’s levied taxes from the Nadus, Nagaras (Towns), Brahmin Gramas, Temples, etc.

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Question 3.
Name major Naduvazhi Swaroopams.
Answer:
Kola Swaroopam (Kolathunadu)
Nediyiruppu Swaroopam (Eranadu)
Perumpadappu Swaroopam (Kochi)
Thrippapur Swaroopam (Venadu)

Question 4.
Analyze the political history of Kerala from the eighth to the eighteenth century.
Answer:
Until the 18th century, the Naduvazhi Swaroopams continued without much change. By the second half of the 18th century is Sultans of Mysore, Hyder Ali and Tipu Sultan led military campaigns which created frenzy among the Naduvahis of northern Kerala. Nediyiruppu, Kola and other smaller Swaroopams quickly came under the Mysore Sultans. Fearing the attack from Mysore Sultans many Naduvazhis and Desavazhis fled to Venado. The Perumpadappu Swaroopam of Kochi soon accepted the suzerainty of the Mysore Sultans. Only Travancore resisted the attacks.

It was during this period that Kerala was divided into three regions namely Travancore, Kochi, and Malabar. By the close of the 18th century, the East India Company defeated Tipu Sultan. As a result of this, the Malabar region which was under Tipu Sultan came completely under the British rule. With this, the independent rule of the Naduvazhis of Malabar came to an end. Travancore and Kochi continued to be princely states.

Question 5.
Which were the 3 types of lands based on ownership rights?
Answer:

  1. Cherikkal,
  2. Brahmswam
  3. Devaswam

Question 6.
Match the following

A B
Cherikkal Temples
Brahmswam Brahmins
Devaswam Naduvazhis

Answer:

A B
Cherikkal Naduvazhis
Brahmswam Brahmins
Devaswam Temples

Question 7.
Prepare a note on the system of tax developed by Mysore sultans.
Answer:
The system of tax, the Mysore Sultans had developed was based on the total production from the land, of which a share was fixed to be collected as tax. Later the British conducted a land survey, divided the land in terms of acres and cents, and allotted them survey numbers. Similar land surveys were conducted in Kochi and Travancore. In Kochi, it was known as Kettezhuthu and in Travancore, Kandezhuthu. Accordingly, tax was fixed on the assessed land.

Question 8.
Identify different occupational groups during the medieval period.
Answer:

  • People engaged in agriculture and the making of agricultural equipment.
  • People involved in handicrafts and the making of metal types of equipment.
  • People engaged in trade.
  • People involved in weaving and oil production.
  • People involved in temple rituals.
  • Officials connected to the Naduvazhi Swaroopams.

Question 9.
Discuss how the caste system formed in Kerala.
The descendants of those who were engaged in a particular occupation followed the same occupation. People engaged in the same occupation evolved into one caste. The Adiyalars who used to farm during the medieval times occupied the lowest rung in the caste hierarchy whereas the Brahmins were at the top.

Based on the family occupation, all other castes came in between these two categories. On the basis of the caste, the concept of purity and impurity sprang up.
By the beginning of the 19th century, the population of Malabar, Kochi, and Travancore were officially categorized on the basis of caste. Subsequently, caste came to be decided on the basis of birth irrespective of the occupation.

HSSLive.Guru

Question 10.
Which were the major trade centers in Kerala during the Medieval period?
Answer:

  • Kodungalloor
  • Kozhikode
  • Madayi

Question 11.
Examine different kinds of trade prevailed in the medieval period.
Answer:
Regional Trade:
Chanthas and Angadies were major regional trade centers. Commodities used daily such as paddy, rice, vegetables, betal nut, salt, fish, etc, were the major items exchanged.
Long-distance Trade:
Long-distance trade was mainly with Tamil Nadu, Karnataka, Andhra Pradesh and Orrisa. Tamil Brahmins and Chettis were the main traders. Rice, Chilli, Cotton, other cloth materials, silk, and horses were brought to Kerala. Black pepper and other spices were taken from here.
Foreign Trade:
The arabs, Chinese, Europeans, etc. were the main foreign traders. Black pepper, ginger, cardamom, cinnamon, other spices, coconut, etc. were taken from here. Gold, copper, silver, china clay pottery, silk, etc. were brought to Kerala.

Question 12.
What do you mean by salais?
Answer:
The centres where the vedas were taught in the medieval Kerala were known as ‘Salas’

Question 13.
By 14th Century, books were written in
Answer:
Manipravalam

Question 14.
List the literary works of the missionaries.
Answer:

  • Samkshepavedartham
  • Puthan Pana by Arnos Pathiri
  • Varthamanapusthakam of Paremakkal Thoma Kathanar.

Questions 15.
Prepare a note on the administrative system of medieval Kerala.
Answer:
During the period of Perumals, a centralized rule came into being in Kerala. By the 12th century, the Perumal rule came to an end. The Nadus became independent. The positions of power that developed in the Nadus were known as Naduvazhi Swaroopams. Until the 18th century, the Naduvazhi Swaroopams continued without much change.

HSSLive.Guru

Question 16.
Malayalam language literature, art, forms, and sciences flourished during the medieval period. Substantiate.
Answer:
Influence of Malayalam is more evident in the works after the 12th century. By the 14th century, books were written in Manipravalam. Bhakti literature was present in the 17th century. District art forms developed during the period. During the medieval period, there was progress in the fields of Ayurveda, Mathematics, Astrology, and Architecture.

Question 17.
What were the features of Swaroopams?
Answer:
The positions of power that developed in the Nadus were known as Naduvazhi Swaroopams. Swaroopams were the ruling families with the right of self-rule and the followed matrilineal system of inheritance. The Swaroopams had their own military.

Question 18.
Elucidate what is Anjuvannam and Manigramam?
Answer:
Anjuvannam and Manigramam are the trade guilds existed in medieval Kerala till the 14th century. They were active in both sea and land trade.

Question 19.
List out various Maryadas existed in medieval Kerala.
Answer:

  • Desamaryada
  • Thozhilmaryada
  • Swaroopa maryada
  • Shudramaryada
  • Jathimaryada

Question 20.
Identify the distinct art forms of Kerala that developed during medieval period.
Answer:

  • Mohiniyattam
  • Ottanthullal
  • Padayani
  • Mangamkali
  • Parichamuttukali
  • Chakyarkoothu
  • Kathakali
  • Theyyam
  • Oppana
  • Duffmuttu
  • Koodiyattam
  • Chavittunatakam

HSSLive.Guru

Question 21.
How did Arab-Malayalm develop?
Answer:
Malayalam was influenced by the language of the people who had come through the sea route for trade. Influence of the Arabs led to the development of Arab- Malayalam literature.

Kerala Syllabus 9th Standard Social Science Solutions Chapter 2 The Signature of Time in Malayalam

Students can Download Social Science Part 2 Chapter 2 The Signature of Time 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.

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The Signature of Time Textual Questions and Answers in Malayalam

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Kerala Syllabus 9th Standard Social Science Solutions Chapter 1 Sun: The Ultimate Source in Malayalam

Students can Download Social Science Part 2 Chapter 1 Sun: The Ultimate Source 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.

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Sun: The Ultimate Source Textual Questions and Answers in Malayalam

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Kerala SSLC Malayalam Model Question Paper 1 (Adisthana Padavali)

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Plus Two Botany Notes Chapter 4 Biotechnology: Principles and Processes

Students can Download Chapter 4 Biotechnology: Principles and Processes Notes, Plus Two Botany Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Botany Notes Chapter 4 Biotechnology: Principles and Processes

Principles Of Biotechnology
The important techniques leads to the origin of modern biotechnology are:

(i) Genetic engineering: Techniques to alter the chemistry of genetic material (DNA and RNA), and introduce these into host organisms and changing the phenotype of the host organism.
(ii) Maintaning the microbial contamination-free condition to promote the growth of desired
microbe/eukaryotic cell in large quantities for the manufacture of biotechnological products like antibiotics, vaccines, enzymes, etc.

In traditional hybridisation the new hybrid formed possess undesirable genes along with the desired genes. But the technique of genetic engineering /recombinant DNA creates transgenic organism contains only the desirable genes.

Plus Two Botany Notes Chapter 4 Biotechnology: Principles and Processes

In a chromosome there is a specific DNA sequence called the origin of replication, which is responsible for the initiation of replication. If the foreign (alien) DNA transferred and integrated into the genome of the recipient, it multiply along with the host DNA. This called as cloning (making multiple identical copies of any template DNA).

The concept of linking a gene coding for antibiotic resistance with a plasmid of Salmonella typhimurium was the first step in the construction recombinant DNA. It was the work of Stanley Cohen and Herbert Boyer (1972).
Antibiotic resistant gene was isolated from a plasmid by cutting DNA at specific locations by restriction enzymes. Then the cut piece of DNA is linked with the plasmid DNA (vectors) with the help of enzyme DNA ligase.

In the transfer of the malarial parasite into human body, mosquito acts vector. In the same way, a plasmid can be used as vector to deliver an alien piece of DNA into the host organism. It results in the creation of new combination of circular autonomously replicating DNA, it is known as recombinant DNA.

When this DNA is introduced into Escherichia coli, it could replicate using the new host’s DNA and polymerase enzyme to make multiple copies. The ability of forming multiple copies of antibiotic resistance gene in E.coliis called cloning.
The three basic steps in the creation of GMO are

  1. Identification of DNA with desirable genes;
  2. Introduction of the identified DNA into the host;
  3. Maintenance of introduced DNA in the host and transfer of the DNA to its progeny.

Plus Two Botany Notes Chapter 4 Biotechnology: Principles and Processes

Tools Of Recombinant Dna Technology
Important tools are

  1. restriction enzymes
  2. polymerase enzymes
  3. ligases
  4. vectors and the
  5. host organism.

1. Restriction Enzymes:
The enzymes restricting the growth of bacteriophage in Escherichia coli is used to cut DNA. This is called restriction endonuclease.

Restriction endonuclease- Hind II cut DNA molecules at a particular point by recognising a specific sequence of six base pairs. This is called recognition sequence.

Another restriction endonuclease EcoRI comes from Escherichia coli RY13. In EcoRI, ‘R’ indicates the order in which the enzymes were isolated from that strain of bacteria.

Restriction enzymes belong to nucleases. These are of two kinds; exonucleases and endonucleases. Exonucleases remove nucleotides from the ends of the DNA whereas, endonucleases bind to the DNA and cut each of the two strands of the double helix at specific points in their sugar-phosphate backbones.
Plus Two Botany Notes Chapter 4 Biotechnology Principles and Processes 1
Plus Two Botany Notes Chapter 4 Biotechnology Principles and Processes 2

Plus Two Botany Notes Chapter 4 Biotechnology: Principles and Processes
Each restriction endonuclease recognises a specific palindromic nucleotide sequence in the DNA. Actually palindromic nucleotide sequences is the same as the word, “MALAYALAM,” read in both forward and backward.
Eg- 5′ ——GAATTC ——3′
3′ —— CTTAAG —— 5
After cutting DNA duplex by an enzyme, it leaves single stranded portions at the ends. These are sticky ends on each strand. This stickiness of the ends facilitates the action of the enzyme DNA ligase.

Separation and isolation of DNA fragments:
The cut fragments of DNA separated by a technique known as gel electrophoresis. Negatively charged DNA fragments are separated by forcing them to move towards the anode under an electric field through a agarose matrix.

The DNA fragments separate according to their size in agarose gel. So the smaller the fragment size moves farther.

The separated DNA fragments can be visualised only after staining the DNA with ethidium bromide followed by exposure to UV light. It appears as bright orange coloured bands. The separated bands of DNA are cut out from the agarose gel and extracted from the gel piece.

This step is known as elution. The DNA fragments thus obtained are used in constructing recombinant DNA by joining them with cloning vectors.
Agarose gel electrophoresis showing migration of digested and un digested DNA
Plus Two Botany Notes Chapter 4 Biotechnology Principles and Processes 3

Plus Two Botany Notes Chapter 4 Biotechnology: Principles and Processes

2. Cloning Vectors:
Plasmids and bacteriophages can replicate within bacterial cells independently without chromosomal DNA. Bacteriophages form a high copy numbers of their genome within the bacterial cells. Plasmids form 15-100 copies per cell.

If linking an alien piece of DNA with bacteriophage or plasmid DNA, it can multiply its numbers equal to the copy number of the plasmid or bacteriophage. So, this is helpful in the selection of recombinants from non-recombinants.

The features of artificial cloning vector are
(i) Origin of replication (oril):
It is a sequence of cloning vector in which replication starts when any piece of DNA linked to it. This sequence is responsible for controlling the copy number of the linked DNA.

(ii) Selectable marker:
In addition to ‘ori’, the vector contains selectable marker, which helps in identifying and eliminating non transformants and selectively permitting the growth of the transformants.

The genes coding for antibiotic resistance such as ampicillin, chloramphenicol, tetracycline or kanamycin, etc., are considered as selectable markers for E. coli. The normal E. coli cells do not show the resistance against any of these antibiotics.

(iii) Cloning sites:
For linking the alien DNA into the vector, there must be preferably single recognition sites for the commonly used restriction enzymes because more than one recognition sites within the vector results several fragments.

The ligation of alien DNA is carried out at a restriction site present in one of the two antibiotic resistance genes.
Plus Two Botany Notes Chapter 4 Biotechnology Principles and Processes 4
For example, ligation of a foreign DNA at the Bam H I site of tetracycline resistance gene in the vector pBR322, the recombinant plasmids lose tetracycline resistance due to insertion of foreign DNA.

Recombinants selected from non-recombinant by plating the transformants on ampicillin containing medium. The transformants growing on ampicillin containing medium are then transferred on a medium containing tetracycline.

It could not grow in the medium containing tetracycline. But, non recombinants can grow on both medium Therefore antibiotic resistance gene helps in selecting the transformants.

Another method of selecting recombinants from non-recombinants is their ability to produce colour in the presence of a chromogenic substrate. For this recombinant DNA is inserted within the coding sequence of an enzyme, beta-galactosidase. This results into inactivation of the enzyme, called as insertional inactivation.

If the bacteria does not have an insert, chromogenic substrate present in the medium react with betagalactosidase enzyme gives blue coloured colonies.

Plus Two Botany Notes Chapter 4 Biotechnology: Principles and Processes

If the plasmid have an insert, they do not produce any colour due to insertional inactivation of the gene coding for beta galactosidase, these are identified as recombinant colonies.

(iv) Vectors for cloning genes in plants and animals:
Normally Agrobacterioum tumifaciens (a pathogen of several dicot plants) transfer its ‘T-DNA’to normal plant cells and causes tumor. Similarly retroviruses in animals have the ability to transform normal cells into cancerous cells and they are used as vectors for delivering genes of interest to humans.

For delivering genes of interest to plants tumor inducing (Ti) plasmid of Agrobacterium tumifaciens is modified (disarming) as non pathogenic Similarly the retroviruses are disarmed and used to deliver desirable genes into animal cells.

3. Competent Host
(For Transformation with Recombinant DNA)
DNA is a hydrophilic molecule, it cannot pass through cell membranes. For this, bacterial cells must have to be competent to take up DNA.

This is done by treating them with calcium ions and incubating the cells and recombinant DNA on ice, followed by placing them at 42°C Then putting them back on ice. This helps the bacteria to take up the recombinant DNA.

Another methods:

  1. Micro-injection – recombinant DNA is directly injected into the nucleus of an animal cell.
  2. Biolistics or gene gun -cells are bombarded with high velocity micro-particles of gold or tungsten coated with DNA. It is suitable for plants.

And the last method uses ‘disarmed pathogen’ vectors, which when allowed to infect the cell, transfer the recombinant DNA into the host.

Plus Two Botany Notes Chapter 4 Biotechnology: Principles and Processes

Processes Of Recombinant Dna Technology
Recombinant DNA technology involves several steps. They are

1. Isolation of the Genetic Material (DNA):
Initially the bacterial cells/plant or animal tissue are treated with enzymes such as lysozyme (bacteria), cellulase (plant cells), chitinase (fungus) to open the cell to release DNA along with other macromolecules such as RNA, proteins, polysaccharides, and Other molecules can be removed by appropriate treatments and purified DNA precipitates out afterthe addition of chilled ethanol. It can be observed as collection also lipids.

To get DNA in a pure form and free from other macro-molecules it is treated with enzymes. RNA can be removed by treating with ribonuclease whereas proteins can be removed by treating with protease, of fine threads in the Suspension.

2. Cutting of DNA at Specific Locations:
It is done by incubating purified DNA molecules with the restriction enzyme. Here Agarose gel electrophoresis is used to check the progression of a restriction enzyme digestion. DNA is a negatively charged molecule, hence it moves towards the positive electrode (anode).

After having cut at the source DNA as well as the vector DNA with a specific restriction enzyme, the cut out ‘gene of interest’ from the source DNA and the cut vector with space are mixed and ligase is added. This results in the preparation of recombinant DNA.

3. Amplification of Gene of Interest using PCR:

Polymerase Chain Reaction is helpful to produce multiple copies ( eg-1 billion copies-) of the gene of interest.

It is synthesised in vitro using two sets of primers (chemically synthesised oligonucleotides that are complementary to the regions of DNA) and the enzyme thermostable DNA polymerase (isolated from a bacterium, Thermus aquaticus).

This enzyme extends the primers using the nucleotides provided in the reaction mixture and the genomic DNA as template. If the process of replication of DNA is repeated many times, the segment of DNA (gene of interest) can be amplified. The amplified fragment is used to ligate with vector for further cloning.

Plus Two Botany Notes Chapter 4 Biotechnology: Principles and Processes

(PCR showing denaturation. annealing and extention)
Plus Two Botany Notes Chapter 4 Biotechnology Principles and Processes 5
Plus Two Botany Notes Chapter 4 Biotechnology Principles and Processes 6

4. Insertion of Recombinant DNA into the Host Cell/Orqanism:
Recombinant DNA carry gene resistant to antibiotic (e.g., ampicillin) is transferred into E. coli cells, the host cells become transformed into ampicillin-resistant cells. If spreading the transformed cells on agar plates containing ampicillin, only the transformants grow and untransformed cells die.

So it is helpful to select a transformed cell in the presence of ampicillin. The ampicillin resistance gene in this case is called a selectable marker.

5. Obtaining the Foreign Gene Product:
The main aim of all recombinant technologies is to produce a desirable protein. Here the foreign gene is expressed under appropriate conditions. If it is necessary to produce target protein i.e recombinant protein on a small scale, rDNA transferred into the host and cloned genes of interest must be grown in the laboratory. Then the protein is extracted and purified.

Plus Two Botany Notes Chapter 4 Biotechnology: Principles and Processes

Stirred tank Bioreactor
Plus Two Botany Notes Chapter 4 Biotechnology Principles and Processes 7
A stirred-tank reactor is a cylindrical vessel that helps in the mixing of the reactor contents. The stirrer also facilitates oxygen availability throughout the bioreactor.

It consist of agitator system, an oxygen delivery system and a foam control system, a temperature control system, pH control system and sampling ports, so that small volumes of the culture can be withdrawn periodically.

6. Downstream Processing:
After the desired product formed, it is subjected to a series of processes. These include separation and purification, which are called as downstream processing.

Plus Two Botany Notes Chapter 4 Biotechnology: Principles and Processes

The product is added with preservatives and undergoes clinical trials as in case of drugs. Later the strict quality control testing is done for each product.

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