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You can Download Population, Migration, Settlements Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Social Science Solutions Part 2 Chapter 8 help you to revise complete Syllabus and score more marks in your examinations.

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

Population, Migration, Settlements Textual Questions and Answers

Population Migration Settlements 9th Kerala Syllabus 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

9th Standard Social Science Notes Pdf In English Medium Question 2.
What is meant by density of population?
Answer:
The average population of every square kilometre is called as density of population.

9th Social Science Notes Kerala Syllabus Question 3.
Complete the flow chart

Answer:

9th Class Social Science Malayalam Medium Question 4.
Name the most densely populated state in India.
Answer:
Bihar

9th Standard Social Science Notes Pdf Kerala Syllabus 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.

9th Standard Social Science Notes Pdf Malayalam Medium Question 6.
Define population growth!
Answer:
Population growth in the change in population of any particular place over a particular period.

Social 9th Standard Notes Kerala Syllabus 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%

Hsslive Guru Social Science Kerala Syllabus Question 8.
Point out the factors causing change in population.
Answer:

• Birth rate
• Death rate
• Migration

9th Class Social Science Notes Kerala Syllabus Question 9.
Define Migration.
Answer:
Permanent or temporary shifting of residence of people from one place to another is called migration.

9th Class Social Notes Kerala Syllabus Question 10.
Prepare a flow chart showing different levels of migration.

Answer:

Social Notes 9th Standard Kerala Syllabus 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.

Kerala Syllabus 9th Standard Social Science Notes Question 12.
Find out the pull factors of migration
Answer:

• Employment opportunities
• Higher education facilities
• Better living standards

9th Std Social Science Kerala Syllabus 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

Social Science 9th Standard Kerala Syllabus Question 14.
Name 2 types of settlements
Answer:

1. Nucleated settlements
2. Dispersed settlements

Social 9th Class Guide Kerala Syllabus 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

9th Class Social Science Textbook Kerala Syllabus Question16.
What are the major problems faced by urban centers?
Answer:

• Slums
• Traffic problems
• Pollution

9th Class Social Science Textbook Malayalam Medium 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.

Social 9th Notes Kerala Syllabus Question 18.
What is meant by population?
Answer:
The total number of people living in a definite area is called population.

Population Class 9 Notes Kerala Syllabus 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.

9th Class Social Science Malayalam Medium Notes 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.

9th Social Notes Kerala Syllabus 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.

Social Science Malayalam Medium 9th Standard 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.

Hss Live Guru Social Science Kerala Syllabus Question 23.
How is density of population calculated?
Answer:
density of popuation = \(\frac{\text { Total Population }}{\text { Land Area }}\)

Social 9th Class Notes Kerala Syllabus 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.

Social Science 9th Class Notes Kerala Syllabus 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:

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

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.

Answer:

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.

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.,

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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

Decimal Forms Class 9 Kerala Syllabus 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

Kerala Syllabus 9th Standard Maths Chapter 2 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

3. \(\frac{1}{2}\) + \(\frac { 1 }{ { 2 }^{ 2 } } \) + \(\frac { 1 }{ { 2 }^{ 3 } } \)
Denominators of all the fractions should be 2³

Hss Live Guru 9th Maths Kerala Syllabus 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

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

Kerala Syllabus 9th Standard Maths Notes 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 + \(\frac{1}{3}\))

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 + \(\frac{4}{6}\)

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 \(\frac{10}{9}\)

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 ……

Hsslive Maths Class 9 Kerala Syllabus 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

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 \(\frac{1}{9}\)

b. \(\frac{1}{10}\) part of a number comes closer to its \(\frac{1}{9}\)

c. \(\frac{1}{10}\) part of a number comes closer to its \(\frac{1}{9}\)

d. Since \(\frac{1}{9}\) is closes to \(\frac{1}{10}\)

e. Since \(\frac{1}{9}\) is closes to \(\frac{1}{10}\)

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.

Hsslive Guru 9th Maths Kerala Syllabus 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 \(\frac{10}{11}\)

Continuing like this,
= 0.090909

2. a. \(\frac{2}{11}\) = \(\frac{2×10}{11×10}\) + \(\frac{2}{10}\) x \(\frac{10}{11}\)

Continuing like this,
\(\frac{2}{11}\) = 0.18181818 …….

b. \(\frac{3}{11}\) = \(\frac{3×10}{11×10}\) = \(\frac{3}{10}\) x \(\frac{10}{11}\)

Continuing like this
\(\frac{3}{11}\) = 0.272727…..

3. \(\frac{10}{11}\) = \(\frac{10×10}{11×10}\) = \(\frac{10}{10}\) x \(\frac{10}{11}\)

Continuing like this
\(\frac{10}{11}\) = 0.90909……….

Hss Live Guru 9 Maths Kerala Syllabus 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 Decimal Forms Exam Oriented Question and Answers

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

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:

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 \(\frac{10}{6}\) = \(\frac{1}{10}\)(\(\frac{50}{6}\))

2. \(\frac{3}{11}\) = \(\frac{3}{100}\) x \(\frac{100}{11}\) = \(\frac{3}{100}\)(9 + \(\frac{1}{11}\))

3. \(\frac{23}{11}\) = \(\frac{23}{100}\) x \(\frac{100}{11}\)

4. \(\frac{1}{13}\) = \(\frac{1}{100}\) x \(\frac{100}{13}\)

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

2. \(\frac{2}{9}\) – \(\frac{2}{10}\) = \(\frac{2}{90}\)

(\(\frac{3}{9}\), \(\frac{6}{9}\) These fractions having common factor in the numerator and the denominator)

3. Repeated deciamals.

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:

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:

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…….

Students can Download Kerala SSLC Biology Previous Year Question Paper March 2019 English Medium Pdf, Kerala SSLC Biology Model Question Papers helps you to revise the complete Kerala State Board New Syllabus and score more marks in your examinations.

Kerala SSLC Biology Model Question Paper 4 English Medium

Instructions :

• The first 15 minutes Is the Cool-off time.
• You may use the time to read the questions and plan your answers.
• Answer only on the basis of instructions and questions given.
• Consider score and time while answering.

Time: 1½ Hours
Total Score: 40 Marks

Section – A

Answer any five questions from Q.No. 1 to 6. Each carries on score. (5 × 1 = 5)

Sslc Biology Question Paper 2019 Question 1.
Identify the word pair relation and fill the blanks.
(a) Monkeys: Cercopithecoidea::
Chimpanzee: ………………..
(b) A.I.Oparin: Theory of chemical evolution::
Hugo de Vries: ………………….
Answer:
a) Hominoidea
b) Mutation theory

Sslc 2019 Biology Question Paper Question 2.
Find out the parts that are not related to retina from the following.

Answer:
Conjuctiva
Iris

Sslc Biology Question Paper 2019 Kerala Question 3.
“Myelin sheath accelerates the speed of impulses through axon and provides nutrition to it.”
(a) How does myelin sheath form?
Answer:
Myelin sheath is formed due to the repeated encir- * cling of Schwann cells around the axone.

Biology Sslc Question Paper 2019 Question 4.
Identify the relation in the Indicator (A) and complete (B) accordingly.

Answer:
i. Ribose Sugar
ii. AUGC.

Sslc Question Paper 2019 Kerala Question 5.
Find out the fungal diseases from the following: Malaria, Ringworm, Filariasis, Athelete’s foot
Answer:
Ring worm Athlete’s foot

Sslc Biology Chapter 1 Questions English Medium Question 6.
Complete the statement suitably:
“In …….(a)……………. the specialised part in pancreas two types of cells are found. Of these (b)……. cells produce insulin.”
Answer:
a) Islets of Langerhans
b) Beta cells

Answer any six questions from Q.No. 7 to 13. Each carries on score. (6 × 2 = 12)

Sslc Biology Previous Year Question Papers Question 7.
Make suitable pairs of different white blood cells and the function they perform.

Answer:

March 2019 Biology Question Paper Question 8.
Write the name of pathogens and symptoms of the given diseases:

Answer:
A: Malaria
Pathogens: Protozoa – plasmodium

Symptoms

• High fever with shivering
• Profuse sweating, severe headache
• Vomitting, diarrhoea, anemia

B: Tuberculosis
Pathogens: bacteria: Mycobacterium tuberculosis

Symptoms

• Loss of body weight
• Fatigue
• Continuous dry cough

Biology Question Paper 2019 Question 9.

• Excess blood is lost even though minor, injuries.
• Loss of body balance
• Accumulation of insoluble proteins in the neutral tissues of brain
• Production of dopamine reduces
• Irregular flow of electric charges in brain.

Answer:

Sslc English Previous Year Question Papers Question 10.
Observe the illustration and answer the questions.

(a) Name the Scientists who devised the experi-mental set up shown above.
(b) Which theory of evolution is substantiated by this experiment?
Answer:
a) Stanley L.Miller and Harold C.Urey
b) Theory of chemical evolution

12th Biology Question Paper 2019 State Board Question 11.
Observe the illustration and answer the questions.

Answer:
a. A: Genetic scissors: Restriction endonuclease
B: Genetic glue : Ligase

b. Yes

• They will have the ability to prouce insulin.
• They change that occured in the genetic constitution will be transferred to the next generation too. In case of mutation the gene might loss the capacity for the production.

Question 12.
List out the four major concepts to be included in a blood donation campaign.
Answer:

• Blood cannot be made artificially. So we can save life of a person by donating blood. So do-nate blood and save life.
• A healthy person can donate 300 ml of blood within a period of 6 months.
• If the level of blood decreases beyond a certain level, it may cause the death of the individual. During such situation for the sustenance of the life the blood donation by a person becomes inevitable.
• Before transfusion, blood group matching should be ensured.

Question 13.
Mutation cause variations in organisms. It leads to evolution of species:
(a) What is mutation?
(b) Explain two other factors that cause variations in organisms.
Answer:
a) A sudden heritable change in the genetic con-stitution of an organism is called mutation.
b) Crossing over in chromosomes

Answer any five from Q.No. 14 to 20. Each carries 3 score. (5 x 3 = 15)

Question 14.
Observe the illustration and answer the questions.

(a) Why do the forelimbs of these organisms show differences in external appearance?
(b) What inferences regarding evolution can be drawn from the anatomy of these organs?
(c) Write any two other scientific evidences which proves evolution.
Answer:
a) Difference in their external appearances are their adaptations to live in their own habitats.
b) Anatomical resemblances justify the interference that all organisms evolved from a common ancestor.
c) Biochemistry and Physiology, Molecular Biology and Evidence from fossils

Question 15.
Observe the illustration.

(a) What does R, r denote in the illustration
(b) Which character is expressed in first generation. Why?
Answer:
a) Gametes / Allele
b) Red Flower
Hybridization experiment, the allele that controls the dominant character (Red) that is expressed, and other character remains hidden (recessive character-white) in the offsprings of the first generation.

Question 16.
“Smoking harmfully affects internal organs.”
This is a general statement.
Explain how smoking affects brain, heart and lungs.
Answer:
a) Brain: Stroke, Addiction to nicotine
b) Lungs: Lung cancer, Bronchitis, Emphysema
c) Heart: Hypertension, loss of elasticity of arteries, Decrease in functional efficiency.

Question 17.
There are certain mistakes in the given chart. Find out and correct it.

Answer:

Question 18.
Analyse the statement and answer the questions. “Antibiotics, the miraculous medicines of 20th century helped a lot to bring many diseases under control. But the use of antibiotics without consulting doctor is not advisable.”
(a) Why antibiotics considered as miraculous medi-cines?
(b) Write two side effects of antibiotics.
Answer:
a) Antibiotics are drugs obtained from microorgan isms that are used to destroy the growth of other microorganisms that cause diseases. Antibiotics are biochemical substances extracted from living things like bacteria and fungi which can or prevent the spreading of germs. Antibiotics target microorganisms such as bacteria, fungi and parasites.

b) Side effects of antibiotics:

• Regular use develops immunity in pathogens against antibiotics.
• Destroys useful bacteria in body.
• Reduces the quantity of some vitamins, in body.

Question 19.
Correct mistakes if any in the underlined part of the given statements.
(a) Curvature of lens increases when viewing near objects.
(b) Vitreous humor is formed from blood, and is re-absorbed by blood.
(c) Membraneous labyrinth in the inner ear is filled with Perilymph.
(d) Eustachian tube amplifies and transmits the vi-brations of tympanum to the internal ear.
Answer:
(a) Curvature of lens increases when viewing near objects.
(b) Vitreous humor is formed from blood,and is re-absorbed by blood.
(c) Membraneous labyrinth in the inner ear is filled with endolvmph.
(d) Ear ossicles amplifies and transmits the vibration of tympanum to the internal ear.

Question 20.
Observe the figure and answer the questions:

(a) Identify the partsAand B.
(b) What is the function of ‘A’?
(c) Explain the process that takes place in ‘B’.
Answer:
a) A: mRNA B: Ribosome
b) mRNA:mRNA carries information from DNA to ribosomes and controls protein synthesis.
c) mRNA molecule that carries information from DNA to ribosomes.
mRNA reaches ribosomes. tRNA carry different kinds of amino acids to ribosomes.
Based on the information in mRNA protein is synthesized in ribosomes adding amino acids.

Answer any two from Q.No. 21 to 23. Each careries 4 score. (2 x 4 = 8)

Question 21.
Analyse the given informations related to plant hor mones and answer the questions.
(a) to increase the size of apple
(b) to prevnt dropping of premature fruits.
(c) to increase the production of latex.
(d) to enable flowering of pineapple plants at a time.

(i) Match hormones and their functions properly.
(ii) Write the name and function of any other two hormones occur naturally in plants.
Answer:
i) a) Gibberellins
b) Auxins
c) Ethyphon
d) Ethylene

ii)

Plant hormones	Functions
Cytokinins	Promotes cell division, cell growth and differentiation along with auxin.
Abscisic acid	Control the dormancy of embryo in the seed. Control flowenng Helps to sustair the plant in adverse conditions

Question 22.
Observe the illustration and answer the following questions.

(a) Name the cells (A) and (B).
(b) Explain the role of these cells in making vision possible.
(c) How impulses are generated in these cells when light rays fall on it?
Answer:
a) A. Rod cells B. Cone cells

Photo receptors	Functions
Rod cells	Vision in dim light, black and white vision
Cone cells	Bright light vision, colour vision

b) Working of the cone
When the light falls on cone cells, the photopsin, in them dissociate into retinal and opsin. This chemical change creates impulses.
Working of rod cells :
d cells the pigment rhodopsin in them dissociate into retinal and opsin. This chemical change creates impulses.

c) The sense of vision :
When the pigment photoreceptors dissociate, impulses are forrmed. When get the sense of vision, when these impulses reach the brain through optic nerve.

Question 23.
(a) Redraw the diagram.
Name and label the parts that perform the given functions.

(a) Relay station of impulses.
(b) Controls heartbeat, breathing etc.
(c) Maintains equilibrium of the body.
Answer:

a) Thalamus
b) Medulla oblongata
c) Cerebellum

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

Kerala From Eighth To Eighteenth Century 9th Kerala Syllabus Question 1.
The chief of Kudi was …………..
Answer:
Kudipathi

Examine The Characteristics Of The Rule Of Perumals 9th Kerala Syllabus 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.

List Out The Names Of The Nadus Of Medieval Kerala 9th Kerala Syllabus Question 3.
Name major Naduvazhi Swaroopams.
Answer:
Kola Swaroopam (Kolathunadu)
Nediyiruppu Swaroopam (Eranadu)
Perumpadappu Swaroopam (Kochi)
Thrippapur Swaroopam (Venadu)

Kerala Syllabus 9th Standard Social Science Guide 9th Kerala Syllabus Question 4.
Analyze the political history of Kerala from the eighth to the eighteenth century.
Answer:
Until the 18^th 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 18^th 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.

9th Standard Social Science Notes Pdf Kerala Syllabus Question 5.
Which were the 3 types of lands based on ownership rights?
Answer:

1. Cherikkal,
2. Brahmswam
3. Devaswam

9th Standard Social Science Notes Pdf Malayalam Medium Question 6.
Match the following

A	B
Cherikkal	Temples
Brahmswam	Brahmins
Devaswam	Naduvazhis

Answer:

A	B
Cherikkal	Naduvazhis
Brahmswam	Brahmins
Devaswam	Temples

Scert Class 9 Social Science Notes Pdf Kerala Syllabus 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.

Kerala Syllabus 9th Standard History Notes 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.

Kerala Syllabus 9th Standard Social Science Notes 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 19^th 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.

Kerala Syllabus 9th Standard Social Science Notes Pdf Question 10.
Which were the major trade centers in Kerala during the Medieval period?
Answer:

• Kodungalloor
• Kozhikode
• Madayi

Scert Class 9 Social Science Notes Malayalam Medium 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.

Scert Class 9 Social Science Notes Kerala Syllabus Question 12.
What do you mean by salais?
Answer:
The centres where the vedas were taught in the medieval Kerala were known as ‘Salas’

Hsslive Class 9 Social Science Kerala Syllabus Question 13.
By 14^th Century, books were written in
Answer:
Manipravalam

9th Social Science Notes Kerala Syllabus Question 14.
List the literary works of the missionaries.
Answer:

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

9th Std Social Science Notes Kerala Syllabus 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 12^th 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.

9th Class Social Science Notes Kerala Syllabus 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 12^th century. By the 14^th century, books were written in Manipravalam. Bhakti literature was present in the 17^th 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 14^th 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

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.

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 Class 9 Kerala Syllabus 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.

Hss Live Guru 9th Physics Kerala Syllabus 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.

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.

9th Class Physics Notes Kerala Syllabus 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

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.

Hss Live Guru Physics 9th Kerala Syllabus Question 3.
How does the ribbon/paper move?
Answer:
Ribbon/ paper moves up and down

9th Class Physics 7th Chapter Kerala Syllabus Question 4.
In which direction does the wave move?
Answer:
The wave moves forward or horizontally.

9th Class Physics Chapter 7 Kerala Syllabus 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.

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

Hsslive Guru Std 9 Physics Kerala Syllabus Question 6.
In the figure, which are the points of the highest displacement (amplitude)?
Answer:
A, C, E, G, I, K, M

Class 9 Physics Kerala Syllabus Kerala Syllabus Question 7.
How many crests and troughs are there in the figure?
Answer:
4 crests, and 3 troughs.

Class 9 Scert Physics Solutions Kerala Syllabus Question 8.
Whether all the particles are in the same phase of vibration at a particular time?
Answer:
No

Hss Live Guru 9 Physics Kerala Syllabus Question 9.
Which are the particles in the same phase of vibration as that of A?
Answer:
E, I, M

Std 9 Physics Kerala Syllabus  Question 10.
What about C?
Answer:
G, K

Std 9 Physics Kerala Syllabus  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).

Hsslive Guru 9th Physics Kerala Syllabus 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.

9th Standard Physics Textbook Kerala Syllabus  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

Kerala Syllabus 9th Standard Physics Notes Pdf Question 14.
Which wave has a higher wavelength?
Answer:
First wave has a higher wavelength.

Kerala Syllabus 9th Standard Physics Notes Chapter 1 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

9th Class Physics Notes Kerala Syllabus Malayalam Medium 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

Question 17.
Observe the graphic representation of a wave motion given below.

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

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

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

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.

Answer:

Sound

Sound is produced by the vibration of objects. Sound is a longitudinal wave. Sound needs a material medium to travel.

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

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.

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

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.

2. Ultrasonic waves are used to detect cracks and flaws in large metal blocks

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.

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

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.

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.

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.

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

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

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

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?

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.

A person standing at A claps his hands,
a) Is there any change for occurring echo?
Answer:
a) Yes, can hear echo.

Question 8.
Figure shows the distance – displacement graph. The wave is formed in 2s.

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.

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.

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

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.

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.

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

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.

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

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.

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

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

Current Electricity Class 9 Kerala Syllabus Chapter 6 Activity -1

A positively charged electroscope is connected to the earth through a switch using a conductor.

9th Class Physics Chapter 6 Notes Kerala Syllabus Question 1.
What kind of charge is present in this electroscope?
Answer:
Static

Current Electricity Class 9 Solutions Kerala Syllabus Chapter 6 Question 2.
What happens to this charge when the switch is turned on?
Answer:
Charge neutralizes

9th Std Physics Notes Kerala Syllabus Chapter 6 Question 3.
Will the flow of charge sustain in this arrangement?
Answer:
The charge will not sustain in this arrangement

9th Standard Physics Notes Kerala Syllabus Chapter 6 Activity – 2

A circuit with a cell, a bulb, and a switch is given in the figure.

Kerala Syllabus 9th Standard Physics Notes Chapter 6 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.

Current Electricity 9th Class Exercise Kerala Syllabus Chapter 6 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.

Kerala Syllabus 9th Standard Physics Notes Pdf Chapter 6 Question 6.
Complete the table based on different situations as shown in figure.


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.

Kerala Syllabus 9th Standard Physics Notes Malayalam Medium Activity-3

observe the figures

Question 7.
If the value is opened, in which one will there be a flow of water and rotation of the wheel?
Answer:

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

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.

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

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

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

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:

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

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

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.

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

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

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.

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

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:

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 )

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.

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

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:

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 =

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

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

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

It is used to remove insulation of wires while combining insulated electric wires or when they are to be connected to the devices

Pliers

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

While doing the work related to electric power, gloves are worn in the hand as a protection from electric shock.

Multimeter

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

It helps to measure the current in a circuit without connecting wires or devices in the circuit.

Insulation tape

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

It is used for fixing nut and bolt. They are available in different sizes.

Soldering iron

It is used to solder electronic components in a circuit

Hammer

It is used for fixing and removing nails.

Drill machine

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

Answer:

Question 2.
Given below are the diagrams showing the connection of ammeter and voltmeter in a circuit. Of these, which are correct?

Answer:

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

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.

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:

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:

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?

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.

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.

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.

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.

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

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.

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.

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.

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.

(PCR showing denaturation. annealing and extention)

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.

Stirred tank Bioreactor

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.

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.

Students can Download Kerala SSLC Social Science Model Question Paper 1 English Medium Pdf, Kerala SSLC Social Science Model Question Papers helps you to revise the complete Kerala State Board New Syllabus and score more marks in your examinations.

Kerala SSLC Social Science Model Question Paper 1 English Medium

Instructions:

• The first 15 minutes is the cool-off time. You may use the time to read the questions and plan your answers.
• Answer all questions in PART – A. Answer any one from the questions given under each question number in PART – B.

Time: 2½ Hours
Total Score: 80 Marks

Part – A

Answer all the questions

Question 1.
Name the first iron and steel plant established in South India. (1)
Answer:
Visweswarayya Iron and Steel Ltd. (VISL)

Question 2.
Which among the following serves as the banker to the central and state governments in India? (1)
a) State Bank of India
b) Indian Bank
b) Bankof India
d) Reserve Bank of India
Answer:
Reserve Bank of India

Question 3.
Identify the incident that forced Gandhiji to stop non-cooperation movement. (1)
Answer:
Chauri Chaura incident.

Question 4.
“The result of your political inactivity is that you will be ruled by people inferior to you”. Name the political thinker who made this statement. (1)
Answer:
Plato

Question 5.
Identify the regions where laterite soils are formed:(1)
a) Regions made of igneous rocks named Basalt
b) Regions with monsoon rains and intermittent hot seasons
c) Desert regions
d) Plains formed by the river deposition.
Answer:
(b) Regions with monsoon rains and intermittent hot seasons.

Question 6.
Compare the Kharif and Rabi cropping seasons in India. (3)
Answer:
Kharif sowing period June (onset of monsoon) harvesting period Early November (End of monsoon) Major crops rice, maize, millets, cotton, jute, sugar, cane, ground nut.
Rabi Sowing period November (Beginning of winter) harvesting period March (Beginning of summer) Wheat, tobacco, mustard, pulses.

Question 7.
What are the methods of study employed in sociology? (3)
Answer:
Social survey, Interview, Observation and Case study.

Question 8.
How, is electronic banking (E-banking) helpful to customers? (3)
Answer:
Money can be sent and bills can be paid anywhere in the world from home

• Saves time
• Low service charge.

Question 9.
Write a short note on Trans Himalayas. (3)
Answer:
Trans Himalayas include Karakoram, Ladakh, and Zaskar mountain ranges. Mount K2 (8661m) also known as Godwin Austin, the highest peak in India, is in the Karakoram range. The average height of the Trans Himalayas is 6000 metres.

Question 10.
What are the discretionary functions of the state?(3)
Answer:

• Protection of health
• Provide educational facilities
• Implement welfare activities
• Provide transportation facilities.

Question 11.
Link column ‘A’ with appropriate items from column ‘B’. (4)

A	B
Deccan Education Society	Dr. Zakir Hussain
Viswa Bhrthi university	Rabindranath Tagore
Jamia Milia Islamia	Mahadev Govind Ranade
Vallthol Narayana Menon	Kerala Kalmandalm

Answer:

A	B
Deccan Education Society	Mahadev Govind Ranade
Viswa Bhrthi university	Rabindranath Tagore
Jamia Milia Islamia	Dr. Zakir Hussain
Vallthol Narayana Menon	Kerala Kalmandalm

Question 12.
Explain the different levels of human resource development.(4)
Answer:
Individuals take efforts to develop their own skills.

• Family creates an environment for the development of the potential of individuals.
• Various institutions and agencies provide facilities for education and training.
• Nation provides the necessary facilities for its people to develop their skills.

Question 13.
What is ‘Mountbatten Plain ? Mention its proposals.(4)
Answer:
The strategyprepared by Mountbatten Plan.

• To form a separate country in Muslim majority area as per the Muslims wish.
• To divide Punjab and Bengal
• To conduct a referendum to determine whether to add North West Frontier province to Pakistan or not

Question 14.
Explain why the Sepoys and the Kins fought against the British during the First War of Indian Independence, 1857. (4)
Answer:
Poor salary and abuse by the British officers were the major reasons for their resentment. The rumour that the cartridge in the newly supplied Enfield rifles were greased with the fat of cows and pigs provoked them. It wounded the religious sentiments of the Hindu and Muslim soldiers. The soldiers who were unwilling to use the new cartridges were punished by the officers, the British rule had adversely affected the kings too. In addition to the Doctrine of Lapse, the princely states were convicted of inefficient rule and were annexed by the British. This made the kings to lead the rebellion.

Question 15.
Mark and label the following geo-information in the outline map of India provided. (4)
Answer:
For marking the places on the map.

a) River Narmada
b) Karakoram range
c) Eastern Coastal plain
d) Haldia port

Part – B

Question 16.
Evaluate the achievements of India in the fields of missile technology and space mission. (3)
OR
Prepare a short note on Civil Disobedience Movement in Malabar.
Answer:
Indian Space Research Organization (ISRO) was established to lead space research. The first rocket-
1. launching station in India was established in Thumba, near Thiruvananthapuram. As a result of the collective efforts of India’s space research experts, first satellite Aryabhatta was successfully launched in 1975. In addition to satellites, space vehicles and rocket launchers were also developed. Jt was because of the far-sightedness of Jawaharlal Nehru that India became the first developing nation to make and launch satellites. There are several agencies that develop satellites in India now, They are:

• National Remote Sensing Agency (NRSA)
• Physical Research Laboratory (PRL) India has also advanced much in missile technology. Agni and Prithwi are the missiles developed by India. Dr. Raja Ramanna and Dr. A.P.J-. Abdul Kalam led our experiments in the atomic energy sector.

OR

In 1930s, the Civil Disobedience Movement gained momentum in Malabar. People broke the salt law by making salt under the leadership of K. Kelappan and Mohammed Abdu Rahiman at Payyannurand Kozhikode respectively. The British army brutally attacked the satyagrahis and arrested the leaders. Boycott of foreign goods, picketing liquor shops and popularising Khadi were also part of the Civil Disobedience Movement.

Question 17.
Name the global pressure belts between which each of the following wind blows: (3)
a) Westerlies
b) Trade winds
c) Polar winds.
OR
Complete the following table showing the apparent movement of the Sun.

Period	The Apparent Movement
i) 21 Match to 21 June	From the equator to Tropic of Cancer
ii) 22 December to 21 March	From the Tropic of Capricorn to the Equator
iii) 23 September to 22 December	From the equator to Tropic of Capricorn

Answer:
a) 30° Latitude to 60° latitude
b) 30° Latitude to 0° latitude
c) 90° Latitude to 60°latitude

OR

Period	The Apparent Movement
i) 21 Match to 21 June	From the equator to Tropic of Cancer
ii) 22 December to 21 March	From the Tropic of Capricorn to the Equator
iii) 23 September to 22 December	From the equator to Tropic of Capricorn

Question 18.
Explain the factors that led to the reorganization of states on the basis of languages in India. (3)
OR
Prepare a note on the rise df modern industries in Kerala.
Answer:
There were demands from different parts of India for the formation of states on the basis of language. In 1920 the Nagpur session of the Indian National Congress resolved to form its state committees on the basis of language.’After independence, people agitated for the formation of states along linguistic lines. Potti Sriramalu, a freedom fighter, started satyagraha for the formation of Andhra Pradesh for Telugu speaking people, the Nagpur session of the Indian National Congress resolved to form its state committees on the basis of language. After independence, people agitated for the formation of states along linguistic lines.

Potti Sriramalu, a freedom fighter, started satyagraha for the formation of Andhra Pradesh for Telugu speaking people. After 58 days of fasting, his martyrdom and it intensified the mass agitation. Following this, in 1953, the Government of India formed the state of Andhra Pradesh for Telugu speaking people. After this, the demand for linguistic states intensified. The Government of India formed a Commission to reorganise Indian states on the basis of languages,

OR

Modern factories were established in Kerala by the middle of the twentieth century. Majority of them were in Travancore and Kochi. Rulers of Travancore adopted policies promoting modern industries. The British provided technical and financial support to the industries. The establishment of Pallivasal Hydro Electric Project propelled the development of modern industries.

Question 19.
What are the features of bureaucracy? (4)
OR
Write any two problems faced by the society and suggest its solutions.
Answer:

• Hierarchical organisation
• Permanence
• Appointment on the basis of Qualification
• Political Neutrality
• Professionalism

OR

Water scarcity

• • effective utilisation of water Environmental pollution
• Garbage treatment at source Flood
• Shifting the residence in safe zone
• Corruption
• Awareness agiinstforruption (any two)

Question 20.
Explain the goals of the fiscal policy.
OR
Explain the structure and jurisdiction of the district consumer disputes.redressal forum.
Answer:
Attain economic stability

• Create employment opportunities
• Control unnecessary expenditure

OR

Functions at district level – president and two members – at least one woman member After collecting evidence based oh the complaint filed by the consumer, verdicts are given where the compensation claimed does not exceed ₹ 20 lakhs.

Question 21.
How are satellite imageries prepared? What is Spatial Resolution? (4)
OR
Analyze the model.reference grids given and find out the following.

i) Locate the spring using 6 figure grid reference method.
ii) Identify the feature with 6 figure grid reference 682 315.
iii) Locate the settlements in 4 figure grid reference method.
iv) Name the method by which elevation and relief is represented in the grids.
Answer:
The sensors on artificial satellites distinguish objects on the earth’s surface based on their spectral signature and transmit the information in digital format to the terrestrial stations. This is interpreted .with the help of computers and converted in to picture formats. These are called satellite imageries. The size of the smallest object on the earth’s surface that a satellite sensor can distinguish is called the spatial resolution of the sensor.

OR

i) 656325
ii) Tube well
iii) 6830
iv) Contour lines

Question 22.
Explain the Central Vigilance Commission and the Ombudsman.
OR
Expain the role of social science learning in the formulation of civic consciousness.
Answer:
The Central Vigilance Commission is the institution constitute^ at the national level to prevent corruption. It came into effect in 1964. It is formed to prevent corruption in the central government offices. The Central Vigilance Commissioner is the head of the Central Vigilance Commission. In every department there will be a Chief Vigilance Officer. The duty of the commission is to enquire into yigilance cases and take necessary actions.

Elected representatives and bureaucrats are part of public administration. Complaints can be filed against their corruption, nepotism or financial misappropriation or negligence of duties. Ombudsman is constituted for this purpose. A retired Judge of the High Court is appointed as the Ombudsman. People can directly approach the Ombudsman with complaints. On receiving complaints, the Ombudsman has the power to summon anyone and can order enquiry and recommend actions.

OR
Equips the individuals to respect diversity and to behave with tolerance.

• Helps to understand the different contexts of political, social, economic and environmental problems.
• Equips the individual to suggest comprehensive solutions to different problems.
• Disseminate the message of peace and co-operation to the society.
• Makes the individual civic conscious and action oriented by familiarising the ideal models and activities of civic consciousness.

Question 23.
Explain the circunastances inhere the consumers are exploited of cheated. (4)
OR
Explain any two direct taxes in India.
Answer:
When the purchased product is damaged or defective.

• Defective services received from government/on – government/ private institutions.
• Appropriation of price over and above the amount legally fixed or marked on the outer casing.
• Violation of the prevention of adulteration law.
• Sale of products which are harmful to life and safety.
• Loss due to trading methods which lead to unfair practices and limited consumer freedom.
• Giving misleading advertisement for increasing sales.

OR

Personal Income Tax It is the tax imposed on the income of individuals. The rate of tax increases as the income increases. Income tax is applicable to the income-that is above a certain limit. In India the income tax is collected by the central government as per the Income Tax Act 1961, Corporate tax This , is the tax imposed on the net income.

Question 24.
Elucidate Local time standard time and Greenwich mean time. Estimate thelndian’Standard time when the Greenwidwhichmean time is 12 midnight.
OR
What is monsoon? Explain the role of Trade wipds in the formation of South-est monsoon winds.
Answer:

• Local time: Local time is the time calculated based on the position of the sun.
• Standard time; Time observed at the standard meridian of a country.
• Greenwich Time: Time observed at 0° longitude is known as the Greenwich Time 5.30 am

OR

Monsoon winds are winds that change direction due to the shift of the pressure belts.
Sun’s rays fall vertically to the North of the Equator during certain months due to the tilt of the Earth’s axis. This leads to an increase in temperature along the region through which Tropic of Cancer passes. The pressure belts also shift slightly northwards in accordance with this, The southeast trade winds also cross the equator and moves towards the north Sun’s rays fall vertically to the North of the Equator during certain months due to the tilt of the Earth’s axis. This leads to an increase in temperature along the region through which Tr: pic of Cancer passes. The pressure belts also shift slightly northwards in accordance with this. The southeast trade winds also cross the equator and moves towards the north.

Question 25.
Analyze the causes of the First World War. (6)
OR
Explain the factors that led to the February Revolution in Russia.
Answer:
Military Alliances : Germany, – Italy and Austria – Hungary formed the Triple Alliance. England, France and Russia formed Triple Entene. These alliances created tension in Europe.’

Aggressive Nationalism: European nations captured other countries and provinces. Formation of Pan Salv movement, Pan German Movement and the Revenge movement were examplesof aggressive nationalism.

Imperial Crisis : Imperial competition among the European nations led to crisis. Moroccan Crisis, Balkan crisis etc. Assassination of Arch Duke Francis Ferdinand. Crown prince of Austria was assassinated on 28 June 1914 at Sarajevo, capital of Bosnia.

Accusing Serbia for this Austria Hungary declared war on 28th July 1914.
OR

Decision of the Russian emperor czar Nicholas II to participate in the First World War ignoring the opposition of the Duma: Scarcity of food by 1917.Soldiers joined the workers in the agitation at Petrograd.
The city of Petrograd was captured by the workers. With this the emperor abandoned the throne and a temporary government under the leadership of Kerensky assumed power. This is known as the February revolution.

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Board	Kerala Board
Textbook	SCERT, Kerala
Class	SSLC Class 10
Subject	SSLC Malayalam
Chapter	Previous Year Question Papers, Model Papers, Sample Papers
Year of Examination	2020, 2019, 2018, 2017
Category	Kerala Syllabus Question Papers

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Students can Download Chapter 9 Accounts from Incomplete Records Questions and Answers, Plus One Accountancy Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Accountancy Chapter Wise Questions and Answers Chapter 9 Accounts from Incomplete Records

Plus One Accountancy Accounts from Incomplete Records One Mark Questions and Answers

Question 1.
Single entry system is also known as ………….
(a) Imprest system
(b) Merchandise system
(c) Incomplete system
(d) Cash system
Answer:
(c) Incomplete system

Question 2.
Incomplete records are usually maintained by ………………………
(a) Small traders
(b) Society
(c) Company
(d) Government
Answer:
(a) Small traders

Question 3.
Credit purchase can be ascertained as the balancing figure in the ………………
(a) Total Debtor Account
(b) Total Creditor Account
(c) Statement of Affairs
(d) Balance Sheet
Answer:
(b) Total Creditors Account

Question 4.
Cash received from debtors can be had from ………………. the account.
(a) Total Debtor
(b) Cash Book
(c) Statement of affairs
(d) Both a & b
Answer:
(d) both a & b.

Question 5.
If capital comparison method of single entry system, the profit or loss is ascertained by
(a) Preparing a statement of affairs
(b) Preparing trading and profit & loss A/c.
(c) Preparing a statement of profit or loss
(d) Both a & c.
Answer:
(d) Both a and c.

Question 6.
Incomplete record mechanism of bookkeeping is:
(a) Scientific
(b) Unscientific
(c) Unsystematic
(d) Both b and C
Answer:
(d) Both b and c

Question 7.
Locate the odd one.
(a) Incomplete system
(b) Unsystematic system
(c) Double-entry system
(d) Single entry system
Answer:
(c) Double-entry system.

Question 8.
……….. account are not kept under single entry system.
Answer:
Impersonal

Question 9.
………….. account is prepared to ascertain credit sale.
Answer:
Total Debtors Account

Question 10.
Bill receivable from debtors during the year can be obtained from ………… account.
Answer:
Bill Receivable

Question 11.
Statement of affairs is prepared to a certain ……………..
Answer:
Capital

Question 12.
Find the odd one and state the reason.

1. credit sale, sales returns, discount allowed, return outwards.
2. Credit purchase, endorsement of the bill, return inwards, return to suppliers.

Answer:

1. return outwards – affected by creditors account, all others are affected by debtors A/c.
2. return inwards – affected by debtors a/c, all others are affected by creditors A/c.

Question 13.
Match the following.

Answer:

• 1 – e
• 2 – c
• 3 – d
• 4 – b
• 5 – a

Question 14.
What does the missing item of the account represent?
Total Debtors A/c

Answer:
Cash received from Debtors Rs. 32,000

Question 15.
In capital comparison method of single entry system, the profit or loss is ascertained by
(a) Preparing trading and profit and loss A/c.
(b) Preparing statement of affairs.
(c) Preparing statement of profit or loss.
(d) Both b and c.
Answer:
(d) Both b and c

Question 16.
Given the opening and closing balances of bills receivable and cash received on account of bills receivable, balancing bills receivable account will show,
(a) Credit purchase
(b) Credit sales
(c) Bills received during the year
Answer:
(c) Bills received during the year.

Question 17.
Given the opening and closing balance of debtors and the figures of credit sales, the balancing figure of total debtors account will give.
(a) Bills honoured during the year.
(b) Closing balance of bills receivable.
(c) Cash received from debtors.
(d) Cash sales.
Answer:
(c) Cash received from debtors.

Plus One Accountancy Accounts from Incomplete Records Two Mark Questions and Answers

Question 1.
State the meaning of incomplete records.
Answer:
Books of accounts that are not maintained according to the double-entry system are generally referred to as incomplete records. The system is also known as single entry. It is an incomplete, unscientific and unsystematic method of keeping the books of accounts of a trader.

Question 2.
Complete the following table:

Answer:

• 2. Credit purchase – Total creditors account
• 3. Cash sales – Receipt side of cash book
• 4. Credit sales – Total Debtors Account
• 5. Capital – Statement of Affairs

Question 3.
Afire occured in the godown of Mr. Asok who keeps his books under single entry and his goods were partly destroyed. Since the goods were insured, he lodged a claim of Rs. 1,00,000/- to the insurance company, out of which only Rs. 60,000 was admitted. On what ground can the Insurance company’s decision be justified?
Answer:
Since, Mr. Asok maintain incomplete records, it is not reliable and scientific. These accounts are not accepted by the Insurance company. It is one of the limitations of single entry.

Plus One Accountancy Accounts from Incomplete Records Three Mark Questions and Answers

Question 1.
Give any five features of single entry system.
Answer:

1. It is an unscientific, unsystematic and incomplete system.
2. Mainly personal accounts are prepared by ignoring fully or partially the impersonal accounts.
3. It is used by small traders.
4. Profit or loss under this system is only an estimate.
5. True financial position cannot be ascertained.

Question 2.
Calculate profit or loss from the following information . for the year ended 31.12.2005.

Answer:
Statement of profit or loss for the year ended 31.12.05

Question 3.
Prepare Total Debtors Account from the following information:

Answer:
Total Debtors Account

Question 4.
Calculation of credit purchase by preparing Total creditors account.

Answer:

Question 5.
Find.out the capital at the beginning.

Answer:
Calculation of Capital at the beginning

Plus One Accountancy Accounts from Incomplete Records Four Mark Questions and Answers

Question 1.
The single entry system of accounting is crude and unsystematic, still is popular among small businessmen. Give reasons.
Answer:
Some businessmen prefer to keep their books under single entry system due to the following reasons.

1. The system is suitable to small traders which have mainly cash transactions and do not have many assets and liabilities to be recorded in details.
2. The system is economical since lesser number of books are maintained.
3. Lack of knowledge about the double-entry system.
4. Ignorance of businessmen as to the statutory requirements of keeping proper books of accounts.
5. Intentional omission to take advantage of taxation.

Question 2.
What are the difference between Balance Sheet and Statement of Affairs?
Answer:

Balance Sheet	Statement of Affairs
1. It is prepared on the basis of those books which are maintained under the double-entry system.	1. It is prepared on the basis of information from incomplete records
2. It is prepared to show the financial position of the concern.	2. It is usually prepared to find out capital.
3. Value of asset and liabilities in a Balance Sheet are based on ledger balances	3. Value of assets and liabilities in a statement of affairs are based on estimates
4. Omission of assets or liabilities can easily be found out when Balance sheet disagree.	4. It is difficult to locate omission of assets or liabilities in statement of affairs.

Plus One Accountancy Accounts from Incomplete Records Five Mark Questions and Answers

Question 1.
What are the limitations of incomplete records?
Answer:
Following are the limitations of single entry system

1. It is not based on the double-entry system, arithmetical accuracy of books of accounts can not proved.
2. No clear idea about the financial position.
3. Comparison with previous years performance is not possible due to incomplete information.
4. It encourage fraud, misappropriation etc. among employess.
5. In the absense of nominal accounts, it is difficult to determine the exact profit or loss.
6. It is difficult to obtain loans from bank or other financial institution.

Question 2.
Mention the difference between double-entry system and single entry system or incomplete records.
Answer:
The following are the difference between the double-entry system and single entry system.

Single Entry System	Double Entry System
1. Dual aspects of transactions are not recorded.	1. Dual aspects of every transaction are recorded.
2. As trial balance is not prepared, arithmetical accuracy can’t be checked.	2. Trial balance is prepared to check the arithmetical accuracy.
3. Only an estimate of profit can be made	3. Actual net profit can be Calculated
4. Balance sheet can not be prepared to ascertain the financial position	4. Balance sheet can be prepared to ascertain the financial position
5. This system is suitable for sole trader who have a few transaction	5. This is suitable for all types of business all types of business

Question 3.
Final accounts can be prepared from incomplete records. Explain the procedure.
Answer:
Though the records are incomplete, the trader has to ascertain the profit or loss of his business and the position regarding assets and liabilities. Two methods are adopted for ascertainment of profit or loss. They are:

1. Ascertainment of profit or loss by statement of affair method.
2. Preparation of profit and loss account and balance sheet under conversion method.

1. Statement of Affair Method:
Under this method, profit or loss can be ascertained by comparing the capital at the beginning and at the end of the financial period. For this purpose, two statements are prepared.

a. Statement of Affairs:
It is a statement prepared by presenting the assets on one side and liabilities on the other side as in the case of a balance sheet. The difference between the totals of the two sides is known as “owners equity or capital”.
Owner equity or capital = Asset – Liabilities

b. Statement of profit or loss:
The statement prepared to ascertain the profit or loss by comparing the opening capital with closing capital is called statement of profit or loss. If the capital at the end of the year exceeds the capital in the beginning of the year, the difference will be treated as “profit.” On the other hand, If the capital in the beginning of the year is more than that at the end of the year, there is “loss.”

2. Conversion Method:
Under single entry system, nominal accounts and real accounts (other than cash) are not maintained. Hence it is not possible to prepare the profit and loss account and balance sheet under the system. In such a situation, financial statements are to be prepared by converting accounts under single entry to that under double entry. This method of preparing financial statements is called conversion method.

Question 4.
From the following particulars, calculate total sales.

Answer:
Bill Receivable Account

Total Debtors Account

Question 5.
From the following information, calculate the amount total purchase.

Answer:
Bills Payable A/c

Total Creditors A/c

Plus One Accountancy Accounts from Incomplete Records Six Mark Questions and Answers

Question 1.
Sumesh keeps incomplete records. You are required to ascertain the profit or loss for the year ending 3-1.3.2004 from the following information.

He had withdrawn Rs. 5,000 during the year and had introduced Rs. 4,000 from the sale of his personal property.
Answer:
Statement of Affairs as on 01.04.2003

Statement of Affairs as on 31.03.2004

Statement of Profit or Loss for the year ended 31.03.2004.

Plus One Accountancy Accounts from Incomplete Records Eight Mark Questions and Answers

Question 1.
Mr. Murali keeps his books under single entry. He supplies you with the following information from which you are to find out his profit or loss for the year ended 31.3.2007.

He had withdrawn Rs. 3,000 during the year for a private purpose and had introduced fresh capital Rs. 6,000 on 1.10.2006. Bad and doubtful debts provision at 5% is to be made on debtors. Depreciation on plant and machinery at 10% and furniture at 15 % is to be made. Allow 6% interest on capital.
Answer:
Statement of Affairs of Mr. Murali

Statement of profit or loss for the year ended 31.3.07

Question 2.
Anil carries on a retailer business and does not keep his books on double entry basis. The following particulars are obtained from his books.

His cash transations during the year were given

During the year Anil had taken goods from the business for private consumption which amounted to Rs. 850. Prepare profit and loss account for the year ending 30-06-2005 and a balance sheet as on that date after charging depreciation @ 10% p.a. on the machinery.
Answer:
Statement of Affairs as at 1.7.04

Total Debtors A/c

Total Creditors A/c

Trading and profit and loss account for the year ended 30.06.05

Balance sheet as on 30.06.05

Question 3.
Shankar maintains his book of account on single entry system. Prepare his final accounts from the information supplied for the year ended 30.9.2008 as follows.
Cash transactions during the year.

Particulars of assets and liabilities are given below:

Additional information:

1. Credit sales for the year Rs. 18,100.
2. Discount allowed to Debtors Rs. 2,100.
3. Return outwards during the year Rs. 500.
4. Salaries outstanding on 30.9.2008 Rs. 3,000.
5. Provision for doubtful debts is to be created to the extent of Rs. 3,000.
6. 5% depreciation is to be provided on furniture and land & buildings.

Answer:
Total Debtors A/c

Total Creditors A/c

Cash Book

Statement of Affairs as at 01.10.2007

Trading and profit and loss A/c for the year ended 30.9.2008.

Balance sheet as on 30.09.2008

Question 4.
Mr. Giri does not keep his books under the double-entry system. The following are his assets and liabilities as on the opening and closing date of 2005.

His Cashbook for the year ended 31.12.05 as follows.

Discount allowed to debtors is Rs. 1,600 and discount allowed by creditors is Rs. 1,300. Bad debts written off is Rs. 400. Provision for bad debts is required at 5%. Depreciation @ 10% is required on furniture. Interest accrued on investments amounts to Rs. 2,200. Prepare Trading and profit and loss A/c and Balance sheet for 2005.
Statement of Affairs as on 01.01.2005
Answer:

Bills Receivable A/c

Bills Payable A/c

Total Debtors A/c

Total Creditors A/c

Trading and Profit & Loss A/c for the year ended 31.12.2005

Balance sheet as on 31.12.2005

Question 5.
Mr. Binu keeps his books under single entry. From the following information, prepare profit and loss account for the year ended 31^st December 2004 and a balance sheet as on that date.
Cashbook

Other Information

Answer:
Total Debtors A/c

Total Creditors A/c

Trading and Profit and Loss A/c for the year ended 31.12.2004

Balance Sheet as on 31.12.2004

Question 6.
Mrs. Bhavana keeps his books by Single Entry System. You’re required to prepare final accounts of her business for the year ended December 31, 2015. Her records relating to cash receipts and cash payments for the above period showed the following particulars.
Summary of Cash

The following information is also available

All her sales and purchases were on credit. Provide depreciation on plant and building by 10% and machinery by 5%. make a provision for bad debts by 5%.
Answer:
Debtor’s Account

Creditor’s Account

Statements of Affairs as on 31^st December 2015

Trading and Profit & Loss Account as on 31^st December 2015

Balance Sheet as on 31 December 2015