Maharashtra Board Class 11 Geography Important Questions Chapter 4 Climatic Regions

Balbharti Maharashtra State Board Class 11 Geography Important Questions Chapter 4 Climatic Regions Important Questions and Answers.

Maharashtra State Board 11th Geography Important Questions Chapter 4 Climatic Regions

Complete the chain.

Question 1.

A B C D
(1) Mediterranean climate (1) Interior area of Asia (1) Hot summer, very low rainfall (1) Coniferous forest
(2) Equatorial rainforest climate (2) Scandinavian countries and Siberia (2) Cool summer, bitterly cold winters (2) Scrubs and hardwood forest
(3) Tropical desert climate (3) Amazon and Congo basins (3) Moist winter and warm summer (3) Dense and evergreen hardwood forest
(4) Taiga climate (4) Central California and South Africa (4) High temperature and convectional rainfall (4) Xerophytic vegetation

Answer:

A B C D
(1) Mediterranean climate (4) Central California and South Africa (1) Moist winter and warm summer (1) Scrubs and hardwood forest
(2) Equatorial rainforest climate (3) Amazon and Congo basins (2) High temperature and convectional rainfall (2) Dense and evergreen hardwood forest
(3) Tropical desert climate (1) Interior area of Asia (3) Hot summer, very low rainfall (3) Xerophytic vegetation
(4) Taiga climate (2) Scandinavian countries and Siberia (4) Cool summer, bitterly cold winters (4) Coniferous forest

Maharashtra Board Class 11 Geography Important Questions Chapter 4 Climatic Regions

Question 2.

A B C D
(1) Tropical monsoon climate (1) Between 10° and 20° N and S (1) Summer temperatures around 35°C, rainfall averaging 250-1000 mm (1) Tundra vegetation
(2) Tundra climate (2) Between 45° to 65° N and S (2) precipitation 500 to 2500 mm (2) Tali thick grass with scattered trees
(3) Tropical savannah type of climate (3) 65° to 90° N (3) Summer temperature 27° to 32°, orographic rainfall (3) Year-round short green grass, trees shed leaves in winter
(4) West European type (4) Between 10° to 20° N and S (4) Summer temperature 10°C! and winter temperature -20°C to -30°C (4) Deciduous forest

Answer:

A B C D
(1) Tropical monsoon climate (1) Between 10° to 30° N and S (1) Summer temperature 27° to 32°, orographic rainfall (1) Deciduous forest
(2) Tundra climate (2) Between 65° to 90° N (2) Summer temperature 10°C and winter temperature -20°C to -30°C (2) Tundra vegetation
(3) Tropical Savannah type of climate (3) Between 10° to 20° N and S (3) Summer temperatures around 35°C, rainfall averaging 250-1000 mm (3) Tall thick grass with scattered trees
(4) West European type (4) Between 45° to 65° N and S (4) Mild cool summer, precipitation 500 to 2500 mm (4) Year round short green grass, trees shed leaves in winter

Choose the correct option by identifying the correct correlation in the sentences.

Question 1.
Bushmen tribe is found in ……………….
(a) Tropical deserts
(b) Highland
(c) Ice sheet
(d) Tundra
Answer:
(a) Tropical deserts

Question 2.
In Marine West European type climate …………………
(a) trees shed leaves during summers
(b) no vegetation
(c) swamps during melting of show
(d) trees shed leaves during winter
Answer:
(d) trees shed leaves during winter

Maharashtra Board Class 11 Geography Important Questions Chapter 4 Climatic Regions

Question 3.
Tundra climate is observed in the belt from ……………………
(a) near poles in both hemispheres
(b) 65° to 90° North
(c) 20° to 30° in both hemispheres
(d) 0° to 10° in both hemispheres
Answer:
(b) 65° to 90° North

Question 4.
Rainforest dense vegetation is …………………… layered vegetation.
(a) two
(b) three
(c) four
(d) one
Answer:
(b) three

Question 5.
Tropical monsoon climate is found within …………….. North and South of the equator.
(a) 10° to 30°
(b) 0° to 10°
(c) 30° to 60°
(d) 60° to 90
Answer:
(a) 10° to 30°

Question 6.
In India parts of the peninsular plateau and rain-shadow zone in Maharashtra has …………………. type of climate.
(a) Mediterranean
(b) Arid
(c) Tropical Savannah
(d) Mountain
Answer:
(c) Tropical Savannah

Question 7.
Rainfall in winter is a characteristic of …………………. climate.
(a) China type
(b) Marine West European
(c) Taiga
(d) Mediterranean
Answer:
(d) Mediterranean

Question 8.
The tribes of Greenland are ……………….
(a) Aborigines
(b) Inuits
(c) Bushmen
(d) Bedouin
Answer:
(b) Inuits

Maharashtra Board Class 11 Geography Important Questions Chapter 4 Climatic Regions

Question 9.
Terrace cultivation is a characteristic of ………………….. type of climate.
(a) mountain
(b) ice cap
(c) equatorial
(d) arid
Answer:
(a) mountain

Identify the correct correlation.

A : Assertion
R : Reasoning

Question 1.
A – In high mountains, large changes in mean temperatures occur over short distances.
R – Highland climates are governed by topography.
(a) Only A is correct.
(b) Only R is correct.
(c) Both A and R are correct and R is the correct explanation of A.
(d) Both A and R are correct but R is not the correct explanation of A.
Answer:
(c) Both A and R are correct and R is the correct explanation of A.

Question 2.
A – In ice cap climate extremely meagre precipitation in the form of snow is found.
R – Precipitation exceeds evaporation.
(a) Only A is correct.
(b) Only R is correct.
(c) Both A and R are correct and R is the correct explanation of A.
(d) Both A and R are correct but R is not the correct explanation of A.
Answer:
(d) Both A and R are correct but R is not the correct explanation of A.

Question 3.
A – In Tundra region, the winter temperature is not as severe as the Taiga.
R – The Tundra is closer to the sea while Taiga is away from any ocean body.
(a) Only A is correct.
(b) Only R is correct.
(c) Both A and R are correct and R is the correct explanation of A.
(d) Both A and R are correct but R is not the correct explanation of A.
Answer:
(c) Both A and R are correct and R is the correct explanation of A.

Question 4.
A – In southern hemisphere permanent settlements are found around Taiga region.
R – Latitudinal location plays a great role in the climate.
(a) Only A is correct.
(b) Only R is correct.
(c) Both A and R are correct and R is the correct explanation of A.
(d) Both A and R are correct but R is not the correct explanation of A.
Answer:
(b) Only R is correct.

Maharashtra Board Class 11 Geography Important Questions Chapter 4 Climatic Regions

A – Rainforest are dense.
R – Constant high temperature and heavy convectional rainfall evenly distributed over the year.
(a) Only A is correct.
(b) Only R is correct.
(c) Both A and R are correct and R is the correct explanation of A.
(d) Both A and R are correct but R is not the correct explanation of A.
Answer:
(c) Both A and R are correct and R is the correct explanation of A.

Identify the incorrect factor.

Question 1.
Equatorial rainforests are observed in-
(a) Congo and Amazon basin
(b) Indonesia
(c) Malaysia
(d) South East Asia
Answer:
(d) South East Asia

Question 2.
Tropical desert or arid type climate is observed in-
(a) Philippines
(b) Rajasthan
(c) Coastal Chile
(d) Peru
Answer:
(a) Philippines

Question 3.
In China type or humid sub-tropical climate the following crops are grown-
(a) olives
(b) citrus fruits
(c) cotton
(d) tobacco
Answer:
(a) olives

Question 4.
In ice sheet regions the characteristics of climate is-
(a) snowfall
(b) frequent anticyclones
(c) precipitation exceeds evaporation
(d) winter rainfall
Answer:
(c) precipitation exceeds evaporation

Question 5.
The characteristics of the climate in tropical desert regions are-
(a) high humidity
(b) highest diurnal range of temperature
(c) precipitation less than 200 mm
(d) windy conditions
Answer:
(a) high humidity

Maharashtra Board Class 11 Geography Important Questions Chapter 4 Climatic Regions

Question 6.
The characteristics of the Mediterranean climate regions are-
(a) mild, moist winter
(b) warm, dry summer
(c) drizzle
(d) foggy coasts
Answer:
(c) drizzle

Question 7.
The characteristics of the China type climate regions are-
(a) frost in winter
(b) windy condition
(c) summer temperature 10°C
(d) winter temperature 0°C to 18°C
Answer:
(b) windy condition

Question 8.
The Taiga type of climate is observed in-
(a) New Zealand
(b) Scandinavian countries
(c) Northern part of North America
(d) Siberia
Answer:
(a) New Zealand

Question 9.
The Marine North Western type of climate is observed in-
(a) Siberia
(b) Canada
(c) New Zealand
(d) Australia
Answer:
(a) Siberia

Question 10.
The main economic activities of people in Tundra region is-
(a) hunting and fishing
(b) lumbering
(c) animal rearing
(d) mining
Answer:
(b) lumbering

Complete the table.

Maharashtra Board Class 11 Geography Important Questions Chapter 4 Climatic Regions 1
Answer:
Maharashtra Board Class 11 Geography Important Questions Chapter 4 Climatic Regions 2

Identify the correct Group

Question 1.
Tropical Monsoon climate.
(A) Within 0° to 10° N and S of equator
High annual rainy temperature
Rainfall 250 to 2500 mm.
Saline soil

(B) Within 10° to 30° N and S of equator
High annual range of temperature
Rainfall 250 to 2500 mm.
Iron rich soil in high rainfall zones

(C) Between 10° to 20° N and S latitude
High diurnal range of temperature
Rainfall 250 to 1000 mm.
Swamps

(D) Between 20° to 30° N and S latitude
High diurnal temperature
Scanty rainfall
Sandy soil
Answer:
B

Maharashtra Board Class 11 Geography Important Questions Chapter 4 Climatic Regions

Question 2.
Mediterranean climate regions.
(A) Central California
South and South West Australia
Cape Town
Central Part of Chile

(B) Canada
Western Europe
South East Australia
New Zealand

(C) Between 10° to 20° N and S latitude
High diurnal range of temperature
Rainfall 250 to 1000 mm.
Swamps

(D) Between 20° to 30° N and S latitude
High diurnal temperature
Scanty rainfall
Sandy soil
Answer:
A

Give geographical reasons.

Question 1.
Most natural regions are homogeneous ecosystem.
Answer:

  1. A natural region is a region which has common natural factors related to geography, geology and climate such as vegetation, animal resources etc.
  2. A natural region has same flora and fauna. For example, in tropical desert region xerophytes grow everywhere due to low rainfall resulting into shortage of water supply.
  3. From ecological point of view, the naturally available flora and fauna of this region are influenced by geographical as well as geological factors such as temperature, rainfall, soil. etc.
  4. Thus, most of the natural regions are homogeneous ecosystem.

Differentiate between.

Question 1.
Tropical Monsoon and Tropical Desert type of climate.
Answer:

Tropical Monsoon Tropical Desert
Latitudinal extent Within 10° to 30° North and South. Found usually between 20° to 30° latitudes in both hemispheres.
Temperature Summer temperature is around 27°C to 32°C while winter temperature is 15°C to 24°C. Summer temperature around 30°C to 45°C, winter temperature around 20°C to 25°C.
Rainfall Rainfall is between 250 to 2500 mm, excessively wet during rainy season. Precipitation less than 200 mm.
Vegetation Tropical rainforest, ranges from jungles to thorn forest in drier boundaries. Xerophytic plants are found which can withstand extreme temperature.
Animal life Large hooved leaf eaters and large carnivores like tigers. Small, nocturnal, burrowing animals are found.
Human life Agriculture is the main occupation of people of monsoon region. Agriculture is practiced near oasis.
Regions of the world Coastal areas of South West India and South East Asia, South West Africa, North East and South East Brazil, Northern part of Australia and part of Japan come under the monsoon winds. Western coasts of all continents, large parts of Gujarat, Rajasthan and South West Haryana, Iran, interior parts of Asia, Coastal Chile, Peru, South-west Africa, interior Mexico, Baja California, North Africa, Namibia and parts of US.

Equatorial Rainforests and Savannah Climatic Regions.

Taiga region
(7) Monsoon region
(8) Greenland
(9) Equatorial region
Answer:

Maharashtra Board Class 11 Geography Important Questions Chapter 4 Climatic Regions 3

Maharashtra Board Class 11 Geography Important Questions Chapter 4 Climatic Regions

Answer the following question by using the given map.

Maharashtra Board Class 11 Geography Important Questions Chapter 4 Climatic Regions 4
Question 1.
Name the climatic region along Mediterranean Sea.
Answer:
Mediterranean type of climate region.

Question 2.
Name the climate region along NW coast of Europe.
Answer:
West European type of climate region.

Question 3.
Name the island which has Ice sheet.
Answer:
Greenland island.

Question 4.
In which part of China, Chine type of climate is found?
Answer:
South eastern part of China.

Maharashtra Board Class 11 Geography Important Questions Chapter 4 Climatic Regions

Question 5.
Name the countries where Tundra type of climate is found.
Answer:
Northern part of Russia and Northern part of Canada and Alaska.

Write short notes on.

Question 1.
Equatorial Rainforest
Answer:
Equatorial region is located between 0° to 10° in both the hemisphere, that is on either side of the equator.

The Amazon basin of South America, Congo basin of Africa, East coast of Central America, Malaysia, Indonesia, Papua New Guinea, Madagascar, Philippines are included in this region.

This region has a unique climate. Being located along the equator, sunrays are vertical throughout the year. Therefore, the temperature is constantly high throughout the year, Average

The days and nights are of equal duration. This may be the only region in the world where there is no summer or winter seasons. Throughout the year there is high temperature and heavy rainfall, thus it has hot and wet climate.

This region has heavy rainfall almost daily. The ITCZ (Inter-Tropical Convergence Zone) is an area of low pressure in this region due to high temperature and convectional currents. In ITCZ, North East and South East trade winds converge. They bring warm and humid air in this region. Due to convectional currents, moist air rises up, gets condensed and gives heavy rainfall every day. Rainfall is 2500 to 3000 mm, evenly distributed over the year.

Due to hot and wet climate, there is a luxuriant growth of the trees. Therefore, the trees are tall and forest have three layered vegetation. Due to heavy rainfall trees are evergreen. Thus, they are evergreen rainforest and they provide hardwood.

Due to dense forest growth, climbing and jumping animals like monkeys, as well as snakes, crocodiles are observed on marshy land. This region has rich biodiversity.

Due to heavy rainfall chemical weathering is more common than physical weathering, fertile elements of soil are washed away. only iron particles remain in the soil.

Tribal people live in this region with the help of hunting of birds and animals and food gathering activity. For example, pygmies in Congo basin of Africa.

Question 2.
Tundra region
Answer:

  1. In higher latitudes that is near the North pole Tundra climate region is located. It is only on the landmass of Antarctica in the Southern Hemisphere.
  2. The countries bordering Arctic Ocean, that is Northern part of North America, Greenland, Eurasia, Antarctic, Peninsula and islands in the polar region are included in this climatic region.
  3. This region has also unique climate. There is day light continuously for six months and night for six months.
  4. This region has long winters of 10 to 11 months and short summers of 1 to 2 months.
  5. This region is away from the equator, so sunrays are slanting, therefore summer temperature is around 10°C, but winters are severe, winter temperature is -20°C to -30°C. Due to low temperature there is low evaporation.
  6. There is no rainfall but there is snowfall. Average snowfall is 300 to 500 mm.
  7. Polar anticyclones, strong winds, snow storms, coastal fog are common.
  8. Due to low temperature hardly moss, lichen grow. In summer when snow melts some flowering plants grow.
  9. The main source of livelihood of people is hunting of seals, walrus, polar bear for collection of fur and meat, fishing and mining at some places. Main tribe in this region is Inuits.

Maharashtra Board Class 11 Geography Important Questions Chapter 4 Climatic Regions

Question 3.
Marine West European type climate
Answer:

  1. This region as the name suggests is located in the western part of the continent in the latitudinal belt of 45° to 65° N and S.
  2. Main countries included in this region are the countries of Western Europe, South Eastern Australia, New Zealand, Western Coastline of USA and Canada, Southern Chile, Southern Alaska.
  3. This region has mild to cool summer; average summer temperature is 20°C. Average winter temperature is 5°C. Winters are quite warm due to warm ocean current passing along the coastline.
  4. Due to nearness to sea, there is always high cloud cover and high humidity. Average rainfall is 500 to 2500 mm. Fog, frost and drizzle are common.
  5. Temperate deciduous trees grow, they shed leaves in winter. Coniferous trees grow on mountain slopes, short green grass grows throughout the year.
  6. Fishing, agriculture and animal rearing is the main occupation of the people.

Question 4.
Tropical monsoon climate
Answer:

  1. This region extends between 10° and 30° N and S latitudes.
  2. This region includes the countries of South East Asia, South West Africa, as well as coastal areas of South West India, North East and South East Brazil, Northern part of Australia and parts of Japan.
  3. This region has three distinctive seasons, summer, winter and rainy season, due to seasonal winds called monsoon winds.
  4. The summer temperature is 27°C to 32°C, winter temperature is 15°C to 24°C. Annual range of temperature is high.
  5. The average rainfall is 250 to 2500 mm. Most of the places receive orographic rainfall. The rainfall is mainly by South West monsoon winds.
  6. The climate is controlled by summer onshore and winter offshore wind movements related to shifting of ITCZ.
  7. Due to seasonal rainfall, deciduous forest grows, trees shed their leaves in summer. Heavy rainfall areas have equatorial forest, less rainfall areas have thorny vegetation. Herbivores and carnivores live in these forests.
  8. Main occupation of people is agriculture due to sufficient temperature, rainfall and fertile soil.

Answer in detail.

Question 1.
Write in brief about China type of climate.
Answer:

  1. China type of climate or humid sub-tropical climate is found along East coastal areas between 20° and 40° N and S.
  2. The countries of South East USA, South East part of South America, as well as Eastern Australia, South Africa, South china to South Japan and Easter Island in Chile.
  3. Summers are mild with temperature 10°C. Winters are cold. The temperature varies between 0°C to 18°C. There is frost in winter.
  4. Since these regions are located along the coastline, onshore humid air gives rainfall almost throughout the year. Cyclonic storms are in winter.
  5. Due to rainfall throughout the year, there is mixed forest growth. Coniferous forests grow on mountain slopes. Lower slopes of mountains are covered with grass.
  6. Agriculture is the main occupation of the people. Main crops grown are rice, wheat, corn, sugarcane, tobacco, cotton, and citrus fruits.

Maharashtra Board Class 11 Geography Important Questions Chapter 4 Climatic Regions

Question 2.
Explain the Highland or Mountain type climate.
Answer:

  1. This type of climate is found in all latitudes where there are mountains or highlands.
  2. The main areas are mountainous area of Asia, Central Europe, Western North and South America and highlands of Africa.
  3. Climate of these regions depend upon the altitude of mountains, highlands, their location on forward and windward slope, their slopes facing the sun etc.
  4. In general, mountain/highland region has low temperature since temperature decreases with increase in height.
  5. Most of the mountains receive orographic rainfall specially if they are near the coastline. For example, Western Ghats in India receive orographic rainfall by South West monsoon winds.
  6. In general, there is snowfall on higher part or slopes of mountains. There is permanent snow cover on the top of mountain. For example, Himalayas are snow covered in their upper part.
  7. Generally coniferous forests grow on the upper slopes of mountains that is below snow line. The lower slopes are covered with evergreen or deciduous forests depending on amount of rainfall.
  8. Main occupation is animal rearing with the help of pastures on mountain slopes. Due to beautiful scenery of the region tourism is developed in most of the mountainous areas. Agriculture is practiced on terraces slopes.

Maharashtra Board Class 11 Geography Important Questions Chapter 3 Agents of Erosion

Balbharti Maharashtra State Board Class 11 Geography Important Questions Chapter 3 Agents of Erosion Important Questions and Answers.

Maharashtra State Board 11th Geography Important Questions Chapter 3 Agents of Erosion

Complete the chain.

Question 1.

Agent Processes Erosional / Deposition features
(1) River (1) Plucking (1) Hollows
(2) Wind (2) Solution (2) Potholes
(3) Glacier (3) Deflation (3) Sinkholes
(4) Groundwater (4) Drilling (4) Roche moutonnees

Answer:

Agent Processes Erosional / Deposition features
(1) River (1) Drilling (1) Potholes
(2) Wind (2) Deflation (2) Hollows
(3) Glacier (3) Plucking (3) Roche moutonnees
(4) Ground water (4) Solution (4) Sink holes

Maharashtra Board Class 11 Geography Important Questions Chapter 3 Agents of Erosion

Question 2.

Agent Erosional features Depositional features
(1) River (1) Gorge (1) Moraine
(2) Sea waves (2) Cirque (2) Barchan
(3) Wind (3) Sea cliff (3) Sand bars
(4) Glaciers (4) Mushroom rock (4) Food plains

Answer:

Agent Erosional features Depositional features
(1) River (1) Gorge (1) Food plains
(2) Sea waves (2) Sea cliff (2) Sand bars
(3) Wind (3) Mushroom rock (3) Barchan
(4) Glaciers (4) Cirque (4) Moraine

Question 3.

Agent Erosional features Deposition features
(1) Ground water (1) Wave cut platforms (1) Loess
(2) Wind (2) Horn (2) Beach
(3) Glacier (3) Sink holes (3) Stalactites
(4) Sea waves (4) Yardang (4) Eskers

Answer:

Agent Erosional features Deposition features
(1) Ground water (1) Sink holes (1) Stalactites
(2) Wind (2) Yardang (2) Loess
(3) Glacier (3) Horn (3) Eskers
(4) Sea waves (4) Wave cut platforms (4) Beach

Choose the correct option by identifying the correct correlation in the sentences.

Question 1.
In mountainous areas river flows at a higher speed. Here, …………………
(a) bed gets more eroded than banks
(b) banks get more eroded than bed
(c) both bed and bank get eroded
(d) none of the above
Answer:
(a) bed gets more eroded than banks

Question 2.
The particles of sand are blown away by the wind. After removal the portion from where sand blown off appears as depression. These are called ……………….
(a) Potholes
(b) Barchans
(c) Sand dunes
(d) Deflation hallows
Answer:
(d) Deflation hallows

Question 3.
Landforms developed by underground water anywhere in the world are called ……………….
(a) Deltas
(b) Karst
(c) Aeolian
(d) None of the above
Answer:
(b) Karst

Question 4.
Where two tributary glaciers join together, their lateral moraines merge underneath …………………
(a) to form terminal moraines
(b) to form medial moraines
(c) to form lateral moraines
(d) to form ground moraines
Answer:
(b) to form medial moraines

Maharashtra Board Class 11 Geography Important Questions Chapter 3 Agents of Erosion

Question 5.
In this process of transportation, the material acquired by the agents is transported by rolling, pushing and dragging along the surface. The material consists of boulders and big rocks.
(a) Abrasion
(b) Attrition
(c) Traction
(d) Saltation
Answer:
(c) Traction

Question 6.
The process by which moving ice exerts pressure on majority portion of rocks on bed or along the bank is known as ………………..
(a) Abrasion
(b) Plucking
(c) Attrition
(d) Solution
Answer:
(b) Plucking

Question 7.
When rocks and pebbles bump into each other and break up into smaller fragments it is called ………………
(a) Attrition
(b) Abrasion
(c) Drilling
(d) Deflation
Answer:
(a) Attrition

Question 8.
Drumlins are formed by deposition by …………………..
(a) Rivers
(b) Sea waves
(c) Glaciers
(d) Wind
Answer:
(c) Glaciers

Question 9.
Sometimes, waves can erode the softer part and make it hollow enough to be called a ………………..
(a) Cave
(b) Beach
(c) Headland
(d) Sea cliff
Answer:
(a) Cave

Question 10.
………………….. beach at Chennai is the longest beach in India.
(a) Diveagar
(b) Guhagar
(c) Harihareshwar
(d) Marina
Answer:
(d) Marina

Question 11.
The ………………… Lake in Orissa becomes a fresh water lagoon during monsoons.
(a) Dal
(b) Wular
(c) Chilika
(d) Periyar
Answer:
(c) Chilika

Identify the correct correlation.

A : Assertion
R : Reasoning

Question 1.
A – In river beds, gravels and pebbles are often rounded.
R – The rock materials carried in the flow tumble and bounce against one another.
(a) Only A is correct.
(b) Only R is correct.
(c) Both A and R are correct and R is the correct explanation of A.
(d) Both A and R are correct but R is not the correct explanation of A.
Answer:
(c) Both A and R are correct and R is the correct explanation of A.

Maharashtra Board Class 11 Geography Important Questions Chapter 3 Agents of Erosion

Question 2.
A – The alluvial fans are formed at the foothills.
R – When the river enters the plains, there is change in slope and the velocity of the river increases abruptly.
(a) Only A is correct.
(b) Only R is correct.
(c) Both A and R are correct and R is the correct explanation of A.
(d) Both A and R are correct but R is not the correct explanation of A.
Answer:
(a) Only A is correct.

Question 3.
A – Some rivers may not form deltas.
R – They do not have huge load of sediment.
(a) Only A is correct.
(b) Only R is correct.
(c) Both A and R are correct and R is the correct explanation of A.
(d) Both A and R are correct but R is not the correct explanation of A.
Answer:
(c) Both A and R are correct and R is the correct explanation of A.

Question 4.
A – Large waves do not get generated in lagoons as they get separated from the sea.
R – The lagoons becomes fresh water lagoon during monsoon.
(a) Only A is correct.
(b) Only R is correct.
(c) Both A and R are correct and R is the correct explanation of A.
(d) Both A and R are correct but R is not the correct explanation of A.
Answer:
(d) Both A and R are correct but R is not the correct explanation of A.

Question 5.
A – In arid areas, the eroded portion of softer rocks appear like elongated ridges and harder rocks appear as elevated portions.
R – Softer rocks get more eroded faster.
(a) Only A is correct.
(b) Only R is correct.
(c) Both A and R are correct and R is the correct explanation of A.
(d) Both A and R are correct but R is not the correct explanation of A.
Answer:
(c) Both A and R are correct and R is the correct explanation of A.

Identify the incorrect factor.

Question 1.
In the following agent’s abrasion process is used in erosion.
(a) Wind
(b) Sea waves
(c) Glacier
(d) Ground water
Answer:
(d) Ground water

Question 2.
In the following agent’s attrition process is used in erosion,
(a) Ground water
(b) Wind
(c) River
(d) Sea waves
Answer:
(a) Ground water

Maharashtra Board Class 11 Geography Important Questions Chapter 3 Agents of Erosion

Question 3.
Solution process of erosion is used in the following agents.
(a) Ground water
(b) Glacier
(c) River
(d) Sea waves
Answer:
(b) Glacier

Question 4.
Lateral erosion process is found in the following agents.
(a) River
(b) Sea waves
(c) Wind
(d) Glacier
Answer:
(c) Wind

Question 5.
Suspension is the process of transportation used by following agents.
(a) Ground water
(b) River
(c) Wind
(d) Sea waves
Answer:
(a) Ground water

Question 6.
Saltation process of transportation is used by following agents.
(a) River
(b) Sea waves
(c) Glacier
(d) Wind
Answer:
(c) Glacier

Question 7.
The erosional landforms of glaciers are.
(a) Cirque
(b) Horn
(c) Moraine
(d) Arete
Answer:
(c) Moraine

Question 8.
The depositional landforms of glaciers are.
(a) Eskers
(b) Drumlin
(c) Erratic rock
(d) Barchan
Answer:
(d) Barchan

Question 9.
The depositional features of ground water are.
(a) Stalactite
(b) Sea stack
(c) Pillar
(d) Stalagmite
Answer:
(b) Sea stack

Question 10.
The erosional features of wind are.
(a) Loess
(b) Ventifacts
(c) Deflation hollows
(d) Yardang
Answer:
(a) Loess

Maharashtra Board Class 11 Geography Important Questions Chapter 3 Agents of Erosion

Question 11.
The depositional features of wind are.
(a) Longitudinal dunes
(b) Loess
(c) Barchan
(d) Flood plains
Answer:
(d) Flood plains

Question 12.
The erosional features of sea waves are.
(a) Sea arch
(b) Sea caves
(c) Sink holes
(d) Wave cut platforms
Answer:
(c) Sink holes

Question 13.
The depositional features of sea waves are.
(a) Beaches
(b) Sand bars
(c) Lagoon
(d) Delta
Answer:
(d) Delta

Question 14.
The erosional features of rivers are.
(a) U-Shaped valley
(b) Gorge
(c) Waterfall
(d) Potholes
Answer:
(a) U-Shaped valley

Question 15.
The depositional features of rivers are.
(a) Oxbow lake
(b) Flood plains
(c) Moraines
(d) Meanders
Answer:
(c) Moraines

Complete the table.

Question 1.

Agent Erosional landforms Deposition landforms
(1) River
(2) Wind
(3) Sea Waves
(4) Groundwater
(5) Glacier

Answer:

Agent Erosional landforms Deposition landforms
(1) River Gorges, V-shaped valleys, canyons, waterfalls, potholes Alluvial fans, meanders, flood plains, natural levees, deltas
(2) Wind deflation hollows, ventifacts mushroom rock, yardang Sand dunes, loess, barchans
(3) Sea Waves Sea cliffs, cave, sea stack, sea arches, wave cut platform Beach, sand bar, lagoon, sand pits
(4) Groundwater Sink holes, caves Stalactite, stalagmites, pillars
(5) Glacier Roche mountonnees, cirque, arete, horn, U-shaped valley, hanging valley Drumlins, eskers, erratic rock, lateral moraines, end moraines, terminal moraines, medial moraines

Question 2.

Agent Process of erosion Process of deposition
(1) River
(2) Wind
(3) Sea Waves
(4) Groundwater
(5) Glacier

Answer:

Agent Process of erosion Process of deposition
(1) River attrition, solution, drilling, downcutting, headward erosion, lateral erosion traction, saltation, suspension
(2) Wind abrasion, attrition, deflation traction, saltation, suspension
(3) Sea Waves abrasion, attrition, solution, lateral erosion traction, saltation, suspension
(4) Groundwater solution solution
(5) Glacier plucking, abrasion, downcutting, headward erosion, lateral erosion traction

Give geographical reasons.

Question 1.
The coastal zone management warrants serious attention.
Answer:
The coastal zone management warrants serious attention because-

  1. Compared to the other agents of erosion the work of sea waves goes on ceaselessly.
  2. Erosion in some part and deposition in the adjoining part keeps on taking place constantly. The beaches and bars which are formed due to deposition are also eroded.
  3. Coastal regions are always vulnerable to the risk of getting submerged due to increase in sea level.
  4. Coastal regions are also the regions of high population density.

Maharashtra Board Class 11 Geography Important Questions Chapter 3 Agents of Erosion

Question 2.
Mushroom rocks have broad tops.
Answer:
Mushroom rocks have broad tops because-

  1. The high rising rocks in the path of the wind are attacked by the sand that moves with the wind.
  2. Winds and the particles they carry, attack the base of an individual rock.
  3. The larger top part is not eroded as much as the basal part since, the particles carried by the wind being heavy are not uplifted up to the top portion of the rock.
  4. The particles at medium height are smaller but their velocities are high. Hence, their impact is more.
  5. As a result, the portion of rock at medium height is eroded more and the rock as a whole gets the shape of a mushroom.

Question 3.
All rivers do not form delta.
Answer:
All rivers do not form delta because –

  1. Delta is a place near the seashore where the sediments brough down by a river are deposited in triangular form.
  2. Most of the rivers finally join to the sea but every river does not form delta because they do not have the huge load of sediments.
  3. If the rivers are travelling from a long distance, they join many tributaries and they deposit the sediment into the main river. Such long and sluggish rivers can deposit sediment at the river mouth and form delta. For example, The River Ganga has formed a huge delta.
  4. Some rivers are short and swift. They do not receive more tributaries, so such rivers have less sediments. These rivers flow very swiftly so they deposit sediments in the sea water. Therefore, they do not form delta.

Differentiate between.

Question 1.
Headward erosion and Lateral erosion.
Answer:

Headward erosion Lateral erosion
(i) Headward erosion is the backward erosion by river in the source region. (i) Lateral erosion is the erosion that occurs on the sides of valleys of a river or glacier.
(ii) This is carried away by the river. (ii) The valley slopes are eroded by tributary streams.
(iii) Headward erosion causes the river to move backward, because erosion takes place in the opposite direction of the follow of river. (iii) Lateral erosion causes widening of valley, because lateral erosion occurs in the valley sides.

Question 2.
Plucking and Abrasion.
Answer:

Plucking Abrasion
(i) Plucking is the process by which moving ice exerts pressure on rocks on bed or along the bank. (i) Abrasion involves the scratching and polishing of the surface or bedrock by the particles moving on it.
(ii) The pressure on the rock causes uprooting of rock portion that is exposed to the flow. (ii) The rock particles involved in abrasion rub against rock and wear away the surface.
(iii) The eroded bedrock will have rugged surface. (iii) The eroded bedrock will have smooth surface.

Question 3.
Barchan and Longitudinal dunes.
Answer:

Barchan Longitudinal dunes
(i) When there is an obstacle in the path of wind, or due to low velocity of wind, the sand gets dropped at some places. Thus, barchans are formed. (i) When sand is deposited parallel to the direction of wind, longitudinal dunes are formed.
(ii) Barchans are crescent-shaped dunes. The slope that faces the wind is gentle whereas the opposite slope is steep. (ii) They appear parallel to the wind direction. Sometimes they are 100km long.
(iii) Barchans migrate from one place to another. (iii) Longitudinal dunes do not migrate but extend in the wind direction.
(iv) They are seen in the Great Indian Desert of Rajasthan. (iv) They are seen in Rub-al-Khali desert of Saudi Arabia.

Write short notes on.

Question 1.
Plucking
Answer:

  1. It is the process by which moving ice exerts pressure on majority portion of rocks on bed or along the bank.
  2. The pressure on the rock causes uprooting of rock portion that is exposed to the flow. This uprooting of particles is called plucking.
  3. The eroded bedrock will have a rugged surface.
  4. Plucking at the base of a glacier lead to formation of Roche moutonnees.

Question 2.
Drilling
Answer:

  1. Bedload moves along the running water.
  2. As and when this flow encounters an obstacle due to relief on the bed or joints in the bed, the flow tends to develop a circular pattern.
  3. This circular system becomes stable, though the water continues to flow in downstream direction.
  4. This leads to development of a whirl.
  5. The trapped sediments also follow the similar circular motion.
  6. The continuous action of these trapped sediments and the whirl deepens the bed of the river at a given point.
  7. Eventually, it develops into a larger depression assuming a shape of a pot.
  8. Potholes are the features formed by this process. For example, they are found in River Indrayani at Bhegadewadi in Pune district.

Maharashtra Board Class 11 Geography Important Questions Chapter 3 Agents of Erosion

Question 3.
The Process of deposition
Answer:

  1. The velocity at which the agent is moving has a great impact on deposition.
  2. If the stream or wind slows down, the carrying capacity and the particle sizes carried and deposited will decrease.
  3. If a stream flows faster, say during floods or when the river is in the mountains, then the carrying power of the stream and the sizes of particles deposited will increase.
  4. If there is a change in the slope of land or in the direction of flow, deposition may occur there.
  5. The slopes are absent in the plain region. Therefore, the rivers are unable to carry the sediments further and they start depositing.
  6. Due to gravity, small particles settle more slowly than the larger particles. Smaller particles are carried away by the river or wind till the end and their deposition occurs in the later stage.
  7. If an obstacle comes in between the flow of the agent it causes the particles to settle and thus deposition takes place.

Question 4.
Moraine
Answer:

  1. Moraine is a depositional feature of a glacier.
  2. Glaciers generally deposit load of sediments along the side and front of the ice. These deposits are called moraines.
  3. There are four types of moraines depending upon the location where sediments are deposited. They are lateral moraines, medial moraines, terminal moraines and end moraines.
  4. The moraines deposited at the sides of the wall are called lateral moraine.
  5. When two tributary glaciers join together, their lateral moraines merge underneath to form medial moraines..
  6. At the foot of the glacier, sediments are deposited like a heap of all grain sizes forming a curved depositional ridge called end moraines. It contains grains of all sizes.
  7. When sediments are deposited at the snout of glacier, when it melts, it is called terminal moraine.

Draw a neat and labelled diagram for.

Question 1.
Yardangs
Answer:
Maharashtra Board Class 11 Geography Important Questions Chapter 3 Agents of Erosion 1

Question 2.
Sea arch
Answer:
Maharashtra Board Class 11 Geography Important Questions Chapter 3 Agents of Erosion 2

Question 3.
Sea stack
Answer:
Maharashtra Board Class 11 Geography Important Questions Chapter 3 Agents of Erosion 3

Question 4.
Pothole drilling
Answer:
Maharashtra Board Class 11 Geography Important Questions Chapter 3 Agents of Erosion 4

Question 5.
Groundwater
Answer:
Maharashtra Board Class 11 Geography Important Questions Chapter 3 Agents of Erosion 5

Maharashtra Board Class 11 Geography Important Questions Chapter 3 Agents of Erosion

Answer in detail.

Question 1.
Write in detail about the depositional features formed by glaciers.
Answer:
Following are the depositional features formed by glaciers.
Drumlin:

  1. Coarse Material that moves with glaciers gets deposited at different places in the form of heaps.
  2. They appear to have egg like shape. Such heaps are called drumlins.
  3. At times, large number of drumlins get deposited in an area. From a higher elevation, it- appears like a basket of eggs. It is also referred to as ‘basket of eggs topography’.

Eskers:
When the coarse material, moving with the glaciers is deposited in linear and zigzag manner, it is called an esker.

Erratic rock:

  1. In the areas under the influence of glaciers generally at high latitudes, huge rock pieces are found to have been deposited in the area where the local rocks are of different formation.
  2. Such rock appears as erratic ones to the area in which they are deposited.
  3. These can be considered as guest rocks.

Moraines:

  1. Generally, glaciers deposit load of sediments along the side and front of the ice. These deposits are called moraines.
  2. The moraines deposited at the sides of the wall are called lateral moraines.
  3. At the toe or foot of the glacier, sediment is deposited in a jumbled heap of all grain sizes forming a curved depositional ridge called end moraines.
  4. End moraines, that mark the farthest advance of a glacier called terminal moraines.
  5. When two tributary glaciers join together, their lateral moraines merge underneath to form medial moraines.

Question 2.
Explain the types of sand dunes.
Answer:
Barchans:

  1. These are crescent-shaped dunes.
  2. They form where supply of sand is minimum.
  3. Due to an obstacle in the path of the wind or due to the lowering of its velocity, the sand moving with it gets dropped at some places.
  4. The barchan slope that faces the wind is gentle whereas, the opposite slope is steep.
  5. Such barchans are seen in large numbers in the Sahara Desert of Africa or in the Great Indian Desert in Rajasthan.

Longitudinal Dunes:

  1. These are long dunes deposited is the direction of the wind.
  2. They appear parallel to the wind direction.
  3. They do not migrate like the barchans.
  4. They are also called seif dunes.
  5. They are sometimes hundreds of kilometres long. Such dunes can be seen in Rub-al- Khali desert in Saudi Arabia.

Maharashtra Board Class 11 Geography Important Questions Chapter 3 Agents of Erosion

Question 3.
Write about the erosional features formed by river.
Answer:
A river, which is flowing water, erodes rock materials, transports them to newer places and deposits them. In this process, it creates many erosional and depositional landforms. Let us study in details about erosional landforms:

George:

  1. In mountainous areas, river flows at a higher speed.
  2. Here, the bed gets more eroded than its banks. Gorges are formed with steep banks and a narrow bed. For example, gorges of river Ulhas and river Narmada. Deeper gorges are called canyons.

V-shaped Valleys:

  1. Over a period of time, the bed of the river is less eroded.
  2. Erosion along the banks and the slope of the valley increases.
  3. The vertical sides of the valley become wide resembling letter V’. These are V-shaped valleys.

Waterfalls:
When a river runs over alternating layers of hard and soft rock, rapids and waterfalls are formed. Jog falls on the Sharavati river.

Potholes:

  1. They generally originate below waterfalls on where rocks are structurally weak.
  2. Swirling whirlpool motions of the river water causes stones at the bottom to grind the bedrock and enlarge the potholes by drilling while finer sediments are carried away in the current.
  3. Potholes may range from a few centimetres to many meters in diameter and depth. For example, large potholes can be seen in the beds of river Kukadi at Nighoj in Ahmednagar district.

Maharashtra Board Class 11 Geography Important Questions Chapter 2 Weathering and Mass Wasting

Balbharti Maharashtra State Board Class 11 Geography Important Questions Chapter 2 Weathering and Mass Wasting Important Questions and Answers.

Maharashtra State Board 11th Geography Important Questions Chapter 2 Weathering and Mass Wasting

Complete the chain

Question 1.

Types of Mass Wasting Characteristics Regions
(1) Creep (1) Rapid fall (1) Periglacial or alpine region
(2) Rockfall (2) Slowest moving (2) Hillsides in a humid region
(3) Earth flow (3) Slow downhill creep (3) Rocky areas
(4) Solifluction (4) Rapid or slow (4) Tops and basal portions of hills

Answer:

Types of Mass Wasting Characteristics Regions
(1) Creep (1) Slowest moving (1) Tops and basal portions of hills
(2) Rock fall (2) Rapid fall (2) Rocky areas
(3) Earth flow (3) Rapid or slow (3) Hill sides in humid region
(4) Solifluction (4) Slow downhill creep (4) Periglacial or alpine region

Maharashtra Board Class 11 Geography Important Questions Chapter 2 Weathering and Mass Wasting

Question 2.

Types of weathering Factors Resultant
(1) Physical (1) Heat (1) Oxidation
(2) Chemical (2) Explosion (2) Granular weathering
(3) Anthropological (3) Water (H2O) (3) Exfoliation
(4) Pressure (4) Mining

Answer:

Types of weathering Factors Resultant
(1) Physical (1) Heat Pressure (1) Exfoliation, Granular weathering
(2) Chemical (2) Water (H2O) (2) Oxidation
(3) Anthropological (3) Explosion (3) Mining

Choose the correct option by identifying the correct correlation in the sentences.

Question 1.
Rocks can break because of …………….
(a) water, pressure, heat, etc.
(b) gravity
(c) deposition
(d) sedimentation
Answer:
(a) water, pressure, heat, etc.

Question 2.
In rocks such as granite which have joints, heat can cause weathering ………………….
(a) by breaking the rock into two halves
(b) by breaking the rock along the joints into blocks
(c) by breaking the rock into pieces
(d) by shattering
Answer:
(b) by breaking the rock along the joints into blocks

Question 3.
Chemical weathering is effective and rapid in …………………
(a) cold climate
(b) hot climate
(c) humid climate
(d) dry climate
Answer:
(c) humid climate

Maharashtra Board Class 11 Geography Important Questions Chapter 2 Weathering and Mass Wasting

Question 4.
Erosion involves the application of ………………
(a) kinetic energy to the surface along which the material moves
(b) gravitational force to the surface along which the material moves
(c) tensional force to the surface along which the material moves
(d) compressional force to the surface along which the material moves
Answer:
(a) kinetic energy to the surface along which the material moves

Question 5.
The salts of calcium, sodium etc., present in rocks have a tendency to expand, this leads to ………………..
(a) breaking of rocks
(b) dissolving the salts
(c) formation of cracks in rocks
(d) crystallization of salts
Answer:
(d) crystallization of salts

Question 6.
Water molecules react with minerals present in the rocks. This process is called ………………..
(a) solution
(b) hydrolysis
(c) oxidation
(d) carbonation
Answer:
(b) hydrolysis

Question 7.
The calcium present in ………………. reacts with the water and air to form carbonic acid.
(a) limestone
(b) marble
(c) granite
(d) slate
Answer:
(a) limestone

Question 8.
During ………………., the calcium and carbonate in limestone detach from each other, thereby decomposing the limestone.
(a) oxidation
(b) solution
(c) desilication
(d) carbonation
Answer:
(d) carbonation

Question 9.
When rocks break along joints due to heat, it is called ……………….
(a) block disintegration
(b) cracks
(c) shattering
(d) granular weathering
Answer:
(a) block disintegration

Question 10.
Mining, blasting of hills and ridges is a type of …………….. weathering.
(a) physical
(b) chemical
(c) biological
(d) anthropogenic
Answer:
(d) anthropogenic

Question 11.
Mass movements occur continuously on all …………………
(a) plains
(b) plateau
(c) slopes
(d) rivers
Answer:
(c) slopes

Maharashtra Board Class 11 Geography Important Questions Chapter 2 Weathering and Mass Wasting

Question 12.
The ……………….. plays a big role in stabilizing slopes.
(a) gravity
(b) water
(c) slope
(d) vegetation
Answer:
(d) vegetation

Question 13.
Slow downhill creep of soil in periglacial or alpine regions is known as ………………..
(a) rock fall
(b) solifluction
(c) earth flow
(d) landslide
Answer:
(b) solifluction

Identify the correct correlation

Question 1.
A – Mass wasting is effective on slopes that are made up of clay and shale.
R – Mass wasting determines the material of the slope.
(a) Only A is correct.
(b) Only R is correct.
(c) Both A and R are correct and R is the correct explanation of A.
(d) Both A and R are correct but R is not the correct explanation of A.
Answer:
(c) Both A and R are correct and R is the correct explanation of A.

Question 2.
A – Microscopic organisms like algae, moss, lichens cause biological weathering.
R – They produce chemicals on the outer layer of the rocks.
(a) Only A is correct.
(b) Only R is correct.
(c) Both A and R are correct and R is the correct explanation of A.
(d) Both A and R are correct but R is not the correct explanation of A.
Answer:
(c) Both A and R are correct and R is the correct explanation of A.

Question 3.
A – Quartzite, a metamorphic rock, is harder than steel.
R – It does not easily react physically but is easily fractured by chemical weathering.
(a) Only A is correct.
(b) Only R is correct.
(c) Both A and R are correct and R is the correct explanation of A.
(d) Both A and R are correct but R is not the correct explanation of A.
Answer:
(d) Both A and R are correct but R is not the correct explanation of A..

Question 4.
A – In mass movements, the materials come down the slope without the aid of a transporting medium like running water, ice or wind.
R – Anthropogenic weathering is responsible for mass movement.
(a) Only A is correct.
(b) Only R is correct.
(c) Both A and R are correct and R is the correct explanation of A.
(d) Both A and R are correct but R is not the correct explanation of A.
Answer:
(a) Only A is correct

Question 5.
A – Steeper slope will also have higher possibility of slow mass movement.
R – Gentle slopes will have slow mass movement.
(a) Only A is correct.
(b) Only R is correct.
(c) Both A and R are correct and R is the correct explanation of A.
(d) Both A and R are correct but R is not the correct explanation of A.
Answer:
(b) Only R is correct.

Maharashtra Board Class 11 Geography Important Questions Chapter 2 Weathering and Mass Wasting

Question 6.
A – Mass movement is based on how quickly the rock and weathered material moves due to friction, gravity, load of material.
R – The rock and material move because of running water.
(a) Only A is correct.
(b) Only R is correct.
(c) Both A and R are correct and R is the correct explanation of A.
(d) Both A and R are correct but R is not the correct explanation of A.
Answer:
(d) Both A and R are correct but R is not the correct explanation of A.

Identify the incorrect factor.

Question 1.
The factors that affect mechanical weathering.
(a) Water
(b) Heat
(c) Pressure
(d) Microscopic organism
Answer:
(d) Microscopic organism

Question 2.
The factors that affect biological weathering.
(a) Mining
(b) Microscopic organism
(c) Barrowing animals
(d) Roots of trees
Answer:
(a) Mining

Question 3.
The reasons for anthropogenic weathering.
(a) Mining
(b) Pressure
(c) Road construction
(d) Quarrying for building material
Answer:
(b) Pressure

Question 4.
The factors responsible for mass movements.
(a) Gravity
(b) Water
(c) Heat
(d) Relief and slope
Answer:
(c) Heat

Question 5.
The mass movements can be divided into four main classes.
(a) Creep
(b) Fall or topple
(c) Flows
(d) Weak material and structure
Answer:
(d) Weak material and structure

Maharashtra Board Class 11 Geography Important Questions Chapter 2 Weathering and Mass Wasting

Question 6.
The types of mass wasting.
(a) Fall or topple
(b) Landslide
(c) Solifluction
(d) Rock fall
Answer:
(a) Fall or topple

Identify the correct group

Question 1.

A B C D
(1) Algae (1) Rock fall (1) Relief and slope (1) Shattering
(2) Roots and plants (2) Landslide (2) Gravity (2) Block disintegration
(3) Lichens (3) Earth flow (3) Water (3) Exfoliation
(4) Moss (4) Exfoliation (4) Vegetation (4) Burrowing

Answer:
A

Give geographical reasons

Question 1.
Many times, rocks break.
Answer:
Rocks break because-

  1. Rocks which are exposed are affected by the climate of that region in various ways.
  2. Rocks can break because of water, pressure, heat, etc. in various ways.
  3. Those rocks which have joints or layers, break easily than rocks which are homogeneous.
  4. The rocks which have vertical strata are easily loosened and break, due to changes in temperature, water, wind or frost action.
  5. If there are larger grains in the rocks, there is ample space in the rocks for water to enter and they get weathered easily and break.
  6. Some rocks are very hard and resistant to chemical weathering but they are easily fractured by physical weathering.
  7. Generally, sedimentary rocks break easily than igneous rocks and metamorphic rocks.

Question 2.
Water plays an important role in weathering.
Answer:
Water plays an important role in weathering because-

  1. In areas where diurnal range of temperature is high, water inside the crack’s freezes. During daytime, it again becomes water.
  2. When water freezes, its volume increases due to the pressure exerted on the walls of the rock, the cracks in the rocks widen and finally rock breaks down due to freeze and thaw weathering.
  3. Water also breaks down the rocks chemically. The water molecules react with minerals present in the rocks. This is called hydrolysis.
  4. Another way by which water can cause chemical weathering is by solution. Minerals like calcium, magnesium dissolves in water.
  5. Oxygen present in air and water reacts with iron or aluminium present in rock and as a result rock rusts and ultimately breaks.

Question 3.
Biological weathering is the disintegration of rocks as a result of action by living organisms.
Answer:
Biological weathering is the disintegration of rocks as a result of action by living organisms because-

  1. Roots of trees and other plants penetrate into the soil and grow in size. Thus, they exert pressure on rocks and make the cracks wider and deeper.
  2. Some plants grow within fissures and lead to disintegration.
  3. Microscopic organisms such as algae, moss, lichens and bacteria grow on rock surface and produce chemicals, they react on outer layers the rocks and break them.
  4. Man known for his economic and technological development becomes the most powerful weathering and erosion agent. Mining, blasting, construction of roads, railways etc., result in fast disintegration of rocks.
  5. There are some barrowing animals like rats, when they barrow, rocks are broken.

Question 4.
Vegetation plays a big role in stabilizing slopes.
Answer:
Vegetation plays a big role in stabilizing slopes because-

  1. The strong root system of trees and other plants help to hold the soil in place.
  2. When these trees are removed, the soil becomes loose and can be easily washed away.
  3. This is why, when there is dense forest cover there is stabilization of slopes but deforested areas are likely to be sites of mass wasting.

Maharashtra Board Class 11 Geography Important Questions Chapter 2 Weathering and Mass Wasting

Differentiate between.

Question 1.
Physical weathering ad Chemical weathering.
Answer:

Physical weathering Chemical weathering
(i) Physical weathering may be defined as the disintegration or breaking of rocks without changing the properties of the rock. (i) Chemical weathering is the process by which the internal structures of a mineral is altered by the addition or renewal of elements.
(ii) This type of weathering occurs mainly by temperature and pressure changes. (ii) This type of weathering occurs when water activates several types of chemical reactions within the rocks.
(iii) In general, cold dry climate accelerates physical weathering. (iii) Hot wet climate accelerates chemical weathering.
(iv) E.g. Exfoliation (iv) E.g. Oxidation

Question 2.
Landslides and Earth flow.
Answer:

Landslides Earth flow
(i) All types of mass movements of rock, debris, including soils and ice are collectively called landslides. (i) Diagonal down slope movement of rock fragments and fine-grained material along sliding plane with enough water is called earth flow.
(ii) Landslides occur on steep or gentle slope gradient, from mountain range to coastal cliffs. (ii) Earth flow is more common or hillsides or valley sides having rich minerals.
(iii) Gravity is primary force for the landslide to occur, others are slope stability, heavy rainfall or earthquake. (iii) The primary reason for Earth flow is continuous heavy rainfall which saturates the ground and increases the pour water pressure, and earth flow starts.

Question 3.
Rockfall and Solifluction.
Answer:

Rockfall Solifluction
(i) Rapid fall of rock material down or rock travel downwards bouncing and rolling is termed as rock fall. (i) Solifluction is the name for the slow downhill creep of soil.
(ii) Rockfalls occur mainly in rocky areas. (ii) Solifluction occurs mainly in periglacial or alpine regions.
(iii) The size of rockfall blocks can range from one cubic centimetre to ten thousand cubic meters. (iii) The size of material in solifluction is very small since it contains particles of soil.

Write short notes on

Question 1.
Types of Mass Wasting
Answer:
Mass wasting or mass movement is the down slope movement of loose mixture of soil, land and rock particles by force of gravity.
Following are different types of mass wasting:
Creep:

  1. Very slow imperceptible down slope movement of materials is called creep.
  2. It occurs along top and basal portions of hills.

Rock fall:

  1. Rock falls are relatively small landslides confined to the removal of individual and superficial blocks down a cliff base.
  2. It mainly occurs in rocky areas.

Earth flow:

  1. Earth flow is promoted by excessive water received mostly through rainfall so that the materials are oversaturated.
  2. It typically occurs on hill sides in humid regions.

Solifluction:

  1. Solifluction is the name for the slow downhill creep of soil, which occurs in a variety of climatic conditions.
  2. It occurs in periglacial or alpine regions.
  3. It occurs very slowly, so it is measured in millimetres or centimetres.
  4. As permafrost is impermeable to water, soil overlying may become oversaturated and the slides slope down under the pull of gravity.

Landslide:

  1. It may be mentioned that generally all types of mass movements of rock wastes including soils and ice are collectively called landslides.
  2. It occurs on moderately steep slopes.
  3. Gravity is primary force for the landslide to occur, others are slope stability, heavy rainfall or earthquake.

Maharashtra Board Class 11 Geography Important Questions Chapter 2 Weathering and Mass Wasting

Question 2.
Biological Weathering:
Answer:
Flora : Roots of trees penetrate into the soil in search of minerals. Sometimes the plants grow within the fissures in the rock leading to widening of the fissures and eventually rocks disintegrate.

Fauna : Microscopic organisms such as algae, moss, lichens and bacteria grow on rock surface and produce chemicals, which react with the rock surface and create weathering. Burrowing animals like moles, mice, rats, reptiles, like snakes etc., dig holes in the soil and speed up disintegration.

Draw a neat diagram and labelled for:

Question 1.
Exfoliation
Answer:
Maharashtra Board Class 11 Geography Important Questions Chapter 2 Weathering and Mass Wasting 1

Answer in Details

Question 1.
Explain different types of mass wasting.
Answer:
Mass wasting is the down slope movement of loose mixture of soil, land and rock particles by force of gravity.
Following are different types of mass wasting:

Creep:

  1. Very slow imperceptible down slope movements of materials is called creep.
  2. It occurs along tops and basal portions of hills.

Rock fall:

  1. Rock falls are relatively small landslides confined to the removal of individual and superficial blocks from cliff base.
  2. It mainly occurs in rocky areas.

Earth flow:

  1. Earth flow is promoted by excessive water received mostly through rainfall so that the materials are oversaturated.
  2. It typically occurs on hill slides in humid regions.

Solifluction:

  1. Solifluction is the name for the slow downhill creep of soil, which occurs in variety of climatic conditions.
  2. It occurs in periglacial or alpine regions.
  3. As permafrost is impermeable to water, soil overlying may become oversaturated and slide slope down under the pull of gravity.

Landslide:

  1. It may be mentioned that generally all types of mass movements of rock wastes including soils and ice are collectively called landslides.
  2. It occurs on moderately steep slopes.

Question 2.
What is weathering? Explain different types of weathering.
Answer:
Weathering refers to the break down or disintegration and decomposition of rocks in situ through mechanical and chemical changes in the rocks and their minerals affected by water temperature, wind, different atmospheric gases and organisms. Following are the types of weathering-
Maharashtra Board Class 11 Geography Important Questions Chapter 2 Weathering and Mass Wasting 2
Let us study the types in details:
Physical Weathering:
(i) Water:

  1. In areas where diurnal temperature range is high, water inside the crack’s freezes during night and melts during daytime.
  2. Due to this continuous freezing and thawing, the cracks widen and disintegration of rocks takes place. This is called freeze and thaw weathering.

(ii) Heat:

  1. In hot deserts the temperature increases during the day. The minerals present in the rock react differently to the increased temperature.
  2. The minerals within the rock expand leading to stress and the rock breaks into individual mineral grains. This is called as granular disintegration.
  3. In the similar process, in the deserts during day time the rock surface gets heated and shatters. It may produce a breaking sound. This is called shattering.
  4. In rocks such as granite which have joints, heat can cause weathering by breaking the rocks along the joints into blocks. This is called block disintegration.

Pressure:

  1. Due to overlying rocks, the rocks beneath the surface experience a lot of pressure.
  2. When the underlying rock gets exposed to the surface, they are subjected to low pressure. As a result of pressure differences, the rock mass expands outwards.
  3. The expansion causes outer layer to separate from lower layer. It is called dislodgement which happens due to pressure release. It is mainly observed in igneous rocks which are more homogeneous. The continuous removal of outer layer of the rock is called exfoliation.

Chemical Weathering:

  1. Oxidation: When oxygen from air and water reacts with iron or bauxite present in the rocks, iron and aluminium oxides are formed. The colour of the rock changes and it is called rusting.
  2. Carbonation: When carbon-dioxide reacts with the minerals in the soil, carbonation takes place. The decomposition of dead produces C02, which reacts with minerals in the rocks.
  3. Solution: When some minerals in the rocks reacts with water in the rock or moisture in the air and gets dissolved, it is called solution. For example, chalk dissolves in water.
  4. Salt: The salts of calcium, sodium, magnesium, potassium etc., present in the rocks have a tendency to expand due to their thermal properties. This leads to crystallization of salts and individual grains spilt from main rocks and fall off at the end.

Biological Weathering:
Flora : Roots of trees penetrate into the soil in search of minerals. Sometimes the plants grow within the fissures in the rock leading to widening of the fissures and eventually disintegrating.

Fauna : Microscopic organisms such as algae, moss, lichens and bacteria grow on rock surface and produce chemicals, which react with the rock surface and create weathering. Burrowing animals like moles, mice, rats, reptiles, like snakes etc., dig holes in the soil and speeds up disintegration.

Anthropogenic weathering:
Man is responsible for weathering as he constructs roads, railways, do mining etc.

  1. Man being a biological agent affects weathering.
  2. With economic and technological development, man has become the most powerful weathering and erosion agent.
  3. Mining, blasting of hills and ridges for road and dam construction, quarrying for industrial and building materials, etc result in a fast rate of disintegration of rocks.
  4. This may be accomplished by natural weathering processes in thousands to millions of years.
  5. Man accelerates the rate of weathering on hill slopes through activities like deforestation.

Maharashtra Board Class 11 Geography Important Questions Chapter 2 Weathering and Mass Wasting

Question 3.
Classify and explain mass movement on basis of weathered material.
Answer:
Maharashtra Board Class 11 Geography Important Questions Chapter 2 Weathering and Mass Wasting 3

  1. Fall or topple: It happens when rocks have a free fall and land at the bottom of a slope.
  2. Flows: A mixture of water, rock and weathered material. They move very quickly. Large flows can bury an entire village. Small flows can block roads.
  3. Slide: A slide happens when section of soil or rock suddenly gives way and moves down a slope. The material moves as a single mass along a slippery zone.
  4. Creep: It is a very slow mass movement that goes on for years or even centuries.

Question 4.
What are the factors affecting mass movement?
Answer:
The factors responsible for mass movement are:
Relief and slope:

  1. There should be considerable elevation in an area for down slope movement to occur.
  2. Such movements will not occur in plain areas.
  3. Hilly, mountainous and plateau areas are more vulnerable to such movements.
  4. The steeper the slope, higher are the chances of occurrence of mass movements. Gentle slopes will have slow movements while steep slopes will have rapid movements.

Gravity:

  1. It is the main force responsible for mass movements.
  2. It is a force that acts everywhere on the surface of the earth, pulling everything down.

Water:

  1. Although water is not always directly involved as transporting medium but it plays an important role.
  2. Addition of water from rainfall or snowfall or melting of snow makes the material on the slope heavier.
  3. Water can reduce the friction along a sliding surface.

Weak material and structures:

  1. Some rocks are weaker than others. Rocks which have joints break easily.
  2. In particular, rocks containing calcium or clay minerals tend to have a low strength.
  3. Such areas will be more susceptible to mass wasting.

Maharashtra Board Class 11 Geography Important Questions Chapter 2 Weathering and Mass Wasting

Question 5.
Complete the following flow chart.
Maharashtra Board Class 11 Geography Important Questions Chapter 2 Weathering and Mass Wasting 4
Answer:
Maharashtra Board Class 11 Geography Important Questions Chapter 2 Weathering and Mass Wasting 5

Maharashtra Board Class 11 Geography Important Questions Chapter 1 Earth Movements

Balbharti Maharashtra State Board Class 11 Geography Important Questions Chapter 1 Earth Movements Important Questions and Answers.

Maharashtra State Board 11th Geography Important Questions Chapter 1 Earth Movements

Complete the chain.

Question 1.

A B C
(1) P-waves (1) Ring of Fire (1) Indonesia
(2) Volcanic eruption (2) Shadow zone (2) Italy
(3) Cinder cone (3) Smoke (3) Cauliflower cloud
(4) Tectonic movements (4) Mt. Nuovo (4) Cotopaxi
(5) Circum pacific belt (5) Convergent and divergent margin (5) Between 105° and 140°

Answer:

A B C
(1) P-waves (1) Shadow zone (1) Between 105° and 140°
(2) Volcanic eruption (2) Smoke (2) Cauliflower cloud
(3) Cinder cone (3) Mt. Nuovo (3) Italy
(4) Tectonic movements (4) Convergent and divergent margin (4) Indonesia
(5) Circum pacific belt (5) Ring of Fire (5) Cotopaxi

Maharashtra Board Class 11 Geography Important Questions Chapter 1 Earth Movements

Choose the correct option by identifying the correct correlation in the sentences.

Question 1.
Formation of mountains and distribution of continents are related to ……………
(a) sudden movements
(b) slow movements
(c) vertical movements
(d) horizontal movements
Answer:
(b) slow movements

Question 2.
A fold where one limb lies over the other in horizontal direction is ………………
(a) symmetrical fold
(b) overturned fold
(c) recumbent fold
(d) isoclinal fold
Answer:
(c) recumbent fold

Question 3.
Young fold mountains are of ………………… age.
(a) 10 to 25 million years
(b) over 200 million years
(c) 25 to 40 million years
(d) 30 to 45 million years
Answer:
(a) 10 to 25 million years

Question 4.
The tensional forces may cause subsidence in the central portion of the crust between two adjacent faults forming …………………
(a) rift valleys
(b) lava domes
(c) fold mountain
(d) volcanic mountain
Answer:
(a) rift valleys

Question 5.
P-waves can pass through ………………….
(a) only solid medium
(b) only liquid medium
(c) all mediums
(d) air
Answer:
(c) all mediums

Question 6.
When the lava solidifies in the volcanic neck, it forms …………….
(a) caldera
(b) cinder cone
(c) crater
(d) volcanic plug
Answer:
(d) volcanic plug

Maharashtra Board Class 11 Geography Important Questions Chapter 1 Earth Movements

Question 7.
When a portion of landslides down along the fault plane and when the exposed portion of the plane faces the sky ………….. is formed.
(a) reverse fault
(b) normal fault
(c) thrust fault
(d) tear fault
Answer:
(b) normal fault

Question 8.
Faulting results in the formation of ……………….
(a) volcanic mountains
(b) block mountains
(c) fold mountains
(d) volcanic plateaus
Answer:
(b) block mountains

Question 9.
A specific area where the seismic waves are not reported is called a ……………….
(a) shadow zone
(b) epicentre
(c) focus
(d) fracture zone
Answer:
(a) shadow zone

Question 10.
The describes the intensity of an earthquake based on its observed effects.
(a) Richter scale
(b) Mercalli scale
(c) seismogram
(d) seismograph
Answer:
(b) Mercalli scale

Question 11.
Sometimes, the lava material thrown into the air solidifies into small fragments before falling on the surface; it is called ………………
(a) crock fragments
(b) ash
(c) volcanic dust
(d) volcanic bombs
Answer:
(d) volcanic bombs

Question 12.
The …………….. are symmetrical in shape.
(a) cinder cone
(b) lava domes
(c) volcanic plateau
(d) composite cone
Answer:
(d) composite cone

Question 13.
Earthquakes are recorded with the help of ………………. instrument.
(a) seismogram
(b) seismograph
(c) Richter scale
(d) Mercalli scale
Answer:
(b) seismograph

Maharashtra Board Class 11 Geography Important Questions Chapter 1 Earth Movements

Identify the correct correlation.

Question 1.
A – The origin of Himalayas is due to the collision of the Indian subcontinent with Eurasian landmass.
R – The convergence of plates is often associated with mountain building.
(a) Only A is correct.
(b) Only R is correct.
(c) Both A and R are correct and R is the correct explanation of A.
(d) Both A and R are correct but R is not the correct explanation of A.
Answer:
(c) Both A and R are correct and R is the correct explanation of A.

Question 2.
A – The structure of the interior of the earth is decided with the help of seismic waves.
R – The S-waves travel in solid, liquid and gaseous medium while P-waves travel only through solid medium.
(a) Only A is correct.
(b) Only R is correct.
(c) Both A and R are correct and R is the correct explanation of A.
(d) Both A and R are correct but R is not the correct explanation of A.
Answer:
(a) Only A is correct.

Question 3.
A – Volcanic activity coincides with seismic activity of the world.
R – The seismic cone coincides with plate boundaries.
(a) Only A is correct.
(b) Only R is correct.
(c) Both A and R are correct and R is the correct explanation of A.
(d) Both A and R are correct but R is not the correct explanation of A.
Answer:
(d) Both A and R are correct but R is not the correct explanation of A.

Question 4.
A – Basic lava contains less percentage of silica.
R – It has high melting point.
(a) Only A is correct.
(b) Only R is correct.
(c) Both A and R are correct and R is the correct explanation of A.
(d) Both A and R are correct but R is not the correct explanation of A.
Answer:
(a) Only A is correct.

Identify the incorrect factor.

Question 1.
The types of fold are-
(a) symmetrical
(b) recumbent
(c) Isoclinal
(d) Reverse
Answer:
(d) Reverse

Maharashtra Board Class 11 Geography Important Questions Chapter 1 Earth Movements

Question 2.
The types of fault are-
(a) overturned
(b) reverse
(c) Thrust
(d) Tear
Answer:
(a) overturned

Question 3.
Causes of earthquake are-
(a) tectonic movements
(b) anthropogenic causes
(c) landslides
(d) Volcanicity
Answer:
(c) landslides

Question 4.
The solid material thro wn out of volcano consists of-
(a) soil
(b) volcanic bomb
(c) breccias
(d) Ash
Answer:
(a) soil

Question 5.
The material thrown out in formation of cinder cone are-
(a) ash
(b) cinder
(c) breccias
(d) Dust
Answer:
(d) Dust

Question 6.
Three major belts of earthquake and volcanoes are-
(a) Mid-Atlantic belt
(b) Mid-continental belt
(c) Circum-Pacific belt
(d) Indian Ocean belt
Answer:
(d) Indian Ocean belt

Question 7.
Three types of seismic waves are-
(a) P-waves
(b) S-waves
(c) F-waves
(d) longitudinal waves
Answer:
(c) F-waves

Question 8.
Examples of block mountain are-
(a) Black Forest Mountains in Germany
(b) Satpuras in Maharashtra
(c) Vosges in France
(d) Himalayas in India
Answer:
(d) Himalayas in India

Identify the correct group.

Question 1.

A B C D
(1) Mt. Fujiyama (1) Normal Fault (1) Rock Fragments (1) Crater Lake
(2) Mt. St. Helens (2) Reverse Fault (2) Volcanic Dust (2) Caldera
(3) Pinatubo (3) Tear Fault (3) Inflammable gases (3) Volcanic Plug
(4) Mayon (4) Overturned fold (4) Biotic Material (4) Volcanic Plateau

Answer:
C

Question 2.

A B C D
(1) Asymmetrical Fold (1) Normal Fault (1) Zone I – Very high intensity (1) Limb
(2) Overturned Fold (2) Thrust Fault (2) Zone II – High intensity (2) Anticline
(3) Isoseismal Lines (3) Reverse Fault (3) Zone III – Moderate intensity (3) Syncline
(4) Symmetrical Folds (4) Tear Fault (4) Zone IV – Low intensity (4) Trough

Answer:
C

Maharashtra Board Class 11 Geography Important Questions Chapter 1 Earth Movements

Give geographical reasons.

Question 1.
Faulting results in formation of block mountains.
Answer:
Faulting results in formation of block mountains because-

  1. Earth movements generate tensional forces that tend to pull the crust apart and faults are developed.
  2. If the block enclosed by the faults rises above or the land on either side subsides, the upstanding portion becomes the block mountain.
  3. The Black forest mountain in Germany, Vosges in France, Satpuras in Maharashtra are examples of block mountains.

Question 2.
The crust of the earth shakes.
Answer:
The crust of the earth shakes because –

  1. Movements occurring in the crust of the earth produce a lot of stress in the rock strata.
  2. When this stress accumulates beyond a limit, it tends to get released in the crust.
  3. Release of stress causes release of energy.
  4. The release of energy produces energy waves and this makes the earth’s crust to shake.

Question 3.
A zone between 105° and 140° from the epicentre is identified as the shadow zone for P and S waves.
Answer:
A zone between 105° and 140° from the epicentre is identified as the shadow zone for P and S waves because-

  1. Generally, seismographs located at any distance within 105° from the epicentre, record the arrival of both P-waves and S-waves.
  2. However, the seismographs located beyond 140° from the epicentre; record the arrival of P-waves but not that of S-waves.
  3. The shadow zone results from S-waves being stopped entirely by the liquid core and P-waves being bent (refracted) by the liquid core.

Question 4.
Volcanic eruptions can cause earthquakes.
Answer:
Volcanic eruptions can cause earthquakes because-

  1. Most of the earthquakes are along the edges of tectonic plates. This is where most volcanoes are too.
  2. Most earthquakes directly beneath a volcano are caused by the movement of magma.
  3. The magma exerts pressure on the rocks until it cracks the rock.
  4. Then the magma flows out into the cracks building pressure again. Every time the rock cracks it makes a small earthquake.

Maharashtra Board Class 11 Geography Important Questions Chapter 1 Earth Movements

Question 5.
Most of the volcanoes are found along plate boundaries.
Answer:
Most of the volcanoes are found along plate boundaries because-

  1. Volcanoes are common along convergent and divergent plate boundaries.
  2. They are also associated with meeting zones of continents and oceans.
  3. Volcanoes erupt because mantle rock melts, due to extremely high temperatures, and pressure.
  4. Along the subduction plate boundaries, the crust heats up as it sinks into the mantle, resulting into upwelling of lava.

Differentiate between.

Question 1.
Volcanoes and Earthquakes.
Answer:

Volcanoes Earthquakes
(i) Volcanoes form at the earth’s surface. (i) Earthquakes originate from deeper within the crust.
(ii) Volcanoes are formed by release of gases and magma. (ii) Earthquakes are caused by movement along a fault.
(iii) Volcanoes lead to formation of new landforms like domes, plateaus, cones, caldera etc (iii) Earthquakes simply cause waves which disturb rocks they do not form any landforms.
(iv) It directly produces ash and other volcanic debris. (iv) Earthquake events does not directly produce debris. But debris results by destruction due to the earthquake.
(v) It is possible to predict a volcanic eruption a few weeks to a few days in advance, though the exact time of eruption can’t be predicted with any accuracy. (v) The likelihood of an earthquake can be predicted but it is not possible to determine any time frame of when the earthquake will take place, or how likely it is to happen at some point in future.

Question 2.
Acid lava and Basic lava.
Answer:

Acid lava Basic lava
(i) It contains high percentage of silica. (i) It contains less percentage of silica.
(ii) It has got high melting point. (ii) It has low melting point.
(iii) It is thick, fluid and moves slowly. (iii) It is more fluid and can flow over long distance.

Mark and name the following on the outline map of the world with suitable index.

(1) Ring of Fire
(2) Indo-Australian plate
(3) Eurasian Plate
(4) Pacific Plate
(5) Mt. Cotopaxi
Answer:
Maharashtra Board Class 11 Geography Important Questions Chapter 1 Earth Movements 1

Answer the following Questions by using the given map

Maharashtra Board Class 11 Geography Important Questions Chapter 1 Earth Movements 2

Question 1.
Name the volcanic mountain in Africa.
Answer:
Kilimanjaro

Question 2.
Name the volcanic mountain in Pacific Ocean.
Answer:
Mauna Loa

Question 3.
Name the volcanic mountain in Japan.
Answer:
Fuji

Question 4.
Name volcanic mountain in Europe.
Answer:
Stromboli

Maharashtra Board Class 11 Geography Important Questions Chapter 1 Earth Movements

Question 5.
Name the volcanic mountain along the north eastern coastline of North America.
Answer:
St. Helens

Write short notes on.

Question 1.
Folding
Answer:

  1. Folding occurs when compressional forces are applied to rocks that are ductile or flexible.
  2. Rocks that lie deep within the crust and are therefore under high pressure are generally ductile and particularly susceptible to folding without breaking.
  3. As a result, rocks deep within the crust typically fold rather than break.
  4. Folding is also likely to occurs where compressional forces are applied slowly.
  5. Folding results into formation of fold mountains e.g., the Himalayas, the Alps etc.
  6. There are different parts of folds – limbs, axis of fold, Anticline, syncline, axial plane.
  7. There are different types of fold formed on basis of inclination of limbs – symmetrical fold, asymmetrical fold, overturned fold, recumbent fold and isoclinal fold.
  8. The nature of fold depends on several factors like, the nature of rocks, the nature and intensity of compressive forces, duration of the operation of compressive forces, etc.

Question 2.
Faulting:
Answer:

  1. In the earth’s crust, the forces operating in opposite direction lead to tension.
  2. As a result, rocks develop cracks or fissure.
  3. In the regions where cracks develop rocks get displaced, such displacement can occur in upward downward or horizontal direction.
  4. Faulting can be classified according to displacement of rocks.
  5. Rock layers that are near the earth’s surface and not under high confining pressures are too rigid to bend into folds.
  6. If the tectonics force is large enough, these rocks will break rather than bend. Such breaks may also be called fracture, ruptures or faults.
  7. Faulting results in formation of block mountains and rift valleys.
  8. There are different types of faults – normal faults, reverse faults, tear fault and thrust fault.

Question 3.
Types of faults
Answer:
A fault is a fracture in the crustal rock, formed when the crustal rocks are displaced due to tensional movement caused by endogenic forces. The displacement of rock can occur in upward, downward different or horizontal direction.
The types of faults determined by the direction of motion are:

  1. Normal fault: It results when portion of land slides down along the fault plane and when the exposed portion of the plane faces the sky.
  2. Reverse fault: It results when a portion of the land is thrown upward relative to other side of the land. In such situation, the fault plane faces the ground.
  3. Tear fault : At times, the rock strata on either side of the fault plane do not have vertical displacement. Instead, movement occurs along the plane in horizontal direction.
  4. Thrust fault: When the portion of the land on one side of the fault plane gets detached and moves over the land on the other side. The angle of fault plane is generally less than 45°.

Question 4.
Types of folds
Answer:
Folding occurs when compressional forces are applied to rocks, that are ductile or flexible.
The different types of folds are:
Symmetrical folds:

  1. The axial plane is vertical.
  2. The limbs are inclined at the same angle.

Asymmetrical folds:

  1. The axial plane is inclined.
  2. The limbs are inclined at different angles.

Overturned fold:

  1. One limb lies above the other limb.
  2. Limbs slope unequally in the same direction.

Recumbent fold:

  1. Axial plane is almost horizontal.
  2. One limb lies over the other in horizontal direction.

Isoclinal fold:

  1. The limbs slope in the same direction with same amount.
  2. The axial plane may be vertical, inclined or horizontal.
  3. Slope of some portion of limbs is near vertical.

Maharashtra Board Class 11 Geography Important Questions Chapter 1 Earth Movements

Question 5.
Volcanic landforms
Answer:
A number of landforms are formed due to cooling and solidification of magma. Some important landforms are as follows :
Lava Domes :

  1. Domes are developed when magma comes out and solidifies around its mouth.
  2. The shape of the dome depends upon the fluidity of lava.
  3. High dome with steep slope is developed by acidic lava.
  4. Due to basic lava, broad-based low domes are developed.

Lava Plateaus:

  1. Due to spread of lava in huge quantity from fissure volcano, it covers large areas and plateaus are formed.
  2. The Deccan Trap in India has developed from volcanic eruptions millions of years ago.

Caldera:

  1. At times, the eruption of volcano brings about large quantity of material and relieves lot of pressure.
  2. After the eruption, a large and deep depression remains in that area. This large depression is called caldera.
  3. These can be around 10 km wide and hundreds of metres deep.
  4. They may later turn into lakes. Smaller calderas are known as craters.

Crater Lake:

  1. When the funnel shaped crater of an extinct volcano gets filled with rain water, it forms a crater lake.

Volcanic Plug:

  1. It is formed when the lava solidifies in the volcanic neck.
  2. The surrounding rock material is removed by different agents of erosion; the vent stands predominantly. It is called as volcanic plug.
  3. The diameter of a plug varies between 300 and 600 metres.
  4. The Devil’s Tower in Wyoming, United State of America is the best example.

Cinder Cone:

  1. Solid material is ejected in large quantity.
  2. This material consists of ash, cinder and breccias.
  3. Cinders are half burnt pieces of solid material.
  4. The solid material is deposited around the mouth until a conical hill with steep slopes is formed.
  5. For example, cone of Mt. Nuovo in Italy.

Composite Cone:

  1. Composite cones are built up of alternate layers of lava and cinder.
  2. This cone is composed of two materials and therefore it is called a composite cone.
  3. It is symmetrical in shape.
  4. For example, Mt. St. Helens, USA.

Question 6.
Volcanic Materials
Answer:
There are three main types of material which come out in volcanic eruptions namely liquid, solid and gaseous form.
Liquid material:

  1. It is the molten rock material. When the molten rock material is below the earth’s surface, it is called ‘Magma’.
  2. When it appears on the surface it is called ‘lava’. On the basic of percentage of silica, it is classified as:

Acidic lava:

  1. It contains higher percentage of silica.
  2. It has high melting point.
  3. It is thick, fluid and moves slowly.

Basic lava:

  1. It contains less percentage of silica.
  2. It has low melting point.
  3. It’s thicker, fluid and can flow over longer distance.

Solid material:

  1. It consists of dust particles and rock fragments.
  2. When the material is very fine, it is called volcanic dust.
  3. The small sized solid particles are called ash.
  4. The solid angular fragments are known breccias.
  5. Sometimes, the lava material thrown into the air solidifies into small fragments before falling on the earth’s surface; it is called volcanic bombs.

Gaseous material:

  1. At the time of volcanic eruption, a dark cloud of smoke can be seen over the crater.
  2. On the basis of shape, cloud is called cauliflower cloud.
  3. Various inflammable gases are found in these clouds.
  4. These gases produce flames.

Question 7.
Shadow Cone
Answer:

  1. The waves which are caused by earthquake are called seismic waves. There are three types of waves: P-waves, S-waves and longitudinal waves.
  2. P- waves can pass through all the mediums while S-waves can pass only through solid medium.
  3. Even though P-waves pass through all mediums, they experience refraction as they pass from one medium to the other.
  4. However, there exist some specific areas where the waves of that earthquake are not reported. Such zones are called ‘shadow zones’.
  5. Generally, seismographs located at any distance within 105° from the epicentre, record both P-waves and S-waves. However, beyond 140° P-waves are recorded but not S-waves.
  6. Thus, the zone between 105° and 140° is identified as the shadow zone for both types of waves.
  7. The entire zone beyond 105° does not receive S-waves. The shadow zone of S-waves is much larger than that of P-waves.
  8. The shadow zone of P-waves appears as a band around the earth between 105° and 140° away from the epicentre.

Draw a neat and labelled diagram for

Question 1.
Parts of fold
Answer:
Maharashtra Board Class 11 Geography Important Questions Chapter 1 Earth Movements 3

Maharashtra Board Class 11 Geography Important Questions Chapter 1 Earth Movements

Question 2.
Block Mountain
Answer:
Maharashtra Board Class 11 Geography Important Questions Chapter 1 Earth Movements 4

Question 3.
Rift Valley
Answer:
Maharashtra Board Class 11 Geography Important Questions Chapter 1 Earth Movements 5

Answer in detail.

Question 1.
Give the distribution of earthquakes and volcanic belts in the world.
Answer:
There are three major belts or zones of earthquakes and volcanoes on the earth.
Circum-Pacific belt:

  1. This belt is also called as ‘Ring of Fire’ as it has maximum number of active volcanoes.
  2. It includes the volcanoes of eastern and western coastal areas of the Pacific Ocean.
  3. This belt contains most of the world-famous volcanic mountains and volcanic cones.
  4. The highest volcanic mountain of the world Cotopaxi is located in this belt.
  5. Fujiyama (Japan) Mt. St. Helens (Washington, USA) Pinatubo and Mayon (Philippines) are other significant volcanoes in this belt.

Mid-Atlantic belt:

  1. This belt covers the volcanoes mainly along the Mid-Atlantic ridge.
  2. The most active volcanic area in Mid-Atlantic ridge is Iceland.

Mid-Continental belt:

  1. This belt includes the volcanoes of Alpine mountain chains, the Mediterranean Sea and volcanoes of fault zone of Eastern Africa.
  2. The famous volcanoes in this belt are Stromboli and Mt. Etna.

Question 2.
Explain different types of folds.
Answer:
Folding occurs when compressional forces are applied to rocks which are ductile or flexible. The different types of folds are.
Symmetrical folds:

  1. The axial plane is vertical.
  2. The limbs are inclined at the same angle.

Asymmetrical folds:

  1. The axial plane is inclined.
  2. The limbs are inclined at different angles.

Overturned fold:

  1. One limb lies above the other limb.
  2. Limbs slope unequally in the same direction.

Recumbent fold:

  1. One Axial plane is almost horizontal.
  2. One limb lies over the other in horizontal direction.

Isoclinal fold:

  1. The limbs slope in the same direction with same amount.
  2. The axial plane may be vertical, inclined or horizontal.
  3. Slope of some portion of limbs is near vertical.

Maharashtra Board Class 11 Geography Important Questions Chapter 1 Earth Movements

Question 3.
Explain different types of slow movements.
Answer:
Slow movement can be classified as:
Vertical or Epeirogenic Movements:

  1. Slow movements keep on taking place either towards the centre of the earth or away from it towards the crust.
  2. Due to such movements, an extensive portion of the crust is either raised up or it subsides.
  3. When a portion of the crust is raised up above sea level, it leads to formation of continents. Hence, they are also called ‘continent-building movements.’
  4. Such movement can also form extensive plateaus.
  5. These movements are not related to development of tensions or pressure in the earth’s crust.

Horizontal or Orogenic Movements:

  1. These movements work in horizontal direction.
  2. These movements produce compression or tension in the rock strata.
  3. These movements lead to either folds or cracks in the surface of the earth. These movements give rise to mountains.
  4. Their speed is more than ‘continental-building’ movements.
  5. These movements either produce folds or faults. Consequently, either fold mountains or block mountains are formed.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 14 Biomolecules Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 14 Biomolecules

Question 1.
What are biomolecules? Give examples of biomolecules.
Answer:
Biomolecules: The lifeless, complex organic molecules which combine in a specific manner to produce life or control biological reactions are called biomolecules.

Examples: Carbohydrates, lipids (fats and oils), nucleic acids, enzymes.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 2.
What is the importance of biomolecules?
Answer:
Biomolecules are organic molecules which combine in a particular fashion to give complex substances which help to sustain life and produce identical daughter cells and play an important role in the actions of an organism.

  • Carbohydrates are the major constituents of food and source of energy.
  • Proteins help in proper functioning of living beings. They are important constituents of skin, hair, muscles. Enzymes which catalyse chemical reactions that take place in cells are proteins.
  • Lipids (fats and oils) function as the storehouses of energy.
  • Nucleic acids, the ribonucleic acid (RNA), and deoxyribonucleic acid (DNA) are responsible for genetic characteristics and synthesis of proteins.

Question 3.
What are carbohydrates?
OR
Define the term : Carbohydrates.
Answer:
Carbohydrates : Carbohydrates are optically active polyhydroxy aldehydes or polyhydroxy ketones, or the compounds which on hydrolysis produce polyhydroxy aldehydes or polyhydroxy ketones.

Examples : Glucose, sucrose, fructose.

Question 4.
What is monosaccharide?
Answer:
The basic unit of all carbohydrates which is a simple carbohydrate and cannot be hydrolysed further is known as monosaccharide. The monosaccharide is crystalline and soluble in water. E.g. Glucose, fructose, ribose.

Question 5.
Mention the names of monosaccharides or simple carbohydrates.
Answer:
Monosaccharides are (1) glucose (2) fructose (3) ribose.

Question 6.
State the basic unit of all carbohydrates.
Answer:
The basic unit of all carbohydrates which is a simple carbohydrate and cannot be hydrolysed further is known as monosaccharide.

Question 7.
How are carbohydrates classified?
OR
Classification of carbohydrates with examples.
Answer:
Carbohydrates are classified as monosaccharides oligosaccharides and polysaccharides.
(1) Monosaccharides : These carbohydrates cannot be further hydrolysed into smaller units. They are basic units of all carbohydrates, and are called monosaccharides.

Examples : Glucose, fructose, ribose

(2) Oligosaccharides : An oligosaccharide is a carbohydrate (sugar) which on hydrolysis gives two to ten monosaccharide units.
Depending on the number of monosaccharides produced on hydrolysis, oligosaccharides are further classified as :
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 1

Oligosaccharide is homogeneous. In this, each molecule of oligosaccharide contains the same number of monosaccharide units joined together in the same order as every other molecule of the same oligosaccharide.

(3) Polysaccharides : These are carbohydrates which on hydrolysis give a large number of monosaccharides.

Polysaccharides are tasteless, amorphous, insoluble in water. They are long chain, naturally αcurring polymers of carbohydrates.

Example : Cellulose, starch, glycogen.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 8.
Classify the following carbohydrates into Monosaccharide, Disaccharide, Oligosaccharide, Polysaccharide.
(1) Glucose
(2) Starch
(3) Sucrose
(4) Maltose
(5) Galactose
(6) Lactose
(7) Ribose.
Answer:

Carbohydrates Class
(1) Glucose Monosaccharide
(2) Starch Polysaccharide
(3) Sucrose Disaccharide
(4) Maltose Disaccharide
(5) Galactose Monosaccharide
(6) Lactose Disaccharide
(7) Ribose Monosaccharide

Question 9.
Classify the following carbohydrates.
(1) Cellulose,
(2) Maltose,
(3) Raffinose,
(4) Fructose.
Answer:

Carbohydrates Class
(1)     Cellulose

(2)     Maltose

(3)     Raffinose

(4)     Fructose

Polysaccharide

Disaccharide

Trisaccharide

Monosaccharide

Question 10.
Classify the following into monosaccharides, oligosaccharides and polysaccharides.
(1) Starch
(2) Glucose
(3) Stachyose
(4) Maltose
(5) Raffinose
(6) Cellulose
(7) Sucrose
(8) Lactose.
Answer:

Monosaccharides Glucose
Oligosaccharides Stachyose, maltose, raffinose, sucrose, lactose
Polysaccharides Starch, cellulose

Question 11.
Classify the following into monosaccharides and disaccharides.
Ribose, maltose, galactose, fructose and lactose (~2 mark each)
Answer:

Monosaccharides Ribose, galactose, fructose
Disaccharides Maltose, lactose

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 12.
Give the preparation of glucose from sucrose or cane sugar.
OR
Describe the laboratory method of preparation of glucose.
Answer:
Preparation of glucose from sucrose (cane sugar) : Laboratory method.

Glucose is prepared in the laboratory by hydrolysis of sucrose by boiling it with dilute hydrαhloric acid or dilute sulphuric acid for about two hours. On hydrolysis, sucrose gives one molecule of glucose and one molecule of fructose.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 5
Alcohol is added during cooling to separate glucose and fructose since, glucose is almost insoluble in alcohol, hence it crystallizes out first. Fructose remains in the solution as it is more soluble than glucose.

Crystals of glucose are separated out by filtration and purified by recrystallization.

Question 13.
Give the preparation of glucose from starch.
OR
How is glucose prepared on commercial scale?
Answer:
Commercially, on a large scale, glucose is prepared by hydrolysis of starch with dilute sulphuric acid. Starchy material is mixed with water and dilute sulphuric acid and heated at 393 K under 2 to 3-atm pressure. Starch is hydrolysed to give glucose.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 6

Question 14.
Explain the structure of glucose.
Answer:
Molecular formula of glucose is C6H12O6.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 7
Glucose has an aldohexose structure. In other words, glucose molecule contains one aldehydic, that is, formyl group and the remaining five carbons carry one hydroxyl group (-OH) each. The six carbons in glucose form one straight chain.

Question 15.
Describe the action of following reagents on glucose :
(1) HI
(2) Hydroxyl amine (NH2OH)
(3) Hydrogen cyanide
(4) Bromine water
(5) dil. Nitric acid
(6) Acetic anhydride.
Answer:
(1) Action of HI : Glucose on prolonged heating with HI gives n-hexane, indicates that all the six carbon atoms are linked in straight chain.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 8

(2) Action of hydroxyl amine : Glucose reacts with hydroxyl amine in an aqueous solution to form glucose oxime. This indicates the presence of CHO group in glucose.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 9

(3) Action of hydrogen cyanide : Glucose reacts with hydrogen cyanide to form glucose cyanohydrin.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 10

(4) Action of bromine water : Glucose on oxidation with mild oxidising agent like bromine water gives gluconic acid, which shows that the carbonyl group in glucose is aldehyde group.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 11

(5) Action of dll. nitric acid : Glucose on oxidation with dilute nitric acid forms dicarboxylic acid, saccharic acid. This indicates the presence of a primary alcoholic group (-CH2OH) in glucose.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 12

(6) Action of acetic anhydride : When glucose is heated with acetic anhydride in the presence of catalyst pyridine, glucose penta acetate is formed. It indicates that glucose is a stable compound and contains five hydroxyl groups.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 13

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 16.
Write Fischer projection formulae for
(1) Glucose
(2) Gluconic acid
(3) Saccharic acid.
Answer:
Fischer projection formulae of glucose, gluconic acid and saccharic acid :
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 17

Question 17.
Explain D and L configuration in sugars.
Answer:
The simplest carbohydrates glyceraldehyde is chosen as the standard, to assign D and L configuration to monosaccharides. Glyceraldehyde contains one asymmetric carbon atom and exist in two enantiomeric forms
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 18

The dextro entantiomer is represented as (+) glyceraldehyde and it is referred as D-configuration i.e., D-glyceraldehyde. The laevo enantiomer of glyceraldehyde is represented as ( -) glyceraldehyde and it corelated as L-configuration i.e., L-glyceraldehyde.

In Fischer projection formula, a monosaccharide is assigned D-configuration if the (- OH) hydroxyl group at the last chiral carbon and lies towards right hand side. On the other hand it is assigned L-configuration if the – OH group on the last chiral carbon atom and lies on the left hand side. In monosaccharides, the most oxidised carbon (i.e., -CHO) is at the top.

Examples :
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 19

Question 18.
Write Fischer projection formulae for (a) L-( + )-erythrose (b) L-( +) ribulose.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 23

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 19.
Is the following sugar, D-sugar or L-sugar?
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 24
Answer:
The compound is L-sugar.
The compound is L-sugar.

Question 20.
Assign D/L configuration to the following monosaccharides.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 25
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 26
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 27

Question 21.
Explain ring structure of glucose.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 28
Glucose has two cyclic structures (II and III) which are in equilibrium with each other through the open chain structure (I) in aqueous solution.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

The ring structure of glucose is formed by reaction between the formyl ( – CHO) group and the alcoholic (- OH) group at C – 5. Thus, the ring structure is called a hemiacetal. The two hemiacetal structures (II and III) differ only in the configuration of C – I (Fig.), the additional chiral centre resulting from ring closure. The two ring structures are called α- and β- anomers of glucose and C-l is called the anomeric carbon. The ring of the cyclic structure of glucose contains five carbons and one oxygen. Thus, it is a six membered ring. It is called pyranose structure, in analogy with the six membered heterαyclic compound pyran (IV). Hence glucose is also called glucopyranose.

Question 22.
Write the structures of α-D-( + )-glucopyranose and β-D-( +) glucopyranose.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 29

Question 23.
Explain Haworth formula of glycopyranose.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 30
In the Haworth formula the pyranose ring is considered to be in a perpendicular plane with respect to the plane of paper. The carbons and oxygen in the ring are in the places as they appear in figure. The lower side of the ring is called α-side and the upper side is the β-side. The α-anomer has its anomeric hydroxyl (- OH) group (at C-l) on the α-side, whereas the β-anomer has its anomeric hydroxyl (- OH) group (at C-l) on the β-side. The groups which appear on right side in the Fischer projection formula appear on α-side in the Haworth formula, and the groups which appear on left side in the fischer projection formula appear on a β-side in the Haworth formula.

Question 24.
Explain the structure of fructose.
Answer:
Fructose has molecular formula C6H12O6. It contains ketonic functional group at carbon number 2 and six carbon atoms in straight chain. It belongs to D-series and is a laevo rotatory compound. It is written as D-( – )-fructose. Being an α-hydroxy keto compound fructose is a reducing sugar.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 31

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 25.
Draw mirror images of glucose and fructose.
Answer:
(1) Glucose
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 32
(2) Fructose
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 33

Question 26.
Write the two cyclic structures of α-D-( – )-fructofuranose and β-D-( – )-fructofuranose exist in equilibrium with open chain structure.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 34

Question 27.
Write the Haworth projection formulae for α -D-( -) – Fructofuranose and β – D – ( -) – Fructo- furanose.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 35

Question 28.
Explain the structure of sucrose.
Answer:
Sucrose is a hexasaccharide and has molecular formula C12H22O11. The structure of shcrose contains glycosidic linkage between C – 1 of α-glucose and C – 2 of β-fructose. Since aldehyde and ketone groups of both monosaccharide units are involved in the formation of glycosidic bond, sucrose is a nonreducing sugar.

Sucrose is dextrorotatory, on hydrolysis with dilute acid or an enzyme invertase gives equimolar mixture of dextrorotatory glucose and laevorotatory fructose.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 36
The solution is laevorotatory because laevo rotation of fructose (- 92.4°) is more than dextrorotation of glucose ( + 52.50), hence the sign of rotation is changed from (+) to (-) after hydrolysis, the product is called invert sugar.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 37

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 29.
Explain the structure of maltose.
Answer:
Maltose is another disaccharide obtained by partial hydrolysis of starch or made of two units of D-glucose. In maltose, C-l of one α-D-glucose is linked to C-4 of another α-D-glucose molecule by glycosidic linkage. The glucose ring which uses its hydroxyl group at C-1 is α – 1 → 4 glycosidic linkage. It is a reducing sugar because a free aldehyde group can be produced at C1 of second glucose molecule. Maltose on hydrolysis with dilute acids gives glucose.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 38

Question 30.
Draw a neat diagram for Haworth formula of maltose.

Question 31.
Explain the structure of lactose.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 39
Lactose (C12H22O11) is a disaccharide. It is found in milk, therefore, it is also known as milk sugar. It is formed from two monosaccharide units, namely D – galactose and D – glucose. The glycosidic linkage is formed between C-l of β-D-galactose and C -4 of glucose. Therefore the linkage in lactose is called β – 1,4 – glycosidic linkage. The hemiacetal group at C-l of the glucose unit is not involved in glycosidic linkage but is free. Hence lactose is a reducing sugar. The above figure shows Haworth formula of lactose.

Question 32.
What are the hydrolysis products of (1) lactose (2) sucrose?
Answer:
(1) Lactose on hydrolysis in presence of an acid or enzyme lactase gives one molecule each of glucose and galactose
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 40
(2) Sucrose on hydrolysis in the presence of dii. acid or the enzyme invertase gives one molecule each of glucose and fructose.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 41

Question 33.
Explain the structure of starch.
Answer:
Starch is found in cereal grains, roots, tubers, potatoes, etc. It is a polymer of α-D-glucose and consists of two components, amylose and amylopectin.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Amylose is water soluble component forms blue coloured complex with iodine. It constitutes about 20 % of starch. Amylose contains 200 to 1000 α-D-glucose units linked together by glycosidic linkage between C-l of one unit and C-4 of another unit. i.e. α-1, 4 glycosidic linkages.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 43
Amylopectin is insoluble in water and constitutes about 80 % starch which forms blue-violet coloured complex with iodine. It is a branched chain polymer. In amylopectin, α-D-glucose molecules are linked together by glycosidic linkage between C1 – of one unit and C-4 of another unit to form long chain and branching αcurs by glycosidic linkage between C-l and C6 glycosidic linkage.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 44

Question 34.
What are polysaccharides?
Answer:
A large number of same or different monosaccharides are joined together by glycosidic linkages are called polysaccharides. They have general formula (C6H10O5)n.

Question 35.
Explain the structure of cellulose.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 45
Cellulose mainly αcurs in plants. Cell wall of plant cells is made up of cellulose. It is a long chain polymer. In cellulose, β-D-glucose units are linked by glycosidic linkage between C1-of one unit of glucose and C4 of another glucose unit. Thus cellulose contains 1 → 4β glycosidic linkages like those in cellobiose.

Question 36.
Explain the structure of glycogen.
Answer:
The glucose is stored in animal body in the form of glycogen. It is also known as animal starch because its structure is similar to amylopectin. Glycogen is highly branched. Whenever the body is required glucose, enzymes breaks the glycogen to glucose.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 37.
How is glycogen different from starch?
Answer:
Starch is the main storage molecules of plants whereas glycogen is the main storage molecule of animals. Starch is found in cereals, roots, tubers, etc. Glycogen is present in liver, muscles and brain.

Question 38.
What do you understand by the term glycosidic linkage?
Answer:
The linkage between two monosaccharide units through oxygen atom is called glycosidic linkage.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 46

Question 39.
What is the basic structural difference between starch and cellulose?
Answer:
Starch is a polymer of a-glucose and consists of two components-amylose and amylopectin. In amylose α-D-D-( + )-glucose units held by C,-C4 glycosidic linkage and in amylopectin, α-D-glucose units held by C1-C4 glycosidic linkage whereas branching αcurs by C1-C6 glycosidic linkage. [Refer Question 35 (i) (ii) Fig.] Cellulose is a straight chain polysaccharide composed only of β-D-glucose units held by C1-C4 glycosidic linkage. (Refer Question 37 Fig.)

Question 40.
Define the term : Protein OR What are proteins?
Answer:
Chemically proteins are polyamides which are high molecular weight polymers of the monomer units i.e. α-amino acids. OR It can also be defined as Proteins are the biopolymers of a large number of a-amino acids and they are naturally occurring polymeric nitrogenous organic compounds containing 16% nitrogen and peptide linkages (-CO-NH-).

Question 41.
Write the common sources of protein.
Answer:
Common sources of proteins are milk, pulses, peanuts, eggs, fishes, cheese, cereals, etc. They are also the principal materials of muscle, nerves, tendons, skin, blood, enzymes, many hormones and antibiotics.

Question 42.
What are the products of hydrolysis of proteins?
Answer:
On hydrolysis, proteins give a mixture of α-anlino acids.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 47
The α-carbon in α-amino acids ohtained by hydrolysis of proteins has ‘L’ configuration.

Question 43.
What are the a-amino acids?
Answer:
α-Amino acids are carboxylic acids having an amino (- NH2) group bonded to the α-carbon, i.e. the carbon next to the carboxyl (- COOH) group.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 48
α-amino acids are derivatives of carboxylic acids, obtained by replacing – H atom by amino group. They are bifunctional compounds containing acidic Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 49 and basic – NH2 groups.
Example : Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 50 (where R is an alkyl group or aryl group).

The amino acids are colourless, crystalline, water soluble, high melting solids. These acids in their aqueous solutions behave like salts due to presence of both acidic, Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 51 and basic. (- NH2) groups in the same molecule.

Such a doubly charged ion is known as zwitter ion. Example : Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 52
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 53

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 44.
What are the final products of hydrolysis of proteins?
Answer:
Proteins on hydrolysis with dilute solution of acids, alkalies or enzymes give a mixture of large number of a-amino acids as final products.

For example :
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 54

Question 45.
Write the classification of amino acids, giving examples.
Answer:
The amino acids are of three types : acidic, basic and neutral. The symbol ‘R’ in the structure of a-amino acids represents side chain and may contain additional functional groups.

(1) Acidic amino acids : If ‘R’ contains a carboxyl (- COOH) group the amino acid is acidic amino acid, i.e. If carboxyl groups are more in number than amino groups, then amino acids are acidic in nature.

Examples : Glutamic acid HOOC-CH2-CH2-; Aspartic acid HOO-CH2

(2) Basic amino acids : If ‘R’ contains an amino (1°, 2°, or 3°) group, it is called basic amino acid i.e. If amino groups are more in number than carboxyl groups then amino acids are basic in nature.

Examples : Arginine Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 55

(3) Neutral amino acids : The other amino acids having neutral or no functional group in ‘R’ are called neutral amino acids, i.e. The amino acids having equal number of amino and carboxyl groups are neutral amino acids.

Examples : Alanine CH3-; Valine (CH3)2-CH

Question 46.
What are essential and non-essential amino acids? Give two examples of each.
Answer:
The amino acids, which cannot be synthesised in the body and are supplied through diet are called essential amino acids. Examples : Lysine H2N-(CH2)4-; Valine (CH3)2CH- The amino acids which are synthesized in the body are called non-essential amino acids.

Examples : Glutamic acid HOO-CH2-CH2-; Serine HO-CH2

Question 47.
What is meant by Zwitter ion?
Answer:
An a-amino acid molecule contains both acidic carboxyl ( – COOH) group as well as basic amino (- NH2) group. Proton transfer from acidic group to basic group of amino acid forms a salt, which is a dipolar ion called a zwitterion.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 61

Question 48.
Draw zwitter ion of alanine and other two forms.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 62

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 49.
What is a peptide bond (peptide linkage)?
OR
Define peptide bond.
Answer:
Proteins are the polymers of a-amino acids and they are connected to each other. The bond that connects a-amino acids to each other is called peptide bond (peptide linkage, – CONH -).

Question 50.
How is peptide linkage (dipeptide linkage) formed in proteins? How is tripeptide formed?
Answer:
Peptide linkage is formed by condensation of acidic Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 63 group of one molecule of a-amino acid and basic -NH2 group of other molecule of α-amino acid with elimination of water.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 64

When one more molecule of amino acid combines with dipeptide, it forms tripeptide.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 65
Thus, it forms tetra, penta and finally a polypeptide chain i.e. proteins. Hence, proteins are basically polypeptides.

Question 51.
Write the structures of all possible dipeptides which can be obtained from glycine and alanine.
Answer:
(1) Dipeptide from glycine :
Carboxylic group of glycine reacting with amino group another molecule of glycine to form dipeptide
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 73
(2) Dipeptide from alanine :
Carboxylic goup of alanine reaction with amino goup of another molecule of alamine to form dipeptide
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 74
(3) Dipeptide from glycine and alanine :
Carboxylic group of glycine reacting with amino group another molecule of alanine to form dipeptide
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 75

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 52.
How are proteins classified on the basis of molecular shapes?
Answer:
On the basis of their molecular shapes proteins are classified as :
(1) Fibrous proteins : The proteins in which the polypeptide chains lie parallel (side by side) to form fibre-like structure, are called fibrous proteins. The polypeptide chains held together by hydrogen bonds. These proteins are insoluble in water.

The fibrous proteins are tough and insoluble in water, and dilute acids or bases.

Example : myαin (in muscles), keratin (in hair, nails, skin), fibroin (in silk), collagen (in tendons), etc.

(2) Globular proteins : The proteins have spherical shape. This shape results from coiling around of the polypeptide chain of protein, and have intramolecular hydrogen bonding are called globular proteins.

They are soluble in water and dilute acids or bases.

Example : Haemoglobin (in blood), albumin (in eggs), insulin (in pancreas), etc.

Question 53.
Distinguish between globular and fibrous proteins.
Answer:

Globular proteins Fibrous proteins
(1) The chains of polypeptides of protein coil around to give a spherical shape.
(2) Globular proteins are soluble in water.
(3) They are sensitive to small changes of temperature and pH.
(4) They possess biological activity.
(1) The proteins in which the polypeptide chains lie parallel to form fibre like structure.
(2) Fibrous proteins are insoluble in water.
(3) They are stable to moderate changes of temperature and pH.
(4) They do not possess biological activity.

Question 54.
Draw a neat labelled diagram for the secondary structure of protein.
Answer:
Secondary structure of proteins : The three-dimensional arrangement of lαalized regions of a long polypeptide chain is called the secondary structure of protein. Hydrogen bonding between N-H proton of one amide linkage and C = O oxygen of another gives rise to the secondary structure. There are two different types of secondary structures i.e. α-helix and β-pleated sheet.

α-Helix : In a-helix structure, a polypeptide chain gets coiled by twisting into a right handed or clαkwise spiral known as a-helixn. The characteristic features of α-helical structure of protein are :
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 78
(1) Each turn of the helix has 3.6 amino acids.
(2) A C = O group of one amino acid is hydrogen bonded to N – H group of the fourth amino acid along the chain.
(3) Hydrogen bonds are parallel to the axis of helix while R groups extend outward from the helix core.
Myosin in muscle and a-keratin in hair are proteins with almost entire a-helical secondary structure.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

β-Pleated sheet : In secondary structure, when two or more polypeptide chains (strands) line up side-by-side is called β-pleated sheets. The β-picate sheet structure of protein consists of extended strands of polypeptide chains held together by intermolecular hydrogen bonding. The characteristics of β-pleated sheet structure are :
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 79

  • The C = O and N – H bonds lie in the planes of the sheet.
  • Hydrogen bonding occurs between the N – H and C = O groups of nearby amino acid residues in the neighbouring chains.
  • The R groups are oriented above and below the plane of the sheet.

The β-pleated sheet arrangement is favoured by amino acids with small R groups.

Question 55.
What is denaturation of proteins? How is denaturation brought about?
OR
What is the effect of denaturation on the structure of proteins?
Answer:
The prαess by which the molecular shape of protein changes without breaking the amide / peptide bonds that form the primary structure is called denaturation. OR Proteins gets easily precipitated. It is an irreversible change and the prαess is called denaturation of proteins.

Denaturation uncoils the protein and destroys the shape and thus loses their characteristic biological activity. Denaturation is brought about by heating the protein with alcohol, concentrated inorganic acids or by salts of heavy metals. During denaturation secondary and tertiary and quternary structures are destroyed but primary structure remains intact.

Example : Boiling of egg to coagulate egg white, conversion of milk to curd.

Question 56.
Define : Enzymes
Answer:
All biological reactions are catalysed by bio-catalyst in living organisms called enzymes.

Question 57.
What are enzymes? Explain with suitable example.
Answer:
All biological or bio-catalysts which catalyse the reactions in living organisms are called enzymes. Chemically all enzymes are proteins. They are required in very small quantities as they are catalyst also they reduce the activation energy for a particular reaction.

Example : Enzyme maltase converts maltose to glucose.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 84

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 58.
Explain the catalytic action of enzymes.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 85
Answer:
Mechanism of enzyme catalysis : Action of an enzyme on a substrate is known as lock-and-key mechanism.

Accordingly, the enzyme has active site on its surface. A substrate molecule can attach to this active site only if it has the right size and shape. Once in the active site, the substrate is held in the correct orientation, enzymes provide functional group which will attack the substrate and forms the products of reaction. The products leave the active site and the enzyme is ready to act as catalyst again.

Question 59.
Give examples of industrial application of enzyme catalysis.
Answer:

  • Glucose Isomerase (enzyme) is used in conversion of glucose to sweet-tasting fructose.
  • New antibiotics are manufactured using penicillin acylase (enzyme).
  • Laundry detergentts are manufactured using proteases (enzyme).
  • Esters used in cosmetics are manufactured using genetically engineered enzyme.

Question 60.
Draw a neat diagram for enzyme catalysis.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 85

Question 61.
State the main functions of enzymes.
Answer:
Enzymes are biological catalyst and they are highly specific in nature. The two main functions are as follows :
(1) They lower the requirement of activation energy.
(2) They speed up the rate of reaction.
E.g. Enzyme maltase catalyses maltose to glucose.
\(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_{2} \mathrm{O} \stackrel{\text { Maltase }}{\longrightarrow} 2 \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 62.
What are nucleic acids?
Answer:
Nucleic acids are unbranched polymers of repeating monomers i.e. nucleotides. In other words, nucleic acids have a polynucleotide structure which in turn consists of a base, a pentose sugar and phosphate moiety.

Nucleic acids are biomolecules which are found in the nuclei of all living cells in the form of nucleoproteins or chromosomes.

(Nucleoproteins = Proteins + Nucleic acid)
(prosthetic group)

Question 63.
State the types of nucleic acids.
Answer:
The types of nucleic acids are : Ribonucleic acid (RNA) and deoxy ribonucleic acid (DNA). DNA molecules contain several million nucleotides while RNA molecules contain a few thousand nucleotides.

Question 64.
Explain chemical composition of nucleic acids.
Answer:
Nucleic acids have a polynucleotide structure. Nucleic acids (RNA and DNA) consists of three components :
(1) monosaccharide (sugar)
(2) nitrogen containing base and
(3) phosphate group.

(1) Monosaccharides : Nucleotides of both RNA consist of five membered monosaccharide ring (furanose), called as simply sugar component.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 86
In RNA, the sugar component of nucleotide units is D-ribose and in DNA 2-deoxy-D-ribose.
2 – deoxy means no – OH group at C2 position.

(2) Nitrogen containing base : Total five nitrogen – containing bases are present in nucleic acids. Three bases with one ring (cytosine, uracil and thymine) are derived from the parent compound pyrimidine. Two bases with two rings (adenine and guanine) are derived from the parent compound purine. Each base in designated by a one-letter symbol. Uracil (U) αcurs only in RNA while thymine (T) ocurs only in DNA.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 87

(3) Phosphate group : The sugar units are joined to phosphate through C3 and C5 hydroxyl groups.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 65.
What is meant by nucleosides?
OR
Write the structure of nucleoside. Give examples.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 89
A nucleoside contains two basic components of nucleic acids i.e. a pentose sugar and a nitrogenous base.

A nucleoside is formed when 1 -position of a pyrimidine (cytosine, thymine or uracil) or 9-position of guanine or adenine base is attached to C- l of sugar by β-linkage.

Examples: Formation of nucleoside:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 90

Question 66.
What is meant by nucleotide?
OR
Write the structure of nucleotide. Give example.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 91
A nucleotide contains all three basic components of nucleic acids i.e., a pentose sugar, a phosphoric acid and a nitrogenous base. These are obtained by esterification of \(\mathrm{C}_{5}^{1}-\mathrm{OH}\) group of the pentose sugar by phosphoric acid. Nucleotides are joined together through phosphate ester linkage. Thus, nucleotides are monophosphates of nucleosides. Abridged names of some nucleotides are AMP, dAMP, UMP, dTMP and so on. Here, the first capital letter is derived from the corresponding base. MP stands for monophosphate. Small letter ‘d’ in the beginning indicates deoxyribose in the nucleotide.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Example :
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 92

 

Question 67.
Write the structure of nucleic acids.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 95
Answer:
Nucleic acids, both DNA and RNA, are polymers of nucleotides, formed by joining the 3′ – OH group of one nucleotide with 5′ – phosphate of another nucleotide. Two ends of polynucleotide chain are distinct from each other. One end having free phosphate group of 5′ position is called 5′ end. The other end is 3′ end and has free OH – group at 3′ position.

Question 68.
Draw a schematic representation of polynucleotide structure of nucleic acids.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 96

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 69.
Explain double helix.
OR
State the salient features of the Watson and Crick mode of DNA.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 98
The Salient features are :

  1. DNA is made of two polynucleotide strands that wind into a right-handed double helix.
  2. The two strands run in opposite directions: one from the Y end to the 3’ end, while the other from the 3’ end to the Y end.
  3. Pcrpcndicular to the axis of the helix, the sugar – phosphate backbone lies on the outside of the helix and the bases lic on the inside.
  4. The hydrogen bonding between the hases of the two DNA strands stabilizes the double helix. This gives rise to a ladder-like structure of DNA double helix.
  5. Adenine always forms two hydrogen bonds with thymine and guanine forms three hydrogen bonds with cytosinc. Thus A – T arid C – G arc complementary hase pairs and the Two strands of the double helix arc complementary to each other.

Question 70.
Give scientific reasons :
1. In the preparation of glucose from sucrose, ethyl alcohol is added at the time of cooling.
Answer:
Hydrolysis of sucrose with dilute hydrαhloric acid gives glucose along with fructose.

Ethyl alcohol is added at the time of cooling in the preparation of glucose, to separate glucose from fructose. Glucose being insoluble in alcohol, crystallizes out first, while fructose being more soluble in alcohol, remains in the solution.

Question 71.
Answer in one sentence :

(1) How is glucose stored in the animal body?
Answer:
Glucose is stored in the form of glycogen in the animal body.

(2) Write other term used for carbohydrates.
Answer:
Carbohydrates are often termed as saccharides or sugars.

(3) How many moles of acetic anhydride will be required to form glucose penta acetate from 1 mole of glucose?
Answer:
10 moles of acetic anhydride.

(4) What are reducing sugars?
Answer:
Reducing sugars : Carbohydrates which reduce Fehling solution to red ppt of Cu20 or Tollen’s reagent to shining metallic silver are called reducing sugars. All monosaccharides and oligosaccharides except sucrose are reducing sugars.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

(5) What are non-reducing sugars?
Answer:
Non-reducing sugars : Carbohydrates which do not reduce Fehling solution and Tollen’s reagent are called non-reducing sugars. E.g. sucrose.

(6) Give an example each of reducing and non-reducing sugars.
Answer:
Reducing sugars : Maltose or lactose
Non-reducing sugars : Sucrose.

(7) Name the linkage which joins two monosaccharide units through oxygen atom.
Answer:
The linkage which joins two monosaccharide units through oxygen atom is called glycosidic linkage.

(8) Name the sugar present in DNA.
Answer:
The sugar present in DNA is deoxyribose.

(9) A nucleotide from DNA containing thymine is hydrolysed. What are the products formed?
Answer:
When nucleotide from DNA containing thymine is hydrolysed, 2-deoxy-D-ribose, thymine and phosphoric acid is obtained.

(10) How is zwitterion formed?
Answer:
In aqueous solution, the carboxyl group loses a proton while the amino group accepts it, as a result, a dipolar or zwitter ion is formed.

(11) Name the amino acids which are synthesized in the body.
Answer:
The amino acids which are synthesized in the body are called non-essential amino acids. Examples : Glutamic acid, serine.

(12) Name the four bases present in DNA which of these is not present in RNA.
Answer:
Purines-adenine (A) and guanine (G); Pyrimidines-thymine (T) and cytosine (C), these four bases are present in DNA. Out of these, thymine (T) is not present in RNA.

(13) What are different types of RNA which are found in the cell?
Answer:
There are three different types of RNA found in the cell. (1) The messenger RNA which carries the message to the ribosome (2) Ribosomal RNA where synthesis of protein takes place (3) The transport RNA.

(14) State the functions of RNA and DNA.
Answer:
RNA and DNA are responsible for generic characteristics : DNA preserves the information and uses it by producing duplicate identical DNA molecules. RNA carries messages and transports them.

Multiple Choice Questions

Question 72.
Select and write the most appropriate answer from the given alternatives for each subquestion :

1. Which of the following is not sugar?
(a) Sucrose
(b) Starch
(c) Fructose
(d) Glucose
Answer:
(b) Starch

2. Which of the following is the example of disaccharide?
(a) Glucose
(b) Raffinose
(c) Cellulose
(d) Sucrose
Answer:
(d) Sucrose

3. Fructose is
(a) aldopentose
(b) aldohexose
(c) ketopentose
(d) ketohexose
Answer:
(d) ketohexose

4. Oxidation product of glucose with bromine water is
(a) sorbitol
(b) gluconic acid
(c) glutamic acid
(d) saccharic acid
Answer:
(b) gluconic acid

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

5. The general formula of carbohydrates is
(a) C(H2O)
(b) Cx(H2O)y
(c) Cx(H2O)
(d) Cx(H2O)x
Answer:
(b) Cx(H2O)y

6. Monosaccharides containing aldehyde group are called
(a) aldoses
(b) ketoses
(c) polysaccharides
(d) disaccharides
Answer:
(a) aldoses

7. Which of the following sugars can be used to prepare glucose on a large scale?
(a) Cellulose
(b) Cane sugar
(c) Galactose
(d) Starch
Answer:
(d) Starch

8. Which of the following carbohydrates cannot undergo hydrolysis?
(a) Glucose
(b) Sucrose
(c) Cellulose
(d) Maltose
Answer:
(a) Glucose

9. Glucose differs from fructose in
(a) the functional group
(b) the number of chiral carbon atoms
(c) the number of carbon atoms
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

10. The example of aldopentose is
(a) arabinose
(b) glucose
(c) fructose
(d) sucrose
Answer:
(a) arabinose

11. Dextrose, grape sugar and blood sugar αcurs in
(a) fructose
(b) glucose
(c) sucrose
(d) starch
Answer:
(b) glucose

12. The example of ketopentose is
(a) galactose
(b) ribose
(c) raffinose
(d) maltose
Answer:
(b) ribose

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

13. Cane sugar on hydrolysis gives
(a) glucose and maltose
(b) glucose and lactose
(c) glucose and fructose
(d) only glucose
Answer:
(c) glucose and fructose

14. On commerical scale, glucose is prepared from
(a) starch
(b) potato pulp
(c) sucrose
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

15. The number of monosaccharide units formed on hydrolysis of glucose are
(a) zero
(b) one
(c) two
(d) three
Answer:
(a) zero

16. Which of the following is NOT TRUE about glucose?
(a) It is monosaccharide
(b) It is a polyhydroxy aldehyde
(c) It is polyhydroxy ketone
(d) It contains six carbon atoms
Answer:
(c) It is polyhydroxy ketone

17. Final hydrolysis product of simple protein is
(a) carboxylic acid
(b) α-amino acid
(c) mineral acid
(d) acetic acid
Answer:
(b) α-amino acid

18. Haemoglobin is the example of-
(a) simple protein
(b) derived protein
(c) fibrous protein
(d) conjugated protein
Answer:
(d) conjugated protein

19. Protein are also called
(a) polysaccharides
(b) polypeptides
(c) polyglycerides
(d) polyster
Answer:
(b) polypeptides

20. The simplest amino acid is
(a) glycine
(b) oxalic acid
(c) adipic acid
(d) caprolactam
Answer:
(a) glycine

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

21. Amino acids usually exist in the form of zwitter ion which consist of
(a) the basic group-NH2 and the acidic group -COOH
(b) the acidic group -N+H3 and the basic group COO
(c) the acidic group -COO+ and the acidic group NH3-
(d) acidic or basic group
Answer:
(b) the acidic group -N+H3 and the basic group COO-

22. The water insoluble protein is
(a) casein of milk
(b) albumin
(c) serum albumin
(d) keratin of hair
Answer:
(d) keratin of hair

23. The main structural feature of a protein molecule is the presence of
(a) an ester linkage
(b) an ether linkage
(c) a peptide linkage
(d) all of these
Answer:
(c) a peptide linkage

24. Milk sugar is
(a) sucrose
(b) lactose
(c) maltose
(d) glucose
Answer:
(b) lactose

25. The carbohydrates used for silvering of mirror is
(a) fructose
(b) starch
(c) glucose
(d) cellulose
Answer:
(c) glucose

26. Which one of the following is NOT produced by human body?
(a) DNA
(b) Hormones
(c) Enzymes
(d) Vitamins
Answer:
(c) Enzymes

27. A biological catalyst is essentially
(a) an amino acid
(b) an enzyme
(c) a nitrogen molecule
(d) a carbohydrate
Answer:
(d) a carbohydrate

28. Which one of the following is not a constituent of RNA?
(a) Ribose
(b) Uracil
(c) Thymine
(d) Phosphate
Answer:
(b) Uracil

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

29. DNA is a polymer of units of
(a) sugars
(b) ribose
(c) amino acids
(d) nucleotides
Answer:
(c) amino acids

30. Which one of the following molecules will form zwitter ion?
(a) CH3COOH
(b) CH3CH2NH2
(c) CCl3NO2
(d) NH2CH2COOH
Answer:
(d) NH2CH2COOH

31. In metabolic prαess the maximum energy is given by
(a) carbohydrates
(b) proteins
(c) vitamins
(d) fats
Answer:
(d) fats

32. DNA has a structure of helix was reported by
(a) Herman Fischer
(b) Fedrick Sauger
(c) Andreas Marggraf
(d) Watson and Crick
Answer:
(d) Watson and Crick

33. The secondary structure of a protein is determined by
(a) co-ordinate bond
(b) covalent bond
(c) ionic bond
(d) hydrogen bond
Answer:
(d) hydrogen bond

34. In maltose, glycosidic linkage is present between the two glucose units at positions
(a) 1, 2
(b) 1, 1
(c) 1, 3
(d) 1, 4
Answer:
(d) 1, 4

35. Which of the following amino acids is basic in nature?
(a) Valine
(b) Tyrosine
(c) Arginine
(d) Luecine
Answer:
(c) Arginine

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

36. Sucrose molecules consists of
(a) a glucofuranose and a fructopyranose
(b) a glucofuranose and a fructofuranose
(c) a glucopyranose and a fructopyranose
(d) a glucopyranose and a’ fructofuranose
Answer:
(d) a glucopyranose and a’ fructofuranose

37. Which one of the following statements is not correct about DNA molecule?
(a) It has double helix structure
(b) It serves as hereditary material
(c) The two DNA strands are exactly similar
(d) Its replication is called semi-conservative mode of replication
Answer:
(c) The two DNA strands are exactly similar

38. Glycine on heating forms
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 107
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 108
Answer:
(a)

39. Acidic amino acid is
(a) Glutamine
(b) Glutamic acid
(c) Tyrosine
(d) Lysine
Answer:
(b) Glutamic acid

40. Basic amino acid is
(a) Lysine
(b) Glycine
(c) Cystine
(d) Alanine
Answer:
(a) Lysine

41. Precipitation of protein is referred to as
(a) destruction of proteins
(b) separation of proteins
(c) denaturation of proteins
(d) fragmentation of proteins
Answer:
(c) denaturation of proteins

42. An amino acid containing sulphur is
(a) serine
(b) cysteine
(c) valine
(d) asparagine
Answer:
(b) cysteine

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

43. Rhamnose is a
(a) carbohydrate
(b) protein
(c) lipid
(d) vitamin
Answer:
(a) carbohydrate

44. Lactose on hydrolysis gives
(a) glucose + glucose
(b) glucose + fructose
(c) glucose + galactose
(d) fructose + galactose
Answer:
(c) glucose + galactose

45. Raffinose on hydrolysis gives
(a) glucose + glucose + galactose
(b) glucose + fructose + galactose
(c) glucose + galactose + galactose
(d) fructose + galactose + galactose
Answer:
(b) glucose + fructose + galactose

46. Naturally αcurring glucose is
(a) dextro rotatory
(b) laevo rotatory
(c) racemic mixture
(d) all of these
Answer:
(a) dextro rotatory

47. Amylopectin is
(a) soluble in water and constitutes about 80% of starch
(b) insoluble in water and constitutes about 80% of starch
(c) Soluble in alcohol and constitutes about 60% of starch
(d) in soluble in alcohol and constitutes about 60% of starch
Answer:
(b) insoluble in water and constitutes about 80% of starch

48. Insulin contains
(a) 51 amino acids
(b) 151 amino acids
(c) 15 amino acids
(d) 115 amino acids
Answer:
(a) 51 amino acids

49. Pyranose structure of glucose is
(a) an open chain structure of glucose
(b) a structure of reduction product of glucose
(c) a cyclic six-membered structure of glucose
(d) a four-membered cyclic form of glucose
Answer:
(c) a cyclic six-membered structure of glucose

50. The number of – OH groups present in ribulose is
(a) 3
(b) 4
(c) 6
(d) 5
Answer:
(b) 4

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

51. Peptide linkage is
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 109
Answer:
(d)

52. Stachyose is an example of
(a) monosaccharides
(b) disaccharides
(c) trisaccharides
(d) tetrasaccharides
Answer:
(d) tetrasaccharides

53. How many moles of (CH3CO)2O will be required to form glucose pentaacetate form 2 moles of glucose?
(a) 2
(b) 5
(c) 10
(d) 2.5
Answer:
(c) 10

54. Which of the following NOT present in DNA?
(a) Adenine
(b) Guanine
(c) Thymine
(d) Uracil
Answer:
(d) Uracil

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

55. Maltose is a
(a) polysaccharide
(b) disaccharide
(c) trisaccharide
(d) monosaccharide
Answer:
(b) disaccharide

12th Chemistry Chapter 4 Exercise Chemical Thermodynamics Solutions Maharashtra Board

Class 12 Chemistry Chapter 4

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 4 Chemical Thermodynamics Textbook Exercise Questions and Answers.

Chemical Thermodynamics Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 4 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 4 Exercise Solutions

1. Select the most apropriate option.

Question 1.
The correct thermodynamic conditions for the spontaneous reaction at all temperatures are
(a) ΔH < 0 and ΔS > 0
(b) ΔH > 0 and ΔS < 0
(c) ΔH < 0 and ΔS < 0
(d) ΔH < 0 and ΔS = 0
Answer:
(a) ΔH < 0 and ΔS > 0

Question ii.
A gas is allowed to expand in a well-insulated container against a constant external pressure of 2.5 bar from an initial volume of 2.5 L to a final volume of 4.5 L. The change in internal energy, ΔU of the gas will be
(a) -500 J
(b) +500J
(c) -1013 J
(d) +1013 J
Answer:
(a) -500 J

Question iii.
In which of the following, entropy of the system decreases ?
(a) Crystallisation of liquid into solid
(b) Temperature of crystalline solid is increased from 0 K to 115 K
(c) H2(g) → 2H(g)
(d) 2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g)
Answer:
(a) Crystallisation of liquid into solid

Question iv.
The enthalpy of formation for all elements in their standard states is
(a) unity
(b) zero
(c) less than zero
(d) different elements
Answer:
(b) zero

Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics

Question v.
Which of the following reactions is exothermic ?
(a) H2(g) → 2H(g)
(b) C(s) → C(g)
(c) 2Cl(g) → Cl2(g)
(d) H2O(s) → H2O(l)
Answer:
(c) 2Cl(g) → Cl2(g)

Question vi.
6.24 g of ethanol are vaporized by supplying 5.89 kJ of heat. Enthalpy of vaporization of ethanol will be
(a) 43.4 kJ mol-1
(b) 60.2 kJ mol-1
(c) 38.9 kJ mol-1
(d) 20.4 kJ mol-1
Answer:
(a) 43.4 kJ mol-1

Question vii.
If the standard enthalpy of formation of methanol is -238.9 kJ mol-1 then entropy change of the surroundings will be
(a) -801.7 JK-1
(b) 801.7 JK-1
(c) 0.8017 JK-1
(d) -0.8017 JK-1
Answer:
(b) 801.7 JK-1

Question viii.
Which of the following are not state functions ?
1. Q + W 2. Q 3. W 4. H-TS
(a) 1, 2 and 3
(b) 2 and 3
(c) 1 and 4
(d) 2, 3 and 4
Answer:
(b) 2 and 3

Question ix.
For vaporization of water at 1 bar, ΔH = 40.63 kJ mol-1 and ΔS =108.8 JK-1 mol-1. At what temperature, ΔG = 0?
(a) 273.4 K
(b) 393.4 K
(c) 373.4 K
(d) 293.4 K
Answer:
(c) 373.4 K

Question x.
Bond enthalpies of H – H, Cl – Cl and H – Cl bonds are 434 kJ mol-1, 242 kJ mol-2 and 431 kJ mol-1, respectively. Enthalpy of formation of HCl is
(a) 245 kJ mol-1
(b) -93 kJ mol-1
(c) -245 kJ mol-1
(d) 93 kJ mol-1
Answer:
(b) -93 kJ mol-1

Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics

2. Answer the following in one or two sentences.

Question i.
Comment on the statement: No work is involved in an expansion of a gas in vacuum.
Answer:
(1) When a gas expands against an external pressure Pex, changing the volume from V1 to V2, the work obtained is given by
W = -Pex (V2 – V1).
(2) Hence the work is performed by the system when it experiences the opposing force or pressure.
(3) Greater the opposing force, more is the work.
(4) In free expansion, the gas expands in vaccum where it does not experience opposing force, (P = 0). Since external pressure is zero, no work is obtained.
∴ W = -Pex (V2 – V1)
= -0 × (V2 – V1)
= 0
(5) Since during expansion in vacuum no energy is expended, it is called free expansion.

Question ii.
State the first law of thermodynamics.
Answer:
The first law of thermodynamics is based on the principle of conservation of energy and can be stated in different ways as follows :

  1. Energy can neither be created nor destroyed, however, it may be converted from one form into another.
  2. Whenever, a quantity of one kind of energy is consumed or disappears, an equivalent amount of another kind of energy appears.
  3. The total mass and energy of an isolated system remain constant, although there may be interconservation of energy from one form to another.
  4. The total energy of the universe remains constant.

Question iii.
What is enthalpy of fusion?
Answer:
Enthalpy of fusion (ΔfusH) : The enthalpy change that accompanies the fusion of one mole of a solid into a liquid at constant temperature and pressure is called enthalpy of fusion.
For example,
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 1
This equation describes that when one mole of ice melts (fuses) at 0 °C (273 K) and 1 atmosphere, 6.1 kJ of heat will be absorbed.

Question iv.
What is standard state of a substance?
Answer:
The thermodynamic standard state of a substance (compound) is the most stable physical state of it at 298 K and 1 atmosphere (or 1 bar). The enthalpy of the substance in the standard state is represented as ΔfH0.

Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics

Question v.
State whether ∆S is positive, negative or zero for the reaction 2H(g) → H2(g). Explain.
Answer:
(i) The given reaction, 2H(g) → H2(g) is the formation of H2(g) from free atoms.
(ii) Since two H atoms form one H2 molecule, ∆n = 1 – 2= -1 and disorder is decreased. Hence entropy change ∆S < 0 (or negative).

Question vi.
State second law of thermodynamics in terms of entropy.
Answer:
The second law of thermodynamics states that the total entropy of the system and its surroundings (universe) increases in a spontaneous process.
OR
Since all the natural processes are spontaneous, the entropy of the universe increases.
It is expressed mathematically as
∆ STotal = ∆ Ssystem + ∆Ssurr > 0
∆ SUniverse = ∆ Ssystem + ∆ Ssurr > 0

Question vii.
If the enthalpy change of a reaction is ∆H how will you calculate entropy of surroundings?
Answer:
(i) For endothermic reaction, ∆H > 0. This shows the system absorbs heat from surroundings.
∴ ∆surr H < 0.
∵ Entropy change = ∆surr S = \(\frac{-\Delta_{\text {surr }} H}{T}\)
There is decrease in entropy of surroundings.
(ii) For exothermic reaction, ∆H < 0, hence for surroundings, ∆surr H > 0

∴ ∆surr > 0.

Question viii.
Comment on spontaneity of reactions for which ∆H is positive and ∆S is negative.
Answer:
Since ∆H is +ve and ∆S is -ve, ∆G will be +ve at all temperatures. Hence reactions will be non-spontaneous at all temperatures.

Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics

3. Answer in brief.

Question i.
Obtain the relationship between ∆G° of a reaction and the equilibrium constant.
Answer:
Consider following reversible reaction, aA + bB ⇌ cC + dD
The reaction quotient Q is,
Q = \(\frac{[\mathrm{C}]^{c} \times[\mathrm{D}]^{d}}{[\mathrm{~A}]^{a} \times[\mathrm{B}]^{b}}\)
The free energy change ∆G for the reaction is ∆G = ∆G° + RT in Q
Where ∆G° is the standard free energy change.
At equilibrium
Q = \(\frac{[\mathrm{C}]_{e}^{c} \times[\mathrm{D}]_{e}^{d}}{[\mathrm{~A}]_{e}^{a} \times[\mathrm{B}]_{e}^{b}}=\mathrm{K}\)
∴ ∆G = ∆G° + RT In K
∵ at equilibrium ∆G = 0
∴ 0 = AG° + RT In K
∴ ∆G° = -RT In K
∴ ∆G°= -2.303 RT log10K.

Question ii.
What is entropy? Give its units.
Answer:
(i) Entropy : Being a state function and thermodynamic function it is defined as entropy change (∆S) of a system in a process which is equal to the amount of heat transferred in a reversible manner (Qrev) divided by the absolute temperature (T), at which the heat is absorbed. Thus,
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 2
(ii) Units of entropy are JK-1 in SI unit and cal K-1 in c.g.s. units. It is also expressed in terms of entropy unit (e.u.). Hence 1 e.u. = 1 JK-1.
(iii) Entropy is a measure of disorder in the system. Higher the disorder, more is entropy of the system.

Question iii.
How will you calculate reaction enthalpy from data on bond enthalpies?
Answer:
(i) In chemical reactions, bonds are broken in the reactant molecules and bonds are formed in the product molecules.
(ii) Energy is always required to break a chemical bond while energy is always released in the formation of the bond.
(iii) The enthalpy change of a gaseous reactions (ΔfH0) involving substances with covalent bonds can be calculated with the help of bond enthalpies of reactants and products. (In case of solids we need lattice energy or heat of sublimation while in case of liquids we need heat of evaporation.)
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 3
If the energy required to break the bonds of reacting molecules is more than the energy released in the bond formation of the products, then the reaction will be endothermic and ∆H0 reaction will be positive. On the other hand if the energy released in the bond formation of the products is more than the energy required to break the bonds of reacting molecules then the reaction will be exothermic and ∆H0 reaction will be negative.

Question iv.
What is the standard enthalpy of combustion ? Give an example.
Answer:
Standard enthalpy of combustion or standard heat of combustion : it is defined as the enthalpy change when one mole of a substance in the standard state undergoes complete combustion in a sufficient amount of oxygen at constant temperature (298 K) and pressure (1 atmosphere or 1 bar). It is denoted by ∆cH0.
E.g. CH3OH(l) + \(\frac {3}{2}\) O2(g) = CO2(g) + 2H2O
cH0= -726 kJ mol-1
(∆cH0 is always negative.)
[Note : Calorific value : It is the enthalpy change or amount of heat liberated when one gram of a substance undergoes combustion.
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 4

Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics

Question v.
What is the enthalpy of atomization? Give an example.
Answer:
Enthalpy of atomisation (∆atomH) : it is the enthalpy change accompanying the dissociation of one mole of gaseous substance into its atoms at constant temperature and pressure.
For example : CH4(g) → C(g) + 4H(g)atomH = 1660 kJ mol-1

Question vi.
Obtain the expression for work done in chemical reaction.
Answer:
Consider n1 moles of gaseous reactants A of volume V1 change to n2 moles of gaseous products B of volume V2 at temperature T and pressure P.
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 5
In the initial state, PV1 = n1RT
In the final state, PV2 = n2RT
PV2 – PV1 = n2RT – n1RT = (n2 – n1)RT = ∆nRT
where ∆n is the change in number of moles of gaseous products and gaseous reactants.
Due to net changes in gaseous moles, there arises change in volume against constant pressure resulting in mechanical work, -P∆V.
∴ W = -P∆V = -P(V2 – V1) = – ∆nRT
(i) If n1 – n2, ∆n = 0, W = 0. No work is performed.
(ii) If n2 > n1, ∆n > 0, there is a work of expansion by the system and W is negative.
(iii) If n2 < n1, ∆n < 0, there is a work of compression, hence work is done on the system and W is positive.

Question vii.
Derive the expression for PV work.
Answer:
Consider a certain amount of an ideal gas enclosed in an ideal cylinder fitted with massless, frictionless rigid movable piston at pressure P, occupying volume V1 at temperature T.
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 6
Fig. 4.8 : Work of expansion
As the gas expands, it pushes the piston upward through a distance d against external force F, pushing the surroundings.
The work done by the gas is,
W = opposing force × distance = -F × d
-ve sign indicates the lowering of energy of the system during expansion.
If a is the cross section area of the cylinder or piston, then,
W = \(-\frac{F}{a}\) × d × a
Now the pressure is Pex = \(\frac{F}{a}\)
while volume change is, ΔV = d × a
∴ W = -Pex × ΔV
If during the expansion, the volume changes from V1 and V2 then, ΔV = V2 – V1
∴ W= -Pex(V2 – V1)
During compression, the work W is +ve, since the energy of the system is increased,
W = +Pex(V2 – V1)

Question viii.
What are intensive properties? Explain why density is intensive property.
Answer:
(A) Intensive property : It is defined as a property of a system whose magnitude is independent of the amount of matter present in the system.
Explanation :

  1. Intensive property is characteristic of the system, e.g., refractive index, density, viscosity, temperature, pressure, boiling point, melting point, freezing point of a pure liquid, surface tension, etc.
  2. The intensive properties are not additive.

(B) Density is a ratio of two extensive properties namely, mass and volume. Since the ratio of two extensive properties represents an intensive property, density is an intensive property. It does not depend on the amount of a substance.

Question ix.
How much heat is evolved when 12 g of CO reacts with NO2 ? The reaction is :
4 CO(g) + 2 NO2(g) → 4CO2(g) + N2(g), ∆H0 = -1200 kJ

Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics

4. Answer the following questions.

Question i.
Derive the expression for the maximum work.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 7
Consider n moles of an ideal gas enclosed in an ideal cylinder fitted with a massless and frictionless movable rigid piston. Let V be the volume of the gas at a pressure P and a temperature T.
If in an infinitesimal change pressure changes from P to P – dP and volume increases from V to V + dV. Then the work obtained is, dW = -(P-dP) dV
= -PdV + dPdV
Since dP.dV is negligibly small relative to PdV
dW= -PdV
Let the state of the system change from A(P1, L1) to B (P2, V2) isothermally and reversibly, at temperature T involving number of infinitesimal steps.
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 8
Then the total work or maximum work in the process is obtained by integrating above equation.
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 9
At constant temperature,
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 10
where n, P, V and T represent number of moles, pressure, volume and temperature respectively. For the process,
ΔU = 0, ΔH = 0.
The heat absorbed in reversible manner
Qrev, is completely converted into work.
Qrev = -Wmax.
Hence work obtained is maximum.

Question ii.
Obtain the relatioship between ∆H and ∆U for gas phase reactions.
Answer:
Consider a reaction in which n1 moles of gaseous reactant in initial state change to n2 moles of gaseous product in the final state.
Let H1, U1, P1, V1 and H2, U2, P2, V2 represent enthalpies, internal energies, pressures and volumes in the initial and final states respectively then,
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 11
The heat of reaction is given by enthalpy change ΔH as,
ΔH = H2 – H1
By definition, H = U + PV
∴ H1 = U1 + P1V1 and H2 = U2+ P2V2
∴ ΔH = (U2 + P2V2) – (U1 + P1V1)
= (U2 – U1) + (P2V2 – P1V1)
Now, ΔU = U2 – U1
Since PV = nRT,
For initial state, P1V1= n1RT
For final state, P2V2 = n2RT
∴ P2V2 – P1V1 = n2RT – n1RT
= (n2 – n1) RT
= ΔnRT
where Δn
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 12
∴ ΔH = ΔU + ΔnRT
If QP and QV are the heats involved in the reaction at constant pressure and constant volume respectively, then since QP = ΔH and QV = ΔU.
∴ QP = QV = ΔnRT

Question iii.
State Hess’s law of constant heat summation. Illustrate with an example. State its applications.
Answer:
Statement of law of constant heat summation : It states that, the heat of a reaction or the enthalpy change in a chemical reaction depends upon initial state of reactants and final state of products and independent of the path by which the reaction is brought about (i.e. in single step or in series of steps).
OR
Heat of reaction is same whether it is carried out in one step or in several steps.
Explanation :
Consider the formation of CO2(g).
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 13
Hess’s law treats thermochemical equations mathematically i.e., they can be added, subtracted or multiplied by numerical factors like algebraic equations.

Applications : Hess’s law is used for :

  1. To calculate heat of formation, combustion, neutralisation, ionization, etc.
  2. To calculate the heat of reactions which may not take place normally or directly.
  3. To calculate heats of extremely slow or fast reactions.
  4. To calculate enthalpies of reactants and products.

Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics

Question iv.
Although ∆S for the formation of two moles of water from H2 and O2 is -327 JK-1, it is spontaneous. Explain. (Given ∆H for the reaction is -572 kJ).
Answer:
Given : ΔS= -327 JK-1; ΔH = -572 kJ
ΔG = ΔH – TΔS, and ΔH << ΔS
∴ ΔG < 0, and hence the formation of H2O(l) is spontaneous.

Question v.
Obtain the relation between ∆G and ∆STotal. Comment on spontaneity of the reaction.
Answer:
(i) Gibbs free energy, G is defined as,
G = H – TS
where H is the enthalpy, S is the entropy of the system at absolute temperature T.
Since H, T and S are state functions, G is a state function and a thermodynamic function.

(ii) At constant temperature and pressure, change in free energy ΔG for the system is represented as, ∆G = ∆H – T∆S
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 14
This is called Gibbs free energy equation for ∆G. In this ∆S is total entropy change, i.e., ∆STotal.

(iii) The SI units of ∆G are J or kJ (or Jmol-1 or kJmol-1).
The c.g.s. units of ∆G are cal or kcal (or cal mol-1 or kcal mol-1.)

The second law explains the conditions of spontaneity as below :
(i) ∆Stotal > 0 and ∆G < 0, the process is spontaneous.
(ii) ∆Stotal < 0 and ∆G > 0, the process is nonspontaneous.
(iii) ∆Stotal = 0 and ∆G = 0, the process is at equilibrium.

Question vi.
One mole of an ideal gas is compressed from 500 cm3 against a constant external pressure of 1.2 × 105 Pa. The work involved in the process is 36.0 J. Calculate the final volume.
Answer:
Given : V1 = 500 cm3 = 0.5 dm3;
Pex = 1.2 × 105 Pa = 1.2 bar; W= 36 J;
1 dm3 bar = 100 J; V2 = ?
W = -Pex (V2 – V1)
36 J = – 1.2 (V2 – 0.5) dm3 bar
= -1.2 × 100 (V2 – 0.5) J
∴ V2 – 0.5 = \(\frac{-36}{1.2 \times 100}=-0.3\)
∴ V2 = 0.5 -0.3 = 0.2 dm3 = 200 cm3
Ans. Final volume = 200 cm3.

Question vii.
Calculate the maximum work when 24g of O2 are expanded isothermally and reversibly from the pressure of 1.6 bar to 1 bar at 298 K.
Answer:
Given : W02 = 24 g, P1 = 1.6 bar, P2 = 1 bar
T = 298 K, Wmax = ?
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 15

Question viii.
Calculate the work done in the decomposition of 132 g of NH4NO3 at 100 °C.
NH4NO3(s) → N2O(g) + 2 H2O(g)
State whether work is done on the system or by the system.
Answer:
NH44NO3(s) → N2O(g) + 2 H2O(g)
mNH4NO3 = 132 g; MNH4NO3 = 80 g mol-1
T = 273 + 100 = 373 K; Δn = ?
For the reaction,
Δn = Σn2 gaseous products – Σn1 gaseous reactants
= 3 – 0 = 3 mol
Since there is an increase in number of gaseous moles, the work is done by the system.
nNH4NO3 = \(\frac{m_{\mathrm{NH}_{4} \mathrm{NO}_{3}}}{M_{\mathrm{NH}_{4} \mathrm{NO}_{3}}}\)
= \(\frac{132}{80}\)
= 1.65 mol
For 1 mol NH4NO3(s) Δn = 3 mol
∴ For 1.65 mol NH4NO3(s) Δn = 3 × 1.65 = 4.95 mol
W = -ΔnRT = -4.95 × 8.314 × 373
= – 15350 J
= – 15.35 kJ
Ans. Work is done by the system.
Work done = – 15.35 kJ

Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics

Question ix.
Calculate standard enthalpy of reaction,
Fe2O3(s) + 3CO(g) → 2 Fe(s) + 3CO2(g),
from the following data.
fH0(Fe2O3) = -824 kJ/mol,
fH0(CO) = -110 kJ/mol,
fH0(CO2) = -393 kJ/mol
Answer:
Given : ∆fH0Fe2O3 = -824 kJ/mol-1;
fH0(CO) = – 110 kJ mol-1
fH0(CO2) = – 393 kJ/mol-1; ∆fH0 = ?
Required equation,
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
∆H1 = ? – (I)
Given equations :
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 16
= -(-824) -3 (-110) + 3(-393)
= 824 + 330 – 1179
fH0 = -25 kJ
Ans. ∆fH0 = -25 kJ

Question x.
For a certain reaction ∆H0 =219 kJ and ∆S0 = -21 J/K. Determine whether the reaction is spontaneous or nonspontaneous.
Answer:
Given : ∆H0 = 219 kJ; ∆S0 = -21 J/K, ∆G0 = ?
For standard state, T = 298 K
∆G0 = ∆H0 – T∆S0
= 219 – 298 × (-21) × 10-3
= 219 + 6.258
= 225.3 kJ
Since ∆S < 0 and ∆G0 > 0, the reaction is non-spontaneous.

Question xi.
Determine whether the following reaction is spontaneous under standard state conditions.
2 H2O(l) + O2(g) → 2H2O2(l)
if ∆H0 = 196 kJ, ∆S0 = -126 J/K
Does it have a cross-over temperature?
Answer:
Given : 2H2O(l) + O2(g) → 2H2O2(l)
∆H0 = +196 kJ
∆S0 = -126 JK-1 =0.126 kj K-1
T= 298 K
∆G0 = ?
Cross over temperature = T = ?
∆G0 = ∆H0 – T∆S0
= 196 – 298 (-0.126)
= 196 + 37.55
= + 233.55 kJ
∵ ∆G0 > 0, the reaction is non-spontaneous.
∆H0 > 0, ∆S0 < 0,
Since at all temperatures, ∆G0 > 0, there is no cross over temperature.
Ans. The reaction is non-spontaneous.
There is no cross-over temperature for the reaction.

Question xii.
Calculate ∆U at 298 K for the reaction,
C2H4(g) + HCl(g) → C2H5Cl(g), ∆H = -72.3 kJ
How much PV work is done?
Answer:
Given : C2H4(g) + HCl(g) → C2H5Cl(g)
T = 298 K; ∆H = -72.3 kJ; PV = ?;
∆U = ?
∆n = Σn2gaseous products – Σn1gaseous reactants
= 1 – (1 + 1)= -1 mol
For PV work :
W = -∆nRT
= – (- 1) × 8.314 × 298
= 2478 J = 2.478 kJ
∆H = ∆U + ∆nRT
∴ ∆U = ∆H – ∆nRT
= – 72.3 – (-2.478)
= – 69.82 kJ
Ans. PV work = 2.478 kJ
∆U = -69.82 kJ

Question xiii.
Calculate the work done during synthesis of NH3 in which volume changes from 8.0 dm3 to 4.0 dm3 at a constant external pressure of 43 bar. In what direction the work energy flows?
Answer:
Given : V1 = 8.0 dm3; V2 = 4.0 dm3; Pex = 43 bar
W = ? What direction work energy flows ?
W = -Pex(V2 – V1)
= -43 (4 – 8)
= 172 dm3 bar
= 172 × 100 J
= 17200 J
= 17.2 kJ
In this compression process, the work is done on the system and work energy flows into the system.

Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics

Question xiv.
Calculate the amount of work done in the
(a) oxidation of 1 mole HCl(g) at 200 °C according to reaction.
4HCl(g) + O2(g) → 2 Cl2(g) + 2 H2O(g)
(b) decomposition of one mole of NO at 300 °C for the reaction
2 NO(g) → N2(g) + O2
Answer:
Given :
(a) 4HCl(g) + O2(g) → 2Cl2(g) + 2H2O(g)
nHCl = 1 mol; T = 273 + 200 = 473 K, W = ?
For 4 mol HCl ∆n = (2 + 2) – (4 + 1) = – 1 mol
∴ For 1 mol HCl ∆n = –\(\frac {1}{4}\) = -0.25 mol
W = -∆nRT = – (-0.25) × 8.314 × 473 = 983.11
(b) ∆n = (1 + 1) – 2 = 0 mol
W = -∆nRT = -(0) × 8.314 × 473 = 0
Ans. (a) W = 983.1 J
(b) W = 0.0 J

Question xv.
When 6.0 g of O2 reacts with CIF as per
2CIF(g) + O2(g) → Cl2O(g) + OF2(g)
The enthalpy change is 38.55 kJ. What is standard enthalpy of the reaction ?
Answer:
Given : The given reaction is for 1 mol O2 or 32 g O2.
∵ For 6.0 g O2
∆ H0 = 38.55 kJ
∴ For 32 g O2
∆ H0 = \(\frac{32 \times 38.55}{6}\)
= 205.6 kJ
Ans. ∆H0 = 205.6 kJ

Question xvi.
Calculate the standard enthalpy of formation of CH3OH(l) from the following data:
i. CH3OH(l) + \(\frac {3}{2}\) O2(g) → CO2(g) + 2H2O(l), ∆H0 = -726 kJ mol-1
ii. C (Graphite) + O2(g) → CO2(g), ∆cH0 = -393 kJ mol-1
iii. H2(g) + \(\frac {1}{2}\) O2(g) → H2O(l), ∆fH0 = -286 kJ mol-1
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 17
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 18
∴ ∆H0
= –\(\Delta H_{2}^{0}\) + \(\Delta H_{3}^{0}\) + 2∆\(\Delta H_{4}^{0}\)
= – (- 726) + (- 393) + 2(- 286)
= 726 – 393 – 572
= – 239 kJ mol-1
Ans. Standard enthalpy of formation = ∆fH0= -239 kJ mol-1

Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics

Question xvii.
Calculate ∆H0 for the following reaction at 298 K
H2B4O7(s) + H2O(l) → 4HBO2(aq)
i. 2H3BO3(aq) → B2O3(s) + 3H2O(l), ∆H0 = 14.4 kJ mol-1
ii. H3BO3(aq) → HBO2(aq) + H2O(l), ∆H0 = -0.02 kJ mol-1
iii. H2B4O7(s) → 2B2O3(s) + H2O(l), ∆H0 = 17.3 kJ mol-1
Answer:
Given equations :
i. 2H3BO3(aq) → B2O3(s) + 3H2O(l), ……….(i)
∆H0 = 14.4 kJ mol-1
ii. H3BO3(aq) → HBO2(aq) + H2O(l) ……….(ii)
∆H0 = -0.02 kJ mol-1
iii. H2B4O7(s) → 2B2O3(s) + H2O(l), ……….(iii)
∆H0 = 17.3 kJ mol-1
Required equation :
(iv) H2B4O7(s) + H2O(l) → 4HBO2(aq) ……. (iv)
\(\Delta H_{4}^{0}=?\)
To obtain eq. (iv) add 4 times equation (ii) and eq.
(iii) and subtract 2 times equation (i).
∴ eq. (iv) = 4 eq. (ii) + eq. (iii) – 2eq. (i)
∴ \(\Delta H_{4}^{0}=4 \Delta H_{2}^{0}+\Delta H_{3}^{0}-2 \Delta H_{1}^{0}\)
= 4(-0.02) + 17.3 – 2(14.4)
= -0.08 + 17.3 – 28.8
= -11.58 kJ
∴ Enthalpy change for the reaction
= ∆rH0 = -11.58 kJ
Ans. ∆rH0 for the given reaction = -11.58 kJ

Question xviii.
Calculate the total heat required (a) to melt 180 g of ice at 0 °C, (b) heat it to 100 °C and then (c) vapourise it at that temperature. Given ∆fusH(ice) = 6.01 kJ mol-1 at 0 °C, ∆vapH(H2O) = 40.7 kJ mol-1 at 100 °C specific heat of water is 4.18 J g-1 K-1.
Answer:
Given : Mass of ice = m = 180 g
T1 = 273 + 0 °C = 273 K
T2 = 273 + 100 °C = 373 K
fusH(ice) = ∆fusH(H2O)(s) = 6.01 kJ mol-1
vapHH2O(l) = 40.7 kJ mol-1
Specific heat of water = C = 4.18 J g-1 K-1
For converting 180 g ice into vapour, ∆ HTotal = ?
Number of moles of H2O = \(\frac {180}{18}\) = 10 mol
The total process can be represented as,
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 19
(i) ∆H1 = ∆fusH = 10 mol × 6.01 kJ mol-1
= 60.1 kJ
(ii) When the temperature of water is raised from 0 °C to 100 °C (i.e., 273 K to 373 K), then
∆ H2 = m × C × ∆T
= m × C × (T2 – T1)
= 180 g × 4.18 Jg-1K-1 × (373 – 273) × 10-3 kJ = 75.24 kJ
∆ H3 = ∆vapH = 10 mol × 40.7 kJ mol-1 = 407 kJ
Hence total enthalpy change,
∆ HTotal = ∆H1 + ∆H2 + ∆H3
= 60.1 + 75.24 + 407
= 542.34 kJ
Ans. Total heat required = 542.34 kJ

Question xix.
The enthalpy change for the reaction,
C2H4(g) + H2(g) → C2H6(g)
is -620 J when 100 ml of ethylene and 100 mL of H2 react at 1 bar pressure. Calculate the pressure volume type of work and ∆U for the reaction.
Answer:
Given :
\(\begin{aligned}
&\mathrm{C}_{2} \mathrm{H}_{4(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6(\mathrm{~g})} \\
&100 \mathrm{~mL} \quad 100 \mathrm{ml} \quad 100 \mathrm{ml}
\end{aligned}\)
∆H = – 620 J; VC2H4 = 100 mL; VH2 = 100 mL
Pex= 1 bar; W=?; ∆U = ?
∆V = 100 – (100 + 100) = -100 mL = -0.1 dm3
W = -Pex(V2 – V1)
= -Pex × ∆V
= -1 × (-0.1)
= 0.1 dm3 bar
= 0.1 × 100 J
= +10 J
∆H = ∆U + P∆V
∴ ∆U = ∆H – P∆V = -620 – (+10) = -610 J
Ans. W = +10 J; ∆U = -610 J

Question xx.
Calculate the work done and comment on whether work is done on or by the system for the decomposition of 2 moles of NH4NO3 at 100 °C
NH4NO3(s) → N2O(g) + 2H2O(g)
Answer:
Given : NH4NO3(s) → N2O(g) + 2H2O(g)
nNH4NO3 = 2 mol; T = 273 + 100 = 373 K
W = ? Comment on work = ?
∆nreaction = (1 + 2) – 0 = 3 mol
∵ For 1 mol of NH4NO3 ∆nreaction = 3 mol
∴ For 2 mol of NH4NO3 ∆nreaction = 6 mol
Due to 6 moles of gaseous products from 2 mol NH4NO3, there is work of expansion, hence work is done by the system.
W = -∆nRT
= – 6 × 8.314 × 373 = -18606 J
= -18.606 kJ
Ans. Work is done by the system.
W= -18.606 kJ

Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics

12th Chemistry Digest Chapter 4 Chemical Thermodynamics Intext Questions and Answers

(Textbook page No. 73)

Question 1.
Under what conditions ∆H = ∆U ?
Answer:
(a) ∆H = ∆U + P∆V
when ∆V = 0, ∆H = ∆U
(b) ∆H = ∆U + ∆nRT
when ∆n = 0, ∆H = ∆U

Try this… (Textbook page No. 71)

Question 1.
25 kJ of work is done on the system and it releases 10 kJ of heat. What is ∆U?
Answer:
W = 25 kJ; Q= -10 kJ
∆U = Q + W = -10 + 25
∆U = + 15 kJ

Try this… (Textbook page No. 75)

Question 1.
For KCl, ∆LH = 699 kJ/mol-1 and ∆hydH = -681.8 kJ/mol-1. What will be its enthalpy of solution?
Answer:
Enthalpy of solution :
solnH = ∆LH + ∆hydH
= 699 + (-681.8)
solnH = +17.2 kJ mol-1

Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics

Try this… (Textbook page No. 76)

Question 1.
Given the thermochemical equation,
C2H2(g) + \(\frac {5}{2}\) O2(g) → 2CO2(g)+ H2O(l), ∆rH0 = -1300 kJ
Write thermochemical equations when
i. Coefficients of substances are multiplied by 2.
ii. equation is reversed.
Answer:
(i) 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l)
rH0 = -2 × 1300 kJ
= – 2600 kJ
(ii) 2CO2(g) + H2O(l) → C2H2(g) + \(\frac {5}{2}\)O2(g)
rH0 = +1300 KJ

Try this… (Textbook page No. 78)

Question 1.
(i) Write thermochemical equation for complete oxidation of one mole of H2(g). Standard enthalpy change of the reaction is -286 kJ.
(ii) Is the value -286 kJ, enthalpy of formation or enthalpy of combustion or both? Explain.
Answer:
(i) H2(g) + \(\frac {1}{2}\)O2(g) → H2O(l) ∆cH0 = -286 KJ mol-1
(ii) The value -286 kJ is the standard enthalpy of formation of H2O(l) or standard enthalpy of combustion of H2(g).

Question 2.
Write equation for bond enthalpy of Cl-Cl bond in Cl2 molecule ∆rH0 for dissociation of Cl2 molecule is 242.7 kJ.
Answer:
Equation for bond enthalpy :
Cl2(g) → 2Cl(g)rH0 = 242.7 kJ mol-1
∴ Bond enthalpy of Cl2 = 242.7 kJ mol-1

Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics

Try this… (Textbook page No. 82)

Question 1.
State whether ∆S is positive, negative or zero for the following reactions.
i. 2H2(g) + O2(g) → 2H2O(l)
ii. CaCO3(s) → CaO(s) + CO2(g)
Answer:
(i) 2H2(g) + O2(g) → 2H2O(l)
Since the system is converted from gaseous state to a liquid state, the disorder is decreased, hence ∆S < O (negative).

(ii) CaCO3(s) → CaO(s) + CO2(g)
Since molecules of solid CaCO3 break giving gaseous CO2, disorder is increased hence ∆S > O (positive).

12th Std Chemistry Questions And Answers:

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Balbharti Maharashtra State Board 11th Biology Important Questions Chapter 4 Kingdom Animalia Important Questions and Answers.

Maharashtra State Board 11th Biology Important Questions Chapter 4 Kingdom Animalia

Question 1.
What are grades of organization in animals?
Answer:
Cellular, cell- tissue, tissue-organ are the grades of organization in animals.

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 2.
How are the animals classified based on body cavity?
Answer:
The animals are classified as acoelomates, pseudocoelomates, and coelomates based on body cavity.

Question 3.
Explain in detail the body plan in animals.
Answer:
Animals show three fundamental body plans as follows:
1. Cell aggregate body plan.
2. Blind sac body plan,
3. Tube within tube body plan.

1. Cell aggregate body plan:
a. In this body plan, cells do not form tissues or organs.
b. Differentiation and division of labour among the cells is minimal.
c. Members of phylum Porifera show cell aggregate body plan.

2. Blind sac body plan:
a. In this body plan, the body is sac-like with a single opening. Digestion is carried out in this sac-like structure.
b. The food is ingested and egested through the same
opening.
c. Members of phylum Cnidaria show a blind sac body plan.

3. Tube within tube body plan:
a. Digestive system is present in tube-like body cavity.
b. Mouth and anus are present at two separate ends of the digestive system.
c. Phylum Annelida onwards all phyla show tube within tube body plan.

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 4.
Give the characteristic features of phylum Porifera.
Answer:
Phylum Porifera (Pori = Pores: feron = bearing): Members of the phylum Porifera are also called sponges. Characteristic features of the phylum:

  1. Habitat: They are aquatic, mostly marine but few species are found in fresh water.
  2. Forms: They are sedentary animals (attached to substratum or rock).
  3. Body shape: They have asymmetrical body. Body of these animals consists of many cells with minimal
    division of labour among cells. Hence, their body is considered as a colony of different types of cells.
  4. Body surface: Their body bears minute pores called ‘ostia’ through which water enters the spongocoel (body cavity). Water leaves the body through a large opening called ‘osculum’. Beating of flagella creates water current.
  5. Circulation: Water is circulated in the body through the ‘canal system’. When the water enters the body of poriferans, cells absorb the food, exchange respiratory gases and release excretory products.
  6. Digestive system: The body cavity of sponges (spongocoel) is lined by unique type of flagellated cells called choanocytes or collar cells for digestion.
  7. Endoskeleton: The body of sponges consists of calcareous / siliceous spicules and proteinaceous ‘spongin fibres’.
  8. Reproduction: Sponges reproduce asexually as well as sexually. Asexual reproduction takes place by fragmentation and gemmule formation. Sexual reproduction is by formation of gametes. Fertilization is internal and development is indirect through larval stage.
  9. Sponges have great power of regeneration.
    e.g. Scypha, Euspongia (Bath sponge), Euplectella (Venus’ flower basket).

Question 5.
Given below is a typical sponge body. Identify i, ii, and iii.
Scypha, Euspongia (Bath sponge), Euplectella (Venus’ flower basket).
Answer:

  1. Ostium,
  2. Choanocyte,
  3. Osculum

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 6.
Identify the organism and enlist the general characters of its phylum.
Answer:
The given organism is Euplectella.
For characters: Phylum Porifera (Pori = Pores: feron = bearing): Members of the phylum Porifera are also called sponges. Characteristic features of the phylum:

  1. Habitat: They are aquatic, mostly marine but few species are found in freshwater.
  2. Forms: They are sedentary animals (attached to substratum or rock).
  3. Body shape: They have asymmetrical body. Body of these animals consists of many cells with minimal
    division of labour among cells. Hence, their body is considered as a colony of different types of cells.
  4. Body surface: Their body bears minute pores called ‘ostia’ through which water enters the spongocoel (body cavity). Water leaves the body through a large opening called ‘oscu lum’. Beating of flagella creates water current.
  5. Circulation: Water is circulated in the body through the ‘canal system’. When the water enters the body of poriferans, cells absorb the food, exchange respiratory gases and release excretory products.
  6. Digestive system: The body cavity of sponges (spongocoel) is lined by unique type of flagellated cells called choanocytes or collar cells for digestion.
  7. Endoskeleton: The body of sponges consists of calcareous / siliceous spicules and proteinaceous ‘spongin fibres’.
  8. Reproduction: Sponges reproduce asexually as well as sexually. Asexual reproduction takes place by fragmentation and gemmule formation. Sexual reproduction is by formation of gametes. Fertilization is internal and development is indirect through larval stage.
  9. Sponges have great power of regeneration.
    e.g. Scypha, Euspongia (Bath sponge), Euplectella (Venus’ flower basket).

Question 7.
State the characteristics of members belonging to phylum Cnidaria.
Answer:
Characteristics of members belonging to phylum Cnidaria:

  1. Habitat: They are aquatic, mostly marine and few of them are fresh – water forms.
  2. Forms: They are sessile or free swimming.
  3. Cnidoblasts: Presence of cnidoblasts or stinging cells are present on the tentacles for anchorage, offence and defence.
  4. Body Symmetry: They have radially symmetrical body.
  5. Germ layer: They are diploblastic.
  6. Body cavity: Cnidarians have a central cavity called coelenteron or gastrovascular cavity, which helps in digestion and circulation. They have blind-sac body plan i.e., single pore opening to the exterior in the digestive system.
  7. Body form: Members of this phylum exhibit two body forms. The cylindrical form, known as polyp e.g. Hydra and the umbrella – like form (.Aurelia – jelly fish) is known as medusa.
  8. Digestion: They have extracellular and intracellular digestion.
  9. Reproduction: Cnidarians reproduce asexually and sexually.

Asexual reproduction takes place by budding and regeneration. Sexual reproduction takes place gamete formation. They exhibit metagenesis i.e. alternation of polypoid generation with medusoid generation. Polyps produce medusae asexually and medusae produce polyps sexually, e.g. Obelia
e.g. Hydra, Aurelia (Jellyfish), Physalia (Portuguese man-of-war), Adamsia (Sea anemone), Diploria (Brain coral), Gorgonia (sea fan).

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 8.
Describe the salient features of phylum Ctenophora.
Answer:
The members of this phylum are commonly known as comb jellies and sea walnuts. They are also known as acnidarians as they lack cnidoblasts. The phylum is considered as one of the minor phyla as it is represented by very few members.

Salient features of phylum Ctenophora:

  1. Habitat: They are exclusively marine.
  2. Forms: They are free swimming animals.
  3. Germ layers: Members of this phylum are diploblastic.
  4. Body Symmetry: They are radially symmetrical.
  5. Body plan: The animals of this phylum show blind-sac body plan.
  6. Body organization: They show tissue level organization.
  7. Locomotion: It is earned out by eight rows of ciliated comb plates.
  8. Bioluminescence: It is the characteristic feature of the members of this phylum.
  9. Digestion: It is extracellular and intracellular.
  10. Reproduction: Reproduction is sexual with indirect development.
  11. Colloblasts: These sticky cells are used to capture prey, e.g. Pleurobrachia, Ctenoplana

Question 9.
Draw a neat and labelled diagram:
1. Cnidoblast
Answer:
Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia 1

2. Colloblast
Answer:
Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia 2

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 10.
Mention the unique features of phylum Platyhelminthes.
Answer:
The members of this phylum are called as flatworms.
Unique features of phylum Platyhelminthes:

  1. Body shape: Body of these animals is dorsoventrally flattened.
  2. Coelom: They are acoelomates.
  3. Germ layers: Platyhelminthes are triploblastic.
  4. Body organization: They show organ-system grade of organization.
  5. Forms: Most of them are endoparasites and few are free-living. Parasitic forms have hooks and suckers for attachment to the hosts’ body. In parasitic forms, body is covered by cuticle and in free-living forms it is covered by cilia.
  6. Digestive system: Parasitic forms generally lack digestive system. In free-living forms, the digestive system is incomplete.
  7. Body plan: They show blind-sac body plan.
  8. Excretion and osmoregulation: It occurs by flame cells or protonephridia.
  9. Reproduction: They are mostly hermaphrodite (bisexual).
  10. Self-fertilization is seen. Few animals show high power of regeneration and show polyembryony.
    e.g. Planaria, Taenia (Tapeworm), Fasciola (Liver fluke) are included in this phylum.

Question 11.
Identify the organisms and label their diagrams,

Question 1.
Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia 3
Answer:
The given organism is Taenia or Tapeworm.

Question 2.
Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia 4
Answer:
The given organism is Fasciola or Liver fluke.

Question 3.
Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia 5
Answer:
The given organism is Planaria.

Question 4.
Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia 6
Answer:
The given organism is Wuchereria.

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 12.
Describe the characteristics of Aschelminthes.
Answer:
Phylum Aschelminthes (ascus – sac, helminth – worm) is also called as Nemathelminthes (Nema = thread, helmins = worms).
Characteristics of Aschelminthes:

  1. Forms: These are mostly parasitic. However, few forms are free-living.
  2. Body shape: The body is long, cylindrical, thread-like, circular in cross-section, hence they are known as roundworms.
  3. Body symmetry: These are bilaterally symmetrical.
  4. Coelom: They are pseudocoelomate animals.
  5. Germ layers: These animals are triploblastic.
  6. Body plan: They show tube within a tube-type body plan.
  7. Body covering: The body is covered by tough, resistant cuticle.
  8. Muscles: Body wall has longitudinal muscles, but circular muscles are absent.
  9. Digestive system: Alimentary canal is complete with mouth and anus, at opposite ends.
  10. Excretion: Excretion takes place either by canals or gland cells.
  11. Nervous system: Nervous system consists of a nerve ring and nerves.
  12. Reproduction: Animals are unisexual i.e. sexes are separate.
  13. Fertilization is internal. Development may or may not include larval stages. It shows sexual dimorphism.
    e.g. Ascaris (Roundworm), Wuchereria (filarial worm) and Ancylostoma (hookworm).

Question 13.
Explain the sexual dimorphism in Ascaris.
Answer:
Animals like Ascaris show sexual dimorphism. The male Ascaris is shorter and narrower than the female and has a curved posterior end with a pair of penial setae for copulation. The female Ascaris is relatively longer and broader and has a straight posterior end without penial setae.

Question 14.
Draw a neat and labelled diagram of Ascaris.
Answer:
Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia 7

Question 15.
Enlist the characteristic features of phylum Annelida.
Answer:
Annelids are commonly called as ring worms or segmented worms.
Characteristic feature of phylum Annelida:

  1. Forms: Annelids may be aquatic, ectoparasitic or free – living or burrowing in moist soil.
  2. Body symmetry: They are bilaterally symmetrical.
  3. Body coelom: They are true coelomates.
  4. Segmentation: Body is metamerically segmented and has a special region called clitellum.
  5. Digestive system: Alimentary canal is complete.
  6. Locomotion: Locomotion takes place with the help of setae (earthworm), parapodia (Nereis) or suckers (leech). Well developed longitudinal and circular muscles help in locomotion.
  7. Nervous system: It consists of nerve ring and ventral solid and ganglionated nerve cord.
  8. Reproduction: Mostly are hermaphrodites and few are dioecious (Nereis).
  9. Respiration: Exchange of gases takes place through body wall.
  10. Circulation: Circulatory system is of closed type. Excretion and osmoregulation is carried out with help of nephridia. e.g. Nereis (Aquatic annelid), Pheretima (Earthworm), Hirudinaria (Leech).

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 16.
Which phylum onwards all phyla show tube within tube body plan?
Answer:
Phylum Annelida onwards all phyla show tube within tube body plan.

Question 17.
State the unique features of phylum Mollusca.
Answer:
Mollusca (Mollis: Soft) is the second-largest phylum.
Unique features of phylum Mollusca:

  1. Habitat: They are aquatic or seen in marshy places. Few of them are terrestrial.
  2. Forms: Molluscs are either free-living or sedentary.
  3. Body plan: These are soft bodied and show tube within a tube type of body plan.
  4. Body symmetry: Most of the Molluscs show bilateral symmetry but few are asymmetrical due to torsion (twisting).
  5. Body division: Body consists of head, foot and visceral mass. Visceral mass is enclosed in thick, muscular fold of body wall called mantle. Mantle secretes a hard-calcareous shell, that may be external or internal or absent. Muscular foot is present ventrally.
  6. Digestive system: Digestive system is well-developed and complete with anterior mouth and posterior anus. Buccal cavity has a rasping organ called radula (helps in feeding), which is provided with transverse rows of teeth.
  7. Respiration: In aquatic forms, numerous feather-like gills called ctenidia, help in aquatic respiration. Terrestrial forms may show the presence of lungs.
  8. Circulatory system: Circulatory system is of open type (except in Sepia, where it is of the closed type). Blood contains a copper-containing blue-coloured respiratory pigment called hemocyanin.
  9. Excretion: Excretion occurs by kidney-like structures, also called ‘Organ of Bojanus’.
  10. Nervous system and sense organs; Nervous system has three rjahs of Mf hxese,
    interconnected by commissures and connectives.
  11. Sense organs: Sense organs such as eyes for vision, tentacles for tactile sensation and osphradia for testing purity of water is present.
  12. Sexual reproduction: Sexes are usually separate. Animals of this phylum are mostly oviparous and the development is direct or indirect.
    e.g. Pila, Spisula (Bivalve), Octopus (devil fish), Sepia (cuttle fish), Chaetopleura (Chiton), Pinctada (Pearl oyster), Loligo (Squid), Aplysia (Sea hare), Dentalium (Tusk shell).

Question 18.
Give the economic importance of molluscs.
Answer:
Economic importance of molluscs:

  1. Pearl oyster (Pinctada) gives precious pearls.
  2. Many molluscs are edible.
  3. Molluscan shells are rich source of calcium.

Question 19.
Fill in the blanks.

  1. The stinging cells on the tentacles of cnidarians are known as _______.
  2. Laccifer lacca which produces lac, belongs to phylum __________.
  3. Excretion in molluscs occurs by _________.
  4. The annelid with locomotory structures like setae is ________.

Answer:

  1. cnidoblasts
  2. Arthropoda
  3. Organ of Bojanus
  4. earthworm

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 20.
Identify the phylum to which the given organism belongs to and enlist the characteristics of this phylum.
Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia 8
Answer:
The given organism (Balanoglossus) belongs to phylum Hemichordata.
Characteristics of phylum Hemichordata:

  1. Habitat: Hemichordates are exclusively marine animals, usually living at the bottom of the sea in burrows. These are mostly free – living but animals like Rhabdopleura are sedentary.
  2. Body shape and division: Body is soft and vermiform. It is unsegmented and divided into three parts namely – proboscis, collar and trunk.
  3. Digestive system: Alimentary canal is complete, straight or ‘U’ shaped. Buccal cavity gives rise to a rod-like buccal diverticulum.
  4. Respiration: Respiration is brought about by numerous gills arranged in two longitudinal rows present in the pharyngeal region. Gills open by gill slits.
  5. Circulation: Circulatory system is simple and open type.
  6. Excretion: It takes place with help of with the glomerulus.
  7. Nervous system: Nervous tissue is embedded in epidermis on the dorsal as well as the ventral side.
  8. Reproduction and development: Sexes are separate (sometimes bisexual). Fertilization is external and development is indirect through free swimming larva.

Question 21.
Name the various subphyla of phylum chordata.
Answer:
Subphyla of phylum chordata:

  1. Urochordata
  2. Cephalochordata
  3. Vertebrata

Question 22.
Members of which subphyla are called protochordates?
Answer:
The members of subphyla – Urochordata and Cephalochordata are collectively called protochordates.

Question 23.
Which subphylum includes the tunicates or ascidians?
Answer:
Subphylum Urochordata includes tunicates or Ascidians.

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 24.
Give the distinguishing features of Tunicata.
Answer:
Distinguishing features of Tunicata or Urochordata:

  1. Habitat: They are exclusively marine.
  2. Body covering: Soft body is covered by ‘test’ or ‘tunic’ which is made up of tunicine.
  3. Notochord: Notochord is present only in the tail of the larva and is lost during metamorphosis. Hence, the name Urochordata.
  4. Respiration: Pharynx with many gill slits for respiration.
  5. Circulation: Closed circulatory system is present.
  6. Reproduction: Development is indirect, e. g. Herdmania, Salpa, Doliolum, Ascidia.

Question 25.
Write a short note on lancelets.
Answer:

  1. Cephalochordates are also known as lancelets and are small fish-like animals that rarely exceed 5 cm in length.
  2. Lancelets are exclusively marine and live partly buried in soft marine sediments.
  3. Notochord extends throughout entire length of the body and persists throughout life.
  4. Myotomes (muscle blocks) are present.
  5. Post anal tail is present.
  6. Circulatory system is closed type. Blood lacks pigment, e.g. Branchiostoma

Question 26.
Classify Branchiostoma.
Answer:
Classification of Branchiostoma:
Kingdom: Animalia Phylum: Chordata Subphylum: Cephalochordata

Question 27.
In some chordates, the notochord is replaced by cartilaginous or bony vertebral column. Name the chordates which possess this character.
Answer:
Vertebrates

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 28.
Explain in brief the divisions of sub-phylum Vertebrata.
Answer:
Sub-phylum vertebrata is divided into two divisions: Agnathostomata (lacks jaw) and Gnathostomata (bears jaw) on the basis of presence/absence of jaws.
1. Division Agnathostomata:
This division consist of the lowest or most primitive vertebrates that lack jaws.
They include only one class of living vertebrates, the Cyclostomata.
2. Division Gnathostomata:
This division includes animals with jaws.
It is divided into two superclasses: Pisces (bear fins) and Tetrapoda (bear four limbs)
[Note: Students can scan the given Q.R code for understanding the characteristics of vertebrates.]

Question 29.
Mention the characteristic feature of class Cyclostomata.
Answer:
Characteristic features of class Cyclostomata (Cyclos: Circular, stoma-mouth) Lat/Grk

  1. Members of class Cyclostomata are jaw-less and eel like organisms.
  2. Their skin is devoid of scales, soft and smooth, containing unicellular mucus glands.
  3. Median fms are present but paired fins are absent.
  4. They are ectoparasites on fishes.
  5. They have sucking circular mouth, without jaws.
  6. Cranium and vertebral column are made up of cartilage.
  7. Their digestive system lacks stomach.
  8. Respiration occurs by 6 – 15 pairs of gill slits. Gills slits are without operculum.
  9. Heart is two chambered with one auricle and one ventricle.
  10. Gonad is single, large and without gonoduct.
  11. Fertilization is external. They are anadromous as they migrate for spawning to fresh – water from marine habitat.
  12. After spawning, they die within few days. Larvae metamorphosize and migrate to ocean.
    e.g Petromyzon (Lamprey), Myxine (Hagfish).

Question 30.
Mention the important features of superclass Pisces.
Answer:
Important features of superclass Pisces:

  1. Habitat: These are aquatic animals and are present in fresh, marine and brackish waters.
  2. Body temperature: Pisces are poikilothermic animals i.e., cold blooded animals, in which body temperature changes according to the change in the surrounding temperature.
  3. Sensory organs:They have lateral line system which shows the presence of rheoreceptors for the detection of water current.
  4. Locomotion: Locomotion is by body muscles and fins. Caudal fin helps in steering wheel.
  5. Skeleton: Exoskeleton is made up of dermal scales. It is either bony or cartilaginous.
  6. Body shape: Body is streamlined and boat-shaped and helps to overcome resistance during swimming.
  7. Respiration: Respiration is by gills.
  8. Circulation: It shows single and closed circulation. Heart is two-chambered and ventral in position. Heart of fishes is described as venous heart (presence of deoxygenated blood).
  9. Nervous system: They have a well-developed brain with large olfactory lobes.
  10. Reproduction: Sexes are separate. Most of the fishes are oviparous, however some are viviparous.

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 31.
Distinguish between Chondrichthyes and Osteichthyes.
Answer:

Chondrichthyes Osteichthyes
1. Endoskeleton is made of cartilage. Endoskeleton is made of bones.
2. Exoskeleton is made of minute scales called placoid scales. Exoskeleton is made of large, flat and overlapping cycloid or ctenoid scales.
3. Mouth is ventral in position. Mouth is mostly terminal in position.
4. 5-7 pairs of gill slits without operculum are present. Four pairs of gills covered by operculum are present.
5. Caudal fin is heterocercal. Caudal fin is homocercal.
6. Males have copulatory organs called claspers located between the pelvic fins. Males lack claspers.
7. Air bladder is absent. Air bladder is present to maintain buoyancy. Thus, these fishes do not need to swim constantly.
8. Fertilization is internal. Fertilization is external.
9. Many of them are viviparous animals. They are oviparous animal
10. Scolidon (Dogfish), Pristis (Sawfish), Electric ray, Common skate, Hammer headed shark, Carcharodon (Great white shark), Trygon (Stingray), Anoxypristis. Exocoetus (Flying fish), Hippocampus (Sea horse), Labeo rohita, Pomphret (Rohu), Catla (Katla), Clarius (Magur), Pterophyllum (Angle fish), Bombay duck, Lung fishes (Protopterus, Lepidosireri), Aquarium fishes like Betta (Fighting fish).

Question 32.
Match the columns:

Column I (Organism) Column II (Characteristic Feature)
1. Euplectella (a) Exoskeleton formed of placoid scales
2. Periplaneta (b) Presence of mantle
3. Sepia (c) Water canal system
4. Scoliodon (d) Jointed appendages
5. Clarias (e) Exoskeleton formed of cycloid scales

Answer:

Column I (Organism) Column II (Characteristic Feature)
1. Euplectella (c) Water canal system
2. Periplaneta (d) Jointed appendages
3. Sepia (b) Presence of mantle
4. Scoliodon (a) Exoskeleton formed of placoid scales
5. Clarias (e) Exoskeleton formed of cycloid scales

Question 33.
What are tetrapods?
Answer:
Tetrapods are group of vertebrates that includes amphibians, reptiles, birds and mammals. It includes animals that bear two pairs of appendages (with some exceptions e.g. Snakes are limbless, etc.)

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 34.
Write a short note on amphibians.
Answer:

  1. These animals live on land as well as in water (freshwater only).
  2. Amphibians are poikilothermic animals.
  3. Body is differentiated into head, and trunk. Neck and tail are absent in many adults with few exceptions.
  4. Two pairs of limbs arise from the pectoral and pelvic girdles respectively, which help in locomotion.
  5. Skin is moist and glandular with mucous glands.
  6. Exoskeleton is absent.
  7. Eyelids are present. Tympanum represents the ear.
  8. Excretory products, digestive products and gametes are released through the common chamber cloaca.
  9. Circulatory system is of closed type. Heart is three-chambered and ventral. RBCs are biconvex and nucleated.
  10. Respiration is by skin, lungs and bucco-pharynx.
  11. Nervous system is well developed.
  12. Sexes are separate. Amphibians are oviparous. Fertilization is external and development is indirect through aquatic larval stage.
  13. They exhibit metamorphosis. e.g. Rana (Frog), Bufo (Toad), Salamandra (Salamander), Ichthyophis, Hyla (Tree frog), etc.

Question 35.
Name the limbless amphibian.
Answer:
Ichthyophis is a limbless amphibian.

Question 36.
Complete the table.

Phylum/Class Excretory organ Circulation Respiratory organ
Arthropoda Lungs/Gills/Tracheal system
Nephridia Closed Skin/Parapodia
Organ of Bojanus Open
Amphibia Closed Lung

Answer:

Phylum/Class Excretory organ Circulation Respirators organ
Arthropoda Malpighian tubule Open Lungs/Gills/Tracheal system
Annelida Nephridia Closed Skin/Parapodia
Mollusca Organ of Bojanus Open Ctenidia
Amphibia Kidneys Closed Lung

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 37.
Give the diagnostic characters of Reptilia.
Answer:
Diagnostic characters of Reptilia:

  1. Habitat: They are crawling animals. They are the first true terrestrial vertebrates. Few may be aquatic or semi- aquatic and are also found in marshy areas.
  2. Locomotion: Locomotion occurs by limbs in most animals. The limbs are pentadactyl with clawed digits, which help the animal to walk, creep or crawl. Snakes are limbless and crawl on their belly.
  3. Body temperature: They are poikilotherms.
  4. Exoskeleton: Skin is dry, non-glandular and covered by an exoskeleton of epidermal scales or scutes, shields or plates. Lizards and snake shed their skin periodically.
  5. Ear: Tympanum is present
  6. Circulatory system: It has two complete auricles but the ventricles are incompletely partitioned. Therefore, the heart of reptiles is not perfectly four chambered (except in crocodile the heart is four chambered).
  7. Nervous system: The brain is well developed. The olfactory lobes and cerebellum are better developed as compared to amphibians.
  8. Reproduction: Sexes are separate and exhibit prominent sexual dimorphism. Fertilization is internal and the animals are oviparous (exception – viper, it is viviparous). They show little parental care.
  9. e.g. Naja naja (Cobra), Hemidactylus (Wall lizard), Chelonia (Turtle), Crocodilus (Crocodile), Testudo (Tortoise), Chameleon (Tree lizard), Bangarus (Krait), Vipera (Viper).

Question 38.
Enlist the salient features of class Aves.
Answer:
The salient features of class Aves:

  1. Habitat: These animals are aerial in habitat.
  2. Locomotion: Forelimbs are modified into wings for flying. Hind limbs are used for walking, clasping tree branches and running. Aquatic birds have webbed toes. This helps in swimming, e.g. Duck.
  3. Body division: Body is differentiated into head, neck, trunk and tail.
  4. Body shape: Body is streamlined (boat-shaped) to reduce resistance during flight.
  5. Body temperature: These are warm-blooded animals (homeotherms) i.e., keep the body temperature constant irrespective of fluctuations in environmental temperature.
  6. Exoskeleton: Exoskeleton is made up of feathers. Scales are present on hind-limbs. Skin is thin, dry and non-glandular except oil gland at the base of tail (uropygial gland).
  7. Endoskeleton: Bones are hollow (pneumatic) with air cavities to reduce body weight.
  8. Digestion: Jaws are modified into beaks. Teeth are absent. Special structures such as crop and gizzard are present.
  9. Circulatory system: They show double circulation. Blood is red in colour due to presence of biconvex and nucleated RBCs. Heart is perfectly four-chambered, with two auricles and two ventricles.
  10. Respiration: Respiration occurs by lungs. Presence of air sacs increases the buoyancy.
  11. Nervous system: Brain is enlarged with a well developed cerebellum for equilibrium.
  12. Reproduction: Sexes are separate and the animals exhibit prominent sexual dimorphism.
  13. The female shows presence of only left ovary and left oviduct.

This helps to reduce body weight during flying. Fertilization is internal. Avians are oviparous. Parental care is very well developed.
e.g. Columba (Pigeon), Psittacula (Parrot), Struthio (Ostrich), Kiwi, Aptenodytes (Penguin), Corvus (Crow), Neophron (Vulture), Passer (Sparrow), Pavo (Peacock), etc.

Question 39.
Name the flightless bird.
Answer:
Ostrich

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 40.
Give important features of class Mammalia.
Answer:
Features of class Mammalia (mammae: breasts, nipple):

  1. Special feature: Presence of mammary glands (milk-producing glands) for the nourishment of young ones. Mammary glands are modified sweat glands.
  2. Habitat: Mammals are omnipresent (present everywhere). These are mostly terrestrial, some are aquatic and few are aerial and arboreal (living on trees).
  3. Locomotion: Limbs are the organs of locomotion and are modified for walking, climbing, burrowing, swimming, etc.
  4. Body division: Body is differentiated into head, neck, trunk and tail. They have external ear (pinna).
  5. Body temperature: Mammals are homeotherms or warm-blooded animals.
  6. Exoskeleton: It is in the form of hair, fur, nails, hooves, horns, etc.
  7. Skin: Skin is glandular and has sweat glands and sebaceous (oil) glands.
  8. Mouth cavity: Mammals show heterodont dentition (various types of teeth like incisors, canines, premolars and molars).
  9. Circulation: Heart is ventral in position, four chambered with two auricles and two ventricles. RBCs are biconcave and enucleated (except camel). Blood is red in colour.
  10. Respiration: Respiration takes place by lungs.
  11. Nervous system: Brain is highly developed. Cerebrum shows a transverse band called corpus callosum.
  12. Reproduction and development: Only few mammals are oviparous, e.g. Duck billed platypus. Some have pouches for development of immature young ones. These are called marsupials, e.g. Kangaroo. Most of the mammals are placental and viviparous.

Question 41.
Give examples of animals belonging to class Mammalia.
Answer:
Bat, Rattus(rat), Macaca (monkey), Camelus (camel), Whale, Human being, Canis (dog), Elephas (elephant), Equus (horse), Pteropus (flying fox), Ornithorhynchus (platypus), Macropus (kangaroo), Trachypithecus.

Question 42.
Distinguish between Reptilia, Amphibia and Aves.
Answer:

Reptilia Amphibia Aves
Members of Reptilia are terrestrial, with few exceptions. Members of Amphibia live on land as well as in water. Aves are terrestrial and aquatic.
They are poikilothermic. They are poikilothermic. They are homeothermic.
All reptiles have three chambered heart, except for crocodiles. They have three chambered heart. They have four-chambered heart.
Olfactory lobes and cerebellum are better developed than those of amphibians. Olfactory lobes and cerebellum are less developed as compared to reptiles. Cerebellum is well developed for equilibrium.
Skin is dry, non-glandular and covered by scales and plates. Skin is moist, glandular with mucous glands. Skin is thin, dry, non-glandular except oil gland at the base of tail.
Digits bear claws. Digits do not bear claws. Digits bear claws.
Exoskeleton bears epidermal scales or scutes, shields or plates. Exoskeleton is absent. Exoskeleton is made up of feathers.

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 43.
Name the phyla to which the following animals belong:

Question 1.
Diploria
Answer:
Cnidaria

Question 2.
Ancyclostoma
Answer:
Aschelminthes

Question 3.
Nereis
Answer:
Annelida

Question 4.
Hottentotta
Answer:
Arthropoda

Question 5.
Chaetopleura
Answer:
Mollusca

Question 6.
Ophiothrix
Answer:
Echinodermata

Question 7.
Rhabdopleura
Answer:
Hemichordata

Question 8.
Exocoetus
Answer:
Chordata

Question 9.
Lepidosiren
Answer:
Chordata

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 44.
Apply Your Knowledge:

Question 1.
A college conducted an inter-college quiz competition. During a round in the quiz, the students were asked to identify animals with respect to their characteristic features given below.

  1. A limbless reptile
  2. Gastrovascular cavity in Hydra
  3. The phylum which includes ringworms or segmented worms
  4. An oviparous mammal
  5. The phylum which includes comb jellies

Answer:

  1. Snake
  2. Coelenteron
  3. Phylum Annelida
  4. Duck-billed platypus
  5. Phylum Ctenophora

Question 45.
An organism has long cylindrical thread-like body. Its body wall has longitudinal muscles but no circular muscles. It is a pseudocoelomate. Identify the phylum to which it belongs.
Answer:
Aschelminthes

Question 46.
Classify the given animals in their respective groups.
Macropus, Struthio, Equus, Bufo, Anura, Salamander, Naja naja, Hippocampus, Bombay duck, Lamprey, Hagfish, Doliolum, Aplysia, Wuchereria, Physalia, Euplectella, Krait, Scypha, Ctenoplana, Brain coral, Obelia, Loligo.
Answer:

Phylum / Subphylum / Class Animals
1. Porifera Euplectella, Scypha
2. Cnidaria Physalia, Brain Coral, Obelia
3. Ctenophora Ctenoplana
4. Mollusca Aplysia, Loligo
5. Aschelminthes Wuchereria
6. Osteichthyes Hippocampus, Bombay duck
7. Cyclostomata Lamprey, Hagfish
8. Urochordata Doliolum
9. Amphibia Bufo, Anura, Salamander
10. Reptilia Naja naja, Krait
11. Aves Struthio
12. Mammalia Macropus, Equus

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 47.
Match the Column.
Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia 9
Answer:
1 – b,
2 – c,
3 – d,
4 – a

Quick Review:

Classification of animals at a glance:

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia 10

Question 48.
Exercise:

Question 1.
Give the difference between diploblastic and triploblastic animals?
Answer:
Number of germ layers:
(a) When an organism shows only two germ layers, they are called diploblastic animals. In this case, the outer ectoderm is separated from the inner endoderm by a non-living substance called mesoglea.
(b) When an organism shows three germinal layers, they are called triploblastic animals. The three layers are namely – outer ectoderm, middle mesoderm and inner endoderm.

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 2.
Name the superclass under the division Gnathostomata.
Answer:
Division Gnathostomata:
This division includes animals with jaws.
It is divided into two superclasses: Pisces (bear fins) and Tetrapoda (bear four limbs)
[Note: Students can scan the given Q.R code for understanding the characteristics of vertebrates.]

Question 3.
In which group, notochord is present only in the tail of larva.
Answer:
Notochord: Notochord is present only in the tail of the larva and is lost during metamorphosis. Hence, the name Urochordata.

Question 4.
Comment on respiration in Aves.
Answer:
Respiration: Respiration occurs by lungs. Presence of air sacs increases the buoyancy.

Question 5.
Aves and Pisces have stream-lined body.
What is the significance of this type of body.
Answer:
Exoskeleton: Exoskeleton is made up of feathers. Scales are present on hind-limbs. Skin is thin, dry and non-glandular except oil gland at the base of tail (uropygial gland).

Question 6.
Which type of circulation occurs in Aves?
Answer:
Circulatory system: They show double circulation. Blood is red in colour due to presence of biconvex and nucleated RBCs. Heart is perfectly four chambered, with two auricles and two ventricles.

Question 7.
Which group of chordates possess sucking and circular mouth without jaws?
Answer:
Characteristic features of class Cyclostomata (Cyclos: Circular, stoma-mouth) Lat/Grk

  1. Members of class Cyclostomata are jaw-less and eel like organisms.
  2. Their skin is devoid of scales, soft and smooth, containing unicellular mucus glands.
  3. Median fms are present but paired fins are absent.
  4. They are ectoparasites on fishes.
  5. They have sucking circular mouth, without jaws.
  6. Cranium and vertebral column are made up of cartilage.
  7. Their digestive system lacks stomach.
  8. Respiration occurs by 6 – 15 pairs of gill slits. Gills slits are without operculum.
  9. Heart is two chambered with one auricle and one ventricle.
  10. Gonad is single, large and without gonoduct.
  11. Fertilization is external. They are anadromous as they migrate for spawning to fresh – water from marine habitat.
  12. After spawning, they die within few days. Larvae metamorphosize and migrate to ocean.e.g Petromyzon (Lamprey), Myxine (Hagfish).

Question 8.
Give any four characteristic features of sponges.
Answer:
Phylum Porifera (Pori = Pores: feron = bearing): Members of the phylum Porifera are also called sponges. Characteristic features of the phylum:

  1. Habitat: They are aquatic, mostly marine but few species are found in freshwater.
  2. Forms: They are sedentary animals (attached to substratum or rock).
  3. Body shape: They have asymmetrical body. Body of these animals consists of many cells with minimal
    division of labour among cells. Hence, their body is considered as a colony of different types of cells.
  4. Body surface: Their body bears minute pores called ‘ostia’ through which water enters the spongocoel (body cavity). Water leaves the body through a large opening called ‘osculum’. Beating of flagella creates water current.
  5. Circulation: Water is circulated in the body through the ‘canal system’. When the water enters the body of poriferans, cells absorb the food, exchange respiratory gases and release excretory products.
  6. Digestive system: The body cavity of sponges (spongocoel) is lined by unique type of flagellated cells called choanocytes or collar cells for digestion.
  7. Endoskeleton: The body of sponges consists of calcareous / siliceous spicules and proteinaceous ‘spongin fibres’.
  8. Reproduction: Sponges reproduce asexually as well as sexually. Asexual reproduction takes place by fragmentation and gemmule formation. Sexual reproduction is by formation of gametes. Fertilization is internal and development is indirect through larval stage.
  9. Sponges have great power of regeneration.
    e.g. Scypha, Euspongia (Bath sponge), Euplectella (Venus’ flower basket).

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 9.
Mention the role of cnidoblasts.
Answer:
Cnidoblasts: Presence of cnidoblasts or stinging cells are present on the tentacles for anchorage, offence and defence.

Question 10.
Name the oil gland present at the base of tail in Aves.
Answer:
Exoskeleton: Exoskeleton is made up of feathers. Scales are present on hind-limbs. Skin is thin, dry and non-glandular except oil gland at the base of tail (uropygial gland).

Question 11.
Why are cyclostomes termed as anadromous?
Answer:
Nervous system: Brain is enlarged with a well developed cerebellum for equilibrium.

Question 12.
Write a short note on Urochordates.
Answer:
Distinguishing features of Tunicata or Urochordata:

  1. Habitat: They are exclusively marine.
  2. Body covering: Soft body is covered by ‘test’ or ‘tunic’ which is made up of tunicine.
  3. Notochord: Notochord is present only in the tail of the larva and is lost during metamorphosis. Hence, the name Urochordata.
  4. Respiration: Pharynx with many gill slits for respiration.
  5. Circulation: Closed circulatory system is present.
  6. Reproduction: Development is indirect, e. g. Herdmania, Salpa, Doliolum, Ascidia.

Question 13.
Enlist the characters of second largest phylum of animal kingdom.
Answer:
Mollusca (Mollis: Soft) is the second largest phylum.
Unique features of phylum Mollusca:

  1. Habitat: They are aquatic or seen in marshy places. Few of them are terrestrial.
  2. Forms: Molluscs are either free-living or sedentary.
  3. Body plan: These are soft bodied and show tube within a tube type of body plan.
  4. Body symmetry: Most of the Molluscs show bilateral symmetry but few are asymmetrical due to torsion (twisting).
  5. Body division: Body consists of head, foot and visceral mass. Visceral mass is enclosed in thick, muscular fold of body wall called mantle. Mantle secretes a hard-calcareous shell, that may be external or internal or absent. Muscular foot is present ventrally.
  6. Digestive system: Digestive system is well-developed and complete with anterior mouth and posterior anus. Buccal cavity has a rasping organ called radula (helps in feeding), which is provided with transverse rows of teeth.
  7. Respiration: In aquatic forms, numerous feather-like gills called ctenidia, help in aquatic respiration. Terrestrial forms may show the presence of lungs.
  8. Circulatory system: Circulatory system is of open type (except in Sepia, where it is of the closed type). Blood contains a copper containing blue coloured respiratory pigment called haemocyanin.
  9. Excretion: Excretion occurs by kidney like structures, also called ‘Organ of Bojanus’.
  10. Nervous system and sense organs; Nervous system has three rjahs of Mf hxese t&e,
    interconnected by commissures and connectives.
  11. Sense orgAnswer:Sense organs such as eyes for vision, tentacles for tactile sensation and osphradia for testing purity of water is present.
  12. Sexual reproduction: Sexes are usually separate. Animals of this phylum are mostly oviparous and the development is direct or indirect.
  13. e.g. Pila, Spisula (Bivalve), Octopus (devil fish), Sepia (cuttle fish), Chaetopleura (Chiton), Pinctada (Pearl oyster), Loligo (Squid), Aplysia (Sea hare), Dentalium (Tusk shell).

Question 14.
Members of which phylum are known as segmented or ring worms?
Answer:
Annelids are commonly called as ring worms or segmented worms.
Characteristic feature of phylum Annelida:

  1. Forms: Annelids may be aquatic, ectoparasitic or free – living or burrowing in moist soil.
  2. Body symmetry: They are bilaterally symmetrical.
  3. Body coelom: They are true coelomates.
  4. Segmentation: Body is metamerically segmented and has a special region called clitellum.
  5. Digestive system: Alimentary canal is complete.
  6. Locomotion: Locomotion takes place with the help of setae (earthworm), parapodia (Nereis) or suckers (leech). Well developed longitudinal and circular muscles help in locomotion.
  7. Nervous system: It consists of nerve ring and ventral solid and ganglionated nerve cord.
  8. Reproduction: Mostly are hermaphrodites and few are dioecious (Nereis).
  9. Respiration: Exchange of gases takes place through body wall.
  10. Circulation: Circulatory system is of closed type. Excretion and osmoregulation is carried out with help of nephridia. e.g. Nereis (Aquatic annelid), Pheretima (Earthworm), Hirudinaria (Leech).

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 15.
Mention the unique features of ctenophores.
Answer:
The members of this phylum are commonly known as comb jellies and sea walnuts. They are also known as acnidarians as they lack cnidoblasts. The phylum is considered as one of the minor phyla as it is represented by very few members.

Salient features of phylum Ctenophora:

  1. Habitat: They are exclusively marine.
  2. Forms: They are free swimming animals.
  3. Germ layers: Members of this phylum are diploblastic.
  4. Body Symmetry: They are radially symmetrical.
  5. Body plan: The animals of this phylum show blind-sac body plan.
  6. Body organization: They show tissue level organization.
  7. Locomotion: It is earned out by eight rows of ciliated comb plates.
  8. Bioluminescence: It is the characteristic feature of the members of this phylum.
  9. Digestion: It is extracellular and intracellular.
  10. Reproduction: Reproduction is sexual with indirect development.
  11. Colloblasts: These sticky cells are used to capture prey, e.g. Pleurobrachia, Ctenoplan

Question 16.
What is ecdysis?
Answer:
Exoskeleton: Body is covered by a tough, non – living chitinous exoskeleton. As the exoskeleton does not allow body growth, arthropods shed off their exoskeleton periodically during growth. This process is called moulting or ecdysis.

Question 17.
Give the characteristic features of class Cephalochordata.
Answer:

  1. Cephalochordates are also known as lancelets and are small fish-like animals that rarely exceed 5 cm in length.
  2. Lancelets are exclusively marine and live partly buried in soft marine sediments.
  3. Notochord extends throughout entire length of the body and persists throughout life.
  4. Myotomes (muscle blocks) are present.
  5. Post anal tail is present.
  6. Circulatory system is closed type. Blood lacks pigment, e.g. Branchiostoma

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 18.
Give an example of:

Question 1.
Animals whose body is covered by shell.
Answer:
Mollusca (Mollis: Soft) is the second largest phylum.
Unique features of phylum Mollusca:

  1. Habitat: They are aquatic or seen in marshy places. Few of them are terrestrial.
  2. Forms: Molluscs are either free-living or sedentary.
  3. Body plan: These are soft bodied and show tube within a tube type of body plan.
  4. Body symmetry: Most of the Molluscs show bilateral symmetry but few are asymmetrical due to torsion (twisting).
  5. Body division: Body consists of head, foot and visceral mass. Visceral mass is enclosed in thick, muscular fold of body wall called mantle. Mantle secretes a hard-calcareous shell, that may be external or internal or absent. Muscular foot is present ventrally.
  6. Digestive system: Digestive system is well-developed and complete with anterior mouth and posterior anus. Buccal cavity has a rasping organ called radula (helps in feeding), which is provided with transverse rows of teeth.
  7. Respiration: In aquatic forms, numerous feather-like gills called ctenidia, help in aquatic respiration. Terrestrial forms may show the presence of lungs.
  8. Circulatory system: Circulatory system is of open type (except in Sepia, where it is of the closed type). Blood contains a copper containing blue coloured respiratory pigment called haemocyanin.
  9. Excretion: Excretion occurs by kidney like structures, also called ‘Organ of Bojanus’.
  10. Nervous system and sense organs; Nervous system has three rjahs of Mf hxese,
    interconnected by commissures and connectives.
  11. Sense organs: Sense organs such as eyes for vision, tentacles for tactile sensation and osphradia for testing purity of water is present.
  12. Sexual reproduction: Sexes are usually separate. Animals of this phylum are mostly oviparous and the development is direct or indirect.
  13. e.g. Pila, Spisula (Bivalve), Octopus (devil fish), Sepia (cuttle fish), Chaetopleura (Chiton), Pinctada (Pearl oyster), Loligo (Squid), Aplysia (Sea hare), Dentalium (Tusk shell).

Question 2.
Animals with organs of Bojanus for excretion.
Answer:
Mollusca (Mollis: Soft) is the second largest phylum.
Unique features of phylum Mollusca:

  1. Habitat: They are aquatic or seen in marshy places. Few of them are terrestrial.
  2. Forms: Molluscs are either free-living or sedentary.
  3. Body plan: These are soft bodied and show tube within a tube type of body plan.
  4. Body symmetry: Most of the Molluscs show bilateral symmetry but few are asymmetrical due to torsion (twisting).
  5. Body division: Body consists of head, foot and visceral mass. Visceral mass is enclosed in thick, muscular fold of body wall called mantle. Mantle secretes a hard-calcareous shell, that may be external or internal or absent. Muscular foot is present ventrally.
  6. Digestive system: Digestive system is well-developed and complete with anterior mouth and posterior anus. Buccal cavity has a rasping organ called radula (helps in feeding), which is provided with transverse rows of teeth.
  7. Respiration: In aquatic forms, numerous feather-like gills called ctenidia, help in aquatic respiration. Terrestrial forms may show the presence of lungs.
  8. Circulatory system: Circulatory system is of open type (except in Sepia, where it is of the closed type). Blood contains a copper-containing blue coloured respiratory pigment called haemocyanin.
  9. Excretion: Excretion occurs by kidney like structures, also called ‘Organ of Bojanus’.
  10. Nervous system and sense organs; Nervous system has three rjahs of Mf hxese,
    interconnected by commissures and connectives.
  11. Sense organs: Sense organs such as eyes for vision, tentacles for tactile sensation and osphradia for testing purity of water is present.
  12. Sexual reproduction: Sexes are usually separate. Animals of this phylum are mostly oviparous and the development is direct or indirect.
  13. e.g. Pila, Spisula (Bivalve), Octopus (devil fish), Sepia (cuttle fish), Chaetopleura (Chiton), Pinctada (Pearl oyster), Loligo (Squid), Aplysia (Sea hare), Dentalium (Tusk shell).

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 19.
Distinguish between Arthropoda and Mollusca
Answer:
Arthropoda (Arthros: Joint, Podos: leg): Arthropoda forms the largest phylum of kingdom Animalia. Characteristics of Arthropoda:
a. Habitat: Arthropods are omnipresent.
b. Forms: Solitary or colonial, most of them are free-living. Barnacles are sedentary. Few are parasitic and sanguivorous, (e.g. Female mosquito, bed bug.)
c. Body symmetry: Body is bilaterally symmetrical.
d. Germ layers: They are triploblastic.
e. Body cavity: Arthropods are eucoelomates.
f. Body plan: They show tube within tube body plan.
g. Level of body organization: They show organ system level of organization.
h. Special features: The members of this phylum have jointed appendages. Hence, they are known as arthropods. Some insects like honey bee, ants, termites, etc. exhibit polymorphism.
i. Exoskeleton: Body is covered by a tough, non – living chitinous exoskeleton. As the exoskeleton does not allow body growth, arthropods shed off their exoskeleton periodically during growth. This process is called moulting or ecdysis.
j. Body division: Body is divided into head, thorax and abdomen.
k. Segmentation: Body shows metameric segmentation.
l. Digestion: Digestive system is complete and divided into foregut, midgut and hindgut.
m. Circulation: Circulatory system is of open type wherein, blood flows through body cavity called haemocoel.
n. Respiration: Respiration occurs through respiratory organs like gills, trachea, book lungs or book
gills.
o. Excretion: Excretion takes place by green glands, Malpighian tubules or coxal glands.
p. Nervous system: Nervous system consists of nerve ring and double, ventral ganglionated nerve cord.
q. Sense organs:Arthropods have well developed sense organs in the form of antennae, simple or compound eye and various receptors.
r. Sexual reproduction: Sexes are generally separate in arthropods with distinct sexual dimorphism.
s. Significance:
Beneficial arthropods: Some arthropods are of economic importance. For example, Honey bees (Apis) are important for their honey and wax, silk worms for the production of silk. Lobsters, prawns, crabs are edible. Harmful arthropods: Some arthropods are harmful and act as vectors to spread various diseases, e.g., Mosquitoes. Locusta (locust) is a gregarious pest. Limulus (King crab) is a living fossil.
Other examples: Cockroach (Periplaneta), butterfly, scorpion (Hottentotta) and millipede (Archispirostreptus) prawn.
Mollusca (Mollis: Soft) is the second largest phylum.

Unique features of phylum Mollusca:

  1. Habitat: They are aquatic or seen in marshy places. Few of them are terrestrial.
  2. Forms: Molluscs are either free-living or sedentary.
  3. Body plan: These are soft bodied and show tube within a tube type of body plan.
  4. Body symmetry: Most of the Molluscs show bilateral symmetry but few are asymmetrical due to torsion (twisting).
  5. Body division: Body consists of head, foot and visceral mass. Visceral mass is enclosed in thick, muscular fold of body wall called mantle. Mantle secretes a hard-calcareous shell, that may be external or internal or absent. Muscular foot is present ventrally.
  6. Digestive system: Digestive system is well-developed and complete with anterior mouth and posterior anus. Buccal cavity has a rasping organ called radula (helps in feeding), which is provided with transverse rows of teeth.
  7. Respiration: In aquatic forms, numerous feather-like gills called ctenidia, help in aquatic respiration. Terrestrial forms may show the presence of lungs.
  8. Circulatory system: Circulatory system is of open type (except in Sepia, where it is of the closed type). Blood contains a copper containing blue coloured respiratory pigment called haemocyanin.
  9. Excretion: Excretion occurs by kidney like structures, also called ‘Organ of Bojanus’.
  10. Nervous system and sense organs; Nervous system has three rjahs of Mf hxese,
    interconnected by commissures and connectives.
  11. Sense organs: Sense organs such as eyes for vision, tentacles for tactile sensation and osphradia for testing purity of water is present.
  12. Sexual reproduction: Sexes are usually separate. Animals of this phylum are mostly oviparous and the development is direct or indirect.
  13. e.g. Pila, Spisula (Bivalve), Octopus (devil fish), Sepia (cuttle fish), Chaetopleura (Chiton), Pinctada (Pearl oyster), Loligo (Squid), Aplysia (Sea hare), Dentalium (Tusk shell).

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 20.
Name the phylum that forms connecting link between Chordates and Non-chordates.
Answer:

  1. Hemiehordata was earlier considered as sub phylum of Chordata because the buccal diverticulum was considered as notochord. It is now placed as a separate phylum under Non-Chordata.
  2. It possesses certain characteristics of both Chordates and Non-chordates.
  3. Absence of notochord worm-like body, heart located on the dorsal side are the Non-chordate like characteristics seen in Hemiehordata.
  4. Presence of nerve chord, pharyngeal gill slits are some of the Chordate-like characters seen in Hemiehordata.
  5. Hence, Hemiehordata is considered as a connecting link between Non-chordata and Chordata.

Question 21.
Comment on the reproduction of the members of the phylum Platyhelminthes.
Answer:
The members of this phylum are called as flatworms.
Unique features of phylum Platyhelminthes:

  1. Body shape: Body of these animals is dorsoventrally flattened.
  2. Coelom: They are acoelomates.
  3. Germ layers: Platyhelminthes are triploblastic.
  4. Body organization: They show organ-system grade of organization.
  5. Forms: Most of them are endoparasites and few are free-living. Parasitic forms have hooks and suckers for attachment to the hosts’ body. In parasitic forms, body is covered by cuticle and in free-living forms it is covered by cilia.
  6. Digestive system: Parasitic forms generally lack digestive system. In free-living forms, the digestive system is incomplete.
  7. Body plan: They show blind-sac body plan.
  8. Excretion and osmoregulation: It occurs by flame cells or protonephridia.
  9. Reproduction: They are mostly hermaphrodite (bisexual). Self fertilization is seen.
  10. Few animals show high power of regeneration and show polyembryony. e.g. Planaria, Taenia (Tapeworm), Fasciola (Liver fluke) are included in this phylum.

Question 22.
Give a list of aerial adaptations shown by birds.
Answer:

  1. In birds, the forelimbs are modified into wings for flying.
  2. They possess stream-lined body to reduce resistance during flight.
  3. Bones are hollow or pneumatic to reduce body weight.
  4. In order to reduce body weight, urinary bladder is absent. Also, females possess only left ovary and oviduct.
  5. Body is covered by feathers to facilitate flying.

Question 23.
Write a short note on Superclass Pisces. Give one example.
Answer:
Important features of superclass Pisces:

  1.  Habitat: These are aquatic animals and are present in fresh, marine and brackish waters.
  2. Body temperature: Pisces are poikilothermic animals i.e., cold blooded animals, in which body temperature changes according to the change in the surrounding temperature.
  3. Sensory orgAnswer:They have lateral line system which shows the presence of rheoreceptors for the detection of water current.
  4. Locomotion: Locomotion is by body muscles and fins. Caudal fin helps in steering wheel.
  5. Skeleton: Exoskeleton is made up of dermal scales. It is either bony or cartilaginous.
  6. Body shape: Body is streamlined and boat-shaped and helps to overcome resistance during swimming.
  7. Respiration: Respiration is by gills.
  8. Circulation: It shows single and closed circulation. Heart is two-chambered and ventral in position. Heart of fishes is described as venous heart (presence of deoxygenated blood).
  9. Nervous system: They have a well-developed brain with large olfactory lobes.
  10. Reproduction: Sexes are separate. Most of the fishes are oviparous, however some are viviparous.

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 24.
Give any six salient features of class Cyclostomata.
Answer:
Characteristic features of class Cyclostomata (Cyclos: Circular, stoma-mouth) Lat/Grk

  1. Members of class Cyclostomata are jaw-less and eel like organisms.
  2. Their skin is devoid of scales, soft and smooth, containing unicellular mucus glands.
  3. Median fms are present but paired fins are absent.
  4. They are ectoparasites on fishes.
  5. They have sucking circular mouth, without jaws.
  6. Cranium and vertebral column are made up of cartilage.
  7. Their digestive system lacks stomach.
  8. Respiration occurs by 6 – 15 pairs of gill slits. Gills slits are without operculum.
  9. Heart is two chambered with one auricle and one ventricle.
  10. Gonad is single, large and without gonoduct.
  11. Fertilization is external. They are anadromous as they migrate for spawning to fresh – water from marine habitat.
  12. After spawning, they die within few days. Larvae metamorphosize and migrate to ocean.
    e.g Petromyzon (Lamprey), Myxine (Hagfish).

Question 25.
Describe salient features of Phylum Echinodermata.
Answer:
Salient features of phylum Echinodermata (Echinus – spines, derma – skin)

  1. Habitat: These are exclusively marine.
  2. Forms: Members of this phylum are solitary, sedentary or free-living and gregarious, benthic.
  3. Body symmetry: These animals are radially symmetrical with pentamerous symmetry.
  4. Shape: Members of Echinodermata are spherical, elongated or star – shaped.
  5. Body: The endoskeleton is made up of calcareous ossicles. Spines are formed on the body. Hence, they are known as echinoderms. The body has two sides oral and aboral and lacks definite divisions. Mouth is ventrally present on oral surface and anus on aboral surface.
  6. Water vascular system: Presence of water vascular system is the peculiar character of echinoderms.
  7. MadrepOrite is the opening of water vascular system through which water enters. Water vascular system is useful in locomotion, food capturing, respiration.
  8. Digestion: Digestive system is complete.
  9. Respiration: Peristomial gills, papillae, respiratory tree, etc. are used for respiration.
  10. Circulatory and excretory systems: Absent in echinoderms.
  11. Nervous system: Nervous system is simple with a nerve ring around the mouth and radial nerves in arms.
  12. Reproduction and development: Sexes are separate (sometimes bisexual). Fertilization is external.
  13. Development is indirect, i.e. through larval stages. They show high power of regeneration.
    e.g. Sea lily (Antedon), Sea star (Asterias), Sea cucumber (Cucumaria), Brittle star (Ophiothrix), Sea urchin (Echinus).

Question 26.
Explain in brief the characteristic features of Phylum Hemichordata.
Answer:
The given organism (Balanoglossus) belongs to phylum Hemichordata.
Characteristics of phylum Hemichordata:

  1. Habitat: Hemichordates are exclusively marine animals, usually living at the bottom of the sea in burrows. These are mostly free – living but animals like Rhabdopleura are sedentary.
  2. Body shape and division: Body is soft and vermiform. It is unsegmented and divided into three parts namely – proboscis, collar and trunk.
  3. Digestive system: Alimentary canal is complete, straight or ‘U’ shaped. Buccal cavity gives rise to a rod-like buccal diverticulum.
  4. Respiration: Respiration is brought about by numerous gills arranged in two longitudinal rows present in the pharyngeal region. Gills open by gill slits.
  5. Circulation: Circulatory system is simple and open type.
  6. Excretion: It takes place with help of with the glomerulus.
  7. Nervous system: Nervous tissue is embedded in epidermis on the dorsal as well as the ventral side.
  8. Reproduction and development: Sexes are separate (sometimes bisexual). Fertilization is external and development is indirect through free swimming larva.

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 27.
Write the distinguishing features of class Reptilia.
Answer:
Diagnostic characters of Reptilia:

  1. Habitat: They are crawling animals. They are the first true terrestrial vertebrates. Few may be aquatic or semi- aquatic and are also found in marshy areas.
  2. Locomotion: Locomotion occurs by limbs in most animals. The limbs are pentadactyl with clawed digits, which help the animal to walk, creep or crawl. Snakes are limbless and crawl on their belly.
  3. Body temperature: They are poikilotherms.
  4. Exoskeleton: Skin is dry, non-glandular and covered by an exoskeleton of epidermal scales or scutes, shields or plates. Lizards and snake shed their skin periodically.
  5. Ear: Tympanum is present
  6. Circulatory system: It has two complete auricles but the ventricles are incompletely partitioned. Therefore, the heart of reptiles is not perfectly four chambered (except in crocodile the heart is four chambered).
  7. Nervous system: The brain is well developed. The olfactory lobes and cerebellum are better developed as compared to amphibians.
  8. Reproduction: Sexes are separate and exhibit prominent sexual dimorphism. Fertilization is internal and the animals are oviparous (exception – viper, it is viviparous). They show little parental care.
  9. e.g. Naja naja (Cobra), Hemidactylus (Wall lizard), Chelonia (Turtle), Crocodilus (Crocodile), Testudo (Tortoise), Chameleon (Tree lizard), Bangarus (Krait), Vipera (Viper).

Question 28.
Mention the unique features of Phylum Cnidaria.
Answer:
Characteristics of members belonging to phylum Cnidaria:

  1. Habitat: They are aquatic, mostly marine and few of them are fresh – water forms.
  2. Forms: They are sessile or free swimming.
  3. Cnidoblasts: Presence of cnidoblasts or stinging cells are present on the tentacles for anchorage, offence and defence.
  4. Body Symmetry: They have radially symmetrical body.
  5. Germ layer: They are diploblastic.
  6. Body cavity: Cnidarians have a central cavity called coelenteron or gastrovascular cavity, which helps in digestion and circulation. They have blind-sac body plan i.e., single pore opening to the exterior in the digestive system.
  7. Body form: Members of this phylum exhibit two body forms. The cylindrical form, known as polyp e.g. Hydra and the umbrella – like form (.Aurelia – jelly fish) is known as medusa.
  8. Digestion: They have extracellular and intracellular digestion.
  9. Reproduction: Cnidarians reproduce asexually and sexually. Asexual reproduction takes place by budding and regeneration.
  10. Sexual reproduction takes place gamete formation. They exhibit metagenesis i.e. alternation of polypoid generation with medusoid generation. Polyps produce medusae asexually and medusae produce polyps sexually, e.g. Obelia
  11. e.g. Hydra, Aurelia (Jellyfish), Physalia (Portuguese man-of-war), Adamsia (Sea anemone), Diploria (Brain coral), Gorgonia (sea fan).

Question 29.
Describe salient features of phylum Arthropoda.
Answer:
Arthropoda (Arthros: Joint, Podos: leg): Arthropoda forms the largest phylum of kingdom Animalia. Characteristics of Arthropoda:
a. Habitat: Arthropods are omnipresent.
b. Forms: Solitary or colonial, most of them are free-living. Barnacles are sedentary. Few are parasitic and sanguivorous, (e.g. Female mosquito, bed bug.)
c. Body symmetry: Body is bilaterally symmetrical.
d. Germ layers: They are triploblastic.
e. Body cavity: Arthropods are eucoelomates.
f. Body plan: They show tube within tube body plan.
g. Level of body organization: They show organ system level of organization.
h. Special features: The members of this phylum have jointed appendages. Hence, they are known as arthropods. Some insects like honey bee, ants, termites, etc. exhibit polymorphism.
i. Exoskeleton: Body is covered by a tough, non – living chitinous exoskeleton. As the exoskeleton does not allow body growth, arthropods shed off their exoskeleton periodically during growth. This process is called moulting or ecdysis.
j. Body division: Body is divided into head, thorax and abdomen.
k. Segmentation: Body shows metameric segmentation.
l. Digestion: Digestive system is complete and divided into foregut, midgut and hindgut.
m. Circulation: Circulatory system is of open type wherein, blood flows through body cavity called haemocoel.
n. Respiration: Respiration occurs through respiratory organs like gills, trachea, book lungs or book
gills.
o. Excretion: Excretion takes place by green glands, Malpighian tubules or coxal glands.
p. Nervous system: Nervous system consists of nerve ring and double, ventral ganglionated nerve cord.
q. Sense organs: Arthropods have well developed sense organs in the form of antennae, simple or compound eye and various receptors.
r. Sexual reproduction: Sexes are generally separate in arthropods with distinct sexual dimorphism.
s. Significance:
Beneficial arthropods: Some arthropods are of economic importance. For example, Honey bees (Apis) are important for their honey and wax, silk worms for the production of silk. Lobsters, prawns, crabs are edible. Harmful arthropods: Some arthropods are harmful and act as vectors to spread various diseases, e.g., Mosquitoes. Locusta (locust) is a gregarious pest. Limulus (King crab) is a living fossil.
Other examples: Cockroach (Periplaneta), butterfly, scorpion (Hottentotta) and millipede (Archispirostreptus) prawn.

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 30.
Name the following.

Question 1.
Pores on the body of sponges through which the water enters.
Answer:
Body surface: Their body bears minute pores called ‘ostia’ through which water enters the spongocoel (body cavity). Water leaves the body through a large opening called ‘osculum’. Beating of flagella creates water current.

Question 2.
Brain coral belongs to this phylum.
Answer:
Characteristics of members belonging to phylum Cnidaria:

  1. Habitat: They are aquatic, mostly marine and few of them are fresh – water forms.
  2. Forms: They are sessile or free swimming.
  3. Cnidoblasts: Presence of cnidoblasts or stinging cells are present on the tentacles for anchorage, offence and defence.
  4. Body Symmetry: They have radially symmetrical body.
  5. Germ layer: They are diploblastic.
  6. Body cavity: Cnidarians have a central cavity called coelenteron or gastrovascular cavity, which helps in digestion and circulation. They have blind-sac body plan i.e., single pore opening to the exterior in the digestive system.
  7. Body form: Members of this phylum exhibit two body forms. The cylindrical form, known as polyp e.g. Hydra and the umbrella – like form (.Aurelia – jelly fish) is known as medusa.
  8. Digestion: They have extracellular and intracellular digestion.
  9. Reproduction: Cnidarians reproduce asexually and sexually. Asexual reproduction takes place by budding and regeneration. Sexual reproduction takes place gamete formation.
  10. They exhibit metagenesis i.e. alternation of polypoid generation with medusoid generation. Polyps produce medusae asexually and medusae produce polyps sexually, e.g. Obelia
  11. e.g. Hydra, Aurelia (Jellyfish), Physalia (Portuguese man-of-war), Adamsia (Sea anemone), Diploria (Brain coral), Gorgonia (sea fan).

Question 3.
Annelid with parapodia.
Answer:
Locomotion: Locomotion takes place with the help of setae (earthworm), parapodia (Nereis) or suckers (leech). Well developed longitudinal and circular muscles help in locomotion.

Question 4.
Groups under phylum Chordata which include poikilotherms?
Answer:
Important features of superclass Pisces:

  1. Habitat: These are aquatic animals and are present in fresh, marine and brackish waters.
  2. Body temperature: Pisces are poikilothermic animals i.e., cold blooded animals, in which body temperature changes according to the change in the surrounding temperature.
  3. Sensory organs: They have lateral line system which shows the presence of rheoreceptors for the detection of water current.
  4. Locomotion: Locomotion is by body muscles and fins. Caudal fin helps in steering wheel.
  5. Skeleton: Exoskeleton is made up of dermal scales. It is either bony or cartilaginous.
  6. Body shape: Body is streamlined and boat-shaped and helps to overcome resistance during swimming.
  7. Respiration: Respiration is by gills.
  8. Circulation: It shows single and closed circulation. Heart is two-chambered and ventral in position. Heart of fishes is described as venous heart (presence of deoxygenated blood).
  9. Nervous system: They have a well-developed brain with large olfactory lobes.
  10. Reproduction: Sexes are separate. Most of the fishes are oviparous, however some are viviparous.

Question 5.
The phenomenon of alternation of generation between asexual and sexual reproduction in cnidarians.
Answer:
Reproduction: Cnidarians reproduce asexually and sexually. Asexual reproduction takes place by budding and regeneration. Sexual reproduction takes place gamete formation. They exhibit metagenesis i.e. alternation of polypoid generation with medusoid generation. Polyps produce medusae asexually and medusae produce polyps sexually, e.g. Obelia

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 31.
Why was phylum Elemichordata earlier considered as a sub phylum of Chordata?
Answer:

  1. Hemiehordata was earlier considered as sub phylum of Chordata because the buccal diverticulum was considered a notochord. It is now placed as a separate phylum under Non-Chordata.
  2. It possesses certain characteristics of both Chordates and Non-chordates.
  3. Absence of notochord worm-like body, heart located on the dorsal side are the Non-chordate like characteristics seen in Hemiehordata.
  4. Presence of nerve chord, pharyngeal gill slits are some of the Chordate-like characters seen in Hemiehordata. Hence, Hemiehordata is considered as a connecting link between Non-chordata and Chordata.

Question 32.
Distinguish between Platyhelminthes and Nemathelminthes.
Answer:
The members of this phylum are called as flatworms.
Unique features of phylum Platyhelminthes:

  1. Body shape: Body of these animals is dorsoventrally flattened.
  2. Coelom: They are acoelomates.
  3. Germ layers: Platyhelminthes are triploblastic.
  4. Body organization: They show organ-system grade of organization.
  5. Forms: Most of them are endoparasites and few are free-living. Parasitic forms have hooks and suckers for attachment to the hosts’ body. In parasitic forms, body is covered by cuticle and in free-living forms it is covered by cilia.
  6. Digestive system: Parasitic forms generally lack digestive system. In free-living forms, the digestive system is incomplete.
  7. Body plan: They show blind-sac body plan.
  8. Excretion and osmoregulation: It occurs by flame cells or protonephridia.
  9. Reproduction: They are mostly hermaphrodite (bisexual). Self fertilization is seen. Few animals show high power of regeneration and show polyembryony.
  10. e.g. Planaria, Taenia (Tapeworm), Fasciola (Liver fluke) are included in this phylum.
    Phylum Aschelminthes (ascus – sac, helminth – worm) is also called as Nemathelminthes (Nema = thread, helmins = worms).

Characteristics of Aschelminthes:

  1. Forms: These are mostly parasitic. However, few forms are free-living.
  2. Body shape: The body is long, cylindrical, thread-like, circular in cross-section, hence they are known as roundworms.
  3. Body symmetry: These are bilaterally symmetrical.
  4. Coelom: They are pseudocoelomate animals.
  5. Germ layers: These animals are triploblastic.
  6. Body plan: They show tube within a tube type body plan.
  7. Body covering: The body is covered by tough, resistant cuticle.
  8. Muscles: Body wall has longitudinal muscles, but circular muscles are absent.
  9. Digestive system: Alimentary canal is complete with mouth and anus, at opposite ends.
  10. Excretion: Excretion takes place either by canals or gland cells.
  11. Nervous system: Nervous system consists of a nerve ring and nerves.
  12. Reproduction: Animals are unisexual i.e. sexes are separate. Fertilization is internal. Development may or may not include larval stages. It shows sexual dimorphism.
  13. e.g. Ascaris (Roundworm), Wuchereria (filarial worm) and Ancylostoma (hookworm).

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 33.
Give one word for the following.

Question 1.
The gastrovascular cavity in cnidarians
Answer:
Body cavity: Cnidarians have a central cavity called coelenteron or gastrovascular cavity, which helps in digestion and circulation. They have blind-sac body plan i.e., single pore opening to the exterior in the digestive system.

Question 2.
Animals known as acnidarians.
Answer:
The members of this phylum are commonly known as comb jellies and sea walnuts. They are also known as acnidarians as they lack cnidoblasts. The phylum is considered as one of the minor phyla as it is represented by very few members.

Salient features of phylum Ctenophora:

  1. Habitat: They are exclusively marine.
  2. Forms: They are free-swimming
  3. Germ layers: Members of this phylum are diploblastic.
  4. Body Symmetry: They are radially symmetrical.
  5. Body plan: The animals of this phylum show blind-sac body plan.
  6. Body organization: They show tissue-level organization.
  7. Locomotion: It is earned out by eight rows of ciliated comb plates.
  8. Bioluminescence: It is the characteristic feature of the members of this phylum.
  9. Digestion: It is extracellular and intracellular.
  10. Reproduction: Reproduction is sexual with indirect development.
  11. Colloblasts: These sticky cells are used to capture prey, e.g. Pleurobrachia, Ctenoplana

Question 3.
The largest phylum of kingdom Animalia.
Answer:
Arthropoda (Arthros: Joint, Podos: leg): Arthropoda forms the largest phylum of kingdom Animalia. Characteristics of Arthropoda:
a. Habitat: Arthropods are omnipresent.
b. Forms: Solitary or colonial, most of them are free-living. Barnacles are sedentary. Few are parasitic and sanguivorous, (e.g. Female mosquito, bed bug.)
c. Body symmetry: Body is bilaterally symmetrical.
d. Germ layers: They are triploblastic.
e. Body cavity: Arthropods are eucoelomates.
f. Body plan: They show tube within tube body plan.
g. Level of body organization: They show organ system level of organization.
h. Special features: The members of this phylum have jointed appendages. Hence, they are known as arthropods. Some insects like honey bee, ants, termites, etc. exhibit polymorphism.
i. Exoskeleton: Body is covered by a tough, non – living chitinous exoskeleton. As the exoskeleton does not allow body growth, arthropods shed off their exoskeleton periodically during growth. This process is called moulting or ecdysis.
j. Body division: Body is divided into head, thorax and abdomen.
k. Segmentation: Body shows metameric segmentation.
l. Digestion: Digestive system is complete and divided into foregut, midgut and hindgut.
m. Circulation: Circulatory system is of open type wherein, blood flows through body cavity called haemocoel.
n. Respiration: Respiration occurs through respiratory organs like gills, trachea, book lungs or book
gills.
o. Excretion: Excretion takes place by green glands, Malpighian tubules or coxal glands.
p. Nervous system: Nervous system consists of nerve ring and double, ventral ganglionated nerve cord.
q. Sense organs:Arthropods have well developed sense organs in the form of antennae, simple or compound eye and various receptors.
r. Sexual reproduction: Sexes are generally separate in arthropods with distinct sexual dimorphism.
s. Significance:
Beneficial arthropods: Some arthropods are of economic importance. For example, Honey bees (Apis) are important for their honey and wax, silk worms for the production of silk. Lobsters, prawns, crabs are edible. Harmful arthropods: Some arthropods are harmful and act as vectors to spread various diseases, e.g., Mosquitoes. Locusta (locust) is a gregarious pest. Limulus (King crab) is a living fossil.
Other examples: Cockroach (Periplaneta), butterfly, scorpion (Hottentotta) and millipede (Archispirostreptus) prawn.

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 4.
Copper containing respiratory pigment in blood of molluscs.
Answer:
Circulatory system: Circulatory system is of open type (except in Sepia, where it is of the closed type). Blood contains a copper-containing blue-coloured respiratory pigment called haemocyanin.

Question 39.
What is the role of radula in mollusca?
Answer:
Digestive system: Digestive system is well-developed and complete with anterior mouth and posterior anus. Buccal cavity has a rasping organ called radula (helps in feeding), which is provided with transverse rows of teeth.

Question 40.
What are choanocytes?
Answer:
Digestive system: The body cavity of sponges (spongocoel) is lined by unique type of flagellated cells called choanocytes or collar cells for digestion.

Question 41.
Name the worm which causes filariasis.
Answer:
e.g. Nereis (3 annelid), Pheretima (Earthworm), Hirudinaria (Leech).

Question 42.
Multiple-choice Questions

Question 1.
Blind sac body plan occurs in
(a) Cnidaria
(b) Arthropoda
(c) Echinodermata
(d) Hemichordata
Answer:
(a) Cnidaria

Question 2.
Physalia belongs to phylum
(a) Platyhelminthes
(b) Cnidaria
(c) Nemathelminthes
(d) Arthropoda
Answer:
(b) Cnidaria

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 3.
Flame cells are found in phylum
(a) Porifera
(b) Coelenterata
(c) Platyhelminthes
(d) Arthropoda
Answer:
(c) Platyhelminthes

Question 4.
___________ is commonly known as hookworm.
(a) Wuchereria
(b) Ancyclostoma
(c) Ascaris
(d) Nereis
Answer:
(b) Ancyclostoma

Question 5.
Which of the following is bilaterally symmetrical?
(a) Pleurobrachia
(b) Cucumaria
(c) Aurelia
(d) Pheretima
Answer:
(d) Pheretima

Question 6.
Malpighian tubules or coxal glands are organs of excretion found in
(a) molluscs
(b) arthropods
(c) hemichordates
(d) platyhelminthes
Answer:
(b) arthropods

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 7.
________ is known as living fossil.
(a) Limulus
(b) Locusta
(c) Laccifer
(d) Loligo
Answer:
(a) Limulus

Question 8.
The member of second largest phylum is
(a) Lobster
(b) Squid
(c) Saccoglossus
(d) Antedon
Answer:
(b) Squid

Question 9.
Excretory system of these molluscs is of open type except,
(a) Sea hare
(b) Pila
(c) Octopus
(d) Sepia
Answer:
(d) Sepia

Question 10.
_________ are exclusively marine animals.
(a) Cnidarians
(b) Echinoderms
(c) Molluses
(d) Arthropoda
Answer:
(b) Echinoderms

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 11.
The peculiar character of echinoderms is
(a) Presence of mantle cavity
(b) Presence of water vascular system
(c) Presence of jointed appendages
(d) Presence of ostia and osculum
Answer:
(b) Presence of water vascular system

Question 12.
Which one of the following belongs to Subphylum Cephalochordata?
(a) Amphioxus
(b) Herdmania
(c) Petromyzon
(d) Ascidia
Answer:
(a) Amphioxus

Question 13.
Complete the analogy:
Salpa: Tunicata : : Myxine : ________ .
(a) Cyclostomata
(b) Chondrichthyes
(c) Cephalochordata
(d) Amphibia
Answer:
(a) Cyclostomata

Question 14.
Members of class Reptilia
(a) are limbless except for Salamander
(b) have moist, glandular skin
(c) have better developed olfactory lobes and cerebellum than amphibians
(d) have four chambered heart except for crocodile
Answer:
(c) have better developed olfactory lobes and cerebellum than amphibians

Question 15.
Which of the following are the first true terrestrial vertebrates?
(a) Mammals
(b) Amphibians
(c) Reptiles
(d) Both (b) and (c)
Answer:
(c) Reptiles

Question 16.
________ is an oviparous mammal.
(a) Macaca
(b) Pteropus
(c) Macropus
(d) Duck billed platypus
Answer:
(d) Duck billed platypus

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 43.
Competitive Corner:

Question 1.
Which of the following animals are TRUE coelomates with bilateral symmetry?
(a) Annelids
(b) Adult Echinoderms
(c) Aschelminthes
(d) Platyhelminthes
Answer:
(a) Annelids

Question 2.
Consider following features:
1. Organ system level of organization
2. Bilateral symmetry
3. True coelomates with segmentation of body Select the correct option of animal groups which possess all the above characteristics.
(a) Arthropoda, Mollusca and Chordata
(b) Annelida, Mollusca and Chordata
(c) Annelida, Arthropoda and Chordata
(d) Annelida, Arthropoda and Mollusca
Hint: In Annelida, Arthropoda and Chordata true segmentation occurs.
Answer:
(c) Annelida, Arthropoda and Chordata

Question 3.
Identify the vertebrate group of animals characterized by crop and gizzard in its digestive system.
(a) Aves
(b) Reptilia
(c) Amphibia
(d) Osteichthyes
Hint: In Aves, crop is associated with storage of food grains and gizzard is used to crush food grain.
Answer:
(a) Aves

Question 4.
Match the following organisms with their respective Characteristics.

1. Pila (P) Flame cells
2. Bomby (q) Comb plates
3. Pleurobrachia (r) Radula
4. Taenia (s) Malpighian tubules

Select the correct option from the following:
(a) i – q, ii – s, iii – r, iv – p
(b) i – r, ii – q, iii – s, iv – p
(c) i – r, ii – q, iii – p, iv – s
(d) i – r, ii – s, iii – q, iv – p
Answer:
(d) i – r, ii – s, iii – q, iv – p

Maharashtra Board Class 11 Biology Important Questions Chapter 4 Kingdom Animalia

Question 5.
An important characteristic that Hemichordates share with Chordates is
(a) absence of notochord
(b) ventral tubular nerve cord
(c) pharynx with gill slits
(d) pharynx without gill slits
Answer:
(c) pharynx with gill slits

Question 6.
Which among these is the CORRECT combination of aquatic mammals?
(a) Seals, Dolphins, Sharks
(b) Dolphins, Seals, Trygon
(c) Whales, Dolphins, Seals
(d) Trygon, Whales, Seals
Hint: Shark and Trygon (sting ray) are cartilaginous fishes. They belong to class Chondrichthyes. While Dolphins, Seals and Whales are aquatic mammals.
Answer:
(c) Whales, Dolphins, Seals

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

Balbharti Maharashtra State Board 12th Commerce Book Keeping & Accountancy Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts Important Questions and Answers.

Maharashtra State Board 12th Commerce BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

I. Objective Questions:

A. Select the most appropriate alternatives from the following & rewrite the sentences:

Question 1.
When dates of drawings are not given, interest on drawings is charged for _____________ months.
(a) three
(b) six
(c) nine
(d) twelve
Answer:
(b) six

Question 2.
A debit balance of the partner’s current account will appear on the _____________ side of the Balance Sheet.
(a) assets
(b) liabilities
(c) debit
(d) credit
Answer:
(a) assets

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

Question 3.
The interest on partner’s capital is credited to _____________ Account.
(a) Trading
(b) Profit and Loss
(c) Capital
(d) Cash
Answer:
(c) Capital

Question 4.
Under fixed capital method, salary or commission to partner is credited to _____________ Account.
(a) Partner’s Capital
(b) Partner’s Current
(c) Partner’s Drawings
(d) Partner’s Salary
Answer:
(b) Partner’s Current

Question 5.
If fixed capital method is adopted, net divisible profit is transferred to _____________ Account.
(a) Partner’s Current
(b) Partner’s Capital
(c) Profit and Loss
(d) Trading
Answer:
(a) Partner’s Current

Question 6.
A statement showing financial position of a business is called a _____________
(a) Balance Sheet
(b) Trial Balance
(c) Capital A/c
(d) Trading A/c
Answer:
(a) Balance Sheet

Question 7.
Wages paid for Installation of machinery should be debited to _____________ Account.
(a) Machinery
(b) Wages
(c) Trading
(d) Profit and Loss
Answer:
(a) Machinery

Question 8.
All indirect expenses are debited to _____________ Account.
(a) Trading
(b) Capital
(c) Profit and Loss
(d) Current
Answer:
(c) Profit and Loss

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

Question 9.
Return outwards are deducted from _____________
(a) Purchase
(b) Sales
(c) Capital
(d) Debtors
Answer:
(a) Purchase

Question 10.
Debit balance of Trading Account indicates _____________
(a) Gross Profit
(b) Gross Loss
(c) Net Profit
(d) Net Loss
Answer:
(b) Gross Loss

Question 11.
Credit balance of Profit and Loss Account indicates _____________
(a) Gross Profit
(b) Gross Loss
(c) Net Profit
(d) Net Loss
Answer:
(c) Net Profit

Question 12.
Income received in advance is shown on _____________ side of the Balance Sheet.
(a) Debit
(b) Credit
(c) Assets
(d) Liabilities
Answer:
(d) Liabilities

Question 13.
Amount irrecoverable from debtors is known as _____________
(a) discount
(b) bad debts
(c) allowance
(d) none of these
Answer:
(b) bad debts

Question 14.
Trading Account is prepared on the basis of _____________ expenses.
(a) indirect
(b) direct
(c) revenue
(d) capital
Answer:
(b) direct

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

Question 15.
Royalty paid on production is shown in the _____________
(a) Balance Sheet
(b) Trading A/c
(c) Profit and Loss A/c
(d) Partner’s Current Account
Answer:
(b) Trading A/c

Question 16.
Prepaid expenses are shown on the _____________ side of the Balance Sheet.
(a) Assets
(b) Liabilities
(c) Debit
(d) Credit
Answer:
(a) Assets

Question 17.
Advertisement expenditure to be written off yet will appear on the _____________ side of Balance Sheet.
(a) Debit
(b) Liabilities
(c) Assets
(d) Credit
Answer:
(c) Assets

Question 18.
_____________ is the list of all ledger balances.
(a) Balance Sheet
(b) Trial Balance
(c) Trading A/c
(d) Profit and Loss A/c
Answer:
(b) Trial balance

Question 19.
Final accounts are prepared on the basis of _____________ and adjustments.
(a) Trial balance
(b) Trading A/c
(c) Profit and Loss A/c
(d) Capital A/c
Answer:
(a) Trial balance

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

Question 20.
The personal medical bill of a partner paid from the business is known as _____________ of the partner.
(a) capital
(b) profit
(c) cash
(d) drawings
Answer:
(d) drawings

B. Write the word/phrase/term, which can substitute each of the following sentences.

Question 1.
The capital method in which the partner’s Current Account is opened.
Answer:
Fixed Capital Method

Question 2.
The capital method in which the partner’s Current Account is not opened.
Answer:
Fluctuating Capital Method

Question 3.
Method of Capital Account in which capital balances of partners change every year.
Answer:
Fluctuating Capital Method

Question 4.
Expenses that are due but not paid at the end of the year.
Answer:
Outstanding/Unpaid expenses

Question 5.
A provision that is created on sundry debtors for likely bad debts.
Answer:
Reserve for Doubtful Debts

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

Question 6.
Income is received before it is due.
Answer:
Income received in advance

Question 7.
The stock is valued at cost price or market price whichever is less.
Answer:
Closing stock

Question 8.
Reduction in the value of fixed assets due to its continuous use.
Answer:
Depreciation

Question 9.
The transport expenses incurred to carry the goods purchased by the firm.
Answer:
Carriage Inward

Question 10.
Income due but not received.
Answer:
Accrued income

Question 11.
Account prepared on the basis of direct expenses and incomes.
Answer:
Trading Account

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

Question 12.
Account prepared on the basis of indirect expenses and incomes.
Answer:
Profit and Loss Account

Question 13.
The transport expenses are paid to Railway, the Airways company, or the Shipping company.
Answer:
Freight

C. State whether the following statements are True or False with reasons:

Question 1.
The partnership agreement must be in written form.
Answer:
This statement is Raise.
A partnership agreement can be in oral or written form. It is advisable to have a partnership agreement in written form, to avoid future conflicts and disputes among the partners. However, it is not compulsory.

Question 2.
There is no limit to a maximum number of partners in a firm.
Answer:
This statement is Raise.
Minimum two persons are required to form the partnership firm. As per the provisions made under the Companies Act 2013 (amended in 2014) the maximum number of partners in a firm is restricted to 50.

Question 3.
Partners are entitled to get a salary or commission.
Answer:
This statement is False.
In Partnership Deed when it is clearly mentioned that all partners or specific partners are entitled to salary or commission then only partners are entitled to get salary or commission. When partnership deed remains silent on salary or commission, then partners are not able to get any salary or commission.

Question 4.
Closing stock is always valued at market price.
Answer:
This statement is False.
As per the conservatism concept, the closing stock is always valued at cost price or market price whichever is less. If the market price of closing stock is greater than its cost then closing stock is recorded at cost.

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

Question 5.
The trial balance is the basis of the Final Account.
Answer:
This statement is True.
Based on the Trial balance and other adjustments, one can prepare Final Accounts. So, the Trial balance is the basis of Final Accounts.

Question 6.
Return inward is deducted from purchases.
Answer:
This statement is False.
Return inward means Sales return and it is to be deducted from sales, not from purchase. Return outward is deducted from purchases.

Question 7.
Discount allowed to Debtors is called as Bad debts.
Answer:
This statement is False.
Discount allowed to Debtors is an expense for the business while Bad debts mean irrecoverable amount from debtors and is a loss to the business. Thus, both have different meanings so, we cannot say that discount allowed to debtors is called as Bad debt.

D. Complete the Sentences.

Question 1.
Documentation charges paid for purchasing a building is debited to _____________ Account.
Answer:
Building

Question 2.
Credit balance of Trading Account indicates _____________
Answer:
Gross Profit

Question 3.
Receivable income is shown on _____________ side of the Balance Sheet.
Answer:
Assets

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

Question 4.
Trademark, Copyright, Patents are the examples of _____________ asset.
Answer:
intangible

Question 5.
Balance Sheet is a _____________ but it is not an _____________
Answer:
Statement, Account

Question 6.
Profit and Loss Account is a _____________ Account.
Answer:
Nominal

Question 7.
The income which is due but not yet received is called _____________ income.
Answer:
accrued/receivable

Question 8.
The statement showing list of all ledger balances is known as _____________
Answer:
Trial balance

Question 9.
Debit balance of Profit and Loss Account means _____________
Answer:
Net Loss

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

Question 10.
Interest on capital is an _____________ for the partner.
Answer:
income

Question 11.
Income accured but not received is _____________ for firm.
Answer:
an asset

Question 12.
Sale of scrap is recorded on _____________ side of _____________ Account.
Answer:
Credit, Profit and Loss

Question 13.
General reserve is recorded in _____________ side of _____________
Answer:
Liability, Balance Sheet

Question 14.
Provision for doubtful debts recorded in _____________ side of _____________ when it is given in the Trial balance only.
Answer:
Liability, Balance Sheet

Question 15.
Provident fund amount is a _____________ for the firm.
Answer:
Liability

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

E. Answer in one sentence only:

Question 1.
Define: ‘Partnership’ as per the Indian Partnership Act 1932.
Answer:
As per the Indian Partnership Act 1932, “partnership is the relation between persons who have agreed to share the profits of a business carried on by all or any one of them acting for all”.

Question 2.
State the two-fold capacities of each partner who works in a business.
Answer:
Each partner works in two-fold capacities viz. Principal and Agent in a business.

Question 3.
As per Income Tax Act 1961, write the dates for Financial or Accounting Year.
Answer:
As per Income Tax Act, 1961, the Financial or Accounting year starts from 1st April of the current year to 31st March of next year. [e.g. 01/04/2019 to 31/03/2020].

Question 4.
How many effects for the hidden adjustment given in the Trial balance are to be passed?
Answer:
Two effects for every hidden adjustment, given in the Trial balance are to be passed, though no special instruction is given in the problem.

F. Do you agree/disagree with the following statements.

Question 1.
A profit and Loss Account is a Real Account.
Answer:
Disagree

Question 2.
Carriage outward means carriage on sales.
Answer:
Agree

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

Question 3.
Adjustments are recorded in Partners Current Account in Fluctuating Capital method.
Answer:
Disagree

Question 4.
Outstanding incomes are treated as an asset.
Answer:
Agree

Question 5.
The balance Sheet is an account.
Answer:
Disagree

Question 6.
R.D.C. is created on creditors.
Answer:
Agree

Question 7.
Depreciation is calculated on fixed assets.
Answer:
Agree

Question 8.
Copyright is a visible asset.
Answer:
Disagree

Question 9.
Interest on drawings is an income for the firm.
Answer:
Agree

Question 10.
Interest on a Partner’s Loan to the firm is always to be allowed.
Answer:
Agree

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

Question 11.
All indirect expenses are debited to Trading Account.
Answer:
Disagree

Question 12.
Capital Account always shows a credit balance.
Answer:
Disagree

Question 13.
Trading Account is prepared to know the profit or loss of the firm.
Answer:
Disagree

Question 14.
Final Accounts are prepared on the basis of Trial Balance and adjustments given.
Answer:
Agree

Question 15.
Royalty paid on production is shown in the Trading Account.
Answer:
Agree

Solved Problems

Question 1.
From the following Trial Balance and Adjustments prepare Trading and Profit and Loss Account and Balance Sheet as on 31st March 2019 for Mr. A and B.
Trial Balance as of 31st March, 2019
Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts Q1
Adjustments:
1. Closing stock: Cost price ₹ 60,000, Market price ₹ 52,500.
2. Interest on fixed deposit ₹ 1800 is still outstanding.
3. Provide R.D.D. at 2.5 % on sundry debtors.
4. Depreciate furniture by 5 %.
5. Goods of ₹ 12,000 were destroyed by fire and the insurance company accepted the claim of ₹ 9,000 only.
Solution:
In the books of Mr. A and Mr. B
Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts Q1.1
Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts Q1.2
Balance Sheet as of 31st March 2019
Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts Q1.3
Working Notes:
1. Closing stock is to be considered at a cost price or market price whichever is less. It is valued at ₹ 52,500.

2. Goods of ₹ 12,000 were destroyed and the insurance company accepted a claim of ₹ 9,000, which means ₹ 3,000 is a loss for the firm. The insurance company accepted the claim, (not yet paid the amount) therefore, it is recorded on the Asset side of the Balance Sheet.

3. Advertisement expense ₹ 45,000 is given for 3 years means for one year, we have to take 45000/3 = ₹ 15,000 as advt. exp. and ₹ 30,000 (45,000 – 15,000) is to be taken as prepaid advt. exp. (Asset side)

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

Question 2.
From the following Trial Balance of Parth and Zalak and given Adjustments, prepare Final Accounts for the year ending on 31st March 2019.
Trial Balance as of 31st March 2019
Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts Q2
Adjustments:
1. Closing stock is valued at ₹ 99,000.
2. Write off ₹ 3,000 as further bad debts and maintain 5% R.D.D. on debtors.
3. Depreciate Plant and Machinery by 10%, Motor car by 15%, Patents by 20%.
4. Furniture costing ₹ 12,000 sold for ₹ 7,500 was wrongly included in sales and the remaining furniture is valued at ₹ 33,000.
5. Outstanding expenses are Wages ₹ 8,100, Salaries ₹ 6,750. The insurance premium is paid for the year ended 31st December 2019.
6. Goods worth ₹ 67,500 were destroyed by fire and the insurance company accepted the claim for only ₹ 57,000.
7. Sale of goods of ₹ 15,000 was wrongly considered as the sale of machinery.
Solution:
In the books of Parth and Zalak
Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts Q2.1
Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts Q2.2
Balance Sheet as of 31st March 2019
Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts Q2.3
Working Notes:
1. Interest on loan @18 % is calculated for 3 months, (i.e. 1/1/2019 to 31/3/2019)
I = \(\frac{\text { PRN }}{100}\)
= 30,000 × \(\frac{18}{100} \times \frac{3}{12}\)
= ₹ 1,350

2. Net loss by fire = 67,500 – 57,000 = ₹ 10,500

3. Depreciation on furniture = Book value – Value given in adj.
= 45,000 – 33,000
= ₹ 12,000

4. Rent is paid for 10 months i.e. 2 months rent is outstanding.

5. Insurance premium is paid upto 31st Dec., 2019. i.e. 9 months insurance premium is prepaid.
= 5,400 × \(\frac{9}{12}\)
= ₹ 4,050

6. Loss on sale of furniture = Cost of furniture sold – Sale proceeds of furniture
= 12,000 – 7,500
= ₹ 4,500

Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts

Question 3.
From the following trial balance and adjustments of Rushabh and Yesha, you are required to prepare final accounts as of 31st March 2019. The profit and Loss sharing ratio of partners is their capital ratio.
Trial Balance as of 31st March 2019
Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts Q3
Adjustments:
1. Closing stock is ₹ 1,10,000. Its market value is 20% more than its value.
2. Calculate interest on capital @ 6% p.a.
3. Interest on drawings to be charged from partners: Rushabh ₹ 900, Yesha ₹ 600
4. Provision for doubtful debts is to be kept at 5%.
5. Outstanding expenses at the end of the year: Rent ₹ 300, Salary ₹ 950.
6. Provide depreciation at 10% on machinery and at 5% on furniture.
7. Write off ₹ 4,000 from leasehold building.
Solution:
In the books of Rushabh and Yesha
Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts Q3.1
Balance Sheet as of 31st March 2019
Maharashtra Board 12th BK Important Questions Chapter 1 Introduction to Partnership and Partnership Final Accounts Q3.2
Working Notes:
1. Consider closing stock value ₹ 1,10,000 as its market value is 20% more.

2. Interest in drawings: Record it on the Cr. side of P & L A/c subtract it from the capital. (As shown)

3. Interest on an 8 % loan is calculated for 5 months (i.e. 1/11/18 to 31/3/19)
I = \(\frac{\mathrm{PRN}}{100}\)
= 30,000 × \(\frac{8}{100} \times \frac{5}{12}\)
= ₹ 1,000

4. In the Trial Balance, the following balances have credit balance:
Bills payable, Bank Overdraft, Provision for doubtful debts, 8 % Loan, etc.

5. O/s wages – Cr. bal. – write it on the Liability side of the Balance Sheet.

6. Prepaid insurance – Dr. bal. write it on the Asset side of the Balance Sheet.

Maharashtra Board Class 5 Hindi Solutions पुनरावर्तन ४

Balbharti Maharashtra State Board Class 5 Hindi Solutions Sulabhbharati पुनरावर्तन ४ Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Hindi Sulabhbharati Solutions पुनरावर्तन ४

5th Standard Hindi Digest पुनरावर्तन ४ Textbook Questions and Answers

1. सुनो, और दोहराओ:

प्रश्न 1.
सुनो, और दोहराओ:
Maharashtra Board Class 5 Hindi Solutions पुनरावर्तन ४ 1
उत्तर:
(अ) 1. मैं जा रह्य / रही हूँ।
2. हम जा रहे हैं / रही हैं।
3. तू जा रहा है / रही है।
4. तुम जा रहे हो / रही हो।
5. आप जा रहे हैं। रही हैं।

(ब) 1. यह जा रहा है। रही है।
2. ये जा रहे हैं / रही हैं।
3. वह जा रहा है / रही है।
4. वे जा रहे हैं / रही हैं।

Maharashtra Board Class 5 Hindi Solutions पुनरावर्तन ४

2. तुम्हारे मित्र के पास कॉपी खरीदने के लिए पैसे नहीं हैं । तुम्हारी माँ ने मेला देखने के लिए तुम्हें तीस रुपये दिए हैं। इस स्थिति में तुम्हारे मन में कौन से भाव जागते हैं ? बताओ।

प्रश्न 1.
तुम्हारे मित्र के पास कॉपी खरीदने के लिए पैसे नहीं हैं । तुम्हारी माँ ने मेला देखने के लिए तुम्हें तीस रुपये दिए हैं। इस स्थिति में तुम्हारे मन में कौन से भाव जागते हैं ? बताओ।
उत्तर:
उन तीस रुपयों में से अपने मित्र को कॉपी खरीदकर दूंगा तथा बचे हुए पैसों से मेला घूम लूँगा।

3. पढ़ो, और समझो:

प्रश्न 1.
पढ़ो, और समझो:
उत्तर:

  1. जहाँ विद्यार्थी पढ़ते हैं – विद्यालय
  2. जादू दिखाने वाला – जादूगर
  3. खेल खेलने वाला – खिलाड़ी
  4. सच बोलनेवाला – सत्यवादी
  5. चित्र बनाने वाला – चित्रकार
  6. मूर्तियाँ बनाने वाला – मूर्तिकार
  7. सोने के गहने बनाने वाला – सुनार
  8. जो कभी न हारा हो – अजेय
  9. जो साग-सब्ज़ी खाता हो – शाकाहारी
  10. जहाँ अनाथ रहते हैं – अनाथालय
  11. जो रोगियों का इलाज करता है – डॉक्टर
  12. जो लकड़ी का काम करता है – बढ़ई
  13. जो मिट्टी के बरतन बनाता है – कुम्हार
  14. जो जूते सिलता है – मोची
  15. जो कपड़े सिलता है – दर्जी

Maharashtra Board Class 5 Hindi Solutions पुनरावर्तन ४

4. सड़क पार करते समय तुम क्या सावधानी बरतते हो? लिखो।

प्रश्न 1.
सड़क पार करते समय तुम क्या सावधानी बरतते हो? लिखो।
Maharashtra Board Class 5 Hindi Solutions पुनरावर्तन ४ 2
उत्तर:
सड़क पार करते समय जेब्रा क्रॉसिंग पर से जाना चाहिए।

5. पढ़ाई करने के बाद भी तुम्हें परीक्षा से डर लग रहा है, तो तुम क्या करोगे? बताओ।

प्रश्न 1.
पढ़ाई करने के बाद भी तुम्हें परीक्षा से डर लग रहा है, तो तुम क्या करोगे? बताओ।
(क) बड़ों से बातचीत करोगे ।
(ख) खेलने जाओगे।
(ग) गाना सुनोगे अथवा गाओगे ।
(घ) आराम करोगे ।
उत्तर:
(क) बड़ों से बातचीत करेंगे।

Maharashtra Board Class 5 Hindi Solutions पुनरावर्तन ४

Hindi Sulabhbharati Class 5 Solutions पुनरावर्तन ४ Additional Important Questions and Answers

प्रश्न 1.
चित्रों को देखकर नीचे दी गई शब्द – पहेली बूझो; (एक अक्षर का एक से अधिक बार उपयोग कर सकते हैं।)
उत्तर:
फूल: कमल, हरसिंगार, गुलाब, गुड़हल, डहलिया, रजनीगंधा, बेला, सूरजमुखी, सदाबहार।
फल: केला, अमरूद, अनार, चीकू, अंगूर, सेब, पपीता, आम, अंजीर, संतरा, सीताफल, अनन्नास।
अन्य शब्द: नीम, पीला, ताला, अंदर, चीता, कान, पतला, खीर, गेंद, जल, दानी

Maharashtra Board Class 5 Hindi Solutions पुनरावर्तन ३

Balbharti Maharashtra State Board Class 5 Hindi Solutions Sulabhbharati पुनरावर्तन ३ Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Hindi Sulabhbharati Solutions पुनरावर्तन ३

5th Standard Hindi Digest पुनरावर्तन ३ Textbook Questions and Answers

1. अनुस्वारवाले (-) शब्दों को सुनो, समझो और दोहराओ:

अङ्क-अंक, चञ्चल-चंचल, झण्डा-झंडा, सुन्दर-सुंदर, मुम्बई-मुंबई; टंकी, पंख, पतंग, कंघी, कंचा, पंछी, अंजीर, झंझावात, घंटी, कंठ, डंडा, पंढरपुर, संत, पंथ, बंदर, कंधा, पंप, गुंफन, कंबल, खंभा ।

Maharashtra Board Class 5 Hindi Solutions पुनरावर्तन ३

2. रास्ते में घायल पक्षी को देखकर तुम्हारे मन में कौन-से भाव आते हैं? बताओ?

प्रश्न 1.
रास्ते में घायल पक्षी को देखकर तुम्हारे मन में कौन-से भाव आते हैं? बताओ?
उत्तर:
रास्ते में घायल पक्षी को देखकर मुझे उस पर दया आ जाती है। मैं उसे उठाकर उसका उपचार करती / करता हूँ, ताकि वह फिर से आकाश में उड़ सके।

3. पढ़ो, समझो और रेखांकित शब्दों पर चर्चा करो और दोनों अक्षरों से नए वाक्य बनाओ

प्रश्न (क)
मैं पाँचवी कक्षा में पढ़ रहा हूँ।
उत्तर:
‘मैं’ का उपयोग हम अपने लिए करते हैं। ‘में’ का उपयोग किसी ओर वस्तु के बारे में करते हैं।

प्रश्न (ख)
मृणाल ने पूछा “उदय! कहाँ गए थे?” उदय ने कहा, “मैं भोपाल गया था।”
उत्तर:
कहाँ – किसी जगह के बारे में पूछते हैं।
कहा – मतलब ‘बोला’ या ‘बताया’।
(अ) तुम कहाँ जा रहे हो?
(आ) राम ने कहा, ‘मैं बीमार हूँ।

प्रश्न (ग)
रौनक बिल्ली की ओर लपका और वह भाग गई।
उत्तर:
ओर – उस तरफ (की तरफ)
और – दो शब्दों अथवा दो वाक्यों को जोड़ने वाला शब्द
(अ) सीमा पाठशाला की ओर जा रही है।
(आ) सीता और गीता जुड़वा बहनें हैं।

Maharashtra Board Class 5 Hindi Solutions पुनरावर्तन ३

प्रश्न (घ)
राजू खेल रहा है। उसके साथी बैठे हैं।
उत्तर:
है – केवल एक व्यक्ति के लिए लगता है।
हैं – बहुवचन या अनेक के लिए लगता है।
(अ) राम बाहर खड़ा है।
(आ) बच्चे खेल रहे हैं।

प्रश्न (ङ)
सोहन की बिल्ली इतनी प्यारी है कि सब उसे उठा लेते हैं।
उत्तर:
की – पहले शब्द की व्याख्या करता है।
कि – दो वाक्यों को जोड़ने का काम करता है।
(अ) राम की बहन सीता है।
(आ) पवन ने कहा कि मैं बीमार हूँ।

4. उचित शब्द बनाकर लिखो:

प्रश्न 1.
उचित शब्द बनाकर लिखो:

  1. ख ओं
  2. र पै
  3. न का
  4. ठ हों
  5. ज ग का
  6. ट ना घु
  7. र द बं
  8. त र भा
  9. ई र पा चा
  10. व ली दी पा
  11. ला शा ठ पा
  12. वा ल री फु

उत्तर:

  1. आँख
  2. पैर
  3. कान
  4. होंठ
  5. कागज
  6. घुटना
  7. बंदर
  8. भारत
  9. चारपाई
  10. दीपावली
  11. पाठशाला
  12. फुलवारी

Maharashtra Board Class 5 Hindi Solutions पुनरावर्तन ३

5. वर्ण के अवयवों का उपयोग करते हुए अपने मन से चित्र बनाओ।

Maharashtra Board Class 5 Hindi Solutions पुनरावर्तन ३ 1

Hindi Sulabhbharati Class 5 Solutions पुनरावर्तन ३ Additional Important Questions and Answers

प्रश्न 1.
आपकी पाठशाला में मनाए गए किसी एक कार्यक्रम पर पाँच वाक्य लिखिए:
उत्तर:

  1. हमारी पाठशाला में हिन्दी दिवस मनाया गया।
  2. सभी बच्चों ने इसमें भाग लिया।
  3. बच्चों ने भाषण, कविता, नाटक प्रस्तुत किए।
  4. पहली कक्षा की एक छात्रा ने बहुत अच्छी कहानी सुनाई।
  5. अंत में सभी को धन्यवाद दिया गया।

Maharashtra Board Class 5 Hindi Solutions पुनरावर्तन ३