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Balbharti Maharashtra State Board Class 11 Geography Solutions Chapter 5 Global Climate Change Textbook Exercise Questions and Answers.

Maharashtra State Board Class 11 Geography Solutions Chapter 5 Global Climate Change

1. Complete the chain.

Question 1.

Answer:

2. Identify the incorrect factor.

Question 1.
The causes of global warming-
(a) Emission of greenhouse gases
(b) Deforestation
(c) Apparent movement of the sun
(d) Industrialization
Answer:
(c) Apparent movement of the sun

Question 2.
The indicators of climate change-
(a) Retreat of glaciers
(b) Increase in the frequency of floods
(c) Increase in frequency of cyclones
(d) Increase in the minimum and maximum temperature
Answer:
(d) Increase in the minimum and maximum temperature

Question 3.
The tools for studying climate change
(a) Ice cores
(b) Coral reef
(c) Tree rings
(d) Ancient forts
Answer:
(d) Ancient forts

Question 4.
Measures to combat climate change –
(a) Banning the use of pesticides and insecticides
(b) Promoting afforestation and banning deforestation
(c) Banning public transport
(d) Banning fossil fuels
Answer:
(c) Banning public transport

3. Give geographical reasons.

Question 1.
It is important to study climate change.
Answer:
It is important to study climate change because –

1. The earth’s climate is changing faster primarily as a result of human activities.
2. Number of flash floods are increasing, especially in urban areas.
3. Due to warming of oceans the rate of evaporation is high, which is the root cause of both flooding and droughts.
4. Due to increase in temperature, there is melting of snow on mountains and sea level is rising.
5. Human activities, especially emissions of heat-trapping greenhouse gases from fossil fuel combustion, deforestation and land use changes are the primary drivers of the climate changes observed in the industrial era.

Question 2.
There is a great possibility that we may not see Maldives on the world map in the future.
Answer:
There is a great possibility that we may not see Maldives on the world map in the future because-

1. Sea level continues to rise at a rate of about 3 mm/ per year, leaving no ground surface higher than 3 m.
2. Due to global warming the rising of sea level is likely to worsen.
3. Higher sea levels may lead to severe cyclones and periodic flooding of coastal areas.
4. Since, Maldives is a flat country there is possibility that it may be submerged under sea water if the rise in sea level continues.

Question 3.
The snowline is retreating.
Answer:
The snowline is retreating because –

1. A snowline is a boundary between a snow-covered and snow-free surface. Above the snowline there is permanent snow cover.
2. When glaciers and icebergs melt at an alarming rate, it is a cause of concern. This is known as retreating glaciers.
3. Due to changes in the climate there is less snowfall as compared to the previous years.
4. This means that there is lesser ice formation each year than its rate of melting.
5. Thus, due to melting of glaciers and icebergs as well as less snowfall the snowline is retreating.

Question 4.
There is an increase in the frequency of droughts and cyclones.
Answer:
There is an increase in the frequency of droughts and cyclones because-

1. In addition to triggering more rainfall, global warming could also increase the occurrence of drought.
2. The roots of both flooding and drought lie in the physical process known as evaporation.
3. As global warming heats the world’s oceans, the water molecules near the sea surface become more energetic and tend to evaporate into the atmosphere more readily. Thus, water vapour is formed.
4. Similarly, due to global warming, the number of cyclones in a year and their intensities have also increased in the tropical regions.

4. Write short notes on.

Question 1.
Bleaching of coral reefs
Answer:

1. When temperature changes, corals throw out the algae living in their tissues.
2. These algae are responsible for their colour.
3. An increase of 1°C – 2°C in ocean temperatures for a long time can lead to bleaching, turning corals white.
4. If corals are bleached for prolonged periods, they eventually die.
5. Coral bleaching leads to death of large amounts of corals.

Question 2.
Flash floods
Answer:

1. A flash flood is a term that describes floods, which is usually observed in the low-lying areas.
2. Floods are caused due to extremely heavy rainfall.
3. Flash floods generally occur after the collapse of natural ice, debris dam or a man-made dam, etc.
4. Flash foods also occur due to very high precipitation in one day (Mumbai in 2005, Kedarnath in 2013) or because of changed weather conditions like cyclones (Chennai, 2015).
5. Flash foods are different from the regular floods by having a time scale of fewer than six hours between rainfall and the onset of flooding.

Question 3.
Tools to study Paleoclimatology
Answer:
Paleoclimatology is the study of the weather and climate of the earth’s past. Scientists and meteorologists have been using instruments to measure climate and weather for the past 140 years. However, millions of years ago historical evidence called proxy data, which includes coral reefs, tree rings, ice cores, etc., were used to tell about the climate.

Coral reefs:

1. Coral reefs are very sensitive to changes in climate. –
2. Corals form skeletons by extracting calcium carbonate from the ocean waters.
3. When the water temperature changes, densities of calcium carbonate in the skeletons also change.
4. Coral formed in the summer has a different density than coral formed in the winter.
5. This creates seasonal growth rings on the corals.
6. Scientists can study these rings to determine the temperature of the water and the season in which the coral grew.

Tree rings:

1. There are rings on the stem of the tree.
2. Variation in these rings is due to variation in the environmental conditions when they were formed.
3. Thus, studying this variation leads to improved understanding of past environmental conditions.

Ice cores:

1. These are samples of ice taken from the inner sides of the ice sheets.
2. Throughout each year, layers of snow fall over the ice sheets in Greenland and Antarctica.
3. Each layer of snow is different.
4. Summer snow differs from winter snow.
5. Each layer gives scientists a lot of information about the climate each year.

Question 4.
Greenhouse gases
Answer:

1. Greenhouse gases are the primary gases, such as, water vapour carbon dioxide, nitrous oxide, methane and ozone in the earth’s atmosphere that absorbs and traps energy from the sun.
2. Of all greenhouse gases, carbon dioxide absorbs more heat.
3. Without greenhouse gases, the average temperature of the earth’s surface would be about -18°C rather than the present average of 14°C.
4. Carbon dioxide emissions comes from many human activities such as combustion of fossil fuels, oil and natural gas, with additional contributions coming from deforestation, changes in land use, soil erosion and agriculture (including live stock).

5. Answer in detail.

Question 1.
Climate change has not always been anthropogenic in nature. Explain.
Answer:
Though the human activities are a major cause behind climate change, there are some natural causes for climate change. They are as follows:
Amount of energy output by the sun:

1. Low energy output by the sun can reduce amount of insolation received by the earth.
2. This can lead to cooling of the earth.

Milankovitch Oscillation:

1. The earth’s rotation around its own axis and revolution around the sun is responsible for the changes in climate.
2. Their changes alter the amount of insolation received from the sun, which in turn might affect its climate.
3. It can cause the earth to be closer to the sun (warmer) and farther from the sun (colder).
4. The ice ages can occur when we are farther from the sun.

Volcanism:

1. Volcanic eruptions throw lots of particulates and aerosols into the atmosphere.
2. These aerosols remain in the atmosphere for a considerable period of time.
3. The winds spread it around the world, thus reducing the sun’s radiation reaching the earth’s surface.

Goldilocks Zone:

1. The earth lies in the habitable zone called the Goldilocks Zone.
2. Earth experiences cooler climate, when it was located in the outer reaches of this zone.
3. As the size of the sun increases the zone moves outward over time.
4. Such changes in the zone causes earth to warm or cool.

Question 2.
What measures will you suggest to combat climate change in your village/city.
Answer:
Following measures can be taken to combat climate change in our village / city.

1. Afforestation should be encouraged in and around play areas, hill sides, schools, etc.
2. Deforestation should be avoided strictly.
3. The use of public transport should be enabled to lessen the vehicular pollution. Thus, emission of harmful gases such as CO₂, etc., can be prevented.
4. Use of natural manure can be encouraged instead of using fertilizers.
5. In villages, Liquefied Petroleum Gas (LPG) should be used instead of using coal or wood as fuel.

11th Geography Digest Chapter 5 Global Climate Change Intext Questions and Answers

Can you tell? (Textbook Page No. 58)

1. The graph in the figure shows the difference between the global average temperature of the twentieth century and global monthly temperatures from 1985 to 2015. Answer the questions given below.

Question 1.
In which year is the difference the least?
Ans.
Difference in the temperature is least in the year 1985.

Question 2.
What is the difference between the mean temperatures of the twentieth century and temperature in 2015?
Answer:
The difference between the mean temperature of twentieth century (1995-2000) and 2015 is 0.8°C.

Question 3.
Why do the temperatures differ in different months?
Answer:
The temperature differs in different months because of the occurrence of different seasons.

2. Look at the graph and answer the questions. (Textbook Page No. 63)

Question 1.
What does this graph show?
Ans.
The graph shows CO₂ concentration (ppm) and its increased level over the period of time from 1900 to 2017.

Question 2.
What does ppm mean?
Answer:
The abbreviation ppm means ‘parts per million’ of carbon concentration in the atmosphere.

Question 3.
Since which year has the increase been phenomenal?
Answer:
After 1980, there has been a phenomenal increase observed.

Question 4.
Can you think of the reasons behind the increase in carbon dioxide?
Answer:
The reasons behind the increase in carbon dioxide is due to combustion of fossil fuels and large- scale deforestation.

Try this. (Textbook Page No. 59)

1. Look at the following graph. Answer the following question.

Question 1.
Which of these gases has the highest contribution?
Answer:
Water vapour, shows the highest contribution.

Question 2.
Which of these gases came from natural and man-made sources?
Answer:
Water vapour, methane, NO₂ and miscellaneous gases are naturally occurring gases while CO₂ is man-made.

Question 3.
Which activities are responsible for their emission?
Answer:
Burning of fossil fuels like coal, etc., and deforestation are responsible for emission of CO₂.

Question 4.
Out of these, whose emission can be controlled by humans?
Ans.
The emission of CO₂ can be controlled by humans.

Question 2.
Make a list of things you need to change in your lifestyle. (Textbook Page No. 67)
Answer:

1. Use of public transport.
2. Less use of wood but more use of alternatives to wood.
3. Regular maintenance of vehicles.
4. Avoid wastage of water.
5. Use of energy efficient devices.
6. Ban on plastic goods.
7. Minimum use of paper in home and office.
8. Purchase of only required goods.
9. Use of Eco-friendly goods.
10. Celebration of festivals without harming the environment.

Find out! (Textbook Page No. 62)

Question 1.
Use internet and reference books on climate change and make a list of species vulnerable to climate change.
Answer:
Species vulnerable to climate change are: corals, polar bear and frogs.

Question 2.
Make a list of movies based on ice age and climate change. (Textbook Page No. 65)
Answer:

1. Interstellar
2. Beasts of the Southern Wild
3. Chasing Coral
4. Snowpiercer
5. An Inconvenient Truth
6. Are you ready to fight? (Vattaram)

Question 3.
With the help of internet, find out the details of the National Action Plan on Climate Change (NAPCC), 2008 and its missions. (Textbook Page No. 66)
Answer:
NAPCC is a government document that includes eight ambitious goals set for the country to achieve.

1. National Solar Mission
2. National Mission for Enhanced Energy Efficiency
3. National Mission on Sustainable Habitat
4. National Water Mission
5. National Mission for Sustaining the Himalayan Ecosystem
6. National Mission for a “Green India”
7. National Mission for Sustainable Agriculture
8. National Mission on Strategic Knowledge for Climate Change

1. Look at the graph and answer the following questions. (Textbook Page No. 60)

Question 1.
What does the graph show?
Answer:
The graph shows an increase in the global sea level from 1880 till date.

Question 2.
In which year is the change around 225 mm?
Answer:
The change around 225 mm is observed is the year 2018.

Question 3.
What conclusions can you draw by seeing the graph?
Answer:
With the continued global warming the sea levels are likely to rise.

Question 4.
What correlation can be seen in this graph and the graph of rising temperature?
Answer:
With the rising temperature the sea level is rising. We can conclude this from both the graphs.

Question 2.
Compare both the satellite images in fig 5.3 (A) and (B). (Textbook Page No. 61)

Answer:
Fig. 5.3 (A) shows the snow-covered Gangotri glacier area in Dec. 1984, whereas in Fig. 5.3 (B) the snow-covered region of the Gangotri glacier has retreated by 2018.

Balbharti Maharashtra State Board Class 11 Geography Solutions Chapter 4 Climatic Regions Textbook Exercise Questions and Answers.

Maharashtra State Board Class 11 Geography Solutions Chapter 4 Climatic Regions

1. Write the names of climatic regions according to the factors that dominate their characteristics.

Answer:

2. Choose the correct alternative

Question 1.
Monsoon region
(a) annual average temperature around 27° C
>2500 mm annual precipitation
Indonesia
Hard-wood evergreen trees

(b) Average temperature in Summers around 35° C
< 2500 mm annual rainfall
South East Asia
Hard wood deciduous trees

(c) Temperatures in summer around 35° C
1000 mm annual rainfall
Continental part of Indian peninsula
Tall and thick grass

(d) Average temperatures in summer around 27° C
1000 mm rainfall in winter
South Africa
Hard–wooded, waxy, evergreen forests
Answer:
(b) Average temperature in Summers around 35° C
< 2500 mm annual rainfall
South East Asia
Hard wood deciduous trees

Question 2.
The region with high diurnal range of temperature
(a) Tropical rainforests
(b) Tropical grasslands
(c) Tropical desert regions
(d) Tropical Monsoon regions
Answer:
(c) Tropical desert regions

Question 3.
Lumbering flourished as an occupation from Newfoundland to Alaska in North America because
(a) Tundra Climatic Region
(b) Taiga Climatic Region
(c) West European Climatic Region
(d) China-type climatic region
Answer:
(b) Taiga Climatic Region

Question 4.
The main reason behind the months of precipitation in the graphs of Monsoon climatic regions being different is
(a) ITCZ
(b) orographic rainfall
(c) hemispheres are different
(d) apparent movement of the sun
Answer:
(a) ITCZ

3. Give geographical reasons.

Question 1.
In monsoon climate region, rainfall occurs in specific season.
Answer:
In monsoon climate region, rainfall occurs in specific season because-

1. The differential heating and cooling of land and water creates low pressure on the land while the sea experiences high pressure.
2. This is strongly related to shifting of the ITCZ. During the summer the ITCZ moves north to the latitudes of 20°-25°.
3. Several months later, the moisture laden summer monsoon is replaced by dry north-east monsoon.
4. By this time, the ITCZ has shifted to the southern hemisphere.
5. In northern hemisphere, the winds move from sea to land bringing moisture along with them in summers and gives rainfall.
6. In southern hemisphere same conditions prevail when there are winters in the northern hemisphere.

Question 2.
Taiga region is not found in Southern Hemisphere.
Answer:
Taiga region is not found in Southern Hemisphere because-

1. There is not much land at higher altitude in the southern hemisphere (until inside the Antarctic circle) where it is permanently ice covered.
2. Most of the land is fairly close to oceans and thus tempered by marine warming.
3. Whereas, Taiga is found throughout the high Northern latitudes between Tundra and the temperate forest, characterised by coniferous forests.
4. The strong oceanic influence in the southern hemisphere reduces the severity of the winter, thus, affecting the temperature and precipitation.

Question 3.
The diurnal range of temperature is more in desert areas.
Answer:
The diurnal range of temperature is more in desert areas because-

1. Deserts get hot during the day, but cool down rapidly during night. The main reasons for the temperature fluctuation between day and night are the lack of humidity and cloud cover.
2. Because of water’s relatively high specific heat, humid air takes longer to warm or cool, while air on land warms or cools more quickly.
3. Because desert areas receive a large amount of sunlight, the temperatures can get very high, but during night the dry air cools quickly.
4. The lack of cloud cover overhead allows the warmer air to rise into the atmosphere while cooler air settles in its place.
5. Basically, desert temperature varies so much because there is less moisture and cloud to keep the heat there.

Question 4.
There is no concept of season in equatorial regions.
Answer:
There is no concept of season in equatorial regions because-

1. The equatorial region lies between 0° to 10° latitudes in both the hemispheres.
2. The sun is exactly overhead in the equatorial regions throughout the year giving maximum insolation.
3. Thus, constant high temperatures are experienced throughout the year, which gives rise to heavy convectional precipitation.
4. Days and nights are almost of equal length and thus, the concept of summer and winter as being hot and cold seasons do not exist in these parts.

Question 5.
The Savannah region is prone to droughts.
Answer:
The Savannah region is prone to droughts because-

1. The region lies between 10° to 20° N and S latitudes, where the sun’s rays at noon are never far from overhead and thus the insolation is maximum and temperature is constantly high here.
2. Distinct wet and dry seasons of relatively equal duration are found. Most of the regions annual rainfall is experienced during the wet season and very little precipitation falls during the dry season.
3. A lengthy dry season and a relatively short wet season is the characteristic of Savannah region.
4. Thus, in Savannah region the dry season becomes more severe and often drought conditions prevail during the course of the year.

Question 6.
Though Mussoorie and Dehradun are located on the same latitude, why is their climate different?
Answer:
Though Mussoorie and Dehradun are located on the same latitude, their climate is different because-

1. Dehradun lies in the humid sub-tropical climatic region where the summers are long, hot and humid. Whereas, Mussoorie is a hill station in Dehradun district of Uttarakhand which lies in sub-tropical climatic regions that are very wet.
2. Dehradun lies at an elevation of average 450 m above mean sea level whereas Mussoorie is at an elevations of 1880 m above mean sea level.
3. In Dehradun during monsoon season there is often heavy and protracted rainfall, whereas in Mussoorie an average of 660 mm orographic rainfall is experienced.
4. The winter temperature is around 1°C and 20°C in Dehradun with fog commonly experienced. Mussoorie usually receives few spells of snowfall in December, January and February.

4. Differentiate between.

Question 1.
Equatorial Rainforests and Savannah Climatic Regions.
Answer:

Question 2.
Taiga Region and Tundra Region.
Answer:

Question 3.
Monsoon and Mediterranean Regions.
Answer:

5. Answer in detail.

Question 1.
Explain, with examples, the effects of latitude on a place’s climate.
Answer:
Latitudes affect the climate of a place in different ways-
Places close to the equator receive more sunlight:

1. The equatorial region lying between 0° to 5° North and South latitudes receive maximum heat as the sunrays are directly falling on the equator.
2. Due to which the region receives heavy rainfall throughout the year.
3. As a result, evergreen forests are found is such regions.
4. Example : Amazon basin.

Places close to the tropical region receives less sunlight:

1. Temperature gradually decreases with increasing distance from the equator, as the angle of sun’s rays decreases from it.
2. In tropical areas the sun rays become slanting and thus it gives less heat but covers more area.
3. Example regions lying between Tropic of Cancer 231/2°N to Tropic of Capricorn 23/4°S. Similarly, in the temperate regions 66X/20N and S and the polar regions the climate become extremely cool as the intensity of heat decreases. These regions are thus covered by snow.

Question 2.
Explain with examples, how winds affect the climate of a place.
Answer:

1. The winds affect the climate because they create the hot and cold air which moves around.
2. Prevailing winds affect the climate of an area. The direction of local winds is determined by the daily temperature of variations.
3. When winds blow from warm areas, they carry higher temperatures while winds blowing from cold areas carry lower temperatures.
4. Thus, winds bring in heat/ cold from the direction it has been blown from and thus changes the temperature of the place which affects the climate. For e.g., cold wind waves from the Himalayas carry coldness to the parts of the Deccan plateau.
5. Winds that blow from the sea often carry moisture and give rain to the coast. For e.g.. south-west monsoon winds give heavy rainfall in the western coastline of India.
6. Some winds are warm and dry, they increase the temperature during winter season. For example, winds that blow to Britain from warm areas such as Africa are warm and dry. Such warm and dry winds increase the summer temperature in some areas.

Question 3.
Russia is larger than Chile in area but does not experience climatic diversity as Chile. Explain.
Answer:

1. The latitude and longitude of Russia is 61.5240°N and 105.3188°E respectively. It is located in northern Asia, between Europe on one side and North Pacific Ocean on the other side.
2. Chile is situated along western margin of South American continent. Its latitude and longitude are 35.6751°S and 71.5430°W respectively.
3. Russia has continental location since it is surrounded to south and west by huge landmass, and to the north lies Arctic ocean which is frozen, so it is like snow covered land.
4. Thus, due to continental location there are dry summer and very cold winters with temperatures of -30° and less and sometimes heavy snowfall.
5. The winter is mostly dry, snow covers the ground from the end of October to mid-March.
6. In Chile the coastal location and the east-west extent is very less.
7. There is a lot of climatic diversity in Chile. There is Atacama Desert in the north, Mediterranean climate at the central part, extreme cold climate in the east.
8. The most important factors that control the climate of Chile are Pacific anticyclone, cold Humboldt ocean current and mountain ranges along the coastline.
9. Thus, though Russia is larger than Chile in area but does not experience climatic diversity as Chile.

Question 4.
Explain the factors affecting climate of a place giving examples.
Answer:

Latitude : Latitude is the primary factor influencing distribution of atmospheric temperature. Generally, latitudes of high degrees are less warm than the low degrees of latitudes. Thus, sun’s rays making a high angle to the ground, experience high temperature throughout the year. Temperature gradually decreases with increasing distance from the equator as the angle of the sun’s rays decreases away from it. Hence, the polar region experiences very low temperature.

Altitude : The atmosphere is indirectly heated by terrestrial radiation from surface of the earth. Therefore, the place near the sea level experiences higher temperature than the places situated at higher altitude. In other words, the temperature generally decreases with increase in height. The rate of decrease of temperature with increase in height is known as Normal Lapse rate. It is 6.4°C per 1000 meters altitude.

Distance from the Sea : The location of place with respect to the sea also determines the temperature of that place. Though, the sun provides same amount of heat over land and water bodies, there is difference in their heating and cooling capacity. Compared to land, the sea (water) gets heated slowly and loses heat slowly. On the other hand, land heats up and cools down faster. Therefore, the variation in temperature over the water bodies is less as compared to land masses. The coastal areas come under the moderating influence of the sea and land breezes and experience moderate or equable climate while interior parts of the continents experience extreme climate.

Prevailing winds : Winds affect the temperature and rainfall of any area. Winds carry with them the temperature of their source regions. Hot winds increase the temperature while cold winds decrease the temperature, e.g., cold wind waves from Himalayas carry low temperature to parts of the Deccan Plateau. Thus, south west monsoon winds in India reduce the temperature after the onset of monsoon in June all over the country.

Some winds carry moisture and give rain to the area they visit. For example, South-west monsoon winds give rainfall to western coastline of India.

Ocean currents : Ocean currents considerably influence temperature of the adjacent land area. Warm currents raise the temperature of the coastal areas, whereas cold currents reduce the temperature near the coast where they flow.

Aspect of slope : In the northern hemisphere the south facing slopes and in the southern hemisphere the north facing slopes always face the sun. Hence, sun facing slopes are warmer and the opposite slopes are cooler.

6. On a world map, show the following areas

(1) Equatorial Rainforests and Savannah Climatic Regions.
(1) Savannah climatic region in Africa
(2) Highland climatic region in India
(3) Chile and Russia
(4) Ice cap climatic region
(5) Desert
Answer:

11th Geography Digest Chapter 4 Climatic Regions Intext Questions and Answers

Use vour brain power!

Question 1.
(i) Have you ever thought why there is difference in the skin colour of various people in the world?
(ii) Why all the people in the world do not eat same food?
(iii) Why there is a variety in clothing pattern and types too? Even our houses are different.
(iv) How come flora and fauna are restricted to a particular region?
(v) Why different fruits are found in different places? (Textbook Page No. 44)
Answer:
(i) There is variation in skin colour of people depending on the latitudes. For. e.g., people near the equator are dark since the sunrays are vertical and hence there is too much heat, while people in temperate or polar region are fair since sunrays in these regions are slanting due to increase in distance from sun. Hence, there is low temperature and colour of skin is fair.

(ii) There is variation in food because food depends on geographical factors like climate, soil, location etc. For example, in extreme cold climate areas soil is covered with snow, so agriculture is not possible, people eat fish and meat of fur bearing animals. The countries located along the coastline eat fish and rice because fishing is carried out at the coastal areas and rice is grown on fertile delta of the river along the coastline.

(iii) People prefer to use clothes according to climate. For example, Inuits in Polar region wear fur clothes due to extreme cold climate, Europeans wear woollen cloths in winter due to cold climate, Asians mostly wear cotton clothes except in winter.

The pattern of house depends upon the climate and availability of material in the surrounding area. For example, in heavy rainfall areas houses have sloping roofs, in hot climatic areas houses are made up of stones or thick walls to prevent the effect of outside hot air.

People use material available in the surrounding area for construction of houses. For example, Inuits in Polar region use snow to build houses, since it is available in plenty. These peculiar houses are called Igloos.

(iv) Particular temperature and rainfall are essential for the growth of flora. For example, coniferous forest grows in cool climate and snowfall. Flora provides food and shelter to fauna, therefore wherever there is thick flora, fauna is bound to be there. In grassland region grass eating animals like dear, antelope, rabbits are seen but the camel is seen only in the desert.

(v) Every fruit crop requires particular type of climate; therefore, different fruits grow in different areas depending upon the climate. For example, apples are grown in Jammu and Kashmir, mangoes, cashew nuts and jackfruits are grown in the Konkan region due to suitable climate.

Question 2.
(i) What would be the annual range of temperature in this region?
(ii) Where is this type of climate found in India? (Textbook Page No. 46)
Answer:
(i) In equatorial region both diurnal and annual range of temperature will be low as it experiences high temperature all year round. The annual range of temperature may be as low as 3°C.

(ii) This type of tropical rainforest climate is found in north eastern states of India, western coast of West Bengal, and Andaman and Nicobar Islands.

Question 3.
Comment upon the type of weathering which will occur in this region. (Textbook Page No. 47)
Answer:
Mechanical and chemical weathering is predominant in this region.

Question 4.
(i) Which agricultural crops are produced here?
(ii) Why are longitudes not given in geographical distribution? (Textbook Page No. 48)
Answer:
(i) Maize and rice are grown all over Savannah region. Sorghum and millets are grown in northern Savannah. Stray crops and vegetables are grown in all regions.

(ii) Temperature is the main element of climate. The temperature varies with latitude and not with longitudes. Geographical distribution is the natural arrangement of the various forms of plants and animals in different regions and localities of the earth. This distribution varies with the latitude as we go away from the equator towards the poles. Thus, longitudes are not given.

Question 5.
Comment upon the rate of weathering in this climate. (Textbook Page No. 49)
Answer:
Mechanical or physical weathering is predominant in arid regions. The rate of weathering is very slow in this region.

Question 6.
Why do people in Europe use olive oil for cooking? (Textbook Page No. 50)
Answer:
Most olives are grown in the southern part and is used for almost every application. It is a healthy type of oil which makes it great for cooking.

Question 7.
What factors make this region agriculturally productive? [Textbook Page No. 51]
Answer:
Year-round precipitation and fertile soil along the rivers make the region agriculturally productive.

Question 8.
(i) Why does Chile recur frequently in examples of geographical distribution?
(ii) Why has fishing developed here? (Textbook Page No. 52)
Answer:
(i) The geography of Chile is extremely diverse as the country extends from a latitude 17° South to Cape Horn at 56° and ocean on the west to Andes on the east. It borders Pacific Ocean towards south and small part of the south is towards the Atlantic Ocean.

(ii) Due to long indented coastlines and cool climate, fishing is done on a large scale. Presence of South Pacific Ocean and small part of South Atlantic Ocean are major fishing regions.

Question 9.
(i) What would be the annual range of temperature in this climate? What could be the occupational activities carried out by humans here?
(ii) What type of weathering will be prominent here? (Textbook Page No. 53)
Answer:
(i) The annual range of temperature in Taiga is 60°, the summer temperature can be 10°C and average winter temperature is -60°C. Hunting and lumbering can be carried out in this region.

(ii) Since the region is covered by snow weathering is slow. Mechanical weathering is prominent here but is very slow. Alternate freezing and thawing takes place here.

Question 10.
What kinds of mass movement may happen here? [Textbook Page No. 54]
Answer:
Slow mass movement takes place here in the form of slumping where snow particles moves short distance down the slope.

Question 11.
(i) What kind of activities will bring people from other regions to this climate?
(ii) What could be the occupations followed here? (Textbook Page No. 55)
Answer:
(i) Since this region is severely covered by snow, scientific research and exploration can be done or fishing and hunting can be done but on a very smaller scale.

(ii) Hunting of aquatic life is the dominant occupation followed here.

Question 12.
(i) What type of human activities will develop in this region?
(ii) In what ways might high latitudes be different from high altitudes? (Textbook Page No. 56)
Answer:
(i) Terrace farming, animal rearing and tourism are the types of human activities that might develop in this region.

(ii) High latitudes are the regions away from the equator. The intensity of sunlight goes on decreasing as we move away from the equator towards the poles. The equatorial and tropical regions will have precipitation in the form of rainfall.

High altitude regions are the regions at an elevation from the mean sea level. In high altitude regions temperature decreases with increase in height therefore they are covered by snow as precipitation is in the form of snowfall.

Can you tell?

Question 1.
Make a list of the human activities you think that are not influenced by climatic elements. (Textbook Page No. 44)
Answer:
There are five major activities of man.
Primary: Agriculture, fishing, lumbering, hunting and gathering.
Secondary: Industries and manufacturing.
Tertiary: Services like transport and communication, teachers, doctors, etc.
Quaternary: Research and development etc.
Quinary: Highly intelligent activities.

2. Read the graphs and answer the following questions. (Textbook Page No. 46)

Question 1.
In which months there is no rainfall?
Answer:
In both Belem and Singapore it rains throughout the year.

Question 2.
In which month is the temperature highest?
Answer:
The temperatures is highest in the month of October in Belem and in the month of May in Singapore.

Question 3.
In which month is the temperature lowest?
Answer:
The temperature is lowest in the months of January and February in Belem and in the month of December and January in Singapore.

Question 4.
What could be the factors which influence the climate of these places?
Answer:
Both Belem and Singapore lie in the equatorial region, i.e., at 1° 27′ S to 48° 30′ W and 1° 17′ N and 103° 51′ E respectively. Thus, in both Belem and Singapore, the climate is characterised by uniform temperature and pressure, high humidity and abundant rainfall as the sunrays are perpendicular on the equator throughout the year.

Question 5.
Write a concluding statement about the climate of both the places based on the questions above.
Answer:
Both cities experiences rainfall throughout the year. These is no dry season month. Average temperature varies little throughout. There is no distinct season, uniform temperature, high humidity and abundant rainfall. They have equatorial type of climate.

3. Read the graph and answer the following questions. (Textbook Page No. 47)

Question 1.
Name the months of highest and lowest rainfall. What is the difference between the values of rainfall?
Answer:
In Mangaluru highest rainfall month is July with around 1143 mm rainfall and lowest rainfall was found in the month of January which is the driest month. Thus, the difference between the values of rainfall is 1143 mm.

In Cairns highest rainfall is 486 mm in the month of February and least amount of rainfall occurs in August around 37 mm. Thus, the difference between the values of rainfall is 449 mm.

Question 2.
Name the months of highest and lowest temperature.
Answer:
Highest and Lowest temperature in Mangaluru is in April around 32.4°C and in January with 21.6°C respectively. Highest and lowest temperature in Cairns is in January at around 27.5° C and July being coldest month 16.4°C respectively.

Question 3.
Are the months of rainfall same in both the places? If not, why?
Answer:
No, the months of rainfall are not same in both the places. It rains in Mangaluru from May to November whereas, in Cairns rainfall is observed in the months from November to May.

Question 4.
What could be the factors which influence the climate of these places?
Answer:
The major factor which influences the climate of Mangaluru and Cairo is that Mangaluru lies in Northern hemispheres and Cairns in Southern hemispheres.

Question 5.
What difference do you find in the graphs in fig and these?
Answer:
There is variation in rainfall and temperature graphs of Mangaluru and Cairo as these lies in different hemispheres while the graphs of Fig. represent same temperature and rainfall throughout the year at Belem and Singapore. Both lie in the equatorial region, so not much variation is observed.

Question 6.
Write a concluding statement about the climate of both the places.
Answer:
The climate of Mangaluru and Cairo is tropical. There is significant rainfall in most of the months of the year with short dry season. In Mangaluru, January is the driest month with 00 mm. of precipitation. Most of the precipitation is in July. While, in Cairo, the least amount of rainfall occurs in August. In February, precipitation is the highest.

4. Read the graph and answer the following questions. (Textbook Page No. 48)

Question 1.
In which months is there no rainfall?
Answer:
It rains throughout the year in both Lagos and Brasilia.

Question 2.
In which month is the temperature highest?
Answer:
Temperature is highest in the month of March in Lagos and in the month of September in Brasilia.

Question 3.
In which month is the temperature lowest?
Answer:
The temperature is lowest in Lagos in the month of August and that in the month of June in Brasilia.

Question 4.
Are the months of rainfall same in both the places? If not, why?
Answer:
It is wettest in June in Lagos and January is the wettest month in Brasilia, Lagos is situated near the equator. Thus, there is significant precipitation difference between the rainy season and summer season.

Brasilia is located at the top of Brazilian highlands. It has milder climate due to elevation and with two distinct seasons, the rainy season and dry season.

Question 5.
What could be the factors influencing the climate of these places?
Answer:
Lagos being near the equator and Brasilia located on the Brasilia highland, influences the climate of these places.

Question 6.
What difference do you find in the previous and these graphs?
Answer:
The total monthly precipitation is light in Mangaluru in the month of June, July, August and that of Cairns in the month of January, February, March whereas in equatorial region precipitation is experienced throughout the year. From this the graph of monsoon climate, we find that maximum precipitation is during the months of June to September.

Question 7.
Write a concluding statement about the climate of both the places.
Answer:
The rainfall is much more in summer than the winters in Lagos. The least amount of rainfall occurs in December while greater amount of precipitation occurs in June. The temperature is higher on an average in March, whereas lowest temperature occurs in August. With regards to Brasilia here too the rainfall is much more in summer than the winters. Precipitation is lowest in June and highest in January, September is the hottest month and lowest temperature is in June.

5. Read the graph and answer the following questions. (Textbook Page No. 48)

Question 1.
Name the months of highest and lowest rainfall. What is the difference between the values of rainfall?
Answer:
In Alice Springs, September is the driest month with around just 10 mm rainfall. Whereas the most precipitation falls in the month of January around 38 mm. The difference between the value is around 28 mm.
Similarly, In Cairo, the driest month is May with no precipitation at all whereas it rains about 5 mm in the month of January. The difference between the values is just 5 mm.

Question 2.
Name the months of highest and lowest temperature.
Answer: The warmest month in Alice Springs is January with an average temperature of 38.7°C while July experiences lowest temperature of about 11.5°C.
In Cairo, July is the warmest month with 27.6°C and lowest temperatures is in January with around 13.1°C temperature.

Question 3.
Are the months of rainfall same in both the place? If not, why?
Answer:
No, the months of rainfall are not same in both the places because Alice Springs lies in the Southern Hemisphere while Cairo lies in the Northern Hemisphere.

Question 4.
What could be the factors which influence the climate of these places?
Answer:
Cairo in Egypt is located to the north of Tropic of Cancer, and Alice Springs is located just south of Tropic of Capricorn. Therefore, both the areas come under subtropical high-pressure belt. The air becomes dry here. The winds in this region blows out of the region, so there is very low rainfall. The region of Alice Springs is part of central Ranges scrub area of dry scrubby grasslands. Majority of the Egypt’s landscape is desert, hence extreme aridity is experienced.

Question 5.
What difference do you find in the previous and these graphs?
Answer:
In the previous graphs rainfall is depicted for almost all the months. But in the graphs representing climate of Alice Springs and Cairo, rainfall is very low or almost negligible.
Temperature variation can also be depicted in the previous graphs and graphs of Alice Springs and Cairo.

Question 6.
Write a concluding statement about the climate of both the places.
Answer:
Rainfall is very less in Alice Springs whereas almost negligible in Cairo. This is due to the latitudinal extent of both the places and subtropical high-pressure belt. Cairo is in or near the subtropical desert biome whereas Alice Springs is in or near subtropical desert scrub biome. The average annual temperature is 35°C warmer. Average monthly temperature varies by 3.7°C in Cairo.

6. Read the graph and answer the following questions. (Textbook Page No. 50)

Question 1.
Name the months of highest rainfall.
Answer:
Perth experiences highest rainfall in the month of June whereas Rome experiences highest rainfall in November.

Question 2.
Name the months of lowest temperatures.
Answer:
The temperature in Perth is the lowest in July and the temperature in Rome is the lowest in January.

Question 3.
Are the answers to 1 and 2 same?
Answer:
No, the answers are not same.

Question 4.
Are the months of rainfall same in both the places If not, why?
Answer:
No, it does not rain in the same months in Perth and Rome. Perth lies on 31°57′ S to 115°51′ E longitude and Rome lies on 41°53′ N to 12°30′ E longitude. It means Perth is located in the Southern | Hemisphere, whereas Rome is in the Northern Hemisphere. Perth is near the Subtropical dry forest biome, whereas Rome is near the warm temperate moist forest biome.

Question 5.
What could be the factors which influence the climate of a place?
Answer:
The special feature of climate of this region is long dry summers and mild and wet winters. This because of subtropical high pressure in summer and westerly wind movement in winter.

Question 6.
What difference do you find in the previous graphs and these graphs?
Answer:
In Mediterranean regions, summers are long and warm and dry whereas winters are mild and wet. They are different from low-latitudinal climates where the temperature is high throughout the year. Rainfall in winter is the characteristic of this region but previous graph shows very meagre rainfall.

Question 7.
Write a concluding statement about the climate of both the places.
Answer:
Rainfall is highest in the month of June, around 175 mm in Perth, while its lowest in the month of January around just 7 mm. Whereas, November experiences highest rainfall of about 114 mm and lowest of about 17 mm in July in Rome. Temperature reaches its highest in February around 24.8°C and lowest in July around 13.6°C in Perth, while July experiences highest temperature of about 24.4° C and least about 7.7° in January in Rome.

7. Read the graph and answer the following questions. (Textbook Page No. 51)

Question 1.
Name the months of highest and lowest rainfall.
Answer:
In Brisbane, highest rainfall occurs in the months of February, while lowest rainfall is in the month of September. In Shanghai, June experiences highest rainfall, while precipitation is lowest in December.

Question 2.
Name the months of highest and lowest temperatures. Relate them with rainfall months.
Answer:
In Brisbane temperature is highest in January and lowest in July, while Shanghai experiences highest temperature in July and lowest in January.
In Brisbane, temperature is highest in January whereas rainfall is highest in February and high in January, whereas in Perth, July records highest temperature and June records highest rainfall. Similarly, temperature is least in January and rainfall is least in December.

Question 3.
Are the months of rainfall same in both the places? If not why?
Answer:
No, both places have rainfall in different months. This is probably because Brisbane is located is Southern Hemisphere. Brisbane is in the south east corner of Queensland. The region is on the coastal plains, east of Great Dividing Range.
Shanghai is located in the Northern Hemisphere, located on Yangtze River Delta on China’s east coast and has proximity to the Pacific Ocean via East China Sea.

Question 4.
What could be the factors which influence the climate of these places?
Answer:
Brisbane and Shanghai, both lie in the eastern parts of the southern and northern hemispheres respectively. Due to proximity to the coral sea of the Pacific Ocean and warm ocean current, Brisbane’s overall temperature variability is somewhat less.

Shanghai is located in the middle of China’s east coast near the mouth of river Yangtze to the North, the East china sea to the east, Hangzhou Bay to the south and Jiangsu and Zhejiang provinces to the west.

Question 5.
What difference do you find in the previous and these graphs?
Answer:
The major difference between Mediterranean and China type is that, the Mediterranean is found on the western margins of the continents while china type is found is the eastern parts almost in same latitudes.
In Mediterranean type winter rainfall is the characteristics feature, while china type experiences rainfall throughout the year.

Question 6.
Write a concluding statement about the climate of both the places.
Answer:
Both the places receive rainfall from convectional showers. Precipitation is year-round. In Brisbane variation in the precipitation between driest and wettest months is 133 mm. During the year, average temperatures vary by 10.3°C.

8. Read the graph and answer the following questions. (Textbook Page No. 52)

Question 1.
In which months do you find temperature the lowest?
Answer:
Temperature is the lowest in the month of January in Portland and in Melbourne it’s in the month of July.

Question 2.
In which month is the precipitation lowest?
Answer:
Precipitation is lowest in the month of July in Portland and lowest precipitation is found in the month of February in Melbourne.

Question 3.
What are the highest values of precipitation?
Answer:
In Portland highest precipitation is in the month of December with an average of 162 mm. While, highest value of precipitation in Melbourne is in October with an average of 71 mm.

Question 4.
Name the months with no precipitation at all.
Answer:
It rains throughout the year in both Portland and Melbourne.

Question 5.
Write a concluding paragraph about this climatic region.
Answer:
Marine west European type of climate is located between 40° and 65° latitudes in both the hemispheres along the western coasts of the continents. This climatic region is surrounded by Mediterranean climate in the south continental dry climate in the east and semi-arctic climate in the north. The temperatures are affected by marine influences, warm ocean currents and prevailing winds. This climate is characterised by cool summer and wild winters. Average temperature during summer season ranges between 15°C and 21°C. It rains throughout the year.

9. Read the graph and answer the following questions. (Textbook Page No. 53)

Question 1.
Note the values of the axes, how different are these graphs from the earlies ones?
Answer:
The values of Taiga or Sub-Arctic regions show the mean annual temperature below freezing point (0°C) that is from 0° to -35° C. These values are not observed in rest of the graphs.

Question 2.
Note the highest and the lowest temperature and their months.
Answer:
In Whitehorse, the highest temperature recorded is around 20.0°C in the month of July, whereas January is the coldest month with average temperature of -23°C.
In Arkhangelsk, the highest temperature is in the month of July around 20°C and lowest in the month of January around -18°C.

Question 3.
Note the highest and lowest rainfall and their months.
Answer:
Highest rainfall recorded is about 35 to 37 mm in the months of July, August in Whitehorse and lowest is in the month of April with just 9 mm.
In Arkhangelsk, the greatest amount of precipitation occurs in August with an average of 66 mm. The lowest amount of rainfall occurs in February. The average is this month in 27 mm.

Question 4.
Why does not a place from Southern Hemisphere appear here?
Answer:
There isn’t much land at higher latitudes in hemisphere (until inside the Antarctic circle where it is permanently ice covered) and most of that land is fairly close to oceans and it has effect of marine warming.

Question 5.
What factors are responsible for this climate?
Answer:
Located in large Continental landmass between 55° to 65° latitude, the Sub-Arctic climate is removed from any moderating influence of an ocean. It therefore, experiences a very large range in annual temperature. During the summer it is dominated by the westerlies and cyclonic activity the winter it is the polar hight and Easterlies. The Sub-Arctic climate has continental polar air
masses.

10. Read the graph and answer the following questions. (Textbook Page No. 54)

Question 1.
Why is it that both the graphs are from the northern Hemisphere?
Answer:
Tundra is located at the top of the world, near the north pole. Tundra is a Finnish word which means barren land. Thus, Tundra region having least vegetation and polar or arctic climate is found in North America and Eurasia between the southern limit of the permanent ice caps in the north and the northern limit of temperate coniferous forest of taiga climate in the south.

Question 2.
Which are the warmest and the coolest months?
Answer:
In Barrow, the warmest month is July while lowest temperature is in the month of February. Same temperature conditions are observed in Barentsburg.

Question 3.
What is the annual range of temperature?
Answer:
The average annual range of temperature in Barrow is -12.2°C while in Barentsburg it is -5.9° C.

Question 4.
Why does not the duration of day (sometimes more than 24 hours) influence its temperatures or precipitation?
Answer:
This is because the sun’s rays are oblique and little insolation is received being at higher latitude -65° to 90° North.

11. Read the graph and answer the following questions. (Textbook Page No. 55)

Question 1.
Which are the warmest and the coolest months?
Answer:
The warmest month is July and the coolest month is February in Eismitte. While in Scott South Station warmest months are December and January and coolest month is August.

Question 2.
Name the months of highest and lowest rainfall.
Answer:
The month with highest rainfall in Eismitte is December and lowest rainfall is in May. While in Scott South Station highest rainfall is in the month of February, while lowest rainfall is observed in July and August.

Question 3.
In what way do you find similarities of this climate with other climate types of high latitudes?
Answer:
Precipitation is very low in high latitude regions and mostly it is in the form of snowfall. Temperature is below freezing point (CPC) in all the high latitude regions. This is because the sun’s rays are oblique and hence temperature is low. These regions are mostly under permanent snow cover.

Question 4.
What factors influence this type of climate?
Answer:
Little or no insolation during most of the months in a year, oblique sunrays are the factors responsible for this type of climate.

12. Read the graph and answer the following questions. (Textbook Page No. 56)

Question 1.
Which are the warmest and the coolest months?
Answer:
The warmest month is June while coolest month is January in Shimla. November is the hottest month and July is the coolest month in El Alto.

Question 2.
Name the months of highest and lowest precipitation.
Answer:
Most of the precipitation is in the months of July, whereas driest month is November is in Shimla. Precipitation is highest in January and lowest in June in El Alto.

Question 3.
In what ways do you find similarities of this climate with other types of high latitudes?
Answer:
In these regions, precipitation is in the form of snowfall. The regions are covered by snow.
Temperature is very low, as the sunrays are oblique and little insolation is received. Temperature decreases with increasing altitude.

Question 4.
Why are the axis showing temperature different in both the graphs?
Answer:
Shimla is located at 31°6 N to 77°10 and its elevation is 2.276 m whereas El Alto is located on 16°31 S to 68°10 W with an elevation of 4.150 mts. Temperature decreases with increasing altitude. Thus, the axis is different temperature in both the graphs.

Question 5.
What factors influence this type of climate?
Answer:
The higher reaches of mountain, altitude, location of leeward or windward side, precipitation are the factors affecting climates of mountain type of climate.

Find out! (Textbook Page No. 44)

Use internet or reference books to find out about the attempts at classification of climates.
Answer:
Classification of climate was given by Wladimir Peter Koppen, Charles Warren Thornthwaite and Glenn Thomas Trewartha.
[Student are required to research and attempt this question on their own]

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

Maharashtra State Board Class 11 Geography Solutions Chapter 3 Agents of Erosion

1. Compete the Chain.

Question 1.

Answer:

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

Question 1.
Water or snow enters the cracks in the rocks and makes it weak. When the glacier passes on these rocks, it pulls the rocks at the bottom along with it. This process is called
(a) Plucking
(b) Abrasion
(c) Attrition
(d) Transportation
Answer:
(d) Transportation

Question 2.
Sometimes, the river starts erosion upstream. This happens when the head stream gets a lot of water in the early stages of river’s flow.
(a) Downcutting
(b) Headward erosion
(c) Lateral erosion
(d) Vertical erosion
Answer:
(b) Headward erosion

Question 3.
Soft rock erodes beneath the hard rock due to sea waves. This results into landforms which further develop as sea arches. The landform is
(a) Sea cave
(b) Sea stack
(c) Sea cliff
(d) Wave cut platform
Answer:
(a) Sea cave

Question 4.
This landform develops due to depositional work of wind. The windward slope of this landform is gentle.
(a) Loess plains
(b) Barchans
(c) Seif
(d) Sand hills
Answer:
(b) Barchans

Question 5.
River, glacier, wind, sea waves and groundwater are the agents of erosion. Following work in the correct order is responsible to form various landforms.
(a) Disintegration, picking up, transportation, weathering
(b) Picking up, disintegration, deposition, weathering
(c) Deposition, transportation, picking up, disintegration
(d) Disintegration, picking up, transportation, deposition
Answer:
(d) Disintegration, picking up, transportation, deposition

3. Give geographical reasons.

Question 1.
The Eastern coast of India have deltas formed by the rivers but the Western coast has estuaries.
Answer:
The Eastern coast of India have deltas formed by the rivers but the Western coast has estuaries because-

1. Deltas can form at the mouth of those rivers where the sediment supply is high. On the other hand, where the rivers do not have load of sediments, estuaries are formed. Deltas can also be formed where the sea is not very deep.
2. The eastern coast borders the Bay of Bengal.
3. Because of the gentle slope on the eastern coast, rivers flow with low velocities and deposit the sediments brought with them at the coast. As a result, deltas are formed at the mouth.
4. The western coast borders with the Arabian Sea, which is by and large a rocky coast.
5. Its width is also less.
6. Moreover, rivers are short and swift. Hence, they flow with high velocities and thus, does not deposit the sediments brought by them at the mouth. They form estuaries.

Question 2.
There is direct relationship between the velocity of the agents and process of deposition.
Answer:
There is direct relationship between the velocity of the agents and the process of deposition because-

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 flo’ods or when the river is in the mountains, then the carrying power of the stream and the sizes of particles deposited will increase.
4. On the other hand, if there is a change in the slope of land or in the direction of flow, deposition may occur there.
5. If the slope of a region is almost absent like in plain region, the rivers are unable to carry the sediments further and start depositing.

Question 3.
Compared to all the agents, sea waves work ceaselessly.
Answer:
Compared to all the agents, sea waves work ceaselessly because-

1. The movement of sea water carried out by waves is mainly responsible for marine erosion and deposition.
2. The sea waves are dashing on the coastline ceaselessly. The erosion in some parts and deposition in the adjoining parts takes place simultaneously.
3. The beaches and sand bars are formed due to deposition but they are also eroded.
4. The erosive power of waves depends partly on velocity of winds, as waves derive their energy from them and partly on the distance of open ocean over which they are blown.
5. Coastal erosion may be caused by hydraulic action, abrasion, impact and corrosion, which work continuously.
6. Abrasion is the most effective form erosion by waves. The softer rock along the coastline gets eroded first.

Question 4.
One finds many sheep rocks, horns, Aretes and hanging valleys in the Himalayas.
Answer:
One finds many sheep rocks, horns, Aretes and hanging valleys in the Himalayas because-
1. In Himalayas work of glacier is more predominant due to snow covered peaks and slopes. Sheep rock, horns, Aretes and hanging valleys are formed by glacier. Like other agents of erosion, glaciers too carry out erosion, transportation and deposition.

2. In the Himalayas, glaciers remove rock particles from the surface on which they flow by abrasion which leads to formation of sheep rocks.

3. Ice movement, accompanied by weathering and mass wasting has steepened the walls at the head of the glacier. This has deepened into armchair-shaped depression called cirque.

4. In Himalayas, two of more cirques have developed and the area between them is narrowed and formed arete and further the headward erosion of the glaciers has eroded the summit leading to the formation of typical peak called horn.

5. The Himalayan glaciers too have tributaries like rivers. The rate of erosion is different. The main valley gets eroded faster and becomes deeper than the tributary glacial valley. These appear to be hanging when seen from the main valley. Thus, hanging valleys are formed.

Question 5.
Karst landforms are seen concealed under the surface of the earth.
Answer:
Karst landforms are seen concealed under the surface of the earth because-

1. Warmer humid climate has greater amount of vegetation, which supplies CO₂ to sub-surface water.
2. CO₂ is necessary for the acidity of water which makes dissolving easier.
3. Water saturated with dissolved CO₂ should have easy movement.
4. The greater the permeability of the rock, the faster ground water will flow.
5. The fractures and joints in the rocks allow passage of water.
6. In the areas of water-soluble rocks, such as limestone and gypsum the acidic water dissolves the rock and various landform develop underground.
7. Thus, landforms developed by underground water anywhere in the world are called karst landforms.

Question 6.
Snowline decides the limit of glacier work as an agent of erosion.
Answer:
Snowline decides the limit of glacier work as an agent of erosion because-

1. Snowline is the lower limit of permanent snow cover; below which snow does not accumulate.
2. The glacier moves through pre-existing valley and forms erosive action.
3. The glacier performs plucking and abrasion processes on the rocks on its way.
4. Below the snowline is the snow frees surface, where the erosional features are not formed.

4. Write short notes on.

Question 1.
Attrition
Answer:

1. Attrition is when rocks and pebbles bump into each other and break up into smaller fragments.
2. During transit, materials reduce in size.
3. Rock particles become more rounded.
4. It relates to the material that moves.
5. This process of erosion is observed in river, wind and sea waves.

Question 2.
The work of rivers in hilly areas and human activities :
Answer:
Gorges: River from its source, when it starts flowing the process of erosion starts. In mountainous areas, rivers flow at a higher speed due to steep slope. As a result, bed gets eroded more than banks and gorges are formed with steep banks and narrow beds. Such gorges have also become famous tourist places. For example, Gorge of Narmada river at Jabalpur.

In ‘V’ shaped valleys of river, terraced farming is practiced on valley slopes.

In hilly areas, the river has speed, slope and hence downcutting is on a lower scale. Thus, agriculture, agroforestry, animal husbandry and forestry are the major human occupations in these regions.

Waterfalls: Water flowing over a hilly region comes down a cliff, forming alternate bands of eroded soft and hard rocks, such features are called waterfalls. These waterfalls become a site for tourist attraction. For example, Jog falls on Sharavati river, Chuliya falls on Chambal river and Venna falls in Mahabaleshwar. Hotel industries also developed along these features.

Question 3.
Conditions necessary for work of wind
Answer:
Wind is a significant agent of erosion in the deserts. Following conditions are necessary for wind to become effective:

1. Arid areas are essential. In such areas, the rate of evaporation is greater than rate of precipitation. Due to dryness the soil becomes loose and is carried by the wind.
2. Sparse vegetation cover or absence of trees, so that there is no obstacle for wind.
3. Presence of dry loose materials at the surface which carries out erosion.
4. A wind velocity high enough to pick up and move sediments.

5. Differentiate between.

Question 1.
Attrition and Abrasion.
Answer:

Question 2.
U shaped valley and V shaped valley.
Answer:

Question 3.
Stalactite and Stalagmite.
Answer:

Question 4.
Tributaries and Distributaries.
Answer:

6. Answer in detail.

Question 1.
Explain the landforms formed by different agents through the process of abrasion.
Answer:
Abrasion involves the scratching and polishing of the surface or bedrock by the particles which are moving onto it. Let us study different landforms formed by different agents of erosion through the process of abrasion.
Work of Seas Waves:

1. Sea cliffs : Sea cliffs are carved where waves strike directly against rocks. The softer rock along the coastline gets eroded first.
2. Caves : Sometimes, waves can erode the softer part and make it hollow enough to be called a cave.
3. A sea arch : A sea arch forms when waves erode a layer of soft rock underneath a layer of hard rock.
4. Sea stack : If a sea arch collapses, it creates a sea stack, which looks like a large rock in the middle of water.
5. Headland : The part of land projecting into the sea is called headland.
6. Wave-cut platforms : Surfaces at the base of the cliffs are called wave-cut platforms. Extensive platforms are developed where the rocks are least resistant to wave erosion. They are visible at lower water levels, such as at low tide.

Work of Wind:
(i) Ventifacts:
Abrasion carves the windward side of rock into smooth sloping surface. These rocks are called ventifacts.

(ii) Mushroom rocks

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 because the particles are not lifted at a height.
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.

(iii) Yardang:

1. In areas where hard and soft rocks are found, the softer rocks get more eroded faster.
2. The eroded portion of softer rocks appear like elongated ridges and harder rocks appear as elevated portions.
3. A yardang is the remaining part of a ridge where rocks have been eroded.

Work of Glaciers
(i) Roche moutonnee:
They are bedrock hills that are smoothly rounded on the upper side by abrasion and plucking on the lower side.

(ii) Cirque

1. An armchair like feature is formed when ice movement accompanied by weathering and mass wasting steepens the wall at the head of the glacier.
2. It deepens into armchair-shaped depression called cirque.

(iii) Arete

1. Often two or more cirques develop side by side.
2. This leaves the area between any two of them into a narrow wall. This is called arete.

(iv) Horn

1. When three or more cirques are formed, the headward erosion of the glacier erodes the summit.
2. This leads to formation of a characteristic peak which is called a horn. The Matterhorn in the Swiss Alps is an example.

(v) U-Shaped Valleys
When glaciers move ahead, they erode the sides as well as the bottom of the valleys they flow through. This makes the valley broad at the bottom, forming a ‘U’. This is called a U-shaped valley.

Question 2.
Explain how the depositional work done by River Ganga has been beneficial to human activities.
Answer:
The River Ganga and its tributaries have deposited load of sediments in the northern Ganga region. Features formed are:
Alluvial Plain Region : The River Ganga and its tributaries bring loads of sediments – organic sediments like dead remains of plants and animals, skin, hide, bones etc., and inorganic sediments like sand, silt, clay, gravel, etc., and deposits at the foothills of the Himalayan family, Northern Ganga Plain region. It is a fertile land and agriculture is the major human occupation. Along with agriculture river transportation and fishing is also done on a large scale. Fertile plains are in Uttar Pradesh, Bihar, Punjab and Haryana.

Delta: Ganga-Brahmaputra delta, world’s largest delta has been formed at the mouth of the River Ganga. This delta is known as Sundarbans and is located in West Bengal. It is one of the most fertile regions in the world. Most delta is composed of alluvial soil. Thus, agriculture is the occupation followed in this region too.

Question 3.
Which agents of erosion can you see on the cover page of the textbook? Which landforms can you see there? Write the process of formation of any one.
Answer:

1. On the cover page of the textbook. We observe the agents of erosion such as rivers, glaciers and sea waves.
2. We see different landforms like – alluvial fans, meander, oxbow lakes, delta, sea and beach.
3. The river in the middle stage, picks up more material by bank erosion and gets overloaded. The speed of erosion is reduced. The water moves sluggishly in a bed and turns at every minor change of slope and serpentine bends are formed.

Every bend is made more and more pronounced by dashing of water on outer bank which gets eroded while inner banks have deposition. In due course of time they develop into circular loops, they are called meanders.

7. Draw neat and labelled diagrams for

Question 1.
Deflation
Answer:

Question 2.
Wave-cut platform
Answer:

Question 3.
Mushroom rocks
Answer:

11th Geography Digest Chapter 3 Agents of Erosion Intext Questions and Answers

Can you tell? (Textbook Page No. 27)

1. The names of many places may have some relation with the landforms located there. The names of few of them are given below as example. You are expected to find out the particular landforms to which they are associated. Find about them and try to locate them on a map using an atlas. Make a list of similar examples from India.

Answer:

2. Look at the figure below and answer the questions. (Textbook Page No. 32)

Question 1.
What features are formed in the upper course of the river? What processes will play an important role?
Answer:
Features formed in the upper course of the river are V-shaped valleys, gorges and canyons, potholes, waterfalls and rapids.
Processes responsible for the formation of these landforms are – downcutting, drilling.

Question 2.
Can you mark the location where waterfall may form?
Answer:
Waterfall may form at location ‘A’

Question 3.
Why is there a change in slope from A to B? How will it influence the flow of the river?
Answer:
There will be a change in slope from A to B when river enters the plains at the foothills of a mountain. The flow of the river decreases, the velocity reduces abruptly.

Question 4.
Which human activities can be conducted in the region around A and B?
Answer:
Around region A terraced farming can be practised. Around region B there is rich with alluvium brough down by the rivers, the plain is extensive and useful for agriculture. As the river widens around the source B, fishing and transportation is also found.

Question 5.
In which area will the process of deposition overtake erosion?
Answer:
Deposition will be found at B and C.

Question 6.
Alluvial fans and deltas are both features formed due to deposition but at different locations. Identify their regions of formation and reason behind their different locations.
Answer:
Alluvial fans will be formed at source ‘B’ at the foothills of the mountains, as there is change in slope and the velocity of the river reduces abruptly. Now the river is unable to carry heavy load. A delta will be formed at the mouth of the river. In this region, which is an extensively flat region, and where the sediments supply is high but velocity is low. So, the river deposits the sediments and branches out to meet ocean or sea.

3. Look at the figure and identify landforms at A, B, C, D, E, F, G. (Textbook Page No. 34)

Answer:
A – Headland
B – Lagoon
C – Beach
D – Sea Cave
E – Sea Cliff
F – Sea Arch
G – Sea Stack

Let’s recall. (Textbook Page No. 30)

You have already learnt about various landforms formed by the agents of erosion in Class IX. Identify the landforms given in class IX textbook from page no. 30 to 38. Identify the agent which is responsible for their formation. Also, state whether they are erosional or depositional landforms. Complete the table accordingly.

Answer:

Think about it. (Textbook Page No. 30)

Have you ever been to a river and seen its bed? Discuss in the class about your observation about the river, its banks, its bed and its velocity.
Answer:
[Students will discuss their experiences with teachers.]

Find out! (Textbook Page No. 31)

Find out famous examples of gorges and canyons.
Answer:
Following are the famous examples of gorges and canyons of the world.

1. Fish River Canyon, Namibia
2. Grand Canyon, Arizona, USA
3. Tiger Leaping Gorge, China
4. Kali Gandaki Gorge, Nepal
5. Blyden River Canyon, South Africa
6. Gorges du Verdon, France
7. Indus Gorge, Pakistan
8. Antelope Canyon, USA

1. Study figure given below and answer the following questions. (Textbook Page No. 38)

Question 1.
Which rocks are mainly found here?
Answer:
Soluble rock like limestone, a sedimentary rock composed of Calcium Carbonate (CaCO₃) are found.

Question 2.
Identify the spot where stream disappears.
Answer:
Steam will disappear below the sink hole.

Question 3.
Which major erosional process works in this area?
Answer:
Solution is the major erosional process in this area.

Question 4.
Identify the landforms formed by deposition.
Answer:
Stalactite, Stalagmite, columns or pillars are the landforms formed by deposition.

Question 5.
Why do depositional landforms not form on the surface in areas of Karst terrain?
Answer:
In Karst terrain groundwater dissolves minerals like calcium carbonate present in the rocks. The dripping water leaves behind a deposit of calcium carbonate. The water saturated with calcium carbonate dripping on the floor of a cave deposits calcium carbonate on the floor. Thus, depositional landforms do not form on the surface in the areas of Karst terrain.

2. See the figure give below A, B, C. Answer the following questions. (Textbook Page No. 40)

Img 8
Question 1.
What difference do you find in the three figures?
Answer:
Fig A shows the regions before glacier formed.
Fig B showed the regions covered by snow during glaciation.
Fig C shows different erosional and depositional features formed by glaciers after glaciation period.

Question 2.
Identify the landforms formed due to erosion by glaciers.
Answer:
Cirques, horn, U-shaped valleys, hanging valleys are formed due to erosion by glaciers.

Question 3.
Where can U-Shaped valleys be formed?
Answer:
U- Shaped valley is formed in the pre-existing valley.

Question 4.
In which region will deposition occur?
Answer:
Deposition starts generally along the side and front of ice.

Question 5.
Identify the landforms formed by deposition by glaciers.
Answer:
Drumlins, eskers, moraines are the depositional features formed by glaciers.

Use your brain power! (Textbook Page No. 41)

In which diagram of the three will you find end moraines? See fig. 3.5 A, B, C.

Answer:
We will find end moraines in Fig. C

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

Maharashtra State Board Class 11 Geography Solutions Chapter 2 Weathering and Mass Wasting

1. Complete the Chain.

Question 1.

Answer:

2. Identify the correct correlation.

A : Assertion
R : Reasoning
Question 1.
A – In areas of high rainfall, slides are very common.
R – Types of mass wasting movements are dependent on a region’s 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:
(d) Both A and R are correct but R is not the correct explanation of A.

Question 2.
A – Gravity is a major factor in mass wasting.
R – Gravity pulls all things down to the earth’s surface.
(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 – Freeze and thaw weathering is common in desert areas.
R – Water gets into cracks and breaks 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:
(d) Both A and R are correct but R is not the correct explanation of A.

Question 4.
A – Surface water helps solifluction.
R – Water table is responsible for the same.
(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.

3. Identify the correct group.

Question 1.

Answer:
C

4. Give geographical reasons.

Question 1.
Temperature is the main factor behind granular weathering.
Answer:
Temperature is the main factor behind granular weathering because-

1. In hot desert areas, diurnal range of temperature is high.
2. As the temperature increases with the rising sun, the rock gets heated and cools down with the decrease in temperature.
3. The minerals in the rock react differently to the temperature increase due to continuous heating and cooling.
4. Consequently, it leads to development of stress within the rock and molecular or granular disintegration takes place.

Question 2.
Human is an agent of weathering.
Answer:
Human is an agent of weathering because-

1. Man is a biological agent of weathering. Due to economic and technological development, man has become the most powerful agent of weathering and erosion.
2. Mining, blasting of hills and ridges for road and dam construction, quarrying for industrial and building materials, etc., results in a fast rate of disintegration of rocks.
3. Man accelerates the rate of weathering on hill slopes through activities like deforestation.

Question 3.
Slope is a major factor in mass wasting.
Answer:
Slope is a major factor in mass wasting because-

1. Mass wasting is the down-slope movement of loose mixture of soil, land and rock particles by the force of gravity.
2. In mass wasting the materials come down the slope without the aid of transporting medium like running water, ice or wind.
3. Mass wasting occurs continuously on all slopes.
4. Some act very slowly, others very suddenly, often with disastrous results.

Question 4.
Oxidation changes the size and colour of the rocks.
Answer:
Oxidation changes the size and colour of the rocks because-

1. Oxygen in the air and water reacts with certain elements in the minerals inside the rock
2. In this process, the minerals in the rock react with the oxygen present in the air or water.
3. Metals, particularly iron and aluminium, commonly oxidize forming iron or aluminium oxides. It is also called rusting.
4. These oxides are larger in volume than in the original rocks.
5. The iron oxides are red in colour and aluminium oxides are yellow.
6. Thus, oxidation changes size and colour of the rock.

Question 5.
Effect of mass movement will be greater along the western slope of the Sahyadri’s than the eastern slope.
Answer:
Effect of the mass movement will be greater along the western slope of the Sahyadri’s than the eastern slope because-

1. The western slope of the Sahyadri’s are steeper than the eastern slopes.
2. The western slopes receive more rainfall than the eastern slopes as a result mass movement is greater in western slopes than eastern slopes.
3. Also, the rivers as an agent of erosion are short and swift on the western slope and thus rapid mass movement is found as compared to eastern coast.

5. Write short notes on.

Question 1.
Gravity and Solifluction
Answer:
Gravity:

1. It is the main force responsible for mass movements.
2. It is a force that acts everywhere on the earth’s surface, pulling everything down.
3. If the friction on the rock is stronger than gravity for a particular slope, the rock material is likely to stay.
4. But if the gravity is stronger, movement will occur in the direction of the slope.

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. As permafrost is impermeable to water, soil overlying may become oversaturated and slide slopes down under the pull of gravity.

Question 2.
Role of water in mass wasting
Answer:

1. Although water is not always directly involved as a transporting medium but it plays an important role in mass wasting.
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.

Question 3.
Exfoliation
Answer:

1. Because of the overlying rocks, the rocks beneath the surface experience a lot of pressure.
2. The exposed part of the rock heats more while the inner part is comparatively cooler.
3. As a result, the outer layer of the rocks fall apart from the main rock just as we peel off onion.

Question 4.
Weathering and homogeneity in rocks
Answer:

1. Weathering is the physical or chemical breakdown of rocks into small pieces due to various reasons like weather conditions, temperature, and water, living organisms like humans, algae, fungi, etc.
2. Weathering depend upon the texture of the rock. Those rocks which have joints on layers break easily than rocks which are homogeneous. Generally, sedimentary rocks break easily than igneous rocks as sedimentary rocks are formed by layering.
3. Sediments brought down by the river are deposited in layers, thus sedimentary rocks are soft and porous, and can be broken easily.
4. Igneous rocks are formed by cooling and solidification of molten magma. Hence, they are hard and non-porous and cannot be broken so easily.
5. Thus, sedimentary rocks are more subjected to weathering than the igneous rocks.

Question 5.
Carbonation
Answer:

1. The decomposition of dead matter in the soil produces CO₂
2. This CO₂ and the CO₂ in the air reacts with minerals in the rocks.
3. Minerals such as feldspar and carbonates decompose when this happens.
4. In humid climate, water adds to the weathering process.
5. In arid climates, the absence of water in the region leads to carbonate rocks to form cliffs that are resistant.
6. Often, carbonation and solution occur simultaneously.
7. During carbonation, the calcium and carbonate in limestone detach from each other, thereby decomposing the limestone.

6. Draw a neat and labelled diagram for

Question 1.
Freeze and thaw weathering
Answer:

Question 2.
Block disintegration
Answer:

Question 3.
Biological Weathering
Answer:

7. Answer in detail.

Question 1.
Explain with examples the process of weathering happening in Konkan.
Answer:

1. Konkan, also known as the Konkan Coast is a rugged, section of the western coastline of India.
2. The region is divided into North Konkan including Mumbai, Thane and Raigad districts whereas South Konkan include Ratnagiri and Sindhudurg districts.
3. Laterite rocks-occur in the form of plateaus at Konkan strip. High altitude laterite plateaus
   are found is Sitara, Kolhapur, Ratnagiri and Sindhudurg district.
4. In the laterite rocks soft silica matter is present.
5. Because of high rainfall exposed rocks become empty to wash out of soft silica matter.
6. Laterite soil is formed by laterite rock in high elevation. The soil contains has high concentration of ferrous and aluminium. This soil is very stony because of the presence of weathering fragments of iron concentrations.
7. Oxygen in the air and water reacts with certain elements in the minerals in the rock. The metals in the rock particularly iron and aluminium oxidise and form iron and aluminium oxides. Thus, chemical weathering takes place.
8. Laterite soil is found in Mahabaleshwar, Southern parts of Mahabaleshwar, around Bhima Shankar and Matheran. Thus, chemical and mechanical weathering is found.

Question 2.
Explain the correlation between Himalayas and mass movements. Give examples wherever necessary.
Answer:

1. Mass movements occur in hilly, mountainous or plateau areas. The steeper the slope more the occurrences of mass movement.
2. The Himalayas have greatest relief, high potential energy, high seismic activity, steep slopes, strong weathering, great snow fall etc.
3. Due to slope, gradient, massive rockslides, debris flow takes place.
4. Steep gradient of precipitation and temperature produce mass movement that loads glacier surface and chokes, rivers with sediments.
5. For example, in Kosi river basin of Himalayas of central Nepal, because of deep weathering, high seismic activity, river undercutting, deforestation and heavy precipitation, especially during the summer months, mass movements are most frequent in highly jointed, sheared, intensely folded and fault rocks.
6. Thus, in short, in the Himalayan region, due to steep slope, melting of glaciers, precipitation, seismic activity, mass movement takes place on a larger scale.

11th Geography Digest Chapter 2 Weathering and Mass Wasting Intext Questions and Answers

Let’s recall (Textbook Page No. 15)

1. Study the diagram in fig 2.1 and answer the following questions.

Question 1.
Identify the types of rocks shown in the diagram.
Answer:
Igneous, sedimentary and metamorphic rocks are shown in the diagram.

Question 2.
Arrange the rocks according to their chronology of origin.
Answer:
Igneous, sedimentary and metamorphic rocks.

Question 3.
Explain how sedimentary rocks are formed.
Answer:
Sedimentary rocks are formed from layering upon layering of all the organic (dead remains of plants and animals) and in organic (sand, silt, clay, gravel, etc.) material in a depression or on low lying area. If there are cementing material like limestone, hardening and compaction takes place, then sedimentary rocks are formed.

Question 4.
Think of all the factors which may break the rocks into smaller pieces.
Answer:
Heat, pressure, water, wind, animals, plants etc., can break rocks into smaller pieces.

Question 5.
Which type of rock will break easily as compared to others? Why?
Answer:

1. Sedimentary rocks will break easily as they are formed by layering of sediments and thus, they are soft and porous and are broken easily.
2. Igneous rocks are formed from cooling and solidification of molten magma; hence they are hard and non- porous.
3. The metamorphic rocks too cannot be broken easily as they are formed from heat and pressure.

Think about it.

Question 1.
In which regions will freeze-thaw weathering not be effective? (Textbook Page No. 17)
Answer:
Freeze-thaw weathering will not be effective on the cold polar regions as the soil cover and rocks are very little on the polar areas and moreover, there is snow everywhere.

Question 2.
Besides climatic factors, rock type and structure, can you think of some more factors that affect weathering? (Textbook Page No. 19)
Answer:
Plants, animals, micro-organisms, humans are some more factors that affect weathering.

Question 3.
Can tectonic forces be responsible for mass movement? (Textbook Page No. 25)
Answer:

1. Plate tectonics are responsible for uplift and mountain building that creates and maintains slopes.
2. Mass wasting is common in tectonically active regions.
3. Plate tectonics causes earthquakes that can trigger landslides and cause sediment to lose its strength through liquefaction.

Question 4.
There is a shift of materials in mass movement as well as in transportation from one place to the other. So, why can’t both not be treated as one and the same? (Textbook Page No. 25)
Answer:
Mass movement is caused by abrupt movement and freefall of loosened rock particles because of gravity and friction falling towards the surface, whereas transportation is carrying of rock materials with the help of agents of erosion such as water, wind, air, ice, etc. Thus, they are not same.

Use your brain power!

Question 1.
Can animals and plants also influence weathering? Will that be physical or chemical weathering? Which type of weathering does stone quarrying cause? (Textbook Page No. 19)
Answer:
Yes, burrowing animals widen the fissures in the rocks. Rocks become weak and disintegrate.

The roots of the trees and other plants penetrate in the soil, they grow in size, exert pressure on rocks, widen cracks in the rocks and rocks break. Many microscopic organisms such as algae, lichens, bacteria, moss etc produce chemicals and they break down the outer layer of the rock. These chemicals are responsible for physical and chemical weathering of rocks. The stone quarrying causes the anthropogenic weathering.

2. A region is having an annual mean temperature of 5° C and an annual rainfall of 1000 mm. Can you comment upon the weathering and the type with the help of following questions? (Textbook Page No. 20)

Question 1.
Which type of weathering will be dominant here?
Answer:
Physical weathering will be dominant here.

Question 2.
Where will such a region be found?
Answer:
Such a region will be found in permafrost conditions, alpine and periglacial region.

Question 1.
Complete the table by using the words: intense, moderate, slight and very slight or no weathering. (Textbook Page No. 20)
Answer:
Rate of Physical Weathering:

Rate of Chemical Weathering:

Can you tell? (Textbook Page No. 21)

1. See the diagram given in fig 2.10 and answer the following questions.

Question 1.
Which rock layer has experienced the most weathering?
Answer:
Rock layer C has experienced the most weathering.

Question 2.
Which rock layer has experienced the least weathering?
Answer:
Rock layer B has experienced the least weathering.

Question 3.
What could be the reason behind difference in weathering?
Answer:
Rock C has lot of fractures and joints so it got weathered easily.
Rock B might have been a hard rock, more resistant, so weathering process is slow.

2. The satellite images given in fig. 2.11 A and B belong to the same location but different timeline. Study the images and answer the following questions. (Textbook Page No. 21)

Question 1.
Compare the images and tell what differences do you find in these images?
Answer:
Image 2.11 (A) shows maximum area under vegetation. Image 2.11 (B) shows development in form of settlements and transportation routes.

Question 2.
In 2019, what does the patch of land going from north-west to south-east signify? Why was it not there in 2011 image?
Answer:
The sweeping portion from north-west to south-east is visible in satellite image of 2019. This is because the village is located at the foothills. The rain and slope were responsible for the mudslide. Heavy rainfall and absence of vegetation aggravated the situation. In 2011, vegetation was thick but in 2019 deforestation has been done extensively for farming practices.

Question 3.
To what extent is the climate of a place responsible for this disaster?
Answer:
Climate plays an important role for the disaster. Heavy rainfall makes soil to move from the surface of the mountain towards the foothills.

Question 4.
Which other factors are responsible for the disaster?
Answer:
Deforestation and absence of vegetation are responsible for this.

Give it a try.

1. Study the following schematic diagram. It shows the relationship between speed of material and moisture content. Read the index, and answer the following questions. (Textbook Page No. 23)

(A) Fast (B) Slow (C) Dry (D) Wet
(1) Creep (2) Slide (3) Fall (4) Flow

Question i.
What will happen when the weather conditions are dry?
Answer:
There will not be moisture content hence speed of material will be low.

Question ii.
When will a flow occur?
Answer:
Flow will occur when the moisture content as well as speed of material will be high.

Question iii.
When will a creep occur?
Answer:
Creep will occur when speed of material will be slow.

Question iv.
Now can you enumerate the factors which affect mass wasting?
Answer:
Factors such as gravity, slope of land, climate of region, amount of water, material and structure of the rock affect mass wasting.

Question 2.
On the basis of given points, differentiate between weathering and erosion. (Textbook Page No. 25)
Answer:
Table

Think a little. (Textbook Page No. 24)

Think of the reason why landslides should be more frequent in foothill zone of the Himalayas and Western Ghats region. Why do landslides not occur in Marathwada in Maharashtra or Maidan area in Karnataka?
Answer:
Himalayas are one of the youngest fold mountains of the world. They are formed due to convergent movement of the Indian plate and erosion plate. They are still rising in height. These tectonic movements cause frequent earthquakes in the region resulting into landslides whereas western Ghats lie in the stable Deccan shield less prone to landslides.

Himalayas are greater in height than the western Ghats. The slopes are comparatively steep and hence landslides are common in Himalayas whereas, the western Ghats are much lesser in height than the Himalayas hence less prone to landslides as compared to Himalayas.

The perennial rivers in Himalayas carry lots of sediments during rainfall and due to melting of glaciers in the summer thus leading to landslides, whereas, in Western Ghats many rivers are non-perennial and hence less amount of silt and debris in carried as compared to Himalayan rivers. Thus, there is reduction in the chances of landslides, only during the rainy season, landslides occur in Western Ghats region.

Try this. (Textbook Page No. 24)

Question 1.
Different types of materials flow down the slope. Types of mass wasting depend on their speed. Observe the pictures given in figure 2.12. Match the explanation given below with the diagrams. Identify them as slow or rapid movements.

Answer:
A – Earth flow – Rapid or slow movement
B – Creep – Slowest movement
C – Land slide – Rapid movement
D – Solifluction – Slow movement
E – Rock fall – Rapid movement

Question 2.
Complete the following flow chart. (Textbook Page No. 25)

Answer:

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

Maharashtra State Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics

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) H_2(g) → 2H_(g)
(d) 2NaHCO_3(s) → Na₂CO_3(s) + CO_2(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

Question v.
Which of the following reactions is exothermic ?
(a) H_2(g) → 2H_(g)
(b) C(s) → C_(g)
(c) 2Cl_(g) → Cl_2(g)
(d) H₂O_(s) → H₂O_(l)
Answer:
(c) 2Cl_(g) → Cl_2(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

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 = -P_ex (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 = -P_ex (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 (Δ_fusH) : 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,

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 Δ_fH⁰.

Question v.
State whether ∆S is positive, negative or zero for the reaction 2H_(g) → H_2(g). Explain.
Answer:
(i) The given reaction, 2H_(g) → H_2(g) is the formation of H_2(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
∆ S_Total = ∆ S_system + ∆S_surr > 0
∆ S_Universe = ∆ S_system + ∆ S_surr > 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.

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 (Q_rev) divided by the absolute temperature (T), at which the heat is absorbed. Thus,

(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 (Δ_fH⁰) 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.)

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) + \(\frac {3}{2}\) O_2(g) = CO_2(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.

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 : CH_4(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.

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.

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 P_ex = \(\frac{F}{a}\)
while volume change is, ΔV = d × a
∴ W = -P_ex × ΔV
If during the expansion, the volume changes from V₁ and V₂ then, ΔV = V₂ – V₁
∴ W= -P_ex(V₂ – V₁)
During compression, the work W is +ve, since the energy of the system is increased,
W = +P_ex(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 NO_2(g) → 4CO_2(g) + N_2(g), ∆H⁰ = -1200 kJ

4. Answer the following questions.

Question i.
Derive the expression for the maximum work.
Answer:

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.

Then the total work or maximum work in the process is obtained by integrating above equation.

At constant temperature,

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
Q_rev, is completely converted into work.
Q_rev = -W_max.
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,

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

∴ ΔH = ΔU + ΔnRT
If Q_P and Q_V are the heats involved in the reaction at constant pressure and constant volume respectively, then since Q_P = ΔH and Q_V = ΔU.
∴ Q_P = Q_V = 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 CO_2(g).

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.

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 ∆S_Total. 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

This is called Gibbs free energy equation for ∆G. In this ∆S is total entropy change, i.e., ∆S_Total.

(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) ∆S_total > 0 and ∆G < 0, the process is spontaneous.
(ii) ∆S_total < 0 and ∆G > 0, the process is nonspontaneous.
(iii) ∆S_total = 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;
P_ex = 1.2 × 10⁵ Pa = 1.2 bar; W= 36 J;
1 dm³ bar = 100 J; V₂ = ?
W = -P_ex (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, W_max = ?

Question viii.
Calculate the work done in the decomposition of 132 g of NH₄NO₃ at 100 °C.
NH₄NO_3(s) → N₂O_(g) + 2 H₂O_(g)
State whether work is done on the system or by the system.
Answer:
NH4₄NO_3(s) → N₂O_(g) + 2 H₂O_(g)
m_NH4NO3 = 132 g; M_NH4NO3 = 80 g mol^-1
T = 273 + 100 = 373 K; Δn = ?
For the reaction,
Δn = Σn_{2 gaseous products} – Σn_{1 gaseous reactants}
= 3 – 0 = 3 mol
Since there is an increase in number of gaseous moles, the work is done by the system.
n_NH4NO3 = \(\frac{m_{\mathrm{NH}_{4} \mathrm{NO}_{3}}}{M_{\mathrm{NH}_{4} \mathrm{NO}_{3}}}\)
= \(\frac{132}{80}\)
= 1.65 mol
For 1 mol NH₄NO_3(s) Δn = 3 mol
∴ For 1.65 mol NH₄NO_3(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

Question ix.
Calculate standard enthalpy of reaction,
Fe₂O_3(s) + 3CO_(g) → 2 Fe_(s) + 3CO_2(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,
Fe₂O_3(s) + 3CO_(g) → 2Fe_(s) + 3CO_2(g)
∆H₁ = ? – (I)
Given equations :

= -(-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) + O_2(g) → 2H₂O_2(l)
if ∆H⁰ = 196 kJ, ∆S⁰ = -126 J/K
Does it have a cross-over temperature?
Answer:
Given : 2H₂O_(l) + O_2(g) → 2H₂O_2(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,
C₂H_4(g) + HCl_(g) → C₂H₅Cl_(g), ∆H = -72.3 kJ
How much PV work is done?
Answer:
Given : C₂H_4(g) + HCl_(g) → C₂H₅Cl_(g)
T = 298 K; ∆H = -72.3 kJ; PV = ?;
∆U = ?
∆n = Σn_{2gaseous products} – Σn_{1gaseous 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; P_ex = 43 bar
W = ? What direction work energy flows ?
W = -P_ex(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.

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) + O_2(g) → 2 Cl_2(g) + 2 H₂O_(g)
(b) decomposition of one mole of NO at 300 °C for the reaction
2 NO_(g) → N_2(g) + O₂
Answer:
Given :
(a) 4HCl_(g) + O_2(g) → 2Cl_2(g) + 2H₂O_(g)
n_HCl = 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) + O_2(g) → Cl₂O_(g) + OF_2(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) + \(\frac {3}{2}\) O_2(g) → CO_2(g) + 2H₂O_(l), ∆H⁰ = -726 kJ mol^-1
ii. C (Graphite) + O_2(g) → CO_2(g), ∆_cH⁰ = -393 kJ mol^-1
iii. H_2(g) + \(\frac {1}{2}\) O_2(g) → H₂O_(l), ∆_fH⁰ = -286 kJ mol^-1
Answer:


∴ ∆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

Question xvii.
Calculate ∆H0 for the following reaction at 298 K
H₂B₄O_7(s) + H₂O_(l) → 4HBO_2(aq)
i. 2H₃BO_3(aq) → B₂O_3(s) + 3H₂O_(l), ∆H⁰ = 14.4 kJ mol^-1
ii. H₃BO_3(aq) → HBO_2(aq) + H₂O_(l), ∆H⁰ = -0.02 kJ mol^-1
iii. H₂B₄O_7(s) → 2B₂O_3(s) + H₂O_(l), ∆H⁰ = 17.3 kJ mol^-1
Answer:
Given equations :
i. 2H₃BO_3(aq) → B₂O_3(s) + 3H₂O_(l), ……….(i)
∆H⁰ = 14.4 kJ mol^-1
ii. H₃BO_3(aq) → HBO_2(aq) + H₂O_(l) ……….(ii)
∆H⁰ = -0.02 kJ mol^-1
iii. H₂B₄O_7(s) → 2B₂O_3(s) + H₂O_(l), ……….(iii)
∆H⁰ = 17.3 kJ mol^-1
Required equation :
(iv) H₂B₄O_7(s) + H₂O_(l) → 4HBO_2(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
∆_vapH_H2O(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, ∆ H_Total = ?
Number of moles of H₂O = \(\frac {180}{18}\) = 10 mol
The total process can be represented as,

(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,
∆ H_Total = ∆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,
C₂H_4(g) + H_2(g) → C₂H_6(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; V_C2H4 = 100 mL; V_H2 = 100 mL
P_ex= 1 bar; W=?; ∆U = ?
∆V = 100 – (100 + 100) = -100 mL = -0.1 dm³
W = -P_ex(V₂ – V₁)
= -P_ex × ∆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
NH₄NO_3(s) → N₂O_(g) + 2H₂O_(g)
Answer:
Given : NH₄NO_3(s) → N₂O_(g) + 2H₂O_(g)
n_NH4NO3 = 2 mol; T = 273 + 100 = 373 K
W = ? Comment on work = ?
∆n_reaction = (1 + 2) – 0 = 3 mol
∵ For 1 mol of NH₄NO₃ ∆n_reaction = 3 mol
∴ For 2 mol of NH₄NO₃ ∆n_reaction = 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

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

Try this… (Textbook page No. 76)

Question 1.
Given the thermochemical equation,
C₂H_2(g) + \(\frac {5}{2}\) O_2(g) → 2CO_2(g)+ H₂O_(l), ∆_rH⁰ = -1300 kJ
Write thermochemical equations when
i. Coefficients of substances are multiplied by 2.
ii. equation is reversed.
Answer:
(i) 2C₂H_2(g) + 5O_2(g) → 4CO_2(g) + 2H₂O_(l)
∆_rH⁰ = -2 × 1300 kJ
= – 2600 kJ
(ii) 2CO_2(g) + H₂O_(l) → C₂H_2(g) + \(\frac {5}{2}\)O_2(g)
∆_rH⁰ = +1300 KJ

Try this… (Textbook page No. 78)

Question 1.
(i) Write thermochemical equation for complete oxidation of one mole of H_2(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) H_2(g) + \(\frac {1}{2}\)O_2(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 H_2(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 :
Cl_2(g) → 2Cl_(g) ∆_rH⁰ = 242.7 kJ mol^-1
∴ Bond enthalpy of Cl₂ = 242.7 kJ mol^-1

Try this… (Textbook page No. 82)

Question 1.
State whether ∆S is positive, negative or zero for the following reactions.
i. 2H_2(g) + O_2(g) → 2H₂O_(l)
ii. CaCO_3(s) → CaO_(s) + CO_2(g)
Answer:
(i) 2H_2(g) + O_2(g) → 2H₂O_(l)
Since the system is converted from gaseous state to a liquid state, the disorder is decreased, hence ∆S < O (negative).

(ii) CaCO_3(s) → CaO_(s) + CO_2(g)
Since molecules of solid CaCO₃ break giving gaseous CO₂, disorder is increased hence ∆S > O (positive).

Balbharti Maharashtra State Board Class 11 Geography Solutions Chapter 1 Earth Movements Textbook Exercise Questions and Answers.

Maharashtra State Board Class 11 Geography Solutions Chapter 1 Earth Movements

1. Complete the chain.

Question 1.

Answer:

2. Identify the correct correlation.

A : Assertion R : Reasoning
Question 1.
A – Faulting leads to development of fold mountains.
R – Faulting occurs when tensional forces move away from each other.
(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.

Question 2.
A – Intensity of an earthquake is a measurement of the energy released during an earthquake.
R – Mercallis scale is used to measure intensity of an earthquake.
(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 – South-East Asia, Japan and islands in the Pacific Ocean are most vulnerable to earthquakes and volcanic eruptions.
R – They are located in ‘Ring of Fire’.
(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.

3. Identify the correct group.

Question 1.

Answer:
C

4. Give geographical reasons.

Question 1.
Extinct conical volcanoes often form crater lakes.
Answer:
Extinct conical volcanoes often form crater lakes because-

1. An extinct volcano had no volcanic eruption for at least 10,000 years and is not expected to erupt again in future.
2. After the eruption a large and deep depression remains in that area. Smaller depressions are called craters.
3. This crater eventually gets filled with water and crater lakes are formed.
4. The water may come from precipitation, ground water circulation or melted ice.

Question 2.
People living in the Himalayas are more vulnerable to earthquakes.
Answer:
People living in the Himalayas are more vulnerable to earthquakes because-

1. Himalayan mountain ranges are formed as a result of collision of Indo-Australian plate and Eurasian plate.
2. According to plate tectonic theory, Indo-Australian plate (on which India lies) is moving towards north and is subducting under Eurasian plate.
3. This movement obviously cause friction between two plates.
4. Thus, these regions are more prone to earthquakes.

Question 3.
L-waves do not have a shadow zone.
Answer:
L-waves do not have a shadow zone because-

1. The shadow zone is the area of the earth from angular distance of 105° to 140° from a given earthquake for both P-waves as well as S-waves.
2. Generally, seismographs located at any distance withinl05° from epicentre record the arrival of S-waves and beyond 140° from epicentre record the arrival of P-waves.
3. L-waves are typically generated when the source of the earthquake is close to the earth’s surface.

Question 4.
Soft rocks form folds while hard rocks form faults.
Answer:
Soft rocks form folds while hard rocks form faults because –

1. Soft and elastic rocks are affected by compressional forces.
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. Whereas rock layers that are near the earth’s surface are too rigid to bend into folds.
4. If the tectonic force is large enough, these rocks will break rather than bend and faulting on rupture occur.

Question 5.
Folds depend on the strength of rocks and intensity of forces.
Answer:
Folds depend on the strength of rocks and intensity of forces because-

1. The elasticity of rocks largely affects the nature and magnitude of folding process.
2. The softer and more elastic rocks are subjected to intense folding while rigid and less elastic rocks are only moderately folded.
3. The compressional forces push two portions of the crustal rocks together and tend to shorten and thicken the crust.
4. Rocks that lie deep within the crust and are therefore under high pressure are generally ductile and particularly, susceptible to folding without breaking

5. Answer in detail.

Question 1.
Explain different 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 or horizontal direction.
The different types of faults determined by the direction of motion are:
Normal fault: It results when a portion of land slide down along the fault plane and when the exposed portion of the plane faces the sky

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.

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.

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 2.
Explain with examples, different types of landforms produced by volcanic eruption.
Answer:
A number of landforms are formed due to cooling and solidification of magma. Some important landforms are:
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 slopes 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:
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 States 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 lave 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 3.
Explain the concept of shadow zone.
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-wave 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 zone is called ‘shadow zone’.
5. Generally, seismographs located at any distance within 105° from the epicentre, record both P-waves and S-wave. However, beyond 140° P-wave 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 round the earth between 105° and 140° away from the epicentre.

Question 4.
Write a note on 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:
3. Acidic lava : (a) It contains higher percentage of silica, (b) It has high melting point, (c) It is thick, fluid and moves slowly.
4. Basic lava : (a) It contains less percentage of silica, (b) It has low melting point, (c) 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.

6. Differentiate between.

Question 1.
Folding and Faulting.
Answer:

Question 2.
Normal fault and Reverse fault.
Answer:

Question 3.
Syncline and Anticline.
Answer:

Question 4.
Asymmetrical fold and Symmetrical fold.
Answer:

Question 5.
Mercalli scale and Richter scale.
Answer:

Question 6.
Slow movements and Sudden movements.
Answer:

7. Draw a neat and labelled diagram.

Question 1.
Types of folds
Answer:

Question 2.
Types of Faults
Answer:

Question 3.
Shadow zone
Answer:

Question 4.
Volcanic landforms
Answer:

11th Geography Digest Chapter 1 Earth Movements Intext Questions and Answers

Let’s recall (Textbook Page No. 1)

1. Observe the following pictures in fig 1.1 and discuss the questions.

Question 1.
What might be the reasons behind buildings collapsing in photo 1?
Answer:
The major reason behind the buildings collapsing in photo 1 is the occurrence of earthquakes.

Question 2.
Which event in depicted in photo 2? What impact does it bring about in the surroundings?
Answer:
The volcanic eruption and flowing of lava is depicted in photo 2. Heavy loss of life and property is experienced when the lava flows to the surrounding area.

Question 3.
In photo 3, what could be the reason behind the bend in the rock strata?
Answer:
Due to compression of sediments within the earth’s interior and pressure from overlying layers the rock strata seems to be bended.

Question 4.
What could be the reason behind the difference in altitudes of the land and the steepness of the slope in photo 4?
Answer:
The reason could be displacement of rock due to vertical movement. It may form plateau on block mountains.

Question 5.
Classify the events in the photos into sudden and slow movements.
Answer:
Sudden movements are earthquake and volcanoes in photo 1 and 2. Slow movements are formation of folds and formation of plateaus or block mountains in photo 3 and 4.

Question 6.
Example of which of these movements is not likely to be found in the mainland of Indian sub- continent?
Answer:
The volcanic eruptions are not found in mainland of India sub-continent.

2. On 19th August, 2018, around 300 people died in Indonesia. Many buildings collapsed. Many roads broke apart. A tsunami was generated. (Textbook Page No. 6)

Question 1.
What was the cause behind these events?
Answer:
Collision of tectonic plates was the causes behind the events.

Question 2.
What actually happened during this natural event?
Answer:
A very high intensity earthquake struck north eastern parts of Indonesian coastline, where subsequent landslides into sea triggered tsunamis.

Question 3.
Name the energy waves involved in this natural event.
Answer:
Tsunamis were involved in this natural event where the tectonic plates moved horizontally.

Question 4.
Observe the diagram in fig. 1.9 and label the boxes.

Answer:

Think about it.

Question 1.
How will compressional forces affect brittle rocks? (Textbook Page No. 3)
Answer:
The compressional forces will break the brittle rocks into pieces.

Question 2.
A mountain never remains a mountain’. Can you relate this idiom with the mountain building process? (Textbook Page No. 4)
Answer:
This happens because of continuous erosional activities taking place on the mountain sides. Running water glaciers, wind, volcanic activities keep on changing the surface of a mountain.

Question 3.
When can faults form is fold mountains? (Textbook Page No. 6)
Answer:
If compression takes place along the faults, folds can be formed.

Question 4.
Can folds form into block mountains? Find the reason and discuss. (Textbook Page No. 6)
Answer:
No, folds cannot form into block mountains because folding takes place due to compression of elastic rocks and block mountains are formed due to displacement of rocks owing to tension.

Question 5.
Look at fig.1.10 and answer following questions. During an earthquake, do you think the seismic waves reach entire portion of the earth. Is there any region on the earth’s surface where a given earthquake will not be reported? (Textbook Page No. 6)
Answer:
It’s the shadow zone of P-waves and S-waves where the seismic waves do not reach. No there is no region on the earth’s surface where a given earthquake is not reported.

Question 6.
Why has the shadow zone for L-waves not been shown in fig 1.10? (Textbook Page No. 7)
Answer:
L-waves travel through the circumference of the earth. Thus, they do not been form shadow zone.

Try this.

Question 1.
Observe the diagrams in fig. 1.5. Try to understand the different types of fold shown in the diagram. Write the name of the fold. (Textbook Page No. 4)

Answer:

1. Symmetrical fold.
2. Isoclinal fold
3. Asymmetrical fold
4. Recumbent fold
5. Overturned fold

Question 2.
Observe the diagram in fig. 1.6 and read the explanation regarding the characteristics of various faults given. Identify the faults and match each of them with their characteristics. (Textbook Page No. 5)

Answer:
A – Normal fault B – Reverse fault C – Thrust fault D – Tear fault

Can you tell? (Textbook Page No. 7)

In fig 1.10 textbook page 6 A, B, C are three points on the earth’s surface. Analyse their location with respect to epicentre and shadow zones
Answer:
A, B, C are the points on the surface of the earth.
(i) ‘A’ point is located with 105° from the epicentre, therefore P waves and S waves reach at ‘A’ point. The seismograph at !A’ point records both waves, therefore ‘A’ point is not in shadow zone.

(ii) ‘B’ point is in shadow zone because both the waves do not reach there. P waves pass through all mediums, they experience refraction as they pass on one medium to another. The seismograph records P waves, which is located away from ‘B’ point. But P waves are not recorded in an area between 105 and 140° from epicentre. The S waves also do not reach here. Therefore, it is called ‘Shadow zone’ of Earthquake waves.

(iii) ‘C’ point is in the area of the shadow zone of ‘S’ waves, that in within 140°. P waves reach beyond 140°. Since ‘S’ waves do not reach here, it is called ‘Shadow zone of S waves’.

Question 1.
On the basis of the intensity of damage risk, India is classified into five risk zone. Use the given weblink http://www.bmtpc.org/DataFiles/CMS/file/map%20of%20india/eq-india. pdf and complete the table given. (Textbook Page No. 9)
Answer:

Read the following passage about Krakatoa volcanic eruption and answer the following questions. (Textbook Page No. 10)

There is an island known as Krakatoa between the islands of Java and Su matra in Indonesia. There were frequent volcanic eruptions here. From May 1883, massive explosions began. The eruption that took place at about 10 in the morning on 28 August 1883 was the largest ever recorded explosion. As a result of this explosion, the entire island disappeared. During this eruption, rock particles and dust thrown up in the atmosphere was about 25 km³. The column of this dust-ash rose as high as 80 km.

The discharge of Krakatoa threw into the air nearly 21 km3 of rock fragments, and large quantities of ash fell over an area of some 800,000 km². Near the volcano, masses of floating pumice, were so thick that ships had to halt. The surrounding region was plunged into darkness for two and a half days because of ash in the air. For some years after this, cloud kept moving round the earth. About 36,000 people died in these eruptions and the tsunami waves created by it.

In 1927, volcanic eruptions began at the same place and a new island rose in place of the Krakatoa island that had vanished. It was named Anak Krakatoa’ or ‘Child Krakatoa’. The volcano here constantly emits ash and steam. This new island has now become a laboratory for geologists and biologists.

Question 1.
Make a list of materials that came out during eruptions.
Answer:
Rock, particles, dust, ash, floating pumice, steam were the materials that came out during eruption.

Question 2.
Classify them into liquid, solid and gaseous forms.
Answer:

Find out! (Textbook Page No. 9)

Find out examples of active, dormant and extinct volcanoes.
Answer:
Examples of active volcanoes are – Mt. Etna (Italy), Stromboli (Italy), Mt. Merapi (Indonesia), Erta Ale (Ethiopia), Mt. Erebus (Antarctica)
Examples of dormant volcanoes – Mt. Kilimanjaro (Africa), Mt. Katmai (Alaska)
Examples of extinct volcanoes – Mt. Egmont (New Zealand), Chimborazo (Ecuador), Mt. Kulal (Kenya)

Give it a try (Textbook Page No. 8)

Take the given hypothetical data in the table. The data shows the time of arrival of P-waves and S-waves at 3 seismograph stations. Assume the scale of the map as 1 cm : 18 km. See fig. 1.11.

Answer:
Speed of waves = 08 kms per second
Scale of the map = 1 cm
= 18 kms
First step : Find out the difference in time for waves at Jalana recording centre.
Difference in time of waves = Time of S-wave – Time of P-wave
11 : 06 : 19 (Time of S-wave)
11 : 06 : 06 (Time of P-wave)
= (Time of S-wave) – (Time of P-wave)
= 11 : 06 : 19 – 11 : 06 : 06
= 00 : 00 : 13
Therefore difference in time for waves at Jalana recording centre is 13 seconds.

Second Step : Find out distance on land between Jalana and Epicentre
Distance on land = Difference in time of waves × speed of waves per second
= 13 × 8
= 104
Therefore distance on land between Jalana and Epicentre is 104 km.

Third Step : Find out radius for Jalana Station.
Jalana – Radius of circle = \(\frac {Distance on land}{Scale of the map}\)
= \(\frac {104}{18}\)
= 5.7 cm.
Therefore radius of circle around Jalana station is 5.7 cm.
Now find out radium for remaing stations.
Jalana – 5.7 cm, Washim – 7.5 cm, Aurangabad – 8 cm.

Fouth step : Now draw circles around Jalana, Washim and Aurangad with the help of radius calculated.
All arcs of circle will intersect one another at a particular point. That point is the location of Epicentre.
In the map given below Epicentre is to the south of Mudgal.

Balbharti Maharashtra State Board Class 5 Hindi Solutions Sulabhbharati Chapter 14 मैं सड़क हूँ Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Hindi Sulabhbharati Solutions Chapter 14 मैं सड़क हूँ

5th Standard Hindi Digest Chapter 14 मैं सड़क हूँ Textbook Questions and Answers

1. अंत्याक्षरी खेलो:
सड़क….किनारे….रास्ता….तकनीक….काम….मकान….निकिता…. ताँगा….गाजर…..रश्मि।

2. सड़क कैसे साफ-सुथरी रख सकते हैं, बताओ।

Hindi Sulabhbharati Class 5 Solutions Chapter 14 मैं सड़क हूँ Additional Important Questions and Answers

निम्नलिखित प्रश्नों के उत्तर एक वाक्य में दीजिए:

प्रश्न 1.
जिस सड़क पर चलते हुए पैर मिट्टी से सन जाएँ, वह सड़क कैसी होती है?
उत्तर:
कच्ची सड़क।

प्रश्न 2.
पगडंडी किसे कहते हैं?
उत्तर:
असमान, संकरी और कच्ची डगर को पगडंडी कहते हैं।

प्रश्न 3.
सड़क किन चीजों से बनती है?
उत्तर:
सड़क सीमेंट, रेत, गिट्टी, सलिया एवं कोलतार से बनती है।

प्रश्न 4.
कच्ची सड़क कहाँ पाई जाती है?
उत्तर:
गाँवों में।

प्रश्न 5.
हम सड़क को साफ-सुथरा कैसे रख सकते हैं? बताओ।
उत्तर:
सड़क पर कचरा ना फेंकें। यहाँ-वहाँ ना थूकें। जगह जगह गड्ढे न खोदें।

निम्नलिखित प्रश्नों के उत्तर लिखिए:

प्रश्न 1.
सड़क पर क्या-क्या चलता है?
उत्तर:
सड़क पर मानव, ताँगे, मोटर, बस, साइकिल इत्यादि चलते हैं।

प्रश्न 2.
लोग सड़क के किनारे क्या बनवा लेते हैं?
उत्तर:
लोग सड़क के किनारे मकान-दुकान बनवा लेते हैं।

प्रश्न 3.
चौराहे पर वर्दी पहनकर कौन खड़ा रहता है?
उत्तर:
चौराहे पर वर्दी पहनकर पुलिस का सिपाही खड़ा रहता है।

प्रश्न 4.
लोग अपनी ज़रूरत से सड़क को कैसा बना देते हैं?
उत्तर:
लोग अपनी ज़रूरत से सड़क को टेढ़ी, कभी नीची, कभी ऊँची बना देते हैं।

प्रश्न 5.
ज्यादातर सड़क किस चीज़ से बनाई जाती है?
उत्तर:
ज्यादातर सड़क कोलतार डालकर बनाई जाती है।

प्रश्न 6.
यह आत्मकथा किसकी है?
उत्तर:
यह आत्मकथा सड़क की है।

प्रश्न 7.
सामान बेचनेवाले अपनी दुकान कहाँ लगाते हैं?
उत्तर:
सामान बेचनेवाले सड़क के दोनों तरफ अपनी दुकान लगाते हैं।

प्रश्न 8.
चौराहे पर पुलिस का सिपाही क्या काम करता है?
उत्तर:
चौराहे पर पुलिस का सिपाही यातायात को अनुशासित रखने का काम करता है।

प्रश्न 9.
सड़क को दुख कब होता है?
उत्तर:
जब लोग उसे गंदा कर देते हैं, तब सड़क को दुख होता है।

प्रश्न 10.
सड़क का क्या फायदा है?
उत्तर:
सड़क के कारण देश में दूर-दूर रहने वाले लोग एक दूसरे से आसानी से मिल पाते हैं।

प्रश्न 11.
सड़क सदा कैसी रहना चाहती है?
उत्तर:
सड़क सदा साफ-सुथरी रहना चाहती है।

प्रश्न 12.
सड़क क्या चाहती है?
उत्तर:
सड़क हर गाँव को एक-दूसरे से जोड़ना और लोगों में मेल-जोल बढ़ाना चाहती है।

प्रश्न 13.
सड़क कहाँ खुशहाली लाना चाहती है?
उत्तर:
सड़क पूरे देश में खुशहाली लाना चाहती है।

निम्नलिखित प्रश्नों के उत्तर एक शब्द में लिखिए:

प्रश्न 1.
सड़क, घर और विद्यालय के बीच क्या है?
उत्तर:
सेतु।

प्रश्न 2.
अब कैसी तकनीक आ गई है?
उत्तर:
नई तकनीक।

प्रश्न 3.
नई तकनीक से सड़क किससे बनाई जाती है?
उत्तर:
सीमेंट।

प्रश्न 4.
गड्ढों के कारण क्या होता है?
उत्तर:
दुर्घटनाएँ।

प्रश्न 5.
सड़क कैसी रहना चाहती है?
उत्तर:
साफ-सुथरी।

प्रश्न 6.
सही विकल्प चुनकर रिक्त स्थानों की पूर्ति कीजिए:
(वर्दी, सड़क, गड्ढे, गंदा, निश्चित)

1. मैं ………………. हूँ।
2. सिपाही ………………. पहने खड़ा होता है।
3. जब लोग मुझे …………….. कर देते हैं।
4. मेरे शरीर पर बहुत से ……………. हो जाते हैं।
5. एक दिन मेरी इच्छा ……………… ही पूरी होगी।

उत्तर:

1. सड़क
2. वर्दी
3. गंदा
4. गड्ढे
5. निश्चित

प्रश्न 7.
रिक्त स्थानों की पूर्ति कीजिए:

1. …………………… , …………………….., …………………….. सभी के काम आती हूँ।
2. लोग मेरे किनारे ………………. बना लेते हैं।
3. लोग अपनी ज़रूरत के अनुसार कभी मुझे ……………… , कभी ………………. और कभी ……………… बना देते हैं।
4. मेरे शरीर पर बहुत ………………. हो जाते हैं।
5. इन ………………………. के कारण अनेक ……………………… होती हैं।
6. मैं सदा …………………… …………………… चाहती हूँ।
7. मैं जहाँ जाती हूँ, वह ………………. विकसित हो जाता है।
8. पूरे देश में ……………… लाऊँ।

उत्तर:

1. बालक, जवान, बूढ़े
2. मकान-दुकान
3. टेढ़ी, नीची, ऊँची
4. घाव
5. गड्ढ़ों, दुर्घटनाएँ
6. साफ -सुथरी
7. परिसर
8. खुशहाली

व्याकरण:

प्रश्न 1.
समान अर्थ वाले शब्द लिखिए:

1. जवान
2. घर
3. विद्यालय
4. सेतु
5. दोष

उत्तर:

1. युवक
2. गृह
3. स्कूल
4. पुल
5. गलती

प्रश्न 2.
विपरीत अर्थ वाले शब्द लिखिए:

1. इधर
2. आना
3. जवान
4. ऊँची
5. अच्छा
6. सुख
7. देश
8. विकसित
9. इच्छा
10. निश्चित

उत्तर:

1. उधर
2. जाना
3. बूढ़ा
4. नीची
5. बुरा
6. दुख
7. विदेश
8. अविकसित
9. अनिच्छा
10. अनिश्चित

प्रश्न 3.
लिंग बदलिए:

1. आदमी
2. बालक
3. बूढ़ा

उत्तर:

1. औरत
2. बालिका
3. बूढ़ी

प्रश्न 4.
वचन बदलिए:

1. बूढ़ा
2. बैलगाड़ी
3. साइकिल
4. ताँगा
5. गड्ढा

उत्तर:

1. बूढ़े
2. बैलगाड़ियाँ
3. साइकिलें
4. ताँगे
5. गड्ढे

प्रश्न 5.
नदी की आत्मकथा लिखिए:
उत्तर:
मैं नदी हूँ। पहाड़ों से उतर कर नीचे ज़मीन पर बहती हूँ। गाँवों, खेतों के पास से होकर जाती हूँ। बच्चे मेरे पानी में डुबकी लगाते हैं। मैं सभी के काम आती हूँ। कुछ लोग मुझे गंदा भी करते हैं। इससे मैं दुखी हो जाती हूँ। फिर भी सब कुछ सहती हुई आगे बढ़ती – रहती हूँ। मैंने रुकना नहीं सीखा। मेरा काम है हमेशा बहते रहना। आज के युग में यदि मेरा ध्यान न रखा गया तो मैं एक दिन समाप्त हो जाऊँगी। मुझे हमेशा बहते देखना चाहते हो तो मेरा सही उपयोग करो, क्योंकि मैं हमेशा सभी को आगे बढ़ने की प्रेरणा देती हूँ।

प्रश्न 6.
उद्यान की आत्मकथा लिखिए:
उत्तर:
मैं उद्यान हूँ। जब मुझे नया-नया बनाया गया था, तब बहुत-से पेड़ लगाए गए थे। बैठने के लिए बेंचें थीं। बच्चों के लिए झूले थे। लोगों का आना – जाना लगा रहता था। बच्चों की आवाजें आती थीं। परंतु आज कल अधिकतर कोई नहीं आता, क्योंकि मेरा सही तरीके से ध्यान नहीं रखा गया। बेंचें टूट गईं, झूले टूट गए, पेडों के सभी पत्ते झड़ गए; इसलिए मैं उदास हूँ। कोई मेरी तरफ ध्यान देगा तो मैं फिर से पहले की तरह बन जाऊँगा। मैं सभी को खुशी देना चाहता हूँ। सबके साथ मिल-जुलकर रहना चाहता हूँ।

मैं सड़क हूँ Summary in Hindi

पाठ का सारांश:

प्रस्तुत पाठ एक सड़क की आत्मकथा है। सड़क खुद अपने बारे में बता रही है कि किस तरह लोग उसका उपयोग करते हैं।

शब्दार्थ:

1. महामार्ग – हाइवे (highway)
2. सेतु – पुल (bridge)
3. द्रुतगति मार्ग – चौड़ा रास्ता (express way)
4. किनारे – आसपास (surround)
5. चौराहा – चार रास्ता (crossroads)
6. वर्दी – गणवेश (uniform)
7. यातायात – परिवहन (transportation)
8. अनुशासन – शिष्टता (discipline)
9. कोलतार – डामर (tar)
10. ताँगा – घोड़ा गाड़ी (tonga)
11. पगडंडी – चलने के लिए बना कच्चा रास्ता (crooked)

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 4 Kingdom Animalia Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Biology Solutions Chapter 4 Kingdom Animalia

1. Choose correct option

Question (A)
Which of the following belongs to a minor phylum?
(a) Comb jelly
(b) Jelly fish
(c) Herdmania
(d) Salpa
Answer:
(a) Comb jelly

Question (B)
Select the animal having venous heart.
(a) Crocodile
(b) Salamander
(c) Rohu
(d) Toad
Answer:
(c) Rohu

Question (C)
In Ascaris, _______ .
(a) mesoglea is present
(b) endoderm is a discontinuous layer
(c) mesoderm is present in patches
(d) body cavity is absent
Answer:
(c) mesoderm is present in patches

Question (D)
Which of the following is INCORRECT in case of birds?
(a) Presence of teeth
(b) Presence of scales
(c) Nucleated RBCs
(d) Hollow bones
Answer:
(a) Presence of teeth

Question (E)
Chitinous exoskeleton is a characteristic of ________ .
(a) Dentalium
(b) Antedon
(c) Millipede
(d) Sea urchin
Answer:
(c) Millipede

2. Answer the following questions.

Question (A)
Reptiles are known for having three chambered heart. Which animal shows a near four chambered condition in reptiles?
Answer:
Crocodiles have a four chambered heart.

Question (B)
The circulatory system has evolved from open to closed type in Animal Kingdom. Which Phylum can be called first to represent closed circulation?
Answer:
Phylum Annelida is the first phylum to represent closed circulation.

Question (C)
Pinna is part of external ear and it is found in mammals. Do Aves and Reptiles show external ear in any form?
Answer:
No, Aves and Reptiles do not show external ear in any form. They possess tympanum which represents the ear.

Question (D)
Fish and frog can respire in water. Can they respire through their skin? If yes, why do they have gills?
Answer:
1. Yes, fishes and frogs can respire through their skin.
2. The larval stage of frog i. e. tadpole respires through gills. During metamorphosis, tadpoles lose their gills and develop lungs.
3. Frogs do not have scales and breathe through their skin underwater.
4. Fishes respire primarily via gills. The body of fishes is covered with scales which limits cutaneous respiration in them.

Question (E)
Birds need to keep their body light to help in flying. Hence, they show presence of some organs only on one side. How their skeleton helps in reducing their weight?
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 (F)
Cnidarians and Ctenophorans are both diploblastic. Which other character do they have in common, which is not found in other phyla?
Answer:
Cnidarians and ctenophorans show tissue level of body organization. They have blind sac body plan and radially symmetrical body.

Question (G)
Crab and Snail both have a protective covering. Is it made up of the same material?
Answer:
No, the protective covering is not made up of same material in crab and snail. The protective covering of crabs is made up of chitin and that of snails is made up of calcium carbonate.

Question (H)
Sponge and sea star show calcareous protective material. Do they belong to the same Phylum?
Answer:
No, they do not belong to same phylum. Sponges belong to phylum Porifera and sea star belongs to phylum Echinodermata.
1. Adult echinoderms are radially symmetrical but larval forms are bilaterally symmetrical.
2. Larvae of echinoderms are free-swimming.

Question (I)
Fish and snake both have scales. How do these scales differ from each other?
Answer:
Fishes have dermal scales covering the body surface whereas snakes have epidermal scales or scutes.

Question (J)
Lower Phyla like Arthropods and Cnidarians show metamorphosis. Is it also found in any class of Phylum Chordata?
Answer:
Yes, it is also found in class Amphibia of phylum Chordata.

Question 3.
Draw neat labelled diagram.
A. Sycon
B. Aurelia
C. Amphioxus
D. Catla
E. Balanoglossus
F. Scolidon
Answer:
A. Sycon

B. Aurelia

C. Amphioxus

D. Catla

E. Balanoglossus

F. Scolidon

Question 4.
Match the following.

Answer:

5. Identify the animals given in pictures and write features of its phylum/class.

Question 1.

Answer:
The organism in the given picture is Comb jelly (Red midwater Comb jelly) and it belongs to phylum Ctenophora.

Question 2.

Answer:
The organism in the given picture is Eel and it belongs to phylum Chordata.

Question 3.

Answer:
The given organism in the given picture is Dolphin and it belongs to class Mammalia.

Question 4.

Answer:
The given organism is Snake and it belongs to class Reptilia

Question 5.

Answer:
The given organism is Sea urchin and belongs to phylum Echinodermata.

Question 6.

Answer:
The given organism is flying lizard and belongs to class Reptilia.

Question 7.

Answer:
The organism is Herdmania and belongs to Phylum Chordata (Subphylum Urochordata).

Question 8.

Answer:
The organism in the given picture is Nautilus and it belongs to phylum Mollusca.

Question 9.

Answer:
The organism in the given picture is Amphioxus and it belongs to Phylum Chordata (Subphylum Cephalochordata).

6. Observe and identify body symmetry of given animals.

Question 1.
Observe and identify body symmetry of given animals.

Answer:
Fig i. represents asymmetry
Fig ii. represents radial symmetry
Fig iii. represents bilateral symmetry

Practical/Project:

Question 1.
Study different animals in kingdom Animalia and prepare the chart with detail scientific information.
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).

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).

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

11th Biology Digest Chapter 4 Kingdom Animalia Intext Questions and Answers

Can you recall? (Textbook Page No. 29)

(i) What is the basis for classification?
Answer:
Grades of organization, body symmetry, body cavity, germ layers and segmentation form the basis for classification.

(ii) Who proposed Five kingdom classification system?
Answer:
Robert Whittaker proposed the five kingdom system of classification.

(iii) What is the need and importance of classification?
Answer:
Need and importance for classification:
a. Classification facilitates the identification of animals with great accuracy.
b. The study of animals becomes convenient.
c. It helps in understanding the relationship of animals with other living organisms.
d. It helps to understand the habitat of each animal along with its role in nature.
e. By studying few animals from a group, we can gain a better understanding about the entire group.
f. It helps in understanding different adaptations shown by animals.
g. It gives an idea about evolution of animals.

Observe and discuss. (Textbook Page No. 29)

Discuss the criteria of classification.
Answer:
1. The given diagrams represents the number of germ layers and body symmetry used as criteria for animal classification.
2. 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.
3. Body symmetry:
Body symmetry implies to the similarity in shape, size and number of parts on the opposite sides of a median line when body is divided into two halves by an imaginary line along different plane. Animals may be asymmetrical, radially symmetrical or bilaterally symmetrical.
(a) Asymmetrical animals:
An animal is said to be asymmetrical when its body cannot be divided into two identical halves in any plane.
(b) Radially symmetrical animals:
In certain animals, body can be cut or divided into two similar halves in a number of planes wherein, all the cuts (planes) pass through the centre. This type of symmetry is called radial symmetry.
(c) Bilaterally symmetrical animals:
In this type, the body of the animal can be bisected or divided in two equal or identical halves by a single median or vertical plane.

Internet my friend. (Textbook Page No. 30)

Which are the larval stages of Porifera.
Answer:
Larval stages of Porifera:
Parenchymula – Flagellate larvae of calcinean sponges
Amphiblastula – Free swimming larval stage of Sycon and many other calcareous sponges. Rhagon— Larval stage which give rise to the leuconoid condition in demospongiae.
[Students are expected to find more information about the larval stages of Porifera on internet.]

Find out. (Textbook Page No. 31)

Information about coral reefs and sea fan.
Answer:
Coral reefs:

1. A coral reef is an underwater ecosystem characterized by reef building corals.
2. Coral reefs constitute 25% of all marine species on the planet.
3. They belong to phylum Cnidaria.
4. There are three main types of coral reefs – fringing, barrier and atoll. Coral reefs provide ecosystem services for tourism, fisheries and shoreline protection.
5. They cannot survive in high temperatures, thus due to climate change there is a sharp decline in their population.

Sea fan or Gorgonia:

1. It is a soft coral composed of numerous polyps – cylindrical, sessile (attached) forms that grow together in a flat, fan-like pattern.
2. It belongs to phylum Cnidaria.
3. It does not produce calcium carbonate skeletons.
[Students can find out more information about coral reefs and sea fan using internet ]

Can you tell? (Textbook Page No. 32)

(i) State the parasitic adaptations in Liver fluke and Ascaris.
Answer:
Parasitic adaptations in Liver fluke:
a. Presence of hooks and suckers
b. Body covered with cuticle
c. Lacks digestive system
d. They are hermaphrodites

Parasitic adaptations in Ascaris:
a. Presence of muscular pharynx for sucking the food.
b. Body covered by tough, thick and resistant cuticle.
c. Secretes enzymes against the enzymes secreted by the host.
d. Respiration is anaerobic.
e. Reproductive system is highly developed.

(ii) Give example of free living platyhelminth.
Answer:
Planaria

Find out. (Textbook Page No. 33)

What are the merits and demerits of hermaphroditism?
Answer:
Hermaphroditism is the condition in which an organism possesses reproductive organs of both the sexes.

Merits of hermaphroditism:
a. Assured fertilization which reduces the risk of a species to become extinct due to unavailability of mating partner.
b. Energy required for searching out mating partner is conserved.
c. Frequency of mating is maximized.

Demerits of hermaphroditism:
a. More energy is required to maintain both the reproductive systems.
b. Limited gene diversity.
[Source: http://floydbiology. blogspot. com/2012/06/httpmattc-thinks. html]
[Students are expected to find more information using the internet.]

Why are leeches used in Ayurveda?
Answer:
a. Leeches are used in blood purification therapy to treat many diseases as they suck impure blood from the affected site of the patient’s body.
b. The anticoagulant – hirudin present in saliva of leech, inhibits the coagulation of blood and makes blood thinner. This dissolves the clots found in vessels and facilitates the blood supply.

What is the role of earthworms in agriculture? What is vermicompost?
Answer:
Role of earthworms in agriculture:
a. Earthworms loosen the soil by burrowing deep into it, thus they help to aerate the soil.
b. This continuous digging of soil also helps the water to reach the roots quickly.
c. Earthworms can decompose the organic matter from the soil and convert it into rich manure.
d. This helps in increasing the fertility of soil which ultimately increases the crop production.
e. Earthworm castings are rich in nutrients which act as natural fertilizer.
Vermicompost:
Vermicompost is the product of vermicomposting. It is organic manure produced as vermicast by earthworm feeding on biological waste material and plant residues.

Can you tell? (Textbook Page No. 34)

(i) Explain the term metameric segmentation.
Answer:
In some animals, body consists of many segments arranged along the length of the body. When the external segmentation coincides with the internal segmentation, it is called as metameric segmentation and the phenomenon is called metamerism.

(ii) Give characteristics of 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.

(iii) Enlist the harmful Arthropods.
Answer:
Significance:
Beneficial arthropods: Some arthropods are of economic importance. For example, Honey bees (Apis) are important for their honey and wax, silkworms 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.

Find out. (Textbook Page No. 34)

(i) Why is phylum Arthropoda considered as most successful phylum?
Answer:
Phylum Arthropoda is considered as most successful phylum because of the following reasons:
a. Phylum Arthropoda is the largest phylum of kingdom Animalia. It includes various forms like lobsters, prawns, crabs, insects, millipedes, locust, honeybees, etc.
b. They are omnipresent (present everywhere). Arthropods show great variety of adaptations as their habitat varies from terrestrial to aquatic habitat.
c. Several others factors also contribute to the success of the phylum which includes:
1. The exoskeleton of arthropods is made up of tough chitinous exoskeleton. This enables them to survive on lands in almost all environment and is a great defense against predators.
2. They possess jointed appendages which allow complex movements.
3. They exhibit moulting or eedysis.
4. They have metamerically segmented body helping in movement around diverse environments.

(ii) What do we mean by parthenogenesis?
Answer:
Development of an egg into a complete individual without fertilization is known parthenogenesis. It is found in many non-vertebrates such as bees, rotifers and even some lizards and birds (turkey).

(iii) What do we mean by living fossil?
Answer:
A member of a living animal or plant species that is almost identical to species known from the fossil record (not the recent fossil record), i.e. they have changed very little over a long period.
[Source:https://www. encyclopedia, com/earth-and-environment/ecology-and- environmentalism/environmental-studies/living-fossil]

(iv) How the bees produce honey?
Answer:
a. Bees produce honey using the nectar of flowering plants. A bee sucks the nectar and stores it in a honey sac until it returns to the hive.
b. The nectar is then transferred to worker bees in the hive who suck the nectar from the honey sac through their proboscis. This nectar contains 70% water and 20% honey. Honeybees get rid of excess water by swallowing and regurgitating the nectar again and again. They also fan their wings over filled cells of honeycomb.

When most of the water has evaporated from the honeycomb, the bee seals the comb with a secretion of liquid from its abdomen which eventually hardens into beeswax. This is how the honey bees use nectar to produce a thick, sticky and sweet honey.

(v) What will happen if arthropods do not moult?
Answer:
a. Moulting or eedysis is a periodic shedding of the outer cuticle layer of body in arthropods.
b. The outer layer of body of arthropods is formed of tough, non-living chitinous substance.
c. If arthropods do not moult, they cannot grow and mature into adult forms

Can you tell? (Textbook Page No. 34)

Why do Molluscs have shell?
Answer:
Molluscs are soft-bodied animals. Thus, the calcareous shell provides supports and protects the organisms from predators.

Can you tell? (Textbook Page No. 36)

Give 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. The madrepOrite is the opening of water vascular system through which water enters. Water vascular system is useful in locomotion, food capturing, respiration.
7. Digestion: Digestive system is complete.
8. Respiration: Peristomial gills, papillae, respiratory tree, etc. are used for respiration.
9. Circulatory and excretory systems: Absent in echinoderms.
10. Nervous system: Nervous system is simple with a nerve ring around the mouth and radial nerves in arms.
11. Reproduction and development: Sexes are separate (sometimes bisexual). Fertilization is external.
12. 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).

Can you tell? (Textbook Page No. 36)

Can you tell? (Textbook Page No. 36)
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. Hence, Hemiehordata is considered as a connecting link between Non-chordata and Chordata.

Find out. (Textbook Page No. 36)

Why Balanoglossus is considered as connecting link between Non-chordates and chordates?
Answer:
Balanoglossus belongs to phylum Hemiehordata. For Explanation:

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. Hence, Hemiehordata is considered as a connecting link between Non-chordata and Chordata.

Observe and discuss. (Textbook Page No. 36)

Compare and contrast between Non-Chordates and Chordates.

Answer:

Can you tell? (Textbook Page No. 37)

Herdmania is called a Chordate. Explain.
Answer:
1. Herdmania belongs to phylum Urochordata.
2. It is called a chordate as it shows the following features:
a. Presence of notochord at least in early embryonic life. (In Herdmania, notochord is present in the tail of the larval forms).
b. Presence of hollow, dorsal nerve chord, running throughout the length of the body.
c. Presence of pharyngeal gill slits.
d. Presence of post-anal tail.

Can you tell? (Textbook Page No. 37)

Give characteristics of Petromyzon. Comment on its mode of nutrition.
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).

Can you tell? (Textbook Page No. 38)

(i) What is the lateral line system?
Answer:
a. Lateral line system is the system with mechanoreceptors called neuromasts, for the detection of watei current.
b. These neuromasts are arranged in an interconnected network along the head and body.
c. Lateral line system also known as lateralis system.

(ii) Why Piscian heart is called a venous heart?
Answer:
a. Pisces have two-chambered heart. They have single and closed circulation.
b. Heart of Pisces receives blood only from veins and thus always shows presence of deoxygenated blood which it pumps directly to the gills for oxygenation.
Thus, the heart of Pisces is called a venous heart.

Can you tell? (Textbook Page No. 40)

Amphibians do not have exoskeleton. Give reason.
Answer:
1. Amphibians live in both water and on land.
2. They perform cutaneous respiration (i. e. gaseous exchange across the skin or outer integument.) under water and when on land, they respire through lungs.
Thus, to facilitate cutaneous respiration, amphibians do not have exoskeleton.

Can you tell? (Textbook Page No. 40)

Why are amphibians and reptilians called poikilotherms?
Answer:
Amphibians and reptilians are called poikilotherms as they cannot maintain a constant body temperature. Their body temperature changes according to the change in surrounding temperature.

Can you tell? (Textbook Page No. 41)

Give adaptations in Aves for flying.
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.

Can you tell? (Textbook Page No. 41)

(i) Aves and mammals are homeotherms. Give reason.
Answer:
a. Aves and mammals can generate heat to maintain their body temperature.
b. They keep their body temperature constant, irrespective of fluctuations in environmental temperature. Thus, Aves and mammals are homeotherms.

(ii) How mammals differ from other groups of animals?
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.

Do yourself. (Textbook Page No. 41)

Observe different animals in your surrounding, write detailed classification and write down the characteristics of animals in following format.
Answer: