Bhagya

Students can download 10th Social Science Geography Chapter 4 Resources and Industries Questions and Answers, Notes, Samacheer Kalvi 10th Social Science Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Social Science Solutions Geography Chapter 4 Resources and Industries

Samacheer Kalvi 10th Social Science Resources and Industries Text Book Back Questions and Answers

I. Choose the correct answer

Question 1.
Manganese is used in:
(a) Storage batteries
(b) Steel Making
(c) Copper smelting
(d) Petroleum Refining
Answer:
(b) Steel Making

Question 2.
The Anthracite coal has
(a) 80 to 95% Carbon
(b) Above 70% Carbon
(c) 60 to 7% Carbon
(d) Below 50% Carbon
Answer:
(a) 80 to 95% Carbon

Question 3.
The most important constituents of petroleum are hydrogen and:
(a) Oxygen
(b) Water
(c) Carbon
(d) Nitrogen
Answer:
(c) Carbon

Question 4.
The city which is called the Manchester of South India is
(a) Chennai
(b) Salem
(c) Madurai
(d) Coimbatore
Answer:
(d) Coimbatore

Question 5.
The first Jute mill of India was established at:
(a) Kolkata
(b) Mumbai
(c) Ahmedabad
(d) Baroda
Answer:
(a) Kolkata

Question 6.
The first Nuclear Power station was commissioned in
(a) Gujarat
(b) Rajasthan
(c) Maharashtra
(d) Tamil Nadu
Answer:
(c) Maharashtra

Question 7.
The most abundant source of energy is:
(a) Biomass
(b) Sun
(c) Coal
(d) Oil
Answer:
(b) Sun

Question 8.
The famous Sindri Fertilizer Plant is located in
(a) Jharkhand
(b) Bihar
(c) Rajasthan
(d) Assam
Answer:
(a) Jharkhand

Question 9.
The nucleus for the development of the chotanagpur plateau region is:
(a) Transport
(b) Mineral Deposits
(c) Large demand
(d) Power Availability
Answer:
(b) Mineral Deposits

Question 10.
One of the shore-based steel plants of India is located at
(a) Kolkata
(b) Tuticorin
(c) Goa
(d) Visakhapatnam
Answer:
(d) Visakhapatnam

II. Match the following

Question 1.
Match the Column I with Column II.

Answer:
A. (ii)
B. (i)
C. (iv)
D. (v)
E. (iii)

Question 2.
Match the Column I with Column II.

Answer:
A. (v)
B. (iv)
C. (i)
D. (ii)
E. (iii)

III. Answer the following questions briefly

Question 1.
Define the resource and state its types.
Answer:
Any matter or energy derived from the environment that is used by living things including humans is called a natural resource.
Types of Natural Resources are:

1. Renewable and
2. Non – renewable resources.

Question 2.
Name the states that lead in the production of Iron ore in India.
Answer:
Jharkhand is the leading producer of Iron ore (25%), other states are Odisha (21%), Chattisgarh (18%), Karnataka (20%), Andhra Pradesh (5%) and also the state of Tamil Nadu.

Question 3.
What are the minerals and its types?
Answer:
A mineral is a natural substance of organic or inorganic origin with definite chemical and physical properties.
Types of Minerals are:

1. Metallic minerals
2. Non – metallic minerals.

Question 4.
State the uses of Manganese.
Answer:

1. Manganese is a silvery grey element always available with iron, laterite and other minerals.
2. It is very hard and brittle. So it is used for making iron and steel to give strength.
3. It also serves as raw materials for alloying.
4. It is also used in the manufacture of bleaching powder, insecticides, paints and batteries.

Question 5.
What is natural gas?
Answer:
It usually accompanies petroleum accumulations.
[OR]
Natural gas is an important clean energy resources found in association with or without petroleum. It is used as a source of energy as well as an industrial raw material in the petrochemical industry. It is considered an eco-friendly fuel because of low carbondioxide emissions.

Question 6.
Name the different types of coal with their carbon content.
Answer:
Based on carbon content coal is classified into

1. Anthracite – contains 80% to 90% carbon.
2. Bituminous – contains 60% to 80% carbon.
3. Lignite – contains 40% to 60% carbon
4. Peat – less than 40% carbon

Question 7.
Mention the major areas of jute production in India.
Answer:
West Bengal, Andhra Pradesh, Bihar, Uttar Pradesh, Assam, Chattisgarh, Odisha are the major jute producing area.

Question 8.
Name the important oil producing regions of India.
Answer:
West coast: Mumbai high, Gujarat coast, Basseim, Aliabet (South of Bhavanagar), Ankleshwar, Cambay – Luni region, Ahmedabad -Kaloi region and Punjab – Haryana.

East Coast: Brahmaputra valley, Digboi, Nahoratitya, Moran-Hugrijan, Rudrasagar-Lawa (Assam region), Surrma Valley. Andaman and Nicobar, Gulf of Mannar, Punjab. Haryana, Baleshwar coast.

IV. Distinguish between

Question 1.
Renewable and non-renewable resources
Answer:
Renewable resources:

1. These resources can be replenished after utilization.
2. Abundantly available in nature.
3. Eg: Sun energy, Wind etc.

Non-Renewable resources:

1. These resources cannot be regained after utilization.
2. Limited stock take millions of years for formation.
3. Eg: Minerals

Question 2.
Metallic and non-metallic minerals
Answer:
Metallic minerals:

1. These minerals contain one or more metallic elements.
2. They can be further classified into ferrous and Non-ferrous minerals.
3. Eg: Iron-ore, Manganese

Non-Metallic minerals:

1. These minerals do not contain metal in them.
2. They can be classified into energy fuels and construction minerals.
3. Eg: Coal, Petroleum, lime stone, Gypsum.

Question 3.
Agra based industry and mineral based industry.
Answer:
Agro based industry:

1. These industries obtain raw materials from agricultural products.
2. Eg: Cotton textile industry, Jute mills, Silk industry, Sugar industry etc.
3. Labour intensive

Mineral based industry:

1. These industries obtain raw materials from metallic and non- metallic minerals.
2. Eg: Coal industry, Iron and Steel industry etc.
3. Capital intensive

Question 4.
Jute industry and sugar industry.
Answer:
Jute industry:

1. Second largest textile industry.
2. Jute fibre is the basic raw material.
3. Gunny bags, Canvas, pack sheets,Cordage are some of the Jute products.
4. Concentrated in and around West Bengal.

Sugar industry:

1. Second largest agro based industry.
2. Sugarcane is the basic raw material.
3. Sugar Jaggery, Khandsari are some of the products of sugar industry.
4. Uttar Pradesh have more than 50% of the sugar industries.

Question 5.
Conventional energy and non- conventional energy
Answer:
Conventional energy:

1. Non renewable energy.
2. Mainly energy is produced by burning fossil fuels.
3. Production of energy till the stock of mineral availability.
4. Thermal power – Coal, petroleum and Natural gas and Nuclear minerals.

Non-Conventional energy:

1. Renewable energy
2. Mainly energy is produced by harnessing power from nature.
3. Continuous flow of energy production is possible.
4. Solar energy, Wind energy. Bio mass energy, tidal and wave energy.

V. Answer the following in a paragraph

Question 1.
Write about the distribution of cotton textile industries in India.
Answer:
The cotton textile industries contribute about 7% of industrial output, 2% of India’s GDP and 15% of the country’s export earnings. It is one of the largest sources of employment generation in the country. At present there are 1,719 textile mills in the country. Out of which 188 mills are in public sector, 147 in cooperative sector and 1,284 in private sector. Currently, India is the third largest producer of cotton and has the largest loom arc and ring spindles in the world. At present, cotton textile industry is the largest organized modem industry of India. About 16% of the industrial capital, 14% of industrial production and over 20% of the industrial labour of the country are engaged in this industry.

The higher concentration of textile mills in and around Mumbai, makes it as “Manchester of India”. Presence of black cotton soil in Maharashtra, humid climate, presence of Mumbai port, availability of hydro power, good market and well developed transport facility favour the cotton textile industries in Mumbai. The major cotton textile industries are concentrated in the states of Maharashtra, Gujarat, West Bengal, Uttar Pradesh and Tamil nadu. Coimbatore is the most important centre in Tamil nadu with 200 mills out of its 435 and called as “Manchester ‘ of South India”. Erode, Tirupur, Karur, Chennai, Thirunelveli, Madurai, Thoothukudi, Salem and Virudhunagar are the other major cotton textiles centres in the state.

Question 2.
Explain the factors responsible for the concentration of jute industries in the Hoogly region.
Answer:
The following factors are responsible for the concentration of Jute industries in the Hoogly region in West Bengal.

1. Raw materials: West Bengal is the largest producer of Jute. Availability of raw Jute for production.
2. Processing: Jute require fresh water for processing. Abundant water supply is available by the riverines and continuous supply of fresh water is ensured due to Perennial nature.
3. Transport: Well connected by the network of water ways, road ways and railways.
4. Cheap labour: West Bengal is one of the densely populated area . So cheap labour is available.
5. Market: Kolkatta being one of the textile centre great demand for the product as well as port facilities available for export.

Question 3.
Write an account on the major iron and steel industries of India.
Answer:
Iron and Steel industry is the basic industry. The raw materials are obtained both from Metallic and Non-Metallic minerals.

Iron and Steel industries are located in close proximity to the coal fields or Iron ore mines.

VI. On the outline map of India mark the following.

Question 1.
iron ore production centres.
Answer:

Question 2.
Centres of Petroleum and Natural Gas production.
Answer:

Question 3.
Coal mining centres.
Answer:

Question 4.
Areas of cultivation of cotton.
Answer:

Question 5.
Iron and Steel industries.
Answer:

Samacheer Kalvi 10th Social Science Resources and Industries Additional Important Questions and Answers

I. Choose the correct answer

Question 1.
The ………………. is called as mineral oil.
(a) Petroleum
(b) Coal
(c) Natural gas
(d) Mica
Answer:
(a) Petroleum

Question 2.
Manchester of India is ……..
(a) Delhi
(b) Mumbai
(c) Chennai
Answer:
(b) Mumbai

Question 3.
The ………………. is the largest oil field in India producing 65% of oil.
(a) Ankaleshwar
(b) Mumbai high
(c) Kalol
(d) Surma valley
Answer:
(b) Mumbai high

Question 4.
Chotta Nagpur plateau is noted for ……..
(a) Natural vegetation
(b) Mineral resource
(c) Cotton cultivation
Answer:
(b) Mineral resource

Question 5.
TamilNadu produces about ………………. % of the total thermal electricity produced in India.
(a) 5
(b) 20
(c) 18
(d) 90
Answer:
(a) 5

Question 6.
In India most of the Iron and Steel industries are located in the …… plateau.
(a) Chott Nagpur
(b) Deccan
(c) Malwa
Answer:
(a) Chott Nagpur

Question 7.
Areas near ………………. district has the largest concentrations of wind farm capacity at a single location in the world.
(a) Ramanathapuram
(b) Tuticorin
(c) Thiruvallur
(d) Kanyakumari
Answer:
(d) Kanyakumari

Question 8.
The resources that can be reproduced again and again is called ……
(a) Mineral resources
(b) Renewable resource
(c) Natural resource
Answer:
(b) Renewable resource

Question 9.
Bysinosis is an occupational lung disease caused by exposure to:
(a) natural gas
(b) cotton dust
(c) coal power
(d) automobile
Answer:
(b) cotton dust

Question 10.
Lignite is extracted in Tamil Nadu …….
(a) Kadaloor
(b) Neyveli
(c) Madurai
Answer:
(b) Neyveli

Question 11.
National News Print and Papermills (NEPA) is in ………………. state.
(a) Odisha
(b) West Bengal
(c) TamilNadu
(d) Madhya Pradesh
Answer:
(d) Madhya Pradesh

Question 12.
Sugar bowl of India is …….
(a) West Bengal
(b) Uttar pradesh and Bihar
(c) Mumbai
Answer:
(b) Uttar pradesh and Bihar

Question 13.
The ………………. is the largest producer of electronic goods in India.
(a) Bengaluru
(b) Mysuru
(c) Delhi
(d) Jaipur
Answer:
(a) Bengaluru

Question 14.
The leading producer of electronic goods is ……..
(a) Bangalore
(b) Coimbatore
(c) Hyderabad
Answer:
(a) Bangalore

Question 15.
………………. is an aluminium ore.
(a) Manganese
(b) Magnesium
(c) Bauxite
(d) Anthracite
Answer:
(c) Bauxite

II. Match the following

Question 1.
Match the Column I with Column II.

Answer:
A. (iv)
B. (iii)
C. (i)
D. (v)
E. (ii)

Question 2.
Match the Column I with Column II.

Answer:
A. (iii)
B. (v)
C. (ii)
D. (i)
E. (iv)

Question 3.
Match the Column I with Column II.

Answer:
A. (iii)
B. (iv)
C. (v)
D. (ii)
E. (i)

III. Answer the following questions briefly

Question 1.
What are the features of copper and aluminium?
Answer:

• Malleability
• Expandability
• Good conductor of heat and energy (electricity)

Question 2.
Where was the first hydro-electric power station in India established?
Answer:
The first hydro-electric power station in India was established at “Darjeeling” in 1897.

Question 3.
In which type of rocks is limestone found?
Answer:
Limestone is found in sedimentary rocks.

Question 4.
What is the origin of the word petroleum?
Answer:
The word petroleum is derived from two Latin words petro (Rock) and Oleum (oil) thus petroleum is oil obtained from rocks of the earth. It is also called as mineral oil.

Question 5.
Name any five software centres.
Answer:

1. Chennai
2. Bangalore
3. Mysore
4. Hyderabad
5. Coimbatore

Question 6.
When was the Nuclear power programme initiated in India and where was the first Atomic Power station was set up?
Answer:

1. Nuclear Power programme was initiated in the 1940s, when “ Tata Atomic Research Commission was incorporated in August 1948.
2. The 1st nuclear power station was set up at Tarapur near Mumbai in 1969.

Question 7.
What are the raw materials for the paper industry?
Answer:
Woodpulp, bamboo, salai, Sabai grasses, waste paper and bagasse

Question 8.
Write a note on NPCIL.
Answer:

1. The Nuclear Power Corporation of India Limited ( NPCIL) is wholly owned by the Government of India.
2. It is a Public sector undertaking.
3. Responsible for the generation of nuclear power for electricity.
4. Administered by the Department of Atomic Energy (DAE). It is responsible for designing, constructing and operating the Nuclear power stations in India.

Question 9.
What are the major centres of the automobile industry?
Answer:
Mumbai, Chennai, Kolkata, New Delhi, Pune, Ahmedabad, Lucknow, Satara and Mysore.

Question 10.
Where are the centres of IT parks located in India?
Answer:
Centres of IT parks in India are located in Chennai, Coimbatore, Thiruvananthapuram, Bengaluru, Mysuru, Hyderabad, Vishakapatnam, Mumbai, Pune, Indore, Gandhinagar, Jaipur, Noida, Mohali and Srinagar.

IV. Distinguish between

Question 1.
Paper industry and Iron and Steel industry.
Answer:
Paper industry

1. Forest-based industry.
2. Serve as an index for education and literacy.
3. Wood pulp, Bamboo, bagasse are some of the basic raw materials.

Iron and Steel industry:

1. Mineral-based industry.
2. Key industry for industrial development.
3. Iron ore and Manganese are the main raw materials.

Question 2.
Solar and Wind energy.
Answer:
Solar energy:

1. Conversion of sunlight into electricity.
2. Photo voltoic cells or use of lenses, mirror and tracking system is used.
3. Installation cost is more and different applications is required as per the need.
4. Occupy more space.

Wind energy:

1. Conversion of energy is from the flow of wind.
2. Wind turbines are used.
3. Installation cost is comparatively less.
4. Occupy less space.

Question 3.
Difference between Manganese and Mica.
Answer:
Manganese:

1. It is a Metallic mineral.
2. Very hard and brittle in nature.
3. Raw material for iron and steel basic raw material for alloying.

Mica:

1. It is a Non-Metallic mineral.
2. Translucent, splitable into thin sheets, elastic and incompressible.
3. Exclusively used in electrical goods and also for making lubricants, paints, varnishes etc.

Question 4.
Iron and Steel industry and Software industry.
Answer:
Iron and Steel industry:

1. Age old industry.
2. Depend upon minerals.
3. Large scale industry.
4. Began in 1907.

Software industry:

1. Recently developed industry.
2. Depend upon skill and knowledge (technical).
3. Small and medium scale industry.
4. Began in 1970.

V. Answer the following in a paragraph

Question 1.
Enumerate the automobile industry as the fast-growing industry of India.
Answer:
The first automobile industry was established at the Premier Automobiles Ltd at Kurla (Mumbai) in 1947 and the Hindustan Motor Ltd at Uttarpara (Kolkata) in 1948. At present India is the 7th largest producer of automobile manufactures which includes two wheelers, commercial vehicles, passenger cars, Jeep, scooty, scooters, motor cycles mopeds and three wheelers.

Among the production of two wheelers, motorcycles are manufactured at Faridabad, Haryana and Mysore. While scooters are manufactured at Lucknow, Satara, Akudi (Pune), Panki (Kanpur) and Odhai (Ahmedabad). The cars produced at Haryana, Kolkata, Mumbai and Chennai are Maruti, Ambassador, Fiat, Ford and Hyndai etc. Presenceof foreign car companies such as Mercedes Benz, Fiat, General Motors, Toyota and the recent entry of passenger cars manufacturers BMW, Audi, Volks Wagen and Volvo makes the Indian automobile sector a special one.

Several new joint venture agreements for the manufacture of cars have recently been signed by the Indian companies and renowned car manufacturers of the world. The Indian auto industry is said to take a big leep in the near fixture. This expected to provide much more competitive environment to the industry and a wide choice of ultra modem cars to the consumers. This is the fast growing industry in India.

Question 2.
What are the different forms of Iron ore and their nature?
Answer:
Iron ores are rocks and minerals from which metallic iron can be economically extracted.

The different forms of Iron ore are:

Nature: The ores are usually rich in iron oxides and vary in colour from dark grey, bright yellow or deep purple to rusty red.

Question 3.
Write about the uses of Coal and its areas of distribution.
Answer:

1. Coal is a non-renewable resource. It is available in the form of sedimentary rocks.
2. It has close association with the industrial development of any country.

Uses: Coal is an important source of energy in India with its varied and innumerable uses

1. It is converted into gas, oil and thermal power electricity.
2. Besides it forms a basic raw materials for the production of chemicals, dyes, fertilizers, paints, synthetic and explosives.
3. Distribution: Indian coal is mostly associated with Gondwana region and is primarily found in peninsular India.
4. The States of Jharkhand, Odisha, West Bengal and Madhya Pradesh alone account for nearly 90% of coal reserves of the country. About 2% of India’s coal is of tertiary type and is mostly in Assam and Jammu and Kashmir.
5. Jharkhand is the largest coal producing state in the country. Indian lignite (brown coal) deposits occur particularly in Tamil Nadu ( Neyveli), Puducherry and Kerala.

Question 4.
Give an account on Automobile industry in India.
Answer:
Automobile industry’ is one of the most dynamic industrial groups in India. India is the 7th largest producer of automobile manufacturers.

Distribution: The automobile industries are found in four clusters viz: Delhi, Gurgaon and Manesar in North India.

West India: Pune, Nasik, Halol and Aurangabad East India: Jamshedpur and Kolkatta South India: Chennai, Bengaluru and Hosur.

Major-Indian companies which manufacture commercial vehicles: Tata motors, Ashok Leyland, Mahindra and Mahindra, Eicher motors and Ford motors.

Foreign companies which manufacture commercial vehicles: Hyundai, Mercedes Benz , ITEC, MAN.

Automobile industries in India manufactures two wheelers passenger car, Jeep, Scooty, Scooter, Mopeds, Motorcycles and three wheelers.

Passenger car manufacturers: Tata motors, Maruti Suzuki , Mahindra and Mahindra and Hindustan motors are Indian companies.

Foreign car companies in India: Mercedez Benz, Fiat, General motors, Toyota, recent entry BMW, Audi, Volkswagon and Volvo.

Two wheeler manufacturing is dominated by Indian companies like Hero, Bajaj Auto and TVS.

Major centres of Automobile industries: Mumbai, Chennai, Jamshedpur, Jabalpur, Kolkata, Pune, New Delhi, Kanpur, Bengaluru, Satara, Lucknow • and Mysuru. Chennai is named as “Detroit of Asia” due to the presence of major automobile manufacturing units and allied industries around the city.

Question 5.
What are the major challenges faced by the Indian industries?
Answer:
Some major challenges faced by industries in India are:

1. Non-availability of large blocks of land.
2. Shortage and fluctuation in power supply.
3. Non-availability of cheap labourers.
4. Poor access to credit.
5. High rate of interest for borrowed loan.
6. Lack of technical and vocational training for employees.
7. In appropriate living conditions nearby industrial areas.

VI. On the outline map of India mark the following

Question 1.
Any two places producing limestone
Answer:

Question 2.
Any two Nuclear power stations
Answer:

Question 3.
Any two Thermal power stations in Tamil Nadu
Answer:

Question 4.
Wind farm places (any 2)
Answer:

Students can Download Tamil Nadu 12th Physics Model Question Paper 4 English Medium Pdf, Tamil Nadu 12th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Physics Model Question Paper 4 English Medium

General Instructions:

• The question paper comprises of four parts.
• You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
• All questions of Part I, II, III, and IV are to be attempted separately.
• Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
  These are to be answered by choosing the most suitable answer from the given four
  alternatives and writing the option code and the corresponding answer
• Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered
  in about one or two sentences.
• Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered
  in about three to five short sentences.
• Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered
  in detaiL Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 70

Part – I

Answer all the questions. Choose the correct answer. [15 x 1 = 15]

Question 1.
Which charge configuration produces a uniform electric field?
(a) point charge
(b) infinite uniform line charge
(c) uniformly charged infinite plane
(d) uniformly charged spherical shell
Answer:
(c) uniformly charged infinite plane

Question 2.
The work done in carrying a charge Q, once round a circle of radius R with a charge Q₂ at the centre is  ………….
(a) \(\frac{\mathrm{Q}_{1} \mathrm{Q}_{2}}{4 \pi \varepsilon_{0} \mathrm{R}^{2}}\)
(b) Zero
(c)  \(\frac{\mathrm{Q}_{1} \mathrm{Q}_{2}}{4 \pi \varepsilon_{0} \mathrm{R}} \)
(d) infinite
Answer:
(b) Zero
Hint: The electric field is conservative. Therefore, no work is done in moving a charge around a closed path in a electric field.

Question 3.
The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 Ω is……………………..
(a) 0.2 Ω
(b) 0.5 Ω
(c) 0.8 Ω
(d) 1.0 Ω
Answer:
(b) 0.5 Ω

Question 4.
An electron moves straight inside a charged parallel plate capacitor of uniform charge density σ. The time taken by the electron to cross the parallel plate capacitor when the plates of the capacitor are kept under constant magnetic field of induction \(\overrightarrow{\mathrm{B}}\) is ……….


Answer:
(d)

Question 5.
In a series resonant RLC circuit, the voltage across 100Ω resistor is 40 V. The resonant frequency co is 250 rad/s. If the value of C is 4 pF, then the voltage across L is……………….
(a) 600 V
(b) 4000 V
(c) 400 V
(d) IV
Answer:
(c) 400 V

Question 6.
During the propagation of electromagnetic waves in a medium:
(a) electric energy density is double of the magnetic energy density
(b) electric energy density is half of the magnetic energy density
(c) electric energy density is equal to the magnetic energy density
(d) both electric and magnetic energy densities are zero
Answer:
(c) electric energy density is equal to the magnetic energy density

Question 7.
First diffraction minimum due to a single slit of width 1.0 x 10^-5 cm is at 30°. Then wavelength of light used is, …………………
(a) 400 Å
(b) 500 Å
(c) 600 Å
(d) 700 Å
Answer:
(b) 500 Å
Hint. For diffraction minima, d sin θ = nλ
\(\lambda=\frac{d \sin \theta}{n}=\frac{1 \times 10^{-5} \times 10^{-2} \times \sin 30^{\circ}}{1}=0.5 \times 10^{-7}\)
λ = 500 Å

Question 8.
The sky would appear red instead of blue if
(a) atmospheric particles scatter blue light more than red light
(b) atmospheric particles scatter all colours equally
(c) atmospheric particle scatter red light more than blue light
(d) the sun was much hotter
Answer:
(c) atmospheric particle scatter red light more than blue light

Question 9.
Kinetic energy of emitted electron depends upon
(a) frequency
(b) intensity
(c) nature of atmosphere surrounding the electron
(d) none of these
Answer:
(a) frequency
Hint: Kinetic energy of emitted electron depends on the frequency of incident radiation.

Question 10.
The ratio between the first three orbits of hydrogen atom is…………….
(a) 1:2:3
(b) 2:4:6
(c) 1:4:9
(d) 1:3:5
Answer:
(c) 1:4:9
Hint :
E_n = \(\frac{-13.6 \times z^{2}}{n^{2}}\)
n = 1; E₁ =- 13.6 eV/ atom
n = 2; E₂ = – 3.4 eV/ atom
n = 3; E₃ = -1.51 eV/atom                             ’
The ratio of three orbits E₁ : E₂ : E₃ = 13.6 : 3.4 : 1.51 = 1 : 4 : 9

Question 11.
Bohr’s theory of hydrogen atom did not explain fully
(a) diameter of H-atom
(b) emission spectra
(c) ionisation energy
(d) the fine structure of even hydrogen spectrum
Answer:
(d) the fine structure of even hydrogen spectrum
Hint: Bohr theory could not explain the five structure of hydrogen spectrum.

Question 12.
If a half-wave rectified voltage is fed to a load resistor, which part of a cycle the load current will flow?
(a) 0° – 90°
(b) 90° – 180°
(c) 0° – 180°
(d) 0° – 360°
Answer:
(c) 0° – 180°

Question 13.
Diamond is very hard because ……………..
(a) it is covalent solid
(b) it has large cohesive energy
(c) high melting point
(d) insoluble in all solvents
Answer:
(b) it has large cohesive energy

Question 14.
The internationally accepted frequency deviation for the purpose of FM broadcasts.
(a) 75 kHz
(b) 68 kHz
(c) 80 kHz
(d) 70 kHz
Answer:
(a) 75 kHz

Question 15.
The blue print for making ultra durable synthetic material is mimicked from
(a) Lotus leaf
(b) Morpho butterfly
(c) Parrot fish
(d) Peacock feather
Answer:
(c) Parrot fish

Part – II

Answer any six questions in which Q. No 17 is compulsory.   [6 x 2 = 12]

Question 16.
Define ‘electric flux’.
Answer:
The number of electric field lines crossing a given area kept normal to the electric field lines  is called electric flux. Its unit is N m² C^-1. Electric flux is a scalar quantity.

Question 17.
Determine the number of electrons flowing per second through a conductor, when a current of 32 A flows through it.
Answer:
I = 32A, t= 1 s
Charge of an electron, e = 1.6 x 10^-19 C
The number of electrons flowing per second, n =?

Question 18.
Define magnetic flux.
Answer:
The number of magnetic field lines crossing per unit area is called magnetic flux φ_B
\(\phi_{\mathrm{B}}=\overrightarrow{\mathrm{B}} \cdot\overrightarrow{\mathrm{A}}=\mathrm{B} \mathrm{A} \cos \theta=\mathrm{B} \perp \mathrm{A}\)

Question 19.
Give any one definition of power factor.
Answer:
The power factor is defined as the ratio of true power to the apparent power of an a.c. circuit.
It is equal to the cosine of the phase angle between current and voltage in the a.c. circuit.

Question 20.
State the laws of reflection.
Answer:

• The incident ray, reflected ray and normal to the reflecting surface all are coplanar (ie. lie in the same plane).
• The angle of incidence i is equal to the angle of reflection r. i = r

Question 21.
Why do metals have a large number of free electrons?
Answer:
In metals, the electrons in the outer most shells are loosely bound to the nucleus. Even at room temperature, there are a large number of free electrons which are moving inside the metal in a random manner.

Question 22.
The radius of the 5th orbit of hydrogen atom is 13.25 A. Calculate the wavelength of the electron in the 5th orbit.
Answer:
2πr = nλ
2 x 3.14 x 13.25 Å = 5 x λ
.’. λ = 16.64 Å

Question 23.
Draw the output waveform of a full wave rectifier.
Answer:

Question 24.
Explain centre frequency or resting frequency in frequency modulation.
Answer:
When the frequency of the baseband signal is zero (no input signal), there is no change in the frequency of the carrier wave. It is at its normal frequency and is called as centre frequency or resting frequency.

Part – III

Answer any six questions in which Q.No. 28 is compulsory. [6 x 3 = 18]

Question 25.
What is corona discharge?
Answer:
The electric field near the edge is very high and it ionizes the surrounding air. The positive ions are repelled at the sharp edge and negative ions are attracted towards the sharper edge.
This reduces the total charge of the conductor near the sharp edge. This is called action at points or corona discharge.

Question 26.
What is electric power and electric energy?
Answer:
Electric power: It is the rate at which an electric appliance converts electric energy into other forms of energy. Or, it is the rate at which work is done by a source of emf in maintaining an electric current through a circuit.
\({ P }=\frac { { W } }{ t } ={ VI }={ I }^{ 2 }{ R }=\frac { { V }^{ 2 } }{ { R } } \)
Electric energy: It is the total workdone in maintaining an electric current in an electric circuit for a given time.
W = Vt = VIr joule = I²R? joule.

Question 27.
A bar magnet having a magnetic moment \(\overrightarrow{\mathrm{M}}\) is cut into four pieces i.e., first cut in two pieces along the axis of the magnet and each piece is further cut into two pieces. Compute the magnetic moment of each piece.
Answer:
Consider a bar magnet of magnetic moment \(\overrightarrow{\mathrm{M}}\). When a bar magnet first cut in two pieces along the axis, their magnetic moment is \(\frac{\overrightarrow{\mathrm{M}}}{2}\)

Question 28.
The current in an inductive circuit is given by 0.3 sin (200t – 40°) A. Write the equation for the voltage across it if the inductance is 40 mH.
Answer:
L = 40 x 10^-3H; i = 0.1 sin (200 t – 40°); X_L = ωL = 200 x 40 x 10^-3 = 8 Ω
V_m = I_m X_L = 0.3 x 8 = 2.4 V
In an inductive circuit, the voltage leads the current by 90° Therefore,
υ = V_m sin (ωt + 90°)
υ = 2.4 sin (200f – 40°+ 90°)
υ = 2.4 sin (200f +50°)volt

Question 29.
Writedown the integral form of modified Ampere’s circuital law.
Answer:
This law relates the magnetic field around any closed path to the conduction current and displacement current through that path.

Question 30.
Two light sources have intensity of light as I_o. What is the resultant intensity at a point where the two light waves have a phase difference of π/3?
Answer:
Let the intensities be I_o
The resultant intensity is, I = 4 I_o cos² (φ/2)
Resultant intensity when, ϕ = π/ 3, is I = 4I₀ cos² (π / 6) I
= 4I₀(√3 / 2)² =3I₀

Question 31.
Write the properties of cathode rays.
Answer:

• Cathode rays possess energy and momentum and travel in a straight line with high speed of the order of 10⁷m s^-1
• It can be deflected by application of electric and magnetic fields.
• When the cathode rays are allowed to fall on matter, they produce heat.
• They affect the photographic plates and also produce fluorescence when they fall on   certain crystals and minerals.
• When the cathode rays fall on a material of high atomic weight, x-rays are produced.
• Cathode rays ionize the gas through which they pass.
• The speed of cathode rays is up \(\left( \frac { 1 }{ 10 } \right) ^{ th }\) of the speed of light.

Question 32.
Distinguish between wireline and wireless communication.
Answer:

Question 33.
What is the difference between Nano materials and Bulk materials?
Answer:

• The solids are made up of particles. Each of the particle has a definite number of atoms, which might differ from material to material. If the particle of a solid is of size less than 100 nm, it is said to be a ‘nano solid’.
• When the particle size exceeds 100 nm, it is a ‘bulk solid’. It is to be noted that nano and bulk solids may be of the same chemical composition.
• For example, ZnO can be both in bulk and nano form.
• Though chemical composition is the; same, nano form of the material shows strikingly different properties when compared to its bulk counterpart.

Part – IV

Answer all the questions.  [5 x 5 = 25]

Question 34.
(a) Calculate the electric field due to a dipole on its equatorial plane.
Answer:
Electric field due to an electric dipole at a point on the equatorial plane. Consider a point C at a distance r from the midpoint O of the dipole on the equatorial plane as shown in Figure.

Since the point C is equi­distant from +q and -q, the magnitude of the electric fields of +q and -q are the same. The direction of \(\overrightarrow{\mathrm{E}}_{+}\) is along BC \(\overrightarrow{\mathrm{E}}_{-} \) and the direction of E is along CA. \(\overrightarrow{\mathrm{E}}_{+} \) and \(\overrightarrow{\mathrm{E}}_{-} \) and E are resolved into two components; one component parallel to the dipole axis and the other perpendicular to it. The perpendicular components \(\left|\overrightarrow{\mathrm{E}}_{+}\right| \sin \theta \) and \(\left|\overrightarrow{\mathrm{E}}_{-}\right| \sin \theta \) are oppositely directed and cancel each other. The magnitude of the total electric field at point C is the sum of the parallel components of E₊ and E and its direction is \(\overrightarrow { { E } } _{ + }\quad and\quad \overrightarrow { { E } } _{ – }\) and its direction is along \(-\hat{p}\)

(b) How the emf of two cells are compared using potentiometer?
Answer:
Comparison of emf of two cells with a potentiometer: To compare the emf of two cells, the circuit connections are made as shown in figure. Potentiometer wire CD is connected to a battery Bf and a key K in series. This is the primary circuit. The end C of the wire is connected to the terminal M of a DPDT (Double Pole Double Throw) switch and the other terminal N is connected to a jockey through a galvanometer G and a high resistance HR. The cells whose emf ξ₁ and ξ₂ and to be compared are connected to the terminals M₁, N₁ and M₂, N₂, of the DPDT switch. The positive terminals of Bf ξ₁ and ξ₂ should be connected to the same end C.

The DPDT switch is pressed towards M₁, N₁ so that cell ξ₁, is included in the secondary circuit and the balancing length l₁, is found by adjusting the jockey for zero deflection. Then the second cells ξ₂, is included in the circuit and the balancing length l₂, is determined. Let r be the resistance per unit length of the potentiometer wire and I be the current flowing through the wire.
we have
ξ₁ = Irl₁ ……………….. (1)
ξ₂ = Irl₂ ……………….. (2)
By dividing equation (1) by (2)
\(\frac{\xi_{1}}{\xi_{2}}=\frac{l_{1}}{l_{2}}\)
By including a rheostat (Rh) in the primary circuit, the experiment can be repeated several times by changing the current flowing through it.

Question 35.
(a) Discuss the working of cyclotron in detail.
Answer:
Cyclotron : Cyclotron is a device used accelerate the charged particles to gain large kinetic energy. It is also called as high energy accelerator. It was invented by Lawrence and Livingston in 1934.

Principle : When a charged particle moves normal to the magnetic field, it experiences magnetic Lorentz force.

Construction : The particles are allowed to move in between two semicircular metal containers called Dees (hollow D – shaped objects). Dees are enclosed in an evacuated chamber and it is kept in a region with uniform magnetic field controlled by an electromagnet. The direction of magnetic field is normal to the plane of the are enclosed in an evacuated chamber and it is kept in a region with uniform magnetic field controlled by an electromagnet. The direction of magnetic field is normal to the plane of the Dees. The two Dees are kept separated with a gap and the source S (which ejects the particle to be accelerated) is placed at the center in the gap between the Dees. Dees are connected to high frequency alternating potential difference.

Working: Let us assume that the ion ejected from source S is positively charged. As soon as ion is ejected, it is accelerated towards a Dee (say, Dee – 1) which has negative potential at that time. Since the magnetic field is normal to the plane of the Dees, the ion undergoes circular path. After one semi-circular path in Dee-1, the ion reaches the gap between Dees. At this time, the polarities of the Dees are reversed so that the ion is now accelerated towards Dee-2 with a greater velocity. For this circular motion, the centripetal force of the charged particle q is provided by Lorentz force.

\(\frac { mv^{ 2 } }{ r } =qv{ B }\Rightarrow r=\frac { m }{ q{ B } } v\Rightarrow r\propto v\)
From the equation, the increase in velocity increases the radius of circular path. This process continues and hence the particle undergoes spiral path of increasing radius. Once it reaches near the Very important condition in cyclotron operation is the resonance condition. It happens when the frequency { at which the positive ion circulates in the magnetic field must be equal to the constant frequency of the electrical oscillator f_osc From equation
\(f_{\text {osc }}=\frac{q \mathrm{B}}{2 \pi m} \Rightarrow \mathrm{T}=\frac{1}{f_{\text {osc }}}\)
The time Period of oscillation is \(\mathrm{T}=\frac{2 \pi m}{q \mathrm{B}}\)
The kinetic energy of the charged particle is \(\mathrm{KE}=\frac{1}{2} m v^{2}=\frac{q^{2} \mathrm{B}^{2} r^{2}}{2 m}\)

Limitations of cyclotron

• the speed of the ion is limited
• electron cannot be accelerated
• uncharged particles cannot be accelerated

[OR]

(b) Give the uses of Foucault current.
Answer:
Though the production of eddy current is undesirable in some cases, it is useful in some other cases. A few of them are

(1) Induction stove : Induction stove is used to cook the food quickly and safely with less energy consumption. Below the cooking zone, there is a tightly wound coil of insulated wire. The cooking pan made of suitable material, is placed over the cooking zone. When the stove is switched on, an alternating current flowing in the coil produces high frequency alternating magnetic field which induces very strong eddy currents in the cooking pan. The eddy currents in the pan produce so much of heat due to Joule heating which is used to cook the food.

(2) Eddy current brake : This eddy current braking system is generally used in high speed trains and roller coasters. Strong electromagnets are fixed just above the rails. To stop the train, electromagnets are switched on. The magnetic field of these magnets induces eddy currents in the rails which oppose or resist the movement of the train. This is Eddy current linear brake.

In some cases, the circular disc, connected to the wheel of the train through a common shaft, is made to rotate in between the poles of an electromagnet. When there is a relative motion between the disc and the magnet, eddy currents are induced in the disc which stop the train. This is Eddy current circular brake.

(3) Eddy current testing : It is one of the simple non-destructive testing methods to find defects like surface cracks and air bubbles present in a specimen. A coil of insulated wire is given an alternating electric current so that it produces an alternating magnetic field. When this coil is brought near the test surface, eddy current is induced in the test surface. The presence of defects causes the change in phase and amplitude of the eddy current that can be detected by some other means. In this way, the defects present in the specimen are identified.

(4) Electro magnetic damping : The armature of the galvanometer coil is wound on a soft iron cylinder. Once the armature is deflected, the relative motion between the soft iron cylinder and the radial magnetic field induces eddy current in the cylinder. The damping force due to the flow of eddy current brings the armature to rest immediately and then galvanometer shows a steady deflection. This is called electromagnetic damping.

Question 36.
(a) Write down the properties of electromagnetic waves.
Answer:
Properties of electromagnetic waves:

• Electromagnetic waves are produced by any accelerated charge.
• Electromagnetic waves do not require any medium for propagation. So electromagnetic wave is a non-mechanical wave.
• Electromagnetic waves are transverse in nature. This means that the oscillating electric field vector, oscillating magnetic field vector and propagation vector (gives direction of propagation) are mutually perpendicular to each other.
• Electromagnetic waves travel with speed which is equal to the speed of light in vacuum
  or free space, \(c=\frac { 1 }{ \sqrt { \varepsilon _{ 0 }\mu _{ 0 } } } =3\times 10^{ 8 }{ ms }^{ -1 }\)
• The speed of electromagnetic wave is less than speed in free space or vacuum, that is, v < c. In a medium of refractive index,

[OR]

(b) Explain the Young’s double slit experimental setup and obtain the equation for path difference.
Answer:
Experimental setup
Wavefronts from S₁ and S₂ spread out and overlapping takes place to the right side of double slit. When a screen is placed at a distance of about 1 meter from the slits, alternate bright and dark fringes which are equally spaced appear on the screen. These are called interference fringes or bands.

S₁ and S₂ travel equal distances and arrive in-phase. These two waves constructively interfere and bright fringe is observed at O. This is called central bright fringe.

The fringes disappear and there is uniform illumination on the screen when one of the slits is covered. This shows clearly that the bands are due to interference.

Equation for path difference
Let d be the distance between the double slits S₁ and S₂ which act as coherent sources of wavelength λ. A screen is placed parallel to the double slit at a distance D from it. The mid-point of S₁ and S₂ is C and the midpoint of the screen O is equidistant from S₁ and S₂. P is any point at a distance y  from O.

The waves from S₁ and S₂ meet at P either inphase or out-of-phase depending upon the path difference between the two waves. The path difference S between the light waves from S₁ and S₂ to the point P is, δ = S₂P and S₁P
A perpendicular is dropped from the point S₁, to the line S₂P at M to find the path difference more precisely.
δ = S₂P – MP = S₂M
The angular position of the point P from C is θ. ∠OCP = θ.
From the geometry, the angles ∠OCP and ∠S₂S₁ M are equal.
∠OCP = ∠S₂S₁ M = θ
In right angle triangle ΔS₁S₂M, the path difference, S₂M = d sin θ
δ = d sin θ
If the angle θ is small, sin θ ≈ tan θ ≈ θ From the right angle triangle ΔOCP, tan θ = y/D. The path difference, δ =dy/D

Question  37.
(a) Give the construction and working of photo emissive cell.
Answer:
Photo emissive cell:
Its working depends on the electron emission from a metal cathode due to irradiation of light or other radiations.

• It consists of an evacuated glass or quartz bulb in which two metallic electrodes – that is, a cathode and an anode are fixed.
• The cathode C is semi-cylindrical in shape and is coated with a photo sensitive material. The anode A is a thin rod or wire kept along the axis of the semi-cylindrical cathode.
• A potential difference is applied between the anode and the cathode through a galvanometer G.

Working:

• When cathode is illuminated, electrons are emitted from it. These electrons are attracted by anode and hence a current is produced which is measured by the galvanometer
• For a given cathode, the magnitude of
  the current depends on the intensity to incident radiation and
  the potential difference between anode and cathode.

[OR]

Question 37.
(b) Explain the J.J. Thomson experiment to determine the specific charge of electron.
Answer:
In 1887, J. J. Thomson made remarkable improvement in the scope of study of gases in discharge tubes. In the presence of electric and magnetic fields, the cathode rays are deflected. By the variation of electric and magnetic fields, mass normalized charge or the specific charge (charge per unit mass) of the cathode rays is measured.

A highly evacuated discharge tube is used and cathode rays (electron beam) produced at cathode are attracted towards anode disc A. Anode disc is made with pin hole in order to allow only a narrow beam of cathode rays. These cathode rays are now allowed to pass through the parallel metal plates, maintained at high voltage. Further, this gas discharge tube is kept in between pole pieces of magnet such that both electric and magnetic fields are perpendicular to each other. When the cathode rays strike the screen, they produce scintillation and hence bright spot is observed. This is achieved by coating the screen with zinc sulphide.

(i) Determination of velocity of cathode rays
For a fixed electric field between the plates, the magnetic field is adjusted such that the cathode rays (electron beam) strike at the original position O. This means that the magnitude of electric force is balanced by the magnitude of force due to magnetic field. Let e be the charge of the cathode rays, then eE = eBv
\(v=\frac{E}{B}\) …………….. (1)

(ii) Determination of specific charge
Since the cathode rays (electron beam) are accelerated from cathode to anode, the potential energy of the electron beam at the cathode is converted into kinetic energy of the electron beam at the anode. Let V be the potential difference between anode and cathode, then the potential energy is eV. Then from law of conservation of energy,
\(e \mathrm{V}=\frac{1}{2} m v^{2} \Rightarrow \frac{e}{m}=\frac{v^{2}}{2 \mathrm{V}}\)
Substituting the value of velocity from equation (1) , we get
\(\frac{e}{m}=\frac{1}{2 \mathrm{V}} \frac{\mathrm{E}^{2}}{\mathrm{B}^{2}}\) ………..(2)
Substituting the value of E ,B and V, the specific charge vam be determined as
\(\frac{e}{m}=1.7 \times 10^{11} \mathrm{Ckg}^{-1}\)

(iii) Deflection of charge only due to uniform electric field
When the magnetic field is turned off, the deflection is only due to electric field. The
deflection in vertical direction is due to the electric force.
F_e = eE ………………….. (3)
Let m be the mass of the electron and by applying Newton’s second law of motion,
acceleration of the electron is
\(a_{e}=\frac{1}{m} \mathrm{F}_{e}\) ………………… (4)
Substituting equation (4) in equation (3),
\(a_{e}=\frac{1}{m} e E=\frac{e}{m} E\)
Let y be the deviation produced from original position on the screen. Let the initial upward velocity of cathode ray be μ = 0 before entering the parallel electric plates. Let l be the time taken by the cathode rays to travel in electric field. Let l be the length of one of the plates, then the time taken is
\(t=\frac{l}{v}\) …………(5)

Hence, the deflection y’ of cathod rays is( note : u = 0 and \(a_{e}=\frac{e}{m} \mathrm{E}\))

Therefore, the deflection y on the screen is
y α y’ ⇒ y = Cy’
where C is proportionality constant which depends on the geometry of the discharge tube and substituting y’ value in equation (6), we get

Substituting the values on RHS, the value of specific charge is calculated as
e/m = 1.7 x 10¹¹Ckg^-1

Question 38.
(a) What is an LED? Give the principle of operation with a diagram.
Answer:
Light Emitting Diode (LED): LED is a p-n junction diode which emits visible or invisible light when it is forward biased. Since, electrical energy is converted into light energy, this process is also called electroluminescence. The cross-sectional view of a commercial LED is shown in figure (b). It consists of a p-layer, n-layer and a substrate. A transparent window is used to allow light to travel in the desired direction. An external resistance in series with the biasing source is required to limit the forward current through the LED. In addition, it has two leads; anode and cathode.

When the p-n junction is forward biased, the conduction band electrons on n-side and valence band holes on p-side diffuse across the junction. When they cross the junction, they become excess minority carriers (electrons in p-side and holes in n-side). These excess minority carriers recombine with oppositely charged majority carriers in the respective regions, i.e. the. electrons in the conduction band recombine with holes in the valence band as shown in the figure (c). During recombination process, energy is released in the form of light (radiative) or heat (non-radiative). For radiative recombination, a photon of energy hv is emitted. For non- radiative recombination, energy is liberated in the form of heat.

The colour of the light is determined by the energy band gap of the material. Therefore, LEDs are available in a wide range of colours such as blue (SiC), green (AlGaP) and red (GaAsP). Now a days, LED which emits white light (GalnN) is also available.

[OR]

(b) Give the applications of ICT in mining and agriculture sectors.
Answer:
(i) Agriculture
The implementation of information and communication technology (ICT) in agriculture sector enhances the productivity, improves the living standards of farmers and overcomes the Challenges and risk factors.

• ICT is widely used in increasing food productivity and farm management.
• It helps to optimize the use of water, seeds and fertilizers etc.
• Sophisticated technologies that include robots, temperature and moisture sensors, aerial images, and GPS technology can be used.
• Geographic information systems are extensively used in farming to decide the suitable place for the species to be planted.

(ii) Mining

• ICT in mining improves operational efficiency, remote monitoring and disaster locating system.
• Information and communication technology provides audio-visual warning to the trapped underground miners.
• It helps to connect remote sites.

Students can download 10th Science Chapter 10 Types of Chemical Reactions Questions and Answers, Notes, Samacheer Kalvi 10th Science Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions

Samacheer Kalvi 10th Science Types of Chemical Reactions Text Book Back Questions and Answers

I. Choose the correct answer:

Question 1
H_2(g) + Cl_29(g) → 2HCl_(g)
(a) Decomposition Reaction
(b) Combination Reaction
(c) Single Displacement Reaction
(d) Double Displacement Reaction
Answer:
(a) Decomposition Reaction

Question 2.
Photolysis is a decomposition reaction caused by ______.
(a) heat
(b) electricity
(c) light
(d) mechanical energy
Answer:.
(c) light
Hint:
\(2 \mathrm{AgBr}_{(\mathrm{s})} \stackrel{\mathrm{Light}}{\longrightarrow} 2 \mathrm{Ag}_{(\mathrm{s})}+\mathrm{Br}_{2(\mathrm{g})}\)

Question 3.
A reaction between carbon and oxygen is represented by C_(s) + O_2(g) → CO_2(g) + Heat. In which of the type(s), the above reaction can be classified?
(i) Combination Reaction
(ii) Combustion Reaction
(iii) Decomposition Reaction
(iv) Irreversible Reaction
(a) (i) and (ii)
(b) (i) and (iv)
(c) (i), (ii) and (iii)
(d) (i), (ii) and (iv)
Answer:
(d) (i), (ii) and (iv)

Question 4.
The chemical equation \(\mathrm{Na}_{2} \mathrm{SO}_{4(\mathrm{aq})}+\mathrm{BaCl}_{2(\mathrm{aq})} \rightarrow \mathrm{BaSO}_{4(\mathrm{s})} \downarrow+2 \mathrm{NaCl}_{(\mathrm{aq})}\) represents which of the following types of reaction?
(a) Neutralisation
(b) Combustion
(c) Precipitation
(d) Single displacement.
Answer:
(c) Precipitation
Hint: This reaction involves the precipitation of white BaSO₄ by mixing of Na₂SO₄ (aq) and BaCl₂ (aq). Hence it belongs to precipitation reaction.

Question 5.
Which of the following statements are correct about a chemical equilibrium?
(i) It is dynamic in nature
(ii) The rate of the forward and backward reactions are equal at equilibrium
(iii) Irreversible reactions do not attain chemical equilibrium
(iv) The concentration of reactants and products may be different
(a) (i), (ii) and (iii)
(b) (i), (ii) and (iv)
(c) (ii), (iii) and (iv)
(d) (i), (iii) and (iv)
Answer:
(a) (i), (ii) and (iii)

Question 6.
A single displacement reaction is represented by \(\mathrm{X}_{(\mathrm{s})}+2 \mathrm{HCl}_{(\mathrm{aq})} \rightarrow \mathrm{XCl}_{2(\mathrm{aq})}+\mathrm{H}_{2(\mathrm{g})}\). the following(s) could be X?
(i) Zn
(ii) Ag
(iii) Cu
(iv) Mg.
Choose the best pair.
(a) i and ii
(b) ii and iii
(c) iii and iv
(d) i and iv.
Answer:
(d) i and iv.
Hint:
Zn + 2HCl → ZnCl₂ + H₂
Mg + 2HCl → MgCl₂ + H_2.

Question 7.
Which of the following is not an “element + element → compound” type reaction?
(a) C_(s) + O_2(g) → CO_2(g)
(b) 2K_(s) + Br_2(l) → 2KBr_(s)
(c) 2CO_(g) + O_2(g) → 2CO_2(g)
(d) 4Fe_(s) + 3O_2(g) → 2Fe₂O_3(s)
Answer:
(c) 2CO_(g) + O_2(g) → 2CO_2(g)

Question 8.
Which of the following represents a precipitation reaction?
(a) A_(s) + B_(s) → C_(s) + D_(s)
(b) A_(s) + B_(aq) → C_(aq) + D_(l)
(c) A_(aq) + B_(aq) → C_(s) + D_(aq)
(d) A_(aq) + B_(s) → C_(aq) + D_(l)
Answer:
(c) A_(aq) + B_(aq) → C_(s) + D_(aq)

Question 9.
The pH of a solution is 3. Its [OH^–] concentration is ______.
(a) 1 × 10^-3 M
(b) 3 M
(c) 1 × 10^-11 M
(d) 11 M.
Answer:
(c) 1 × 10^-11 M
Hint: pH = 3
It means [H⁺] = 10^-3
[H⁺] [OH^–] = 10^-14
[10^-3] [OH^–] = 10^-14
[OH^–] = 10^-11

Question 10.
Powdered CaCO₃ reacts more rapidly than flaky CaCO₃ because of :
(a) large surface area
(b) high pressure
(c) high concentration
(d) high temperature
Answer:
(a) large surface area

II. Fill in the blanks:

1. A reaction between an acid and a base is called ………..
2. When zinc metal is placed in hydrochloric acid, ………. gas is evolved.
3. The equilibrium attained during the meiting of ice is known as ………..
4. The pH of a fruit juice is 5.6. If you add slaked lime to this juice, its pH ……….
5. The value of ionic product of water at 25 °C is ………..
6. The normal pH of human blood is …………
7. Electrolysis is type of ……….. reaction.
8. The number of products formed in a synthesis reaction is ………..
9. Chemical volcano is an example for ……….. type of reaction.
10. The ion formed by dissolution of H⁺ in water is called …………
Answer:
1. neutralization
2. H₂
3. physical equilibrium
4. increases to ‘7’
5. 1 × 10^-14 mol² dm^-6
6. 7.4
7. decomposition
8. 1
9. decomposition
10. hydronium ion

III. Match the following:

Question 1.
Identify the types of reaction:

Answer:
A. (iii)
B. (i)
C. (iv)
D. (ii)

IV. True or False: (If false give the correct statement)

1. Silver metal can replace hydrogen gas from nitric acid.
2. The pH of rain water containing dissolved gases like SO₃, CO₂, NO₂ will be less than 7.
3. At the equilibrium of a reversible reaction, the concentration of the reactants and the products will be equal.
4. Periodical removal of one of the products of a reversible reaction increases the yield.
5. On dipping a pH paper in a solution, it turns into yellow. Then the solution is basic.

Answer:

1. False – Silver cannot displace H₂ from HNO₃ acid, since it is placed below hydrogen in the activity series.
2. True
3. False – At equilibrium the concentration of the reactants and products do not change it remains constant, but the concentration of the reactants and the products will not be equal.
4. True
5. False – The solution is neutral if the solution is basic it will be green in colour.

V. Short answer questions:

Question 1.
When an aqueous solution of potassium chloride is added to an aqueous solution of silver nitrate, a white precipitate is formed. Give the chemical equation of this reaction.
Answer:

Question 2.
Why does the reaction rate of a reaction increase in raising the temperature?
Answer:
On increasing temperature heat is supplied to the reactant. This energy breaks more bonds and thus speed up the chemical reaction. Foods kept at room temperature spoils faster than that kept in the refrigerator.

Question 3.
Define combination reaction. Give one example for an exothermic combination reaction.
Answer:
A combination reaction is a reaction in which two or more reactants combine to form a compound.
Eg: C_(s) + O_2(g) → CO_2(g) + heat

Question 4.
Differentiate reversible and irreversible reactions.
Answer:

VI. Answer in detail:

Question 1.
What are called thermolysis reactions?
Answer:
Thermal decomposition reactions are called ‘thermolysis’ reaction. In this type of reaction, the reactant is decomposed by applying heat. There are two types of thermolysis reactions. They are:
(i) Compound to element / element decomposition:
A compound is decomposed into two elements.
Eg:

(ii) Compound to compound / compound decomposition:
A compound is decomposed into two compounds.
Eg:

Question 2.
Explain the types of double displacement reactions with examples.
Answer:
There are two major classes of double displacement reactions. They are,
(i) Precipitation Reactions: When aqueous solutions of two compounds are mixed, if they react to form an insoluble compound and a soluble compound, then it is called precipitation reaction.
\(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2(\mathrm{aq})}+2 \mathrm{KI}_{(\mathrm{aq})} \rightarrow \mathrm{PbI}_{2(\mathrm{s})} \downarrow+2 \mathrm{KNO}_{3(\mathrm{aq})}\)

(ii) Neutralisation Reactions: Another type of displacement reaction in which the acid reacts with the base to form a salt and water. It is called ‘neutralisation reaction’ as both acid and base neutralize each other.
\(\mathrm{NaOH}_{(\mathrm{aq})}+\mathrm{HCl}_{(\mathrm{aq})} \rightarrow \mathrm{NaCl}_{(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}\).

Question 3.
Explain the factors influencing the rate of a reaction.
Answer:
Important factors that affect rate of a reaction are:

1. Nature of the reactants
2. Concentration of the reactants
3. Temperature
4. Catalyst
5. Pressure
6. Surface area of the reactants

1. Nature of the reactants : The reaction of sodium with hydrochloric acid is faster than that with acetic acid, because Hydrochloric acid is a stronger acid than acetic acid and thus more reactive. So, the nature of the reactants influence the reaction rate.
2Na_(s) + 2HCl_(aq) → 2NaCl_(aq) + H_2(g) (fast)
2Na_(s) + 2CH₃COOH_(aq) → 2CH₃COONa_(aq) + H_2(g) (slow)

2. Concentration of the reactants : Changing the amount of the reactants also increases the reaction rate. More the concentration, more particles per volume exist in it and hence faster the reaction. Granulated zinc reacts faster with 2M hydrochloric acid than 1M hydrochloric acid.

3. Temperature : Most of the reactions go faster at higher temperature. Because adding heat to the reactants provides energy to break more bonds and thus speed up the reaction. Calcium carbonate reacts slowly with hydrochloric acid at room temperature. When the reaction mixture is heated the reaction rate increases.

4. Pressure : If the reactants are gases, increasing their pressure increases the reaction rate. This is because, on increasing the pressure the reacting particles come closer and collide frequently.

5. Catalyst : A catalyst is a substance which increases the reaction rate without being consumed in the reaction. In certain reactions, adding a substance as catalyst speeds up the reaction. For example, on heating potassium chlorate, it decomposes into potassium chloride and oxygen gas, but at a slower rate. If manganese dioxide is added, it increases the reaction rate.

6. Surface area of the reactants : Powdered calcium carbonate reacts more readily with hydrochloric acid than marble chips. Because, powdering of the reactants increases the surface area and more energy is available on collision of the reactant particles. Thus, the reaction rate is increased.

Question 4.
How does pH play an important role in everyday life?
Answer:

• The pH of blood is almost 7.4. Any increase or decrease in this value leads to diseases
• Citrus fruits require slightly alkaline soil, while rice requires acidic soil and sugarcane requires neutral soil.
• If the pH of rainwater becomes less than 7, it becomes acid rain which is harmful in day-to-day life.
• pH changes cause tooth decay.
• During indigestion, the stomach produces too much acid and this causes pain and irritation.

Question 5.
What is chemical equilibrium? What are its characteristics?
Answer:
Chemical equilibrium is a state of a reversible chemical reaction where the,
Rate of forward reaction = Rate of backward reaction.
No change in the amount of the reactants and products takes place.
Characteristics of equilibrium:

1. In a chemical equilibrium, the rates of the forward and backward reactions are equal.
2. The observable properties such as pressure, concentration, colour, density, viscosity etc., of the system remain unchanged with time.
3. The chemical equilibrium is a dynamic equilibrium, because both the forward and backward reactions continue to occur even though it appears static externally.
4. In physical equilibrium, the volume of all the phases remain constant.

VII. HOT Questions:

Question 1.
A solid compound ‘A’ decomposes on heating into ‘B’ and a gas ‘C’ On passing the gas ‘C’ through water, it becomes acidic. Identify A, B and C.
Answer:
A – CaCO₃, solid compound
‘A’ decomposes on heating into ‘B’ and a gas ‘C’.

On passing the gas CO₂ through water, it becomes acidic.

A – CaCO₃, Calcium carbonate
B – CaO, Calcium oxide
C – CO₂, Carbondioxide gas

Question 2.
Can a nickel spatula be used to stir copper sulphate solution? Justify your answer.
Answer:
No, nickel spatula cannot be used to stir the copper sulphate solution. Actually, on the basis of activity series, nickel is more reactive than copper, so nickel will displace copper from its solution and copper will be deposited on nickel spatula.

VIII. Solve the following problems:

Question 1.
Lemon juice has a pH 2, what is the concentration of H⁺ ions?
Answer:
pH = – log [H⁺]
[H⁺] = antilog of [-pH]
= antilog [-2]
[H⁺] = 10^-2 M
[OR]
PH = – log [H⁺]
[H⁺] = 10^-pH
[H⁺] = 10^-2M

Question 2.
Calculate the pH of 1.0 × 10^-4 molar solution of HNO₃.
Answer:
pH = – log [H⁺]
HNO3 → H⁺ + NO₃^–
pH = -log [1 × 10^-4]
= -(-4)log₁₀ 10 = 4
pH = 4

Question 3.
What is the pH of 1.0 x 10^-5 molar solution of KOH?
Answer:
KOH → K⁺ + OH^–
pOH = -log[OH^–]
= -log [1 × 10^-5]
pOH = 5
pH + pOH = 14
∴ pH of KOH = 14 – 5 = 9
pH = 9

Question 4.
Laundry detergent has a pH 8.5, What is the concentration of H⁺ ions?
Answer:
pH = 8.5
pH = – log [H⁺]
[H⁺] = 10^-pH
[H⁺] = 10^-8.5
[H⁺] = 3.16 × 10^-9 M

Question 5.
The hydroxide ion concentration of a solution is 1 × 10^-11M. What is the pH of the solution?
Answer:
[OH^–] = 1 × 10^-11 M
pOH = – log[OH^–]
= – log[1 × 10^-11]
= -log₁₀1 – log₁₀10^-11
= -(-11) log₁₀10 = 11
pOH = 11
pH + pOH = 14
pH = 14 – 11
pH = 3

Samacheer Kalvi 10th Science Types of Chemical Reactions Additional Important Questions and Answers

I. Choose the correct answer:

Question 1.
The unit of rate of a reaction is:
(a) dm³/mol
(b) dm^-3
(c) mol dm^-3
(d) mol
Answer:
(c) mol dm^-3

Question 2.
As the molecule is dissociated by the absorption of heat it is otherwise called as ______.
(a) Thermolysis
(b) Photolysis
(c) Electrolysis
(d) None of these.
Answer:
(a) Thermolysis

Question 3.
The chemical formula of marble is:
(a) CaCO₃
(b) MgCO₃
(c) Na₂CO₃
(d) PbCO₃
Answer:
(a) CaCO₃

Question 4.
As the decomposition is caused by light, this kind of reaction is called ______.
(a) Thermolysis
(b) Photolysis
(c) Electrolysis
(d) None of these.
Answer:
(b) Photolysis

Question 5.
Fluorine will displace the following halide ion from the solution:
(a) chloride
(b) bromide
(c) iodide
(d) all the above
Answer:
(d) all the above

Question 6.
The decomposition of AgBr into grey coloured silver metal is an example of ……… reaction.
(a) compound to element/element
(b) compound to compound/compound
(c) combination
(d) neutralization
Answer:
(a) compound to element/element

Question 7.
The Metathesis reaction among the following is:
(a) C₃H_8(g) +5O_2(g) → 3CO_2(g) + 4H₂O + heat
(b) Zn_(s) + CuSO_4(aq) → ZnSO_4(aq) + Cu_(s)
(c) HNO_3(aq) + NH₄OH_(aq) → NH₄NO_3(aq) + H₂O_(l)

Answer:
(c) HNO_3(aq) + NH₄OH_(aq) → NH₄NO_3(aq) + H₂O_(l)

Question 8.
KI and Pb(NO₃)₂ solutions are mixed to give a precipitate. What is the colour of the precipitate?
(a) White
(b) Brown
(c) Red
(d) Yellow.
Answer:
(d) Yellow

Question 9.
The pH of rain water is approximately:
(a) 7
(b) 8
(c) 4
(d) 14
Answer:
(a) 7

Question 10.
Most of the reactions go faster at ______.
(a) low temperature
(b) moderate temperature
(c) 0°C
(d) high temperature.
Answer:
(d) high temperature

II. Fill in the blanks.

1. A chemical equation provides information on the ……….. of the substances and the reaction condition.
2. The symbol ‘aq’ in a chemical equation represent the physical state of the substance as ……….
3. 2Na_(s) + Cl_2(g) → 2NaCl_(s) represent the combination reaction between a ……… and ……
4. After white washing with a solution of slaked lime a thin layer of ……….. is formed.
5. Reactions in which heat is absorbed is called ……….. reactions.
6. Electrolytic refining of copper is based on ……….. reaction.
7. When 10^-6 mole of a monobasic strong acid is dissolved in water, the pH of the solution is ………..
8. When pH of a solution is 2, the [H⁺] in mol/L is ……….
9. Combustion of coal is an example of ………. reaction.
10. [ ] represents the concentration of either the reactant or product in ……….
Answer:
1. physical state
2. aqueous solution
3. metal, non-metal
4. CaCO₃
5. endothermic
6. electrolytic decomposition
7. 6
8. 1 × 10^-2
9. irreversible
10. mol/Lit

III. Match the following.

Question 1.
Match the following table:

Answer:
A. (v)
B. (iii)
C. (iv)
D. (ii)
E. (i)

Question 2.
Match the following table:

Answer:
A. (v)
B. (iv)
C. (i)
D. (iii)
E. (ii)

Question 3.
Match the following table:

Answer:
A. (iii)
B. (v)
C. (iv)
D. (ii)
E. (i)

Question 4.
Match the following table:

Answer:
A. (iv)
B. (v)
C. (i)
D. (ii)
E. (iii)

IV. True or False: (if false give the correct statement)

1. Formation of calcium silicate from silica and calcium oxide is a combination reaction.
2. Most of the combination reactions are endothermic in nature.
3. Decomposition of mercuric oxide into mercury and O₂ is an example of . photolysis.
4. Chlorine can displace fluoride ion from its aqueous solution.
5. Magnesium is more reactive than iron.

Answer:

1. True
2. False – Most of the combination reactions are exothermic in nature.
3. False – Decomposition of mercuric oxide into mercury and O₂ is an example of thermolysis.
4. False – Chlorine is less reactive than Fluorine, so it cannot displace fluoride ion from its aqueous solution.
5. True

V. Short answer questions:

Question 1.
Write a note on double displacement reaction with an example.
Answer:
When two compounds react, if their ions are interchanged, then the reaction is called double displacement reaction.
Eg: NaOH_(aq) + HCl_(aq) → NaCl_(aq) + H₂O

Question 2.
Identify the wrong statements and correct them.

1. Sodium benzoate is used in food preservative.
2. Nitric acid is not used as fertilizer in agriculture.
3. Sulphuric acid is called the king of chemicals.
4. The pH of acid is greater than 7.
5. Acetic acid is used in aerated drinks.

Answer:

1. Correct statement.
2. Wrong statement. Nitric acid is used as a fertilizer in agriculture.
3. Correct statement.
4. Wrong statement. The pH of the acid is lesser than 7.
5. Wrong statement. Carbonic acid is used in aerated drinks.

Question 3.
Why a combustion reaction may be called as an exothermic oxidation?
Answer:
In a combustion reaction heat is evolved, it is an exothermic reaction. As oxygen is added, it is also an oxidation. So, combustion may be called as an exothermic oxidation.

Question 4.
Take two conical flasks. Label them as I and II. Take a small amount of copper sulphate solution in the first conical flask. Take a small amount of granulated zinc in the second conical flask. Allow the copper sulphate solution to react with the zinc.

1. Name the type of reaction.
2. Say whether the metal zinc is more reactive or less reactive.
3. Write a complete and balanced reaction.
4. Say whether this change is reversible or irreversible.

Answer:

1. The reaction taken place is displacement reaction.
2. Metal zinc is more reactive.
3. Balanced chemical equation.
   \(\mathrm{Zn}_{(\mathrm{s})}+\mathrm{CuSO}_{4(\mathrm{aq})} \rightarrow \mathrm{ZnSO}_{4(\mathrm{aq})}+\mathrm{Cu}_{(\mathrm{s})} \downarrow\)
4. This change is an irreversible change.

Question 5.
What is an irreversible reaction? Give an example.
Answer:
The reaction that cannot be reversed is called irreversible reaction. The irreversible reactions are unidirectional, i.e., they take place only in the forward direction. Consider the combustion of coal into carbon dioxide and water.

Question 6.
Define the rate of a reaction.
Answer:
“Rate of a reaction is the change in the amount or concentration of any one of the reactants or products per unit time”.
Consider the following reaction,
A → B
The rate of this reaction is given by
Rate = –\(\frac{d[A]}{dt}\) = +\(\frac{d[B]}{dt}\)
Where,
[A] – Concentration of A
[B] – Concentration of B
The negative sign indicates the decrease in the concentration of A with time. The positive sign indicates the increase in the concentration of B with time.

Question 7.
What is meant by combination reaction? Give an example.
Answer:
A reaction in which a single product is formed from two or more reactants is known combination reaction.
2Mg + O₂ → 2MgO.

Question 8.
What is a catalyst?
Answer:
A catalyst is a substance which increases the reaction rate without being consumed in the reaction.

Question 9.
Define Displacement reaction. Give an example.
Answer:
The reaction in which a more reactive element displaces a less reactive element from its compound is called displacement reaction.
\(\mathrm{Pb}+\mathrm{CuCl}_{2} \longrightarrow \mathrm{PbCl}_{2}+\mathrm{Cu} \downarrow\)
Lead displaces copper from copper chloride solution.

Question 10.
When a aerated soft drink bottle is kept open it will go flat. Why?
Answer:
(i) In the sealed aerated soft drink bottle, the dissolved CO₂, in the form of carbonic acid and gaseous CO₂ are in equilibrium.
(ii) When we open the bottle the gaseous CO₂ will escape and the dissolved CO₂ begins to undissolve to the gas phase to maintain the equilibrium. So when we keep the bottle open for a long time it will go flat with all the dissolved CO₂ gone.

Question 11.
Can copper displace zinc or lead from their salt solutions?
Answer:
No, copper cannot displace zinc or lead from their salt solutions. Because copper is less reactive than zinc and lead.

Question 12.
What is called as acid rain?
Answer:
The pH of rain water is approximately ‘7’ but when the air is polluted with oxides of S and N, they get dissolved in the rain water and make its pH less than 7, then it is called acid rain.

Question 13.
Write the differences between combination and decomposition reaction.
Answer:

VI. Answer in detail:

Question 1.
Two acids ‘A’ and ‘B’ were kept in beakers. Acid ‘A’ undergoes partial dissociation in water, whereas acid ‘B’ undergoes complete dissociation in water.

1. Of the two acids ‘A’ and ‘B’ which is weak acid and which is strong acid?
2. What is a weak acid?
3. What is a strong acid?
4. Give one example of each.

Answer:

1. Of the two acids ‘A’ and ‘B’, ‘A’ is a weak acid and ‘B’ is a strong acid.
2. A weak acid is the one which ionises partially when dissolved in water.
3. A strong acid is the one which ionises completely when dissolved in water.
4. Weak acid – Acetic acid(CH₃COOH)
   Strong acid – Sulphuric acid (H₂SO₄).

Question 2.
Sodium hydroxide and HCl acid react as shown in this equation
NaOH_(aq)4 + HCl_(aq) → NaCl_(aq) + H₂O
(i) Which type of chemical reaction is this?
Answer:
Neutralization reaction

(ii) The reaction is exothermic. Explain what that means?
Answer:
When heat is evolved during a chemical reaction it is called exothermic.

(iii) Differentiate exothermic and endothermic reaction.
Answer:
Exothermic :

1. Heat is evolved.
2. Temperature increases.

Endothermic :

1. Heat is absorbed.
2. Temperature decreases.

(iv) What happens to the temperature of the solution as the chemicals react?
Answer:
Temperature of the solution increases.

Question 3.
Take two conical flasks. Label them as I and II. Take a small amount of CuSO₄ in the I conical flask and small amount of granulated Zinc in the II conical flask. Allow the CuSO₄ solution to react with Zinc.
(i) Name the type of reaction.
Answer:
Displacement reaction.

(ii) Say whether the metal Zn is more reactive or less reactive.
Answer:
Zinc is more reactive than Copper.

(iii) Write the complete and balanced reaction.
Answer:
Zn_(s)+ CuSO_4(aq) → ZnSO_4(aq) + Cu_(s)↓

(iv) Say whether this change is reversible or irreversible
Answer:
Irreversible.

Question 4.
Suggest a reason for each observation given below.

1. In fireworks, powdered magnesium is used rather than magnesium ribbon.
2. Zinc and dilute H₂SO₄ react much more quickly when a few drops of copper sulphate solutions are added.
3. The reaction between magnesium carbonate and dilute hydrochloric acid speeds up when some concentrated HCl is added.

Answer:

1. In fireworks, powdered magnesium is used because it has more surface area than magnesium ribbon. Because of more surface area, powdered magnesium reacts faster than magnesium ribbon.
2. Zinc and dilute H2SO4 react much more quickly when a few drops of copper sulphate solution is added. Copper sulphate acts as a catalyst which increases the rate of the reaction.
3. When the concentration of the reactants increases, the rate of the reaction also increases. So the reaction between magnesium carbonate and dilute hydrochloric acid speeds up by the addition of some concentrated HCl.

VII. Hot Questions:

Question 1.

Answer:
According to the rate Law for the reaction.
2N₂O₅ → 4NO₂ + O₂

2K₁ = K₂ = 4K₃ [neglecting negative and positive signs]

Question 2.
When solutions of silver nitrate and potassium bromide are mixed, a pale yellow precipitate is formed.
The ionic equation for the reaction is Ag⁺ + Br^– → AgBr

1. (a) What is the name of the pale yellow precipitate?
   (b) Is it soluble or insoluble?
2. Is the formation of silver bromide precipitate, a result of redox reaction or not? Justify your answer.
3. What is this type of reaction called?

Answer:

1. (a) The pale yellow precipitate is silver bromide.
   (b) Silver bromide is sparingly soluble.
2. Yes, the formation of silver bromide precipitate is due to redox reaction. Ag⁺ gains electron (reduction) and Br^– loses electron (oxidation). So it is a redox reaction.
3. This reaction is a double displacement reaction.

VIII. Numericals:

Question 1.
The hydrogen ion concentration of a fruit juice is 3.3 × 10^-2M. What is the pH of the juice? Is it acidic or basic?
Answer:
[H⁺] = 3.3 × 10^-2 M
Formula:
pH = -log[H⁺]
pH = -log[3.3 × 10^-2]
= -[log 3.3 + log 10^-2]
= -[0.5185 – 2 log₁₀10]
= -0.5185 + 2 log₁₀10 [log₁₀10 = 1]
= -0.5185 + 2
pH = 1.4815
Since pH is less than 7, the solution is acidic.

Question 2.
If a solution has a pH of 7.41, determine its H⁺ concentration.
Answer:
pH = 7.41
Formula:
[H⁺] = antilog[-pH]
= antilog[-7.41] = 10^-7.41
= 3.89 × 10^-8M

Question 3.
The pH of a solution is 5.5 at 25°C. Calculate its [OH^–].
Answer:
pH = 5.5
Since pH + pOH = 14
pOH = 14 – pH
= 14 – 5.5 = 8.5
[OH^–] = 10^-pOH
= 10^-8.5
= 3.16 × 10^-9M
(OR)
[OH^–] = antilog[-8.5]
= 3.16 × 10^-9M
[OH^–] = 3.16 × 10^-9M

Question 4.
Calculate the pH of 0.001 M HCl solution.
Answer:
Since HCl is a strong acid, it dissociates
HCl → H⁺ + Cl^–
[H⁺] = [HCl] = 0.001 M
[H⁺] = 1 × 10^-3M
pH = -log[H⁺]
= -log[1 × 10^-3]
pH = 3

Question 5.
The hydrogen ion concentration of a solution is 1 × 10^-8M.
(i) What is the pH of the solution?
Answer:
pH = -log[1 × 10^-8]
pH = 8

(ii) What is the pOH of the solution?
Answer:
pOH = 14 – 8
= 6

(iii) Is the given solution acidic or basic?
Answer:
Acidic

Question 6.
Calculate the pH of 0.02 M Ba(OH)₂, Ba(OH)₂ solution is a strong electrolyte.
Answer:
[Ba(OH)₂] = 0.02 M
Ba(OH)₂ → Ba²⁺ + 2OH^–
[OH^–] = 2[Ba(OH)₂]
= 2 × 0.02 = 0.04 M
pOH = – log[0.04]
= -log[4 × 10^-2]
= -[log 4 + log 10^-2]
= -[0.6020 – 2 log 10]
pOH = -0.6020 + 2 × 1
= 1.398
pH = 14 – 1.398
= 12.602

Students can download 10th Social Science Geography Chapter 3 Components of Agriculture Questions and Answers, Notes, Samacheer Kalvi 10th Social Science Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Social Science Solutions Geography Chapter 3 Components of Agriculture

Samacheer Kalvi 10th Social Science Components of Agriculture Text Book Back Questions and Answers

I. Choose the correct answer

Question 1.
The soil which is rich in iron oxides is
(a) Alluvial
(b) Black
(c) Red
(d) Alkaline
Answer:
(c) Red

Question 2.
Which of the following organization has divided the Indian soils into 8 major groups?
(a) Indian Council of Agricultural Research
(b) Indian Meteorological Department
(c) Soil Survey of India
(d) Indian Institute of Soil Science
Answer:
(a) Indian Council of Agricultural Research

Question 3.
The soils formed by the rivers are:
(a) Red soils
(b) Black soils
(c) Desert soils
(d) Alluvial soils india.
Answer:
(d) Alluvial soils india.

Question 4.
……….. dam is the highest gravity in India.
(a) Hirakud dam
(b) Bhakra Nangal dam
(c) Mettur dam
(d) Nagaijuna Sagar dam
Answer:
(b) Bhakra Nangal dam

Question 5.
……………… is a cash crop.
(a) Cotton
(b) Wheat
(c) Rice
(d) Maize
Answer:
(a) Cotton

Question 6.
Black soils are also called as ………..
(a) Arid soils
(b) Saline soils
(c) Regur soils
(d) Mountain soils
Answer:
(c) Regur soils

Question 7.
The longest dam in the world is:
(a) Mettur dam
(b) Kosi dam
(c) Hirakud dam
(d) Bhakra Nangal dam
Answer:
(c) Hirakud dam

Question 8.
The leading producer of rice in India is ……….
(a) Punjab
(b) Maharashtra
(c) Uttar Pradesh
(d) West Bengal
Answer:
(c) Uttar Pradesh

Question 9.
Which crop is called as “Golden Fibre” in India?
(a) Cotton
(b) Wheat
(c) Jute
(d) Tobacco
Answer:
(c) Jute

Question 10.
The state which leads in the production of coffee is ……….
(a) West Bengal
(b) Karnataka
(c) Odisha
(d) Punjab
Answer:
(b) Karnataka

II. Consider the given statements and choose the right option given below

Question 1.
Assertion (A): Horticulture involves the cultivation of fruits, vegetables, and flowers.
Reason (R): India ranks first in the world in the production of mango, banana, and citrus fruits.
(a) Both (A) and (R) are true and (R) explains (A)
(b) Both (A) and (R)are true: (R) does not explain (A)
(c) (A) is correct (R) is false
(d) (A) is false (R) is true
Answer:
(c) (A) is correct (R) is false

Question 2.
Assertion(A): Alluvial soil is formed by the deposition of eroded and decayed materials brought by the rivers.
Reason(R): Paddy and wheat are grown well in the soil.
(a) Both (A) and (R) are true and (R) explains (A)
(b) Both (A) and (R)are true and (R) does not explain (A)
(c) (A) is correct (R) is false
(d) (A) is false (R) is true
Answer:
(b) Both (A) and (R)are true and (R) does not explain (A)

III. Pick the odd one out

Question 1.
(a) Wheat
(b) Rice
(c) Millets
(d) Coffee
Answer:
(d) Coffee

Question 2.
(a) Khadar
(b) Bhangar
(c) Alluvial soil
(d) Black soil
Answer:
(d) Black soil

Question 3.
(a) Inundation canals
(b) Perennial canals
(c) Tanks
(d) Canals
Answer:
(c) Tanks

IV. Match the following

Answer:
A. (iv)
B. (iii)
C. (v)
D. (i)
E. (ii)

V. Answer in brief

Question 1.
Define soil.
Answer:
Soil is the uppermost layer of the land surface, usually composed of minerals, organic matter, living organisms, air and water. Its formation is mainly related to the parent rock material, surface relief, climate and natural vegetation.

Question 2.
Name the types of soil found in India.
Answer:
There are 8 major groups of soil found in India classified by the Indian Council of Agricultural Research. They are

1. Alluvial soil
2. Black soil
3. Red soil
4. Forest and Mountain soil
5. Arid and Desert soil
6. Laterite soil
7. Saline and Alkaline soil
8. Peaty and Marshy soil

Question 3.
State any two characteristics of black cotton soil.
Answer:

1. This soil is rich in calcium carbonate, magnesium, potash, lime and iron but deficient in phosphorous. It is clayey and impermeable which has great capacity to retain moisture for a long time.
2. It becomes sticky when wet but develops cracks during dry summer season. The soil is suited for dry farming due to its high moisture retentivity.

Question 4.
What is Multipurpose project?
Answer:

1. Multipurpose projects are the River valley projects.
2. Dams are constructed across rivers aims to serve many purposes such as irrigation, power generation, water supply, controlling floods, navigation, development of fisheries etc.

Question 5.
Define Agriculture.
Answer:
Agriculture is the process of producing food for people, fodder for cattle, fiber and many other desired products by the cultivation of certain plants and the raising of domesticated animals

Question 6.
State the types of agriculture practices in India?
Answer:

1. Owing to physical environment and culture the following cultivation systems prevail in India.
2. Subsistence farming
3. Intensive farming
4. Dry farming
5. Mixed farming agriculture
6. Terrace cultivation

Question 7.
Name the seasons of agriculture in India.
Answer:

1. Kharif Season
2. Rabi Season
3. Zaid Season

Question 8.
Mention the plantation crops of India.
Answer:

1. Plantation crops are mainly cultivated for the purpose of exports.
2. These crops are cultivated in large estates on hilly slopes.
3. Tea, coffee, rubber and spices are the main plantation crops of India.

Question 9.
What do you mean by livestock?
Answer:
Livestocks is defined as farm animals who are raised to generate a profile. It is an integral component of the farming system in India.

Question 10.
Write a brief note on the categories of fisheries in India?
Answer:
In India fisheries are categorised into two.They are marine or sea fisheries and inland or fresh water fisheries.

Marine or Sea fisheries:

1. It includes coastal off-shore and deep sea fishing mainly on continental shelf upto a depth of 200 mts.
2. Kerala leads in Marine fish production.

Inland or Fresh water fisheries:

1. Rivers, lakes, canals, reservoirs, ponds tanks etc are the sources of fresh water fisheries,
2. Andhra Pradesh is leading in the fresh water fisheries.
3. About 50% of the country’s total fish production comes from inland fisheries.

VI. Give reasons

Question 1.
Agriculture is the backbone of India.
Answer:
Agriculture provide food for the entire population. It supplies raw materials to the agro-based industries. It contributes to export trade. So, agriculture is the backbone of India.

Question 2.
Rain water harvesting is necessary.
Answer:
India experiences Tropical monsoon climate. Rainfall in India is seasonal, irregular and uneven and highly eractic. Hence it is necessary’ to save rain water when it is available and use in times of need. In order to prevent surface . run off Rainwater harvesting is needed.

Question 3.
Small farms are predominant in India.
Answer:
The predominant type of Indian agriculture is subsistence farming. In this agriculture land holding is small and half of the production is used for family consumption and the rest is sold in the nearby markets. The farmers concentrate on stage food crops like rice and wheat. As the farmers are poor, they can’t apply the modem inputs which cost more.

VII. Distinguish between the following

Question 1.
Rabi and Kharif crop seasons.
Answer:
Rabi crop season:

1. Rabi crop season is from October – March.
2. Seeds are sown in winter and harvested before summer.
3. Major crops of this season are mainly Wheat, Mustard, Maize etc.

Kharif crop season:

1. Kharif crop season is from June – September.
2. Seeds are sown after the summer monsoon and harvested before winter.
3. Major crops are Rice, Cotton, Groundnut, Turmeric etc.

Question 2.
Inundation canal and perennial canal.
Answer:
Inundation canal:

1. Undependable source of irrigation.
2. Operational only during flood in rivers and not have weir system to regulate water.

Perennial canal:

1. Dependable source of irrigation.
2. Have weir system through barrage to regulate water from perennial rivers or dams.

Question 3.
Marine fishing and Inland fishing.
Answer:
Marine fishing:

1. Marine fishing includes coastal, off¬shore and deep sea fishing mainly on the continental shelf upto a depth of 200m.
2. Kerala is the leading producer in marine fishing.

Inland fishing:

1. Inland fishing is done in rivers, lakes, canals, ponds, tanks, reservoirs etc.
2. Andhra Pradesh, West Bengal, Gujarat, Kerala and TamilNadu are the leading states.

Question 4.
Alluvial soils and Black soils.
Answer:
Alluvial soil:

1. Formed by the sediments deposited by the rivers and streams.
2. Silt clay, sandy and loamy in nature.
3. Found in the river valleys and plains.
4. Rice, wheat, sugarcane grow well in this soil.

Black soil:

1. Formed by the disintegration of basalt rocks.
2. Sticky when wet develops cracks when dry.
3. Found in plateau region especially in Deccan trap.
4. Cotton, sugarcane and tobacco grow well in this soil.

VIII. Answer in a paragraph

Question 1.
State the types of soil in India and explain the characteristics and distribution of soil.
Answer:

Question 2.
Write about any two multipurpose projects of India.
Answer:
A comprehensive river valley project which serves a number of purposes simultaneously is called a “Multi purposes project”.

1. The Bhakra Nangal Project: India’s biggest multipurpose river valley project is ‘Bhakra Nangal Project’. It has been built at a strategic point where two hills on either side of the Sutlej are very close to each other. It is the highest gravity dam in the world. Its length is 226 metres from the river bed. The canals taken out are 1100 kilometres long. The ‘Nangal Power Plant’ on the Sutlej produces electricity, and serves the states of Himachal Pradesh, Punjab, Haryana, Rajasthan and New Delhi. The distributaries are 8400 km in length. It irrigates an area of 1.4 million hectares.

2. Indira Gandhi Project: This project is an ambitious scheme to bring new areas under irrigation so that more areas could be cultivated. The waters of the River Beas and the Ravi are diverted to the River Sutlej. The ‘Pong’ Dam on the River Beas has been constructed to divert the Beas water into the Sutlej in a regulated manner. So that ‘Rajasthan canal’, the longest irrigation canal in the world can irrigate Gandhi Nagar, Bikaner and Jaisalmer districts of North West Rajasthan, (i.e) a part of Thar desert. The main canal now called ‘Indira Gandhi Canal’ is 468 km long runs entirely in Rajasthan, Western of Sutlej, Beas and Ravi are now being fully utilised for irrigating thirsty lands of South Western parts of our country.

Question 3.
Bring out the characteristics of intensive and Plantation farming.
Answer:
Characteristics of Intensive farming:

1. Farming is done with intensification and mechanisation system to maximize yields from available land.
2. In this farming heavy use of pesticides and chemical fertilizers have been applied.
3. It has also been applied to the raising of live stock being held indoor as factory farms.
4. Plantation fanning.
5. Single crop raised on a large area testates).
6. Crops are grown mainly for export purpose,
7. The plantations are mostly owned by companies.

Question 4.
Examine the geographical conditions favourable for the cultivation of rice and wheat.
Answer:
Rice:
Rice is an indigenous crop. India is the second-largest producer of rice in the world. It is a tropical crop which is grown well in alluvial plains and river deltas. It requires a mean monthly temperature of 24°C and an average rainfall of 150 cm and deep fertile alluvial soil for its growth. It also needs an abundant supply of cheap labour. In areas of less rainfall particularly in Punjab and Haryana, it is grown with the help of irrigation.
Rice in India is sown in three ways:

1. Broadcasting,
2. Ploughing or drilling, and
3. Transplanting.

Due to increased use of High Yielding Variety (HYV) seeds (CR Dhan 205, AR Dhan 306, CRR 451 etc.), many of the indigenous varieties were disappeared. In 2016, the first 10 leading rice-producing states are West Bengal (First in India) Uttar Pradesh, Punjab, Tamil Nadu, Andhra Pradesh, Bihar, Chhattisgarh, Odisha, Assam, and Haryana.

Wheat:
Wheat is a temperate crop. Its cultivation is mainly confined to the northwestern part of India. India produces both winter wheat and spring wheat. It is the second most important food crop of the country, after rice.

It requires 10-15°C at the time of sowing and 20-25°C at the time of ripening of grains. Over 85% of India’s wheat production comes from 5 states namely Uttar Pradesh, Punjab, Haryana, Rajasthan and Madhya Pradesh. Apart from these regions, the black soil tract of the Deccan covering parts of Maharashtra and Gujarat also contribute a major wheat production.

IX. HOTS Questions

Question 1.
Can you imagine a world without agriculture?
Answer:
Agriculture is the art and science of cultivating the soil, growing crops and raising livestock. It includes the preparation of plant and animal products for people to use and their distribution to markets.

Agriculture provides most of the world’s food and fabrics. Cotton, wool and leather are all agricultural products. It also provides wood for construction and paper products.

A world without agriculture would be very different compared to the world we live in today. It is easy for us as humans to take for granted things when we don’t really know how they are grown or produced. Without agriculture, we may not get food and clothing. Without timber, we may not be able to build houses and furniture. We may also be deprived of some everyday-products like soap, paper and ethanol, which are made up of some sort of agricultural by-product. Modern medicine also depends on agriculture. Without agriculture, we wouldn’t be here. We must always be thankful for this industry and for those involved in the various activities related to it.

Question 2.
Can you give solutions for the prevailing water disputes in South India (construction of dams / raising of dams / cleaning of tanks}?
Answer:

1. Construction of check dams: In order to prevent surface run off water • during heavy rains. Check dams has to be constructed to regulate the flow of water.
2. Raising of dams: Improves the storage capacity, Strengthening the dams is a must.
3. Cleaning of tanks: Preventing encroachment by strict legal action, desilting the tanks and deepening it in the local areas will provide water supply to the locality, as well as to maintain the areas of ground water level cleaning of tanks is a must

X. Map exercise

Question 1.
Demarcate the major tracts of alluvial soils.
Answer:

Question 2.
Delineate the main regions of black soil.
Answer:

Question 3.
Locate the Hirakud dam, Mettur dam and Damodar dam.
Answer:

Question 4.
Shade the regions of jute cultivation.
Answer:

Question 5.
Mark any three tea and coffee growing areas.
Answer:

Question 6.
Demarcate the regions of desert soil.
Answer:

Question 7.
Locate the fishing hubs: Tuticorin, Chennai,Cochin, Mumbai, Machiiipatnam
Answer:

Question 8.
Demarcate: Cauvery delta, Godavari delta.
Answer:

TB. PNo: 115
Activity 1:

Question 1.
Soil Texture (sand, silt, clay) influence on some properties of soils including water holding capacity. Find out water holding capacity of soils which given above based on following table.
Answer:

TB. PNo: 131
Activity 2:

Question 1.
Complete the following table by your day to day life experience.

Answer:

Samacheer Kalvi 10th Social Science Components of Agriculture Additional Important Questions and Answers

I. Choose the correct answer

Question 1.
……………… soil is formed by the process of leaching.
(a) Alluvial
(b) Black
(c) Laterite
(d) Arid
Answer:
(c) Laterite

Question 2.
Cotton is a ………
(a) food crop
(b) cash crop
(c) dry crop
Answer:
(b) cash crop

Question 3.
……………… is one of the method of soil conservation.
(a) Deforestation
(b) Irrigation
(c) Water logging
(d) Afforestation
Answer:
(d) Afforestation

Question 4.
The “rice bowl of Tamil nadu” is ………..
(a) Madurai
(b) Chennai
(c) Thanjavur
Answer:
(c) Thanjavur

Question 5
……………… are useful for the diversion of flood water from the rivers during rainy season.
(a) Perennial canals
(b) Inundation canals
(c) Open wells
(d) Tube wells
Answer:
(b) Inundation canals

Question 6
The type of fanning which is practised in Punjab and Haryana is ………
(a) subsistence farming
(b) commercial farming
(c) wet farming
Answer:
(b) commercial farming

Question 7.
……………… project is constructed on the river Kaveri in TamilNadu.
(a) Tehri Dam
(b) Kosi Dam
(c) Mettur Dam
(d) Horakud Dam
Answer:
(c) Mettur Dam

Question 8.
In the regions with abundant rainfall ……. is grown.
(a) millet
(b) wheat
(c) rice
Answer:
(c) rice

Question 9.
The traditional farming method that results in low productivity is:
(a) mixed farming
(b) shifting agriculture
(c) intensive farming
(d) subsistence farming
Answer:
(d) subsistence farming

Question 10.
One of the important zaid crops is ………….
(a) rice
(b) wheat
(c) watermelon
Answer:
(c) watermelon

Question 11.
First Live-stock census was conducted in India in the year:
(a) 1819
(b) 1919
(c) 1618
(d) 1981
Answer:
(b) 1919

Question 12.
The National Research Center on Plant Biotechnology was established in ………
(a) 1985
(b) 1980
(c) 1990
Answer:
(a) 1985

II. Consider the given statements and choose the right option given below

Question 1.
Assertion (A): Indian agriculture is largely dominated by food crops. Reason (R): India has a large population.
(a) Both (A) and (R) are true and (R) explains (A)
(b) Both (A) and (R)are true: (R) does not explain (A)
(c) (A) is correct (R) is false
(d) (A) is false (R) is true
Answer:
(c) (A) is correct (R) is false

Question 2.
Assertion (A): Forest and mountain soils differ from region to region. Reason (R): Due to the absence of vegetative cover.
(a) Both (A) and (R) are true and (R) explains (A)
(b) Both (A) and (R)are true: (R) does not explain (A)
(c) (A) is correct (R) is false
(d) (A) is false (R) is true
Answer:
(c) (A) is correct (R) is false
Hint: Reason is due to mechanical weathering caused by snow, rain, temperature variation depending on climate.

Question 3.
Assertion (A): Terrace farming method is practiced on hill and mountain slopes. Reason (R): The availability of land is limited and it checks soil erosion.
(a) Both (A) and (R) are true and (R) explains (A)
(b) Both (A) and (R)are true: (R) does not explain (A)
(c) (A) is correct (R) is false
(d) (A) is false (R) is true
Answer:
(a) Both (A) and (R) are true and (R) explains (A)

Question 4.
Assertion (A): Pulses are usually rotated with other crops.
Reason (R): They are used as human food and feeding cattle.
(a) Both (A) and (R) are true and (R) explains (A)
(b) Both (A) and (R) are true and (R) does not explain (A)
(c) (A) is correct (R) is false
(d) (A) is false (R) is true
Answer:
(c) (A) is correct (R) is false
Hint: They are mostly leguminous in nature and fixes atmospheric nitrogen helps soil to regain its fertility.

III. Pick the odd one out

1. (a) Tea, (b) Sugarcane, (c) Cotton, (d) Ragi
2. (a) Khandsari, (b) Herrings, (c) Mackerels, (d) Eels
3. (a) Damodhar, (b) Mahanadi, (c) Kaveri, (d) Raingun
4. (a) Water wheel, (b) Sprinkler, (c) Open well, (d) poly house
5. (a) Blue Revolution, (b) soil erosion, (c) Green Revolution, (d) Soil erosion
6. (a) Fishing, (b) Kharif, (c) Rabi, (d) Zaid

Answer:

1. (d) Ragi
2. (a) Khandsari
3. (d) Raingun
4. (c) Open well
5. (d) Soil erosion
6. (a) Fishing

IV. Match the following

Question 1.
Match the Column I with Column II.

Answer:
A. (v)
B. (i)
C. (ii)
D. (iii)
E. (iv)

Question 2.
Match the Column I with Column II.

Answer:
A. (ii)
B. (iv)
C. (v)
D. (i)
E. (iii)

Question 3.
Match the Column I with Column II.

Answer:
A. (vi)
B. (iii)
C. (ii)
D. (i)
E. (iv)
F. (v)

V. Answer in brief

Question 1.
Why is tank irrigation more prevalent in South India than North India?
Answer:
The undulated topography of peninsular region forms large depressions to collect water which become tanks. So the tank irrigation is more prevalent in South India than North India.

Question 2.
How laterite soils are formed?
Answer:
Laterite soils are formed in the regions where alternate wet and dry conditions prevail.
It is formed by the process of leaching.

Question 3.
Which is an acute problem in India? How?
Answer:
Soil degradation is an acute problem in India. According to a 2015 report of the Indian institute of remote sensing (IIRS). The estimated the amount of soil .erosion that occurred in India was 147 million hectares.
The main problems of the Indian soils are

1. Soil erosion
2. Degradation of Soil
3. Water-logging
4. Saline and Alkaline and
5. Salt Flats.

Question 4.
Draw a flow chart showing the sources of irrigation and their classification.
Answer:

Question 5.
Write few lines about Rain Gun.
Answer:
It is used to spread water like rain and used to serve for crops which used to grow up to 4 feet or high also but we have to adjust sprinklers height as per crop size, typical usage of Rain guns are in sugarcane, maize crops.

Question 6.
Name the food crops grown in India.
Answer:
Food crops that are grown in India include cereals and pulses amongst which rice, wheat, jowar, bajra, maize, barely, ragi, gram and tur are important.

Question 7.
Give short notes about Terrace cultivation.
Answer:
This is practised especially in hilly areas, where lands are of sloping nature. The hill and mountains slopes are cut to form terraces and the land is used in the same way as in permanent agriculture.

Question 8.
Name the five leading states that contribute 85% of India’s wheat production.
Answer:
Uttar Pradesh, Punjab, Haryana, Rajasthan and Madhya Pradesh are the leading states in wheat production in India.

Question 9.
What are the products of Sugarcane?
Answer:
Sugarcane provides raw material for the sugar industry. Besides providing sugar, gur and Khandsari it supplies molasses for alcohol industry and bagasse for paper industry.

Question 10.
How jute is useful for trade?
Answer:

1. Jute fibre provides raw material for Jute industry.
2. It is used for manufactruing of gunny bags, carpets, hessian, ropes and strings, rugs, clothes, tarpaulins, upholstery etc.

Question 11.
What is the other name of ‘Shifting Agriculture1? Who practice shifting agriculture? and how it is called in different regions?
Answer:
‘Shifting Agriculture’ is also called as ‘Slash and bum’ Agriculture. The tribal people follow this type of agriculture. It is called by different names in different regions in India as follows.

Question 12.
What is dry farming?
Answer:
Dry farming is the type of farming practiced in arid areas where there is lack of irrigation facilities. Crops grown in these areas can withstand dry conditions.

Question 13.
What are called cash crops? Give examples.
Answer:
Crops which are cultivated for commercial purpose are called cash crops, eg: Sugarcane, tobacco, fibre crops and oil seeds.

Question 14.
Which is the largest oil seeds producing state? What is the position of India in oil seeds production in the world?
Answer:
Gujarat is India’s largest oil seeds producing state. India is the second largest producer of oil seeds in the world next to China.

Question 15.
Define Green Revolution.
Answer:
The process of improving and increasing the production of food crops using modem techniques is referred as Green Revolution.

Question 16.
Mention the different breeds of cattle population in India.
Answer:
Cattle population in India belongs to different breeds. They are

1. Milch breed
2. Draught breed
3. Mixed or General breed

VI. Give Reasons

Question 1.
Tank irrigation is popular in peninsular India.
Answer:
The undulating relief, absence of perennial rivers, Impermeable rock structure and natural depression are the reasons for having tank irrigation most popular in Peninsular India.

Question 2.
Why tea is grown on the hill slopes?
Answer:
Tea plants require high rainfall but its roots cannot tolerate water logging. Frost condition is a must for tea plants. So it is grown on the hill slopes.

Question 3.
Inundation canals are not dependable source of irrigation.
Answer:
Inundation canals are operational only during rainy season for the diversion of flood water directly from the rivers. Hence these canals are not dependable source of irrigation.

Question 4.
Shifting agriculture is also called as slash and burn agriculture.
Answer:
Once the forest land piece is cleared by tribal people they grow crops for two to three years and bum the stumps after harvest and abandoning the lands to regain its fertility and then move to new areas. Thus, shifting agriculture is also called ‘slash and bum’ agriculture.

Question 5.
Only little surplus is left in subsistence farming.
Answer:
Mainly in subsistence farming crops are grown for family consumption leaving little surplus to sell in the market. Preference will be given to food crops as per their needs.

Question 6.
Indian agriculture is mainly dominated by the food crops.
Answer:
As the population increases the need for food also increase. Due to the need and to supply to the large population Indian agriculture is mainly dominated by the food crops.

VII. Distinguish between the following

Question 1.
Laterite and Red soil.
Solution:
Laterite soil:

1. Femed by the process of leaching under wet and hot dry conditions.
2. Mainly composed of hydrated oxides of iron and aluminium.
3. Found mostly on hilly areas.
4. Suitable for tea, coffee, rubber and cashewnut.

Red soil:

1. Formed by the decomposition of ancient crystalline rocks.
2. Rich in minerals like iron and magnesium.
3. Found in plateau regions.
4. Suitable for pulses and sugarcane.

Question 2.
Arid and desert soil and Peaty and marshy soil.
Answer:
Arid and Desert soil:

1. Formed due to dry climate and high temperature.
2. Poor in organic matter and nitrogen rich in salt content.
3. Found in Rajasthan Northern Gujarat and Southern Punjab.
4. Millets, barley and pulses grow with irrigation.

Peaty and Marshy soil:

1. Formed in humid regions from organic matter.
2. Poor in potash and phosphate rich in vegetable organic matter.
3. Found in coastal areas and Sunderban Deltaic region, Kerala, Odisha, West Bengal.
4. Ideal for paddy and Jute cultivation.

Question 3.
Open well and Tube well irrigation.
Answer:
Open well irrigation:

1. Need sufficient ground water availability.
2. Practiced in areas of Ganga plains and river Deltaic regions.

Tube well irrigation:

1. Can be found in areas of low water table and soft geological subsurface.
2. Need sufficient water supply. Predominant in the states of Gujarat, Maharashtra, Madhya Pradesh and TamilNadu.

Question 4.
Bhakra Nangal and Hirakud project.
Answer:
Bhakra Nangal Project:

1. Constructed on the river Sutlej.
2. Highest gravity dam in the world.
3. Punjab, Haryana and Rajasthan states are benefitted.

Hirakud Project:

1. Constructed on the river Mahanadhi.
2. Longest dam in the world.
3. State of Odhisha is benefitted by this project.

Question 5.
Subsistence and Mixed farming.
Answer:
Subsistence farming:

1. Crops is mainly cultivated for family consumption.
2. Traditional farming methods are used due to small land holdings.
3. Mainly concentrated on food crops.

Mixed farming:

1. Integrated farming to satisfy many needs of the farmers.
2. Modem techniques is used for crop cultivation done in a vast area.
3. System include crop production along with raising live stock, poultry, fisheries, bee keeping etc.

Question 6.
Food and Cash crops.
Answer:
Food crops:

1. Mainly grown for consumption purpose.
2. Food crops include rice, wheat, pulses etc.
3. Apart from consumption used as raw materials for agro based industries.

Cash crops:

1. Crops are cultivated for commercial purpose.
2. Sugarcane, cotton, jute, oil seeds are the main cash crops. .
3. Mainly produced for raw materials for industries and earn valuable foreign exchange.

Question 7.
Rice and Wheat.
Answer:
Rice:

1. Mainly Tropical crop.
2. Need abundant supply of water till harvest.
3. West Bengal, Uttar Pradesh, Punjab, TamilNadu, Andhra Pradesh are some of the leading states in Rice production.

Wheat:

1. Sub tropical or Temperate crop.
2. Need moderate water supply.
3. Uttar Pradesh, Punjab, Haryana, Rajasthan and Madhya Pradesh are the leading states in wheat production.

VIII. Answer in a paragraph

Question 1.
What is a multipurpose project and explain its purpose?
Answer:
A comprehensive river valley project which serves a number of purposes simultaneously is called a “Multipurpose project”. Multipurpose projects serve the following purposes.

1. They help to store water, that can be utilised, when water, is in great demand both for agricultural and domestic purpose.
2. They check floods and famines.
3. Afforestation is undertaken in the catchment areas of river, which helps conservation of water, soil and wildlife. Thus it keeps ecosystem intact.
4. Production of hydro electricity is also another purpose of these projects. It is a pollution free of energy and is renewable energy.
5. They attract tourists and develops tourism industry.
6. Soil conservation and land relamation are other purpose of these projects.

Question 2.
What do you mean by Irrigation? Why it is necessary in India?
Answer:

1. Watering to agricultural lands by artificial means for cultivation is called Irrigation.
2. India is a tropical monsoon country.
3. 75% of rainfall is received within a span of four months.
4. Indian rainfall from monsoons are seasonal, uneven, irregular and erratic in nature.
5. So always there is a need for irrigation to carry out agricultural activities during dry period.
6. Besides erratic rainfall prevalence of high temperature, cultivation of annual crops and hydrophytes.
7. Commercial farming and porous soil make irrigation an essential one for agriculture in India.

Question 3.
What are the different sources of irrigation used in different parts of India?
Answer:
The main sources of irrigations are
1. Canal irrigation
2. Well irrigation
3. Tank irrigation

1. Canal irrigation:
It is the second most important sources of irrigation in India. The Canals are of two types.
i. Inundation canals
ii. Perennial canals

i. Inundation canals:
In this, water is taken directly from the rivers without making any kind of barrage or dam. Such canals are useful for the diversion of flood water from the rivers and remains operational during rainy season.

ii. Perennial canals:
These are developed from perennial work by constructing barrage to regulate the flow of water. These canals are useful for irrigation.

2. Well irrigation:
A Well is a hole or trough usually vertical excavated in the earth for brining groundwater to the surface. It contributes about 62% of net irrigated area in India. Wells are of two types.
i. Open well
ii. Tube well

i. Open well:
This type of irrigation is widely practiced in the areas where groundwater is sufficiently available. The areas are in Ganga plains, the deltaic region of Mahanadi, Godavari, Krishna, Cauvery and parts of Narmada and Tapti valleys.

ii. Tube well:
Tube wells are developed in the areas of low water table, sufficient power supply and soft subsurface geological units. Tube wells are predominant in the states of Gujarat, Maharashtra, Punjab, MP, and Tamil Nadu.

3. Tank irrigation:
A tank is a natural or man-made hollow on the surface developed by constructing a small bund around it across a stream It is used to collect and store water for irrigation and other purpose. It also includes irrigation from taken and ponds.

Question 4.
How is Live-stock census conducted in TamilNadu?
Answer:
State Government is conducting Live stock census with the help of:

1. Department of Animal Husbandary at State level and
2. Regional Joint Director at District level under the guidelines of Government of India ministry of Agriculture and farmers welfare,
3. Department of Animal husbandary, Dairying and Fisheries conducted once in 5 years.

Question 5.
Explain about the major issues faced by the Indian farmers?
Answer:
We can divide the problems faced by the Indian agriculture and by the farmers into two Natural and Man made.
Natural problems:

1. Soil erosion: Large tracts of land suffer from soil erosion by wind and water.
2. Infertile soil: Growing crops for many years without replenishing led to the exhaution of soil and its depletion.
3. Lack of Irrigation: Only some areas of the cropped falls under irrigation.

Man made problems:

1. Small land holdings and fragmented land: Poor status made the farmers to have small land and also some farmers possesses share from their ancestral property cannot afford to apply mechanism.
2. High costs of Inputs: Good quality of seeds are out of reach for many small and marginal farmers due to their high price.
3. Agricultural marketing: Absence of sound marketing, facility, interference of local traders and middlemen for the disposal of their produce, fluctuations in the price.
4. In adequate transport: Lakhs of villages are not w7ell connected with main roads or with market centres. Lack of cheap and efficient means of transportation is not available.
5. Scarcity of capital: Huge capital is needed to purchase advanced farm machineries and equipments which the poor farmers cannot afford to buy.

Question 6.
List out some of the Agricultural Revolutions in India.
Answer:

Question 7.
“Per Drop more Crop”. Explain this statement.
Answer:

1. It is the scheme introduced by the Government of India with the objective to enhance water use efficiency.
2. This micro irrigation scheme comes under “Pradhan Mantri Krishi Sinchayee Yojana (PMKSY) and centrally sponsored scheme on micro irrigation.
3. It promotes appropriate technological interventions like drop and sprinkler irrigation technologies in agriculture and encourage farmers to use water saving and conservation technique.
4. Following five states progressed 78% under micro irrigation scheme
   • Andhra Pradesh
   • Karnataka
   • Gujarat
   • Maharashtra
   • TamilNadu

IX. Map Questions
Mark the following on the outline map of India.

Question 1.
Areas of Red soil and Mountain soil.
Answer:

Question 2.
Demarcate the states producing the highest for the following crops.
Answer:
(i) Paddy

(ii) Sugarcane

(iii) Cotton

(iv) Wheat

Question 3.
Kosi project, Tungabadra, Periyar Dam, Nagarjuna Sagar.
Answer:

Students can Download Tamil Nadu 12th Physics Model Question Paper 5 English Medium Pdf, Tamil Nadu 12th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Physics Model Question Paper 5 English Medium

General Instructions:

• The question paper comprises of four parts.
• You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
• All questions of Part I, II, III, and IV are to be attempted separately.
• Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
  These are to be answered by choosing the most suitable answer from the given four
  alternatives and writing the option code and the corresponding answer
• Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered
  in about one or two sentences.
• Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered
  in about three to five short sentences.
• Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered
  in detaiL Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 70

Part – I

Answer all the questions. Choose the correct answer. [15 x 1 = 15]

Question 1.
If voltage applied on a capacitor is increased from V to 2V, Choose the correct conclusion,
(a) Q remains the same, C is doubled
(b) Q is doubled, C doubled
(c) C remains same, Q doubled
(d) Both Q and C remain same
Answer:
(c) C remains same, Q doubled

Question 2.
The electric field in the region between two concentric charged spherical shells
(a) is zero
(b) increases with distance from centre
(c) is constant
(d) decreases with distance from centre
Answer:
(d) decreases with distance from centre

Question 3.
In India electricity is supplied for domestic use at 220 V. It is supplied at 110 V in USA. If the resistance of a 60W bulb for use in India is R, the resistance of a 60W bulb for use in USA will be …………
(a) R
(b) 2R
(c) \(\frac{\mathbf{R}}{4}\)
(d) \(\frac{\mathbf{R}}{2}\)
Answer:
(c) \(\frac{\mathbf{R}}{4}\)

Question 4.
The vertical component of Earth’s magnetic field at a place is equal to the horizontal component. What is the value of angle of dip at this place?
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer:
(b) 45°

Question 5.
A step-down transformer reduces the supply voltage from 220 V to 11 V and increase the current from 6 A to 100 A. Then its efficiency is
(a) 1.2
(b) 0.83
(c) 0.12
(d) 0.9
Answer:
(b) 0.83

Question 6.
Faraday’s law of electromagnetic induction is related to the …………….
(a) Law of conservation of charge
(b) Law of conservation of energy
(c) Third law of motion
(d) Law of conservation of angular momentum
Answer:
(b) Law of conservation of energy

Question 7.
The dimension of \(\frac{1}{\mu_{0} \varepsilon_{0}}\) is ……….
(a) [L T^-1]
(b) [L² T^-2]
(c) [L^-1 T]
(d) [L^-2 T²]
Answer:
(b) [L² T^-2]

Question 8.
In a Young’s double-slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to,…………..
(a) 2D
(b) \(\frac{\mathrm{D}}{2}\)
(c) \(\sqrt{2} \mathrm{D}\)
(d) \(\frac{\mathrm{D}}{\sqrt{2}}\)
Answer:
(a) 2D

Question 9.
The speed of light in an isotropic medium depends on,
(a) its intensity
(b) its wavelength
(c) the nature of propagation
(d) the motion of the source w.r.to medium
Answer:
(b) its wavelength

Question 10.
The mean wavelength of light from sun is taken to be 550 nm and its mean power is 3.8 x 10²⁶ W. The number of photons received by the human eye per second on the average from sunlight is of the order of ……………..
(a) 10⁴⁵
(b) 10⁴²
(c) 10⁵⁴
(d) 10⁵¹
Answer:
(a) 10⁴⁵
Hint:

Question 11.
As the intensity of incident light increases
(a) kinetic energy of emitted photoelectrons increases
(b) photoelectric current decreases
(c) photoelectric current increases
(d) kinetic energy of emitted photoelectrons decreases
Answer:
(c) photoelectric current increases
Hint: As the intensity of incident light increases, photoelectric current increases.

Question 12.
In J.J. Thomson e/m experiment, a beam of electron is replaced by that of muons (particle with same charge as that of electrons but mass 208 times that of electrons). No deflection condition is achieved only if
(a) B is increased by 208 times
(b) B is decreased by 208 times
(c) B is increased by 14.4 times
(d) B is decreased by 14.4 times
Answer:
(c) B is increased by 14.4 times
Hint: In the condition of no deflection \(\frac{e}{m}=\frac{\mathrm{E}^{2}}{2 v \mathrm{B}^{2}}\)
If m is increased by 208 times then B should be increased \(\sqrt{208}\) = 14.4 times

Question 13.
A forward biased diode is treated as
(a) An open switch with infinite resistance
(b) A closed switch with a voltage drop of 0V
(c) A closed switch in series with a battery voltage of 0.7V
(d) A closed switch in series with a small resistance and a battery.
Answer:
(d) A closed switch in series with a small resistance and a battery.

Question 14.
The variation of frequency of carrier wave with respect to the amplitude of the modulating
signal is called …………
(a) Amplitude modulation
(b) Frequency modulation
(c) Phase modulation
(d) Pulse width modulation
Answer:
(b) Frequency modulation

Question 15.
Who is the father of the modem robotics industry formed the world’s first robotic company in 1956?
(a) Joliot
(b) Cormark
(c) Engelberger
(d) Edward purcell
Answer:
(c) Engelberger

Part – II

Answer any six questions in which Q. No 17 is compulsory.

Question 16.
What is the general definition of electric dipole moment?
Answer:
The electric dipole moment vector lies along the line joining two charges and is directed from  – q to +q. The SI unit of dipole moment is coulomb meter (cm).
\(\vec{p}=q a \hat{i}-q a(-\hat{i})=2 q a \hat{i}\)

Question 17.
Calculate the electric flux through the rectangle of sides 5 cm and 10 cm kept in the region of a uniform electric field 100 NC^-1. The angle 0 is 60°. Suppose 0 becomes zero, what is the electric flux?
Answer:

Question 18.
What is meant by hysteresis?
Answer:
The phenomenon of lagging of magnetic induction behind the magnetising field is called hysteresis. Hysteresis means Tagging behind’.

Question 19.
What is meant by wattles current?
Answer:
The component of current (I_pMS sin φ), which has a phase angle of \(\frac{\pi}{2}\) with the voltage is called reactive component. The power consumed is zero. So that it is also known as ‘Wattless’ current.

Question 20.
What is Snell’s window?
Answer:
When light entering the water from outside is seen from inside the water, the view is restricted to a particular angle equal to the critical angle i_c. The restricted illuminated circular area is called Snell’s window.

Question 21.
Define work function of a metal. Give its unit.
Answer:
The minimum energy needed for an electron to escape from the metal surface is called work function of that metal. It’s unit is electron volt (eV).

Question 22.
What is meant by radioactivity?
Answer:
The phenomenon of spontaneous emission of highly penetrating radiations such as α, β and γ rays by an element is called radioactivity.

Question 23.
What is the angular momentum of an electron in the third orbit of an atom?
Answer:
Here n = 3; h = 6.6 x 10^-34 Js
Angular momentum.

Question 24.
Explain the current flow in a NPN transistor
Answer:

• The conventional flow of current is based on the direction of the motion of holes
• In NPN transistor, current enters from the base into the emitter.

Part – III

Answer any six questions in which Q.No. 27 is compulsory. [6×3 = 18]

Question 25.
Define ‘capacitance’. Give its unit.
Answer:
The capacitance C of a capacitor is defined as the ratio of the magnitude of charge on either of the conductor plates to the potential difference existing between the conductors.
\(C=\frac{Q}{V} \text { or } Q \propto V\)
The SI unit of capacitance is coulomb per volt or farad (F).

Question 26.
Distinguish between drift velocity and mobility
Answer:

Question 27.
A coil of a tangent galvanometer of diametre 0.24 m has 100 turns. If the horizontal component of Earth’s magnetic field is 25 x 10^-6 T then, calculate the current which gives a deflection of 60°.
Answer:
The diameter of the coil is 0.24 m. Therefore, radius of the coil is 0.12 m.
Number of turns is 100 turns. Earth’s magnetic field is 25 x 10^-6 T

Question 28.
Mention the ways of producing induced emf.
Answer:
Induced emf can be produced by changing magnetic flux in any of the following ways:

• By changing the magnetic field B
• By changing the area A of the coil and
• By changing the relative orientation 0 of the coil with magnetic field

Question 29.
A 400 mH coil of negligible resistance is connected to an AC circuit in which an effective current of 6 mA is flowing. Find out the voltage across the coil if the frequency is 1000 Hz.
Answer:
L = 400 x 10^-3 H; I_eff = 6 x 10^-3A; f= 1000 Hz
Inductive reactance, X_L = Lω = L x 2πf = 2 x 3.14 x 1000 x 0.4 = 2512Ω
Voltage across L, V – IX_L = 6 x 10^-3 x 2512 =15.072 V (RMS)

Question 30.
What are the sign conventions followed for lenses?
Answer:
The sign conventions for thin lenses differ only in the signs followed for focal lengths.
(a) The sign of focal length is not decided on the direction of measurement of the focal length from the pole of the lens as they have two focal lengths, one to the left and another to the right (primary and secondary focal lengths on either side of the lens).

(b) The focal length of the thin lens is taken as positive for a converging lens and negative for a diverging lens.

Question 31.
Write down the draw backs of Bohr atom’model.
Answer:
Limitations of Bohr atom model
The following are the drawbacks of Bohr atom model

• Bohr atom model is valid only for hydrogen atom or hydrogen like-atoms but not for
  complex atoms.
• When the spectral lines are closely examined, individual lines of hydrogen spectrum is
  accompanied by a number of faint lines. These are often called fine structure. This is not explained by Bohr atom model.
• Bohr atom model fails to explain the intensity variations in the spectral lines.
• The distribution of electrons in atoms is not completely explained by Bohr atom model.

Question 32.
What do you mean by leakage current in a diode?
Answer:
The leakage current in a diode is the current that the diode will leak when a reverse voltage is applied to it. Under the reverse bias, a very small current in μA, flows across the junction. This is due to the flow of the minority charge carriers called the leakage current or reverse saturation current.

Question 33.
Distinguish between wireline and wireless communication.
Answer:

Part – IV

Answer all the questions. [5 x 5 = 25]

Question 34.
(a) Derive an expression for the torque experienced by a dipole due to a uniform electric field.
Answer:
Torque experienced by an electric dipole in the uniform electric field: Consider an electric dipole of dipole moment \(\vec{p}\) placed in a uniform electric field \(\vec{E}\) whose field lines are equally spaced and point in the same direction. The charge +q will experience a force q\(\vec{E}\)  in the direction of the field and charge -q will experience a force -q \(\vec{E}\) in a direction opposite to the field. Since the external field \(\vec{E}\) is uniform, the total force acting on the dipole is zero. These two forces acting at different points will constitute a couple and the dipole experience a torque. This torque tends to rotate the dipole. (Note that electric field lines of a uniform field are equally spaced and point in the same direction). The total torque on the dipole about the point O.

Using right-hand corkscrew rule, it is found that total torque is perpendicular to the plane of the paper and is directed into it.
The magnitude of the total torque


where θ is the angle made by \(\vec{p} \text { with } \overrightarrow{\mathrm{E}}\) .
Since p = 2aq, the torque is written in terms of the vector product as \(\vec{\tau}=\vec{p} \times \vec{E}\)
The magnitude of this torque is τ = pE sin θ and is maximum
when θ =90°.
This torque tends to rotate the dipole and align it with the electric field [/latex]\overrightarrow{\mathrm{E}}[/latex] . Once \(\overrightarrow{\mathrm{p}}\) is aligned with \(\overrightarrow{\mathrm{E}}\), the total torque on the dipole becomes zero.

[OR]

(b) Explain the equivalent resistance of a parallel resistor network.
Answer:
Resistors in parallel: Resistors are in parallel when they are connected across the same potential difference as shown in fig. (a).
In this case, the total current I that leaves the battery in split into three separate paths. Let I₁, I₂ and I₃ be the current through the resistors R_{1 ,}R₂ and R₃ respectively. Due to the conservation of charge, total current in the circuit I is equal to sum of the currents through each of the three resistors.
I = I_{1 +} I₂ + I₃ ………………… (1)

Since the voltage across each resistor is the same, applying Ohm’s law to each resistor, we have

Substituting these values in equation (1) we get.

Here R_p is the equivalent resistance of the parallel combination of the resistors. Thus, when a number of resistors are connected in parallel, the sum of the reciprocal of the values of resistance of the individual resistor is equal to the reciprocal of the effective resistance of the combination as shown in the fig. (b).
Note: The value of equivalent resistance in parallel connection will be lesser than each individual resistance.

Question 35.
(a) Obtain a relation for the magnetic induction at a point along the axis of a circular coil ‘ carrying current.
Answer:
Magnetic field produced along the axis of the current carrying circular coil: Consider a current carrying circular loop of radius R and let I be the current flowing through the wire in the direction. The magnetic field at a point P on the axis of the circular coil at a distance z from its center of the coil O. It is computed by taking two diametrically opposite line elements of the coil each of length \(d \bar{l} \) at C and D Let r be the vector joining the current element (I \(d \bar{l}\)) at C to the point P.
PC = PD = r = \(r=\sqrt{\mathrm{R}^{2}+\mathrm{Z}^{2}}\)
PC = PD = r = \(r=\sqrt{\mathrm{R}^{2}+\mathrm{Z}^{2}}\)
angle ∠CPO = ∠DPO = θ
According to Biot-Savart’s law, the magnetic field at P due to the current element \(d \bar{l} \) is


The magnitude of magnetic field due to current element I \(d \bar{l} \) at C and D are equal because of equal distance from the coil. The magnetic field dB due to each current element I dl is resolved into two components; dB sin θ along y-direction and dB cos θ along the z-direction. Horizontal components of each current element cancels out while the vertical components (dB cos θ \(\hat{k}\) ) alone contribute to total magnetic field at the point P.
If we integrate \(d \bar{l} \) around the loop, \(d \bar{B} \) sweeps out a cone, then the net magnetic field \(\overrightarrow{\mathrm{B}}\) at point P is

Using Pythagorous theorem r² = R² + Z² and integrating line element from 0 to 2πR, we get

Note that the magnetic field \(\bar{B} \) points along the direction from the point O to P. Suppose if the current flows in clockwise direction, then magnetic field points in the direction from the point P to O.

[OR]

(b) Explain the working of a single-phase AC generator with necessary diagram.
Answer:
Single phase AC generator: In a single phase AC generator, the armature conductors are connected in series so as to form a single circuit which generates a single-phase alternating emf and hence it is called single-phase alternator.

The simplified version of a AC generator is discussed hire. Consider a stator core consisting of 2 slots in which 2 armature conductors PQ and RS are mounted to form single-turn rectangular loop PQRS. Rotor has 2 salient poles with field windings which can be magnetized by means of DC source.

Working: The loop PQRS is stationary and is perpendicular to the plane of the paper. When field windings are excited, magnetic field is produced around it. The direction of magnetic field passing through the armature core. Let the field magnet be rotated in clockwise direction by the prime mover. The axis of rotation is perpendicular to the plane of the paper.

Assume that initial position of the field magnet is horizontal. At that instant, the direction of magnetic field is perpendicular to the plane of the loop PQRS. The induced emf is zero. This is represented by origin O in the graph between induced emf and time angle.

When field magnet rotates through 90°, magnetic field becomes parallel to PQRS. The induced emfs across PQ and RS would become maximum. Since they are connected in series, emfs are added up and the direction of total induced emf is given by Fleming’s right hand rule.

Care has to be taken while applying this rule; the thumb indicates the direction of the motion of the conductor with respect to field. For clockwise rotating poles, the conductor appears to be rotating anti-clockwise. Flence, thumb should point to the left. The direction of the induced emf is at right angles to the plane of the paper. For PQ, it is downwards and for RS upwards. Therefore, the current flows along PQRS. The point A in the graph represents this maximum emf.

For the rotation of 180° from the initial position, the field is again perpendicular to PQRS and the induced emf becomes zero. This is represented by point B. The field magnet becomes again parallel to PQRS for 270° rotation of field magnet. The induced emf is maximum but the direction is reversed. Thus the current flows along SRQP. This is represented by point C.

On completion of 360°, the induced emf becomes zero and is represented by the point D. From the graph, it is clear that emf induced in PQRS is alternating in nature.

Therefore, when field magnet completes one rotation, induced emf in PQRS finishes one cycle. For this construction, the frequency of the induced emf depends on the speed at which the field magnet rotates.

Question 36.
(a) What is emission spectra? Give their types.
Answer:
Emission spectra: When the spectrum of self luminous source is taken, we get emission spectrum. Each source has its own characteristic emission spectrum. The emission spectrum _ can be divided into three types:

(i) Continuous emission spectra (or continuous spectra): If the light from incandescent lamp filament bulb) is allowed to pass through prism (simplest spectroscope), it splits into seven colours. Thus, it consists of wavelengths containing all the visible colours ranging from violet to red. Examples: spectrum obtained from carbon arc, incandescent solids, liquids gives continuous spectra.

(ii) Line emission spectrum (or line spectrum): Suppose light from hot gas is allowed to pass through prism, line spectrum is observed. Line spectra are also known as discontinuous spectra. The line spectra are sharp lines of definite wavelengths or frequencies. Such spectra arise due to excited atoms of elements. These lines are the characteristics of the element which means it is different for different elements. Examples: spectra of atomic hydrogen, helium, etc.

(iii) Band emission spectrum (or band spectrum): Band spectrum consists of several number of very closely spaced spectral lines which overlapped together forming specific bands which are separated by dark spaces, known as band spectra. This spectrum has a sharp edge at one end and fades out at the other end. Such spectra arise when the molecules are excited. Band spectrum is the characteristic of the molecule hence, the structure of the molecules can be studied using their band spectra. Examples, spectra of hydrogen gas, ammonia gas in the discharge tube etc.

[Or]

(b) Describe the Fizeau’s method to determine speed of light.
Answer:
Fizeau’s method to determine speed of light:
Apparatus: The light from the source S was first allowed to fall on a partially silvered glass plate G kept at an angle of 45° to the incident light from the source. The light then was allowed to pass through a rotating toothed-wheel with N teeth and N cuts of equal widths whose speed of rotation could be varied through an external mechanism.

The light passing through one cut in the wheel will, get reflected by a mirror M kept at a long distance d, about 8 km from the toothed wheel. If the toothed wheel was not rotating, the reflected light from the mirror would again pass through the same cut and reach the eyes of the observer through the partially silvered glass plate.

Working: The angular speed c rotation of the toothed wheel was increased from zero to a value co until light passing through one cut would completely be blocked by the adjacent tooth. This is ensured by the disappearance of light while looking through the partially silvered glass plate.

Expression for speed of light: The speed of light in air v is equal to the ratio of the distance the light travelled from the toothed wheel to the mirror and back 2d to the time taken t.
\(v=\frac{2 d}{t}\) ………………… (1)

The distance d is a known value from the arrangement. The time taken t for the light to travel the distance to and fro is calculated from the angular speed ω of the toothed wheel. The angular speed ω of the toothed wheel when the light disappeared for the first time is,
\(\omega=\frac{\theta}{t}\) ………………….. (2)

Here, θ is the angle between the tooth and the slot which is rotated by the toothed wheel within that time t

Rewriting the above equation for t
\(t=\frac{\pi}{\mathrm{N} \omega}\) ……………….. (3)
Substituting t from equation (3) in equation (1)
\(v=\frac{2 d}{\pi / \mathrm{N} \omega}\)
\(v=\frac{2 d \mathrm{N} \omega}{\pi}\) …………….(4)
Fizeau had some difficulty to visually estimate the minimum intensity blocked by the adjacent tooth, and his value for speed of light was very value.

Question 37.
(a) List out the laws of photoelectric effect.
Answer:
Laws of photoelectric effect:

• For a given frequency of incident light, the number of photoelectrons emitted is directly proportional to the intensity of the incident light. The saturation current is also directly proportional to the intensity of incident light.
• Maximum kinetic energy of the photo electrons is independent of intensity of the incident light.
• Maximum kinetic energy of the photo electrons from a given metal is directly proportional to the frequency of incident light.
• For a given surface, the emission of photoelectrons takes place only if the frequency of incident light is greater than a certain minimum frequency called the threshold frequency.
• There is no time lag between incidence of light and ejection of photoelectrons.

[OR]

(b) Describe the working of nuclear reactor with a block diagram.
Answer:
Nuclear reactor:

• Nuclear reactor is a system in which the nuclear fission takes place in a self-sustained controlled manner and the energy produced is used either for research purpose or for power generation.
• The main parts of a nuclear reactor are fuel, moderator and control rods. In addition to this, there is a cooling system which is connected with power generation set up.

Fuel:

• The fuel is fissionable material, usually uranium or plutonium. Naturally occurring uranium contains only \(0.7 \% \text { of }_{92}^{235} \mathrm{U} \text { and } 99.3 \% \text { are only }^{238} \mathrm{g}_{2} \mathrm{U}\) . So the \(_{ 92 }^{ 235 }{ U }\) must be enriched such that it contains at least 2 to 4% of \(_{ 92 }^{ 235 }{ U }\) .
• In addition to this, a neutron source is required to initiate the chain reaction for the first time. A mixture of beryllium with plutonium or polonium is used as the neutron source. During fission of \(_{ 92 }^{ 235 }{ U }\), only fast neutrons are emitted but the probability of initiating fission by it in another nucleus is very low. Therefore, slow neutrons are preferred for sustained nuclear reactions.

Moderators:

• The moderator is a material used to convert fast neutrons into slow neutrons. Usually the moderators are chosen in such a way that it must be very light nucleus having mass comparable to that of neutrons. Hence, these light nuclei undergo collision with fast neutrons and the speed of the neutron is reduced
• Most of the reactors use water, heavy water (D₂0) and graphite as moderators. The blocks of uranium stacked together with blocks of graphite (the moderator) to form a large pile.

Control rods:

• The control rods are used to adjust the reaction rate. During each fission, on an average 2.5 neutrons are emitted and in order to have the controlled chain reactions, only one neutron is allowed to cause another fission and the remaining neutrons are absorbed by the control rods.
• Usually cadmium or boron acts as control rod material and these rods are inserted into the uranium blocks. Depending on the insertion depth of control rod into the uranium, the average number of neutrons produced per fission is set to be equal to one or greater than one.
• If the average number of neutrons produced per fission is equal to one, then reactor is said to be in critical state. In fact, all the nuclear reactors are maintained in critical state by suitable adjustment of control rods. If it is greater than one, then reactor is said to be in super-critical and it may explode sooner or may cause massive destruction.

Shielding:

• For a protection against harmful radiations, the nuclear reactor is surrounded by a concrete wall of thickness of about 2 to 2.5 m.

Cooling system:

• The cooling system removes the heat generated in the reactor core. Ordinary water, heavy water and liquid sodium are used as coolant since they have very high specific heat capacity and have large boiling point under high pressure.
• This coolant passes through the fuel block and carries away the heat to the steam generator through heat exchanger. The steam runs the turbines which produces electricity in power reactors.

Question 38.
(a) Explain the working principle of a solar cell. Mention its applications.
Answer:
Solar cell:
A solar cell, also known as photovoltaic cell, converts light energy directly into electricity or electric potential difference by photovoltaic effect. It is basically a p-n junction which generates emf when solar radiation falls on the p-n junction. A solar cell is of two types: p-type and n-type.

Both types use a combination of p-type and n-type Silicon which together forms the p-n junction of the solar cell. The difference is that p-type solar cells use p-type Silicon as the base with an ultra-thin layer of n-type Silicon as shown in Figure, while n-type solar cell uses the opposite combination. The other side of the p-Silicon is coated with metal which forms the back electrical contact. On top of the n-type Silicon, metal grid is deposited which acts as the front electrical contact. The top of the solar cell is coated with anti-reflection coating and toughened glass.

In a solar cell, electron-hole pairs are generated due to the absorption of light near the junction. Then the charge carriers are separated due to the electric field of the depletion region. Electrons move towards n-type Silicon and holes move towards p-type Silicon layer.

The electrons reaching the n-side are collected by the front contact and holes reaching p-side are collected by the back electrical contact. Thus a potential difference is developed across solar cell. When an external load is connected to the solar cell, photocurrent flows through the load.

Many solar cells are connected together either in series or in parallel combination to form solar panel or module. Many solar panels are connected with each other to form solar arrays. For  high power applications, solar panels and solar arrays are used.

Applications:

• Solar cells are widely used in calculators, watches, toys, portable power supplies, etc.
• Solar cells are used in satellites and space applications
• Solar panels are used to generate electricity.

(b) Elaborate on the basic elements of communication system with the necessary block diagram.
Answer:
Elements of an electronic communication system
Information (Baseband or input signal): Information can be in the form of a sound signal like speech, music, pictures, or computer data which is given as input to the input transducer.

Input transducer
A transducer is a device that converts variations in a physical quantity (pressure, temperature, sound) into an equivalent electrical signal or vice versa. In communication system, the transducer converts the information which is in the form of sound, music, pictures or computer data into corresponding electrical signals. The electrical equivalent of the original information is called the baseband signal. The best example for the transducer is the microphone that converts sound energy into electrical energy.

Transmitter:
It feeds the electrical signal from the transducer to the communication channel. It consists of circuits such as amplifier, oscillator, modulator, and power amplifier. The transmitter is located at the broadcasting station.

Amplifier:
The transducer output is very weak and is amplified by the amplifier. Oscillator: It generates high-frequency carrier wave (a sinusoidal wave) for long distance transmission into space. As the energy of a wave is proportional to its frequency, the carrier wave has very high energy.

Modulator:
It superimposes the baseband signal onto the carrier signal and generates the modulated signal.

Power amplifier:
It increases the power level of the electrical signal in order to cover a large distance.

Transmitting antenna:
It radiates the radio signal into space in all directions. It travels in the form of electromagnetic waves with the velocity of light (3 ^x 10⁸ ms^-1).

Communication channel:
Communication channel is used to carry the electrical signal from transmitter to receiver with less noise or distortion. The communication medium is basically of two types: wireline communication and wireless communication.

Noise:
It is the undesirable electrical signal that interfaces with the transmitted signal. Noise attenuates or reduces the quality of the transmitted signal. It may be man-made (automobiles, welding machines, electric motors etc .) or natural (lightening, radiation from sun and stars and environmental effects). Noise cannot be completely eliminated. However, it can be reduced using various techniques.

Receiver:
The signals that are transmitted through the communication medium are received with the help of a receiving antenna and are fed into the receiver. The receiver consists of electronic circuits like demodulator, amplifier, detector etc. The demodulator extracts the baseband signal from the carrier signal. Then the baseband signal is detected and amplified using amplifiers. Finally, it is fed to the output transducer.

Repeaters:
Repeaters are used to increase the range or distance through which the signals are sent. It is a combination of transmitter and receiver. The signals are received, amplified, and retransmitted with a carrier signal of different frequency to the destination. The best example is the communication satellite in space.

Output transducer:
It converts the electrical signal back to its original form such as sound, music, pictures or data. Examples of output transducers are loudspeakers, picture tubes, computer monitor, etc.

Attenuation:
The loss of strength of a signal while propagating through a medium is known as attenuation.

Range:
It is the maximum distance between the source and the destination up to which the signal is received with sufficient strength.

Students can download 10th Science Chapter 9 Solutions Questions and Answers, Notes, Samacheer Kalvi 10th Science Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 9 Solutions

Samacheer Kalvi 10th Science Solutions Text Book Back Questions and Answers

I. Choose the best answer:

Question 1.
A solution is a mixture.
(a) homogeneous
(b) heterogeneous
(c) homogeneous and heterogeneous
(d) non-homogeneous
Answer:
(a) homogeneous

Question 2.
The number of components in a binary solution is ______.
(a) 2
(b) 3
(c) 4
(d) 5.
Answer:
(a) 2

Question 3.
Which of the following is the universal solvent?
(a) Acetone
(b) Benzene
(c) Water
(d) Alcohol
Answer:
(c) Water

Question 4.
A solution in which no more solute can be dissolved in a definite amount of solvent at a given temperature is called ______.
(a) Saturated solution
(b) Un saturated solution
(c) Supersaturated solution
(d) Dilute solution.
Answer:
(a) Saturated solution

Question 5.
Identify the non-aqueous solution.
(a) sodium chloride in water
(b) glucose in water
(c) copper sulphate in water
(d) sulphur in carbon-di-sulphide
Answer:
(d) sulphur in carbon-di-sulphide

Question 6.
When pressure is increased at a constant temperature the solubility of gases in liquid ______.
(a) No change
(b) increases
(c) decreases
(d) no reaction.
Answer:
(b) increases

Question 7.
Solubility of NaCl in 100 ml water is 36 g. If 25 g of salt is dissolved in 100 ml of water how much more salt is required for saturation:
(a) 12 g
(b) 11 g
(c) 16 g
(d) 20 g
Answer:
(b) 11 g

Question 8.
A 25% alcohol solution means ______.
(a) 25 ml of alcohol in. 100 ml of water
(b) 25 ml of alcohol in 25 ml of water
(c) 25 ml of alcohol in 75 ml of water
(d) 75 ml of alcohol in 25 ml of water.
Answer:
(c) 25 ml of alcohol in 75 ml of water

Question 9.
Deliquescence is due to:
(a) Strong affinity to water
(b) Less affinity to water
(c) Strong hatred to water
(d) Inertness to water
Answer:
(a) Strong affinity to water

Question 10.
Which of the following is hygroscopic in nature?
(a) ferric chloride
(b) copper sulphate pentahydrate
(c) silica gel
(d) none of the above.
Answer:
(c) silica gel

II. Fill in the blanks:

1. The component present in lesser amount, in a solution is called ……..
2. Example for liquid in solid type solution is ……….
3. Solubility is the amount of solute dissolved in ……… g of solvent.
4. Polar compounds are soluble in ……… solvents.
5. Volume percentage decreases with increases in temperature because ………

Answer:

1. solute
2. amalgam
3. 100
4. Polar
5. of expansion of liquid

III. Match the following:

Answer:
A. (iii)
B. (i)
C. (iv)
D. (ii)

IV. True or False: (If false give the correct statement):

1. Solutions which contain three components are called binary solution.
2. In a solution the component which is present in lesser amount is called solvent.
3. Sodium chloride dissolved in water forms a non-aqueous solution.
4. The molecular formula of green vitriol is MgSO₄. 7H₂O
5. When Silica gel is kept open, it absorbs moisture from the air, because it is hygroscopic in nature.

Answer:

1. False – Solutions which contain two components are called binary solution.
2. False – In a solution the component which is present in lesser amount is called solute.
3. False – Sodium chloride dissolved in water forms an aqueous solution.
4. False – The molecular formula of green vitriol is FeSO₄. 7H₂O
5. True

V. Short Answer Questions:

Question 1.
Define the term: Solution
Answer:
A solution is a homogeneous mixture of two or more substances.

Question 2.
What is mean by the binary solution?
Answer:
A solution must at least be consisting of two components. Such solutions which are made of one solute and one solvent are called binary solutions.
E.g., On adding CuSO₄ crystals to water.

Question 3.
Give an example each

1. gas in liquid;
2. solid in liquid;
3. solid in solid;
4. gas in gas.

Answer:

1. Gas in liquid – CO₂ in water
2. Solid in liquid – NaCl in water
3. Solid in solid – Alloys
4. Gas in gas – He – O₂ gas

Question 4.
What is the aqueous and non-aqueous solution? Give an example.
Answer:
Aqueous solution: The solution in which water act as a solvent is called aqueous solution. In general, ionic compounds are soluble in water and form aqueous solutions more readily than covalent compounds. E.g. Common salt in water.

Non – Aqueous solution: The solution in which any liquid, other than water act as a solvent is called non-aqueous solution. Alcohols, benzene, ethers, etc., are used as non – aqueous solvents. E.g. Sulphur dissolved in carbon disulphide.

Question 5.
Define Volume percentage.
Answer:
Volume percentage is defined as the percentage by volume of solute (in ml) present in the given volume of solution.

Question 6.
The aquatic animals live more in a cold region. Why?
Answer:
Aquatic animals live more in cold regions because the solubility of oxygen is more in cold water (at low temperature). Therefore, aquatic animals are more comfortable in cold water.

Question 7.
Define Hydrated salt.
Answer:
Ionic substances which crystallise out from their saturated aqueous solution with a definite number of molecules of water are called hydrated salts.

Question 8.
A hot saturated solution of copper sulphate forms crystals as it cools. Why?
Answer:
The capability of a solution to maintain a certain concentration of solute is temperature-dependent. When a saturated solution of copper sulphate at above room temperature is allowed to cool, the solution becomes supersaturated and in the absence of stirring or the return of the previous solution temperature, the solute starts to precipitate out. i.e., crystal formation occurs.

Question 9.
Classify the following substances into deliquescent, hygroscopic. Cone. Sulphuric acid, Copper sulphate penta hydrate, Silica gel, Calcium chloride and Gypsum salt.
Answer:

VI. Long answer questions:

Question 1.
Write notes on?

1. saturated solution
2. unsaturated solution

Answer:

1. Saturated solution: A solution in which no more solute can be dissolved in a definite amount of the solvent at a given temperature is called saturated solution, e.g. 36 g of sodium chloride in 100 g of water at 25°C forms a saturated solution.
2. Unsaturated solution: Unsaturated solution is one that contains less solute than that of the saturated solution at a given temperature, e.g. 10 g or 20 g or 30 g of Sodium chloride in 100 g of water at 25°C forms an unsaturated solution.

Question 2.
Write notes on various factors affecting solubility.
Answer:
There are three main factors which affects the solubility of a solute. They are

1. Nature of the solute and solvent
2. Temperature
3. Pressure

1. Nature of the solute and solvent : The nature of the solute and solvent plays an important role in solubility. Even though water is Universal solvent, all substances do not dissolve in water. Dissolution occurs when similarities exist between the solvent and the solute.

Ionic compounds are soluble in polar solvent like water and covalent compounds are soluble in non-polar solvents like ether, benzene, alcohol etc.

2. Effect of Temperature :
Solubility of solid in liquid : Generally solubility of a solid solute in a liquid increases with increase in temperature.
In Endothermic process : Solubility increases with increase in temperature.
In Exothermic process : Solubility decreases with increase in temperature.
Solubility of Gases in liquid : Solubility of gases in liquid decreases with increase in temperature.

3. Effect of Pressure : Effect of pressure is observed only in the case of solubility of a gas in a liquid. When the pressure is increased, the solubility of a gas in liquid increases.

Question 3.
(a) What happens when MgSO₄.7H₂O is heated? Write the appropriate equation
(b) Define solubility.
Answer:
(a) When Epsom salt MgSO₄.7H₂O crystals are gently heated, it loses seven water molecules and becomes anhydrous MgSO₄.

(b) Solubility is defined as the amount of solute in grams that can be dissolved in 100 g of the solvent to form its saturated solution at a given temperature and pressure.

Question 4.
In what way hygroscopic substances differ from deliquescent substances.
Answer:

Question 5.
A solution is prepared by dissolving 45 g of sugar in 180 g of water. Calculate the mass percentage of solute.
Answer:
Mass of the solute (sugar) = 45 g
Mass of the solvent (Water) = 180 g
Formula:
Mass percentage of solute (sugar)

The mass percentage of solute = 20%

Question 6.
3.5 litres of ethanol is present in 15 litres of aqueous solution of ethanol. Calculate volume percent of ethanol solution.
Answer:
Volume of ethanol = 3.5 lit = 3500 ml
Volume of water = 15 lit = 15000 ml
Formula:

The volume percentage of ethanol solution = 18.92

VII. HOT Questions

Question 1.
Vinu dissolves 50 g of sugar in 250 ml of hot water, Sarath dissolves 50 g of same sugar in 250 ml of cold water. Who will get a faster dissolution of sugar? and Why?
Answer:
Vinu will get a faster dissolution of sugar. Because generally solubility of a solid solute in a liquid solvent increases with increase in temperature. Therefore Vinu dissolves 50 g of sugar in 250 ml of hot water than Sarath dissolves 50 g of sugar in 250 ml of cold water.

Question 2.
‘A’ is a blue coloured crystalline salt. On heating it loses blue colour and to give ‘B’ When water is added, ‘B’ gives back to ‘A’. Identify A and B, write the equation.
Answer:
Since ‘A’ is a blue coloured crystalline salt, it is CuSO₄. 5H₂O (Blue vitriol). On heating it loses all five water molecules and becomes colourless anhydrous CuSO₄.

When water is added ‘B’ gives back A.

Question 3.
Will the cool drinks give more fizz at top of the hills or at the foot? Explain.
Answer:
At hilltops, the temperature will become less and pressure also decreases. Because temperature and pressure are directly proportional to each other. At low-pressure carbonate, cool drinks will give less fizz and give more fizz at the foot.

Samacheer Kalvi 10th Science Solutions Additional Important Questions and Answers

I. Choose the correct answer:

Question 1.
The dissolution of sugar and salt in water results in a solution.
(a) Binary
(b) Ternary
(c) Quaternary
(d) Saturated
Answer:
(b) Ternary

Question 2.
In a solution, the component which is present in a lesser amount is called ______.
(a) solvent
(b) dissolution
(c) solute
(d) mole.
Answer:
(c) solute

Question 3.
The supersaturated solution of NaCl in 100 g of water at 25°C contains:
(a) 40 g of NaCl
(b) 10 g of NaCl
(c) 20 g of NaCl
(d) 30 g of NaCl
Answer:
(a) 40 g of NaCl

Question 4.
How many component(s) present in binary solution?
(a) 1
(b) 2
(c) 3
(d) 4.
Answer:
(b) 2

Question 5.
Formalin is an aqueous solution of:
(a) formic acid
(b) ammonia
(c) formaldehyde
(d) carbon tetrachloride
Answer:
(c) formaldehyde

Question 6.
The effect of pressure on the solubility of a gas in liquids is given by:
(a) Boyle’s Law
(b) Charle’s Law
(c) Henry’s Law
(d) Avogadro’s Law
Answer:
(c) Henry’s Law

Question 7.
Which one of the following is an example of an aqueous solution?
(a) Sugar in water
(b) Sulphur in carbon disulphide
(c) Iodine dissolved in carbon tetrachloride
(d) Benzoic acid in ethers.
Answer:
(a) Sugar in water

Question 8.
The type of solution when CO₂ is dissolved in water:
(a) solid/liquid
(b) liquid in liquid
(c) gas in liquid
(d) liquid in solid
Answer:
(c) gas in liquid

Question 9.
Tin amalgam is an example of ……… solution.
(a) solid in solid
(b) liquid in solid
(c) solid in liquid
(d) liquid in liquid
Answer:
(b) liquid in solid

Question 10.
In which case solubility decreases with increase in temperature?
(a) Endothermic process
(b) Exothermic process
(c) Both (a) and (b)
(d) None of these.
Answer:
(b) Exothermic process

Question 11.
Fat is soluble in:
(a) water
(b) alcohol
(c) CCl₄
(d) ether
Answer:
(d) ether

Question 12.
The deliquescent substance among the following is:
(a) con.H₂SO₄
(b) P₂O₅
(c) CaCl₂
(d) SiO₂
Answer:
(c) CaCl₂

Question 13.
Mass percentage is expressed as ______.
(a) v/v
(b) w/w
(c) v/w
(d) w/v.
Answer:
(b) w/w

Question 14.
Hygroscopic substances are used as ……… agents.
(a) foaming
(b) drying
(c) oxidising
(d) reducing
Answer:
(b) drying

Question 15.
The molecular formula of Epsom salt is ______.
(a) CuSO₄.5H₂O
(b) FeSO₄.7H₂O
(c) MgSO₄.7H₂O
(d) ZnSO₄.7H₂O.
Answer:
(c) MgSO₄.7H₂O

II. Fill in the blanks:

1. A true solution is a ……… mixture of solute and solvent.
2. Soil cannot store more nitrogen than it can hold because soil is said to be in a state of ………
3. In the dissolution of NaOH in water, the solubility …….. with increase in temperature.
4. Aquatic animals are more comfortable in cold water because as the temperature is less the solubility of dissolved oxygen ………
5. Hydrated salts contain ……… of crystallization.
6. He-O₂ mixture is a binary solution of …….. in ………. solution.
7. The solvent used for dissolving Sulphur is ……….
8. The solubility of NaOH at 25°C is ……….
9. According to Henry’s Law, the solubility of a gas in liquid is ………. proportional to the pressure of the gas over the solution at definite temperature.
10. Anhydrous Calcium chloride is a ………. substance.
11. ……… substances absorb enough water from the atmosphere and get completely dissolved.
12. When 90g of sodium bromide is dissolved in 100 g of water at 25°C it forms a ………. solution.
13. ………. is an example of a binary solution with liquid in Gas.
14. Air and sea water are important ……… solution.
15. A quaternary solution contains ……….. components.
16. The primary factor which determines the characteristic of a solution is ………..
Answer:
1. Homogeneous
2. saturation
3. decreases
4. increases
5. water
6. Gas, Gas
7. CS₂ (or) Carbon disulphide
8. 80 g
9. directly
10. Hygroscopic
11. deliquescent
12. Unsaturated
13. Cloud
14. Homogeneous
15. four
16. Physical state

III. Match the following

Question 1.
Match the column I with column II.

Answer:
A. (iv)
B. (iii)
C. (v)
D. (ii)
E. (i)

Question 2.
Match the column I with column II.

Answer:
A. (iii)
B. (iv)
C. (v)
D. (ii)
E. (i)

Question 3.
Match the column I with column II.

Answer:
A. (v)
B. (iii)
C. (ii)
D. (i)
E. (iv)

Question 4.
Match the column I with column II.

Answer:
A. (iii)
B. (iv)
C. (v)
D. (i)
E. (ii)

Question 5.
Match the column I with column II.

Answer:
A. (iv)
B. (iii)
C. (i)
D. (v)
E. (ii)

IV. True or False: (If false give the correct statement)

1. In an aqueous solution of copper sulphate, the solvent is copper sulphate.
2. A solution containing sugar and salt in water is a binary solution.
3. An example of a solid solution is alloy.
4. The difference between concentrated and dilute solution can be observed by means of colour (or) density.
5. A saturated solution contains 91 g of Glucose in 100 g of water at 25°C.
6. Fat is dissolved in the aqueous solvent ether.
7. The solubility of a gas in a liquid is inversely proportional to the pressure of the gas at a definite temperature.
8. Mass percentage of a solution is expressed as .
9. The white vitriol is represented by the formula ZnSO₄ . 7H₂O.
10. Ferric chloride is a Hygroscopic substance.
Answer:
1. False – In an aqueous solution of copper sulphate, the solvent is water.
2. False – A solution containing sugar and salt in water is a ternary solution.
3. True
4. True
5. True
6. False – Fat is dissolved in the non-aqueous solvent ether.
7. False – The solubility of a gas in a liquid is directly proportional to the pressure of the gas at a definite temperature.
8. True
9. True
10. False – Ferric chloride is a deliquescent substance.

V. Short answer questions:

Question 1.
(i) Which gas is dissolved in soft drinks?
(ii) What will you do to increase the solubility of this gas?
Answer:
(i) Carbon-di-oxide (CO₂) is dissolved in soft drinks.
(ii) An increase in pressure will increase the solubility of CO₂ gas.

Question 2.
Identify the type of binary solution given below.
Answer:

1. Alloys
2. Amalgam
3. Ethyl alcohol in water
4. Aerated drinks

Answer:

1. Solid in solid
2. liquid in solid
3. liquid in liquid
4. Gas in liquid

Question 3.
Explain why Nitrogen in soil is called a saturated solution in nature?
Answer:
Nitrogen in soil is an example of a saturated solution in nature. Soil cannot store more Nitrogen than it can hold.

Question 4.
Define Mass percentage.
Answer:
Mass percentage of a solution is defined as the percentage by mass of the solute present in the solution.

Question 5.
Define the term Molarity (M).
Answer:

Question 6.
Define the term Molality (m).
Answer:

Question 7.
Define the supersaturated solution.
Answer:
A supersaturated solution is one that contains more solute than the saturated solution at a given temperature.
E.g. 40 g of sodium chloride in 100 g of water at 25°C.

Question 8.
Justify the following statements with an explanation.
(i) Solubility of NH₄Cl increases with increase in temperature.
Answer:
Solubility of NH₄Cl increases with increase in temperature because it is an endothermic process.

(ii) Solubility of NaOH decreases with increase in temperature.
Answer:
Solubility of NaOH decreases with increase in temperature because it is an exothermic process.

Question 9.
Calculate the molarity of a solution containing 4 g of NaOH in 500 ml of water.
Answer:
Mass of NaOH = 4 g
Volume of solution = 500 ml

= 0.1 × 2
= 0.2 M

Question 10.
Calculate the molality of a solution containing 3 g of urea (molecular mass = 60) in 750 g of water.
Answer:
Mass of urea (solute) = 3 g
Mass of water (solvent) = 750 g
Formula:

Question 11.
Define dissolution.
Answer:
The process of uniform distribution of solute into solvent is called dissolution.

VI. Long answer questions:

Question 1.
Answer the blanks given in the table.
Answer:

Question 2.
Write a note on the type of solution based on the type of solvent.
Answer:
(i) Aqueous solution : The solution in which water acts as a solvent is called aqueous solution. In general, ionic compounds are soluble in water and form aqueous solutions more readily than covalent compounds.
Eg: Common salt in water, Sugar in water, Copper sulphate in water etc..

(ii) Non-Aqueous solution : The solution in which any liquid, otter than water, acts as a solvent is called non-aqueous solution. Solvent other than water is referred to as non-aqueous solvent. Generally, alcohols, benzene, ethers, carbon disulphide, acetone, etc., are used as non- aqueous solvents.
Eg: Sulphur dissolved in carbon disulphide, Iodine dissolved in carbon tetrachloride.

Question 3.
Justify the following statements with an explanation.

1. The solubility of calcium oxide decreases with increase in temperature,
2. What happens to the solubility in the exothermic process with regard to temperature?
3. In the endothermic process, solubility increases with increase in temperature.
4. At a given temperature, an increase in pressure increases the solubility of the gas

Answer:

1. In an exothermic process, the solubility decreases. When calcium oxide dissolves in water, an exothermic reaction takes place, and so the solubility of calcium oxide decreases
2. In an exothermic process, the solubility decreases with the increase in temperature, as there is already an evolution of heat and it is observed.
3. In an endothermic process, the solubility increases. The solubility of KNO₃ in water is an endothermic reaction and so solubility increases with the increase of temperature.
4. At a given temperature, an increase in pressure increases the solubility of gas according to Henry’s law. e.g. (CO₂ in soft drinks)

VII. Hot Questions.

Question 1.
50 ml tincture of benzoin, antiseptic solution contains 10 ml of benzoin. Calculate the volume of percentage of benzoin.
Answer:
Volume of the solute, Benzoin = 10 ml
Volume of the solution, tincture of benzoin = 50 ml

= 20% (v/v)

Question 2.
Neomycin, the antibiotic cream contains 300 mg of neomycin sulphate the active ingredient in 30 g of ointment base. Calculate the mass percentage of neomycin.
Answer:
Mass of neomycin sulphate(solute) = 300 mg
Mass of the ointment (solution) = 30 g
Formula:

VIII. Numerical problems:

Question 1.
Calculate the molality of the solution containing 18 g of Glucose (Molecular mass 180) in 2 kg of water.
Answer:
Mass of Glucose = 18 g
Molecular mass of Glucose = 180

= 0. 05 m

Question 2.
Calculate the molarity of a solution containing 5.85 g of sodium chloride in 500 ml of the water. (Molecular mass = 58.5)
Answer:
Mass of the solute = 5.85 g
Volume of the solution = 500
No. of moles of NaCl = \(\frac{5.85}{58.5}\) = 0.1

= \(\frac{0.1}{500}\) × 1000 = 0.2 M

Students can Download Tamil Nadu 12th Physics Model Question Paper 2 English Medium Pdf, Tamil Nadu 12th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Physics Model Question Paper 2 English Medium

General Instructions:

• The question paper comprises of four parts.
• You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
• All questions of Part I, II, III, and IV are to be attempted separately.
• Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
  These are to be answered by choosing the most suitable answer from the given four
  alternatives and writing the option code and the corresponding answer
• Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered
  in about one or two sentences.
• Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered
  in about three to five short sentences.
• Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered
  in detaiL Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 70

Part – I

Answer all the questions. Choose the correct answer. [15 x 1 = 15]

Question 1.
Three capacitors are connected in triangle as shown in the figure. The equivalent capacitance between points A and C is……………………………..
(a) 1 μF
(b) 2 μF
(c) 3 μF
(d) 1/4 μF

Answer:
(b) 2 μF

Question 2.
If the electric field in a region is given by \(\overrightarrow{\mathrm{E}}=5 \hat{i}+4 \hat{j}+9 \hat{k}\), then the electric flux through a surface of area 20 units lying in the
y – z plane will be …………………
(a) 20 units
(b) 80 units
(c) 100 units
(d) 180 units
Answer:
(c) 100 units
Hint. The area vector
\(\overrightarrow{\mathrm{A}}=20 \hat{i} ; \overrightarrow{\mathrm{E}}=(5 \hat{i}+4 \hat{j}+9 \hat{k})\)
Flux ( φ)\(\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}}\) = 5 x 20 = 100 units

Question 3.
A wire of resistance 2 ohms per meter is bent to form a circle of radius 1 m. The equivalent resistance between its two diametrically opposite points, A and B as shown in the figure is

(a) π Ω
(b) \(\frac{\pi}{2} \Omega\)
(c) 2π Ω
(d) \(\frac{\pi}{4} \Omega\)
Answer:
(b) \(\frac{\pi}{2} \Omega\)

Question 4.
A non-conducting charged ring of charge q, mass m and radius r is rotated with constant angular speed co. Find the ratio of its magnetic moment with angular momentum is …………..
(a) M
(b) \(\frac{3}{\pi} \mathrm{M}\)
(c) \(\frac{2}{\pi} \mathrm{M}\)
(d) \(\frac{1}{2} \mathrm{M}\)

Answer:
(b) \(\frac{3}{\pi} \mathrm{M}\)

Question 5.
A proton enters a magnetic field of flux density 1.5 Wb/m² with a speed of 2 x 10⁷ m/s at angle of 30° with the field. The force on the proton will be ……………….
(a) 0.24 x 10^-12 N
(b) 2.4 x 10 ^-12 N
(c) 24 x 10^-12 N
(d) 0.024 x 10^-12 N
Answer:
(b) 2.4 x 10 ^-12 N
Hint: F = Bqv sin θ = 1.5 x 1.6 x 10^-19 x 2 x 10⁷ x sin 30°= 2.4 x 10^-12 N

Question 6.
In an electrical circuit, R, L, C and AC voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage and current in the circuit is \(\frac{\pi}{3}\) Instead, if C is removed from the circuit, the phase difference is again \(\frac{\pi}{3}\) . The power factor of the of the circuit is ……………
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{\sqrt{2}}\)
(c) 1
(d) \(\frac{\sqrt{3}}{2}\)
Answer:
(c) 1

Question 7.
The inductance of a coil is proportional to…………………………………….
(a) its length
(b) the number of turns
(c) the resistance of the coil
(d) square of the number of turns
Answer:
(d) square of the number of turns

Question 8.
The electric and magnetic fields of an electromagnetic wave are……………….
(a) in phase and perpendicular to each other
(b) out of phase and not perpendicular to each other
(c) in phase and not perpendicular to each other
(d) out of phase and perpendicular to each other
Answer:
(a) in phase and perpendicular to each other

Question 9.
One of the of Young’s double slits is covered with a glass plate as shown in figure. The position of central maximum will ……………
(a) get shifted downwards
(b) get shifted upward
(c) will remain the same
(d) data insufficient to conclude

Answer:
(b) get shifted upward

Question 10.
The wavelength λ_e of an electron and λ_p of a photon of same energy E are related ………..

Answer:
(d) \(\lambda_{p} \propto \lambda_{e}^{2}\)

Question 11.
A system consists of N_o nucleus at t = 0. The number of nuclei remaining after half of a half-life (that  is, at time \(t=\frac{1}{2} \mathrm{T}_{\frac{1}{2}}\)
Answer:

Hint:

Question 12.
In a pure semiconductor crystal, if current flows due to breakage of crystal bonds, then the semiconductor is called……………………………….
(a) acceptor
(b) donor
(c) intrinsic semiconductor
(d) extrinsic semiconductor
Answer:
(c) intrinsic semiconductor
Hint: Pure semiconductors are called intrinsic semiconductors.

Question 13.
The light emitted in an LED is due to……………………………
(a) Recombination of charge carriers
(b) Reflection of light due to lens action
(c) Amplification of light falling at the junction
Answer:
(a) Recombination of charge carriers

Question 14.
The frequency range of 3 MHz to 30 MHz is used for………………………………..
(a) Ground wave propagation
(b) Space wave propagation
(c) Sky wave propagation
(d) Satellite communication
Answer:
(c) Sky wave propagation

Question 15.
The materials used in Robotics are……………………..
(a) Aluminium and silver
(b) Silver and gold
(c) Copper and gold
(d) Steel and aluminium
Answer:
(d) Steel and aluminium

Part – III

Answer any six questions. Question No. 20 is compulsory.   [6×2 = 12]

Question 16.
Define ‘Electric dipole’
Answer:
Two equal and opposite charges separated by a small distance constitute an electric dipole.

Question 17.
Define current density.
Answer:
The current density (J) is defined as the current per unit area of cross section of the conductor
\(\mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}\)
The S.I. unit of current density. \(\frac{\mathrm{A}}{\mathrm{m}^{2}}(\text { or }) \mathrm{Am}^{-2}\)

Question 18.
What is magnetic susceptibility?
Answer:
It is defined as the ratio of the intensity of magnetisation \((\overrightarrow{\mathrm{M}})\) induced in the material due to magnetising field \((\overrightarrow{\mathrm{H}})\)
\(\chi_{m}=\left|\frac{\overrightarrow{\mathrm{M}}}{\overrightarrow{\mathrm{H}}}\right|\)

Question 19.
What is meant by electromagnetic induction?
Answer:
Whenever the magnetic flux linked with a closed coil changes, an emf (electromotive force) is induced and hence an electric current flows in the circuit.

Question 20.
A coil of 200 turns carries a current of 0.4 A. If the magnetic flux of 4 mWb is linked with the coil, find the inductance of the coil.
Answer:
Number of turns, N = 200; Current, I = 0.4 A
Magnetic flux linked with coil, φ = 4 mWb = 4 x 10^-3 Wb
Induction of the coil , L
\(\mathrm{L}=\frac{\mathrm{N} \phi}{\mathrm{I}}=\frac{200 \times 4 \times 10^{-3}}{0.4}=\frac{800 \times 10^{-3}}{0.4}=2 \mathrm{H}\)

Question 21.
Why do stars twinkle?
Answer:
The stars actually do not twinkle. They appear twinkling because of the movement of the atmospheric layers with varying refractive indices which is clearly seen in the night sky.

Question 22.
How many photons of frequency 10¹⁴ Hz will make up 19.86 J of energy?
Answer:
Total energy emitted per second = Power x time
19.863 = Power x is
∴ Power 19.86 W
Number of photons =

Question 23.
Define curie.
Answer:
One curie was defined as number of decays per second in 1 g of radium and it is equal to 3.7 x 10¹⁰ decays/s

Question 24.
A transistor having α =0.99 and V_BE = 0.7V, is given in the circuit. Find the value of the collector current.
Answer:

Part – III

Answer any six questions. Question No. 26 is compulsory.   [6 x 3 = 18]

Question 25.
Write a short note on ‘electrostatic shielding’.
Answer:
Consider a cavity inside the conductor. Whatever the charges at the surfaces and whatever the electrical disturbances outside, the electric field inside the cavity is zero. A sensitive electrical instrument which is to be protected from external electrical disturbance is kept inside this cavity. This is called electrostatic shielding.

Question 26.
A conductor of linear mass density 0.2 g m^-1 suspended ,by two flexible wire as shown in figure. Suppose the tension in the supporting wires is zero when it is kept inside the magnetic field of 1 T whose direction is into the x page. Compute the current inside the conductor and also the direction of the current. Assume g = 10 m s^-2 = 111.87.
Answer:

Question 27.
How is Eddy current produced? How do they flow in a conductor?
Answer:
Even for a conductor in the form of a sheet or plate, an emf is induced when magnetic flux linked with it changes. But the difference is that there is no definite loop or path for induced current to flow away. As a result, the induced currents flow in concentric circular paths. As these electric currents resemble eddies of water, these are known as Eddy currents. They are also called Foucault currents.

Question 28.
Explain the concept of intensity of electromagnetic waves.
Answer:
The energy crossing per unit area per unit time and perpendicular to the direction of propagation of electromagnetic wave is called the intensity.
Intensity, I = (u)c

Question 29.
If the focal length is 150 cm for a glass lens, what is the power of the lens?
Answer:
Given: focal length,f = 150 cm (or) f= 1.5 m
Equation for power of lens is, P = 1/f
Substituting the values,
\(P = \frac{1}{1.5}\)= 0.067 diopter
As the power is positive, it is a converging lens.

Question 30.
A proton and an electron have same kinetic energy. Which one has greater de Broglie wavelength. Justify.
Answer:
de-Broglic wavelength of the particle is \(\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m \mathrm{K}}}\) \(\text { i.e. } \lambda \propto \frac{1}{\sqrt{m}}\)
As m_e <<m_p ,so λ_e >>λ_P
Hence protons have greater de- broglic wavelength

Question 31.
Distinguish between avalanche and zener breakdown.
Answer:

Question 32.
Explain centre frequency or resting frequency in frequency modulation.
Answer:
When the frequency of the baseband signal is zero (no input signal), there is no change in the frequency of the carrier wave. It is at its normal frequency and is called as centre frequency or resting frequency.

Question 33.
What are black holes?
Answer:
Black holes are end stage of stars Which are highly dense massive object. Its mass ranges from 20 times mass of the sun to 1 million times mass of the sun. It has very strong gravitational force such that no particle or even light can escape from it. The existence of black holes is studied when the stars orbiting the black hole behave differently from the other starts. Every galaxy has black hole at its center. Sagittarius A* is the black hole at the center of the Milky Way galaxy.

Part – IV

Answer all the questions.  [5 x 5 = 25]

Question 34.
(a) How do we determine the electric field due to a continuous charge distribution? Explain. Electric field due to continuous charge distribution
Answer:
The electric charge is quantized microscopically. The expressions of Coulomb’s Law, superposition principle force and electric field are applicable to only point charges. While dealing with the electric field due to a charged sphere or a charged wire etc., it is very difficult to look at individual charges in these charged bodies.

Therefore, it is assumed that charge is distributed continuously on the charged bodies and the discrete nature of charges is not considered here. The electric field due to such continuous charge distributions is found by invoking the method of calculus.

Consider the following charged object of irregular shape. The entire charged object is divided into a large number of charge elements
Δq₁, Δq₂, Δq₃ ….. Δq_n ……… and each charge element Δq is taken as a point charge.
The electric field at a point P due to a charged object is approximately given by the sum of the fields at P due to all such change elements

Here Δ qi is the i^th charge element, r is the distance of the point P from the z^th charge element and \(\hat{r}_{i \mathrm{P}} \) is the unit vector from ith charge element to the point P.

However, the equation is only an approximation. To incorporate the continuous distribution of charge, we take the limit Δq → 0(= dq). In this limit, the summation in the equation becomes an integration and takes the following form
\(\overrightarrow{\mathrm{E}}=\frac{1}{4 \pi \varepsilon_{0}} \int \frac{d q}{r^{2}} \hat{r}\)

Here r is the distance of the point P from the infinitesimal charge dq and \(\hat{r}\) is the unit vector from dq to point P. Even though the electric field for a continuous charge distribution is difficult to evaluate, the force experienced by some test charge q in this electric field is still given by \(\overrightarrow{\mathrm{F}}=q \overrightarrow{\mathrm{E}}\)

(a) Line charge distribution: If the charge Q is uniformly distributed along the wire of length L, then linear charge density (charge per unit length) is λ = Q/L. Its unit is coulomb per meter (Cm^-1). The charge present in the infinitesimal length dl is dq = λ dl
The electric field due to the line of total charge Q is given by

(b) Surface charge distribution: If the charge Q is uniformly distributed on. a surface of area A, then surface charge density (charge per unit area) is \(\lambda=\frac{Q}{L}\). Its unit is coulomb per square meter (Cm^-2 ). The charge present in the infinitesimal area dA is dq = adA. The electric field due to a of total charge Q is given by

(c) Volume charge distribution: If the charge Q is uniformly distributed in a volume V, then volume charge density (charge per unit volume) is given by \(\rho=\frac{Q}{V}\) . Its unit is coulomb per cubic meter (Cm^-3 ). The charge present in the infinitesimal volume element dV is dq = ρdV.
The electric field due to a volume of total charge Q is given by

[OR]

(b) Obtain the macroscopic form of Ohm’s law from its microscopic form and discuss its limitation.
Answer:
Ohm’s law: The Ohm’s law can be derived from the equation J = σE. Consider a segment of wire of length l and cross sectional area A.

When a potential difference V is applied across the wire, a net electric field is created in the wire which constitutes the current. For simplicity, we assume that the electric field is uniform in the entire length of the wire, the potential difference (voltage V) can be written as
V = El
As we know, the magnitude of current density
\(\mathrm{J}=\sigma \mathrm{E}=\sigma \frac{\mathrm{V}}{l}\)
But \(\mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}\),so we write the equation as
\(\frac{\mathrm{I}}{\mathrm{A}}=\sigma \frac{\mathrm{V}}{l}\)
By rearranging the above equations, we get
\(\mathrm{V}=\mathrm{I}\left(\frac{l}{\sigma \mathrm{A}}\right)\)

The quantity \(\frac{l}{\sigma \mathrm{A}}\) is called resistance of the conductor and it is denoted as R. Note that the resistance is directly proportional to the length of the conductor and inversely proportional to area of cross section.
Therefore, the macroscopic form of Ohm’s law can be stated as
V = IR …………….. (3)

Question 35.
(a) Calculate the magnetic held inside and outside of the long solenoid using Ampere’s circuital law.
Answer:
Magnetic field due to a long current carrying solenoid: Consider a solenoid of length L having N turns. The diametre of the solenoid is assumed to be much smaller when compared to its length and the coil is wound very closely.

In order to calculate the magnetic field at any point inside the solenoid, we use Ampere’s circuital law. Consider a rectangular loop abed. Then from Ampere’s circuital law.
\(\oint_{C} \overrightarrow{\mathrm{B}} \cdot d \vec{l}=\mu_{0} \mathrm{I}_{\text {enclosed }}\) = μ₀  x (total current enclosed by Amperian loop)

Since the elemental lengths along be and da are perpendicular to the magnetic field which is along the axis of the solenoid, the integrals.

Since the magnetic field outside the   solenoid is zero, the integral
\(\int_{c}^{d} \overrightarrow{\mathrm{B}} \cdot d \vec{l}=0\)
For the path along ab, the integral is

where the length of the loop ab is h. But the choice of length of the loop ab is arbitrary. We can take very large loop such that it is equal to the length of the solenoid L. Therefore the integral is

Let NI be the current passing through the solenoid of N turns, then

The number of turns per unit length is given by \(\frac{\mathrm{NI}}{\mathrm{L}}=n\) then
\(\mathrm{v}_{p}=\varepsilon_{p}=-\mathrm{N}_{p} \frac{d \Phi_{\mathrm{B}}}{d t}\)
Since n is a constant for a given solenoid and μ_o is also constant. For a fixed current I, the magnetic field inside the solenoid is also a constant.

[OR]

(b) Explain the construction and working of transformer.
Answer:
Construction and working of transformer:
Principle: The principle of transformer is the mutual induction between two coils. That is, when an electric current passing through a coil changes with time, an emf is induced in the neighbouring coil.

Construction: In the simple construction of transformers, there are two coils of high mutual inductance wound over the same transformer core. The core is generally laminated and is made up of a good magnetic material like silicon steel. Coils are electrically insulated but magnetically linked via transformer core.

The coil across which alternating voltage is applied is called primary coil P and the coil from which output power is drawn out is called secondary coil S. The assembled core and coils are kept in a container which is filled with suitable medium for better insulation and cooling purpose.

Working: If the primary coil is connected to a source of alternating voltage, an alternating magnetic flux is set up in the laminated core. If there is no magnetic flux leakage, then whole of magnetic flux linked with primary coil is also linked with secondary coil. This means that rate at which magnetic flux changes through each turn is same for both primary and secondary coils.

As a result of flux change, emf is induced in both primary and secondary coils. The emf induced in the primary coil ε_p is almost equal and opposite to the applied voltage υ_p and is given by
\(\mathrm{v}_{p}=\varepsilon_{p}=-\mathrm{N}_{p} \frac{d \Phi_{\mathrm{B}}}{d t}\)
The frequency of alternating magnetic flux in the core is same as the frequency of the applied voltage. Therefore, induced emf in secondary will also have same frequency as that of applied voltage. The emf induced in the secondary coil ε_s is given by
\(\varepsilon_{\mathrm{s}}=-N_{s} \frac{d \Phi_{\mathrm{B}}}{d t}\) ………………… (1)
where N_p and N_s are the number of turns in the primary and secondary coil, respectively. If the secondary circuit is open, then  ε_s = υ_s where υ_s is the voltage across secondary coil.
\(v_{s}=\varepsilon_{s}=-\mathrm{N}_{s} \frac{d \Phi_{\mathrm{B}}}{d t}\) ……….. (2)

From equations (1) and (2),
\(\frac{v_{s}}{v_{p}}=\frac{N_{s}}{N_{p}}=K\) …………………… (3)

This constant K is known as voltage transformation ratio. For an ideal transformer,
Input power υ_p i_p= Output power υ_si_s
where i_pand i_s are the currents in the primary and secondary coil respectively. Therefore,
\(\frac{v_{s}}{v_{p}}=\frac{N_{s}}{N_{p}}=\frac{i_{p}}{i_{s}}\) ………………….. (4)

Equation (4) is written in terms of amplitude of corresponding quantities,
\(\frac{V_{s}}{V_{p}}=\frac{N_{s}}{N_{p}}=\frac{I_{p}}{I_{s}}=K\)

(i) If N_s> N_p ( or K > 1), ∴ V_s > V_p and I_s < I_p. This is the case of step-up transformer in which voltage is increased and the corresponding current is decreased.

(ii) If N_s < N_p (or K < 1) , ∴ V_s < V_p and I_s > I_p . This is step-down transformer where voltage is decreased and the current is increased.

Question 36.
(a) Discuss the source of electromagnetic waves.
Answer:
Sources of electromagnetic waves: Any stationary source charge produces only electric field. When the charge moves with uniform velocity, it produces steady current which gives rise to magnetic field (not time dependent, only space dependent) around the conductor in which charge flows. If the charged -particle accelerates, in addition to electric field it also produces magnetic field. Both electric and magnetic fields are time varying fields. Since the electromagnetic waves are transverse waves, the direction of propagation of electromagnetic waves is perpendicular to the plane containing electric and magnetic field vectors.

Any oscillatory motion is also an accelerating motion, so, when the charge oscillates (oscillating molecular dipole) about their mean position, it produces electromagnetic waves. Suppose the electromagnetic field in free space propagates along z direction, and if the electric field vector points along y axis then the magnetic field vector will be mutually perpendicular to both electric field and the propagation vector direction, which means
E_y =E₀ sin (kz-ωt)
B_r = B₀ sin(kz – ωt) where, E_o and B_o are amplitude of oscillating electric and magnetic field,\(\hat{k} \) is a wave number, ω is the angular frequency of the wave and k (unit vector, here it is called propagation vector) denotes the direction of propagation of electromagnetic wave.

Note that both electric field and magnetic field oscillate with a frequency (frequency of electromagnetic wave) which is equal to the frequency of the source (here, oscillating charge is the source for the production of electromagnetic waves). In free space or in vacuum, the ratio between E_o and B_o is equal to the speed of electromagnetic wave, which is equal to speed of light c.
\(c=\frac{E_{0}}{B_{0}}\)

In any medium, the ratio of E_o and B_o is equal to the speed of electromagnetic wave in that medium, mathematically, it can be written as
\(v=\frac{E_{0}}{B_{0}}<c\)
Further, the energy of electromagnetic waves comes from the energy of the oscillating charge.

[OR]

(b) Explain about compound microscope and obtain the equation for magnification.
Answer:
Compound microscope:
The lens near the object, called the objective, forms a real, inverted, magnified image of the object. This serves as the object for the second lens which is the eyepiece. Eyepiece serves as a simple microscope that produces finally an enlarged and virtual image. The first inverted image formed by the objective is to be adjusted close to, but within the focal plane of the eyepiece so that the final image is formed nearly at infinity or at the near point. The final image is inverted with respect to the original object. We can obtain the magnification for a compound microscope.

Magnification of compound microscope
From the ray diagram, the linear magnification due to the objective is,
\(m_{0}=\frac{h^{\prime}}{h}\)
from the figure ,\(\tan \beta=\frac{h}{f_{0}}=\frac{h^{\prime}}{L} \), then
\(\frac{h^{\prime}}{h}=\frac{L}{f_{0}} ; m_{o}=\frac{L}{f_{o}}\)

Here, the distance L is between the first focal point of the eyepiece to the second focal point of the objective. This is called the tube length L of the microscope as f_o and f_e are comparatively smaller than L. If the final image is formed at P (near point focusing), the magnification m_e of the eyepiece is,
\(m_{e}=1+\frac{D}{f_{e}}\)

The total magnification m in near point focusing is,
\(m=m_{o} m_{e}=\left(\frac{L}{f_{o}}\right)\left(1+\frac{D}{f_{e}}\right)\)

If the final image is formed at infinity (normal focusing), the magnification m_e of the eyepiece is
\(m_{e}=\frac{D}{f_{e}}\)

The total magnification m in normal focusing is,
\(m=m_{o} m_{e}=\left(\frac{L}{f_{o}}\right)\left(\frac{D}{f_{e}}\right)\)

Question 37.
(a) Briefly explain the principle and working of electron microscope.
Answer:
Electron Microscope:
Principle:

• This is the direct application of wave nature of particles. The wave nature of the electron is used in the construction of microscope called electron microscope.
• The resolving power of a microscope is inversely proportional to the wavelength of the radiation used for illuminating the object under study. Higher magnification as well as higher resolving power can be obtained by employing the waves of shorter wavelengths.
• De Broglie wavelength of electron is very much less than (a few thousands less) that of the visible light being used in optical microscopes.
• As a result, the microscopes employing de Broglie waves of electrons have very much higher resolving power than optical microscope.
• Electron microscopes giving magnification more than 2,00,000 times are common in research laboratories.

Working:

• The electron beam passing across a suitably arranged either electric or magnetic fields undergoes divergence or convergence thereby focussing of the beam is done.
• The electrons emitted from the source are accelerated by high potentials. The beam is made parallel by magnetic condenser lens. When the beam passes through the sample whose magnified image is needed, the beam carries the image of the sample.
• With the help of magnetic objective lens and magnetic projector lens system, the magnified image is obtained on the screen. These electron microscopes are being used in almost all branches of science.

[OR]

(b) Discuss the process of nuclear fission and its properties.
Answer:

• When uranium nucleus is bombarded with a neutron, it breaks up into two smaller nuclei of comparable masses with the release of energy.
• The process of breaking up of the nucleus of a heavier atom into two smaller nuclei with the release of a large amount of energy is called nuclear fission.
• The fission is accompanied by the release of neutrons. The energy that is released in the nuclear fission is of many orders of magnitude greater than the energy released in chemical reactions.
• Uranium undergoes fission reaction in 90 different ways. The most common fission reactions of
  
• Here Q is energy released during the decay of each uranium nuclei. When the slow neutron is absorbed by the uranium nuclei, the mass number increases by one and goes to an excited state. \(_{ 92 }^{ 236 }{ U }\) . But this excited state does not last longer than 10^-12s and decay into two daughter nuclei along with 2 or 3 neutrons. From each reaction, on an average, 2.5 neutrons are emitted.

Question 38.
(a) Transistor functions as a switch. Explain.
Answer:
The transistor in saturation and cut-off regions functions like an electronic switch that helps to. turn ON or OFF a given circuit by a small control signal.

Presence of dc source at the input (saturation region):
When a high input voltage (V. = +5V) is applied, the base current (I ) increases and in turn increases the collector current. The transistor will move into the saturation region (turned ON). The increase in collector current (I_c) increases the voltage drop across R_c .thereby lowering the output voltage, close to zero. The transistor acts like a closed switch and is equivalent to ON condition.

Absence of dc source at the input (cut-off region):
A low input voltage (V_in = OV), decreases the base current (I_B) and in turn decreases the collector current (I_c ). The transistor will move into the cut-off region (turned OFF). The decrease in collector current (I_c) decreases the drop across, thereby increasing the output voltage, dose to +5 V. The transistor acts as an open switch which is considered as the OFF condition.

It is manifested that, a high input gives a low output and a low input gives a high output. In addition, we can say that the output voltage is opposite to the applied input voltage. Therefore, a transistor can be used as an inverter in computer logic circuitry

[OR]

(b) What is modulation? Explain the types of modulation with necessary diagrams.
Answer:
Modulation: For long distance transmission, the low frequency baseband signal (input signal) is superimposed onto a high frequency radio signal by a process called modulation. There are 3 types of modulation based on which parameter is modified.
They are

1. Amplitude modulation,
2. Frequency modulation, and
3. Phase modulation.

1. Amplitude Modulation (AM): If the amplitude of the carrier signal is modified according to the instantaneous amplitude of the baseband signal, then it is called amplitude modulation. Here the frequency and the phase of the carrier signal remain constant. Amplitude modulation is used in radio and TV broadcasting.

The signal shown in figure
(a) is the message signal or baseband signal that carries information, figure
(b) shows the high-frequency carrier signal and figure
(c) gives the amplitude modulated signal. We can see clearly that the carrier wave is modified in proportion to the amplitude of the baseband signal.

(ii) Frequency Modulation (FM):
The frequency of the carrier signal is modified according to the instantaneous amplitude of the baseband signal in frequency modulation. Here the amplitude and the phase of the carrier signal remain constant. An increase in the amplitude of the ‘ baseband signal increases the frequency of the carrier signal and vice versa. This leads to compressions and rarefactions in the frequency spectrum of the modulated wave.

Louder signal leads to compressions and relatively weaker signals to rarefactions. When the amplitude of the baseband signal is zero in Figure (a), the frequency of the modulated signal is the same as the carrier signal. The frequency of the modulated wave increases when the amplitude of the baseband signal increases in the positive direction (A, C). The increase in amplitude in the negative half cycle (B, D) reduces the frequency of the modulated wave (Figure (c)).

(iii) Phase Modulation (PM)
The instantaneous amplitude of the baseband signal modifies the phase of the carrier signal keeping the amplitude and frequency constant is called phase modulation. This modulation is used to generate frequency modulated signals. It is similar to frequency modulation except that the phase of the carrier is varied instead of varying frequency.

The carrier phase changes according to increase or decrease in the amplitude of the baseband signal. When the modulating signal goes positive, the amount of phase lead increases with the amplitude of the modulating signal. Due to this, the carrier signal is compressed or its frequency is increased.

On the other hand, the negative half cycle of the baseband signal produces a phase lag in the carrier signal. This appears to have stretched the frequency of the carrier wave. Hence similar to frequency modulated wave, phase modulated wave also comprises of compressions and rarefactions. When the signal voltage is zero (A, C and E) the carrier frequency is unchanged.

Students can download 10th Science Chapter 8 Periodic Classification of Elements Questions and Answers, Notes, Samacheer Kalvi 10th Science Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 8 Periodic Classification of Elements

Samacheer Kalvi 10th Science Periodic Classification of Elements Text Book Back Questions and Answers

I. Choose the best answer:

Question 1.
The number of periods and groups in the periodic table are:
(a) 6, 16
(b) 7, 17
(c) 8, 18
(d) 7, 18
Answer:
(d) 7, 18

Question 2.
The basis of modem periodic law is ____.
(a) atomic number
(b) atomic mass
(c) isotopic mass
(d) number of neutrons.
Answer:
(a) atomic number

Question 3.
……….. group contains the member of the halogen family.
(a) 17th
(b) 15th
(c) 18th
(d) 16th
Answer:
(a) 17th

Question 4.
_______ is a relative periodic property.
(a) atomic radii
(b) ionic radii
(c) electron affinity
(d) electronegativity.
Answer:
(b) ionic radii

Question 5.
Chemical formula of rust is:
(a) Fe0.xH₂O
(b) Fe04.xH₂O
(c) Fe₂O₃. xH₂O
(d) FeO
Answer:
(c) Fe₂O₃. xH₂O

Question 6.
In the aluminothermic process the role of Al is:
(a) oxidizing agent
(b) reducing agent
(c) hydrogenating agent
(d) sulphurising agent
Answer:
(b) reducing agent

Question 7.
The process of coating the surface of the metal with a thin layer of zinc is called ____.
(a) painting
(b) thinning
(c) galvanization
(d) electroplating.
Answer:
(c) galvanization

Question 8.
Which of the following inert gas has electrons in the outermost shell?
(a) He
(b) Ne
(c) Ar
(d) Kr
Answer:
(a) He

Question 9.
Neon shows zero electron affinity due to ____.
(a) stable arrangement of neutrons
(b) stable configuration of electrons
(c) reduced size
(d) increased density.
Answer:
(b) stable configuration of electrons

Question 10.
……….. is an important metal to form amalgam.
(a) Ag
(b) Hg
(c) Mg
(d) Al
Answer:
(b) Hg

II. Fill in the blanks:

1. If the electronegativity difference between two bonded atoms in a molecule is greater than 1.7, the nature of bonding is ………..
2. …………. is the longest period in the periodical table.
3. ………… forms the basis of modern periodic table.
4. If the distance between two Cl atoms in Cl₂ molecule is 1.98 Å, then the radius of Cl atom is ………..
5. Among the given species A^– A⁺, and A, the smallest one in size is ……….
6. The scientist who propounded the modern periodic law is …………
7. Across the period, ionic radii ………… (increases,decreases).
8. ……….. and ………… are called inner transition elements.
9. The chief ore of Aluminium is ………….
10. The chemical name of rust is ………….
Answer:
1. ionic
2. 6th (sixth) period
3. Atomic number
4. 0.99 Å
5. A⁺
6. Dimitri Mendeleev
7. decreases
8. Lanthanides, Actinides
9. bauxite
10. hydrated ferric oxide

III. Match the following:

Answer:
A. (ii)
B. (v)
C. (iv)
D. (iii)
E. (i)

IV. True or False: (If false give the correct statement)

1. Moseley’s periodic table is based on atomic mass.
2. Ionic radius increases across the period from left to right.
3. All ores are minerals; but all minerals cannot be called as ores.
4. Aluminium wires are used as electric cables due to their silvery white colour.
5. An alloy is a heterogenous mixture of metals.
Answer:
1. False – Moseley’s periodic table is based on atomic number.
2. True
3. True
4. False – Aluminium wires are used as electric cables because it is a good conductor of heat and electricity.
5. False – An alloy is an homogeneous mixture of metals.

V. Assertion and Reason:

Answer the following questions using the data given below:
Question 1.
Assertion: The nature of bond in HF molecule is ionic.
Reason: The electronegativity difference between H and F is 1.9.
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(a) Assertion and Reason are correct, Reason explains the Assertion.

Question 2.
Assertion: Magnesium is used to protect steel from rusting.
Reason: Magnesium is more reactive than iron.
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(a) Assertion and Reason are correct, Reason explains the Assertion.

Question 3.
Assertion: An uncleaned copper vessel is covered with greenish layer. Reason: copper is not attacked by alkali.
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.

VI. Short answer questions:

Question 1.
A is a reddish brown metal, which combines with O₂ at < 1370 K gives B, a black coloured compound. At a temperature > 1370 K, A gives C which is red in colour. Find A, B and C with reaction.
Answer:
(A) is a reddish brown metal – Copper

A – Copper; B – Cupric oxide; C – Cuprous oxide.

Question 2.
A is a silvery white metal. A combines with O₂ to form B at 800°C, the alloy of A is used in making the aircraft. Find A and B.
Answer:
A – Silvery white metal – Aluminium

The alloys of aluminium, Duralumin and Magnalium are used in making the aircraft.
A – Aluminium; B – Aluminium oxide.

Question 3.
What is rust? Give the equation for the formation of rust.
Answer:
When iron is exposed to moist air, it forms a layer of brown hydrated ferric oxide on its surface. This compound is known as rust and the phenomenon of formation of rust is known as rusting.
4Fe + 3O₂ + xH₂O → 2Fe₂O₃. xH₂O (Rust).

Question 4.
State two conditions necessary for rusting of iron.
Answer:
(i) The presence of water and oxygen is essential for the rusting of iron.
(ii) Impurities in the iron, the presence of water vapour, acids, salts and CO₂ speeds up rusting.

VII. Long answer questions:

Question 1.
(a) State the reason for addition of caustic alkali to bauxite ore during purification of bauxite.
Answer:
Caustic alkali is added to bauxite, to dissolve bauxite ore and obtain a solution of sodium aluminate.

(b) Along with cryolite and alumina, another substance is added to the electrolyte mixture. Name the substance and give one reason for the addition.
Answer:
CaF₂ (Fluorspar) is added along with cryolite and alumina. It is added to reduce the high melting point of the electrolyte.

Question 2.
The electronic configuration of metal A is 2, 8, 18, 1.
The metal A when exposed to air and moisture forms B a green layered compound. A with con. H₂SO₄ forms C and D along with water. D is a gaseous compound. Find A, B, C and D.
Answer:
Metal (A) with electronic configuration- 2, 8, 18, 1 is copper.
2Cu + O₂ + CO₂ + H₂O → CuCO₃. Cu(OH)₂ (B)
Copper carbonate (Green layer)

(A) – Copper (Cu)
(B) – Copper Carbonate (CuCO₃. Cu(OH)₂)
(C) – Copper Sulphate (CuSO₄)
(D) – Sulphur dioxide (SO₂)

Question 3.
Explain the smelting process.
Answer:
The roasted ore of copper is mixed with powdered coke and sand and is heated in a blast furnace to obtain matte (Cu₂S + FeS) and slag. The slag is removed as waste.
2 FeS + 3O₂ → 2 FeO + 2 SO₂
2 Cu₂S + 3O₂ → 2 Cu₂O + 2SO₂
Cu₂O + FeS → Cu₂S + FeO
FeO + SiO₂ → FeSiO₂ (slag)

VIII. HOT questions:

Question 1.
Metal A belongs to period 3 and group 13. A in red hot condition reacts with steam to form B. A with strong alkali forms C. Find A, B and C with reactions.
Answer:
Metal A belongs to Period 3 and Group 13. So metal ‘A’ is aluminium.
(A) in red hot condition reacts with steam to form ‘B’.

‘A’ with strong alkali forms ‘C’

(A) – Aluminium
(B) – Aluminium oxide
(C) – Sodium meta aluminate

Question 2.
Name the acid that renders aluminium passive. Why?
Answer:
Dilute or concentrated nitric acid (HNO₃) renders aluminium passive. Because nitric acid does not attack aluminium but it renders aluminium passive due to the formation of an oxide film on its surface.

Question 3.
(a) Identify the bond between H and F in HF molecule.
Answer:
Ionic, because the electronegativity difference is more than 1.7.

(b) What property forms the basis of identification?
Answer:
Electronegativity.

(c) How does the property vary in periods and in groups?
Answer:
In a period, from left to right the electronegativity increases because of the increase in the nuclear charge.
In a Group, from top to bottom, the electronegativity decreases because of the increase in size of the elements.

Samacheer Kalvi 10th Science Periodic Classification of Elements Additional Important Questions and Answers

I. Choose the correct answer:

Question 1.
The shortest period in the periodic table contains elements.
(a) 18
(b) 8
(c) 2
(d) 32
Answer:
(c) 2

Question 2.
Group number of carbon family is _____.
(a) 13
(b) 15
(c) 17
(d) 14.
Answer:
(d) 14.

Question 3.
The ore forming elements, chalcogens are present in ……… group of the modern periodic table.
(a) 18th
(b) 1st
(c) 2nd
(d) 16th
Answer:
(d) 16th

Question 4.
Valency of all the alkali metals is _____.
(a) 1
(b) 2
(c) 3
(d) 4.
Answer:
(a) 1

Question 5.
The largest atom in the 2nd period of the periodic table is:
(a) Li
(b) Be
(c) F
(d) Ne
Answer:
(a) Li

Question 6.
The covalent radii of Hydrogen, if the distance between the Hydrogen nuclei of the molecule is 0.74 Å is:
(a) 1.58 Å
(b) \(\frac{0.74}{4}\) Å
(c) 0.37 Å
(d) 0.74 Å
Answer:
(c) 0.37 Å

Question 7.
Pick out the correct ionic radii in increasing order for the following species – Na, Cl, Na⁺, Cl^– _____.
(a) Na > Cl > Na⁺ > Cl^–
(b) Cl > Na > Na⁺ > Cl^–
(c) Cl^– > Na > Na⁺ > Cl
(d) Cl < Na⁺ < Cl^– < Na.
Answer:
(d) Cl < Na⁺ < Cl^– < Na.
Hint:
Na = 186 pm,
Cl = 91 pm,
Na⁺ = 116 pm,
Cl^– = 167 pm.

Question 8.
In the third period, the first ionization potential is of the order:
(a) Na > Al > Mg > Si > P
(b) Mg > Na > Si > P > Al
(c) Na < Al < Mg < Si < P
(d) Na < Al < Mg < Si < P
Answer:
(c) Na < Al < Mg < Si < P

Question 9.
Which one of the following is the least electronegative element?
(a) Bromine
(b) Chlorine
(c) Iodine
(d) Hydrogen
Answer:
(d) Hydrogen

Question 10.
Which is a widely used a scale to determine the electronegativity?
(a) Pauling scale
(b) Moseley scale
(c) Mendeleev scale
(d) none of these.
Answer:
(a) Pauling scale

Question 11.
Which one of the following orders of ionic radii is correct?
(a) H^– > H⁺ > H
(b) Na⁺ > F^– > O^2-
(c) F > O^2- > Na⁺
(d) None of these
Answer:
(d) None of these

Question 12.
The percentage of carbon in Pig iron is:
(a) < 0.25%
(b) 0.25 – 2%
(c) 2 – 4.5%
(d) > 5%
Answer:
(c) 2 – 4.5%

Question 13.
The chemical formula of clay is _____.
(a) Al₂O₃
(b) Al₂O₃.2H₂O
(c) Al₂O₃. 2SiO₂.2H₂O
(d) Al₂O₃. 2SiO₂.H₂O.
Answer:
(c) Al₂O₃. 2SiO₂.2H₂O

Question 14.
The temperature in the combustion zone is maintained at:
(a) 1500°C
(b) 400°C
(c) 1000°C
(d) 1380°C
Answer:
(a) 1500°C

Question 15.
Oil used in Froth floatation method is _____.
(a) pine oil
(b) natural oil
(c) crude oil
(d) Synthetic oil.
Answer:
(a) pine oil

Question 16.
The first most abundant metal present in the Earth crust is:
(a) Iron
(b) Aluminium
(c) Zinc
(d) Copper
Answer:
(b) Aluminium

Question 17.
……….. metal is used for making calorimeters.
(a) Copper
(b) Tin
(c) Mercury
(d) Iron
Answer:
(a) Copper

Question 18.
More reactive metal is _____.
(a) Zn
(b) Fe
(c) Ag
(d) Na.
Answer:
(d) Na.

Question 19.
The chief ore of Iron is:
(a) Magnesite
(b) Galena
(c) Cinnabar
(d) Haematite
Answer:
(d) Haematite

Question 20.
The metal which melts at room temperature is:
(a) Zinc
(b) Lead
(c) Gallium
(d) Tin
Answer:
(c) Gallium

Question 21.
Conversion of bauxite into alumina is _____.
(a) Hall’s process
(b) Alumino thermic process
(c) Baeyer’s process
(d) Bessemerisation process.
Answer:
(c) Baeyer’s process

Question 22.
………. metal can be cut with knife.
(a) Potassium
(b) Gallium
(c) Mercury
(d) Gold
Answer:
(a) Potassium

Question 23.
………. is not a good conductor of heat and electricity.
(a) Silver
(b) Tungsten
(c) Copper
(d) Aluminium
Answer:
(b) Tungsten

Question 24.
Electrolyte used in Hall’s process ______.
(a) Pure alumina + molten cryolite + fluorspar
(b) Pure alumina + molten bauxite + fluorspar
(c) Pure bauxite + molten cryolite + fluorspar
(d) Pure bauxite + molten Haematite + fluorspar.
Answer:
(a) Pure alumina + molten cryolite + fluorspar

Question 25.
The foaming agent used for froth floatation process is:
(a) Coconut oil
(b) Pine oil
(c) Sodium chloride
(d) Groundnut oil
Answer:
(b) Pine oil

Question 26.
Three elements A, B and C are having the electronic configuration Is² 2s¹, Is² 2s² and Is² 2s² 2p¹ respectively. Which element will have the lowest ionization energy?
(a) A
(b) B
(c) C
(d) B and C
Answer:
(a) A

Question 27.
Metal used in household utensils is ______.
(a) Al
(b) Co
(c) Fe
(d) Na.
Answer:
(a) Al

Question 28.
Which one of the following pair is a metalloid?
(a) Na and K
(b) F and Cl
(c) Cu and Hg
(d) Si and Ge
Answer:
(d) Si and Ge

Question 29.
The highly metallic element will have the configuration of:
(a) 2, 8, 7
(b) 2, 8, 8, 5
(c) 2, 8, 8, 1
(d) 2, 8, 2
Answer:
(c) 2, 8, 8, 1

Question 30.
The metal used in electroplating is ______.
(a) Cu
(b) Al
(c) Fe
(d) Co.
Answer:
(a) Cu

Question 31.
The flux which is used when the gangue present in the ore is acidic:
(a) Silica
(b) Calcium oxide
(c) Calcium silicate
(d) Cuprous sulphide
Answer:
(b) Calcium oxide

Question 32.
Matte is a mixture of:
(a) Cu₂O + Cu₂S
(b) Cu₂O + FeS
(c) Cu₂S + FeS
(d) Cu₂O + PbS
Answer:
(c) Cu₂S + FeS

Question 33.
Fe reacts with dilute nitric acid in cold condition to give ______.
(a) Ferrous nitride
(b) Ferrous nitrate
(c) Ferric nitride
(d) Ferric nitrate.
Answer:
(b) Ferrous nitrate

Question 34.
In the brass alloy, which is solvent?
(a) Zn
(b) Co
(c) Ag
(d) Cu.
Answer:
(d) Cu.

II. Fill in the blanks:

1. The major component of the matte is ………….
2. The modern periodic table is based on ………..
3. The valency of alkali metals is …………
4. The unreactive elements are present in group ………..
5. In the 2nd period, the smallest atom is ……….
6. The size of a cation is ………… than the neutral atom.
7. ……….. is the unit of ionization energy.
8. The ionization energy ……… down the group in the periodic table.
9. The electron affinities of noble gases are …………
10. ………. is the process of extracting the ore from the Earth’s crust.
11. Galena is the ore of …………..
12. The silvery white metal which is a good conductor of heat and electricity is …………
13. The slag formed during Bessemerisation process is ………….
14. Blister copper contains ………. percentage of copper.
15. Haematite ore is concentrated by ……….. washing.
Answer:
1. Cu₂S
2. atomic number
3. one
4. 18
5. Neon
6. smaller
7. KJ/mol
8. decreases
9. zero
10. Mining
11. lead
12. aluminium
13. Iron silicate or FeSiO₃
14. 98%
15. hydraulic

III. Match the following:

Question 1.
Match the column I with column II.

Answer:
A. (iv)
B. (v)
C. (i)
D. (ii)
E. (iii)

Question 2.
Match the column I with column II.

Answer:
A. (ii)
B. (iv)
C. (v)
D. (iii)
E. (i)

Question 3.
Match the column I with column II.

Answer:
A. (iii)
B. (iv)
C. (v)
D. (i)
E. (ii)

Question 4.
Match the column I with column II.

Answer:
A. (v)
B. (iv)
C. (ii)
D. (iii)
E. (i)

IV. True or false. (If false give the correct statement)

1. Alkali metals are generally extracted by the electrolysis of their ores in fused state.
2. Every mineral is an ore but every ore is not a mineral.
3. Slag is a product formed during smelting by combination of flux and impurities.
4. Reactive metals occur in native state.
5. Malachite is a sulphide ore of copper.
6. Lanthanides are present in the 6th group of the periodic table.
7. Atomic radius increases as we go across the period due to increase in size.
8. As the positive charge increases, the size of the cation decreases.
9. If the difference in electronegativity is greater than 1.7, the bond is considered to be covalent.
10. Siderite is the carbonate ore of calcium.
Answer:
1.True
2. True
3. True
4. False – Reactive metals always occur in the combined state.
5. False – Malachite is the carbonate ore of copper.
6. False – Lanthanides are present in the 6th period of the periodic table.
7. False – Atomic radius increases as we go across the period due to decrease in size.
8. True
9. False – If the difference in electronegativity is greater than 1.7, the bond is considered to be ionic.
10. False – Siderite is the carbonate ore of Iron.

V. Assertion and Reason:

Answer the following questions using the data given below:
Question 1.
Assertion: Zinc blende is concentrated by Froth floatation process.
Reason: Zinc blende is a sulphide ore.
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(a) Assertion and Reason are correct, Reason explains the Assertion.

Question 2.
Assertion: In thermite welding, aluminium powder and Fe₂O₃ are used.
Reason: Aluminium powder is a strong reducing agent.
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(a) Assertion and Reason are correct, Reason explains the Assertion.

Question 3.
Assertion: To design the body of an aircraft, aluminium alloys are used.
Reason: Aluminium becomes passive when it is treated with dil or con.HNO₃
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.

Question 4.
Assertion: Tinstone and the impurity wolframite are seperated by magnetic separation.
Reason: Tinstone is magnetic and wolframite is non-magnetic in nature.
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(a) Assertion and Reason are correct, Reason explains the Assertion.

Question 5.
Assertion: Bauxite is purified by leaching.
Reason: Bauxite undergoes thermal decomposition.
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(b) Assertion is correct, Reason is wrong.

VI. Short answer questions:

Question 1.
State the modern periodic law.
Answer:
The physical and chemical properties of the elements are the periodic function of their atomic number.

Question 2.
‘X’ is a silvery white metal. X reacts with O₂ to form Y. The same compound is obtained from the metal on reaction with steam with the liberation of hydrogen gas. Identify X and Y.
Answer:
(i) Silvery white metal ‘X’ is Aluminium.
(ii) It reacts with O₂ to form ‘Y’

(iii) Y can also be obtained on reaction with steam with the liberation of H₂.

Question 3.
Write any four characteristics of periods.
Answer:

• In a period, the electrons are filled in the same valence shell of all elements.
• As the electronic configuration changes along the period, the chemical properties of the elements also change.
• The atomic size of the elements in a period decreases from left to right
• In a period, die metallic character of the element decreases, while their non-metallic character increases.

Question 4.
Write the Principle of Hydraulic washing.
Answer:
The difference in the densities or specific gravities of the ore and the gangue is the main principle behind this method.

Question 5.
What are coinage metals?
Answer:
Copper, silver and gold are called coinage metals, as they are used in making coins and jewellery.

Question 6.
How will you separate tinstone from wolframite?
Answer:
Magnetic separation method. Tinstone is magnetic in nature.
Method: The crushed ore is placed over a conveyer belt which rotates around two metal wheels, one of which is magnetic. The magnetic particles are attracted to the magnetic wheel and fall separately apart from the non¬magnetic particles.

Question 7.
What are ores?
Answer:
The mineral from which a metal can be readily and economically extracted on a large scale is said to be ore.
eg. Bauxite Al₂O₃.2H₂O is the ore of Aluminium

Question 8.
Define electronegativity.
Answer:
It is the tendency of an element in a covalent bond to attract the shared pair of electrons towards itself. It is a relative property.

Question 9.
In what period and group will an element with z = 118 will be present?
Answer:
Elements with z = 118 will be present in Period number ‘7’ and Group number 18.

Question 10.
Why flux is added during metallurgy?
Answer:
Flux is the substance added to the ore to reduce the fusion temperature and to remove impurities.
e.g. CaO, SiO₂

Question 11.
State the trends in the electronegativity in a Group and period.
Answer:
In a Group: Electronegativity decreases in a group because of the increased number of energy levels.
In a Period: The electronegativity increases because the increase in the nuclear charge.

Question 12.
Write a note about smelting.
Answer:
Smelting is a process of reducing the roasted metallic oxide to metal in a molten condition. In this process, impurities are removed by the addition of flux as slag.

Question 13.
Write the formula of the ores of Aluminium.
Answer:

Question 14.
Explain the action of Aluminium with air.
Answer:
On heating at 800°C, aluminium bums in the air very brightly forming its oxide and nitride.
4Al + 3O₂ → 2Al₂O₃ (Aluminium oxide)
2Al + N₂ → 2AlN (Aluminium nitride).

Question 15.
What happen when Aluminium reacts with steam?
Answer:
When steam is passed over red hot aluminium, H₂ gas is evolved.

Question 16.
Write the reaction of Aluminium with Sodium hydroxide?
Answer:

Question 17.
Explain the electrolytic refining of copper.
Answer:
Cathode: A thin plate of pure copper metal.
Anode: A block of impure copper metal.
Electrolyte: Copper sulphate solution + dilute H₂SO₄
When an electric current is passed through the electrolytic solution, pure copper gets deposited at the cathode and the impurities are settled at the bottom of the anode as anode mud.

Question 18.
Mention the uses of Aluminium.
Answer:
Aluminium is used in

1. household utensils
2. electrical cable industry
3. making aeroplanes and other industrial machine parts.

Question 19.
What are the methods employed to make an alloy?
Answer:

1. By fusing the metals together. Eg: Brass is made by melting zinc and copper.
2. By compressing finely divided metals. Eg: wood mexai.

Question 20.
Write the components of wood metal.
Answer:
Wood metal is an alloy of Lead, Tin, Bismuth and Cadmium.

Question 21.
What are the uses of copper?
Answer:

• Copper is used in manufacturing electric cables and other electric appliances.
• Copper is used for making utensils, containers, calorimeters and coins.
• Copper is used in electroplating.
• Copper is alloyed with gold and silver for making coins and jewels.

Question 22.
Give example for non-ferrous copper and aluminium alloys. Non-ferrous copper alloys: Brass (Cu, Zn), Bronze (Cu, Sn)
Answer:
Non-ferrous aluminium alloys: Duralumin (Al, Mg, Cu, Mn), Magnalium (Al, Mg)

Question 23.
How is rust formed?
Answer:
When iron is exposed to moist air, it forms a layer of brown hydrated Ferric oxide on its surface. This compound is known as rust.
4Fe + 3O₂ + xH₂O → 2Fe₂O₃. xH₂O (Rust).

Question 24.
Why are the alloys prepared?
Answer:

1. To modify appearance and colour.
2. To modify chemical activity.
3. To lower the melting point.
4. To increase hardness and tensile strength.
5. To increase resistance to electricity.

Question 25.
Define corrosion.
Answer:
Corrosion is the gradual destruction of metals by chemical or electrochemical reaction with the environment.

Question 26.
What are alloys? How are they prepared?
Answer:

• An alloy is a homogeneous mixture of a metal with other metals or with non-metals that are fused together. e.g. Brass is an alloy of zinc (solute) in copper (solvent)
• Alloys are prepared by fusing the metals together.
• Alloys are prepared by compressing finely divided metals one over the other.

Question 27.
Which is known as Wet corrosion or Electrochemical corrosion?
Answer:
The corrosive action in the presence of moisture is called wet corrosion. It occurs as a result of electrochemical reaction of metal with water or aqueous solution of salt or acids or bases.

Question 28.
Write a note on Cathodic protection.
Answer:
It is the method of controlling corrosion of a metal surface protected is coated with the metal which is easily corrodible. The easily corrodible metal is called Sacrificial metal to act as anode ensuring cathodic protection.

Question 29.
What are the methods used to prevent corrosion?
Answer:
Corrosion of metals is prevented

• by coating with paints
• by coating with oil and grease
• by alloying with other metals
• by the process of galvanization
• by electroplating
• by sacrificial protection

Question 30.
A reddish brown metal ‘A’ reacts with dil.HCl in the presence of O₂ and forms the compound ‘B’. ‘B’ can also be prepared by heating the metal A with Cl₂. Identify A and B.
Answer:
Reddish brown metal ‘A’ is copper.
(A) reacts with dil.HCl in the presence of O₂ and forms CuCl₂ which is ‘B’.

(B) can also prepared by the action of Cl₂.

Question 31.
Write the uses of copper.
Answer:

1. It is extensively used in manufacturing electric cables and other electric appliances.
2. It is used for making utensils, containers, calorimeters and coins.
3. It is used in electroplating.
4. It is alloyed with gold and silver for making coins and jewels.

Question 32.
Write the name and formula of the ores of iron.
Answer:

Question 33.
Define periodicity.
Answer:
The electronic configurations of elements help us to explain the periodic recurrence of physical and chemical properties. Anything which repeats itself after a regular interval is called periodic and this behaviour is called periodicity.

Question 34.
What happens in the combustion zone during the extraction of iron.
Answer:
The temperature in the combustion zone is 150°C. In this region coke bums 02 to form CO₂, when the charge comes in contact with a hot blast of air.

Question 35.
Explain the reactions taking place in the reduction zone.
Answer:
In the upper region of reduction zone, the temperature is at 400°C. In this region CO reduces ferric oxide to form spongy iron.

Question 36.
Define Metallic radius.
Answer:
It is defined as half the distance between the nuclei of adjacent metal atoms.

Question 37.
Complete the following reactions.

1. 4Fe + 10HNO₃ → 4Fe(NO₃)₂ + ………. + 3H₂O
2. 2Fe + 6H₂SO₄ → Fe₂(SO₄)₃ + ………. + 6H₂O

Answer:

1. NH₄NO₃
2. 3SO₂

Question 38.
What happens when steam is passed over red hot iron?
Answer:
When steam is passed over red hot iron magnetic oxide is formed.
3Fe + 4H₂O (steam) → Fe₃O₄ + 4H₂↑

Question 39.
Define Electron affinity.
Answer:
Electron affinity is the amount of energy released when a gaseous atom gains an electron to form its anion. It is also measured in kJ / mol.

Question 40.
Complete the table.
Answer:

Question 41.
Define Metallurgy.
Answer:
Metallurgy is a science of extracting metals from their ores and modifying the metals into alloys for various uses, based on their physical and chemical properties and their structural arrangement of atoms.

Question 42.
Write a short note on leaching or chemical process.
Answer:
This method is employed when the ore is in a very pure form. The ore is treated with a suitable reagent such that the ore is soluble in it but the impurities are not. The impurities are removed by filtration. The solution of the ore, ie., the filtrate is treated with a suitable reagent which precipitates the ore.
E.g. Bauxite Al₂O₃.2H₂O, the ore of aluminium.

Question 43.
Relate all the four columns of the table with their unique properties.
Answer:

Question 44.
Guess Who I am?
(i) I am preserved in Kerosene.
Answer:
Sodium

(ii) My ore is leached with NaOH.
Answer:
Aluminium

(iii) I sacrifice myself to protect my friend Iron.
Answer:
Magnesium

(iv) I am being used in propellers
Answer:
Nickel steel

Question 45.
Explain the method of making alloys.
Answer:

• By fusing the metals together. E.g. Brass is made by melting zinc and copper.
• By compressing finely divided metals. E.g. Wood metal: an alloy of lead, tin, bismuth and cadmium powder is a fusible alloy.

Question 46.
Write the differences between a mineral and a ore.
Answer:

VII. Long answer questions:

Question 1.
Write the reactions taking place during Bessemerisation of copper.
Answer:
2 FeS + 3O₂ → 2FeO + 2SO₂↑

2Cu₂S + 3O₂ → 2 Cu₂O + 2 SO₂↑
2Cu₂O + Cu₂S → 6 Cu + SO₂↑

Question 2.
How do electronegativity values help to find out the nature of bonding between atoms?
Answer:

• If the difference in electronegativity between two elements is 1.7, the bond has 50% ionic character and 50 % covalent character.
• If the difference is less than 1.7, the bond is considered to be covalent.
• If the difference is greater than 1.7, the bond is considered to be ionic.

Question 3.
Explain Froth floatation with diagram.
Answer:
Principle: This process depends on the preferential wettability of the ore with oil (pine oil) and the gangue particles by water. Lighter ores, such as sulphide ores, are concentrated by this method. Eg: Zinc blende (ZnS).

Question 4.
Explain the Baeyer’s process of conversion of Bauxite into alumina.
Answer:
(i) Bauxite ore is finely ground and heated under pressure with a solution of concentrated caustic soda solution at 150°C to obtain sodium meta aluminate.
(ii) On diluting sodium meta aluminate with water, a precipitate of aluminium hydroxide is formed.
(iii) The precipitate is filtered, washed, dried and ignited at 1000°C to get alumina.

Question 5.
Explain the Hall’s Process of electrolytic reduction of alumina with diagram.
Answer:
Hall’s Process:
Aluminium is produced by the electrolytic reduction of fused alumina (Al₂O₃) in the electrolytic cell.
Cathode : Iron tank linked with graphite
Anode : A bunch of graphite rods suspended in molten electrolyte.
Electrolyte : Pure alumina + molten cryolite + fluorspar (fluorspar lowers the fusion temperature of electrolyte)

Temperature: 900 – 950 °C
Voltage used: 5 – 6 V
Overall reaction: 2 Al₂O₃ → 4 Al + 3O₂↑

Question 6.
Write the reaction involved in the middle region of blast furnace during the extraction of iron.
Answer:
The Middle Region (Fusion Zone): The temperature prevails at 1000°C. In this region, CO₂ is reduced to CO.

Limestone decomposes to calcium oxide and CO₂.

These two reactions are endothermic due to absorption of heat. Calcium oxide combines with silica to form calcium silicate slag.
CaO + SiO₂ → CaSiO₃

Question 7.
What are the three different types of iron? Write their uses.
Answer:
(i) Pig iron (Iron with 2-4.5% of carbon): It is used in making pipes, stoves, radiators, railings, manhole covers and drain pipes.
(ii) Steel (Iron with < 0.25% of carbon): It is used in the construction of buildings, machinery, transmission cables and T. V towers and in making alloys.
(iii) Wrought iron (Iron with 0.25-2% of wraught carbon): It is used in making springs, anchors and electromagnets.

Question 8.
What is corrosion? Write the chemistry behind the formation of rust.
Answer:
(i) The slow and steady destruction of a metal by chemical or electro chemical reaction with the environment.
(ii) When the surface of iron is exposed to moisture and other gases present in the atmosphere, the following chemical reaction takes place.
Fe → Fe²⁺ + 2e^–
O₂ + 2H₂O + 4e^– → 4OH^–
O₂ + 4H⁺ + 4e^– → 2H₂O
The Fe²⁺ ions are oxidised to Fe³⁺ ions.
The Fe³⁺ ions combine OH^– ions to form Fe(OH)₃. This becomes rust which is hydrated ferric oxide with the formula Fe₂O₃.xH₂O. It is a reddish brown substance.

Question 9.
Explain the methods of preventing corrosion.
Answer:
(i) Alloying : The metals can be alloyed to prevent the process of corrosion. Eg: Stainless Steel

(ii) Surface Coating : It involves application of a protective coating over the metal. It is of the following types:
(a) Galvanization: It is the process of coating zinc on iron sheets by using electric current.
(b) Electroplating: It is a method of coating one metal over another metal by passing electric current.
(c) Anodizing: It is an electrochemical process that converts the metal surface into a decorative, durable and corrosion resistant. Aluminium is widely used for anodizing process.
(d) Cathodic Protection: It is the method of controlling corrosion of a metal surface protected is coated with the metal which is easily corrodible. The easily corrodible metal is called sacrificial metal to act as anode ensuring cathodic protection.

Question 10.
Discuss the main featured of Periods in the modern periodic table (or) long form of periodic table.
Answer:
The horizontal rows are called periods.There are seven periods in the periodic table.

1. First period (Atomic number 1 and 2): This is the shortest period. It contains only two elements (Hydrogen and Helium).
2. Second period (Atomic number 3 to 10): This is a short period. It contains eight elements (Lithium to Neon).
3. Third period (Atomic number 11 to 18): This is also a short period. It contains eight elements (Sodium to Argon).
4. Fourth period (Atomic number 19 to 36): This is a long period. It contains eighteen elements (Potassium to Krypton). This includes 8 normal elements and 10 transition elements.
5. Fifth period (Atomic number 37 to 54): This is also a long period. It contains 18 elements (Rubidium to Xenon). This includes 8 normal elements and 10 transition elements.
6. Sixth period (Atomic number 55 to 86): This is the longest period. It contains 32 elements (Caesium to Radon). This includes 8 normal elements, 10 transition elements and 14 inner transition elements (Lanthanides).
7. Seventh period (Atomic number 87 to 118): Like the sixth period, this period also accommodates 32 elements. Recently 4 elements have been included by IUPAC.

Question 11.
Discuss the main feature of Groups in the long form of periodic table.
Answer:
(i) The vertical columns in the periodic table starting from top to bottom are called groups. There are 18 groups in the periodic table.

(ii) Based on the common characteristics of elements in each group, they can be grouped as various families.

(iii) The Lanthanides and Actinides, which form part of Group 3 are called inner transition elements.

(iv) Except ‘group O’, all the elements present in each group have the same number of electrons in their valence shell and thus have the same valency. Eg: all the elements of group 1 have one electron in their valence shells (Is¹). So, the valency of all the alkali metals is ‘ 1’.

(v) As the elements present in a group have identical valence shell electronic configurations, they possess similar chemical properties.

(vi) The physical properties of the elements in a group such as melting point, boiling point and density vary gradually.

(vii) The atoms of the ‘group 0’ elements have stable electronic configuration in their valence shells and hence they are unreactive.

VIII. Hot Questions:

Question 1.
Why noble gases have zero electron affinity value?
Answer:
Noble gases show no tendency to accept electrons because the outers and p orbitals of noble gases are completely filled. No more electrons can be added to them and hence their electron affinities are zero.

Question 2.
Arrange the following ions in order of their increasing ionic radii.
Answer:
Li⁺, Mg²⁺, K⁺ Al³⁺
Al³⁺ < Li⁺ < Mg²⁺ < K⁺

Question 3.
Cationic radius is smaller than its corresponding neutral atom. Why?
Answer:
When a neutral atom lose one or more electrons it forms a cation.
Na → Na⁺ + e^–
The radius of this cation (r_Na+)is decreased than its parent atom (r_Na).
When an atom is charged to cation, the number of nuclear charges becomes greater than the number of orbital electrons. Florence the remaining electrons are more strongly attracted by the nucleus. Hence the cationic radius is smaller than its corresponding neutral atom.

Students can download 10th Science Chapter 11 Carbon and its Compounds Questions and Answers, Notes, Samacheer Kalvi 10th Science Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 11 Carbon and its Compounds

Samacheer Kalvi 10th Science Carbon and its Compounds Text Book Back Questions and Answers

I. Choose the best answer:

Question 1.
The molecular formula of an open-chain organic compound is C₃H₆. The class of the compound is:
(a) alkane
(b) alkene
(c) alkyne
(d) alcohol
Answer:
(b) alkene

Question 2.
The IUPAC name of an organic compound is 3-Methyl butan-1-ol. What type of compound it is?
(a) Aldehyde
(b) Carboxylic acid
(c) Ketone
(d) Alcohol
Answer:
(d) Alcohol

Question 3.
The secondary suffix used in IUPAC nomenclature of an aldehyde is:
(a) – ol
(b) – oic acid
(c) – al
(d) – one
Answer:
(c) – al

Question 4.
Which of the following pairs can be the successive members of a homologous series?
(a) C₃H₈ and C₄H₁₀
(b) C₂H₂ and C₂H₄
(c) CH₄ and C₃H₆
(d) C₂H₅OH and C₄H₈OH.
Answer:
(a) C₃H₈ and C₄H₁₀
Hint: Two successive members of a homologous series must have a difference of -CH₂ in the molecular formula.
\(\mathrm{C}_{3} \mathrm{H}_{8} \stackrel{+\mathrm{CH}_{2}}{\longrightarrow} \mathrm{C}_{4} \mathrm{H}_{10}\).

Question 5.
C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O is a:
(a) Reduction of ethanol
(b) Combustion of ethanol
(c) Oxidation of ethanoic acid
(d) Oxidation of ethanal
Answer:
(b) Combustion of ethanol

Question 6.
Rectified spirit is an aqueous solution which contains about ______ of ethanol.
(a) 95.5 %
(b) 75.5 %
(c) 55.5 %
(d) 45.5 %.
Answer:
(a) 95.5 %
Rectified spirit is a mixture of 95.5 % of ethanol and 4.5 % of water.

Question 7.
Which of the following are used as anaesthetics?
(a) Carboxylic acids
(b) Ethers
(c) Esters
(d) Aldehydes
Answer:
(b) Ethers

Question 8.
TFM in soaps represents _____ content in soap.
(a) mineral
(b) vitamin
(c) fatty acid
(d) carbohydrate.
Answer:
(c) fatty acid
Hint: TFM – Total Fatty Matter. It corresponds the fatty acid (oil).

Question 9.
Which of the following statements is wrong about detergents?
(a) It is a sodium salt of long chain fatty acids
(b) It is sodium salts of sulphonic acids
(c) The ionic part in a detergent is – SO₃ ^–Na⁺
(d) It is effective even in hard water.
Answer:
(a) It is a sodium salt of long-chain fatty acids

II. Fill in the blanks:

1. An atom or a group of atoms which is responsible for chemical characteristics of an organic compound is called …………
2. The general molecular formula of alkynes is ………..
3. In IUPAC name, the carbon skeleton of a compound is represented by ………. (root word / prefix / suffix)
4. (Saturated / Unsaturated) ……….. compounds decolourize bromine water.
5. Dehydration of ethanol by cone. Sulphuric acid forms ………. (ethene/ ethane)
6. 100 % pure ethanol is called ………..
7. Ethanoic acid turns ………… litmus to …………
8. The alkaline hydrolysis of fatty acids is termed as …………
9. Biodegradable detergents are made of …………. (branched / straight) chain hydrocarbons.
Answer:
1. Functional group
2- C_nH_2n-2
3. root word
4. unsaturated
5. ethene
6. absolute alcohol
7. Blue, red
8. Saponification
9. straight

III Match the following:

Question 1.
Match the Column I and Column II.

Answer:
A. (iii)
B. (iv)
C. (v)
D. (ii)
E. (i)

IV. Assertion and Reason:

Answer the following questions using the data given below:
Question 1.
Assertion: Detergents are more effective cleansing agents than soaps in hard water.
Reason: Calcium and magnesium salts of detergents are water soluble.
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(a) Assertion and Reason are correct, Reason explains the Assertion.

Question 2.
Assertion: Alkanes are saturated hydrocrabons.
Reason: Hydrocarbons consits of covalnet bonds.
(a) Assertion and Reason are correct, Reason explains the Assertion .
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.

V. Short answer questions:

Question 1.
Name the simplest ketone and give its structural formula.
Answer:
The simplest ketone is Propanone.
It’s structural formula:

Question 2.
Classify the following compounds based on the pattern of carbon chain and give their structural formula:
(i) Propane
(ii) Benzene
(iii) Cyclo butane
(iv) Furan.
Answer:

Question 3.
How is ethanoic acid prepared from ethanol? Give the chemical equation.
Answer:
Ethanol is oxidized to ethanoic acid with alkaline KMnO₄ or acidified K₂Cr₂O₇.

Question 4.
How do detergents cause water pollution? Suggest remedial measures to prevent this pollution?
Answer:

• Some detergents having a branched hydrocarbon chain are not fully biodegradable by microorganisms present in water. So, they cause water pollution.
• They have straight hydrocarbon chains, in biodegradable detergents, which can be easily degraded by bacteria.

Question 5.
Differentiate soaps and detergents.
Answer:

VI. Long answer questions.

Question 1.
What is called a homologous series? Give any three of its characteristics?
Answer:
A homologous series is a group or a class of organic compounds having the same general formula and similar chemical properties in which the successive members differ by a – CH₂ group.
Characteristics of homologous series:

• All members of a homologous series contain the same elements and functional group.
• All members of a homologous series can be prepared by a common method.
• Chemical properties of the members of a homologous series are similar.

Question 2.
Arrive at, systematically, the IUPAC name of the compound: CH₃-CH₂– CH₂-OH.
Answer:
Step 1: The parent chain consists of 3 carbon atoms. The root word is ‘Prop’.
Step 2: There are single bonds between the carbon atoms of the chain. So, the primary suffix is ‘ane’.
Step 3: Since, the compound contains – OH group, it is an alcohol. The carbon chain is numbered from the end which is closest to – OH group. (Rule 3)

Step 4: The locant number of – OH group is l and thus the secondary suffix is ‘l-ol’.
The name of the compound is Prop + ane + (l – ol) = Propan-l-ol

Question 3.
How is ethanol manufactured from sugarcane?
Answer:
Ethanol is manufactured in industries by the fermentation of molasses, which is a by-product obtained during the manufacture of sugar from sugarcane. Molasses is a dark coloured syrupy liquid left after the crystallization of sugar from the concentrated sugarcane juice. Molasses contain about 30% of sucrose, which cannot be separated by crystallization. It is converted into ethanol by the following steps:
(i) Dilution of molasses : Molasses is first diluted with water to bring down the concentration of sugar to about 8 to 10 percent.

(ii) Addition of Nitrogen source : Molasses usually contains enough nitrogenous matter to act as food for yeast during the fermentation process. If the nitrogen content of the molasses is poor, it may be fortified by the addition of ammonium sulphate or ammonium phosphate.

(iii) Addition of YeastrThe solution obtained in step (ii) is collected in large ‘fermentation tanks’ and yeast is added to it. The mixture is kept at about 303K for a few days. During this period, the enzymes invertase and zymase present in yeast, bring about the conversion of sucrose into ethanol.

The fermented liquid is technically called wash.

(iv) Distillation of ‘Wash’: The fermented liquid (i.e., wash), containing 15 to 18 percent alcohol, is now subjected to fractional distillation. The main fraction drawn is an aqueous solution of ethanol which contains 95.5% of ethanol and 4.5% of water. This is called rectified spirit. This mixture is then refluxed over quicklime for about 5 to 6 hours and then allowed to stand for 12 hours. On distillation of this mixture, pure alcohol (100%) is obtained. This is called absolute alcohol.

Question 4.
Give the balanced chemical equation of the following reactions:
(i) Neutralization of NaOH with ethanoic acid.
Answer:

(ii) Evolution of carbon dioxide by the action of ethanoic acid with NaHCO₃.
Answer:

(iii) Oxidation of ethanol by acidified potassium dichromate.
Answer:

(iv) Combustion of ethanol.
Answer:

Question 5.
Explain the mechanism of cleansing action of soap.
Answer:
A soap molecule contains two chemically distinct parts that interact differently with water. It has one polar end, which is a short head with a carboxylate group (-COONa) and one non – polar end having the long tail made of the hydrocarbon chain.

The polar end is hydrophilic (Water-loving) in nature and this end is attracted towards the water. The non – polar end is hydrophobic (Water hating) in nature and it is attracted towards dirt or oil on the cloth, but not attracted towards the water. Thus, the hydrophobic part of the soap molecule traps the dirt and the hydrophilic part makes the entire molecule soluble in water.

When soap or detergent is dissolved in water, the molecules join together as clusters called ‘micelles’. Their long hydrocarbon chains attach themselves to the oil and dirt. The dirt is thus surrounded by the non-polar end of the soap molecules. The charged carboxylate end of the soap molecules makes the micelles soluble in water. Thus, the dirt is washed away with the soap.

VII. Hot Questions.

Question 1.
The molecular formula of an alcohol is C₄H₁₀O. The locant number of its -OH group is 2.
(i) Draw its structural formula.
Answer:

(ii) Give its IUPAC name.
Answer:
Butan-2-ol

(iii) Is it saturated or unsaturated?
Answer:
Saturated

Question 2.
An organic compound ‘A’ is widely used as a preservative and has the molecular formula C₂H₄O₂. This compound reacts with ethanol to form a sweet smelling compound ‘B’.
(i) Identify the compound ‘A’.
Answer:
Organic Compound ‘A’ with the molecules formula C₂ H₄ O₂ which is a preservative is acetic acid (or) ethanoic acid.
A – CH₃COOH

(ii) Write the chemical equation for its reaction with ethanol to form compound ‘B’.
Answer:

B – Ethyl ethanoate (ester – sweet smelling)

(iii) Name the process.
Answer:
Esterification

Samacheer Kalvi 10th Science Carbon and its Compounds Additional Important Questions and Answers

I. Choose the best answer.

Question 1.
The general formula of alkane series is:
(a) C_nH_2n
(b) C_nH_2n-1
(C) C_nH_2n+2
(d) C_nH_2n-2
Answer:
(C) C_nH_2n+2

Question 2.
Organic compounds are _____ in nature.
(a) flammable
(b) inflammable
(c) heavy
(d) light.
Answer:
(b) inflammable

Question 3.
The Heterocyclic compound is:
(a) Benzene
(b) Pyridine
(c) Naphthalene
(d) Camphor
Answer:
(b) Pyridine

Question 4.
Pick out the unsaturated compound from the following ______.
(a) CH₃ – CH₂ – CH₃
(b) CH₃ – CH = CH₂
(c) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}\)
(d) both (b) and (c).
Answer:
(d) both (b) and (c).

Question 5.
……….. will decolourise Br₂/H₂O.
(a) Methane
(b) Pent-2-ene
(c) Ethyl alcohol
(d) Ethanal
Answer:
(b) Pent-2-ene

Question 6.
The simplest alkane is ______.
(a) Ethane
(b) Ethyne
(c) Propane
(d) Methane.
Answer:
(d) Methane.

Question 7.
……….. is the prefix used for -NH₂ Group.
(a) Fluoro
(b) Methyl
(c) Amino
(d) Nitro
Answer:
(c) Amino

Question 8.
The IUPAC name of an organic compound is Pentan-2-one. The secondary suffix is:
(a) Pentan
(b) an
(c) -one
(d) -2-
Answer:
(c) -one

Question 9.
Molasses is fortified with ………… to increase the nitrogen content.
(a) (NH₄)₂SO₄ (or) (NH₄)₃PO₄
(b) (NH₄)₂CO₃ (or) NH₄Cl
(c) (NH₄)₂CO₃ (or) NH₄OH
(d) None of the above
Answer:
(a) (NH₄)₂SO₄ (or) (NH₄)₃PO₄

Question 10.
Which one of the following is a general formula for alkyne?
(a) C_nH_2n
(b) C_nH_2n+2
(c) C_nH_2n-2
(d) C_nH_n
Answer:
(c) C_nH_2n-2

II. Fill in the blanks.

1. The common difference between the successive member of the homologous series if ……….. group.
2. Denatured spirit is a mixture of ethanol and …………
3. Methane gas is produced when the sodium salt of ethanoic acid is ………… with soda lime.
4. For coagulating rubber from latex ………… is used.
5. ………… is added to prevent the caking of the detergent powder.
6. When a soap or detergent is added to water the moleculer cluster together to form ……….
7. The terminal functional group among aldehydes and ketones is ………..
8. On dehydrogenation of ethanol with Cu/573 K it gives ………… gas.
9. ……… converts glucose into ethanol and carbondi-oxide.
10. The structural formula of Pentanoic acid is …………
Answer:
1. -CH₂
2. pyridine
3. decarboxylated
4. ethanoic acid
5. Na₂SO₄ or Sodium sulphate
6. micelles
7. aldehyde
8. H₂
9. Zymase
10. CH₃CH₂CH₂CH₂COOH

III Match the following:

Question 1.
Match the column I and column II.

Answer:
A. (ii)
B. (iii)
C. (v)
D. (i)
E. (iv)

Question 2.
Match the column I and column II.

Answer:
A. (v)
B. (iv)
C. (i)
D. (iii)
E. (ii)

IV Assertion and Reason.

Answer the following questions using the data given below:
Question 1.
Assertion: Alkynes decolourise bromine water.
Reason: Alkynes are unsaturated compounds.
(a) Assertion and Reason are correct, Reason explains Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(a) Assertion and Reason are correct, Reason explains Assertion.

Question 2.
Assertion: Denaturation of ethanol makes it unfit for drinking purpose.
Reason: Ethanol is mixed with Pyridine for denaturation.
(a) Assertion and Reason are correct, Reason explains Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(a) Assertion and Reason are correct, Reason explains Assertion.

Question 3.
Assertion: Organic compounds contains covalent bond.
Reason: Organic compounds have low melting and boiling points.
(a) Assertion and Reason are correct, Reason explains Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.

Question 4.
Assertion: Due to catenation a large number of carbon compounds are formed.
Reason: Carbon compounds show the property of allotropy.
(a) Assertion and Reason are correct, Reason explains Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(b) Assertion is correct, Reason is wrong.

V. Short answer questions:

Question 1.
Define isomerism with an example.
Answer:
Isomerism is a phenomenon in which organic compounds having the same molecular formula will have different structural formula.
Eg: for the molecular formula C₂H₆O we can write
CH₃CH₂OH – ethanol
CH₄O CH₃ – Methoxy methane

Question 2.
What are root words?
Answer:
Root words are the basic unit which describes the carbon skeleton. It gives the number of carbon atoms present in the parent chain of the compound and the pattern of their arrangement.

Question 3.
Write The functional group and the secondary suffix of the following compounds.
Answer:

Question 4.
What happens when ethanol is dehydrated with con.H₂SO₄ at 443K?
Answer:
Dehydration (Loss of water): When ethanol is heated with cone. H₂SO₄ at 443K, it loses a water molecule i.e., dehydrated to form ethene.

Question 5.
Write a note on esterification.
Answer:
The reaction of an alcohol with a carboxylic acid gives a compound having fruity odour. This compound is called an ester and the reaction is called esterification. Ethanol reacts with ethanoic acid in the presence of cone. H₂SO₄ to form ethyl ethanoate, an ester.

Question 6.
Write tests to identify the presence of ethanoic acid.
Answer:

1. Ethanoic acid turns blue litmus paper to red.
2. Ethanoic acid gives brisk effervescence when treated with Na₂CO₃
3. Ethanoic acid gives a sweet-smelling compound called ester when treated with ethanol.

Question 7.
Write a note on decarboxylation reaction.
Answer:
Decarboxylation (Removal of CO₂): When a sodium salt of ethanoic acid is heated with soda lime (solid mixure of 3 parts of NaOH and 1 part of CaO), methane gas is formed.

Question 8.
What is hard soap?
Answer:
Soaps, which are prepared by the saponification of oils or fats with caustic soda (sodium hydroxide), are known as hard soaps. They are usually used for washing purposes.

Question 9.
Why ordinary soap is not suitable for using with hard water?
Answer:
Ordinary soaps when treated with hard water, precipitate as salts of calcium and magnesium. They appear at the surface of the cloth as sticky grey scum. Thus, the soaps cannot be used conveniently in hard water.

Question 10.
What are the advantages of detergents over soaps?
Answer:
Detergents are better than soaps because of they:

1. can be used in both hard and soft water and can clean more effectively in hard water than soap.
2. can also be used in saline and acidic water.
3. do not leave any soap scum on the tub or clothes.
4. dissolve freely even in cool water and rinse freely in hard water.
5. can be used for washing woollen garments, whereas soap cannot be used.
6. have a linear hydrocarbon chain, which is biodegradable.
7. are active emulsifiers of motor grease.
8. do an effective and safe cleaning, keeping even synthetic fabrics brighter and whiter.

VI. Long answer questions.

Question 1.
Write the IUPAC name of the following compounds
Answer:

Question 2.
Write the characteristics of organic compounds.
Answer:

1. Organic compounds have a high molecular weight and a complex structure.
2. They are mostly insoluble in water but soluble in organic solvents such as ether, carbon tetrachloride, toluene, etc.
3. They are highly inflammable in nature.
4. Organic compounds are less reactive compared to inorganic compounds. Hence, the reactions involving organic compounds proceed at slower rates.
5. Mostly organic compounds form covalent bonds in nature.
6. They have a lower melting point and boiling point when compared to inorganic compounds
7. They exhibit the phenomenon of isomerism, in which a single molecular formula represents several organic compounds that differ in their physical and chemical properties
8. They are volatile in nature.
9. Organic compounds can be prepared in the laboratory.

Question 3.
List the advantages of detergents over soaps.
Answer:

1. Can be used in both hard and soft water and can clean more effectively in hard water than soap.
2. Can also be used in saline and acidic water.
3. Do not leave any soap scum on the tub or clothes.
4. Dissolve freely even in cool water and rinse freely in hard water.
5. Can be used for washing woollen garments, where as soap cannot be used.
6. Have a linear hydrocarbon chain, which is biodegradable.
7. Are active emulsifiers of motor grease.
8. Do an effective and safe cleansing, keeping even synthetic fabrics brighter and whiter.

Question 4.
Draw the schematic diagram for the classification of organic compounds based on the pattern of carbon chain with example.
Answer:

Question 5.
Explain the manufacture of soap.
Answer:
Manufacture of soap by Kettle Process This is the oldest method. But, it is still widely used in the small scale preparation of soap. There are mainly, two steps to be followed in this process.

1. Saponification of oil: The oil, which is used in this process, is taken in an iron tank (kettle). The alkaline solution (10%) is added into the kettle, a little in excess. The mixture is boiled by passing steam through it. The oil gets hydrolysed after several hours of boiling. This process is called Saponification.

2. Salting out of soap: Common salt is then added to the boiling mixture. Soap is finally precipitated in the tank. After several hours the soap rises to the top of the liquid as a ‘curdy mass’. The neat soap is taken off from the top. It is then allowed to cool down.

Question 6.
Ethanol is heated with excess con.H₂SO₄ at 443K.
(a) Name the reaction that occurs and explain it.
Answer:
Dehydration (Loss of water).
When ethanol is heated with conc.H₂SO₄ at 443K, it loses a water molecule i.e., dehydrated to form ethene.

Answer:

(c) What is the product formed? What happen when this gas is passed through Br₂/H₂O.
Answer:
Ethene gas.
When it is passed through Br₂/H₂O it gets decolourised because it is an unsaturated compound.

(d) Why does no decolonization occurs when ethanol is treated with Br₂/H₂O?
Answer:
Ethanol is a saturated compound. Therefore no decolourisation occurs.

VII. Hot Questions.

Question 1.
Organic compounds A and B are isomers with the molecular formula C₂H₆O (A) liberates H2 gas when it reacts with metallic sodium whereas (B) does not. Compound (A) reacts with ethanoic acid and forms a fruity smelling compound (C). Identify A, B and C and explain the reactions.
Answer:
(A) and (B) are compounds with same molecular formula C₂H₆O.
(A) reacts with metallic Na and liberates H₂↑
∴ (A) is ethanol [CH₃CH₂OH]
(B) is methoxy methane [CH₃OCH₃]

(A) reacts with ethanoic acid and forms ester.

(A) Ethanol, CH₃CH₂OH
(B) Methoxy methane, CH₃OCH₃
(C) Ethyl ethanoate, CH₃COOC₂H₅

Question 2.
Write the IUPAC names of the following compounds.
Answer:

Students can download Maths Chapter 1 Set Language Ex 1.4 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.4

Question 1.
If P = {1, 2, 5, 7, 9}, Q = {2, 3, 5, 9, 11}, R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8} then find
(i) (P∪Q)∪R
(ii) (P∩Q)∩S
(m) (Q∩S)∩R
Solution:
P = {1, 2, 5, 7, 9}; Q = {2, 3, 5, 9, 11}; R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8}
(i) P∪Q = {1, 2, 5, 7, 9} ∪ {2, 3, 5, 9, 11}
= {1, 2, 3, 5, 7, 9, 11}
(P∪Q)∪R = {1, 2, 3, 5, 7, 9, 11} ∪ {3, 4, 5, 7, 9}
= {1, 2, 3, 4, 5, 7, 9, 11}

(ii) P∩Q = {1, 2, 5, 7, 9} ∩ {2, 3, 5, 9, 11}
= {2, 5, 9}
(P∩Q)∩S = {2, 5, 9} ∩ {2, 3, 4, 5, 8}
= (2, 5}

(iii) Q∩S = {2, 3, 5, 9, 11} ∩ {2, 3, 4, 5, 8}
= {2, 3, 5}
(Q∩S)∩R = {2, 3, 5} ∩ {3, 4, 5, 7, 9}
= {3, 5}

Question 2.
Test for the commutative property of union and intersection of the sets
P = {x : x is a real number between 2 and 7} and
Q = {x : x is an irrational number between 2 and 7}
Solution:
P is a real number set
Q is a set of irrational number
∴ Q⊂P
P∪Q= Q∪P = P
∴ Union of sets is commutative.
P∩Q = Q∩P = Q
∴ Intersection of sets is commutative.

Question 3.
If A = {p, q, r, s}, B = {m, n, q, s, t} and C = {m, n, p, q, s}, then verify the associative property of union of sets.
Solution:
When union of sets is associative
A∪(B∪C) = (A∪B)∪C
(B∪C) = {m, n, q, s, t) ∪ {m, n, p, q, s}
= {m, n, p, q, s, t}
A∪(B∪C) = {p, q, r, s} ∪ {m, n, p, q, s, t}
= {m, n, p, q, r, s, t} ……..(1)
(A∪B) = {p, q, r, s} ∪ {m, n, q, s, t}
= {m, n, p, q, r, s, t}
(A∪B)∪C = {m, n, p, q, r, s, t} ∪ {m, n, p, q, s}
= {m, n, p, q, r, s, t} ……….(2)
From (1) and (2) it is verified that A∪(B∪C) = (A∪B)∪C

Question 4.
Verify the associative property of intersection of sets for A = {-11, √2, √5, 7},
B = {√3, √5, 6, 13} and C = {√2, √3, √5, 9}.
Solution:
When intersection of sets is associative
A∩(B∩C) = (A∩B)∩C
(B∩C) = {√3, √5, 6, 13} ∩ {√2, √3, √5, 9}
= {√3, √5}
A∩(B∩C) = {-11, √2, √5, 7} ∩ {√3, √5}
{√5} ………(i)
(A∩B) = {-11, √2, √5, 7} ∩ {√3, √5 ,6, 13}
= {√5}
(A∩B)∩C = {√5} n {√2, √3, √5, 9}
= {√5}……..(2)
From (1) and (2) it is verified that A∩(B∩C) = (A∩B)∩C

Question 5.
If A={ x : x = 2ⁿ, n ∈ W and n < 4}, B = {x : x = 2n, n ∈ N and n ≤ 4} and C = {0, 1, 2, 5, 6}, then verify the associative property of intersection of sets.
Solution:
A = {x : x = 2ⁿ, n ∈ W and n < 4}
A = {1, 2, 4, 8}
B = {x : x = 2n, n ∈ N and n ≤ 4}
B = {2, 4, 6, 8}
C ={0, 1, 2, 5, 6}
When intersection of sets is associative
A∩(B∩C) = (A∩B)∩C
(B∩C) = {2, 4, 6, 8} ∩ {0, 1, 2, 5, 6}
= {2, 6}
A∩(B∩C)= {1 ,2, 4, 8} ∩ {2, 6}
= {2}……….(1)
(A∩B) = {1, 2, 4, 8} ∩ {2, 4, 6, 8}
= {2, 4, 8}
(A∩B)∩c= {2, 4, 8} n {0, 1, 2, 5, 6}
= {2}………..(2)
From (1) and (2) we get A∩(B∩C) = (A∩B)∩C

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