Bhagya

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:

Avalanche Breakdown	Zener Breakdown
It occurs injunctions which are lightly and have wide depletion widths.	It occurs in junctions which are heavily doped and have narrow depletion widths.
It occurs at higher reverse voltages when thermally generated electrons get enough kinetic energy to produce more electrons by collision.	It occurs due to rupture of covalent bonds by strong electric fields set up in depletion region by the reverse voltage.
At reverse voltage above 6V breakdown is due to avalanche effect.	At reverse voltage below 6V breakdown is due to zener effect.
Electric field produced is weak in nature.	A strong electric field is produced
Charge carriers obtain energy from the applied potential.	Zener current is independent of applied voltage.

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

శ్రీ సాయి కాకడ ఆరతి

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 2 Integral Calculus I Ex 2.8 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.8

Using second fundamental theorem, evaluate the following:

Question 1.
\(\int_{0}^{1}\) e^2x dx
Solution:

Question 2.
\(\int_{0}^{1/4}\) \(\sqrt { 1 -4x}\) dx
Solution:

Question 3.
\(\int_{0}^{1}\) \(\frac { xdx }{x^2+1}\)
Solution:

Question 4.
\(\int_{0}^{3}\) \(\frac { e^xdx }{1+e^x}\)
Solution:
\(\int_{0}^{3}\) \(\frac { e^xdx }{1+e^x}\)
= {log |1 + ex|}\(_{0}^{3}\)
= log |1 + e³| – log |1 + e°|
= log |1 + e³| – log |1 + 1|
= log |1 + e³| – log |2|
= log |\(\frac { 1+e^3 }{2}\)|

Question 5.
\(\int_{0}^{1}\) xe^x² dx
Solution:

Question 6.
\(\int_{1}^{e}\) \(\frac { dx }{x(1+logx)^3}\)
Solution:

Question 7.
\(\int_{-1}^{1}\) \(\frac { 2x+3 }{x^2+3x+7}\) dx
Solution:

Question 8.
\(\int_{0}^{π/2}\) \(\sqrt { 1 +cosx} \) dx
Solution:

Question 9.
\(\int_{1}^{2}\) \(\frac { x-1 }{x^2}\) dx
Solution:

Evaluate the following

Question 10.
\(\int_{1}^{4}\) f(x) dx where f(x) = \(\left\{\begin{array}{l}
4 x+3,1 \leq x \leq 2 \\
3 x+5,2 \end{array}\right.\)
Solution:
\(\int_{1}^{4}\) f(x) dx
= \(\int_{1}^{2}\) f(x) dx + \(\int_{2}^{4}\) f(x) dx
= \(\int_{1}^{2}\) (4x + 3) dx + \(\int_{2}^{4}\) (3x + 5) dx

(8 + 6) – [5] + [24 + 20] – [6 + 10]
= 14 – 5 + 44 – 16
= 58 – 21
= 37

Question 11.
\(\int_{0}^{2}\) f(x) dx where f(x) = \(\left\{\begin{array}{ll}
3-2 x-x^{2}, & x \leq 1 \\
x^{2}+2 x-3, & 1<x \leq 2
\end{array}\right.\)
Solution:
\(\int_{0}^{2}\) f(x) dx
= \(\int_{0}^{1}\) f(x) dx + \(\int_{1}^{2}\) f(x) dx
= \(\int_{0}^{1}\) (3 – 2x – x²) dx + \(\int_{1}^{2}\) (x² + 2x – 3) dx

Question 12.
\(\int_{-1}^{1}\) f(x) dx where f(x) = \(\left\{\begin{array}{ll}
x, & x \geq 0 \\
-x, & x<0
\end{array}\right.\)
Solution:
\(\int_{-1}^{1}\) f(x) dx
\(\int_{-1}^{0}\) f(x) dx + \(\int_{0}^{1}\) f(x) dx
= \(\int_{-1}^{0}\) (-x) dx + \(\int_{0}^{1}\) x dx

Question 13.
f(x) = \(\left\{\begin{array}{l}
c x, \quad 0<x<1 \\
0, \text { otherwise }
\end{array}\right.\) find ‘c’ if \(\int_{0}^{1}\) f(x) dx = 2
Solution:
Given
f(x) = \(\left\{\begin{array}{l}
c x, \quad 0<x<1 \\
0, \text { otherwise }
\end{array}\right.\)
⇒ \(\int_{0}^{1}\) f(x) dx = 2
⇒ \(\int_{0}^{1}\) cx dx = 2
c[ \(\frac { x^2 }{ 2 }\) ]\(_{0}^{1}\) = 2
c[ \(\frac { 1 }{ 2 }\) – 0 ] = 2
\(\frac { 1 }{ 2 }\) = 2
⇒ c = 4

Must Read:

MARICO Pivot Point Calculator

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 2 Integral Calculus I Ex 2.9 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.9

Evaluate the following using properties of definite integrals:

Question 1.
\(\int_{-π/4}^{π/4}\) x³ cos³ x dx
Solution:
Let f(x) = x³cos³x
f(-x) = (-x)³ cos³(-x)
= -x³ cos³x
f(-x) = -f(x)
⇒ f(x) is an odd function
∴ \(\int_{-π/4}^{π/4}\) x³ cos³ x dx = 0

Question 2.
\(\int_{-π/2}^{π/2}\) sin² θ dθ
Solution:
Let f(θ)= sin² θ
f(-θ) = sin² (-θ) = [sin (-θ)]²
= [-sin θ]² = sin² θ
f(-θ) = f(θ)
∴ f(θ) is an even function

Question 3.
\(\int_{-1}^{1}\) log(\(\frac { 2-x }{2+x}\)) dx
Solution:

Question 4.
\(\int_{0}^{π/2}\) \(\frac { sin^7x }{sin^7x+cos^7x}\) dx
Solution:
Using the property
\(\int_{0}^{a}\) f(x) dx = \(\int_{0}^{a}\) f(a – x) dx

Question 5.
\(\int_{0}^{1}\) log (\(\frac { 1 }{x}\) – 1) dx
Solution:

Question 6.
\(\int_{0}^{1}\) \(\frac { x }{(1-x)^{3/4}}\) dx
Solution:
Let I = \(\int_{0}^{1}\) log \(\frac { x }{(1-x)^{3/4}}\) dx
Using the property
\(\int_{0}^{a}\) f(x) dx = \(\int_{0}^{a}\) f(a – x) dx

Read More:

CUB Pivot Point Calculator

Students can download Maths Chapter 3 Bill, Profit and Loss Ex 3.2 Questions and Answers, Notes, Samacheer Kalvi 6th 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.

ASTRAL Pivot Point Calculator

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 3 Bill, Profit and Loss Ex 3.2

Miscellaneous Practice Problems

Question 1.
A Shopkeeper buys three articles for Rs 325, Rs 450, and Rs 510. He is able to sell them for Rs 350, Rs 425, and Rs 525 respectively. Find the gain or loss to the shopkeeper on the whole.
Solution:
C.P of three articles = 325 + 450 + 510 = ₹ 1285
S.P of three articles = 350 + 425 + 525 = ₹ 1,300
Here S.P > C.P
Profit = S.P – C.P = 1,300 – 1285 = ₹ 15
The shopkeeper gained = ₹ 15

Question 2.
A stationery shop owner bought a scientific calculator for ₹ 750. He had put a battery worth ₹ 100 in it. He had spent ₹ 50 for its outer pouch. He was able to sell it at ₹ 850. Find his profit or loss.
Solution:
Cost of the scientific calculator = ₹ 750
Cost of its battery = ₹ 100
Cost of outer pouch = ₹ 50
Cost Price of the calculator = ₹ 750 + ₹ 100 + ₹ 50 = ₹ 900
S.P = ₹ 850
Here S.P > C.P
Loss = C.P – S.P = 900 – 850 = ₹ 50
Loss = ₹ 50

Question 3.
Nathan paid Rs 800 and bought 10 bottles of honey from a village vendor. He sold them in a gain for Rs 100 per bottle. Find his profit or loss.
Solution:
C.P of 10 bottles of honey = ₹ 800
C.P of 1 bottle honey = 800/10 = ₹ 80
S.P of a bottle honey = ₹ 100
Here S.P > C.P
Profit per bottle = ₹ 100 – ₹ 80 = ₹ 20
Profit for 10 bottles = 20 × 10 = ₹ 200
Profit = ₹ 200

Question 4.
A man bought 400 metres of cloth for ₹ 60,000 and sold it at the rate of ₹ 400 per metre. Find his profit or loss.
Solution:
C.P of 400 metres of cloth = ₹ 60,000
S.P per metre = ₹ 400
S.P of 400 metres of cloth = 400 × 400 = ₹ 1,60,000
Here S.P > C.P
Profit = C.P – S.P = 1,60,000 – 60,000 = ₹ 1,00,000

Challenge Problems

Question 5.
A fruit seller bought 2 dozen bananas at Rs 20 a dozen and sold them at Rs 3 per banana. Find his gain or loss.
Solution:
Cost of one dozen banana = ₹ 20
Cost of 2 dozen bananas = ₹ 20 × 2 = ₹ 40
C.P = ₹ 40
S.P per banana = ₹ 3
S.P for 2 dozen banana = ₹ 3 × 24 = ₹ 72
Here S.P > C.P
Profit = S.P – C.P = 72 – 40 = 32
Profit = ₹ 32

Question 6.
A store purchased pens at ₹ 216 per dozen. He paid ₹ 58 for conveyance and sold the pens at the discount of n per pen and made an overall profit of ₹ 50. Find the M.P of each pen.
Solution:
Cost of a dozen pens = ₹ 216
Paid for conveyance = ₹ 58
Cost price of 12 pens = 216 + 58 = ₹ 274 [∵ 1 dozen = 12]
Profit of 12 pens = ₹ 50
Profit = S.P – C.P
⇒ 50 = S.P – 274
⇒ S.P = 50 + 274 = ₹ 324
Also discount allowed per pen = ₹ 2
Discount for 12 pens = 2 × 12 = ₹ 24
S.P = M.P – Discount
⇒ 324 = M.P – 24
⇒ M.P = 324 + 24 = ₹ 348
Marked price for 12 pens = ₹ 348
M.P of a pen = \(\frac { 348 }{ 12 }\) = ₹ 29
M.P per pen = ₹ 29

Question 7.
A Vegetable vendor buys 10 kg of tomatoes per day at Rs 10 per kg, for the first three days of a week. 1 kg of tomatoes got smashed every day for those 3 days. For the remaining 4 days of the week, he buys 15 kg of tomatoes daily per kg. If for the entire week he sells tomatoes at Rs 20 per kg, then find his profit or loss for the week.
Solution:
Total tomatoes bought for 3 days = 3 × 10 = 30 kg
Cost of 1 kg = ₹ 10
Cost of 30kg tomatoes = 30 × 10 = ₹ 300
Total tomatoes bought for other 4 days = 4 × 15 = 60 kg
Cost of 1 kg = ₹ 8
Cost of 60 kg tomatoes = 60 × 8 = ₹ 480
Total cost of 90 kg tomatoes = 300 + 480 = ₹ 780
C.P = ₹ 780
Tomatoes smashed = 3 kg
Total kg of Tomatoes for sale = 90 – 3 = 87 kg
S.P of 1 kg tomatoes = ₹ 20
S.P of 87 kg tomatoes = 87 × 20 = ₹ 1740
Here S.P > C.P
Profit = S.P – C.P = 1740 – 780 = ₹ 960
Profit = ₹ 960

Question 8.
An electrician buys a used T.V for ₹ 12,000 and a used Fridge for ₹ 11,000. After spending ₹ 1000 on repairing the T.V and ₹ 1500 on painting the Fridge, he fixes up the M.P of T.V as ₹ 15,000 and that of the Fridge as ₹ 15,500. If he gives each ₹ 1000 discount oh each find his profit or loss.
Solution:
(i) Cost of a T.V = ₹ 12,000
Paid for repair = ₹ 1,000
C.P of the T.V = 12,000 + 1000 = ₹ 13,000
M.P of the T.V = ₹ 15,000
Discount on a TV = ₹ 1000
S.P = M.P – Discount = 15,000 – 1000 = ₹ 14,000
Here S.P > C.P
Profit = S.P – C.P = 14,000 – 13,000 = ₹ 1,000
Profit on the T.V = ₹ 1,000
(ii) Cost of the Fridge = ₹ 11,000
Painting charge = ₹ 1500
C.P of the Fridge = 11000 + 1500 = ₹ 12,500
M.P of the Fridge = ₹ 15,500
Discount allowed = ₹ 1000
S.P = M.P – Discount = ₹ 15,500 – ₹ 1000 = ₹ 14,500
Here also S.P > C.P
Profit = ₹ 14,500 – ₹ 12,500 = ₹ 2000
Total profit = Profit on T.V + Profit on Fridge = ₹ 1000 + ₹ 2000 = ₹ 3000
Profit = ₹ 3000

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 2 Integral Calculus I Ex 2.10 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.10

Evaluate the following:

Question 1.
(i) \(\Upsilon\) (4)
Solution:
Γ(4) = Γ(3 + 1) = 3! = 6

(ii) \(\Upsilon\) (\(\frac { 9 }{2}\))
Solution:

(iii) \(\int_{0}^{∞}\) e^-mx x⁶ dx
Solution:
W.K.T \(\int_{0}^{∞}\) xⁿ e^-ax dx = \(\frac { n! }{a^{n+1}}\)
∴ \(\int_{0}^{∞}\) e^-mx x⁶ dx = \(\frac { 6! }{3^{6+1}}\) = \(\frac { 6! }{m^7}\)

(iv) \(\int_{0}^{∞}\) e^-4x x⁴ dx
Solution:

(v) \(\int_{0}^{∞}\) e^-x/2 x⁵ dx
Solution:

Question 2.
If f(x) = \(\left\{\begin{array}{l}
x^{2} e^{-2 x}, x \geq 0 \\
0, \text { otherwise }
\end{array}\right.\), then evaluate \(\int_{0}^{∞}\) f(x) dx
Solution:
Given

Read More:

• ABCAPITAL Pivot Point Calculator

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 3 Integral Calculus II Ex 3.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 3 Integral Calculus II Ex 3.4

Choose the most suitable answer from the given four alternatives:

Question 1.
Area bounded by the curve y = x (4 – x) between the limits 0 and 4 with x-axis is
(a) \(\frac { 30 }{3}\) sq.unit
(b) \(\frac { 31 }{2}\) sq.unit
(c) \(\frac { 32 }{3}\) sq.unit
(d) \(\frac { 15 }{3}\) sq.unit
Solution:
(c) \(\frac { 32 }{3}\) sq.unit
Hint:

Question 2.
Area bounded by the curve y = e^-2x between the limits 0 < x < ∞ is
(a) 1 sq.units
(b) \(\frac { 1 }{2}\) sq.units
(c) 5 sq.units
(d) 2 sq.units
Solution:
(b) \(\frac { 1 }{2}\) sq.units
Hint:

Question 3.
Area bounded by the curve y = \(\frac { 1 }{x}\) between the limits 1 and 2 is
(a) log 2 sq.units
(b) log 5 sq.units
(c) log 3 sq.units
(d) log 4 sq.units
Solution:
(a) log 2 sq.units
Hint:
Area = \(\int_{1}^{2} \frac{1}{x} d x\)
= \((\log x)_{1}^{2}\)
= log 2 – log 1
= log 2 (Since log 1 = 0)

Question 4.
If the marginal revenue function of a firm is MR = e^{\(\frac { -x }{10}\)} then revenue is
(a) 1 – e^-x/10
(b) e^-x/10 + 10
(c) 10(1 – e^-x/10)
(d) -10e^-x/10
Solution:
(c) 10(1 – e^-x/10)
Hint:
MR = e^{\(\frac { -x }{10}\)} then R = ∫MR dx
R = ∫e^-x/10 dx = \(\frac { e^{-x/10} }{(-1/10)}\) + k
R = -10e^-x/10 + k when x = 0, R = 0
⇒ 0 = -10e⁰ + k
0 = -10(1) + k
∴ k = 10
R = -10e^-x/10 + 10 = 10(1 – e^-x/10)

Question 5.
If MR and MC denotes the marginal revenue and marginal cost functions, then the profit functions is
(a) P = ∫(MR – MC) dx + k
(b) P = ∫(R – C) dx + k
(c) P = ∫(MR + MC)dx + k
(d) P = ∫(MR) (MC) dx + k
Solution:
(a) P = ∫(MR – MC) dx + k
Hint:
Profit = Revenue – Cost

Question 6.
The demand and supply functions are given by D(x) = 16 – x² and S(x) = 2x² + 4 are under perfect competition, then the equilibrium price x is
(a) 2
(b) 3
(c) 4
(d) 5
Solution:
(a) 2
Hint:
D(x) =16 – x² and S(x) = 2x² + 4
Under perfect competition D(x) = S(x)
16 – x² = 2x² + 4; 16 – 4 = 2x² + x²
3x² = 12 ⇒ x² = \(\frac { 12 }{3}\) = 4
∴ x = ± 2, x cannot be in negative
∴ x = 2

Question 7.
The marginal revenue and marginal coast functions of a company are MR = 30 – 6x and MC = -24 + 3x where x is the product, profit function is
(a) 9x² + 54x
(b) 9x² – 54x
(c) 54x – \(\frac { 9x^2 }{2}\)
(d) 54x – \(\frac { 9x^2 }{2}\) + k
Solution:
(d) 54x – \(\frac { 9x^2 }{2}\) + k
Hint:
Profit = ∫(MR – MC) dx + k
= ∫(30 – 60) – (-24 + 3x) dx + k
= ∫(54 – 9x) dx + k
= 54x – \(\frac{9 x^{2}}{2}\) + k

Question 8.
The given demand and supply function are given by D(x) = 20 – 5x and S(x) = 4x + 8 if they are under perfect competition then the equilibrium demand is
(a) 40
(b) \(\frac { 41 }{2}\)
(c) \(\frac { 40 }{3}\)
(d) \(\frac { 41 }{5}\)
Solution:
(c) \(\frac { 40 }{3}\)
Hint:
Under perfect competition D(x) = S(x)
20 – 5x = 4x + 8
20 – 8 = 4x + 5x ⇒ 9x = 12
x = \(\frac { 4 }{3}\)
when x = \(\frac { 4 }{3}\); D(x) = 20 – 5(\(\frac { 4 }{3}\)) = 20 – \(\frac { 20 }{3}\)
= \(\frac { 40 }{3}\)

Question 9.
If the marginal revenue MR = 35 +7x – 3x², then the average revenue AR is.
(a) 35x + \(\frac { 7x^2 }{2}\) – x³
(b) 35x + \(\frac { 7x }{2}\) – x²
(c) 35x + \(\frac { 7x }{2}\) + x²
(d) 35x + 7x + x²
Solution:
(c) \(\frac { 40 }{3}\)
Hint:
R = ∫MR dx = ∫(35 + 7x – 3x²) dx

Question 10.
The profit of a function p(x) is maximum when
(a) MC – MR = 0
(b) MC = 0
(c) MR = 0
(d) MC + MR = 0
Solution:
(a) MC – MR = 0
Hint:
P = Revenue – Cost
P is maximum when \(\frac{d p}{d x}\) = 0
\(\frac{d p}{d x}\) = R'(x) – C'(x) = MR – MC = 0

Question 11.
For the demand function p(x), the elasticity of demand with respect to price is unity then.
(a) revenue is constant
(b) a cost function is constant
(c) profit is constant
(d) none of these
Solution:
(a) Revenue is constant

Question 12.
The demand function for the marginal function MR = 100 – 9x² is
(a) 100 – 3x²
(b) 100x – 3x²
(c) 100x – 9x²
(d) 100 + 9x²
Solution:
(a) 100 – 3x²
Hint:
R = ∫(MR) dx + c₁
R = ∫(100 – 9x²) dx + c₁
R = 100x – 3x³ + c₁
When R = 0, x = 0, c₁ = 0
R = 100x – 3x³
Demand function is \(\frac{R}{x}\) = 100 – 3x²

Question 13.
When x₀ = 5 and p₀ = 3 the consumer’s surplus for the demand function p_d = 28 – x²
(a) 250 units
(b) \(\frac { 250 }{3}\) units
(c) \(\frac { 251 }{2}\) units
(d) \(\frac { 251 }{3}\) units
Solution:
(b) \(\frac { 250 }{3}\) units
Hint:

Question 14.
When x₀ = 2 and P₀ = 12 the producer’s surplus for the supply function P₀ = 2x² + 4 is
(a) \(\frac { 31 }{5}\) units
(b) \(\frac { 31 }{2}\) units
(c) \(\frac { 32 }{2}\) units
(d) \(\frac { 30 }{7}\) units
Solution:
(c) \(\frac { 32 }{2}\) units
Hint:
Producer’s Surplus

Question 15.
Area bounded by y = x between the lines y = 1, y = 2 with y = axis is
(a) \(\frac { 1 }{2}\) sq units
(b) \(\frac { 5 }{2}\) sq units
(c) \(\frac { 3 }{2}\) sq units
(d) 1 sq units
Solution:
(c) \(\frac { 3 }{2}\) sq units
Hint:

Question 16.
The producer’s surplus when supply the function for a commodity is p = 3 + x and x₀ = 3 is
(a) \(\frac { 1 }{2}\)
(b) \(\frac { 9 }{2}\)
(c) \(\frac { 3 }{2}\)
(d) \(\frac { 7 }{2}\)
Solution:
(b) \(\frac { 9 }{2}\)
Hint:
p = 3 + x and x₀ = 3
then p₀ = 3 + 3 = 6

Question 17.
The marginal cost function is MC = 100√x find AC given that TC = 0 when the out put is zero is
(a) \(\frac { 200 }{3}\) x^1/2
(b) \(\frac { 200 }{3}\) x^3/2
(c) \(\frac { 200 }{3x^{3/2}}\)
(d) \(\frac { 200 }{3x^{1/2}}\)
Solution:
(a) \(\frac { 200 }{3}\) x^1/2
Hint:
TC = ∫MC dx = ∫100√x dx = 100 ∫(x)^1/2 dx

Question 18.
The demand and supply function of a commodity are P(x) = (x – 5)² and S(x) = x² + x + 3 then the equilibrium quantity x₀ is
(a) 5
(b) 2
(c) 3
(d) 10
Solution:
(b) 2
Hint:
At equilibrium, P(x) = S(x)
⇒ (x – 5)² = x² + x + 3
⇒ x² – 10x + 25 = x² + x + 3
⇒ 11x = 22
⇒ x = 2

Question 19.
The demand and supply function of a commodity are D(x) = 25 – 2x and S(x) = \(\frac { 10+x }{2}\) then the equilibrium price P₀ is
(a) 2
(b) 2
(c) 3
(d) 10
Solution:
(a) 2
Hint:
At equilibrium, D(x) = S(x)
25 – 2x = \(\frac{10+x}{4}\)
⇒ 100 – 8x = 10 + x
⇒ x = 10
That is x₀ = 10
P₀ = 25 – 2(x₀) = 25 – 20 = 5

Question 20.
If MR and MC denote the marginal revenue and marginal cost and MR – MC = 36x – 3x² – 81, then maximum profit at x equal to
(a) 3
(b) 6
(c) 9
(d) 10
Solution
(c) 9
Hint:
Profit P = ∫(MR – MC) dx = ∫(36x – 3x² – 81) dx
P = [\(\frac { 36x^2 }{2}\) – \(\frac { 3x^3 }{3}\) – 81x] = 18x² – x³ – 81x
when p = 0; 18x² – x³ – 81x = 0 ⇒ x² – 18x + 81 = 0
(x – 9)² = 0 ⇒ x – 9 = 0
∴ x = 9

Question 21.
If the marginal revenue of a firm is constant, then the demand function is
(a) MR
(b) MC
(c) C(x)
(d) AC
Solution:
(a) MR
Hint:
MR = k (constant)
Revenue function R = ∫(MR) dx + c₁
= ∫kdx + c₁
= kx + c₁
When R = 0, x = 0, ⇒ c₁ = 0
R = kx
Demand function p = \(\frac{R}{x}=\frac{k x}{x}\) = k constant
⇒ p = MR

Question 22.
For a demand function p, if ∫\(\frac { dp }{p}\) = k ∫\(\frac { dx }{x}\) then k is equal to
(a) n_d
(b) -n_d
(c) \(\frac { -1 }{n_d}\)
(d) \(\frac { 1 }{n_d}\)
Solution:
(c) \(\frac { -1 }{n_d}\)

Question 23.
The area bounded by y = e^x between the limits 0 to 1 is
(a) (e – 1) sq.units
(b) (e + 1) sq.units
(c) (1 – \(\frac { 1 }{e}\)) sq.units
(d) (1 + \(\frac { 1 }{e}\)) sq.units
Solution:
(a) (e – 1) sq.units
Hint:
Area A = \(\int_{a}^{b}\)ydx = \(\int_{0}^{1}\)e^xdx = [e^x]\(_{0}^{1}\)
= [e^x – e°] = [e – 1]

Question 24.
The area bounded by the parabola y² = 4x bounded by its latus rectum is
(a) \(\frac { 16 }{3}\) sq units
(b) \(\frac { 8 }{3}\) sq units
(c) \(\frac { 72 }{3}\) sq units
(d) \(\frac { 1 }{3}\) sq units
Solution:
(b) \(\frac { 8 }{3}\) sq units
Hint:

y² = 4x ⇒ y = \(\sqrt { 4x}\) 2√x = 2(x)^1/2
In this parabola 4a = 4 ⇒ a = 1 and vertex V(0, 0)

Question 25.
The area bounded by y = |x| between the limits 0 and 2 is
(a) 1 sq.units
(b) 3 sq.units
(c) 2 sq.units
(d) 4 sq.units
Solution:
(c) 2 sq.units
Hint:
Area A = \(\int_{a}^{b}\)ydx = \(\int_{0}^{2}\)x dx = [ \(\frac { x^2 }{2}\) ]\(_{0}^{2}\)
= \(\frac { (2)^2 }{2}\) – (0) = \(\frac { 4 }{2}\) = 2

Must Read:

Icici Bank Pivot Point