Important Questions

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 16 Chemistry in Everyday Life Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 16 Chemistry in Everyday Life

Very Short Answer Type Questions

Question 1.
Where are receptors located? [NCERT Exemplar]
Answer:
Receptors are embedded in cell membrane.

Question 2.
Which site of an enzyme is called allosteric site? [NCERT Exemplar]
Answer:
Sites different from active site of enzyme where a molecule can bind and affect the active site is called allosteric site.

Question 3.
What is the harmful effect of hyperacidity? [NCERT Exemplar]
Answer:
Ulcer development in stomach.

Question 4.
Write the name of an antacid which is often used as a medicine.
Answer:
Ranitidine (Zantac).

Question 5.
What is the medicinal use of narcotic drugs? [NCERT Exemplar]
Answer:
Since narcotic drugs relieve pain and produce sleep, these are chiefly used for the relief of post-operative pain, cardiac pain and pain of terminal cancer and in childbirth.

Question 6.
Which type of drugs come under antimicrobial drugs? [NCERT Exemplar]
Answer:
Antiseptics, antibiotics and disinfectants.

Question 7.
What is the mode of action of antimicrobial drugs? [NCERT Exemplar]
Answer:
Antimicrobial drugs can kill the microorganism such as bacteria, virus, fungi or other parasites. They can, alternatively, inhibit the pathogenic action of microbes.

Question 8.
Which one of the following drugs is an antibiotic? Morphine, Equanil, Chloramphenicol, Aspirin
Answer:
Chloramphenicol.

Question 9.
What is meant by ‘narrow-spectrum antibiotics’?
Answer:
Antibiotics which are mainly effective against Gram-positive or Gram-negative bacteria are known as narrow-spectrum antibiotics. For , example, penicillin G.

Question 10.
Define the limited spectrum antibiotics.
Answer:
Antibiotics which are mainly effective against a single organism or disease, are called as limited spectrum antibiotics.

Short answer type questions

Question 1.
Why are certain drugs called enzyme inhibitors?[NCERT Exemplar]
Answer:
Enzymes have active sites that bind the substrate for effective and quick chemical reactions. The functional groups present at the active site of enzyme interact with functional groups of substrate via ionic bonding, hydrogen bonding, van der Waal’s interaction, etc. Some drugs interfere with this interaction by blocking the binding site of enzyme and prevent the binding of actual substrate with enzyme. This inhibits the catalytic activity of the enzyme, therefore, these are called inhibitors.

Question 2.
Sodium salts of some acids are very useful as food preservatives. Suggest a few such acids. [NCERT Exemplar]
Answer:
A preservative is naturally occurring or synthetically produced substance that is added to foods to prevent decomposition by microbial growth or by undesirable chemical changes. Sodium salts of some acids are very useful as food preservatives.

Some examples of such acids are as follows :

• Benzoic acid in the form of its sodium salts constitutes one of the most common food preservatives. Sodium benzoate is a common preservative in acid or acidified foods such as fruit, juices, pickles etc. Yeasts are inhibited by benzoate to a greater extent than are moulds and bacteria.
• Sorbic acid and its salts (sodium, potassium, and calcium) also have preservative activities but the applications of -sodium sorbate (C₆H₇NaO₂) are limited compared to that for potassium salt.
• Sodium erythorbate (C₆H₇NaO₆) is a food additive used predominate in meats, poultry and soft drinks.
• Sodium propanoate[Na(C₂H₅COO)] is used in bakery products as mould inhibitor.

Question 3.
What is the side product of soap industry? Give reactions showing soap formation. [NCERT Exemplar]
Answer:
Soaps are sodium or potassium salts of long-chain fatty acids such as stearic acid, oleic acid and palmitic acid. Soaps containing sodium salts are formed by heating fat (i. e., glyceryl ester of fatty acid) with aqueous sodium hydroxide solution.

This reaction is known as saponification. In this reaction, esters of fatty acids are hydrolyzed and the soap obtained remains in colloidal form. It is precipitated from the solution by adding NaCl. The solution left after removing the soap contains glycerol as side product.

Question 4.
Hair shampoos belong to which class of synthetic detergent? [NCERT Exemplar]
Answer:
Hair shampoos are made up of cationic detergents. These are quaternary ammonium salts of amines with acetates, chlorides or bromides as anions, e. g., cetyltrimethylammonium bromide.

Question 5.
Explain why sometimes foaming is seen in river water near the place where sewage water is poured after treatment?
[NCERT Exemplar]
Answer:
Detergents have long hydrocarbon chains. If their hydrocarbon chain is highly branched, then bacteria cannot degrade this easily. Such detergents are non-biodegradable. Slow degradation of detergents leads to their accumulation.

These non-biodegradable detergents persist in water even after sewage treatment and cause foaming in rivers, ponds and their water get polluted. In order to overcome this issue branching of the hydrocarbon chain is controlled and kept to a minimum.

Long answer type questions

Question 1.
What are enzyme inhibitors? Classify them on the basis of their mode of attachment on the active site of enzymes. With the help of diagrams explain how do inhibitors inhibit the enzymatic activity? [NCERT Exemplar]
Answer:
Enzymes are responsible to hold the substrate molecule for a chemical reaction and they provide functional groups which will attack the substrate to carry out the chemical reaction. Drugs which inhibit any of the two activities of enzymes are called enzyme inhibitors.

Enzyme inhibitors can block the binding site thereby preventing the binding of the substrate to the active site and hence inhibiting the catalytic activity of the enzyme.
Drugs inhibit the attachment of natural substrate on the active site of enzymes in two different ways as explained below :
(i) Drugs which compete with natural substrate for their attachment on the active sites of enzymes are called competitive inhibitors.

(ii) Some drugs, however, do not bind to the active site but bind to a different site of the enzyme which is called allosteric site. This binding of the drug at allosteric site changes the shape of the active site of the enzyme in such a way that the natural substrate cannot recognize it. Such enzymes are called non-competitive inhibitors.

Question 2.
In what respect to prontosil and salvarsan resemble? Is there any resemblance between azo dye and prontosil? Explain. [NCERT Exemplar]
Answer:
Prontosil, also called sulfamide chrysoidine, trade name of the first synthetic drug used in the treatment of general bacterial infections in humans. Prontosil resulted from research, directed by German chemist and pathologist Gerhard Domagk, on the antibacterial action of azo dyes. A red azo dye of low toxicity, prontosil was shown by Domagk to prevent mortality in mice infected with Streptococcus bacteria.

The dye was also effective in controlling staphylococcus infections in rabbits. Within a relatively short period, it was demonstrated that prontosil was effective not only in combating experimental infections in animals but also against Streptococcal disease in humans, including meningitis and puerperal sepsis. Structural formula of prontosil is as follows :

From the structure of prontosil, it is very clear that it has -N = N- linkage. It was discovered that the part of the structure of prontosil molecule shown inbox, i.e., p-amino benzene sulphonamide has antibacterial activity.
Salvarsan is also known as arsphenamine. It was introduced at the beginning of 1910s as the first effective treatment for syphilis. It is an organoarsenic molecule and has -As = As- double bond.

Salvarsan and prontosil show similarity in their structure. Both of these drugs are antimicrobials. Salvarsan contains -As = As- linkage whereas prontosil has —N = N— linkage.

Prontosil (a red azo dye) and azo dye both have -N = N- linkage.

Question 3.
Ashwin observed that his friend Shubhain was staying aloof, not playing with friends, and becoming easily irritable for some weeks. Ashwin told his teacher about this, who, in turn, called Shubham’s parents and advised them to consult a doctor. The doctor after examining Shubham prescribed antidepressant drugs for him.
After reading the above passage, answer the following questions:
(i) Name two antidepressant drugs.
(ii) Mention the values shown by Ashwin.
(iii) How should Shubham’s family help him other than providing medicine?
(iv) What is the scientific explanation for the feeling of depression?
Answer:
(i) Equanil, Iproniazid, phenelzine (any two)
(ii) Empathetic, caring, sensitive.
(iii) They should talk to him, be a patient listener, can discuss the matter with the psychologist.
(iv) If the level of noradrenaline is low, then the signal sending activity becomes low and the person suffers from depression.

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 2 Solutions Important Questions and Answers.

PSEB 12th Class Chemistry Important Questions Chapter 2 Solutions

Very short answer type questions

Question 1.
Define mole fraction.
Answer:
Mole fraction of a component in a solution may be defined as the ratio of moles of that compdnent to the total number of moles of all the components present in the solution.

Question 2.
What is the similarity between Raoult’s law and Henry’s law?
Answer:
The similarity between Raoult’s law and Henry’s law is that both state that the partial vapour pressure of the volatile component or gas is directly proportional to its mole fraction in the solution.

Question 3.
Why is the vapour pressure of a solution of glucose in water lower than that of water? (NCERT Exemplar)
Answer:
This is due to decrease in the escaping tendency of the water molecules from the surface of solution as some of the surface area is occupied by non-volatile solute, glucose particles.

Question 4.
What type of deviation is shown by a mixture of ethanol and acetone? What type of azeotrope is formed by mixing ethanol and acetone?
Answer:
Positive deviation.
Minimum boiling azeotropes.

Question 5.
State how does osmotic pressure vary with temperature.
Answer:
Osmotic pressure increases with increase in temperature.

Question 6.
What are isotonic solutions?
Answer:
The solutions of the same osmotic pressure at a given temperature are called isotonic solutions.

Question 7.
Define van’t Hoff factor.
Answer:
van’t Hoff factor may be defined as the ratio of normal molecular mass to observed molecular mass or the ratio of observed colligative property to normal colligative property.

Question 8.
Why are aquatic species more comfortable in cold water in comparison to warm water? (NCERT Exemplar)
Answer:
Solubility of oxygen in water increases with decrease in temperature. Presence of more oxygen at lower temperature makes the aquatic species more comfortable in cold water.

Question 9.
What is semipermeable membrane? (NCERT Exemplar)
Answer:
Continuous sheets or films (natural or synthetic) which contain a network of submicroscopic holes or pores through which small solvent molecules (water etc.) can pass, but solute molecules of bigger size cannot pass are called semipermeable membrane.

Question 10.
Give an example of a material used for making semipermeable membrane for carrying out reverse osmosis. (NCERT Exemplar)
Answer:
Cellulose acetate.

Short answer type questions

Question 1.
State Henry’s law. Write its one application. What is the effect of temperature on solubility of gases in liquid?
Answer:
It states that the partial pressure of a gas in vapour phase (p) is proportional to the mole fraction of the gas (χ) in the solution.
p ∝ χ or p = K_Hχ where K_H is the
Henry’s constant.
Application of Henry’s law:
To increase the solubility of CO₂ in soft drinks and soda water, the bottle is sealed under high pressure.

Effect of temperature on solubility:
As dissolution is an exothermic process, therefore, according to Le Chatelier’s principle solubility should decrease with rise in temperature.

Question 2.
Why does a solution containing non-volatile solute have higher boiling point than the pure solvent? Why is elevation of boiling point a colligative property?
Answer:
The addition of a non-volatile solute to a volatile solvent lowers its vapour pressure. In order to boil the solution, i.e., to make its vapour pressure equal to atmospheric pressure, the solution has to be heated at a higher temperature. In other words, the boiling point of solution becomes higher than solvent.

As elevation in boiling point depends on the number of moles of solute particles and independent of their nature, therefore, it is a colligative property.

Question 3.
Will the elevation in boiling point be same if 0.1 mol of sodium chloride or 0.1 mol of sugar is dissolved in 1L of water? Explain.
Answer:
No, the elevation in boiling point is not the same. NaCl, being an electrolyte, dissociates almost completely to give Na⁺ and Cl^– ions whereas glucose, being non-electrolyte does not dissociate. Hence, the number of particles in 0.1 M NaCl solution is nearly double than 0.1 M glucose solution. Elevation in boiling point being a colligative property, is therefore, nearly twice for 0.1 M NaCl solution than for 0.1 M glucose solution.

Question 4.
Explain the solubility rule ‘like dissolves like’ in terms of intermolecular forces that exist in solutions. (NCERT Exemplar)
Answer:
If the intermolecular interactions are similar in both constituents, i.e., solute and solvent then solute dissolves in the solvent. e.g., polar solutes dissolve in polar solvents and non-polar solutes in non-polar solvents. Thus, the statement ‘like dissolved like’ proves to be true.

Question 5.
Concentration terms such as mass percentage, ppm, mole fraction and molality are independent of temperature. However, molarity is a function of temperature. Explain.
(NCERT Exemplar)
Answer:
Molarity is defined as the number of moles of solute dissolved per litre of a solution. Since, volume depends on temperature and changes with change in temperature, the molarity will also change with change in temperature. On the other hand, mass does not change with change in temperature, so other concentration terms given in the question also do not do so. According to the definition of all these terms, mass of solvent used for making the solution is related to the mass of solute.

Question 6.
Which of the following solutions has higher freezing point?
0.05 M Al₂(SO₄)₃, 0.1 M K₃ [Fe(CN)₆] Justify.
Answer:
0.05 M Al₂(SO₄)₃ has higher freezing point.

Justification:
For 0.05 MAl₂(SO₄) ₃,i = 5
Number of particles = i × concentration
= 5 × 0.05
= 0.25 moles of ions
For 0.1MK₃ [Fe(CN)₆], i = 4
Number of particles = i × concentration
= 4 × 0.1
= 0.4 moles of ions
We know that, ΔT_f. Number of particles
Hence, 0.05 M Al₂(SO₄)₃ has higher freezing point because it has lower number of particles than 0.1 M K₃ [Fe (CN)₆].

Question 7.
The freezing point of benzene decreases by 2.12 K when 2.5 g of benzoic acid (C₆H₅COOH) is dissolved in 25 g of benzene. If benzoic acid forms a dimer in benzene, calculate the van’t Hoff factor and the percentage association of benzoic acid. (K_f for benzene = 5.12 K kg mol^-1)
Answer:
Given, ΔT_f = 2.12 K, K_f = 5.12 K kg mol^-1
We know that, ΔT_f = i K_f m
2.12 = \(\frac{i \times 5.12 \times 2.5 \times 1000}{122 \times 25}\)
or i = 0.505
For association,
i = 1 – \(\frac{\alpha}{2}\)
0.505 = 1 – \(\frac{\alpha}{2}\)
or α =0.99
Hence, percentage association of benzoic acid is 99%.

Long answer type questions

Question 1.
(a) What type of deviation is shown by a mixture of ethanol and acetone? Give reason.
(b) A solution of glucose (molar mass = 180 g mol^-1) in water is labelled as 10% (by mass). What would be the molality and molarity of the solution?
(Density of solution = 1.2 g mL^-1)
Answer
(a) A mixture of ethanol and acetone shows positive deviation from Raoult’s law.
In pure ethanol hydrogen bond exist between the molecules. On adding acetone to ethanol, acetone molecules get in between the molecules of . ethanol thus breaking some of the hydrogen bonds and weakening molecular interactions considerably. Weakening of molecular interactions leads to increase in vapour pressure resulting in positive deviation from Raoult’s law.

(b) Let the mass of solution = 100 g
∴ Mass of glucose = 10 g
Number of moles of glucose \(=\frac{\text { Mass of glucose }}{\text { Molar mass }}\)
= \(\frac{10 \mathrm{~g}}{180 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 0.056 mol

Question 2.
Discuss biological arid industrial importance of osmosis.
(NCERT Exemplar)
Answer:
The process of osmosis is of great biological and industrial importance as is evident from the following examples:

• Movement of water from soil into plant roots and subsequently into upper portion of the plant occurs partly due to osmosis.
• Preservation of meat against bacterial action by adding salt.
• Preservation of fruits against bacterial action by adding sugar.
  Bacterium in canned fruit loses water through the process of osmosis, shrivels and dies.
• Reverse osmosis is used for desalination of water.

Punjab State Board Syllabus PSEB 12th Class Chemistry Important Questions Pdf in English Medium and Punjabi Medium are part of PSEB Solutions for Class 12.

PSEB 12th Class Chemistry Important Questions in Punjabi English Medium

• Chapter 1 The Solid State Important Questions
• Chapter 2 Solutions Important Questions
• Chapter 3 Electrochemistry Important Questions
• Chapter 4 Chemical Kinetics Important Questions
• Chapter 5 Surface Chemistry Important Questions
• Chapter 6 General Principles and Processes of Isolation of Elements Important Questions
• Chapter 7 The p-Block Elements Important Questions
• Chapter 8 The d-and f-Block Elements Important Questions
• Chapter 9 Coordination Compounds Important Questions
• Chapter 10 Haloalkanes and Haloarenes Important Questions
• Chapter 11 Alcohols, Phenols and Ethers Important Questions
• Chapter 12 Aldehydes, Ketones and Carboxylic Acids Important Questions
• Chapter 13 Amines Important Questions
• Chapter 14 Biomolecules Important Questions
• Chapter 15 Polymers Important Questions
• Chapter 16 Chemistry in Everyday Life Important Questions

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 14 Biomolecules Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 14 Biomolecules

Very short answer type questions

Question 1.
What are oligosaccharides?
Answer:
Carbohydrates which on hydrolysis give two to ten molecules of monosaccharides are called oligosaccharides e. g., sucrose.

Question 2.
How do you explain the presence of all the six carbon atoms in glucose in a straight chain? (NCERT Exemplar)
Answer:
On prolonged heating with HI, glucose gives n-hexane.

Question 3.
Write the product obtained when D-glucose reacts with H₂N—OH.
Answer:

Question 4.
What are anomers?
Answer:
A pair of stereoisomers such as α-D-(+) glucose and β-D-(+) glucose which differ in configuration only around C₁ are called anomers.

Question 5.
What type of linkage is present in proteins?
Answer:

Question 6.
What are Vitamins?
Answer:
Vitamins are generally regarded as organic compounds required in the diet in small amounts to perform specific biological functions for normal maintenance of optimum growth and health of the organism.

Question 7.
Why must vitamin C be supplied regularly in diet? [NCERT Exemplar]
Answer:
Vitamin C is water-soluble hence, they are regularly excreted in urine and can not be stored in our body, so, they are supplied regularly in diet.

Question 8.
Of the two bases named below, which one is present in RNA and which one is present in DNA? Thymine, Uracil.
Answer:

1. Thymine is present in DNA.
2. Uracil is present in RNA.

Question 9.
The activation energy for the acid-catalyzed hydrolysis of sucrose is 6.22 kJ mol^-1, while the activation energy is only 2.15 kJ mol-1 when hydrolysis is catalyzed by the enzyme sucrase. Explain. [NCERT Exemplar]
Answer:
Enzymes, the biocatalysts reduce the magnitude of activation energy by providing alternative paths. In the hydrolysis of sucrose, the enzyme sucrase reduces the activation energy from 6.22 kJ mol^-1 to 2.15 kJ mol^-1.

Question 10.
Name the bases present in RNA. Which one of these is not present in DNA?
Answer:
The bases present in RNA are Adenine (A), Guanine (G), Cytosine (C) and Uracil (U). Uracil is not present in DNA.

Short answer type questions

Question 1.
What is essentially the difference between α-glucose and β-glucose? What is meany by pyranose structure of glucose?
Answer:
α -Glucose and β-Glucose differ only in the configuration of hydroxy group at C₁ and are called anomers and the C₁ carbon is called anomeric carbon. The six-membered cyclic structure of glucose is called pyranose (α-or β), in analogy with pyran. The cyclic structure of glucose is more correctly represented by Haworth structure as given below:

Question 2.
Describe the term D- and L-configuration used for amino acids with examples. [NCERT Exemplar]
Answer:
All naturally occurring a-amino acids (except glycine) are optically active due to the presence of chiral carbon atom. These have either D- or  L-configuration. D-form means that the amino (-NH₂) group is present towards the right-hand side. L-form shows the presence of -NH₂ group on the left-hand side.

Question 3.
Aldopentoses named as ribose and 2-deoxyribose are found in nucleic acids. What is their relative configuration? [NCERT Exemplar]
Answer:
In case of cyclic structure of saccharide, if -OH group present at second last carbon is present at bottom side, then it is considered as D configuration (as shown above).

Question 4.
How do enzymes help a substrate to be attacked by the reagent effectively? [NCERT Exemplar]
Answer:
At the surface of enzyme, active sites are present. These active sites of enzymes hold the substrate molecule in a suitable position so that it can be attacked by the reagent effectively. This reduces the magnitude of activation energy.

Enzymes contains cavities of characteristics shape and possessing active groups known as active centre on the surface. The molecules of the reactant (substrate) having complementary shape, fit into these cavities. On account of these active groups, an activated complex is formed which then decomposes to yield the products.

Question 5.
Amino acids can be classified as α-, β-, γ-, δ- and so on depending upon the relative position of amino group with respect to carboxyl group. Which type of amino acids form polypeptide chain in proteins? [NCERT Exemplar]
Answer:
α-amino acid forms a polypeptide chain by elimination of water molecules.

Question 6.
(i) Winch vitamin deficiency causes rickets?
(ii) Name the base that is found in nucleotide of RNA only.
(iii) Glucose on reaction with acetic acid gives glucose Penta acetate. What does it suggest about the structure of glucose?
Answer:
(i) Vitamin D
(ii) Uracil
(iii) 5-OH groups are present in glucose.

Long answer type questions

Question 1.
Which moieties of nucleosides are involved in the formation of phosphodiester linkages present in dinucleotides? What does the word diester in the name of linkage indicate? Which acid is involved in the formation of this linkage?
Answer:
(i) 5′ and 3′ carbon atoms of pentose sugar.
(ii) Most probably the resemblance of with 2 esters (-COO)² groups joined together.

(iii) Phosphoric acid (H₃PO₄).
Nucleosides are joined together by phosphodiester linkage between 5′ and 3′ carbon atoms of pentose sugar and a dinucleotide with phosphoric acid (CH₃PO₄) is formed.

Question 2.
Write the structures of fragments produced on complete hydrolysis of DNA. How are they linked in DNA molecules? Draw a diagram to show pairing of nucleotide bases in double helix of DNA. [NCERT Exemplar]
Answer:
On complete hydrolysis of DNA, following fragments are formed-pentose sugar (β-D-2-deoxyribose), phosphoric acid (H₃PO₄) and bases (nitrogen-containing heterocyclic compounds).
Structures
(i) Sugar

(ii) Phosphoric acid

(iii) Nitrogen bases: DNA contains four bases.
Adenine (A), Guanine (G), Cytosine (C), and Thymine (T).

A unit formed by the attachment of a base to 1′-position of sugar is called nucleoside. When nucleoside links to phosphoric acid at 5′-position of sugar moiety, a nucleotide is formed. Nucleotides are joined together by phosphodiester linkage between 5′ and 3′ carbon atoms of the pentose sugar. In DNA two nucleic acid chains are wound about each other and held together by hydrogen bonds between pairs of bases.

The two strands are complementary to each other because hydrogen bonds are formed between specific pairs of base adenine form hydrogen bonds with thymine whereas cytosine forms hydrogen bonds with guanine.

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 15 Polymers Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 15 Polymers

Very short answer type questions

Question 1.
Which of the following is a natural polymer? Buna-S, Proteins, PVC
Answer:
Proteins.

Question 2.
Can enzyme be called a polymer? [NCERT Exemplar]
Answer:
Enzymes are biocatalysts which are proteins and are thus polymers.

Question 3.

Answer:
Homopolymer.

Question 4.
Identify the type of polymer -A-B-B-A-A-A-B-A- [NCERT Exemplar]
Answer:
Copolymer.

Question 5.
Out of chain growth polymerisation and step-growth polymerisation, in which type will you place the following: [NCERT Exemplar]

Answer:
Chain growth polymerisation, as there is no loss of small molecules like water; methanol, etc.

Question 6.
What is the role off-butyl peroxide in the polymerisation of ethene?
Answer:
It acts as a free radical generating initiator in the chain initiation step of polymerisation of ethene.

Question 7.
Can nucleic acids, proteins and starch be considered as step growth polymers? [NCERT Exemplar]
Answer:
Yes, step-growth polymers are condensation polymers and they are formed by the loss of simple molecules like water leading to the formation of high molecular mass polymers.

Question 8.
Write the structure of the monomer used for getting the melamine-formaldehyde polymer.
Answer:
Melamine and formaldehyde

Question 9.
How is the following resin intermediate prepared and which polymer is formed by this monomer unit?

Answer:
Melamine and formaldehyde are starting materials for this intermediate. Its polymerisation gives melamine polymer.

Question 10.
Why does cis-polyisoprene possess elastic property? [NCERT Exemplar]
Answer:
cis-polyisoprene is also known’ as natural rubber. Its elastic property is due to the existence of weak van der Waals’ interactions between their various polymer chains.

Short answer type questions

Question 1.
Which of the following polymers soften on heating and harden on cooling? What are the polymers with this property collectively called? What are the structural similarities between such polymers? Bakelite, urea-formaldehyde resin, polythene, polyvinyls, polystyrene. [NCERT Exemplar]
Answer:
Polythene, polyvinyls and polystyrene soften on heating and harden on cooling. Such polymers are called thermoplastic polymers. These polymers are linear or slightly branched long-chain molecules. These possess intermolecular forces whose strength lies between strength of intermolecular forces of elastomers and fibres.

Question 2.
What is the role of benzoyl peroxide in addition polymerisation of alkenes? Explain its mode of action with the help of an example. [NCERT Exemplar]
Answer:
Role of benzoyl peroxide is to initiate the free radical polymerisation reaction which can be easily understood by taking an example of polymerisation of ethene of polythene.
(i) Chain initiation

(ii) Chain propagation

(iii) Chain terminator step

Question 3.
Low-density polythene and high-density polythene, both are polymers of ethene but there is marked difference in their properties. Explain. [NCERT Exemplar]
Answer:
Low density and high-density polythenes are obtained under different conditions. These differ in their structural features. Low-density polythenes are highly branched structures while high-density polythene consists of closely packed linear molecules. Close packing increases the density.

Question 4.
Differentiate between rubbers and plastics on the basis of intermolecular forces. [NCERT Exemplar]
Answer:
Rubber is a natural polymer which possess elastic properties. Natural polymer is a linear polymer of isoprene (2-methyl-1, 3-butadiene).
In natural rubber cis-polyisoprene molecules consists of various chains held together by weak van der Waals’ interaction and has coiling structure. So, it can be stretched like a spring.

Plastics are generally polymers of ethene known as polythene. Polythene is thermoplastic polymer which may be linear (HDP) or branched (LDP) these type of polymers. Possesses intermediate intermolecular forces of attraction. It has linear, structure that can be moulded but can’t be regained on its original shape after stretching.

Question 5.
A natural linear polymer of 2-methyl-l, 3-butadiene becomes hard on treatment with sulphur between 373 to 415 K and -S-S- bonds are formed between chains. Write the structure of the product of this treatment? [NCERT Exemplar]
Answer:
The product is called vulcanised rubber. Its structure is as follows :

Question 6.
Name the type of reaction involved in the formation of the following polymers from their respective monomers
(i) PVC.
(ii) Nylon6.
(iii) PHBV.
Answer:
(i) Addition
(ii) Condensation/Hydrolysis
(iii) Condensation.

Long answer type questions

Question 1.
Explain the following terms giving a suitable example for each:
(i) Elastomers
(ii) Condensation polymers
(iii) Addition polymers
Answer:
(i) Elastomers: These are the polymers having the weakest intermolecular forces of attraction between the polymer chains. The weak forces permit the polymer to be stretched. A few ‘cross links’ are introduced between the chains, which help the polymer to retract to its original position after the force is released as in vulcanised rubber. Elastomers thus possess an elastic character. For example, buna-S, buna-N, neoprene, etc.

(ii) Condensation polymers: The condensation polymers are formed by the repeated condensation reaction between different bi-functional or tri-functional monomer units usually with elimination of small molecules such as water, alcohol, hydrogen chloride, etc. For example, Nylon-6,6, nylon 6, terylene, etc.

(iii) Addition polymers: Addition polymers are formed by repeated addition of same or different monomer molecules. The monomers used are unsaturated compounds. For example, alkenes alkadienes and their derivatives. Polythene is an example of addition polymer.

Question 2.
Phenol and formaldehyde undergo condensation to give a polymer (A) which on heating with formaldehyde gives a thermosetting polymer (B). Name the polymers. Write the reactions involved in the formation of (A). What is the structural difference between two polymers? [NCERT Exemplar]
Answer:
Phenol and formaldehyde undergo condensation to give a polymer novolac (A) which on heating with formaldehyde gives bakelite (B) as a thermosetting polymer.
A sequence of the reaction can be written as follows :

Structural difference in between these two is that novolac is a linear polymer while bakelite is a cross-linked polymer.

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 1 The Solid State Important Questions and Answers.

PSEB 12th Class Chemistry Important Questions Chapter 1 The Solid State

Very short answer type questions

Question 1.
Give an example each of a molecular solid and an ionic solid.
Answer:
Molecular solids: CO₂, I₂, HCl
Ionic solids: NaCl, ZnS, CaF₂

Question 2.
Why does the window glass of the old buildings look milky?
Answer:
It is due to heating during the day and cooling at night, i.e., due to annealing over a number of years, glass acquires crystalline character.

Question 3.
What would be the nature of solid if there is no energy gap between valence band and conduction band?
Answer:
Conductor.

Question 4.
Express the relationship between atomic radius (r) and the edge length (a) in the bcc unit cell,
Answer:
Atomic radius (r) = \(\frac{\sqrt{3}}{4}\) a (edge length of unit cell).

Question 5.
Which point defect in its crystal units increases the density of a solid?
Answer:
Interstitial defect.

Question 6.
What is meant by the term ‘forbidden zone’ in reference to band theory of solids?
Answer:
The energy gap between valence band and conduction band is known as forbidden zone.

Question 7.
‘Crystalline solids are anisotropic in nature.’ What does this statement mean?
Answer:
It means that some of their physical properties like electrical conductivity, refractive index, etc., are diferent in different directions.

Question 8.
Why does the electrical conductivity of semiconductors increase with rise in temperature? [NCERT Exemplar)
Answer:
The gap between conduction band and valence band is small in semiconductors. Therefore, electrons from the valence band can jump to the conduction band on increasing temperature. Thus, they become more conducting as the temperature increases.

Question 9.
Why does table salt NaCl sometimes appear yellow in colour?
(NCERT Exemplar)
Answer:
Yellow colour in NaCl is due to metal excess defect due to which unpaired electrons occupy anionic sites, known as F-centres. These electrons absorb energy from the visible region for the excitation which makes crystal appear yellow.

Question 10.
Why are liquids and gases categorised as fluids? (NCERT Exemplar)
Answer:
Liquids and gases have the tendency to flow, i.e., their molecules can move freely from one place to another. Therefore, they are known as fluids.

Short answer type questions

Question 1.
(i) What type of stoichiometric defect is shown by KC1 and why?
(ii) What type of semiconductor is formed when silicon is doped with As?
(iii) Which one of the following is an example of molecular solid : CO₂ or SiO₂?
Answer:
(i) KCl shows Schottky defect as the cation, K⁺ and anion, Cl^– are of almost similar sizes.
(ii) n-type semiconductor.
(iii) CO₂

Question 2.
Iron has a body centred cubic unit cell with a cell dimension of 286.65 pm. The density of iron is 7.874 g cm^-3. Use this information to calculate Avogadro’s number. (At. mass of Fe = 55.845 u)
Solution:
Given, a =286.65 pm = 286.65 × 10^-10 cm; M = 55.845 g mol^-1;
d = 7.874 g cm^-3
For bee unit cell, z = 2
Substituting the values in the expression, N_A = \(\frac{z \times M}{a^{3} \times d}\), we get
N_A = \(\frac{2 \times 55.845 \mathrm{~g} \mathrm{~mol}^{-1}}{\left(286.65 \times 10^{-10} \mathrm{~cm}\right)^{3} \times 7.874 \mathrm{~g} \mathrm{~cm}^{-3}}\)
N_A = 6.022 × 10²³ mol^-1

Question 3.
An element crystallises in a fee lattice with cell edge of 400 pm. The density of the element is 7 g cm^-3. How many atoms are present in 280 g of the element?
Solution:
Given, a = 400 pm = 400 × 10^-10 cm = 4 x× 10^-8 cm
Volume of the unit cell = a³
= (4 × 10^-8 cm)³ = 6.4 × 10^-23cm³

Since each fee unit cell contains 4 atoms, therefore, the total number of atoms in 280 g = 4 × 6.25 × 10²³ = 2.5 × 10²⁴ atoms

Question 4.
An element crystallises in a fee lattice with cell edge of 400 pm. Calculate the density if 200 g of this element contain 2.5 × 10²⁴ atoms.

Molar mass M = 48.18 g mol^-1
Here, z = 4, M = 48.18 g mol^-1, N_A = 6.022 × 10²³ mol^-1
a = 400 pm = 400 × 10^-10 cm = 4 × 10^-8 cm
Substituting these values in the expression,
d = \(\frac{z \times M}{a^{3} \times N_{A}}\) , we get
d = \(\frac{4 \times 48.18 \mathrm{~g} \mathrm{~mol}^{-1}}{\left(4 \times 10^{-8} \mathrm{~cm}\right)^{3} \times\left(6.022 \times 10^{23} \mathrm{~mol}^{-1}\right)}\) = 5 g cm^-3

Question 5.
Explain why does conductivity of germanium crystals increase on doping with galium?
Answer:
On doping germanium with galium some of the positions of lattice of germanium are occupied be galium. Galium atom has only three valence electrons. Therefore, fourth valency of nearly germanium atom is not satisfied. The place remains vacant. This place is deficient of electrons and is therefore called electron hole or electron vacancy. Electron from neighbouring atom comes and fills the hole, thereby creating a hole in its original position. Under the influence of electric field electrons move towards positively charged plates through these and conduct electricity. The holes appear to move towards negatively charged plates.

Long answer type questions

Question 1.
A sample of ferrous oxide has actual formula Fe_0.93O_1.00In this
sample what fraction of metal ions are Fe²⁺ ions? What type of non-stoichiometric defect is present in this sample? (NCERT Exemplar)
Solution:
Let the formula of the sample be (Fe²⁺ )x (Fe³⁺ )_y O
On looking at the given formula of the compound
x + y = 0.93 ………….. (i)
Total positive charge on ferrous and ferric ions should balance the two units of negative charge on oxygen.
Therefore, 2x + 3y = 2 ……….. (ii)
x + \(\frac{3}{2}\)y = 1 ……………..(iii)
On subtracting equation (i) from equation (iii) we have
\(\frac{3}{2}\) y – y = 1 – 0.93 ⇒ \(\frac{1}{2}\)y = 0.07 ⇒ y = 0.14
On putting the value of y in equation (i), we get
x + 0.14 = 0.93
⇒ x = 0.93 – 0.14 ⇒ x = 0.79
Fraction of Fe2+ ions present in the sample = \(\frac{0.79}{0.93}\) = 0.849
Metal deficiency defect is present in the sample because iron is less in amount than that required for stoichiometric composition.

Question 2.
(i) Following is the schematic alignment of magnetic moments:

Identify the type of magnetism. What happens when these substances are heated?
(ii) If the radius of the octahedral void is ‘r’ and radius of the atoms in close packing is ‘R’ What is the relation between ‘r’ and ‘R’?
(iii) Tungsten crystallises in body centred cubic unit cell. If the edge of the unit cell is 316.5 pm. What is the radius of tungsten atom?
Answer:
(i) The given schematic alignment of magnetic moments shows ferrimagnetism. When these substances are heated they lose ferrimagnetism and become paramagnetic.

(ii) The radius of the octahedral void = r
The radius of the atoms in close packing = R
Relation between r and R is given as :
r = 0.414 R

(iii) Given, a = 316.5 pm
Vs
We know that for body centred cubic unit cell r = \(\frac{\sqrt{3}}{4}\) a
∴ r = \(\frac{\sqrt{3}}{4}\) × 316.5 pm = 136.88 pm

Question 3.
(i) Identify the type of defect shown in the following figure:

What type of substances show this defect?
(ii) A metal crystallises in a body centred cubic structure. If ‘a’ is the edge length of its unit cell, ‘r’ is the radius of the sphere. What is the relationship between ‘r’ and ‘a’?
(iii) An element with molar mass 63 g/ mol forms a cubic unit cell with edge length of 360.8 pm. If its density is 8.92 g/cm 3. What is the nature of the cubic unit cell?
Answer:
(i) The given figure shows Schottky defect. This defect arises when equal number of cations and anions are missing from the lattice. It is a common defect in ionic compounds of high coordination number.
Where born cations and anion are of the same size, e.g., KCl, NaCl, KBr etc.

(ii) Edge length of the unit cell = a
Radius of the sphere = r
For body centred cubic structure ,
r = \(\frac{\sqrt{3}}{4}\)

(iii) We know that,
Density d = \(\frac{z \times M}{a^{3} \times N_{A}}\) or z = \(\frac{d \times a^{3} \times N_{A}}{M}\)
Given,
M = 63 g-mol^-1 6.3 × 10^-2 kg mol^-1
a = 360.8 pm = 360.8 × 10^-12 m = 3.608 × 10^-10m
d = 8.92 g/cm² = 0.892 kgm^-3
N_A = 6.022 × 10²³ mol^-1
On putting the given values in formula,

= 3.97 ≅ 4.
Since, 4 atoms of the element present per unit cell. Hence, the cubic unit cell must be face-centered.

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 6 Electromagnetic Induction Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 6 Electromagnetic Induction

Very short answer type questions

Question 1.
The motion of copper plate is damped when it is allowed to oscillate between the two poles of a magnet. What is the cause of this damping?
Answer:
As the plate oscillate, the changing magnetic flux through the plate produces a strong eddy current in the direction, which opposes the cause. Also, copper being substance, it gets magnetised in the opposite direction, so the plate motion gets damped.

Question 2.
On what factors does the magnitude of the emf induced in the circuit due to magnetic flux depend ?
Answer:
Depends on the time rate of change in magnetic flux (or simply change in Magnetic flux)
\(|\varepsilon|=\frac{\Delta \phi}{\Delta t}\)

Question 3.
A light metal disc on the top of an electromagnet is thrown up as the current is switched on. Why? Give reason.
Answer:
When the current begins to grow through the electromagnet, the magnetic flux through the disc begins to increase. This sets up eddy current in the disc in the same direction as that of the electromagnetic current.

Thus, if the upper surface of electromagnetic acquires AT-polarity, the lower surface of the disc also acquires N-polarity. As, same magnetic poles repel each other, the light metallic disc is thrown up.

Question 4.
State the Faraday’s law’ of electromagnetic induction.
Answer:
On the basis of his experiment, Faraday gave the following two laws:
First Law: Whenever magnetic flux linked with a circuit changes, an emf is induced in it which lasts, so long as change in flux continuous.
Second Law: The emf induced in loop or closed circuit is directly proportional to the rate of change of magnetic flux linked with the loop
i.e., ε ∝ \(\frac{(-) d \phi}{d t}\) or ε = -N \(\frac{d \phi}{d t}\)
where, N= number of turns in the coil. Negative sign indicates the Lenz’s law.

Question 5.
State Lenz’s law. A metallic rod held horizontally along East-West direction, is allowed to fall under gravity. Will there be an emf induced at its ends? Justify your answer.
Answer:
Lenz’s Law: The direction of the induced emf, or the current, in any circuit is such as to oppose the cause that produces it.

Yes, emf will be induced in the rod as there is change in magnetic flux. When a metallic rod held horizontally along East-West direction, is allowed to fall freely under gravity i.e., fall from North to South, the intensity of magnetic lines of the earth’s magnetic field changes through it, i.e., the magnetic flux changes and hence emf induced in it.

Question 6.
How does the mutual inductance of a pair of coils change, when
(i) distance between the coils is increased and
(ii) number of turns in the coils is increased?
Answer:
(i) AΦ = MI, with the increase in the distance between the coils the magnetic flux linked with the secondary coil decreases and hence, the mutual inductance of the two coils will decreases with the increase of separation between them.

(ii) Mutual inductance of two coils can be found out by
M = μ₀N₁N₂ Al i.e.,
M ∝ N₁N₂, SO, with the increase in number of turns mutual inductance increases.

Question 7.
Why is the core of a transformer laminated?
Answer:
The core of a transformer is laminated because of preventing eddy current being produced in the core.

Question 8.
How can the self-inductance of a given coil having N number of turns, area of cross-section A and lengths l be increased?
Answer:
The self-inductance can be increased by the help of electric fields. It does not depend on the current through circuit but depends upon the permeability of material from which the core is made up off.

Question 9.
Consider a magnet surrounded by a wire with an on/off switch S (as shown in figure). If the switch is thrown from the off position (open circuit) to the on position (closed circuit), will a current ?Explain (NCERT Exemplar)

Answer:
No part of the wire is moving and so motional e.m.f. is zero. The magnet is stationary and hence the magnetic field does not change with time. This means no electromotive force is produced and hence no current will flow in the circuit.

Question 10.
A wire in the form of a tightly wound solenoid is connected to a DC source, and carries a current. If the coil is stretched so that there are gaps between successive elements of the spiral coil, will the current increase or decrease? Explain. (NCERT Exemplar)
Answer:
The current will increase. As the wires are pulled apart the flux will leak through the gaps. Lenz’s law demands that induced emf resist this decrease, which can be done by an increase in current.

Question 11.
A solenoid is connected to a battery so that a steady current flows through it. If an iron core is inserted into the solenoid, will the current increase or decrease? Explain. (NCERT Exemplar)
Answer:
The current will decrease. As the iron core is inserted in the solenoid, the magnetic field increases and the flux increases. Lenz’s law implies that induced emf should resist this increase, which can be achieved by a decrease in current. However, this change will be momentarily.

Question 12.
Consider a metal ring kept (supported by a cardboard) on top of a fixed solenoid carrying a current /. The centre of the ring coincides with the axis of the solenoid. If the current in the solenoid is switched off, what will happen to the ring? (NCERT Exemplar)
Answer:
When the current in the solenoid decreases a current flows in the same direction in the metal ring as in the solenoid. Thus there will be a downward force. This means the ring will remain on the cardboard. The upward reaction of the cardboard on the ring will increase.

Short answer type questions

Question 1.
Consider a closed loop C in a magnetic field (figure). The flux passing through the loop is defined by choosing a surface whose edge coincides with the loop and using the formula Φ = B₁ dA₁, B₂ dA₂…. Now, if we choose two different surfaces S₁ and S₂ having C as their edge, would we get the same answer for flux. Justify your answer. (NCERTExemplar)

The magnetic flux linked with the surface can considered as the number of magnetic field lines passing through the surface. So, let dΦ = BdA represents magnetic lines in an area A to B.

By the concept of continuity of lines B cannot end or start in space, therefore the number of lines passing through surface S₁ must be the same as the number of lines passing through the surface S₂. Therefore, in both the cases we gets the same answer for flux.

Question 2.
What are eddy currents? Write their two applications.
Answer:
Eddy Current: Eddy currents are the currents induced in the bulk pieces of conductors when the amount of magnetic flux linked with the conductor changes.

Eddy currents can be minimised by taking laminated core, consists of thin metallic sheet insulated from each other by varnish instead of a single solid mass. The plane of the sheets should be kept perpendicular to the direction of the currents. The insulation provides high resistance hence, eddy current gets minimised.

Applications
(i) Electromagnetic damping
(ii) Induction furnace.

Question 3.
(i) A rod of length l is moved horizontally with a uniform – velocity v in a direction perpendicular to its length through a region in which a uniform magnetic field is acting vertically downward. Derive the expression for the emf induced across the ends of the rod.

(ii) How does one understand this motional emf by invoking the Lorentz force acting on the free charge carriers of the conductor? Explain.
Answer:
(i) Consider a straight conductor moving with velocity v and U shaped conductor placed in perpendicular magnetic field as shown in the figure.

Let conductor shifts from ab to a’ b’ in time dt, then change in magnetic flux
dΦ = B × change in area
= B × (areaa’b’ab)
= B × (l × vdt)
∴  \(\frac{d \phi}{d t}\) Bvl
∴  Induced emf lei \(|\varepsilon|=\frac{d \phi}{d t}\) = Bvl

(ii) During motion, free e^– are shifted at one end due to magnetic force so due to polarisation of rod electric field is produced which applies electric force on free e^– on opposite direction.

At equilibrium of Lorentz force,
F_e + F_m = 0
qE + q(v × B) = 0
E = -v × B = B × v
\(|E|=|B v \sin 90|\)
\(\frac{d v}{d r}\) = Bv
P_D = Bvl

Question 4.
(a) How does the mutual inductance of a pair of coils change when
(i) distance between the coils is increased and
(ii) number of turns in the coils is increased?

(b) A plot of magnetic flux (Φ) versus current (I), is shown in the figure for two inductors A and B. Which of the two has large value of self-inductance?

(c) How is the mutual inductance of a pair of coils affected when
(i) separation between the coils is increased?
(ii) the number of turns in each coil is increased?
(iii) a thin iron sheet is placed between the two coils, other factors remaining the same?
Justify your answer in each case.

Answer:
(a)
(i) Mutual inductance decreases.
(ii) Mutual inductance increases.
Concept
(i) If distance between two coils is increased as shown in figure.

It causes decrease in magnetic flux linked with the coil C₂. Hence induced emf in coil C₂ decreases by relation ε₂ = \(\frac{-d \phi_{2}}{d t}\). Hence mutual inductance decreases.
(ii) From relation M₂₁ = μ₀N₁N₂ Al, if number of turns in one of the coils or both increases, means mutual inductance will increase.

(b) Φ = LI ⇒ \(\frac{\phi}{I}\) = L
The slope of \(\frac{\phi}{I}\) of straight line is equal to self-inductance L. It is larger for inductor A; therefore inductor A has larger value of self inductanc ‘ L’.

(c)
(i) When the relative distance between the coil is increased, the leakage
of flux increases which reduces the magnetic coupling of the coils. So magnetic flux linked with all the turns decreases. Therefore, mutual inductance will be decreased.

(ii) Mutual inductance for a pair of coil is given by
M = K\(\sqrt{L_{1} L_{2}}\)
where, L = \(\frac{\mu N^{2} A}{l}\) and L is called self inductance. Therefore, when the number of turns in each coil increases, the mutual inductance also increases.

(iii) When a thin iron sheet is placed between the two coils, the mutual inductance increases because M ∝ permeability. The permeability of the medium between coils increases.

Question 5.
Consider a metallic pipe with an inner radius of 1 cm. If a cylindrical bar magnet of radius 0.8 cm is dropped through the pipe, It takes more time to come down than It takes for a similar
unmagnetised cylindrical iron bar dropped through the metallic pipe. Explain. (NCERT Exemplar)
Answer:
For the magnet, eddy currents are produced in the metallic pipe. These currents will oppose the motion of the magnet. Therefore magnet’s downward acceleration will be less than the acceleration due to gravity g. On the other hand, an unmagnetised iron bar will not produce eddy
currents and will fall an acceleration g. Thus the magnet will take more time.

Question 6.
A magnetic field B = B₀ sin(ωt) k̂ covers a large region where a wire AB slides smoothly over two parallel conductors separated by a distance d (figure). The wires are in the x-y plane. The wire AB (of length d) has resistance R and the parallel wires have negligible resistance. If AB is moving with velocity y, what is the current in the circuit. What is the force needed to keep the wire moving at constant velocity? (NCERT Exemplar)

Answer:
Let us assume that the parallel wires are at y = 0 i. e., along x-axis and y = d. At t = 0, AB has x = 0, i. e., along y-axis and moves with a velocity v. Let at time t, wire is at x (t) = vt.
Now, the motional emf across AB is
= (B₀sinωt) vd(-ĵ)
emf due to change in field (along OBAC)
= -B₀ωcosωt (t)d
Total emf in the circuit = emf due to change in field (along OBAC) + the motional emf across AB = -B₀d[ωxcos(ωt) + vsin (ωt)]
Electric current in clockwise direction is given by,
= \(\frac{B_{0} d}{R}\) = (ωxcosωt + vsinωt)
The force acting on the conductor is given by F = ilB sin 90° = ilB
Substituting the values, we have
Force needed along i = \(\frac{B_{0} d}{R}\) (ωx cos ωt + vsinωt) × d × B sinωt
= \(\frac{B_{0}^{2} d^{2}}{R}\)(ωx cos ωt + vsinωt) sinωt
This is the required expression for force.

Long answer type questions

Question 1.
(i) How is magnetic flux linked with the armature coil changed in a generator ?
(ii) Derive the expression for maximum value of the induced emf and state the rule that gives the direction of the induced emf.
(iii) Show the variation of the emf generated versus time as the armature is rotated with respect to the direction of the magnetic fields.
Answer:
(i) The direction of flow of current in resistance R get changed alternatively after every half cycle.
Thus, AC is produced in coil.

(ii) Let at any instant total magnetic flux linked with the armature coil is G. and θ = ωt is the angle made by area vector of coil with magnetic field.
Φ = NBA cosθ = NBA cosωt
\(\frac{d \phi}{d t}\) = -NBAω sin ωt
– \(\frac{d \phi}{d t}\) = NBAω sin ωt
By Faraday’s law of emf, e = \(\frac{-d \phi}{d t}\)
Induced emf in coil is given by,
e = NBAω sinωt
e = e₀ sinωt
where, e₀ = NBAω = peak value of induced emf

(iii) The mechanical energy spent in rotating the coil in magnetic field appears in the form of electrical energy.

Question 2.
State the working of AC generator with the help of a labelled diagram.
The coil of an AC generator having N turns, each of area A, is rotated with a constant angular velocity to. Deduce the expression for the alternating emf generated in the coil.
What is the source of energy generation in this device?
Answer:
AC Generator: A dynamo or generator is a device which converts mechanical energy into electrical energy.

Principle: It works on the principle of electromagnetic induction. When a coil rotates continuously in a magnetic field, the effective area of the coil linked normally with the magnetic field lines, changes continuously with time. This variation of magnetic flux with time results in the production of an alternating emf in the coil.

Construction: It consists of the four main parts
(i) Field magnet: It produces the magnetic field. In the case of a low power dynamo, the magnetic field is generated by a permanent magnet, while in the case of large power dynamo, the magnetic field is produced by an electromagnet.

(ii) Armature: It consists of a large number of turns of insulated wire in the soft iron drum or ring. It can revolve an axle between the two poles of the field magnet. The drum or ring serves the two purposes: (a) It serves as a support to coils and (b) It increases the magnetic field due to air core being replaced by an iron core.

(iii) Slip rings: The slip rings R₁ and R₂ are the two metal rings to which the ends of armature coil are connected. These rings are fixed to the shaft which rotates the armature coil so that the rings also rotate along with the armature.

(iv) Brushes: There are two flexible metal plates or carbon rods (B₁ and B₂) which are fixed and constantly touch the revolving rings. The output current in external load RL is taken through these brushes.

Working: When the armature coil is rotated in the strong magnetic field, the magnetic flux linked with the coil changes and the current is induced in the coil, its direction being given by Fleming’s right hand rule. Considering the armature to be in vertical position and as it rotates in anticlockwise direction, the wire ab moves upward and cd downward, so that the direction of induced current is shown in fig. In the external circuit, the current flows along B₁R_lB₂. The direction of current remains unchanged during the first half turn of armature. During the second half revolution, the wire ab moves downward and cd upward, so the direction of current is reversed and in external circuit it flows along B₂R_LB₁ Thus the direction of induced emf and current changes in the external circuit after each half revolution.

Expression for Induced emf: If N is number of turns in coil, f the frequency of rotation, A area of coil and B the magnetic induction, then induced emf
e = – \(\frac{d \phi}{d t}\) = –\(\frac{d}{d t}\) {NBA (cos 2π ft)} dt dt
= 2π NBA f sin 2π ft
Obviously, the emf produced is alternating and hence the current is also alternating.
Current produced by an AC generator cannot be measured by moving coil ammeter; because the average value of AC over full cycle is zero.
The source of energy generation is the mechanical energy of rotation of armature coil.

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 7 Alternating Current Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 7 Alternating Current

Very short answer type questions

Question 1.
Why is the use of AC voltage preferred over DC voltage? Give two reasons.
Answer:
The use of AC voltage is preferred over DC voltage because of

• the loss of energy in transmitting the AC voltage over long distance with the help of step-up transformers is negligible as compared to DC voltage.
• AC voltage can be stepped up and stepped down as per the requirement by using a transformer.

Question 2.
Explain why current flows through an ideal capacitor when it is connected to an AC source, but not when it is connected to a DC source in a steady state.
Answer:
For AC source, circuit is complete due to the presence of displacement current in the capacitor. For steady DC, there is no displacement current, therefore, circuit is not complete.
Mathematically, capacitive reactance
X_C = \(\frac{1}{2 \pi f C}=\frac{1}{\omega C}\)
So, capacitor allows easy path for AC source.
For DC, / = 0, so X_C = infinity.
So, capacitor blocks DC.

Question 3.
Define capacitor reactance. Write its SI units.
Answer:
Capacitor reactance is the resistance offered by a capacitor, when it is connected to an electric circuit. It is given by X_C = \(\frac{1}{\omega C}\)
where, ω = angular frequency of the source
C = capacitance of the capacitor
The SI unit of capacitor reactance is ohm (Ω).

Question 4.
In a series LCR circuit, V_L = V_C ≠ V_R What is the value of power factor for this circuit?
Answer:
Power factor = 1
Since V_L = V_C, the inductor and capacitor will nullify the effect of each other and it will be a resistive circuit.
For Φ =0; power factor cosΦ = 1

Question 5.
The power factor of an AC circuit is 0.5. What is the phase difference between voltage and current in this circuit?
Answer:
Power factor between voltage and current is given by cosΦ, where Φ is phase difference
cosΦ = 0.5 = \(\frac{1}{2}\) ⇒ Φ = cos^-1 (\(\frac{1}{2}\)) = \(\frac{\pi}{3}\)

Question 6.
What is wattless current?
Answer:
When pure inductor and/or pure capacitor is connected to AC source, the current flows in the circuit, but with no power loss; the phase difference between voltage and current is \(\frac{\pi}{3}\) . Such a current is called the wattless current.

Question 7.
An AC source of voltage V = V₀ sin ωt is connected to an ideal inductor. Draw graphs of voltage V and current I versus cat.
Answer:
Graphs of V and I versus ωt for this circuit is shown below:

Question 8.
Define ‘quality factor’ of resonance in series LCR circuit. What is its SI unit?
Answer:
The quality factor (Q) of series LCR circuit is defined as the ratio of the resonant frequency to frequency band width of the resonant curve.
Q = \(\frac{\omega_{r}}{\omega_{2}-\omega_{1}}=\frac{\omega_{r} L}{R}\)

Clearly, smaller the value of R, larger is the quality factor and sharper the resonance. Thus, quality factor determines the nature of sharpness of resonance. It has no units.

Question 9.
What is the function of a step-up transformer?
Answer:
Step-up transformer converts low alternating voltage into high alternating voltage and high alternating current into low alternating current. The secondary coil of step-up transformer has greater number of turns than the primary (N_s > N_p ).

Question 10.
Mention the two important properties of the material suitable for making core of a transformer.
Answer:
Two characteristic properties:

1. Low hysteresis loss
2. Low coercivity

Question 11.
If an LC circuit is considered analogous to a harmonically oscillating spring block system, which energy of the LC circuit would be analogous to potential energy and which one analogous to kinetic energy? (NCERTExemplar)
Answer:
Magnetic energy analogous to kinetic energy and electrical energy analogous to potential energy.

Question 12.
A device ‘X’ is connected to an a.c. source. The variation of voltage, current and power in one complete cycle is shown in the figure.

(a) Which curve shows power consumption over a full cycle?
(b) What is the average power consumption over a cycle?
(c) Identify the device ‘X’. (NCERT Exemplar)
Answer:
(a) A
(b) Zero
(c) L or C or LC

Short answer type questions

Question 1.
Prove that an ideal capacitor in an AC circuit does not dissipate power.
Answer:
Since, average power consumption in an AC circuit is given by
P_av = V_rms × I_rms × cosΦ
But in pure capacitive circuit, phase difference between voltage and current is given by
Φ = \(\frac{\pi}{2}\)
∴ P_av = V_rms × I_rms × cos \(\frac{\pi}{2}\)
⇒ P_av = 0 ( ∵ cos \(\frac{\pi}{2}\) = 0)
Thus, no power is consumed in pure capacitive AC circuit.

Question 2.
A circuit is set up by connecting inductance L 100 mil, resistor R -100 D. and a capacitor of reactance 200 Ω in series. An alternating emf of 150 √2 V, 500/ π Hz is applied across this series combination. Calculate the power dissipated in the resistor.
Answer:
Here, L =100 x 10^-3 H,R =100 Ω,
X_C = 200 Ω,V_rms = 150√2 V
v = \(\frac{500}{\pi}\) HZ

= 225 W

Question 3.
A series L-C-R circuit is connected to an AC source. Using the phasor diagram, derive the expression for the impedance of the circuit. Plot a graph to show the variation of current with frequency of the source, explaining the nature of its variation.
Answer:
Assuming X_L > X_C
⇒ V_L > V_C
∵ Net voltage, V = \(\sqrt{V_{R}^{2}+\left(V_{L}-V_{C}\right)^{2}}\)
where, V_L, V_C and are alternating voltages across L,C and R respectively.


But, V_R= IR,V_L = IX_L,
V_C = IX_C
∴ Net voltage, V = \(\sqrt{(I R)^{2}+\left(I X_{L}-I X_{C}\right)^{2}}\)
\(\frac{V}{I}\) = \(\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}\)
Impedance of LCR circuit,
Z = \(\frac{V}{I}\) = \(\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}\)

Question 4.
In a series LCR circuit connected to an AC source of variable frequency and voltage V = V_m sin ωt, draw a graph showing the variation of current (I) with angular frequency (ω) for two different values of resistance R₁ and R₂ (R₁ > R₂ ). Write the condition under which the phenomenon of resonance occurs. For which value of the resistance out of the two curves, a sharper resonance is produced? Define Q-factor of the circuit and give its significance.
Answer:
Figure shows the variation of i_m with ω in a LCR series circuit for two values of resistance R₁ and R₂ (R₁ > R₂).

The condition for resonance in the LCR circuit is
ω₀ = \(\frac{1}{\sqrt{L C}}\)
We see that the current amplitude is maximum at the resonant frequencyω. Since i_m = v_m / R at resonance, the current amplitude for case R₂ is sharper to that for case R₁.

Quality factor or simply the Q-factor of a resonant LCR circuit is defined as the ratio of voltage drop across the capacitor (or inductor) to that of applied voltage.
It is given by Q = \(\frac{1}{R} \cdot \sqrt{\frac{L}{C}}\)
The Q-factor determines the sharpness of the resonance curve. Less sharp the resonance, less is the selectivity of the circuit while higher is the Q, sharper is the resonance curve and lesser will be the loss in energy of the circuit.

Question 5.
Both alternating current and direct current y, are measured in amperes. But how is the ampere defined for an alternating current? (NCERT Exemplar)

Answer:
An ac current changes direction with the source frequency and the attractive force would average to zero. Thus, the ac ampere must be defined in terms vc of some property that is independent of the direction of current. Joule’s heating effect is such property and hence it is used to define rms value of ac.

Question 6.
Explain why the reactance provided by a capacitor to an alternating current decreases with increasing frequency. (NCERT Exemplar)
Answer:
A capacitor does not allow flow of direct current through it as the resistance across the gap is infinite. When an alternating voltage is applied across the capacitor plates, the plates are alternately charged and discharged. The current through the capacitor is a result of this changing voltage (or charge). Thus, a capacitor will pass more current through it if the voltage is changing at a faster rate, i. e., if the frequency of supply is higher. This implies that the reactance offered by a capacitor is less with increasing frequency, it is given byl/©C.

Question 7.
Explain why the reactance offered by an inductor increases with increasing frequency of an alternating voltage.
(NCERT Exemplar)
Answer:
An inductor opposes flow of current through it by developing an induced emf according to Lenz’s law. The induced voltage has a polarity so as to maintain the current at its present value. If the current is decreasing, the polarity of the induced emf will be so as to increase the current and vice versa. Since the induced emf is proportional to the rate of change of current, it will provide greater reactance to the flow of current if the rate of change is faster, i.e., if the frequency is higher. The reactance of an inductor, therefore, is proportional to the frequency, being given by ωL.

Long answer type questions

Question 1.
(a) An AC source of voltage V = V₀ sin ωt is connected to a series combination of L, C and R. Use the phasor diagram to obtain expression for impedance of the circuit and phase angle between voltage and current. Find the condition when current will be in phase with the voltage. What is the circuit in this condition called?
(b) In a series LR circuit X_L = R and power factor of the circuit is P₁. When capacitor with capacitance C such that X_L = X_C is put in series, the power factor becomes P₂. Calculate \(\frac{P_{1}}{P_{2}}\).
Answer:
(a) Expression for Impedance in LCR Series Circuit : Suppose resistance R, inductance L and capacitance C are connected in series and an alternating source of voltage V = V₀ sin ωt is applied across it. (fig. a) On account of being in series, the current (i) flowing through all of them is the same.

Suppose, the voltage across resistance R isV_R, voltage across inductance L is V_L and voltage across capacitance C is V_C. The voltage V_R and current i are in the same phase, the voltage V_L will lead the current by angle 90° while the voltage V_C will lag behind the current by angle 90° (fig. b). Clearly,V_C and V_L are in opposite directions, therefore their resultant potential difference = V_C – V_L (if V_C >,V_L).

Thus, V_R and (V_C – V_L) are mutually perpendicular and the phase difference between them is 90°. As applied voltage across the circuit is V, the resultant of VR and (V_C – V_L) will also be V. From fig.

Question 2.
(i) What do you understand by sharpness of resonance in a series L-C-R circuit? Derive an expression for Q-factor of the circuit.
Three electrical circuits having AC sources of variable frequency are shown in the figures. Initially, the current flowing in each of these is same. If the frequency of the applied AC source is increased, how will the current flowing in these circuits be affected? Give the reason for your answer.

Answer:

The sharpness of resonance in series LCR circuit refers how quick fall of alternating current in circuit takes place when frequency of alternating voltage shifts away from resonant frequency. It is measured by quality factor (Q-factor) of circuit.

The Q-factor of series resonant circuit is defined as the ratio of the voltage developed across the capacitance or inductance at resonance to the impressed voltage which is the voltage applied.

This is the required expression.

(ii) Let initially I_r current is flowing in all the three circuits. If frequency of applied AC source is increased, then the change in current will occur in the following manner.
(a) Circuit Containing Resistance R Only: There will not be any effect in the current on changing the frequency of AC source.

where,f_i = initial frequency of AC source.
There is no effect on current with the increase in frequency.

(b) AC Circuit Containing Inductance

Only: With the increase of frequency current of AC source inductive reactance increase as
I = \(\frac{V_{r m s}}{X_{L}}=\frac{V_{r m s}}{2 \pi f L}\)
For given circuit,
I ∝ \(\frac{1}{f}\)
Current decreases with the increase of frequency.

(c) AC Circuit Containing Capacitor Only:
X_C = \(\frac{1}{\omega C}=\frac{1}{2 \pi f C}\)
Current, I = \(\frac{V_{r m s}}{X_{C}}\) = \(\frac{V_{r m s}}{\left(\frac{1}{2 \pi f C}\right)}\)
I = 2πfCV_rms
For given circuit, I ∝ f
Current increases with the increase of frequency.

Question 3.
(a) Describe briefly, with the help of a labelled diagram, the working of a step up transformer.
(b) Write any two sources of energy loss in a transformer.
(c) A step up transformer converts a low voltage into high voltage. Does it not violate the principle of conservation of energy? Explain
Or Draw a labelled diagram of a step-down transformer. State the principle of its working. Express the turn ratio in terms of voltages.
Find the ratio of primary and secondary currents in terms of turn ratio in an ideal transformer.
How much current is drawn by the primary of a transformer connected to 220 V supply when it delivers power to a 110 V-550 W refrigerator?
Answer:
(a) Transformer: Transformer is a device by which an alternating voltage may be decreased or increased. It is based on the principle of mutual-induction.

Construction: It consists of laminated core of soft iron, on which two coils of insulated copper wire are separately wound. These coils are kept insulated from each other and from the iron-core, but are coupled through mutual induction. The number of turns in these coils are different. Out of these coils one coil is called primary coil and other is called the secondary coil. The terminals of primary coils are connected to AC mains and the terminals of the secondary coil are connected to external circuit in which alternating current of desired voltage is required. Transformers are of two types:
1. Step-up transformer: It transforms the alternating low voltage to alternating high voltage and in this the number of turns in secondary coil is more than that in primary coil. (i. e.,N_s> N_p).
2. Step-down transformer: It transforms the alternating high voltage to alternating low voltage and in this the number of turns in secondary coil is less than that in primary coil (i. e.N_s < N_p)

Working: When alternating current source is connected to the ends of primary coil, the current changes continuously in the primary coil; due to which the magnetic flux linked with the secondary coil changes continuously, therefore the alternating emf of same frequency is developed across the secondary.

Let N_p be the number of turns in primary coil, N_s the number of turns in secondary coil and Φ the magnetic flux linked with each turn. We assume that there is no leakage of flux so that the flux linked with each turn of primary coil and secondary coil is the same. According to Faraday’s laws the emf induced in the primary coil
ε ₀ = -N_p\(\frac{\Delta \phi}{\Delta t}\) …………….. (1)
and emf induced in the secondary coil
ε _s = -N_p\(\frac{\Delta \phi}{\Delta t}\) ……………… (2)
From eq. (1) and eq, (2)
\(\frac{\varepsilon_{s}}{\varepsilon_{p}}=\frac{N_{s}}{N_{p}}\) …………………. (3)
If the resistance of primary coil is negligible, the emf (ε _p) induced in the primary coil, will be equal to the applied potential difference (V_p) across its ends. Similarly if the secondary circuit is open, then the potential difference Vs across its ends will be equal to the emf (ε _s) induced in it; therefore,
\(\frac{V_{s}}{V_{p}}=\frac{\varepsilon_{s}}{\varepsilon_{p}}=\frac{N_{s}}{N_{p}}\) r(say) …………… (4)
where r = \(\frac{N_{S}}{N_{P}}\) is called the transformation ratio. If i_p and i_s are the instantaneous currents in primary and secondary coils and there is no loss of energy.
For about 100% efficiency,
Power in primary = Power in secondary
V_p i_p = V_si_s
∴ \(\frac{i_{s}}{i_{p}}=\frac{V_{p}^{F}}{V_{s}}=\frac{N_{p}}{N_{s}}=\frac{1}{r}\) ………….. (5)

In step-up transformer, N_s > N_p → r > 1 ;
So V_s > V_p and i_s < i_p
i.e., Step up transformer increases the voltage.

In step down transformer, N_s < N_p → r < 1
So V_s < V_p and i_s > i_p
i.e., step-up down transformer decreases the voltage, but increase the current.

Laminated Core: The core of a transformer is laminated to reduce the energy losses due to eddy currents. So, that its efficiency may remain nearly 100%.
In a transformer with 100% efficiency (say), Input power = output power dV_pI_p = V_sI_s
(b) The sources of energy loss in a transformer are, (i) eddy current losses due to iron core, (ii) flux leakage losses, (iii) copper losses due to heating up of copper wires, (iv) Hysteresis losses due to magnetisation and demagnetisation of core.

(c) When output voltage increases, the output current automatically decreases to keep the power same. Thus, there is no violation of conservation of energy in a step-up transformer.
We have, i_p V_p = i_sV_s = 550 W
V_p 220V
i_p = \(\frac{550}{220}=\frac{5}{2}\) = 2.5A

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 9 Ray Optics and Optical Instruments Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 9 Ray Optics and Optical Instruments

Very short answer type questions

Question 1.
What is reflection?
Answer:
When a light ray incident on a smooth surface bounces back to the same medium, it is called reflection.

Question 2.
State new cartesian sign conventions used for mirrors.
Answer:

• All the distances are measured from the pole of the mirror.
• All the distances measured in the direction of incident ray are taken as positive and the distances measured opposite to the incident ray are taken as – ve.
• All heights measured perpendicular to the principal axis in the upward direction are taken as + ve and those measured in downward direction are taken as – ve.

Note: Direction of incident light is always to be shown falling from left to right. So distance of the object and real image is always -ve while that of virtual image is always + ve, height of real image is always – ve while that of the virtual image and the size of real object are always + ve.

Question 3.
How does focal length of a lens change when red light incident on it is replaced by violet light? Give reason for your answer.
Answer:
The refractive index of the material of a lens increases with the decrease in wavelength of the incident light. So, focal length will decrease with a decrease in wavelength according to the formula.
\(\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
Thus, when we replace red light with violet light then due to increase in wavelength the focal length of the lens will decrease.

Question 4.
Define refraction of light.
Answer:
It is defined as the process of bending of light from its path when it travels from one medium to the another.

Question 5.
State
(a) Laws of reflection.
(b) Laws of refraction.
Answer:
(a) The following are the two laws of reflection :
(i) Angle of incidence is always equal to the angle of reflection.
(ii) The incident ray, reflected ray and normal to the surface at the point of incidence all lie in the same plane.

(b) The following are the two laws of refraction :
(i) The ratio of the sine of angle of incidence to the sine of the angle of refraction is always constant for a given pair of media.
i.e., \(\frac{\sin i}{\sin r}\) = constant = ^aµb
where ^aµb is called relative refractive index of medium b w.r.t. a.
(ii) The incident ray, refracted ray and the normal to the refracting surface at the point of incidence all lie in the same plane.

Question 6.
(i) What is the relation between critical angle and refractive index of a material?
(ii) Does critical angle depend on the colour of light? Explain.
Answer:
(i) Refractive index (µ) = \(\frac{1}{\sin C}\)
where, C is the critical angle.
(ii) Since, refractive index depends upon the wavelength of light, the critical angle for a given pair of media is different for different wavelengths (colours) of light.

Question 7.
Under what condition does a biconvex lens of glass having a certain refractive index act as a plane glass sheet when immersed in a liquid?
Answer:
A biconvex lens will act like a plane sheet of glass if it is immersed in a liquid having the same index of refraction as itself. In this case, the focal length 1/f = 0 or f→ ∞.

Question 8.
A biconvex lens made of a transparent material of refractive index 1.25 is immersed in water of refractive index 1.33. Will the lens behave as a converging lens? Give reason.
Answer:
No, it will behave as a diverging lens.
On Using thin lens maker formula
\(\frac{1}{f_{w}}=\left(\frac{n_{g}}{n_{m}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
On Using sign convention R₁ = +ve, R₂ = -ve and n_g = 1.25 and n_m = 1.33
\(\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right)\) +ve,value and \(\left(\frac{1.25}{1.33}-1\right)\) =-ve value Hence f_w = -ve , so it behaves as a diverging lens.

Question 9.
Define total internal reflection.
Answer:
It is defined as the process of reflection of light that takes place when a ray of light travelling from denser to rarer medium gets incident at the interface of the two media at an angle greater than the critical angle for the given air of media.

Question 10.
State the criteria for the phenomenon of total internal reflection of light to take place.
Answer:
Following are the criteria for total internal reflection

• Light must pass from a denser to a rarer medium.
• Angle of incidence must be greater than critical angle.

Question 11.
Define mirage.
Answer:
It is defined as an optical illusion that occurs in deserts and coal tarred roads appear to be covered with water but on approaching at that place no water is obtained. In deserts thirsty animals observe virtual images of trees on hot sand so expecting a pond of water there but on reaching there, they do not get water pond and hence called optical illusion.

Question 12.
Why diamond sparkles?
Answer:
The critical angle for diamond is low i.e., 23° and its refractive index is 2.47. The faces of diamond are cut in such a way that when a ray of light entering from a face undergoes multiple total internal reflections from its different faces. Due to small value of the critical angle, almost all light rays entering the diamond suffer multiple total internal reflection and thus it shines brilliantly.

Question 13.
What are optical fibres? Give their one use.
Answer:
Optical fibres are thousands of very fine quality fibres of glass or quartz. The diameter of each fibre is of the order of 10^-4 cm having refractive index of material equal to 1.7. These are coated with a thin layer of material having µ = 1.5.
They are used in transmission and reception of electrical signals by converting them first into light signals.

Question 14.
Write the relationship between angle of incidence ‘i’ angle of prism ‘A’ and angle of minimum deviation for a triangular prism.
Answer:
i = \(\frac{A+\delta_{m}}{2}\)
where, δ_m = angle of minimum deviation.

Question 15.
Define dispersion of light. What is its cause?
Answer:
It is defined as the process of splitting up of white light into its constituent colours on passing through a prism.
We know that for small angled prism,
δ = (µ -1)A.
Also according to Cauchy’s formula, we know that µ ∝ \(\frac{1}{\lambda^{2}}\)
Thus µ of the material of prism is different for different colours, so δ is also different for different incident colours.
Thus due to different values of angle of deviation, each colour occupies different direction in emergent beam of light and thus constituent colours of white light get dispersed. λv < λr, so δv > δr.
The violet colour deviates more than the red colour.

Question 16.
Explain the rainbow.
Answer:
The rainbow is an example of the dispersion of sunlight by the water drops in the atmosphere. This is a phenomenon due to combined effect of dispersion, refraction and reflection of sunlight by spherical water droplets of rain. The conditions for observing a rainbow are that the sun should be shining in one part of the sky (say near western horizon) while it is raining in the opposite part of the sky (say eastern horizon). An observer can therefore see a rainbow only when his back is towards the sun.

Question 17.
Why does the Sun look reddish at sunset or sunrise?
Answer:
During sunset or sunrise, the sun is just above the horizon, the blue colour gets scattered most by the atmospheric molecules while red light gets scattered least, hence Sun appears red.
I ∝ \(\frac{1}{2^{4}}\) and λ_B << λ_R.

Question 18.
Will the focal length of a lens for red light be more, same or less than that for blue light? (NCERT Exemplar)
Answer:
As the refractive index for red is less than that for blue, parallel beams of light incident on a lens will be bent more towards the axis for blue light compared to red. Thus the focal length for red light will be more than that for blue.

Question 19.
An unsymmetrical double convex thin lens forms the image of a point object on its axis. Will the position of the image change if the lens is reversed? (NCERT Exemplar)
Answer:
No, the reversibility of the tens makes equation.
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}=(n-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
= -(n-1) \(\left(\frac{1}{R_{2}}-\frac{1}{R_{1}}\right)\)
On reversing the lens, values of R₁ and R₂ are reversed and so their signs.
Hence, for a given position of object (u), position of image (v) remains unaffected.

Question 20.
Why danger signals are of red light?
Answer:
Scattering of light is inversely proportional to the fourth power of wavelength of incident light. As red light has longer wavelength as compared to other visible colours, so its scattering is least and thus red light signals can be seen from a longer distance.

Short answer type questions

Question 1.
Will the focal length of a lens for red light be more, same or less than that for blue light? [NCERT Exemplar]
Answer:
As the refractive index for red is less than that for blue parallel beams of light incident on a lens will be bent more towards the axis for blue light compared to red.
In other words, μ_b > μ_r By lens maker’s formula,
\(\frac{1}{f}=\left(n_{21}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
Therefore, f_b < f_r
Thus, the focal length for blue light will be smaller than that for red.

Question 2.
Define power of a lens. Write its units. Deduce the relation \(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\) for two thin lenses kept in contact coaxially.
Answer:
The power of a lens is equal to the reciprocal of its focal length when it is measured in metre. Power of a lens,
P = \(\frac{1}{f(\text { metre })}\)
Its SI unit is dioptre (D).

Consider two lenses A and B of focal lengths, f₁ and f₂ placed in contact with each other. An object is placed at a point O beyond the focus of the first lens A. .
The first lens produces an image (real image) at I₁ which serves as a virtual object for the second lens B producing the final image at I.

Since, the lenses are thin, we assume the optical centres P of the lenses to be coincident. For the image formed by the first lens A, we obtain
\(\frac{1}{v_{1}}-\frac{1}{u}=\frac{1}{f_{1}}\) ……………………………. (1)
For the image formed by the second lens B, we obtain
\(\frac{1}{v}-\frac{1}{v_{1}}=\frac{1}{f_{2}}\) …………………………………. (2)
Adding eqs. (1) and (2), we obtain
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\) …………………………. (3)

If the two lenses system is regarded as equivalent to a single lens of focal length f, we have
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) ……………………………… (4)
From eqs. (3) and (4), we obtain
\(\frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{1}{f}\) .

Question 3.
(a) Draw a schematic labelled ray diagram of a reflecting type telescope (cassegrain).
(b) The objective of telescope is of larger focal length and of larger aperture (compared to the eyepiece). Why? Given reasons.
(c) State the advantages of reflecting telescope over refracting telescope.
Answer:
(a)

(b) In normal adjustment, magnifying power of the telescope, M = \(\frac{f_{0}}{f_{e}}\)
(i) If focal length of the objective lens is large in comparison to the eyepiece, magnifying power increases.
(ii) Resolving power of the telescope RP = \(\frac{D}{1.22 \lambda}\)
D being the diameter of the objective. To increase the resolving power of the telescope, large aperture of the objective lens is required.

Advantages

• There is no chromatic aberration in a mirror.
• Brighter image.
• High resolving power.
• Large magnifying power.

Question 4.
How is the working of a telescope different from that of a microscope?
Answer:
Difference in working of telescope and microscope

• Objective of telescope forms the image of a very far off object at or within the focus of its eyepiece. The microscope does the same for a small object kept just beyond the focus of its objective.
• The final image formed by a telescope is magnified relative to its size as seen by the unaided eye while the final image formed by a microscope is magnified relative to its absolute size.
• The objective of a telescope has large focal length and large aperture while the corresponding parameters for a microscope have very small ‘ values.

Question 5.
For a glass prism (μ = \(\sqrt{3}\) ) the angle of minimum deviation is equal to the angle of the prism. Find the angle of the prism. (NCERT Exemplar)
Answer:
At minimum deviation μ = \(\frac{\sin \left[\frac{\left(A+\delta_{m}\right)}{2}\right]}{\sin \left(\frac{A}{2}\right)} \)
Given, δ_m = A
∴ μ = \(\frac{\sin A}{\sin \frac{A}{2}}=\frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}}=2 \cos \frac{A}{2}\)
∴ \(\cos \frac{A}{2}=\frac{\sqrt{3}}{2} \text { or } \frac{A}{2}=30\)
⇒ A = 60⁰.

Long Answer Type Questions

Question 1.
(a) Draw a ray diagram for formation of image of a point object by a thin double convex lens having radii of curvature R₁ and R₂. Hence, derive lens maker’s formula for a double convex lens. State the assumptions made and sign convention used.
(b) A convex lens is placed over a plane mirror. A pin is now positioned so that there is no parallax between the pin and its image formed by this lens-mirror combination. How will you use this observation to find focal length of the lens? Explain briefly.
Answer:
(a) Lens Maker’s Formula: Suppose L is a thin lens. The refractive index of the material of lens is n₂ and it is placed in a medium of refractive index n₁. The optical centre of lens is C and X’ X is principal axis. The radii of curvature of the surfaces of the lens are R₁ and R₂ and their poles are P₁ and P₂.

The thickness of lens is t, which is very small. O is a point object on the principal axis of the lens. The distance of O from pole P₁ is u. The first refracting surface forms the image of O at I’ at a distance v’ from P₁.
From the refraction formula at spherical surface, \(\frac{n_{2}}{v^{\prime}}-\frac{n_{1}}{u}=\frac{n_{2}-n_{1}}{R_{1}}\) ……………………………….. (1)
The image I’ acts as a virtual object for second surface and after refraction at second surface, the final image is formed at I.

The distance of I from pole P₂ of second surface is v. The distance of virtual object (I’) from pole P₂ is (v’ -t).
For refraction at second surface, the ray is going from second medium (refractive index n₂) to first medium (refractive index n₁), therefore, from refraction formula at spherical surface

\(\frac{n_{1}}{v}-\frac{n_{2}}{\left(v^{\prime}-t\right)}=\frac{n_{1}-n_{2}}{R_{2}}\) ……………….. (2)
For a thin lens t is negligible as compared to v’, therefore from eq. (2)
\(\frac{n_{1}}{v}-\frac{n_{2}}{v^{\prime}}=-\frac{n_{2}-n_{1}}{R_{2}}\) ……………………………….. (3)
Adding equation (1) and (3),we get

where, ₁n₂ = \(\frac{n_{2}}{n_{1}}\) is refractive index of second medium (te. medium of lens) with respect to first medium. If the object O is at infinity, the image will be formed at second focus i.e., if u = ∞, v = f₂ =f
Therefore, from equation (4)

This formula is called Lens-Maker’s formula. If first medium is air and refractive index of material of lens be n, then ₁n₂ = n, therefore, the modified equation (5) may be written as
\(\frac{1}{f}=(n-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\) ………………………………. (6)

(b) Focal length = distance of the pin from the mirror.
The rays from the object after refraction from lens should fall normally on the plane mirror. So, they retrace their path. Hence, rays must be originating from focus and thus distance of the pin from the plane mirror gives focal length of the lens.

Question 2.
(a) Draw the labelled ray diagram for the formation of image by a compound microscope. Derive an expression for its total magnification (or magnifying power), when the final image is formed at the near point. Why both objective and eyepiece of a compound microscope must have short focal lengths?
(b) Draw a ray diagram showing the image formation by a compound microscope. Hence, obtain expression for total magnification when the image is formed at infinity.
Answer:
(a) Compound Microscope: It consists of a long cylindrical tube, containing at one end a convex lens of small aperture and small focal length. This is called the objective lens (0). At the other end of the tube another co-axial smaller and wide tube is fitted, which carries a convex lens (E) at its outer end. This lens is towards the eye and is called the eyepiece. The focal length and aperture of eyepiece are somewhat larger than those of objective lens. Cross-wires are mounted at a definite distance before the eyepiece. The entire tube can be moved forward and backward by the rack and pinion arrangement.

Adjustment: First of all the eyepiece is displaced backward and forward to focus it on cross-wires. Now the object is placed just in front of the objective lens and the entire tube is moved by rack and pinion arrangement until there is no parallax between image of object and cross wire. In this position, the image of the object appears quite distinct.

Working: Suppose a small object AB is placed slightly away from the first focus F_o‘of the objective lens. The objective lens forms the real, inverted and magnified image A’ B’, which acts as an object for eyepiece. The eyepiece is so adjusted that the image A’B’ lies between the first focus F_e‘ and the eyepiece E. The eyepiece forms its image A”B” which is virtual, erect and magnified. Thus the final image A”B” formed by the microscope is inverted and magnified and its position is outside the objective and eyepiece towards objective lens.

The magnifying power of a microscope is defined as the ratio of angle (β) subtended by final image on the eye to the angle (α) subtended by the object on eye, when the object is placed at the least distance of distinct vision, i.e., Magnifying power,
M = \(\frac{\beta}{\alpha}\)
As object is very small, angles a and 1 are very small and so tan α = α and tan β = β. By definition the object AB is placed at the least distance of distinct vision.

∴ α = tan α = \(\frac{A B}{E A}\)
By sign convention, EA = – D,
∴ α = \(\frac{A B}{-D}\)
and from figure
β = tan β = \(\frac{A^{\prime} B^{\prime}}{E A^{\prime}}\)
If u_e is distance of image A’ B’ from eyepiece E, then by sign convention, EA’ = -u_e
and so, β = \(\frac{A^{\prime} B^{\prime}}{-u_{e}}\)

Hence, magnifying power,
M = \(\frac{\beta}{\alpha}=\frac{A^{\prime} B^{\prime} /\left(-u_{e}\right)}{A B /(-D)}=\frac{A^{\prime} B^{\prime}}{A B} \cdot \frac{D}{u_{e}}\)
By sign conventions, magnification of objective lens
\(\frac{A^{\prime} B^{\prime}}{A B}=\frac{v_{o}}{\left(-u_{o}\right)}\)
∴ M = \(-\frac{v_{o}}{u_{o}} \cdot \frac{D}{u_{e}}\) ………………………………….. (2)

Using lens formula \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\) for eyelens,
(i.e. using f = f_e’ V = -v_e, U = -u_e ) we get
\(\frac{1}{f_{e}}=\frac{1}{-v_{e}}-\frac{1}{\left(-u_{e}\right)}\)
or \(\frac{1}{u_{e}}=\frac{1}{f_{e}}+\frac{1}{v_{e}}\)
Magnifying power,
M = \(-\frac{v_{o}}{u_{o}} D\left(\frac{1}{f_{e}}+\frac{1}{v_{e}}\right)\)

or M = \(-\frac{v_{o}}{u_{o}}\left(\frac{D}{f_{e}}+\frac{D}{v_{e}}\right)\)
When final image is formed at the distance of distinct vision, V_e = D
∴ Magnification,
M= – \(\frac{v_{o}}{u_{o}}\left(1+\frac{D}{f_{e}}\right)\)

For greater magnification of a compound microscope, f_e should be small. As f_o < f_e’ so f₀ is small.
Hence, for greater magnification both f₀ and f_e should be small with f₀ to be smaller of the two.

(b) If image A’B’ is exactly at the focus of the eyepiece, then image A”B” is formed at infinity.
If the object AB is very close to the focus of the objective lens of focal length f₀, then magnification M₀ by the objective lens
M_e = \(\frac{L}{f_{0}}\)

where, L is tube length (or distance between lenses L₀ and L_e) Magnification M_e by the eyepiece
M_e = \(\frac{D}{F_{e}} \)
where, D = Least distance of distinct vision
Total magnification, m = M₀ M_e = \(\left(\frac{L}{f_{o}}\right)\left(\frac{D}{f_{e}}\right)\)

Question 3.
Explain with the help of a labelled ray diagram, how is image formed in an astronomical telescope. Derive an expression for its magnifying power.
Or
Draw a ray diagram showing the image formation of a distant object by a refracting telescope. Define Its magnifying power and write the two important factors considered to increase the magnifying power. Describe briefly the two main limitations and explain how far these can be minimised in a reflecting telescope.
Answer:
Astronomical (Refracti ng) Telescope
Construction: It consists of two co-axial cylindrical tubes, out of which one tube is long and wide, while the other tube is small and narrow. The narrow tube may be moved in and out of the wide tube by rack and pinion arrangement. At one end of wide tube an achromatic convex lens L₁ is placed, which faces the object and is so-called objective (lens). The focal length and aperture of this lens are kept large. The large aperture of objective is taken that it may collect sufficient light to form a bright image of a distant object. The narrow tube is towards eye and carries an achromatic convex lens 12 of small focal length and small aperture on its outer end. This is called eye-lens or eyepiece.

The small aperture of eye lens is taken so that the whole light refracted by it may reach the eye. Cross-wires are fitted at a definite distance from the eye lens. Due to large focal length of objective lens and small focal length of eye lens, the final image subtends a large angle at the eye and hence the object appears large. The distance between the two lenses may be arranged by displacing narrow tube in or out of wide tube by means of rack and pinion arrangement.

Adjustment: First of all the eyepiece is moved backward and forward in the narrow tube and focused on the cross-wires. Then the objective lens is directed towards the object and narrow tube is displaced in or out of wide tube until the image of object is formed on cross-wires and there is no parallax between the image and cross-wires. In this position, a clear image of the object is seen. As the image is formed by refraction of light through both the lenses, this telescope is called the refracting telescope.

Working: Suppose AB is an object whose end A is on the axis of telescope. The objective lens (L₁) forms the image A’B’ of the object AB at its second principal focus F₀.
This image is real, inverted and diminished. This image A’ B’ acts as an object for the eyepiece L₂ and lies between first focus f_e‘ and optical centre C₂ of lens L₂.
Therefore, eyepiece forms its image A” B” which is virtual, erect and magnified.
Thus, the final image A” B” of object AB formed by the telescope is magnified, inverted and lies between objective and eyepiece.

Magnifying Power: The magnifying power of a telescope is measured by the ratio of angle (β) subtended by final image on the eye to the angle (α) subtended by object on the eye. i.e.,
Magnifying power M = \(\frac{\beta}{\alpha}\)
As α and β are very small angles, therefore, from figure.

The angle subtended by final image A” B” on eye.
β = angle subtended by image A’B’ on eye
= tanβ = \(\frac{A^{\prime} B^{\prime}}{C_{2} A^{\prime}}\)
As the object is very far (at infinity) from the telescope, the angle subtended by object at eye is same as the angle subtended by object on objective lens.
∴ α = tan α = \(\frac{A^{\prime} B^{\prime}}{C_{1} A^{\prime}}\)
∴ M = \(\frac{\beta}{\alpha}=\frac{A^{\prime} B^{\prime} / C_{2} A^{\prime}}{A^{\prime} B^{\prime} / C_{1} A^{\prime}}=\frac{C_{1} A^{\prime}}{C_{2} A^{\prime}}\)
If the focal lengths of objective and eyepiece be f₀ and f_e, distance of image A’B’ from eyepiece be u_e, then by sign convention
C₁A’ = +f₀
C₂A’ = – u_e
∴ M = –\(\frac{f_{o}}{u_{e}}\) ……………………………… (1)

If v_e is the distance of A” B” from eye-piece, then by sign convention, f_e is positive, u_e and v_e are both negative. Hence, by lens formula = \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
we have
\(\frac{1}{f_{e}}=\frac{1}{-v_{e}}-\frac{1}{\left(-u_{e}\right)}\)
or
\(\frac{1}{u_{e}}=\frac{1}{f_{e}}+\frac{1}{v_{e}}\)
Substituting this value in eq. (1), we get
M = -f₀ \(\left(\frac{1}{f_{e}}+\frac{1}{v_{e}}\right)\) …………………………. (2)

This is the general formula for magnifying power. In this formula, only numerical values of f₀, f_e and v_e are to be used because signs have already been used.
Length of Telescope : The distance between objective and eyepiece is called the length (L) of the telescope. Obviously,
L = L₁L₂ =C₁C₂ = f₀+u_e …………………… (3)

Now there arise two cases :
(i) When the final image is formed at minimum distance (D) of distinct vision then v_e =D
∴ M = -f₀ \(\left(\frac{1}{f_{e}}+\frac{1}{D}\right)=-\frac{f_{o}}{f_{e}}\left(1+\frac{f_{e}}{D}\right)\) …………………………… (4)
Length of telescope L = f₀ + u_e

(ii) In normal adjustment position, the final image is formed at infinity: For relaxed eye, the final image is formed at infinity. In this state, the image A’B’ formed by objective lens should be at first the principal focus of eyepiece, i.e.,
u_e = f_e and v_e
∴ Magnifying power,
M = – f₀ \(\left(\frac{1}{f_{e}}+\frac{1}{\infty}\right)\) = –\(\frac{f_{o}}{f_{e}}\)
Length of telescope = f₀ + f_e
For large magnifying power, f₀ should be large and f_e should be small. For high resolution of the telescope, diameter of the objective should be large.

Factors for Increasing the Magnifying Power
1. Increasing focal length of objective
2. Decreasing focal length of eyepiece

Limitations
1. Suffers from chromatic aberration
2. Suffers from spherical aberration
3. Small magnifying power
4. Small resolving power

Advantages of Reflecting Telescope
1. No chromatic aberration, because mirror is used.
2. Spherical aberration can be removed by using a parabolic mirror.
3. Image is bright because no loss of energy due to reflection.
4. Large mirror can provide easier mechanical support.

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 8 Electromagnetic Waves Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 8 Electromagnetic Waves

Very short answer type questions

Question 1.
How are radiowaves produced?
Answer:
They are produced by rapid accelerations and deaccelerations of electrons in aerials.

Question 2.
How are microwaves produced?
Answer:
By using a magnetron.

Question 3.
Write two uses of microwaves.
Answer:
Uses of Microwaves

• In RADAR communication.
• In analysis of molecular and atomic structure.

Question 4.
To which part of the electromagnetic spectrum does a wave of frequency 3 × 10¹³ Hz belong?
Answer:
The frequency of 3 × 10¹³ Hz belongs to the infrared waves.

Question 5.
Name the electromagnetic waves, which (i) maintain the Earth’s warmth and (ii) are used in aircraft navigation.
Answer:
(i) Infrared rays
(ii) Microwaves

Question 6.
Welders wear special goggles or face masks with glass windows to protect their eyes from electromagnetic radiation. Name the radiations and write the range of their frequency.
Answer:
Welders wear special goggles or face mask with glass windows to protect their eyes from ultraviolet rays. The range of UV rays is 4 × 10^-7 m (400 nm) to 6 x 10^-10 m (0.6 nm).

Question 7.
How are X-rays produced?
Answer:
X-rays are produced when high energetic electron beam is made incident on a metallic target of high melting point and high atomic weight.

Question 8.
Write two uses of X-rays.
Answer:
Uses of X-rays

• In medical diagnosis as they pass through the muscles not through the bones.
• In detecting faults, cracks, etc. in metal products.

Question 9.
A variable frequency AC source is connected to a capacitor. How will the displacement current change with decrease in frequency? (NCERT Exemplar)
Answer:
On decreasing the frequency, reactance X_C = \(\frac{1}{\omega C}\) will increase which will lead to decrease in conduction current. In this case I_d = I_c, hence displacement current will decrease.

Question 10.
Do electromagnetic waves carry energy and momentum?
Answer:
Yes. Electromagnetic waves carry energy and momentum.

Question 11.
Why is the orientation of the portable radio with respect to broadcasting station important? (NCERT Exemplar)
Answer:
As electromagnetic waves are plane polarised, so the receiving antenna should be parallel to electric/magnetic part of the wave.

Question 12.
The charge on a parallel plate capacitor varies as q = q₀ cos 2πvt. The plates are very large and close together (area = A, separation = d). Neglecting the edge effects, find the displacement current through the capacitor? (NCERT Exemplar)
Answer:
Conduction current I_C = Displacement current I_D
I_C = I_D = \(\frac{d q}{d t}\) = \(\frac{d}{d t}\) (q₀ cos 2π vt) = -2πcq₀vsin2πvt

Question 13.
Professor C.V. Raman surprised his students by suspending freely a tiny light ball in a transparent vacuum chamber by shining a laser beam on it. Which property of electromagnetic waves was he exhibiting? Give one more example of this property. (NCERT Exemplar)
Answer:
Electromagnetic waves exert radiation pressure. Tails of comets are due to solar radiation.

Short answer type questions

Question 1.
Write the generalised expression for the Ampere’s circuital law in terms of the conduction current and the displacement current. Mention the situation when there is
(i) only conduction current and no displacement current,
(ii) only displacement current and no conduction current.
Answer:
Generalised Ampere’s Circuital Law
\(\oint \vec{B} \cdot \overrightarrow{d l}\) = μ₀I_c + μ₀ε₀\(\frac{d \phi_{E}}{d t}\)
Line integral of magnetic field over closed loop is equal to p 0 times sum of conduction current and displacement current.

(i) In case of steady electric field in a conducting wire, electric field does not change with time, conduction current exists in the wire but displacement current may be zero.
So \(\oint \vec{B} \cdot \overrightarrow{d l}\) = μ₀I_c

(ii) In large region of space, where there is no conduction current, but there is only a displacement current due to time varying electric field (or flux).
So Φ \(\vec{B} \cdot \overrightarrow{d l}\) = μ₀ ε₀ \(\frac{d \phi_{E}}{d t}\)

Question 2.
How are infrared waves produced? Why are these referred to as ‘heat waves’? Write their one important use.
Answer:
Infrared waves are produced by hot bodies and molecules. Infrared waves are sometimes referred to as heatwaves. This is because water molecules present in most materials readily absorb infrared waves. After absorption, their thermal motion increases, that is they heat up and heat their surroundings.
Infrared lamps are used in physical therapy and in remote control of devices.

Question 3.
(i) Arrange the following electromagnetic waves in the descending order of their wavelength.
(a) Microwaves
(b) Infrared rays
(c) Ultraviolet radiation
(d) γ-rays
(ii) Write one use each of any two of them.
Answer:
(i) The decreasing order ofwavelength of electromagnetic waves are Microwaves > Infrared > Ultraviolet > y-rays

(ii) Microwaves: They are used in RADAR devices,
γ-rays: It is used in radio therapy.

Question 4.
Write Maxwell’s generalization of Ampere’s Circuital Law. Show that in the process of charging a capacitor, the current produced within the plates of the capacitor is
i = ε₀\(\frac{d \phi_{\boldsymbol{E}}}{d t}\)
Where Φ_E is the electric flux produced during charging of the capacitor plates.
Answer:
Ampere’s circuital law is given by
\(\oint \vec{B} \cdot \overrightarrow{d l}\) = μ₀I_c
For a circuit containing capacitor, during its charging or discharging the current within the plates of the capacitor varies producing displacement current I_d Hence, Ampere’s circuital law is generalised by Maxwell, given as
\(\oint \vec{B} \cdot \overrightarrow{d l}\) = μ₀I_c + μ₀I_d
The electric flux (Φ_E) between the plates of capacitor changes with time, producing current within the plates which is proportional to (\(\frac{d \phi_{E}}{d t}\))
Thus, we get,
I_c = ε₀ \(\frac{d \phi_{E}}{d t}\)ε

Question 5.
How are electromagnetic waves produced? What is the source of energy of these waves? Write mathematical expressions for electric and magnetic fields of an electromagnetic wave propagating along the z-axis. Write any two important properties of electromagnetic waves.
Answer:
Electromagnetic wave produced by oscillating charged particle. Mathematical expression for electromagnetic wave travel along z-axis:
E_x = E₀ sin(kz – ωt) [For electric field]
B_y = B₀ sin(kz – ωt) [For magnetic field]
Properties
(i) Have oscillating electric perpendicular direction.
(ii) Transverse nature.

Question 6.
Electromagnetic waves with wavelength
(i) λ₁ is used in satellite communication.
(ii) λ₂ is used to kill germs in water purifier.
(iii) λ₃ is used to detect leakage of oil in underground pipelines.
(iv) λ₄ is used to improve visibility in runways during fog and mist conditions.
(a) Identify and name the part of electromagnetic spectrum to which these radiations belong.
(b) Arrange these wavelengths in ascending order of their magnitude.
(c) Write one more application of each. (NCERTExemplar)
(a) λ₁ → Microwave, λ₂ → UV
λ₃ → X rays, λ₄ → Infrared

(b) λ₃ < λ₂ <λ₄ < λ₁

(c) Microwave-RADAR
UV-LASIK eye surgery
X-ray-Bone fracture identification (bone scanning)
Infrared-Optical communicatio

Long answer type questions

Question 1.
Draw a labelled diagram of Hertz’s experiment. Explain how electromagnetic radiations are produced using this set-up.
Answer:
Hertz Experiment: Hertz’s experiment was based on the fact that an oscillating electric charge radiates electromagnetic waves and these waves carry energy which is being supplied at the cost of K.E. of the oscillating charge.

Hertz Apparatus: The experimental arrangement used by Hertz for the production and detection of electromagnetic waves in the laboratory, is shown in fig. His experimental arrangement consists of two metal sheets P₁ and P₂. These sheets are connected to a source of very high voltage (i.e. an induction coil, which can supply a potential difference of several thousand volts). S₁ and S₂ are two metal spheres connected to the metal sheets P₁ and P₂ . The distance between the metal sheets is kept nearly 60 cm and that between the sphere is normally from 2 cm to 2.5 cm.

The two plates P₁ and P₂ form a capacitor of very low capacitance (C). The circuit containing P₁ and P₂ (being completed by conducting wire), has also some low value of inductance L. It thus forms an LC circuit. Detector (D) consisting of a coil to the ends of which two other small metal spheres S₁‘and S₂‘ are connected.

Working of Hertz Apparatus: Due to existence of very high voltage, air present in the gap across the plates of spheres S₁ and S₂ gets ionised. Due to presence of the ions or charged particles, the path between the spheres S₁ and S₂ become conducting. As a result of this, very high time- varying current flows across the gap between S₁ and S₂ (as plates P₁ and P₂ form an LC circuit). Due to this a spark is produced. Since, sheets P₁ ,
P₂ form an LC-circuit, hence, electromagnetic waves of frequency f = \(\frac{1}{2 \pi} \sqrt{\frac{1}{L C}}\)

Function of the Detector D: Hertz detected the electromagnetic waves by means of a detector D kept at suitable distance from the conducting spheres S₁, S₂ Detector D is made of two similar conducting spheres S₁‘ and S₂‘ joined to the ends of a coil to form another LC circuit.

The frequency of this LC circuit is made equal to the frequency of electromagnetic waves reaching it. The frequency can be adjusted by changing the diameter of the coil of the detector and by changing the distance between S₁‘ and S₂‘. Hertz placed the detector in such a way that the magnetic lines of force produced by the oscillating electric field across the gap between S₁‘ and S₂‘ are normal to the plane of coil (C). When magnetic lines of force cut the detector coil, an emf is induced in it. Hence, air in the gap between S₁‘ and S₂‘ gets ionised. A conducting path becomes available for the induced current to flow across the gap. Thus, the spark is produced between S₁‘ and S₂‘. Hertz also observed that the spark across S₁‘ and S₂‘ was greatest when the S₁‘ S₂‘ and S₁ S₂ were parallel to each other. This clearly established that electromagnetic waves produced were polarise i.e., \(\vec{E}\) and \(\vec{B}\) always lie in one plane.