Important Questions

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 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 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.

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers Important Questions and Answers.

PSEB 12th Class Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Very Short Answer Type Questions

Question 1.
Write the structures of the products when Butan-2-ol reacts with the following:
(i) CrO₃
(ii) SOCl₂
Answer:

Question 2.
What happens when ethanol reacts with CH₃COCl/pyridine ?
Answer:

Question 3.
When phenol is created with bromine water, while precipitate is obtained. Prove the structure and the name of the compound formed.
Answer:
When phenol is treated with bromine water, white ppt. of 2, 4, 6-tribromophenol is obtained.

Question 4.
Answer the following questions :
(i) Dipole moment of phenol is smaller than that of methanol. Why?
(ii) In Kolbe’s reaction, instead of phenol, phenoxide ion is treated with carbon dioxide. Why ?
Answer:
(i) In phenol, C—O bond is less polar due to electron-withdrawing effect of benzene ring whereas in methanol, C—O bond is more polar due to electron-releasing effect of —CH₃ group.

(ii) Phenoxide ion is more reactive than phenol towards electrophilic aromatic substitution and hence undergoes electrophilic substitution with carbon dioxide which is a weak electrophile.

Question 5.
What is denatured alcohol ?
Answer:
Alcohol is made unfit for drinking by mixing some copper sulphate and pyridine in it. This is called denatured alcohol.

Question 6.
Arrange the following compounds in the increasing order of their acidic strength: p-cresol, p -nitrophenol, phenol
Answer:
p-cresol < phenol < p-nitrophenol

Question 7.
Arrange the following compounds in decreasing order of acidity.
(i) H₂O, ROH, HC ☰ CH
(ii)
(iii) CH₃OH, H₂O, C₆H₆OH
Answer:
(i) H₂O > ROH > HC ☰ CH
(ii)
(iii) C₆H₅OH > H₂O > CH₃OH

Question 8.
Suggest a reagent for conversion of ethanol to ethanal.
Answer:
Ethanol can be oxidises into ethanal by using pyridinium chlorochromate.

Question 9.
Explain why sodium metal can be used for drying diethyl ether but not ethyl alcohol.
Answer:
Due to presence of an active hydrogen atom, ethyl alcohol reacts with sodium metal.
2CH₃ — CH₂ — OH + 2Na → 2CH₃ — CH₂ — ONa + H₂
Diethyl ether, on the other hand, does not have replaceable hydrogen atom therefore does not react with sodium metal hence can be dried by metallic sodium.

Question 10.
Phenol is an acid but does not react with sodium bicarbonate solution. Why?
Answer:
Phenol is a weaker acid than carbonic acid (H₂CO₃) and hence does not liberate CO₂from sodium bicarbonate.

Question 11.
In the process of wine making, ripened grapes are crushed so that sugar and enzyme should come in contact with each other and fermentation should start. What will happen if anaerobic conditions are not maintained during this process?
Answer:
Ethanol will be converted into ethanoic acid.

Short Answer Type Questions

Question 1.
Why is the reactivity of all the three classes of alcohols with cone. HCl and ZnCl₂ (Lucas reagent) different ?
Answer:
The reaction of alcohols with Lucas reagent (cone. HCl and ZnCl₂) follow S_N1 mechanism. S_N1 mechanism depends upon the stability of carbocations (intermediate). More stable the intermediate carbocation, more reactive is the alcohol.

Tertiary carbocations are most stable among the three classes of carbocations and the order of the stability of carbocation is 3° > 2° > 1°. This order, intum, reflects the order of reactivity of three classes of alcohols i. e., 3° > 2° > 1°.

Thus , as the stability of carbocations are different so the reactivity of all the three classes of alcohols with Lucas reagent is different.

Question 2.
Write the mechanism of the following reaction:

Answer:
S_N2 mechanism

Question 3.
Explain a process in which a biocatalyst is used in industrial preparation of a compound known to you.
Answer:
Enzymes are biocatalyst. These biocatalysts (enzymes) are used in the industrial preparation of ethanol. Ethanol is prepared by the fermentation of molasses—a dark brown coloured syrup left after crystallisation of sugar which still contains about 40% of sugar.

The process of fermentation actually involves breaking down of large molecules into simple ones in the presence of enzymes. The source of these enzymes is yeast. The various reactions taking place during fermentation of carbohydrates are :

In wine making, grapes are the source of sugars and yeast. As grapes ripen, the quantity of sugar increases and yeast grows on the outer skin. When grapes are crushed, sugar and the enzyme come in contact and fermentation starts. Fermentation takes place in anaerobic conditions i.e., in absence of air. CO₂ gas is released during fermentation.

The action of zymase is inhibited once the percentage of alcohol ,formed exceeds 14 per cent. If air gets into fermentation mixture, the oxygen of air oxidises ethanol to ethanoic acid which in turn destroys the taste of alcoholic drinks.

Question 4.
Explain why alcohols and ethers of comparable molecular mass have different boiling points ?
Answer:
Boiling point depends upon the strength of intermolecular forces of attraction. Higher these forces of attraction, more will be the boiling point. Alcohols undergo intermolecular hydrogen bonding. So, the molecules of alcohols are held together by strong intermolecular forces of attraction.

But in ethers no hydrogen atom is bonded to oxygen. Therefore, ethers are held together by weak dipole-dipole forces, not by strong hydrogen bond.

Since, lesser amount of energy is required than to break weak dipole-dipole forces in ethers than to break strong hydrogen bonds in alcohol.

Question 5.
Explain why is O ＝ C ＝ O non-polar while R—O—R is polar ?
Answer:
CO₂ is a linear molecule. The dipole moment of two C —O bonds are equal and opposite and they cancel each other and hence the dipole moment of CO₂ is zero and it is a non-polar molecule.

While for ethers, two dipoles are pointing in the same direction. These two dipoles do not cancel the effect of each other. Therefore, there is a finite resultant dipoles and hence R—O—R is a polar molecule.

Question 6.
Give reasons for the following:
(i) p-Nitrophenol is more acidic than o-nitrophenol
(ii) Bond angle C—O—C in ethers is slightly higher than the tetrahedral angle (109°28′).
(iii) (CH₃)₃C—Br on reaction with NaOCH₃ gives an alkene instead of an ether.
Answer:
(i)
Intramolecular H-bonding in o-nitrophenol makes loss of proton difficult. Therefore, p-nitrophenol is more acidic than o-nitrophenol.

(ii) The bond angle in ether is slightly higher than 109 °28′ due to repulsive interaction between the two bulky alkyl groups.

(iii) It is because NaOCH₃ is a strong nucleophile as well as a strong base. Thus, elimination reaction predominates over substitution reaction.

Question 7.
Explain the following behaviours :
(i) Alcohols are more soluble in water than the hydrocarbons of comparable molecular masses.
(ii) Ortho-nitrophenol is more acidic than ortho-methoxyphenol.
(iii) Cumene is a better starting material for the preparation of phenol.
Answer:
(i) Alcohols are more soluble in water than the hydrocarbons of comparable molecular masses because of H-bond formation between alcohol and water molecules.
(ii) Ortho-nitrophenol is more acidic than ortho-methoxyphenol because nitro being the electron with drawing group stabilises the phenoxids ion.
(iii) Cumene is a better starting material for the preparation of phenol because side product formed in this reaction is acetone which is another important organic compound.

Long Answer Type Questions

Question 1.
(a) Name the starting material used in the industrial preparation of phenol.
(b) Write complete reaction for the bromination of phenol in aqueous and non-aqueous medium.
(c) Explain why Lewis acid is not required in bromination of phenol?
Answer:
(a) The starting material used in the industrial preparation of phenol is cumene.

(b) Phenols when treated with bromine water gives polyhalogen derivatives in which all the hydrogen atoms present at ortho and para positions with respect to —OH group are replaced by bromine atoms.

However, in non-aqueous medium such as CS₂, CCl₄, CHCl₃ monobromophenols are obtained.

In aqueous solution, phenol ionises to form phenoxide ion. This ion activates the benzene ring to a very large extent and hence the substitution of halogen takes place at all three positions.

On the other hand, in non-aqueous solution ionisation of phenol is greatly suppressed. Therefore, ring is activated slightly and hence monosubstitution occur.

(c) Lewis acid is an electron deficient molecule. In bromination of benzene, Lewis acid is used-to polarise Br₂ to form Br+ electrophile.

In case of phenol, oxygen atom of phenol itself polarises the bromine molecule to form Br+ ion (electrophile). So, Lewis acid is not required in the bromination of phenol.

Question 2.
Explain the mechanism of the following reactions :
(i) Addition of Grignard’s reagent to the carbonyl group of a compound forming an adduct followed by hydrolysis.
(ii) Acid catalysed dehydration of an alcohol forming an alkene.
(iii) Acid catalysed hydration of an alkene forming an alcohol.
Answer:
(i) Step I : Nucleophilic addition of Grignard reagent to carbonyl group.

Step II : Formation of carbocation : It is the slowest step and hence, the rate determining step.

To drive the equilibrium to the right, ethylene is removed as it is formed.

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes Important Questions and Answers.

PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes

Very Short Answer Type Questions

Question 1.
Out of and , which is an example of allylic halide?
X
Answer:
is an example of allylic halide.

Question 2.
Which of the following reactions is S_N1 type ?

Answer:
Reaction (ii) is S_N1 reaction.

Question 3.
Which one of the following compounds is more easily hydrolysed by KOH and why?
CH₃CHClCH₂CH₃ or CH₃CH₂CH₂CH₂Cl
Answer:
Due to +1 effect of alkyl groups the 2° carbonium ion CH₃—CH—CH₂—CH₃ derived from sec-butyl chloride is more stable than the 1° carbonium ion \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\stackrel{+}{\mathrm{C}} \mathrm{H}_{2}\) derived from n-propyl chloride. Therefore sec-butyl chloride gets hydrolysed more easily than n-propyl chloride under S_N1 conditions.

Question 4.
What is known as a racemic mixture ? Give an example.
Answer:
equimolar mixture of a pair of enantiomers is called racemic mixture. A racemic mixture is optically inactive due to external compensation.

Question 5.
Consider the three types of replacement of group X by group Y as shown here.

This can result in giving compound (A) or (B) or both. What is the process called if
(A) is the only compound obtained ?
(B) is the only compound obtained ?
(A) and (B) are formed in equal proportions ?
Answer:
(i) Retention
(ii) Inversion
(iii) Racemisation.

Question 6.
What is an asymmetric carbon?
Answer:
A carbon which is attached to four different atoms/groups is called asymmetric carbon. For example, the carbon atom in BrCHClI.

Question 7.
What is plane polarized light?
Answer:
A beam of light which has vibration in only one plane is called plane polarized light.

Question 8.
Why iodoform has appreciable antiseptic property ?
Answer:
Iodoform liberate I₂ when it comes in contact with skin. Antiseptic property of iodine is due to the liberation of I₂ not because of iodoform itself.
(responsible for antiseptic property)

Question 9.
How does the ordinary light differ from the plane polarized light?
Answer:
Ordinary light has oscillations in all the directions perpendicular to the path of propagation whereas plane polarised light has all oscillations in the same plane.

Question 10.
What do you .understand by the term optical activity of compounds?
Answer:
The property of certain compounds to rotate the plane of polarzed light in a characteristic way when it is passed through their solutions is called optical activity of compounds.

Short Answer Type Questions

Question 1.
Give reasons for the following:
(i) Ethyl iodide undergoes S_N2 reaction faster than ethyl bromide.
(ii) (±) 2-Butanol is optically inactive.
(iii) C—X bond length in halobenzene is smaller than C—X bond length in CH₃ —X.
Answer:
(i) Since I^– ion is a better leaving group than Br^– ion, hence, CH3I reacts faster than CH₃Br in SN₂ reaction with OH^– ion.

(ii) (±) 2-Butanol is a racemic mixture, i.e., there are two enantiomers in equal proportions. The rotation by one enantiomer will be cancelled by the rotation due to the other isomer, making the mixture optically inactive.

(iii) In CH₃—X the carbon atom is sp²-hybridised while in halobenzene the carbon atom is sp³-hybridised. The sp²-hybridised carbon is more electronegative due to greater s-character and holds the electron pair of C—X bond more tightly than sp³-hybridised carbon with less s-character. Thus, C—X bond length in CH₃—X is bigger than C—X in halobenzene.

Question 2.
Give reasons for the following:
(i) Haloalkanes easily dissolve in organic solvents.
(ii) Halogen compounds used in industry as solvents are alkyl chlorides rather than bromides and iodides.
Answer:
(i) Haloalkanes dissolve in organic solvents because the new intermolecular attractions between haloalkanes and organic solvent molecules have much the same strength as ones being broken in the separate haloalkanes and solvent molecules.
(ii) Because alkyl chlorides are more stable and more volatile than bromides and iodides.

Question 3.
Write the mechanism of the following reaction

Answer:
S_N2 mechanism

Question 4.
Which would undergo S_N1 reaction faster in the following pairs and why ?

Answer:
(i)
Tertiary halide reacts faster than primary halide because of greater stability of 3°-carbocation.

(ii)
Because the secondary carbocation formed in the slowest step is more stable than the primary carbocation.

Question 5.
Give reasons:
(i) n-Butyl bromide has higher boiling point than f-butyl bromide. Racemic mixture is optically inactive.
(ii) The presence of nitro group (—NO₂) at o/p positions increases the reactivity of haloarenes towards nucleophilic substitution reactions.
Answer:
(i) n-butyl bromide being a straight chain alkyl halide has larger surface area than tert butyl bromide. Larger the surface area, larger the magnitude of the van der Waal’s forces and hence higher is the boiling point.

(ii) A racemic mixture contains the two enantiomers d and l in equal •proportions. As the rotation due to one enantiomer is cancelled by equal and opposite rotation of another enantiomer, therefore, it is optically inactive.

(iii) The presence of NO₂ group at o/p position in haloarenes helps in the stabilisation of resulting carbanion by -R and -I effects and hence increases the reactivity of haloarenes towards nucleophilic substitution reactions.

Question 6.
(i) Why are alkyl halides insoluble in water ?
(ii) Why is butan-l-ol optically inactive but butan-2-ol is optically active ?
(iii) Although chlorine is an electron withdrawing group, yet it is ortho, para directing in electrophilic aromatic substitution reactions. Why ?
Answer:
(i) This is due to the inability of alkyl halide molecule to form intermolecular hydrogen bonds with water molecules.
(ii)
due to presence of a chiral carbon butan-2-ol is an optically active compound.

(iii) As the weaker resonance (+R) effect of Cl which stabilise the carbocation formed tends to oppose the stronger inductive (-I) effect of Cl which destabilise the carbocation at ortho and para positions and makes deactivation less for ortho and para position.

Question 7.
(i) Allyl chloride can be distinguished from vinyl chloride by NaOH and silver nitrate test. Comment.
(ii) Alkyl halide reacts with lithium aluminium hydride to give alkane. Name the attacking reagent which will bring out this change.
Answer:
(i) Vinyl chloride does not respond to NaOH and silver nitrate test because of partial double bond character due to resonance.
(ii) Hydride ion (H )

Question 8.
Give the ITJPAC name of the product formed when :
(i) 2-Methyl-l-bromopropane is treated with sodium in the presence of dry ether.
(ii) 1-Methyl cyclohexene is treated with HI.
(iii) Chloroethane is treated with silver nitrite.
Answer:
(i) 2, 5-dimethylhexane
(ii) 1 -Methyl-1 -iodocyclohexane
(iii) Nitroethane

Long Answer Type Questions

Question 1.
Some alkyl halides undergo substitution whereas some undergo elimination reaction on treatment with base. Discuss the structural features of alkyl halides with the help of examples which are responsible for this difference.
Answer:
Primary alkyl halides follow S_N2 mechanism in which a nucleophile attacks at 180° to the halogen atom. A transition state is formed in which carbon is bonded to two nucleophiles and finally halogen atom is pushed out. In S_N2 mechanism, substitution of nucleophile takes place as follows :

Thus, in S_N2 mechanism, substitution takes place. Tertiary alkyl halides follow? S_N1 mechanism, In this case, tert-alkyl halides form 3° carbocation. Now, if the reagent used is a weak base then substitution occur while if it is a strong base then instead of substitution elimination occur.

As alc. KOH is a strong base, so elimination competes over substitution and alkene is formed.

Question 2.
Why are aryl halides less reactive towards nucleophilic substitution reactions than alkyl halides ? How can we enhance the reactivity of aryl halides ?
Answer:
Aryl halides are less reactive towards nucleophilic substitution reaction due to the following reasons :
(i) In haloarenes, the lone pair of electron on halogen are in resonance with benzene ring. So, C—Cl bond acquires partial double bond character which strengthen C—Cl bond. Therefore, they are less reactive towards nucleophilic substitution reaction.

(ii) In haloarenes, the carbon atom attached to halogen is sp²-hybridised. The sp²-hybridised carbon is more electronegative than sp³-hybridised carbon. This sp ²-hybridised carbon in haloarenes can hold the electron pair of C—X bond more tightly and make this C—Cl bond shorter than C—Cl bond haloalkanes.

(iii) Since, it is difficult to break a shorter bond than a longer bond therefore haloarenes are less reactive than haloarenes.
In haloarenes, the phenyl cation will not be stabilised by resonance therefore S_N1 mechanism ruled out.

(iv) Because of the repulsion between the nucleophile and electron rich arenes, aryl halides are less reactive than alkyl halides.

The reactivity of aryl halides can be increased by the presence of an electron withdrawing group (—NO₂) at ortho and para positions. However, no effect on reactivity of haloarenes is observed by the presence of electron withdrawing group at meta-position. Mechanism of the reaction is as depicted with OH^– ion.



From the above resonance, it is very clear that electron density is rich at ortho and para positions. So, presence of EWG will facilitate nucleophilic at ortho and para positions not on meta position.

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 9 Coordination Compounds Important Questions and Answers.

PSEB 12th Class Chemistry Important Questions Chapter 9 Coordination Compounds

Very Short Answer Type Questions

Question 1.
Why is CO a stronger ligand than Cl^– ?
Answer:
CO forms π bonds so it is a stronger ligand than Cl^–.

Question 2.
What is the relationship between observed colour of the complex and the wavelength of light absorbed by the complex?
Answer:
When white light falls on the complex, some part of it is absorbed. Higher the crystal field splitting, lower will be the wavelength absorbed by the complex. The observed colour of complex is the colour generated from the wavelength left over.

Question 3.
How many isomers are there for octahedral complex [CoCl₂ (en) (NH₃)₂]⁺?
Answer:
There will be three isomers: cis and trans isomers. Cis will also show optical isomerism.

Question 4.
Why are low spin tetrahedral complexes not formed?
Answer:
Because for tetrahedral complexes, the crystal field stabilisation energy is lower than pairing energy.

Question 5.
A complex of the type [M(AA)₂X₂]ⁿ⁺ is known to be optically active. What does this indicate about the structure of the complex? Give one example of such complex.
Answer:
An optically active complex of the type [M(AA)₂X₂]ⁿ⁺ indicates cis- octahedral structure, e.g., cis-[Pt(en)₂Cl₂]²⁺ or cis-[Cr(en)₂Cl₂]⁺.

Question 6.
Why is the complex [Co(en)₃]³⁺ more stable than the complex [CoF₆]^3-?
Answer:
Due to chelate effect as the complex [Co(en)₃]³⁺ contains chelating ligand \(\ddot{\mathrm{NH}}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\ddot{\mathrm{NH}}_{2}\).

Question 7.
What do you understand by ‘denticity of a ligand’?
Answer:
The number of coordinating groups present in ligand is called the denticity of ligand. For example, denticity of ethane-1, 2-diamine is 2, as it has two donor nitrogen atoms which can link to central metal atom.

Question 8.
What type of isomerism is shown by the complex [CO(NH₃)₅(SCN)]²⁺?
Answer:
Linkage isomerism.

Question 9.
Arrange the following complex ions in increasing order of crystal field splitting energy △₀ :
[Cr(Cl)₆]^3-, [Cr(CN)₆]^3-, [Cr(NH₃)₆]³⁺
Answer:
[Cr(Cl)₆]^3- < [Cr(NH₃)₆]³⁺ < [Cr(CN)₆]^3-

Question 10.
A coordination compound with molecular formula CrCl₃.4H₂O precipitates one mole of AgCl with AgNO₃ solution. Its molar conductivity is found to be equivalent to two ions. What is the structural formula and name of the compound?
Answer:
[Cr(H₂O)₄Cl₂] Cl
[Tetraaquadichloridochromium (III) chloride]

Short Answer Type Questions

Question 1.
Give the electronic configuration of the following complexes on the basis of crystal field splitting theory.
[CoF₆]^3-, [Fe(CN)₆]^4- and [Cu(NH₃)₆]²⁺
Answer:
[CoF₆]^3-: Co³⁺(d⁶) \(t_{2 g}^{4} e_{g}^{2}\)
[Fe(CN)₆]^4- : Fe²⁺ (d⁶) \(t_{2 g}^{6} e_{g}^{0}\)
[Cu(NH₃)₆]²⁺ : Cu²⁺ (d⁹) \(t_{2 g}^{6} e_{g}^{3}\)

Question 2.
(i) What type of isomerism is shown by [Co(NH₃) ₅ONO]Cl₂?
(ii) On the basis of crystal field theory, write the electronic configuration for d⁴ ion, if △₀ < P.
(iii) Write the hybridisation and shape of [Fe(CN)₆]^3-.
(Atomic number of Fe = 26)
Answer:
(i) Linkage isomerism and the linkage isomer is [Co(NH₃) ₅ONO]Cl₂.
(ii) If △₀ < P, the fourth electron enters one of two e_g orbitals giving the configuration \(t_{2 g}^{3} e_{g}^{1}\).
(iii) Fe³⁺ : 3d⁵ 4s⁰

Question 3.
Explain why [Fe(H₂O)₆]³⁺ 5.92 BM whereas [Fe(CN)₆]^3- has a value of only 1.74 BM.
Answer:
[Fe(CN)₆]^3- involves d²sp³ hybridisation with one unpaired electron and [Fe(H₂O)₆]³⁺ involves sp³d² hybridisation with five unpaired electrons. This difference is due to the presence of strong CN^– and weak ligand H₂O in these complexes.

Question 4.
CuSO₄∙5H₂O is blue in colour while CuSO₄ is colourless. Why?
Answer:
In CuSO₄∙5H₂O, water acts as ligand as a result it causes crystal field splitting. Hence, d-d transition is possible in CuSO₄∙5H₂O and shows colour. In the anhydrous CuSO₄ due to the absence of water (ligand), crystal field splitting is not possible and hence it is colourless.

Question 5.
Why do compounds having similar geometry have different magnetic moment?
Answer:
It is due to the presence of weak and strong ligands in complexes, if CFSE is high, the complex will show low value of magnetic moment and vice versa, e.g., [CoF₆]^3- and [Co(NH₃)₆]³⁺, the former is paramagnetic and the latter is diamagnetic.

Question 6.
A metal ion Mⁿ⁺ having d⁴ valence electronic configuration combines with three bidentate ligands to form a complex compound. Assuming △₀ > P:
(i) Write the electronic configuration of d⁴ ion.
(ii) What type of hybridisation will Mⁿ⁺ ion has?
(iii) Name the type of isomerism exhibited by this complex.
Answer:
(i) \(t_{2 g}^{4} e_{g}^{0}\)
(ii) sp³d²
(iii) Optical isomerism

Long Answer Type Questions

Question 1.
Using crystal field theory, draw energy level diagram, write electronic configuration of the central metal atom/ion and determine the magnetic moment value in the following: [COF₆]^3-, [CO(H₂O)₆]²⁺, [CO(CN)₆]³
Answer:
Magnetic moment, μ = \(\sqrt{n(n+2)}\)
Where, n = Number of unpaired electrons

No unpaired electrons, so it is diamagnetic.

Question 2.
(i) Draw the geometrical isomers of complex [Pt(NH₃)₂Cl₂].
(ii) Write the hybridisation and magnetic behaviour of the complex [Ni(CO)₄].
(Atomic no. of Ni = 28)
Answer:

Geometrical isomers of [Pt(NH₃)₂Cl₂]

(ii) The complex [Ni(CO)₄] involves sp³ hybridisation.

The complex is diamagnetic as evident from the absence of unpaired electrons.

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 4 Chemical Kinetics Important Questions and Answers.

PSEB 12th Class Chemistry Important Questions Chapter 4 Chemical Kinetics

Very Short Answer Type Questions

Question 1.
For which type of reactions, order and molecularity have the same value?
Answer:
If the reaction is an elementary reaction, order is same as molecularity.

Question 2.
Why is the probability of reaction with molecularity higher than three very rare?
Answer:
The probability of more than three molecules colliding simultaneously is very small. Hence, possibility of molecularity being three is very low.

Question 3.
State a condition under which a bimolecular reaction is kinetically first order.
Answer:
A bimolecular reaction may become kinetically of first order if one of the reactants is in excess.

Question 4.
Thermodynamic feasibility of the reaction alone cannot decide the rate of the reaction. Explain with the help of one example.
Answer:
Thermodynamically the conversion of diamond to graphite is highly feasible but this reaction is very slow because its activation energy is high.

Question 5.
Why is it that instantaneous rate of reaction does not change when a part of the reacting solution is taken out?
Answer:
Instantaneous rate is measured over a very small interval of time, hence, it does not change when a part of solution is taken out.

Question 6.
A reaction is 50% complete in 2 hours and 75% complete in 4 hours. What is the order of the reaction?
Solution:
As t_75% = 2t_50%
Therefore, it is a first order reaction.

Question 7.
Define threshold energy of a reaction.
Answer:
Threshold energy is the minimum energy which must be possessed by reacting molecules in order to undergo effective collision which leads to formation of product molecules.

Question 8.
Why does the rate of a reaction increase with rise in temperature?
Answer:
At higher temperatures, larger fraction of colliding particles can cross the energy barrier (i.e., the activation energy), which leads to faster rate.

Question 9.
What is the difference between rate law and law of mass action?
Answer:
Rate law is an experimental law. On the other hand, law of mass action is a theoretical law based on the balanced chemical reaction.

Question 10.
What do you understand by ‘Rate of reaction’?
Answer:
The change in the concentration of any one of the reactants or products per unit time is termed as the rate of reaction.

Question 11.
In the Arrhenius equation, what does the factor e a corresponds to?
Answer:
e^-E_a/RT corresponds to the fraction of molecules that have kinetic energy greater than E_a,

Short Answer Type Questions

Question 1
What do you understand by the ‘order of a reaction’? Identify the reaction order from each of the following units of reaction rate constant:
(i) L^-1mols^-1
(ii) Lmol^-1s^-1.
Solution:
The sum of powers of the concentration of the reactants in the rate law expression is called order of reaction.
For a general reaction: aA + bB → Products
If rate = k[A]^m [B]ⁿ; order of reaction = m + n

(i) General unit of rate constant, k = (mol L^-1 )^1-ns^-1
L^-1mol s^-1 = (mol L^-1 )^1-ns^-1
-1 = -1 + n ⇒ n = 0 ∴ Reaction order = 0

(ii) L mol^-1 s^-1 = (mol L^-1)^1-n s^-1
1 = -1 + n ⇒ n = 2 ∴ Reaction order = 2

Question 2.
The rate constant for the first order decomposition of H₂O₂ is given by the following equation :
log k = 14.2 – \(\frac{1.0 \times 10^{4}}{T}\)K
Calculate E_a for this reaction and rate constant k if its half-life period be 200 minutes.
(Given: R = 8.314 JK^-1 mol^-1)
Solution:
Comparing the equation, log k = 14.2 – \(\frac{1.0 \times 10^{4}}{T}\) K with the equation,
log k = log A = \(\frac{E_{a}}{2.303 R T}\), we get
\(\frac{E_{a}}{2.303 R}\) = 1.0 × 10⁴ K or E_a = 1.0 × 10⁴ K × 2.303 × R
E_a = 1.0 × 10⁴ K × 2.303 × 8.314 JK^-1
= 19.1471 × 10⁴ Jmol^-1
= 191.47 kJ mol^-1
For a first order reaction, t_t/2 = \(\frac{0.693}{k}\) or k = \(\frac{0.693}{t_{1 / 2}}\)
k = \(\frac{0.693}{200 \mathrm{~min}}\) = 3.465 × 10 ^-3min^-1

Question 3.
The reaction, N₂(g) + O₂(g) ⇌ 2NO(g) contributes to air pollution whenever a fuel is burnt in air at a high temperature. At 1500 K, equilibrium constant K for it is 1.0 × 10^-5. Suppose in a case [N₂] = 0.80 mol L^-1 and [O₂] = 0.20 mol L^-1 before any reaction occurs. Calculate the equilibrium concentrations of the reactants and the product after the mixture has been heated to 1500 K.
Solution:

[N₂] = 0.8 – 6.324 × 10⁴ mol L^-1
= 0.799 molL^-1
[O₂] = 0.2 – 6.324 × 10^-4 mol L^-1
= 0.199 mol L^-1

Question 4.
For a general reaction, A → B, plot of concentration of A vs time is given in figure. Answer the following questions on the basis of this graph.

(i) What is the order of the reaction?
(ii) What is the slope of the curve?
(iii) What are the units of rate constant?
Answer:
(i) Zero order
(ii) Slope = – k
(iii) Units of rate constant = mol L^-1 s^-1

Question 5.
For a reaction, A + B → products, the rate law is rate = k [A][B]a^3/2. Can the reaction be an elementary reaction? Explain.
Answer:
During an elementary reaction, the number of atoms or ions colliding to react is referred to as molecularity. Had this been an elementary reaction the order of reaction with respect to B would have been 1, but in the given rate law it is \(\frac{3}{2}\). This indicated that the reaction is not an elementary reaction.

Question 6.
The rate constant for a first order reaction is 60 s^-1. How much time will it take to reduce 1 g of the reactant to 0.0625 g?
Answer:
We know that, t = \(\frac{2.303}{k}\) log \(\frac{[R]_{0}}{[R]}\)
t = \(\frac{2.303}{60}\) log \(\frac{1}{0.0625}\)
t = 0.0462 s

Long Answer Type Questions

Question 1.
For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained:

t/s	0	30	60
[CH3COOCH3]/mol L-1	0.60	0.30	0.15

(i) Show that it follows pseudo first order reaction, as the concentration of water remains constant.
(ii) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(Given log 2 = 0.3010, log 4 = 0.6021)
Solution:

As the value of k is same in both the cases, therefore, hydrolysis of methylacetate in aqueous solution follows pseudo first order reaction.

(ii) Average rate = \(-\frac{\Delta\left[\mathrm{CH}_{3} \mathrm{COOCH}_{3}\right]}{\Delta t}\)
= \(\frac{-[0.15-0.30]}{60-30}\) = \(\frac{0.15}{30}\)
Average rate = 0.005 mol L^-1s^-1

Question 2.
Describe how does the enthalpy of reaction remain unchanged when a catalyst is used in the reaction?
Answer:
A catalyst is a substance which increases the speed of a reaction without itself undergoing any chemical change.
According to “intermediate complex formation theory” reactants first combine with the catalyst to form an intermediate complex which is short-lived and decomposes to form the products and regenerating the catalyst.

The intermediate formed has much lower potential energy than the intermediate complex formed between the reactants in the absence of the catalyst.

Thus, the presence of catalyst lowers the potential energy barrier and the reaction follows a new alternate pathway which require less activation energy.

We know that, lower the activation energy, faster is the reaction because more reactant molecules can cross the energy barrier and change into products.

Enthalpy, △H is a ‘state function. Enthalpy of reaction, i.e., difference in energy between reactants and product is constant, which is clear from potential energy diagram.
Potential energy diagram of catalysed reaction is given as:

Question 3.
All energetically effective collisions do not result in a chemical change. Explain with the help of an example.
Answer:
Only effective collision lead to the formation of products. It means that collisions in which molecules collide with sufficient kinetic energy (called threshold energy = activation energy + energy possessed by reacting species).

And proper orientation lead to a chemical change because it facilitates the breaking of old bonds between (reactant) molecules and formation of the new ones i.e., in products.
e.g., formation of methanol from bromomethane depends upon the orientation of the reactant molecules.

The proper orientation of reactant molecules leads to bond formation whereas improper orientation makes them simply back and no products are formed.

To account for effective collisions, another factor P (probability of steric factor) is introduced K = PZ_ABe^-Ea/RT.

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.