Maharashtra State Board Class 12 Biology Solutions Digest

Maharashtra State Board 12th Std Biology Textbook Solutions Digest

Maharashtra State Board Class 12 Textbook Solutions

Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 1 Solid State Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Chemistry Solutions Chapter 1 Solid State

1. Choose the most correct answer.

Question i.
Molecular solids are
(a) crystalline solids
(b) amorphous solids
(c) ionic solids
(d) metallic solids
Answer:
(b) amorphous solids

Question ii.
Which of the following is n-type semiconductor?
(a) Pure Si
(b) Si doped with As
(c) Si doped with Ga
(d) Ge doped with In
Answer:
(b) Si doped with As

Question iii.
In Frenkel defect
(a) electrical neutrality of the substance is changed.
(b) density of the substance is changed.
(c) both cation and anion are missing
(d) overall electrical neutrality is preserved
Answer:
(d) overall electrical neutrality is preserved

Question iv.
In crystal lattice formed by bcc unit cell the void volume is
(a) 68%
(b) 74%
(c) 32%
(d) 26%
Answer:
(c) 32%

Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State

Question v.
The coordination number of atoms in bcc crystal lattice is
(a) 2
(b) 4
(c) 6
(d) 8
Answer:
(d) 8

Question vi.
Which of the following is not correct ?
(a) Four spheres are involved in the formation of tetrahedral void.
(b) The centres of spheres in octahedral voids are at the a pices of a regular tetrahedron.
(c) If the number of atoms is N the number of octahedral voids is 2N.
(d) If the number of atoms is N/2, the number of tetrahedral voids is N.
Answer:
(c) If the number of atoms is N the number of octahedral voids is 2N.

Question vii.
A compound forms hcp structure. Number of octahedral and tetrahedral voids in 0.5 mole of substance is respectively
(a) 3.011 × 1023, 6.022 × 1023
(b) 6.022 × 1023, 3.011 × 1023
(c) 4.011 × 1023, 2.011 × 1023
(d) 6.011 × 1023, 12.022 × 1023
Answer:
(a) 3.011 × 1023, 6.022 × 1023

Question viii.
Pb has fcc structure with edge length of unit cell 495 pm. Radius of Pb atom is
(a) 205 pm
(b) 185 pm
(c) 260 pm
(d) 175 pm
Answer:
(d) 175 pm

Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State

2. Answer the following in one or two sentences.

Question i.
What are the types of particles in each of the four main classes of crystalline solids?
Answer:
The smallest constituents or particles of various solids are atoms, ions or molecules.

Question ii.
Which of the three types of packing used by metals makes the most efficient use of space and which makes the least efficient use ?
Answer:
fcc has the most efficient packing of particles while scc has the least efficient packing.

Question iii.
The following pictures show population of bands for materials having different electrical properties. Classify them as insulator, semiconductor or a metal.
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 1a
Answer:
Picture A represents metal conductor,
Picture B represents insulator,
Picture C represents semiconductor.

Question iv.
What is a unit cell?
Answer:

  • Unit cell : It is the smallest repeating structural unit of a crystalline solid (or crystal lattice) which when repeated in different directions produces the crystalline solid (lattice).
  • The crystal is considered to consist of an infinite number of unit cells.
  • The unit cell possesses all the characteristics of the crystalline solid.

Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State

Question v.
How does electrical conductivity of a semiconductor change with temperature ? Why?
Answer:

  • Since the energy difference between valence band and conduction band in semiconductor is not large, the electrons from valence band can be promoted to conduction by heating.
  • Hence electrical conductivity of a semiconductor increases with temperature.

Question vi.
The picture represents bands of MOs for Si. Label valence band, conduction band and band gap.
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 2
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 3

Question vii.
A solid is hard, brittle and electrically non-conductor. Its melt conducts electricity. What type of solid is it?
Answer:
A solid crystalline electrolyte like NaCl is hard, brittle and electrically nonconductor. But its melt conducts electricity.

Question viii.
Mention two properties that are common to both hep and ccp lattices.
Answer:
In hcp and ccp crystal lattices coordination number is 12 and packing efficiency is 74%.

Question ix.
Sketch a tetrahedral void.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 4

Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State

Question x.
What are ferromagnetic substances?
Answer:

  1. The substances which possess unpaired electrons and high paramagnetic character and when placed in a magnetic field are strongly attracted and show permanent magnetic moment even when the external magnetic field is removed are said to be ferromagnetic. They can be permanently magnetised.
  2. In the solid state, the metal ions of ferromagnetic substance are grouped together into small regions called domains, where each domain acts as a tiny magnet.

Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 5
For example : Fe, Co, Gd, Ni, CrO2, etc.

3. Answer the following in brief.

Question i.
What are valence band and conduction band?
Answer:
There are two types of bands of molecular orbitals as follows :

  • Valence band : The atomic orbitals with filled electrons from the inner shells form valence bands, where there are no free mobile electrons since they are involved in bonding.
  • Conduction band : Atomic orbitals which are partially filled or empty on overlapping form closely placed molecular orbitals giving conduction bands where electrons are delocalised and can conduct, heat and electricity.

Question ii.
Distinguish between ionic solids and molecular solids.
Answer:

Type/ Property Ionic solids Molecular solids
1. Particles of unit cell Cations and anions Monoatomic or polyatomic molecules
2. Interparticle forces Electrostatic London, dipole-dipole forces and/or hydrogen bonds
3. Hardness Hard and brittle Soft
4. Melting points High 600 °C to 3000 °C Low (-272 °C to 400 °C)
5. Thermal and electrical conductivity Poor electrical conductors in solid state. Good conductors when melted or dissolved in water. Poor conductor of heat and electricity
6. Examples NaCl, CaF2 ice, benzoic acid

Question iii.
Calculate the number of atoms in fcc unit cell.
Answer:
Number of atoms in face-centred cubic (fcc) unit cell :
In this unit cell, there are 8 atoms at 8 corners and 6 atoms at 6 face centres. Each corner contributes 1/8th atom to the unit cell, hence due to 8 corners,
Number of atoms = \(\frac {1}{8}\) × 8 = 1.
Each face centre contributes half of the atom to the unit cell, hence due to 6 face centres,
Number of atoms = \(\frac {1}{2}\) × 6 = 3.
∴ Total number of atoms present in fee unit cell = 1 + 3 = 4.
Hence the volume of the unit cell is equal to the volume of four atoms.
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 6
Face centered unit cell

Question iv.
How are the spheres arranged in first layer of simple cubic close-packed structures? How are the successive layers of spheres placed above this layer ?
Answer:
(i) Stacking of square close packed layers :
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 7
Stacking of square close packed layers

In this arrangement, the two dimensional AAAA type square closed packed layers are placed one over the other in such a way that the crests of all spheres are in contact with successive layers in all directions. All spheres of different layers are perfectly aligned horizontally and vertically forming unit cells having primitive or simple cubic structure. Since all the layers are identical and if each layer is labelled as layer A, then whole three dimensional crystal lattice will be of AAAA… type.

Each sphere is in contact with six surrounded spheres, hence the coordination number of each sphere is six.

(ii) Stacking of two hexagonal close packed layers :
A close packed three dimensional structure can be generated by arranging hexagonal close packed layers in a particular manner.

In this the spheres of second layer are placed in the depression of the first layer.
In this if first layer is labelled as A then second layer is labelled as B since they are aligned differently.

Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 8
Two layers of closed packed spheres

In this, all triangular voids of the first layers are not covered by the spheres of the second layer. The triangular voids which are covered by second layer spheres generate tetrahedral void which is surrounded by four spheres. The triangular voids in one layer have above them triangular voids of successive layers.

The overlapping triangular voids from two layers together form an octahedral void which is surrounded by six spheres.

Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State

Question v.
Calculate the packing efficiency of metal crystal that has simple cubic structure.
Answer:
Step 1 : Radius of sphere : In simple cubic lattice, the atoms (spheres) are present at eight corners and in contact along the edge in the unit cell.
If ‘a’ is the edge length of the unit cell and ‘r’ is the radius of the atom, then
a = 2r or r = a/2
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 9
scc structure

Step 2 : Volume of sphere :
Volume of one particle = \(\frac{4 \pi}{3}\) × r3
= \(\frac{4 \pi}{3}\) × (a/2)3 = \(\frac{\pi a^{3}}{6}\)

Step 3 : Total volume of particles : Since the unit cell contains one particle. Volume occupied by one particle in unit cell = \(\frac{\pi a^{3}}{6}\)

Step 4 : Packing efficiency :
Packing efficiency
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 10
∴ Packing efficiency = 52.36%
Percentage of void space = 100 – 52.36
= 47.64%

Question vi.
What are paramagnetic substances? Give examples.
Answer:
(1) The magnetic properties of a substance arise due to the presence of electrons.
(2) An electron while revolving around the nucleus, also spins around its own axis and generates a magnetic moment and magnetic properties.
(3) If an atom or a molecule contains one or more unpaired electrons spinning in same direction, clockwise or anticlockwise, then the substance is associated with net magnetic moment and magnetic properties. They experience a net force of attraction when placed in the magnetic field. This phenomenon is called paramagnetism and the substance is said to be paramagnetic.
For example, O2, Cu2+, Fe3+ , Cr3+ , NO, etc.

Question vii.
What are the consequences of Schottky defect?
Answer:
Consequences of Schottky defect :

  • Since the number of ions (cations and anions) decreases but volume remains unchanged, the density of a substance decreases.
  • As the number of missing cations and anions is equal, the electrical neutrality of the compound remains same.
  • This defect arises in ionic crystals like NaCl, AgBr, KCl, etc.

Question viii.
Cesium chloride crystallizes in cubic unit cell with Cl ions at the corners and Cs+ ion in the centre of the cube. How many CsCl molecules are there in the unit cell ?
Answer:
Number of Cs+ ion at body centre = 1
Number of Cl ions due to 8 comers = 8 × \(\frac {1}{8}\) = 1
Hence unit cell contains 1 CsCl molecule.

Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State

Question ix.
Cu crystallizes in fee unit cell with edge length of 495 pm. What is the radius of Cu atom ?
Answer:
Given : a = 495 pm
Radius, r = ?
For fee structure,
radius = r = \(\frac{a}{2 \sqrt{2}}=\frac{495}{2 \times \sqrt{2}}\) = 175 cm.
Radius of Cu atom = 175 pm

Question x.
Obtain the relationship between density of a substance and the edge length of unit cell.
Answer:
(1) Consider a cubic unit cell of edge length ‘a’.
The volume of unit cell = a3

(2) If there are ‘n’ particles per unit cell and the mass of particle is ‘m’, then
Mass of unit cell = m × n.

(3) If the density of the unit cell of the substance is p then
Density of unit cell = \(\frac{\text { Mass of unit cell }}{\text { Volume of unit cell }}\)
ρ = \(\frac{m \times n}{a^{3}}\)

Question 4.
The density of iridium is 22.4 g/cm3. The unit cell of iridium is fcc. Calculate the radius of iridium atom. Molar mass of iridium is 192.2 g/mol.
Answer:
Given : Crystal structure of iridium = fcc
Molar mass of iridium = 192.2 gmol-1
Density = ρ = 22.4 gcm-3
Radius of iridium = ?
In fcc structure, there are 8 Ir atoms at 8 comers and 6 Ir atoms at 6 face centres.
∴ Total number of Ir atoms = \(\frac {1}{8}\) × 8 + \(\frac {1}{2}\) × 6
= 1 + 3
= 4
Mass of Ir atom = \(\frac{192.2}{6.022 \times 10^{23}}\)
= 31.92 × 10-23 g
∴ Mass of 4 Ir atoms = 4 × 31.92 × 10-23
= 1.277 × 10-21 g
∴ Mass of unit cell = 1.277 × 10-21 g
Density of unit cell = \(\frac{\text { Mass of unit cell }}{\text { Volume of unit cell }}\)
22.4 = \(\frac{1.277 \times 10^{-21}}{a^{3}}\)
∴ a3 = \(\frac{1.277 \times 10^{-21}}{22.4}\)
= 57 × 10-24 cm3
∴ a = (57 × 10-24)3 = 3.848 × 10-8 cm
If r is the radius of iridium atom, then for fcc structure,
r = \(\frac{a}{2 \sqrt{2}}\)
= \(\frac{3.848 \times 10^{-8}}{2 \times 1.414}\)
= 1.36 × 10-8 cm
= 136 pm
Radius of iridium atom = 136 pm

Question 5.
Aluminium crystallizes in cubic close packed structure with unit cell edge length of 353.6 pm. What is the radius of Al atom ? How many unit cells are there in 1.00 cm3 of Al ?
Answer:
Given : Structure of Al
= Cubic close packed structure
= ccp structure
Edge length of unit cell = a = 353.6 pm
= 3.536 × 10-8 cm
r = ?
Number of unit cells in 1.00 cm3 of Al = ?
Radius of Al atom = r = \(\frac{a}{2 \sqrt{2}}=\frac{353.6}{2 \sqrt{2}}\)
= \(\frac{353.6}{2 \times 1.414}\) = 125 pm
Volume of one unit cell = a3 = (3.536 × 10-8)3
= 4.421 × 10-23 cm3
Number of unit cells = \(\frac{1.00}{4.421 \times 10^{-23}}\)
= 2.26 × 1022
Radius of Al atom = 125 pm
Number of unit cells = 2.26 × 1022

Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State

Question 6.
In an ionic crystalline solid atoms of element Y form hcp lattice. The atoms of element X occupy one third of tetrahedral voids. What is the formula of the compound?
Answer:
In the given hcp lattice, Y atoms are present at 12 corners and 2 face centres.
∴ Number of Y atoms = \(\frac {1}{2}\) × 12 + 2 × \(\frac {1}{2}\) = 3
There are 6 tetrahedral voids, the number of X atoms = \(\frac {1}{3}\) × 6 = 2
∴ Formula of the compound is X2Y3.

Question 7.
How are tetrahedral and octahedral voids formed?
Answer:
Tetrahedral void : The vacant space or void among four constituent particles having tetrahedral arrangement in the crystal lattice is called tetrahedral void.
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 11
The arrangement of four spheres around the void is tetrahedral. A tetrahedral void is formed when a triangular void made by three coplanar spheres is in contact with fourth sphere above or below it.

Octahedral void : The vacant space or void at the centre of six spheres (or atoms) which are placed octahedrally is called octahedral void.
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 12

Question 8.
Third layer of spheres is added to second layer so as to form hcp or ccp structure. What is the difference between the addition of third layer to form these hexagonal close-packed structures?
Answer:

  1. In the formation of hexagonal closed-packed (hcp) structure, the first one dimensional row shows depressions between neighbouring atoms.
  2. When a second row is arranged so that spheres fit in these depressions then a staggered arrangement is obtained. If the first row is A then the second row is B.
  3. When third row is placed in staggered manner in contact with second row then A type arrangement is obtained.
  4. Similarly, the spheres in fourth row can be arranged as B type layer. This results in ABAB… type setting of the layers. This gives hexagonal close packing (hcp) structure.

Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 13
Hexagonal close packing (hcp)

Question 9.
An element with molar mass 27 g/mol forms cubic unit cell with edge length of 405 pm. If density of the element is 2.7 g/cm3, what is the nature of cubic unit cell ? (fcc or ccp)
Answer:
Given : Molar mass = M = 27 g mol-1
Nature of crystal = cubic unit cell
Edge length = a = 405 pm = 4.05 × 10-8 cm
Density = ρ = 2.7 g cm-3
Nature of unit cell = ?
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 14
= 3.997
≅ 4
Hence the nature of unit cell = face-centred cubic unit cell
Radius of Al atom = 125 pm
The nature of cubic unit cell is fcc.

Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State

Question 10.
An element has a bcc structure with unit cell edge length of 288 pm. How many unit cells and number of atoms are present in 200 g of the element? (1.16 × 1024, 2.32 × 1024)

Question 11.
Distinguish with the help of diagrams metal conductors, insulators and semiconductors from each other.
Answer:
Conductor:

  1. A substance which conducts heat and electricity to a greater extent is called conductor.
  2. In this, conduction bands and valence bands overlap or are very closely spaced.
  3. There is no energy difference or very less energy difference between valence bands and conduction bands.
  4. There are free electrons in the conduction bands.
  5. The conductance decreases with the increase in temperature.
  6. E.g., Metals, alloys.
  7. The conducting properties can’t be improved by adding third substance.

Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 15 b

Insulator:

  1. A substance which cannot conduct heat and electricity under any conditions is called insulator.
  2. In this, conduction bands and valence bands are far apart.
  3. The energy difference between conduction bands and valence bands is very large.
  4. There are no free electrons in the conduction bands and electrons can’t be excited from valence bands to conduction bands due to large energy difference.
  5. No effect of temperature on conducting properties.
  6. E.g., Wood, rubber, plastics.
  7. No effect of addition of any substance.

Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 24

Semiconductor:

  1. A substance that has poor electrical conductance at low temperature but higher conductance at higher temperature is called semiconductor.
  2. In this, conduction bands and valence bands are spaced closely.
  3. The energy difference between conduction bands and valence bands is small.
  4. The electrons can be easily excited from valence bands to conduction bands by heating.
  5. Conductance increases with the increase in temperature.
  6. E.g., Si, Ge
  7. By doping, conducting properties improve. E.g. n-type, p-type semiconductors.

Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 25

Question 12.
What are n-type semiconductors? Why is the conductivity of doped n-type semiconductor higher than that of pure semiconductor ? Explain with diagram.
Answer:
n-type semiconductor:

  • n-type semiconductor contains increased number of electrons in the conduction band.
  • When Si semiconductor is doped with 15th group element phosphorus, P, the new atoms occupy some vacant sites in the lattice in place of Si atoms.
  • P has five valence electrons, out of which four are involved in covalent bonding with neighboring Si atoms while one electrons remains free and delocalised.
  • These free electrons increase the electrical conductivity of the semiconductor.
  • The semiconductors with extra non-bonding free electrons are called n-type semiconductors.

Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 16
P atom occupying regular site of Si atom

Question 13.
Explain with diagram. Frenkel defect. What are the conditions for its formation? What is its effect on density and electrical neutrality of the crystal?
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 17

  1. Frenkel defect : This defect arises when an ion of an ionic compound is missing from its regular site and occupies interstitial vacant position between lattice points.
  2. Cations have smaller size than anions, hence generally cations occupy the interstitial sites.
  3. This creates a vacancy defect at its original position and interstitial defect at new position.
  4. Frenkel defect is regarded as the combination of interstitial defect and vacancy defect.

Conditions for the formation of Frenkel defect :

  1. This defect arises in ionic compounds with a large difference between the sizes of cation and anion.
  2. The ionic compounds must have ions with low coordination number.

Consequences of Frenkel defect :

  1. Since there is no loss of ions from the crystal lattice, the density of the solid remains unchanged.
  2. The crystal remains electrically neutral.
  3. This defect is observed in ZnS, AgCl, AgBr, Agl, CaF2, etc.

Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State

Question 14.
What is an impurity defect? What are its types? Explain the formation of vacancies through aliovalent impurity with example.
Answer:
Impurity defect : This defect arises when foreign atoms, that is, atoms different from the host atoms are present in the crystal lattice.

There are two types of impurity defects namely

  1. Substitutional defects and
  2. Interstitial defects.

(1) Substitutional defects : These defects arises when foreign atoms occupy the lattice sites in place of host atoms, due to their displacements.
Examples : Solid solutions of metals (alloys). For example. Brass in which host atoms are of Cu which are replaced by impurity of Zn atoms. In this Zn atoms occupy regular sites while Cu atoms occupy substituted sites.
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 18
Brass

Vacancy through aliovalent impurity :
By addition of impurities of aliovalent ions :
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 19
Vacancy through aliovalent ion

When aliovalent ion like Sr2+ in small amount is added by additing SrCl2 to NaCl during its crystallisation, each Sr2+ ion (oxidation state 2+) removes 2 Na+ ions from their lattice points, to maintain electrical neutrality. Hence one of vacant lattice site is occupied by Sr2+ ion while other site remains vacant.

Interstitial impurity defect :
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 20
Stainless steel

A defect in solid in which the impurity atoms occupy interstitial vacant spaces of lattice structure is called interstitial impurity defect.

For example, in steel, normal lattice sites are occupied by Fe atoms but interstitial spaces are occupied by carbon atoms.

Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State

12th Chemistry Digest Chapter 1 Solid State Intext Questions and Answers

Try this… (Textbook Page No. 1)

Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 21
Observe the above figure carefully. The two types of circles in this figure represent two types of constituent particles of a solid.

Question 1.
Will you call the arrangement of particles in this solid regular or irregular ?
Answer:
The arrangement of particles in this solid is regular.

Question 2.
Is the arrangement of constituent particles in directions \(\overrightarrow{\mathbf{A B}}\), \(\overrightarrow{\mathbf{C D}}\) and \(\overrightarrow{\mathbf{E F}}\) same or different?
Answer:
\(\overrightarrow{\mathbf{A B}}\) represents arrangement of identical particles of one type.
\(\overrightarrow{\mathbf{C D}}\) represents arrangement of identical particles of another type.
\(\overrightarrow{\mathbf{E F}}\) represents regular arrangement of two different particles in alternate positions.

Use your brain power ! (Textbook Page No. 2)

Question 1.
Identify the arrangements A and B as crystalline or amorphous.
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 22
Answer:
Arrangement in image A indicates the substance is crystalline.
Arrangement in image B indicates the substance is amorphous.

Try this… (Textbook Page No. 3)

Question 1.
Graphite is a covalent solid yet soft and good conductor of electricity. Explain.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 23

  1. Each carbon atom in graphite is sp2 hybridised and covalently bonded to other three sp2 hybridised carbon atoms forming σ bonds and the fourth electron in 2pz orbital of each carbon atom is used in the formation of a π bond. This results in the formation of hexagonal rings in two dimensions.
  2. In graphite, the layers consisting of hexagonal carbon network are held together by weak van der Waal’s forces imparting softness.
  3. The electrons in π bonds in the ring are delocalised and free to move in the delocalised molecular orbitals giving good electrical conductance.

Use your brain power ! (Textbook Page No. 13)

Question 1.
Which of the three lattices scc, bcc and fcc has the most efficient packing of particles ? Which one has the least efficient packing ?
Answer:
fcc has the most efficient packing of particles while see has the least efficient packing.

Can you think ? (Textbook Page No. 20)

Question 1.
When ZnO is heated it turns yellow and returns back to original white colour on cooling. What could be the reason ?
Answer:
When colourless ZnO is strongly heated, the metal atoms are deposited on crystal surface and anions O2- migrate to the surface producing vacancies at anion lattice points.

These anions combine with Zn atoms forming ZnO and release electrons.
Zn + O2- → ZnO + 2e

These released electrons diffuse into the crystal and occupy vacant sites of anions and produce F- centres. Due to these colour centres, ZnO turns yellow.

Can you tell ? (Textbook Page No. 23)

Let a small quantity of phosphorus be doped into pure silicon.

Question 1.
Will the resulting material contain the same number of total number of electrons as the original pure silicon ?
Answer:
Total number of electrons in doped silicon will be more than in original silicon.

Question 2.
Will the material be electrically neutral or charged ?
Answer:
Material will be electrically neutral.

Maharashtra State Board Class 12 Chemistry Solutions Digest

Maharashtra State Board 12th Std Chemistry Textbook Solutions Digest

Maharashtra State Board Class 12 Textbook Solutions

Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 5 Oscillations Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 5 Oscillations

1. Choose the correct option.

i) A particle performs linear S.H.M. starting from the mean position. Its amplitude is A and time period is T. At the instance when its speed is half the maximum speed, its displacement x is
(A) \(\frac{\sqrt{3}}{2}\)A
(B) \(\frac{2}{\sqrt{3}}\)
(C) A/2
(D) \(\frac{1}{\sqrt{2}}\)
Answer:
(A) \(\frac{\sqrt{3}}{2}\)A

ii) A body of mass 1 kg is performing linear S.H.M. Its displacement x (cm) at t (second) is given by x = 6 sin (100t + π/4). Maximum kinetic energy of the body is
(A) 36 J
(B) 9 J
(C) 27 J
(D) 18 J
Answer:
(D) 18 J

Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations

iii) The length of second’s pendulum on the surface of earth is nearly 1 m. Its length on the surface of moon should be [Given: acceleration due to gravity (g) on moon is 1/6 th of that on the earth’s surface]
(A) \(\frac{1}{6}\) m
(B) 6 m
(C) \(\frac{1}{36}\) m
(D) \(\frac{1}{\sqrt{6}}\) m.
Answer:
(A) \(\frac{1}{6}\) m

iv) Two identical springs of constant k are connected, first in series and then in parallel. A metal block of mass m is suspended from their combination. The ratio of their frequencies of vertical oscillations will be in a ratio
(A) 1:4
(B) 1:2
(C) 2:1
(D) 4:1
Answer:
(B) 1:2

v) The graph shows variation of displacement of a particle performing S.H.M. with time t. Which of the following statements is correct from the graph?
(A) The acceleration is maximum at time T.
(B) The force is maximum at time 3T/4.
(C) The velocity is zero at time T/2.
(D) The kinetic energy is equal to total energy at time T/4.
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 1
Answer:
(B) The force is maximum at time 3T/4.

2. Answer in brief.

i) Define linear simple harmonic motion.
Answer:
Definition: Linear simple harmonic motion (SHM) is defined as the linear periodic motion of a body, in which the force (or acceleration) is always directed towards the mean position and its magnitude is proportional to the displacement from the mean position.
OR
A particle is said to execute linear SHM if the particle undergoes oscillations about a point of stable equilibrium, subject to a linear restoring force always directed towards that point and whose magnitude is proportional to the magnitude of the displacement of the particle from that point.
Examples : The vibrations of the tines (prongs) of a tuning fork, the oscillations of the needle of a sewing machine.

Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations

ii) Using differential equation of linear S.H.M, obtain the expression for
(a) velocity in S.H.M.,
(b) acceleration in S.H.M.
Answer:
The general expression for the displacement of a particle in SHM at time t is x = A sin (ωt + α) … (1)
where A is the amplitude, ω is a constant in a particular case and α is the initial phase.
The velocity of the particle is
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 10
Equation (2) gives the velocity as a function of x.
The acceleration of the particle is
a = \(\frac{d v}{d t}\) = \(\frac{d}{d t}\) [Aω cos (ωt + α) J at at
∴ a = – ω2 A sin (ωt + α)
But from Eq. (1), A sin (ωt + α) = x
∴ a = -ω2x … (3)
Equation (3) gives the acceleration as a function of x. The minus sign shows that the direction of the acceleration is opposite to that of the displacement.

iii) Obtain the expression for the period of a simple pendulum performing S.H.M.
Answer:
An ideal simple pendulum is defined as a heavy point mass suspended from a rigid support by a weightless, inextensible and twistless string, and set oscillating under gravity through a small angle in a vertical plane.

In practice, a small but heavy sphere, called the bob, is used. The distance from the point of suspension to the centre of gravity of the bob is called the length of the pendulum.

Consider a simple pendulum of length L1 – suspended from a rigid support O. When displaced from its initial position of rest through a small angle θ in a vertical plane and released, it performs oscillations between two extremes, B and C, as shown in below figure. At B, the forces on the bob are its weight \(m \vec{g}\) and the tension \(\overrightarrow{F_{1}}\) in the string. Resolve \(m \vec{g}\) into two components : mg cos θ in the direction opposite to that of the tension and mg sin θ perpendicular to the string.
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 20
mg cos θ balanced by the tension in the string. mg sin θ restores the bob to the equilibrium position.
Restoring force, F = – mg sin θ
If θ is small and expressed in radian,
sin θ \(\approx\) θ = \(\frac{\text { arc }}{\text { radius }}\) = \(\frac{\mathrm{AB}}{\mathrm{OB}}\) = \(\frac{x}{L}\)
∴ F = – mgθ = -mg\(\frac{x}{L}\) …. (1)
Since m, g and L are constant,
F ∝ (-x) …. (2)

Thus, the net force on the bob is in the direction opposite to that of displacement x of the bob from its mean position as indicated by the minus sign, and the magnitude of the force is proportional to the magnitude of the displacement. Hence, it follows that the motion of a simple pendulum is linear SHM.
Acceleration, a = \(\frac{F}{m}\) = –\(\frac{g}{L}\)x … (3)
Therefore, acceleration per unit displacement
= |\(\frac{a}{x}\)| = \(\frac{g}{L}\) ….. (4)
Period of SHM,
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 25
This gives the expression for the period of a simple pendulum.

Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations

iv) State the laws of simple pendulum.
Answer:
The period of a simple pendulum at a given place is
T = 2π\(\sqrt{\frac{L}{g}}\)
where L is the length of the simple pendulum and g is the acceleration due to gravity at that place. From the above expression, the laws of simple pendulum are as follows :

(1) Law of length : The period of a simple pendulum at a given place (g constant) is directly proportional to the square root of its length.
∴ T ∝\(\sqrt{L}\)
(2) Law of acceleration due to gravity : The period of a simple pendulum of a given length (L constant) is inversely proportional to the square root of the acceleration due to gravity.
∴ T ∝ \(\frac{1}{\sqrt{g}}\)
(3) Law of mass : The period of a simple pendulum does not depend on the mass or material of the bob of the pendulum.
(4) Law of isochronism : The period of a simple pendulum does not depend on the amplitude of oscillations, provided that the amplitude is small.

v) Prove that under certain conditions a magnet vibrating in uniform magnetic field performs angular S.H.M.
Answer:
Consider a bar magnet of magnetic moment μ, suspended horizontally by a light twistless fibre in a region where the horizontal component of the Earth’s magnetic field is Bh. The bar magnet is free to rotate in a horizontal plane. It comes to rest in approximately the North-South direction, along Bh. If it is rotated in the horizontal plane by a small
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 30
displacement θ from its rest position (θ = 0), the suspension fibre is twisted. When the magnet is released, it oscillates about the rest position in angular or torsional oscillation.

The bar magnet experiences a torque \(\tau\) due to the field Bh. Which tends to restore it to its original orientation parallel to Bh. For small θ, this restoring torque is
\(\tau\) = – μBh sin θ = – μBhμ …. (1)

where the minus sign indicates that the torque is opposite in direction to the angular displacement θ. Equation (1) shows that the torque (and hence the angular acceleration) is directly proportional in magnitude of the angular displacement but opposite in direction. Hence, for small angular displacement, the oscillations of the bar magnet in a uniform magnetic field is simple harmonic.

Question 3.
Obtain the expression for the period of a magnet vibrating in a uniform magnetic field and performing S.H.M.
Answer:
Definition : Angular SHM is defined as the oscillatory motion of a body in which the restoring torque responsible for angular acceleration is directly proportional to the angular displacement and its direction is opposite to that of angular displacement.
The differential equation of angular SHM is
I\(\frac{d^{2} \theta}{d t^{2}}\) + cθ = 0 ….. (1)
where I = moment of inertia of the oscillating body, \(\frac{d^{2} \theta}{d t^{2}}\) = angular acceleration of the body when its angular displacement is θ, and c = torsion constant of the suspension wire,
∴ \(\frac{d^{2} \theta}{d t^{2}}\) + \(\frac{c}{I}\)θ = 0
Let \(\frac{c}{I}\) = ω2, a constant. Therefore, the angular frequency, ω = \(\sqrt{c / I}\) and the angular acceleration,
a = \(\frac{d^{2} \theta}{d t^{2}}\) = -ω2θ … (2)
The minus sign shows that the α and θ have opposite directions. The period T of angular SHM is
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 80
This is the expression for the period in terms of torque constant. Also, from Eq. (2),
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 81

Question 4.
Show that a linear S.H.M. is the projection of a U.C.M. along any of its diameter.
Answer:
Consider a particle which moves anticlockwise around a circular path of radius A with a constant angular speed ω. Let the path lie in the x-y plane with the centre at the origin O. The instantaneous position P of the particle is called the reference point and the circle in which the particle moves as the reference circle.
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 35
The perpendicular projection of P onto the y-axis is Q. Then, as the particle travels around the circle, Q moves to-and-fro along the y-axis. Line OP makes an angle α with the x-axis at t = 0. At time t, this angle becomes θ = ωt + α.
The projection Q of the reference point is described by the y-coordinate,
y = OQ = OP sin ∠OPQ, Since ∠OPQ = ωt + α, y = A sin(ωt + α)
which is the equation of a linear SHM of amplitude A. The angular frequency w of a linear SHM can thus be understood as the angular velocity of the reference particle.

The tangential velocity of the reference particle is v = ωA. Its y-component at time t is vy = ωA sin (90° – θ) = ωA cos θ
∴ vy = ωA cos (ωt + α)
The centripetal acceleration of the reference particle is ar = ω2A, so that its y-component at time t is ax = ar sin ∠OPQ
∴ ax = – ω2A sin (ωt + α)

Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations

Question 5.
Draw graphs of displacement, velocity and acceleration against phase angle, for a particle performing linear S.H.M. from (a) the mean position
(b) the positive extreme position. Deduce your conclusions from the graph.
Answer:
Consider a particle performing SHM, with amplitude A and period T = 2π/ω starting from the mean position towards the positive extreme position where ω is the angular frequency. Its displacement from the mean position (x), velocity (v) and acceleration (a) at any instant are
x = A sin ωt = A sin\(\left(\frac{2 \pi}{T} t\right)\) (∵ω = \(\frac{2 \pi}{T}\))
v = \(\frac{d x}{d t}\) = ωA cos ωt = ωA cos\(\left(\frac{2 \pi}{T} t\right)\)
a = \(\frac{d v}{d t}\) = -ω2 A sin ωt = – ω2A sin\(\left(\frac{2 \pi}{T} t\right)\) as the initial phase α = 0.
Using these expressions, the values of x, v and a at the end of every quarter of a period, starting from t = 0, are tabulated below.
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 38
Using the values in the table we can plot graphs of displacement, velocity and acceleration with time.
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 39
Conclusions :

  1. The displacement, velocity and acceleration of a particle performing linear SHM are periodic (harmonic) functions of time. For a particle starting at the mean position, the x-t and a-t graphs are sine curves. The v-t graph is a cosine curve.
  2. There is a phase difference of \(\frac{\pi}{2}\) radians between x and v, and between v and a.
  3. There is a phase difference of π radians between x and a.

Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 40

Consider a particle performing linear SHM with amplitude A and period T = 2π/ω, starting from the positive extreme position, where ω is the angular frequency. Its displacement from the mean position (x), velocity (y) and acceleration (a) at any instant (t) are
x = A cos ωt = A cos \(\left(\frac{2 \pi}{T} t\right)\) (∵ω = \(\frac{2 \pi}{T}\))
v = -ωA sin ωt = – ωA sin \(\left(\frac{2 \pi}{T} t\right)\)
a = -ω2A cos ωt = -ω2A cos \(\left(\frac{2 \pi}{T} t\right)\)
Using these expressions, the values of x, y and a at the end of every quarter of a period, starting from t = 0, are tabulated below.
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 41
Using these values, we can plot graphs showing the variation of displacement, velocity and acceleration with time.
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 42
Conclusions :

  1. The displacement, velocity and acceleration of a particle performing linear SHM are periodic (harmonic) functions of time. For a particle starting from an extreme position, the x-t and a-t graphs are cosine curves; the v-t graph is a sine curve.
  2. There is a phase difference of \(\frac{\pi}{2}\) radians between x and v, and between v and a.
  3. There is a phase difference of n radians between x and a.

Explanations :
(1) v-t graph : It is a sine curve, i.e., the velocity is a periodic (harmonic) function of time which repeats after a phase of 2π rad. There is a phase difference of π/2 rad between a and v.
v is minimum (equal to zero) at the extreme positions (i.e., at x = ± A) and v is maximum ( = ± ωA) at the mean position (x = 0).

(2) a-t graph : It is a cosine curve, i.e., the acceleration is a periodic (harmonic) function of time which repeats after a phase of 2π rad. There is a phase difference of π rad between v and a. a is minimum (equal to zero) at the mean position (x = 0) and a is maximum ( = \(\mp\)ω2A) at the extreme positions (x = ±A).
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 43

Question 6.
Deduce the expressions for the kinetic energy and potential energy of a particle executing S.H.M. Hence obtain the expression for total energy of a particle performing S.H.M and show that the total energy is conserved. State the factors on which total energy depends.
Answer:
Consider a particle of mass m performing linear SHM with amplitude A. The restoring force acting r on the particle is F = – kx, where k is the force constant and x is the displacement of the particle from its mean position.
(1) Kinetic energy : At distance x from the mean position, the velocity is
v = ω\(\sqrt{A^{2}-x^{2}}\)
where ω = \(\sqrt{k / m} .\) The kinetic energy (KE) of the particle is
KE = \(\frac{1}{2}\) mv2 = \(\frac{1}{2}\) mω2 (A2 – x2)
= \(\frac{1}{2}\)k(A2 – x2) … (1)
If the phase of the particle at an instant t is θ = ωt + α, where α is initial phase, its velocity at that instant is
v = ωA cos (ωt + α)
and its KE at that instant is
KE = \(\frac{1}{2}\)mv2 = \(\frac{1}{2}\)mω2A2 cos2(ωt + α) ….. (2)
Therefore, the KE varies with time as cos2 θ.

(2) Potential energy : The potential energy of a particle in linear SHM is defined as the work done by an external agent, against the restoring force, in taking the particle from its mean position to a given point in the path, keeping the particle in equilibirum.

Suppose the particle in below figure is displaced from P1 to P2, through an infinitesimal distance dx against the restoring force F as shown.
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 45
The corresponding work done by the external agent will be dW = ( – F)dx = kx dx. This work done is stored in the form of potential energy. The potential energy (PE) of the particle when its displacement from the mean position is x can be found by integrating the above expression from 0 to x.
∴ PE = \(\int\)dW = \(\int_{0}^{x}\) kx dx = \(\frac{1}{2}\) kx2 … (3)
The displacement of the particle at an instant t being
x = A sin (wt + α)
its PE at that instant is
PE = \(\frac{1}{2}\)kx2 = \(\frac{1}{2}\)kA2 sin2(ωt + α) … (4)
Therefore, the PE varies with time as sin2θ.

(3) Total energy : The total energy of the particle is equal to the sum of its potential energy and kinetic energy.
From Eqs. (1) and (2), total energy is E = PE + KE
= \(\frac{1}{2}\)kx2 + \(\frac{1}{2}\)k(A2 – x2)
= \(\frac{1}{2}\)kx2 + \(\frac{1}{2}\)kA2 – \(\frac{1}{2}\)kx2
∴ E = \(\frac{1}{2}\)kA2 = \(\frac{1}{2}\)mω2A2 … (5)
As m is constant, ω and A are constants of the motion, the total energy of the particle remains constant (or its conserved).

Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations

Question 7.
Deduce the expression for period of simple pendulum. Hence state the factors on which its period depends.
Answer:
An ideal simple pendulum is defined as a heavy point mass suspended from a rigid support by a weightless, inextensible and twistless string, and set oscillating under gravity through a small angle in a vertical plane.

In practice, a small but heavy sphere, called the bob, is used. The distance from the point of suspension to the centre of gravity of the bob is called the length of the pendulum.

Consider a simple pendulum of length L1 – suspended from a rigid support O. When displaced from its initial position of rest through a small angle θ in a vertical plane and released, it performs oscillations between two extremes, B and C, as shown in below figure. At B, the forces on the bob are its weight \(m \vec{g}\) and the tension \(\overrightarrow{F_{1}}\) in the string. Resolve \(m \vec{g}\) into two components : mg cos θ in the direction opposite to that of the tension and mg sin θ perpendicular to the string.
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 20
mg cos θ balanced by the tension in the string. mg sin θ restores the bob to the equilibrium position.
Restoring force, F = – mg sin θ
If θ is small and expressed in radian,
sin θ \(\approx\) θ = \(\frac{\text { arc }}{\text { radius }}\) = \(\frac{\mathrm{AB}}{\mathrm{OB}}\) = \(\frac{x}{L}\)
∴ F = – mgθ = -mg\(\frac{x}{L}\) …. (1)
Since m, g and L are constant,
F ∝ (-x) …. (2)

Thus, the net force on the bob is in the direction opposite to that of displacement x of the bob from its mean position as indicated by the minus sign, and the magnitude of the force is proportional to the magnitude of the displacement. Hence, it follows that the motion of a simple pendulum is linear SHM.
Acceleration, a = \(\frac{F}{m}\) = –\(\frac{g}{L}\)x … (3)
Therefore, acceleration per unit displacement
= |\(\frac{a}{x}\)| = \(\frac{g}{L}\) ….. (4)
Period of SHM,
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 25
This gives the expression for the period of a simple pendulum.

The period of a simple pendulum at a given place is
T = 2π\(\sqrt{\frac{L}{g}}\)
where L is the length of the simple pendulum and g is the acceleration due to gravity at that place. From the above expression, the laws of simple pendulum are as follows :

(1) Law of length : The period of a simple pendulum at a given place (g constant) is directly proportional to the square root of its length.
∴ T ∝\(\sqrt{L}\)
(2) Law of acceleration due to gravity : The period of a simple pendulum of a given length (L constant) is inversely proportional to the square root of the acceleration due to gravity.
∴ T ∝ \(\frac{1}{\sqrt{g}}\)

Question 8.
At what distance from the mean position is the speed of a particle performing S.H.M. half its maximum speed. Given path length of S.H.M. = 10 cm. [Ans: 4.33 cm]
Answer:
Data : v = \(\frac{1}{2}\)vmax, 2A = 10 cm
∴ A = 5 cm
v = ω\(\sqrt{A^{2}-x^{2}}\) and vmax = ωA
Since v = \(\frac{1}{2}\)vmax,
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 50
This gives the required displacement.

Question 9.
In SI units, the differential equation of an S.H.M. is \(\frac{d^{2} x}{d t^{2}}\) = -36x. Find its frequency and period. Find its frequency and period.
[Ans: 0.9548 Hz, 1.047 s]
Answer:
\(\frac{d^{2} x}{d t^{2}}\) = -36x
Comparing this equation with the general equation,
\(\frac{d^{2} x}{d t^{2}}\) = -ω2x
We get, ω2 = 36 ∴ ω = 6 rad/s
ω = 2πf
∴ The frequency, f = \(\frac{\omega}{2 \pi}\) = \(\frac{6}{2(3.142)}\) = \(\frac{6}{6.284}\) = 0.9548 Hz
and the period, T = \(\frac{1}{f}\) = \(\frac{1}{0.9548}\) = 1.047 s

Question 10.
A needle of a sewing machine moves along a path of amplitude 4 cm with frequency 5 Hz. Find its acceleration \(\frac{1}{30}\)s after it has crossed the mean position. [Ans: 34.2 m/s2]
Answer:
Data : A = 4 cm = 4 × 10-2 m, f = 5Hz, t = \(\frac{1}{30}\)s
ω = 2πf = 2π (5) = 10π rad/s
Therefore, the magnitude of the acceleration,
|a| = ω2x = ω2A sin ωt
= (10π)2 (4 × 102)
= 10π2 sin \(\frac{\pi}{3}\) = 10 (9.872)(0.866) = 34.20 m/s2

Question 11.
Potential energy of a particle performing linear S.H.M is 0.1 π2 x2 joule. If mass of the particle is 20 g, find the frequency of S.H.M. [Ans: 1.581 Hz]
Answer:
Data : PE = 0.1 π2 x2 J, m = 20 g = 2 × 10-2 kg
PE = \(\frac{1}{2}\)mω2x2 = \(\frac{1}{2}\)m (4π2f2)x2
∴ \(\frac{1}{2}\)m(4π2f2)x2 = 0.1 π2 x2
∴ 2mf2 = 0.1 ∴ f2 = \(\frac{1}{20\left(2 \times 10^{-2}\right)}\) = 2.5
∴ The frequency of SHM is f = \(\sqrt{2.5}\) = 1.581 Hz

Question 12.
The total energy of a body of mass 2 kg performing S.H.M. is 40 J. Find its speed while crossing the centre of the path. [Ans: 6.324 cm/s]
Answer:
Data : m = 2 kg, E = 40 J
The speed of the body while crossing the centre of the path (mean position) is vmax and the total energy is entirely kinetic energy.
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 60

Question 13.
A simple pendulum performs S.H.M of period 4 seconds. How much time after crossing the mean position, will the displacement of the bob be one third of its amplitude. [Ans: 0.2163 s]
Answer:
Data : T = 4 s, x = A/3
The displacement of a particle starting into SHM from the mean position is x = A sin ωt = A sin \(\frac{2 \pi}{T}\) t
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 61
∴ the displacement of the bob will be one-third of its amplitude 0.2163 s after crossing the mean position.

Question 14.
A simple pendulum of length 100 cm performs S.H.M. Find the restoring force acting on its bob of mass 50 g when the displacement from the mean position is 3 cm. [Ans: 1.48 × 10-2 N]
Answer:
Data : L = 100 cm, m = 50 g = 5 × 10-2 kg, x = 3 cm, g = 9.8 m/s2
Restoring force, F = mg sin θ = mgθ
= (5 × 10-2)(9.8)\(\left(\frac{3}{100}\right)\)
= 1.47 × 10-2 N

Question 15.
Find the change in length of a second’s pendulum, if the acceleration due to gravity at the place changes from 9.75
m/s2 to 9.80 m/s2. [Ans: Decreases by 5.07 mm]
Answer:
Data : g1 = 9.75 m/s2, g2 = 9.8 m/s2
Length of a seconds pendulum, L = \(\frac{g}{\pi^{2}}\)
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 62
∴ The length of the seconds pendulum must be increased from 0.9876 m to 0.9927 m, i.e., by 0.0051 m.

Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations

Question 16.
At what distance from the mean position is the kinetic energy of a particle performing S.H.M. of amplitude 8 cm, three times its potential energy? [Ans: 4 cm]
Answer:
Data : A = 8 cm, KE = 3 PE
KE = \(\frac{1}{2}\) (A2 – x2) and PE = \(\frac{1}{2}\)kx2
Given, KE = 3PE.
∴ \(\frac{1}{2}\)k(A2 – x2) = 3\(\left(\frac{1}{2} k x^{2}\right)\)
∴ A2 – x2 = 3x2 ∴ 4x2 = A2
∴ the required displacement is
x = ±\(\frac{A}{2}\) = ±\(\frac{8}{2}\) = ± 4 cm

Question 17.
A particle performing linear S.H.M. of period 2π seconds about the mean position O is observed to have a speed of \(b \sqrt{3}\) m/s, when at a distance b (metre) from O. If the particle is moving away from O at that instant, find the
time required by the particle, to travel a further distance b. [Ans: π/3 s]
Answer:
Data : T = 2πs, v = b\(\sqrt{3}\) m/s at x = b
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 63
∴ Assuming the particle starts from the mean position, its displacement is given by
x = A sin ωt = 2b sin t
If the particle is at x = b at t = t1,
b = 2b sint1 ∴ t1 = sin-1 \(\frac{1}{2}\) = \(\frac{\pi}{6}\)s
Also, with period T = 2πs, on travelling a further distance b the particle will reach the positive extremity at time t2 = \(\frac{\pi}{2}\)s.
∴ The time taken to travel a further distance b from x = b is t2 – t1 = \(\frac{\pi}{2}\) – \(\frac{\pi}{6}\) = \(\frac{\pi}{3}\)s.

Question 18.
The period of oscillation of a body of mass m1 suspended from a light spring is T. When a body of mass m2 is tied to the first body and the system is made to oscillate, the period is 2T. Compare the masses m1 and m2 [Ans: 1/3]
Answer:
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 64
This gives the required ratio of the masses.

Question 19.
The displacement of an oscillating particle is given by x = asinωt + bcosωt where a, b and ω are constants. Prove that the particle performs a linear S.H.M. with amplitude A = \(\sqrt{a^{2}+b^{2}}\)
Answer:
x = asinωt + bcosωt
Let a = A cos φ and b = A sin φ, so that
A2 = a2 + b2 and tan φ = \(\frac{b}{a}\)
∴ x = A cos φ sin ωt + A sin φ cos ωt
∴ x = A sin (ωt + φ)
which is the equation of a linear SHM with amplitude A = \(\sqrt{a^{2}+b^{2}}\) and phase constant φ = tan-1 \(\frac{b}{a}\), as required.

Question 20.
Two parallel S.H.M.s represented by x1 = 5sin (4πt + \(\frac{\pi}{3}\)) cm and x2 = 3sin(4πt + π/4) cm are superposed on a particle. Determine the amplitude and epoch of the resultant S.H.M. [Ans: 7.936 cm, 54° 23′]
Answer:
Data: x1 = 5 sin (4πt + \(\frac{\pi}{3}\)) = A1 sin(ωt + α),
x2 = 3 sin (4πt + \(\frac{\pi}{4}\)) = A2 sin(ωt + β)
∴ A1 = 5 cm, A2 = 3 cm, α = \(\frac{\pi}{3}\) rad, β = \(\frac{\pi}{4}\) rad
(i) Resultant amplitude,
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 65

(ii) Epoch of the resultant SHM,
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 66

Question 21.
A 20 cm wide thin circular disc of mass 200 g is suspended to a rigid support from a thin metallic string. By holding the rim of the disc, the string is twisted through 60° and released. It now performs angular oscillations of period 1 second. Calculate the maximum restoring torque generated in the string under undamped conditions. (π3 ≈ 31)
[Ans: 0.04133 N m]
Answer:
Data: R = 10cm = 0.1 m, M = 0.2 kg, θm = 60° = \(\frac{\pi}{3}\) rad, T = 1 s, π3 ≈ 31
The Ml of the disc about the rotation axis (perperdicular through its centre) is
I = \(\frac{1}{2}\)MR2 = (0.2)(0.1)2 = 10-3 kg.m2
The period of torsional oscillation, T = 2π\(\sqrt{\frac{I}{c}}\)
∴ The torsion constant, c = 4πr2\(\frac{I}{T^{2}}\)
The magnitude of the maximum restoring torque,
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 67

Question 22.
Find the number of oscillations performed per minute by a magnet is vibrating in the plane of a uniform field of 1.6 × 10-5 Wb/m2. The magnet has moment of inertia 3 × 10-6 kgm2 and magnetic moment 3 A m2. [Ans:38.19 osc/min.]
Answer:
Data : B = 1.6 × 10-5 T, I = 3 × 10-6kg/m2,
µ = 3 A.m2
The period of oscillation, T = 2π \(\sqrt{\frac{I}{\mu B_{\mathrm{h}}}}\)
∴ The frequency of oscillation is
f = \(\frac{1}{2 \pi}\)\(\sqrt{\frac{\mu B}{I}}\)
∴ The number of oscillations per minute
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 68
= 38.19 per minute

Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations

Question 23.
A wooden block of mass m is kept on a piston that can perform vertical vibrations of adjustable frequency and amplitude. During vibrations, we don’t want the block to leave the contact with the piston. How much maximum frequency is possible if the amplitude of vibrations is restricted to 25 cm? In this case, how much is the energy per unit mass of the block? (g ≈ π2 ≈ 10 m s-2)
[Ans: nmax = 1/s, E/m = 1.25 J/kg]
Answer:
Data : A = 0.25 m, g = π2 = 10 m/s2
During vertical oscillations, the acceleration is maximum at the turning points at the top and bottom. The block will just lose contact with the piston when its apparent weight is zero at the top, i. e., when its acceleration is amax = g, downwards.
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 69
This gives the required frequency of the piston.
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 70

12th Physics Digest Chapter 5 Oscillations of Waves Intext Questions and Answers

Can you tell? (Textbook Page No. 112)

Question 1.
Why is the term angular frequency (ω) used here for a linear motion ?
Answer:
A linear SHM is the projection of a UCM on a diameter of the circle. The angular speed co of a particle moving along this reference circle is called the angular frequency of the particle executing linear SHM.

Can you tell? (Textbook Page No. 114)

Question 1.
State at which point during an oscillation the oscillator has zero velocity but positive acceleration ?
Answer:
At the left extreme, i.e., x = – A, so that a = – ω2x = – ω2(- A) = ω2A = amax

Question 2.
During which part of the simple harmonic motion velocity is positive but the displacement is negative, and vice versa ?
Answer:
Velocity v is positive (to the right) while displacement x is negative when the particle in SHM is moving from the left extreme towards the mean position. Velocity v is negative (to the left) while displacement x is positive when the particle in SHM is moving from the right extreme towards the mean position.

Can you tell? (Textbook page 76)

Question 1.
To start a pendulum swinging, usually you pull it slightly to one side and release. What kind of energy is transferred to the mass in doing this?
Answer:
On pulling the bob of a simple pendulum slightly to one side, it is raised to a slightly higher position. Thus, it gains gravitational potential energy.

Question 2.
Describe the energy changes that occur when the mass is released.
Answer:
When released, the bob oscillates in SHM in a vertical plane and the energy oscillates back and forth between kinetic and potential, going completely from one form of energy to the other as the pendulum oscillates. In the case of undamped SHM, the motion starts with all of the energy as gravitational potential energy. As the object starts to move, the gravitational potential energy is converted into kinetic energy, becoming entirely kinetic energy at the equilibrium position. The velocity becomes zero at the other extreme as the kinetic energy is completely converted back into gravitational potential energy,
and this cycle then repeats.

Question 3.
Is/are there any other way/ways to start the oscillations of a pendulum? Which energy is supplied in this case/cases?
Answer:
The bob can be given a kinetic energy at its equilibrium position or at any other position of its path. In the first case, the motion starts with all of the energy as kinetic energy. In the second case, the motion starts with partly gravitational potential energy and partly kinetic energy.

Can you tell? (Textbook Page No. 109)

Question 1.
Is the motion of a leaf of a tree blowing in the wind periodic ?
Answer:
The leaf of a tree blowing in the wind oscillates, but the motion is not periodic. Also, its displacement from the equilibrium position is not a regular function of time.

Maharashtra Board Class 12 Secretarial Practice Solutions Chapter 1 Introduction to Corporate Finance

Balbharti Maharashtra State Board Class 12 Secretarial Practice Solutions Chapter 1 Introduction to Corporate Finance Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Secretarial Practice Solutions Chapter 1 Introduction to Corporate Finance

1A. Select the correct answer from the options given below and rewrite the statements.

Question 1.
_____________ is related to money and money management.
(a) Production
(b) Marketing
(c) Finance
Answer:
(c) Finance

Question 2.
Finance is the management of _____________ affairs of the company.
(a) monetary
(b) marketing
(c) production
Answer:
(a) monetary

Maharashtra Board Class 12 Secretarial Practice Solutions Chapter 1 Introduction to Corporate Finance

Question 3.
Corporation finance deals with the acquisition and use of _____________ by business corporation.
(a) goods
(b) capital
(c) land
Answer:
(b) capital

Question 4.
Company has to pay _____________ to government.
(a) taxes
(b) dividend
(c) interest
Answer:
(a) taxes

Question 5.
_____________ refers to any kind of fixed assets.
(a) Authorised capital
(b) Issued capital
(c) Fixed capital
Answer:
(c) Fixed capital

Question 6.
_____________ refers to the excess of current assets over current liabilities.
(a) Working capital
(b) Paid-up capital
(c) Subscribed capital
Answer:
(a) Working capital

Question 7.
Manufacturing industries have to invest _____________ amount of funds to acquire fixed assets.
(a) huge
(b) less
(c) minimal
Answer:
(a) huge

Maharashtra Board Class 12 Secretarial Practice Solutions Chapter 1 Introduction to Corporate Finance

Question 8.
When the population is increasing at a high rate, certain manufacturers find this as an opportunity to _____________ business.
(a) close
(b) expand
(c) contract
Answer:
(b) expand

Question 9.
The sum of all _____________ is gross working capital.
(a) expenses
(b) current assets
(c) current liabilities
Answer:
(b) current assets

Question 10.
_____________ means mix up of various sources of funds in desired proportion.
(a) Capital budgeting
(b) Capital structure
(c) Capital goods
Answer:
(b) Capital structure

1B. Match the pairs:

Question 1.
Maharashtra Board Class 12 Secretarial Practice Solutions Chapter 1 Introduction to Corporate Finance 1B Q1
Answer:

Group ‘A’ Group ‘B’
(a) Capital budgeting (6) Investment decision
(b) Fixed capital (5) Fixed assets
(c) Working capital (1) Sum of current assets
(d) Capital structure (9) Mix up various sources of funds
(e) Corporate finance (2) Deals with acquisition and use of capital

1C. Write a word or term or a phrase that can substitute each of the following statements:

Question 1.
A key determinant of the success of any business function.
Answer:
Finance

Question 2.
The decision of the finance manager ensures that the firm is well-capitalized.
Answer:
Financing decision

Maharashtra Board Class 12 Secretarial Practice Solutions Chapter 1 Introduction to Corporate Finance

Question 3.
The decision of the finance manager to deploy the funds in a systematic manner.
Answer:
Investment decision

Question 4.
Capital is needed to acquire fixed assets that are used for a longer period of time.
Answer:
Fixed capital

Question 5.
The sum of current assets.
Answer:
Gross working capital

Question 6.
The excess of current assets over current liabilities.
Answer:
Networking capital

Question 7.
The process of converting raw material into finished goods.
Answer:
Production cycle

Question 8.
The boom and recession cycle in the economy.
Answer:
Economic Trend

Question 9.
The ratio of different sources of funds in the total capital.
Answer:
Capital Structure

Maharashtra Board Class 12 Secretarial Practice Solutions Chapter 1 Introduction to Corporate Finance

Question 10.
The internal source of financing.
Answer:
Retained earnings

1D. State whether the following statements are True or False:

Question 1.
Finance is related to money and money management.
Answer:
True

Question 2.
The business firm gives a green signal to the project only when it is profitable.
Answer:
True

Question 3.
Corporate finance brings coordination between various business activities.
Answer:
True

Question 4.
Fixed capital is also referred as circulating capital.
Answer:
False

Question 5.
Working capital stays in the business almost permanently.
Answer:
False

Question 6.
The business will require huge funds if assets are acquired on a lease basis.
Answer:
False

Maharashtra Board Class 12 Secretarial Practice Solutions Chapter 1 Introduction to Corporate Finance

Question 7.
The business dealing in luxurious products will require a huge amount of working capital.
Answer:
True

Question 8.
A firm with large-scale operations will require more working capital.
Answer:
True

Question 9.
Liberal credit policy creates a problem of bad debt.
Answer:
True

Question 10.
Financial institutions and banks cater to the working capital requirement of the business.
Answer:
True

1E. Find the odd one.

Question 1.
Land and Building, Plant and Machinery, Cash.
Answer:
Cash

Question 2.
Debenture Capital, Equity Share Capital, Preference Share Capital.
Answer:
Debenture Capital

Maharashtra Board Class 12 Secretarial Practice Solutions Chapter 1 Introduction to Corporate Finance

Question 3.
Fixed Capital, Capital Structure, Working Capital.
Answer:
Capital Structure

1F. Complete the sentences.

Question 1.
Initial planning of capital requirement is made by _____________
Answer:
entrepreneur

Question 2.
When there is boom in economy, sales will _____________
Answer:
increase

Question 3.
The process of converting raw material into finished goods is called _____________
Answer:
production cycle

Question 4.
During recession period sales will _____________
Answer:
decrease

1G. Select the correct option from the bracket.

Question 1.
Maharashtra Board Class 12 Secretarial Practice Solutions Chapter 1 Introduction to Corporate Finance 1G Q1
(To have the right amount of capital, deploy funds in a systematic manner, fixed capital, working capital, capital structure, carry dividend at a fixed rate)
Answer:

Group ‘A’ Group ‘B’
(a) Financing decision (1) To have the right amount of capital
(b) Fixed capital (2) Longer period of time
(c) Investment decision (3) Deploy funds in a systematic manner
(d) Working capital (4) Circulating capital
(e) Combination of various sources of funds (5) capital structure

1H. Answer in one sentence.

Question 1.
Define corporate finance.
Answer:
Corporate finance deals with the raising and using of finance by a corporation.

Question 2.
What is fixed capital?
Answer:
Fixed capital is the capital that is used for buying fixed assets that are used for a longer period of time in the business eg. Capital for plant and machinery etc.

Question 3.
What is working capital?/Define working capital.
Answer:
Working capital is the capital that is used to carry out day-to-day business activities and takes into consideration all current assets of the company.
Eg: for building up inventories.

Maharashtra Board Class 12 Secretarial Practice Solutions Chapter 1 Introduction to Corporate Finance

Question 4.
What is the production cycle?
Answer:
The process of converting raw material into finished goods is called the production cycle.

Question 5.
Define capital structure.
Answer:
Capital structure means to mix up various sources of funds in the desired proportion. To decide capital structure means to decide upon the ratio of different types of capital.

1I. Correct the underlined word and rewrite the following sentences.

Question 1.
Finance is needed to pay dividends to debenture holders.
Answer:
Finance is needed to pay interest to debenture holders.

Question 2.
When there is a recession in the economy sales will increase.
Answer:
When there is a boom in the economy sales will increase.

Question 3.
Share is an acknowledgment of a loan raised by the company.
Answer:
A debenture is an acknowledgment of a loan raised by a company.

Question 4.
Equity shares carry dividends at a fixed rate.
Answer:
Preference shares carry dividends at a fixed rate.

2. Explain the following terms/concepts.

Question 1.
Financing decision
Answer:
A financing decision is a right decision that is made by a finance manager of any corporation ensuring that the firm is well capitalized with the right combination of debt and equity, having access to multiple choices of sources of financing.

Maharashtra Board Class 12 Secretarial Practice Solutions Chapter 1 Introduction to Corporate Finance

Question 2.
Investment decision
Answer:
Investment decisions mean capital budgeting i.e. finding investments and deploying them successfully in the business for greater profits.

Question 3.
Fixed capital
Answer:
Fixed capital is the capital that is used for buying fixed assets that are used for a longer period of time in the business. These assets are not meant for. resale. Examples of fixed capital are capital used for purchasing land and building, furniture, plant, and machinery, etc.

Question 4.
Working Capital
Answer:
Working capital is the capital that is used to carry out day-to-day business activities. It takes into consideration all current assets, of the company. It also refers to ‘Gross Working Capital’.
Examples of working capital are

  • for building up inventories.
  • for financing receivables.
  • for covering day-to-day operating expenses.

3. Study the following case/situation and express your opinion.

1. The management of ‘Maharashtra State Road Transport Corporation’ wants to determine the size of working capital.

Question (a).
Being a public utility service provider will it need less working capital or more?
Answer:
MSRTC being a public utility service provider will need less working capital because of a continuous flow of cash from there, customers thus liabilities are taken care of.

Question (b).
Being a public utility service provider, will it need more fixed capital?
Answer:
Being a public utility service provider MSRTC will need a huge amount of funds to acquire fixed assets thus it will need more fixed capital.

Question (c).
Give one example of a public utility service that you come across on a day-to-day basis.
Answer:
The Indian Railways.

Maharashtra Board Class 12 Secretarial Practice Solutions Chapter 1 Introduction to Corporate Finance

2. A company is planning to enhance its production capacity and is evaluating the possibility of purchasing new machinery whose cost is ₹ 2 crore or has alternative of machinery available on a lease basis.

Question (a).
What type of asset is machinery?
Answer:
Machinery is a Fixed Asset.
A fixed asset may be held for 5, 10 or 20 years and more. But if assets are acquired on a lease or rental basis, then less amount of funds for fixed assets will be needed for business.

Question (b).
Capital used for the purchase of machinery is fixed capital or working capital.
Answer:
Capital used for the purchase of machinery is fixed capital.

Question (c).
Does the size of a business determine the fixed capital requirement?
Answer:
Yes. Where a business firm is set up to carry on large-scale operations, its fixed capital requirements are likely to be high.

4. Distinguish between the following.

Question 1.
Fixed Capital and Working Capital
Answer:

Points Fixed Capital Working Capital
1. Meaning Fixed capital refers to any kind of physical asset, a portion of total capital that is invested in fixed assets. Working capital refers to the sum of current assets or gross working capital.
2. Nature It stays in the business almost permanently. Working capital is circulatory capital. It keeps changing.
3. Purpose It is invested in fixed assets such as land, building, equipment, etc. Working capital is invested in short-term assets such as cash, account receivable, inventory, etc.
4. Sources Fixed capital funding can come from selling shares, debentures, bonds, long-term loans, etc. Working capital can be funded with short-term loans, deposits, trade credit, etc.
5. Objectives of investors Investors invest money in fixed capital hoping to make a future profit. Investors invest money in working capital for getting immediate returns.
6. Risk Investment in fixed capital implies more risk. Investment in working capital is less risky. Eg. Land, building, plant and machinery
7. Decisions Decisions relating to fixed capital investment are generally made by top-level management. Eg. Cash, bills receivable, inventories, cash at the bank Decisions relating to working capital needs are generally made by middle-level or lower-level management.

5. Answer in brief:

Question 1.
Define capital structure and state its components.
Answer:
Definition: R.H. Wessel “The long term sources of funds employed in a business enterprise.”
John H. Hampton “A firm’s capital structure is the relation between the debt and equity securities that make up the firm’s financing of its assets.” Thus, the term capital structure means security mix. It refers to the proportion of different securities raised by a firm for long-term finance.

Components/Parts of Capital Structure:
There are four basic components of capital structure. They are as follows:
(i) Equity Share Capital:

  • It is the basic source of financing activities of the business. Equity share capital is provided by equity shareholders.
  • They buy equity shares and help a business firm to raise necessary funds. They bear the ultimate risk associated with ownership.
  • Equity shares carry dividends at a fluctuating rate depending upon profit.

Maharashtra Board Class 12 Secretarial Practice Solutions Chapter 1 Introduction to Corporate Finance

(ii) Preference Share Capital:

  • Preference shares carry preferential rights as to payment of dividends and have priority over equity shares for return of capital when the company is liquidated.
  • These shares carry dividends at a fixed rate.
  • They enjoy limited voting rights.

(iii) Retained earnings:

  • It is an internal source of financing.
  • It is nothing but ploughing back of profit.

(iv) Borrowed capital: It comprises of the following:

  • Debentures: A debenture is an acknowledgment of a loan raised by the company. The company has to pay interest at an agreed rate.
  • Term Loan: Term loans are provided by the bank and other financial institutions. They carry fixed rate of interest.

Question 2.
State any four factors affecting fixed capital requirements?
Answer:
(i) Nature of business:

  • The nature of business certainly plays a role in determining fixed capital requirements. They need to invest a huge amount of money in fixed assets.
  • e.g. Rail, road, and other public utility services have large fixed investments.
  • Their working capital requirements are nominal because they supply services and not the product.
  • They deal in cash sales only.

(ii) Size of business:
The size of a business also affects fixed capital needs. A general rule applies that the bigger the business, the higher the need for fixed capital. The size of the firm, either in terms of its assets or sales, affects the need for fixed capital.

(iii) Scope of business:
Some business firms that manufacture the entire range of their production would require a huge investment in fixed capital. However, those companies that are labour intensive and who do not use the latest technology may require less fixed capital and vice versa.

(iv) Extent of lease or rent:
Companies who take their assets on a lease basis or on a rental basis will require less amount of funds for fixed assets. On the other side, firms that purchase assets will naturally require more fixed capital in the initial stages.

Question 3.
What are Corporate Finance and State’s two decisions which are basic of corporate finance?
OR
Write short note on Corporate Finance
Answer:
Corporate finance deals with the raising and using of finance by a corporation. It includes various financial activities like capital structuring and making investment decisions, financial planning, capital formation, and foreign capital, etc. Even financial organisations and banks play a vital role in corporate financing.

Henry Hoagland expresses, “Corporate Finance deals primarily with the acquisition and use of capital by the business corporation”.

Following two decisions are the basis of corporate finances:
(i) Financing decision:
Every business firm must carefully estimate its capital needs i.e. working capital and fixed capital. The firm needs to mobilize funds from the right sources also maintaining the right combination of debt capital and equity capital. For this balance, a company may go for or raise equity capital or even opt for borrowed funds by way of debentures, public deposits term loans, etc. to raise funds.

(ii) Investment decision:
Once the capital needs are accessed, the finance manager needs to take correct decisions regarding the use of the funds in a systematic manner, productively, using effective cost control measures to generate high profits. Finding investments through proper decisions and using them successfully in business is called ‘capital budgeting

Maharashtra Board Class 12 Secretarial Practice Solutions Chapter 1 Introduction to Corporate Finance

6. Justify the following statements.

Question 1.
The firm has multiple choices of sources of financing.
Answer:

  • Business firms require finance in terms of working capital and fixed capital.
  • Funds are required at different stages of business.
  • The company can raise funds from various sources i.e. from internal and external sources.
  • Internal sources could be cash inflows on sales turnover, income from investments, and retained earnings.
  • External sources can be obtained for short-term requirements through cash credit, overdraft trade credit, discounting bills of Exchange issues of commercial paper, etc.
  • For long-term needs, a firm can meet its financing needs through the issue of shares, debentures, bonds, public deposits, etc. Thus, it is rightly said that the firm has multiple choices of sources of financing.

Question 2.
There are various factors affecting the requirements of fixed capital.
Answer:

  • Fixed capital being long-term capital is required for the development and expansion of the company.
  • The nature and size of a business have a great impact on fixed capital. Manufacturing businesses require huge fixed capital whereas trading organizations like retailers require less fixed capital.
  • Methods of acquiring assets on rentals or on a lease/installment basis will require less amount of fixed assets.
  • If fixed assets are available at low prices and concessional rates then it would reduce the need for investment in fixed assets.
  • International conditions and economic trends like a boom period will require high investment in fixed assets and a recession will lead to less requirement.
  • Similarly, consumer preferences, competition, and highly demanded goods and services will require a large amount of fixed capital. E.g. Mobile phones. Thus, it is rightly said that there are various factors affecting the requirements of fixed capital.

Question 3.
Fixed capital stays in the business almost permanently.
Answer:
Factors determining fixed capital requirements are:

  • Fixed capital refers to capital invested for acquiring fixed assets.
  • These assets are not meant for resale.
  • Fixed capital is capital used for purchasing land and building, furniture, plant, and machinery, etc.
  • Such cap al is usually required at the time of the establishment of a new company.
  • Existing companies may also need such capital for their expansion and development, replacement of equipment, etc.
  • Modern industrial processes require the increased use of heavy automated machinery. Thus, it is rightly said that fixed capital stays in the business almost permanently.

Question 4.
Capital structure is composed of owned funds and borrowed funds.
Answer:

  • Capital structure means to mix up of various sources of funds in desired proportions.
  • To decide capital structures means to decide upon the ratio of different types of capital.
  • A firm’s capital structure is the relation between the debt and equity securities that make up the firm’s financing of its assets.
  • The capital structure is composed of own funds which include share capital, free serves, and surplus, and borrowed funds which represent debentures, bank loans, and long-term loans provided by financial institutions.
  • Thus capital structure = Equity share capital + preference share capital + reserves + debentures.
  • Thus, it is rightly said that capital structure is composed of owned funds and borrowed funds.

Maharashtra Board Class 12 Secretarial Practice Solutions Chapter 1 Introduction to Corporate Finance

Question 5.
There are various factors affecting the requirement of working capital.
Answer:

  • The nature and size of a business affect the requirement of working capital. Trading or merchandising firms and big retail enterprises need a large amount of capital compared to small firms which need a small amount of working capital.
  • If the period of the production cycle is longer then the firm needs more amount of working capital. If the manufacturing cycle is short, it requires less working capital.
  • During the boom period sales will increase leading to increased investment in stocks, thus requiring additional working capital and during the recession, it is vice versa.
  • Along with the expansion and growth of the firm or company in terms of sales and fixed assets, the requirement of working capital increases.
  • If there is proper coordination, communication, and co-operation between production and sales departments then the requirement of working capital is less.
  • A liberal credit policy increases the possibility of bad debts and in such cases, the requirement of working capital is high, whereas a firm making cash sales requires less working capital.

7. Answer the following questions.

Question 1.
Discuss the importance of Corporate Finance?
Answer:
Corporate finance deals with the raising and utilizing of finance by a corporation. It also deals with capital structuring and making investment decisions, financial planning of capital, and the money market. The finance manager should ensure that:
The firm has adequate finance and it’s being utilized effectively;
Generate minimum return for its owners.

The importance of Corporate Finance are as follows:
(i) Helps in decision making:
Most important decisions of business enterprises are made on the basis of availability of funds, as without finance any function of business enterprise is difficult to be performed independently. Obtaining the funds from the right sources at a lower cost and productive utilization of funds would lead to higher profits. Thus corporate finance plays a significant role in the decision-making process.

(ii) Helps in raising capital for a project:
A new business venture needs to raise capital. Business firms can raise funds by issuing shares, debentures, bonds or even by taking loans from the banks.

(iii) Helps in Research and Development
Research and Development need to be undertaken by firms for growth and expansion of business and to enjoy a competitive advantage. Research and development mostly involve lengthy and detailed technical work for the execution of projects. Through surveys and market analysis etc. companies may have to upgrade old products or develop new products to face competition and attract consumers. Thus the availability of adequate finance helps to generate high efficiency.

Maharashtra Board Class 12 Secretarial Practice Solutions Chapter 1 Introduction to Corporate Finance

(iv) Helps in the smooth running of the business firm:
A smooth flow of corporate finance is important to pay the salaries of employees on time, pay loans, and purchase the required raw materials. At the same time finance is needed for sales promotion of existing products and more so for the launch of new products effectively.

(v) Brings co-ordination between various activities:
Corporate finance plays a significant role in the coordination and control of all activities in an organization. Production activity requires adequate finance for the purchase of raw materials and meeting other day-to-day financial requirements for the smooth running of the production unit. If the production increases, sales will also increase by contributing the income of the concern and profit to increase.

(vi) Promotes expansion and diversification:
Corporate finance provides money for the purchase of modern machines and sophisticated technology. Modern machines and technology help to improve the performance of the firm in terms of profits. It also helps the firm to expand and diversify the business.

(vii) Managing risk:
Companies have to manage several risks such as sudden fall in sales, loss due to natural calamity, loss due to workers strikes, change in government policies, etc. Financial aids help in such situations to manage such risks.

(viii) Replace old assets:
Assets like plants and machinery have become old and outdated over the years. Finance is required to purchase new assets or replace the old assets with new assets having new technology and features.

(ix) Payment of dividend and interest:
Finance is needed to pay the dividend to shareholders, interest to creditors, bank, etc.

(x) Payment of taxes/fees:
The company has to pay taxes to the government such as Income tax, Goods and Service Tax (GST), and fees to the Registrar of Companies on various occasions. Finance is needed for paying these taxes and fees.

Question 2.
Discuss the factors determining working capital requirements?
Answer:
Working Capital = Current Assets – Current Liabilities.
In other words, it is also called ‘Circulating Capital’. Also, refer to ‘GROSS WORKING CAPITAL.’ Management needs to determine the size of working capital with reference to the economic environment and other aspects within the business firm.

Factors determining/influencing working capital requirements are as follows:
(i) Nature of Business:
The working capital requirements are highly influenced by the nature of the business. Trading/ merchandising forms concerned with the distribution of goods require a huge amount of working capital to maintain a large stock of the variety of goods to meet customers’ demands are extend credit facilities to attract them. Whereas public utility concerns have to maintain small working capital because of a continuous flow of cash from their customers.

(ii) Size of business:
The size of a business also affects the requirements of working capital. Size of the firm refers to the scale of operation i.e. a firm with large scale operations will require more working capital and vice versa.

(iii) Volume of Sales:
The volume of sales and the size of the working capital have a direct relationship with each other. If the volume of sales increases there is an increase in the amount of working capital.

Maharashtra Board Class 12 Secretarial Practice Solutions Chapter 1 Introduction to Corporate Finance

(iv) Production cycle:
The process of converting raw material into finished goods is called the ‘production cycle’. If the production cycle period is longer, the firm needs more amount of working capital. If the manufacturing cycle is short, it requires less working capital.

(v) Business cycle:
When there is a boom in the economy, sales will increase resulting in to increase in investment in stock. This will require additional working capital. During a recession period, sales will decline and consequently, the need for working capital will also decrease.

(vi) Terms of purchases and sales:
If credit terms of purchase are favourable and terms of sales are less liberal, then the requirement of cash will be less. Thus, the working capital requirement will be reduced.
A firm that enjoys more credit facilities needs less working capital. On the other hand, if a firm does not get proper credit for purchases and adopts a liberal credit policy for sales if requires more working capital.

(vii) Credit Control:
Credit control includes the factors such as volume of credit sales, the terms of credit sales, the collection policy etc. A firm with a good credit control policy will have more cash flow reducing the working capital requirement. Whereas if the firm’s credit policy is liberal there would be more requirements of the working capital.

(viii) Growth and Expansion:
Those firms which are growing and expanding at a rapid pace need more working capital compared to those firms which are stable in their growth.

(ix) Management ability:
The requirement of working capital is reduced if there is proper coordination in the production and distribution of goods. A firm stocking on heavy inventory calls for a higher level of working capital.

Maharashtra Board Class 12 Secretarial Practice Solutions Chapter 1 Introduction to Corporate Finance

(x) External factors:
If the financial institutions and banks provide funds to the firm as and when required, the need for working capital is reduced.

Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 13 AC Circuits Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 13 AC Circuits

1. Choose the correct option.

i) If the RMS current in a 50 Hz AC circuit is 5A, the value of the current \(\frac{1}{300}\) seconds after its value becomes zero is
(A) 5\(\sqrt {2}\) A
(B) 5\(\sqrt{\frac{3}{2}}\) A
(C) \(\frac{5}{6}\) A
(D) \(\frac{5}{\sqrt{2}}\) A
Answer:
(B) 5\(\sqrt{\frac{3}{2}}\) A

ii) A resistor of 500 Ω and an inductance of 0.5 H are in series with an AC source which is given by V = 100 \(\sqrt {2}\) sin (1000t). The power factor of the combination
(A) \(\frac{1}{\sqrt{2}}\)
(B) \(\frac{1}{\sqrt{3}}\)
(C) 0.5
(D) 0.6
Answer:
(A) \(\frac{1}{\sqrt{2}}\)

Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits

iii) In a circuit L, C & R are connected in series with an alternating voltage of frequency f. the current leads the voltage by 450. The value of C is
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 1
Answer:
(B) \(\frac{1}{2 \pi f(2 \pi f L-R)}\)

iv) In an AC circuit, e and i are given by e = 150 sin (150t) V and i = 150 sin (150 t + \(\frac{\pi}{3}\)) A. the power dissipated in the circuit is
(A) 106W
(B) 150W
(C) 5625W
(D) Zero
Answer:
(C) 5625W

v) In a series LCR circuit the phase difference between the voltage and the current is 45°. Then the power factor will be
(A) 0.607
(B) 0.707
(C) 0.808
(D) 1
Answer:
(B) 0.707

2. Answer in brief.

i) An electric lamp is connected in series with a capacitor and an AC source is glowing with a certain brightness. How does the brightness of the lamp change on increasing the capacitance ?
Answer:
Impedance, Z = \(\sqrt{R^{2}+\frac{1}{\omega^{2} C^{2}}}\), where R is the resistance of the lamp, w is the angular frequency of AC and C is the capacitance of the capacitor connected in series with the AC source and the lamp. When C is increased, \(\) decreases. Hence, Z increases.
Power factor, cos Φ = \(\frac{R}{Z}\)
As Z increases, the power factor decreases.
Now, the average power over one cycle,
Pav = vrms irms cos Φ
= Vrms \(\left(\frac{V_{\mathrm{rms}}}{Z}\right)\) cos Φ
= \(\frac{V_{\mathrm{rms}}^{2}}{\mathrm{Z}} \cos \phi\)
∴ Pav decreases as Z increases and cos Φ decreases.
As the current through the lamp \(\left(\frac{V_{\mathrm{rms}}}{Z}\right)\) decreases, the brightness of the lamp will decrease when C is increased.

ii) The total impedance of a circuit decreases when a capacitor is added in series with L and R. Explain why ?
Answer:
For an LR circuit, the impedance,
ZLR = \(\sqrt{R^{2}+X_{\mathrm{L}}^{2}}\), where XL is the reactance of the inductor.
When a capacitor of capacitance C is added in series with L and R, the impedance,
ZLCR = \(\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}\) because in the case of an inductor the current lags behind the voltage by a phase angle of \(\frac{\pi}{2}\) rad while in the case of a capacitor the current leads the voltage by a phase angle of \(\frac{\pi}{2}\) rad. The decrease in net reactance decreases the total impedance (ZLCR < ZLR).

iii) For very high frequency AC supply, a capacitor behaves like a pure conductor. Why ?
Answer:
The reactance of a capacitor is XC = \(\frac{1}{2 \pi f C}\), where f is the frequency of the AC supply and C is the capacitance of the capacitor. For very high frequency, f, XC is very small. Hence, for very high frequency AC supply, a capacitor behaves like a pure conductor.

Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits

iv) What is wattless current ?
Answer:
The current that does not lead to energy consumption, hence zero power consumption, is called wattless current.
In the case of a purely inductive circuit or a purely capacitive circuit, average power consumed over a complete cycle is zero and hence the corresponding alternating current in the circuit is called wattless current.
[Note : In this case, the power factor is zero.]

v) What is the natural frequency of L C circuit ? What is the reactance of this circuit at this frequency
Answer:
The natural frequency of LC circuit is \(\frac{1}{2 \pi \sqrt{L C}}\) ,
where L is the inductance and C is the capacitance. The reactance of this circuit at this frequency is
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 18

Question 3.
In a series LR circuit XL = R and power factor of the circuit is P1. When capacitor with capacitance C such that XL = XC is put in series, the power factor becomes P2. Calculate P1 / P2 .
Answer:
For a series LR circuit, power factor,
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 17

Question 4.
When an AC source is connected to an ideal inductor show that the average power supplied by the source over a complete cycle is zero.
Answer:
In an AC circuit containing only an ideal inductor, the current i lags behind the emf e by a phase angle of \(\frac{\pi}{2}\) rad. Here, for e = e0 sin ωt, we have, i = i0 sin(ωt – \(\frac{\pi}{2}\))
Instantaneous power, P = ei
= (e0 sin ωt) [i0 (sin ωt cos \(\frac{\pi}{2}\) – cos ωt sin \(\frac{\pi}{2}\))]
= – e0i0 sin ωt cos ωt as cos \(\frac{\pi}{2}\) = 0 and sin \(\frac{\pi}{2}\) = 1.
Average power over one cycle,
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 15
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 16
= erms irms cos Φ = erms irms (\(\frac{R}{Z}\)), where the impedance Z = \(\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}\).
∴ Pav = 0, i.e., the circuit does not dissipate power.

Question 5.
Prove that an ideal capacitor in an AC circuit does not dissipate power
Answer:
In an AC circuit containing only an ideal inductor, the current i lags behind the emf e by a phase angle of \(\frac{\pi}{2}\) rad. Here, for e = e0 sin ωt, we have, i = i0 sin(ωt – \(\frac{\pi}{2}\))
Instantaneous power, P = ei
= (e0 sin ωt) [i0 (sin ωt cos \(\frac{\pi}{2}\) + cos ωt sin \(\frac{\pi}{2}\))]
= – e0i0 sin ωt cos ωt as cos \(\frac{\pi}{2}\) = 0 and sin \(\frac{\pi}{2}\) = 1.
Average power over one cycle, Pav
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 15
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 16
= erms irms cos Φ = erms irms (\(\frac{R}{Z}\)), where the impedance Z = \(\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}\).
∴ Pav = 0, i.e., the circuit does not dissipate power.

Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits

Question 6.
(a) An emf e = e0 sin ωt applied to a series L – C – R circuit derives a current I = I0 sinωt in the circuit. Deduce the expression for the average power dissipated in the circuit.
(b) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain.
Answer:
(a) Instantaneous power,
P = ei
= (e0 sin ωt) [i0 (sin ωt ± Φ)]
= e0i0 sin ωt(sin ωt cos Φ ± cos ωt sin Φ)
= e0i0 sin2 ωt ± e0i0 sin Φ sin ωt cos ωt
Average power over one cycle,
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 15
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 16
= erms irms cos Φ = erms irms (\(\frac{R}{Z}\)), where the impedance Z = \(\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}\).

(b) Pav = erms irms cos Φ
The factor cos Φ is called as power factor. For circuits used for transporting electric power, a low power factor means the power available on transportation is much less than erms irms It means there is significant loss of power during transportation.

Question 7.
A device Y is connected across an AC source of emf e = e0 sinωt. The current through Y is given as i = i0 sin(ωt + π/2)
a) Identify the device Y and write the expression for its reactance.
b) Draw graphs showing variation of emf and current with time over one cycle of AC for Y.
c) How does the reactance of the device Y vary with the frequency of the AC ? Show graphically
d) Draw the phasor diagram for the device Y.
Answer:
(a) The device Y is a capacitor. Its reactance is Xc = \(\frac{1}{\omega C}\),
where ω is the angular frequency of the applied emf and C is the capacitance of the capacitor.

(b)
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 6

(c) XC = \(\frac{1}{\omega C}=\frac{1}{2 \pi f C}\). Thus XC ∝ \(\frac{1}{f}\), where f is the frequency of AC. Suppose C = \(\left(\frac{1000}{2 \pi}\right)\) pF
For f= 100 Hz, XC = 1 × 107Ω = 10MΩ;
for f = 200 Hz, XC = 5 MΩ;
for f = 300 Hz, XC = \(\frac{10}{3}\) MΩ;
for f = 400 Hz, XC = 2.5 MΩ
for f = 500 Hz, XC = 2 MΩ and so on
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 8

(d)
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 9
The phasor representing the peak emf (e0) makes an angle (ωt) in an anticlockwise direction with respect to the horizontal axis. As the current leads the voltage by 90°, the phasor representing the peak current (i0) is turned 90° anticlockwise with respect to the phasor representing emf e0. The projections of these phasors on the vertical axis give instantaneous values of e and i.

Question 8.
Derive an expression for the impedance of an LCR circuit connected to an AC power supply.
Answer:
Figure shows an inductor of inductance L, capacitor of capacitance C, resistor of resistance R, key K and source (power supply) of alternating emf (e) connected to form a closed series circuit.
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 10
We assume the inductor, capacitor and resistor to be ideal. As these are connected in series, at any instant, they carry the same current i = i0 sin ωt. The voltage across the resistor, eR = Ri, is in phase with the current. The voltage across the inductor, eL = XLi, leads the current by \(\frac{\pi}{2}\) rad and that across the capacitor, eC = XCi, lags behind the current by \(\frac{\pi}{2}\) rad. This is shown in the phasor diagram.
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 11
is the effective resistance of the circuit. It is called the impedance.

Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits

Question 9.
Compare resistance and reactance.
Answer:
(1) Resistance is opposition to flow of charges (current) and appears in a DC circuit as well as in an AC circuit.
The term reactance appears only in an AC circuit. It occurs when an inductor and/or a capacitor is used.

(2) In a purely resistive circuit, current and voltage are always in phase.
When reactance is not zero, there is nonzero phase difference between current and voltage.

(3) Resistance does not depend on the frequency of AC.
Reactance depends on the frequency of AC. In case of an inductor, reactance increases linearly with frequency. In case of a capacitor, reactance decreases as frequency of AC increases; it is inversely proportional to frequency.

(4) Resistance gives rise to production of Joule heat in a component.
In a circuit with pure reactance, there is no production of heat.

Question 10.
Show that in an AC circuit containing a pure inductor, the voltage is ahead of current by π/2 in phase.
Answer:
Figure 13.8 shows an AC source, generating a voltage e = e0 sin ωt, connected to a key K and a pure inductor of inductance L to form a closed circuit.
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 2
On closing the key K, an emf is induced in the inductor as the magnetic flux linked with it changes with time. This emf opposes the applied emf and according to the laws of electromagnetic induction by Faraday and Lenz, we have,
e’ = -L\(\frac{d i}{d t}\) ………………. (1)
where e’ is the induced emf and i is the current through the inductor. To maintain the current; e and e’ must be equal in magnitude and opposite in direction.

According to Kirchhoff’s voltage law, as the resistance of the inductor is assumed to be zero, we
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 3
where C is the constant of integration. C must be time independent and have the dimension of current. As e oscillates about zero, i also oscillates about zero and hence there cannot be any time independent component of current.
∴ C = 0. ∴ i = –\(\frac{e_{0}}{\omega L}\)cos ωt = – \(\frac{e_{0}}{\omega L}\)sin(\(\frac{\pi}{2}\) – ωt)
∴ i = \(\frac{e_{0}}{\omega L}\) sin(ωt – \(\frac{\pi}{2}\)) ……………. (3)
as sin (-θ) = – sin θ
From Eq. (3), ipeak = i0 = \(\frac{e_{0}}{\omega L}\)
∴ i = i0 sin(ωt – \(\frac{\pi}{2}\)) ………………. (4)
Comparison of this equation with e = e0 sin ωt shows that e leads i by \(\frac{\pi}{2}\) rad, i.e., the voltage is ahead of current by \(\frac{\pi}{2}\) rad in phase.

Question 11.
An AC source generating a voltage e = e0 sinωt is connected to a capacitor of capacitance C. Find the expression for the current i flowing through it. Plot a graph of e and i versus ωt.
Answer:
Figure 13.12 shows an AC source, generating a voltage e = e0 sin ωt, connected to a capacitor of capacitance C. The plates of the capacitor get charged due to the applied voltage. As the alternating voltage is reversed in each half cycle, the
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 4
capacitor is alternately charged and discharged. If q is the charge on the capacitor, the corresponding potential difference across the plates of the capacitor is V = \(\frac{q}{C}\) ∴ q = CV. q and V are functions of time, with V = e = e0 sin ωt. The instantaneous current in the circuit is i = \(\frac{d q}{d t}=\frac{d}{d t}\)(CV) = C \(\frac{d v}{d t}\) = C \(\frac{d}{d t}\) (e0 sin ωt) = ωC e0 cos ωt
∴ i = \(\frac{e_{0}}{(1 / \omega C)} \sin \left(\omega t+\frac{\pi}{2}\right)=i_{0} \sin \left(\omega t+\frac{\pi}{2}\right)\)
where i0 = \(\frac{e_{0}}{(1 / \omega C)}\) is the peak value of the current.
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 5
Table gives the values of e and i for different values of cot and Fig shows graphs of e and i versus ωt. i leads e by phase angle of \(\frac{\pi}{2}\) rad.
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 6

Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits

Question 12.
If the effective current in a 50 cycle AC circuit is 5 A, what is the peak value of current? What is the current 1/600 sec.after if was zero ?
Answer:
Data : f = 50 Hz, irms = 5 A, t = \(\frac{1}{600}\) s
The peak value of the current,
i0 = irms\(\sqrt {2}\) = (5)(1.414) = 7.07 A
= i0sin (2πft)
= 7.07 sin [2π(5o) (\(\frac{1}{600}\))]
= 7.07 sin (\(\frac{\pi}{6}\)) = (7.07)(0.5)
= 3.535 A
This is the required current.

Question 13.
A light bulb is rated 100W for 220 V AC supply of 50 Hz. Calculate (a) resistance of the bulb. (b) the rms current through the bulb.
Answer:
Data: Power (Vrms irms) = 100 W, Vrms = 220V,
f = 50 Hz
The rms current through the bulb,
irms = \(\frac{\text { power }}{V_{\mathrm{rms}}}=\frac{100}{220}\) = 0.4545 A
The resistance of the bulb,
R = \(\frac{V_{\mathrm{rms}}}{i_{\mathrm{rms}}}=\frac{220}{(100 / 220)}\) = (22) (22) = 484 Ω

Question 14.
A 15.0 µF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak)
in the circuit. If the frequency is doubled, what will happen to the capacitive reactance and the current.
Answer:
Data : C = 15 µF = 15 × 10-6 F, Vrms = 220V, f = 50 Hz,
The capacitive reactance = \(\frac{1}{2 \pi f C}\)
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 12
If the frequency is doubled, the capacitive reactance will be halved and the current will be doubled.

Question 15.
An AC circuit consists of only an inductor of inductance 2 H. If the current is represented by a sine wave of amplitude 0.25 A and frequency 60 Hz, calculate the effective potential difference across the inductor (π = 3.142)
Answer:
Data : L = 2H, i0 = 0.25 A, f = 60 Hz, π = 3.142
ωL = 2πfL = 2(3.142)(60)(2) = 754.1 Ω
The effective potential difference across the inductor = ωLirms = ωL \(\frac{i_{0}}{\sqrt{2}}\)
= \(\frac{(754.1)(0.25)}{1.414}\) = 133.3 V

Question 16.
Alternating emf of e = 220 sin 100 πt is applied to a circuit containing an inductance of (1/π) henry. Write an equation for instantaneous current through the circuit. What will be the reading of the AC galvanometer connected in the circuit?
Answer:
Data: e = 220 sin 100 πt, L = (\(\frac{1}{\pi}\))H
Comparing e = 220 sin 100 πt with
e = e0 sin ωt, we get
ω = 100 π ∴ ωL = (100 π) (\(\frac{1}{\pi}\)) = 100 Ω
∴ The instantaneous current through the circuit
= i = \(\frac{e_{0}}{\omega L}\) sin(100 πt – \(\frac{\pi}{2}\))
= \(\frac{220}{100}\) sin (100 πt – \(\frac{\pi}{2}\)) = 2.2 sin (100 πt – \(\frac{\pi}{2}\)) in ampere [assuming that e is in volt.]
irms = \(\frac{i_{0}}{\sqrt{2}}=\frac{2.2}{1.414}\) = 1.556 A is the reading of the AC galvanometer connected in the circuit.

Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits

Question 17.
A 25 µF capacitor, a 0.10 H inductor and a 25Ω resistor are connected in series with an AC source whose emf is given by e = 310 sin 314 t (volt). What is the frequency, reactance, impedance, current and phase angle of the circuit?
Answer:
Data: C = 25 µF = 25 × 10-6F, L = 0.10H, R = 25 Ω ,
e = 310 sin (314 t) [volt]
Comparing e = 310 sin (314 t) with
e = e0 sin (2πft), we get,
the frequency of the alternating emf as
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 13
cos Φ = \(\frac{R}{Z}=\frac{25}{99.2}\) = 0.2520
∴ The phase angle, Φ = cos-1(0.2520) = 75.40° = 1.316 rad

Question 18.
A capacitor of 100 µF, a coil of resistance 50Ω and an inductance 0.5 H are connected in series with a 110 V-50Hz source. Calculate the rms value of current in the circuit.
Answer:
Data : C = 100 µF = 100 × 10-6 F = 10-4 F,
R = 50 Ω, L = 0.5H, f = 50 Hz, Vrms = 110 V
∴ ωL = 2πfL = 2 (3.142)(50)(0.5) = 157.1 Ω
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 14
2500 + 15700 = 18200 Ω2
∴ Impedance, Z = \(\sqrt {18200}\) Ω = 134.9 Ω
The rms value of the current in the circuit,
irms = \(\frac{V_{\mathrm{rms}}}{\mathrm{Z}}=\frac{110}{134.9} \mathrm{~A}\)
= 0.8154 A

Question 19.
Find the capacity of a capacitor which when put in series with a 10Ω resistor makes the power factor equal to 0.5. Assume an 80V-100Hz AC supply.
Answer:
Data : R = 10 Ω, power factor = 0.5, f = 100 Hz
Power factor = \(\frac{1}{2 \pi f C R}\)
∴ 0.5 = \(\frac{1}{2(3.142)(100) C(10)}\)
∴ C = \(\frac{1}{3.142 \times 10^{3}}\)
= \(\frac{10 \times 10^{-4}}{3.142}\)
= 3.182 × 10-4 F
This is the capacity of the capacitor.

Question 20.
Find the time required for a 50 Hz alternating current to change its value from zero to the rms value.
Answer:
Data : f = 50 Hz, i = \(\frac{i_{0}}{\sqrt{2}}\) ∴ \(\frac{i}{i_{0}}=\frac{1}{\sqrt{2}}\)
i = i0 sinωt
∴ sinωt = \(\frac{i}{i_{0}}=\frac{1}{\sqrt{2}}\)
∴ ωt = \(\frac{\pi}{4}\) rad
∴ 2πft = \(\frac{\pi}{4}\)
∴ t = \(\frac{1}{8 f}=\frac{1}{8(50)}=\frac{1}{400}\)
= \(\frac{1000 \times 10^{-3}}{400}\) = 2.5 × 10-3 s
This is the required time.

Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits

Question 21.
Calculate the value of capacitance in picofarad, which will make 101.4 micro henry inductance to oscillate with frequency of one megahertz.
Answer:
Data : fr = 106 Hz, L = 101.4 × 10-6 H
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 19
= \(\frac{10000 \times 10^{-10}}{4(3.142)^{2}(101.4)}\) = 2.497 × 10-10 F
= 249.7 × 10-12 F = 249.7 picofarad
This is the value of the capacity.

Question 22.
A 10 µF capacitor is charged to a 25 volt of potential. The battery is disconnected and a pure 100 m H coil is connected across the capacitor so that LC oscillations are set up. Calculate the maximum current in the coil.
Answer:
Data: C = 10 µF = 10 × 10-6F = 10-5F,
L = 100mH = 100 × 10-3 H = 10-1 H, V = 25V
For reference, see the solved example (8) above.
\(\frac{1}{2}\)CV2 = \(\frac{1}{2}\)Li2
∴i2 = \(\frac{C}{L} V^{2}=\frac{10^{-5}}{10^{-1}}(25)^{2}\)
∴i = 25 × 10-2 A = 0.25 A
This is the maximum current in the coil.

Question 23.
A 100 µF capacitor is charged with a 50 V source supply. Then source supply is removed and the capacitor is connected across an inductance, as a result of which 5A current flows through the inductance. Calculate the value of the inductance.
Answer:
Data: C = 100 µF = 100 × 10-6 F = 10-4 F,
V = 50V, i = 5A
The energy stored in the electric field in the capacitor
= \(\frac{1}{2}\)CV2
The energy stored in the magnetic field in the inductor = \(\frac{1}{2}\)Li2
Here, \(\frac{1}{2}\)CV2 = \(\frac{1}{2}\)Li2
∴ L = C\(\frac{V^{2}}{i^{2}}\)
∴ L = C\(\left(\frac{V}{i}\right)^{2}=10^{-4}\left(\frac{50}{5}\right)^{2}\) = 10-4 × 102
= 10-2H
This is the value of the inductance.

Maharashtra State Board Class 12 Secretarial Practice Solutions Digest

Maharashtra State Board 12th Std Secretarial Practice Textbook Solutions Digest

Maharashtra State Board Class 12 Textbook Solutions

Maharashtra Board Class 12 Physics Solutions Chapter 4 Thermodynamics

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

Maharashtra State Board 12th Physics Solutions Chapter 4 Thermodynamics

1. Choose the correct option.

i) A gas in a closed container is heated with 10J of energy, causing the lid of the container to rise 2m with 3N of force. What is the total change in the internal energy of the system?
(A) 10J
(B) 4J
(C) -10J
(D) – 4J
Answer:
(B) 4J

ii) Which of the following is an example of the first law of thermodynamics?
(A) The specific heat of an object explains how easily it changes temperatures.
(B) While melting, an ice cube remains at the same temperature.
(C) When a refrigerator is unplugged, everything inside of it returns to room temperature after some time.
(D) After falling down the hill, a ball’s kinetic energy plus heat energy equals the initial potential energy.
Answer:
(B) While melting, an ice cube remains at the same temperature. [Here, ∆u = 0, W = Q]
(C) When a refrigerator is unplugged, everything inside of it returns to room temperature after some time.
(D) After falling down the hill, a ball’s kinetic energy plus heat energy equals the initial potential energy.

Maharashtra Board Class 12 Physics Solutions Chapter 4 Thermodynamics

iii) Efficiency of a Carnot engine is large when
(A) TH is large
(B) TC is low
(C) TH – TC is large
(D) TH – TC is small
Answer:
(A) TH is large
(B) TC is low
(C) TH – TC is large
[η = \(\frac{T_{\mathrm{H}}-T_{\mathrm{c}}}{T_{\mathrm{H}}}\) = 1 – \(\frac{T_{\mathrm{c}}}{T_{\mathrm{H}}}\)]

iv) The second law of thermodynamics deals with transfer of:
(A) work done
(B) energy
(C) momentum
(D) mass
Answer:
(B) energy

v) During refrigeration cycle, heat is rejected by the refrigerant in the :
(A) condenser
(B) cold chamber
(C) evaporator
(D) hot chamber
Answer:
closed tube[See the textbook]

2. Answer in brief.

i) A gas contained in a cylinder surrounded by a thick layer of insulating material is quickly compressed.
(a) Has there been a transfer of heat?
(b) Has work been done?
Answer:
(a) There is no transfer of heat.
(b) The work is done on the gas.

ii) Give an example of some familiar process in which no heat is added to or removed form a system, but the temperature of the system changes.
Answer:
Hot water in a container cools after sometime. Its temperature goes on decreasing with time and after sometime it attains room temperature.
[Note : Here, we do not provide heat to the water or remove heat from the water. The water cools on exchange of heat with the surroundings. Recall the portion covered in chapter 3.]

iii) Give an example of some familiar process in which heat is added to an object, without changing its temperature.
Answer:

  1. Melting of ice
  2. Boiling of water.

iv) What sets the limits on efficiency of a heat engine?
Answer:
The temperature of the cold reservoir sets the limit on the efficiency of a heat engine.
[Notes : (1) η = 1 – \(\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}\)
This formula shows that for maximum efficiency, TC should be as low as possible and TH should be as high as possible.
(2) For a Carnot engine, efficiency
η = 1 – \(\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}\).η → 1 TC → 0.]

v) Why should a Carnot cycle have two isothermal two adiabatic processes?
Answer:
With two isothermal and two adiabatic processes, all reversible, the efficiency of the Carnot engine depends only on the temperatures of the hot and cold reservoirs.
[Note : This is not so in the Otto cycle and Diesel cycle.]

3. Answer the following questions.

i) A mixture of hydrogen and oxygen is enclosed in a rigid insulting cylinder. It is ignited by a spark. The temperature and the pressure both increase considerably. Assume that the energy supplied by the spark is negligible, what conclusions may be drawn by application of the first law of thermodynamics?
Answer:
The internal energy of a system is the sum of potential energy and kinetic energy of all the constituents of the system. In the example stated above, conversion of potential energy into kinetic energy is responsible for a considerable rise in pressure and temperature of the mixture of hydrogen and oxygen ignited by the spark.

ii) A resistor held in running water carries electric current. Treat the resistor as the system
(a) Does heat flow into the resistor?
(b) Is there a flow of heat into the water?
(c) Is any work done?
(d) Assuming the state of resistance to remain unchanged, apply the first law of thermodynamics to this process.
Answer:
(a) Heat is generated into the resistor due to the passage of electric current. In the usual notation, heat generated = I2Rt.
(b) Yes. Water receives heat from the resistor.
Maharashtra Board Class 12 Physics Solutions Chapter 4 Thermodynamics 10
Here, I = current through the resistor, R = resistance of the resistor, t = time for which the current is passed through the resistor, M = mass of the water, S = specific heat of water, T = rise in the temperature of water, P = pressure against which the work is done by the water, ∆u= increase in the volume of the water.

Maharashtra Board Class 12 Physics Solutions Chapter 4 Thermodynamics

iii) A mixture of fuel and oxygen is burned in a constant-volume chamber surrounded by a water bath. It was noticed that the temperature of water is increased during the process. Treating the mixture of fuel and oxygen as the system,
(a) Has heat been transferred ?
(b) Has work been done?
(c) What is the sign of ∆u ?
Answer:
(a) Heat has been transferred from the chamber to the water bath.
(b) No work is done by the system (the mixture of fuel and oxygen) as there is no change in its volume.
(c) There is increase in the temperature of water. Therefore, ∆u is positive for water.
For the system (the mixture of fuel and oxygen), ∆u is negative.

iv) Draw a p-V diagram and explain the concept of positive and negative work. Give one example each.
Answer:
Consider some quantity of an ideal gas enclosed in a cylinder fitted with a movable, massless and frictionless piston.
Maharashtra Board Class 12 Physics Solutions Chapter 4 Thermodynamics 11
Suppose the gas is allowed to expand by moving the piston outward extremely slowly. There is decrease in pressure of the gas as the volume of the gas increases. Below figure shows the corresponding P-V diagram.
Maharashtra Board Class 12 Physics Solutions Chapter 4 Thermodynamics 12
In this case, the work done by the gas on its surroundings,
W = \(\int_{V_{\mathrm{i}}}^{V_{\mathrm{f}}} P d V\) (= area under the curve) is positive as the volume of the gas has increased from Vi to Vf.
Let us now suppose that starting from the same
Maharashtra Board Class 12 Physics Solutions Chapter 4 Thermodynamics 13
initial condition, the piston is moved inward extremely slowly so that the gas is compressed. There is increase in pressure of the gas as the volume of the gas decreases. Figure shows the corresponding P-V diagram.
In this case, the work done by the gas on its surroundings, W = \(\int_{V_{\mathrm{i}}}^{V_{\mathrm{f}}} P d V\) (= area under the curve) is negative as the volume of the gas has decreased from Vi to Vf.
Maharashtra Board Class 12 Physics Solutions Chapter 4 Thermodynamics 15

v) A solar cooker and a pressure cooker both are used to cook food. Treating them as thermodynamic systems, discuss the similarities and differences between them.
Answer:
Similarities :

  1. Heat is added to the system.
  2. There is increase in the internal energy of the system.
  3. Work is done by the system on its environment.

Differences : In a solar cooker, heat is supplied in the form of solar radiation. The rate of supply of heat is relatively low.

In a pressure cooker, usually LPG is used (burned) to provide heat. The rate of supply of heat v is relatively high.

As a result, it takes very long time for cooking when a solar cooker is used. With a pressure cooker, it does not take very long time for cooking.
[Note : A solar cooker can be used only when enough solar radiation is available.]

Question 4.
A gas contained in a cylinder fitted with a frictionless piston expands against a constant external pressure of 1 atm from a volume of 5 litres to a volume of 10 litres. In doing so it absorbs 400 J of thermal energy from its surroundings. Determine the change in internal energy of system. [Ans: -106.5 J]
Answer:
Data : P = 1 atm = 1.013 × 105 Pa, V1 = 5 litres = 5 × 10-3 m3 V2 = 10 litres = 10 × 10-3 m3, Q = 400J.
The work done by the system (gas in this case) on its surroundings,
W = P(V2 – V1)
= (1.013 × 105 Pa) (10 × 10-3 m3 – 5 × 10-3 m3)
= 1.013 (5 × 102)J = 5.065 × 102J
The change in the internal energy of the system, ∆u = Q – W = 400J – 506.5J = -106.5J
The minus sign shows that there is a decrease in the internal energy of the system.

Question 5.
A system releases 130 kJ of heat while 109 kJ of work is done on the system. Calculate the change in internal energy.
[Ans: ∆U = 21 kJ]
Answer:
Data : Q = -130kj, W= – 109kJ
∆u = Q – W = – 130kJ – ( – 109kJ)
= (-130 + 104) kJ = – 26 kj.
This is the change (decrease) in the internal energy.

Question 6.
Efficiency of a Carnot cycle is 75%. If temperature of the hot reservoir is 727ºC, calculate the temperature of the cold reservoir. [Ans: 23ºC]
Data : η = 75% = 0.75, TH = (273 + 727) K = 1000 K
η = 1 – \(\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}\) ∴ \(\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}\) = 1 – η
∴ TC = TC(1 – η) = 1000 K (1 – 0.75)
= 250K = (250 – 273)°C
= -23 °C
This is the temperature of the cold reservoir.

Maharashtra Board Class 12 Physics Solutions Chapter 4 Thermodynamics

Question 7.
A Carnot refrigerator operates between 250K and 300K. Calculate its coefficient of performance. [Ans: 5]
Answer:
Data : TC = 250 K, TH = 300 K
K = \(\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}-T_{\mathrm{C}}}\) = \(\frac{250 \mathrm{~K}}{300 \mathrm{~K}-250 \mathrm{~K}}\) = \(\frac{250}{50}\) = 5
This is the coefficient of performance of the refrigerator.

Question 8.
An ideal gas is taken through an isothermal process. If it does 2000 J of work on its environment, how much heat is added to it? [Ans: 2000J]
Answer:
Data : W = 2000 J, isothermal process
In this case, the change in the internal energy of the gas, ∆u, is zero as the gas is taken through an isothermal process.
Hence, the heat added to it,
Q = ∆ u + W = 0 + W = 200J

Question 9.
An ideal monatomic gas is adiabatically compressed so that its final temperature is twice its initial temperature. What is the ratio of the final pressure to its initial pressure? [Ans: 5.656]
Answer:
Data : Tf = 2Ti, monatomic gas ∴ γ = 5/3
Maharashtra Board Class 12 Physics Solutions Chapter 4 Thermodynamics 16
Maharashtra Board Class 12 Physics Solutions Chapter 4 Thermodynamics 17
This is the ratio of the final pressure (Pf) to the initial pressure (Pi).

Question 10.
A hypothetical thermodynamic cycle is shown in the figure. Calculate the work done in 25 cycles.
Maharashtra Board Class 12 Physics Solutions Chapter 4 Thermodynamics 18
[Ans: 7.855 × 104 J]
Answer:
Maharashtra Board Class 12 Physics Solutions Chapter 4 Thermodynamics 19
Maharashtra Board Class 12 Physics Solutions Chapter 4 Thermodynamics 20
25 cycles
The work done in one cycle, \(\oint\)PdV
= πab = (3.142) (2 × 10-3 m3) (5 × 105 Pa)
= 3.142 × 103J
Hence, the work done in 25 cycles
= (25) (3.142 × 103 J) = 7.855 × 104J

Question 11.
The figure shows the V-T diagram for one cycle of a hypothetical heat engine which uses the ideal gas. Draw the p-V diagram and P-V diagram of the system. [Ans: (a)]
Maharashtra Board Class 12 Physics Solutions Chapter 4 Thermodynamics 21
[Ans: (b)]
Answer:
(a) P-V diagram (Schematic)
Maharashtra Board Class 12 Physics Solutions Chapter 4 Thermodynamics 22
ab: isobaric process,
bc : isothermal process,
cd : isobaric process,
da : isothermal process
\(\frac{P_{\mathrm{a}} V_{\mathrm{a}}}{T_{\mathrm{a}}}\) = \(\frac{P_{\mathrm{b}} V_{\mathrm{b}}}{T_{\mathrm{b}}}\) = \(\frac{P_{\mathrm{c}} V_{\mathrm{c}}}{T_{\mathrm{c}}}\) = \(\frac{P_{\mathrm{d}} V_{\mathrm{d}}}{T_{\mathrm{d}}}\) = nR

(b) P—T diagram (Schematic)
Maharashtra Board Class 12 Physics Solutions Chapter 4 Thermodynamics 23

Question 12.
A system is taken to its final state from initial state in hypothetical paths as shown figure. Calculate the work done in each case.
Maharashtra Board Class 12 Physics Solutions Chapter 4 Thermodynamics 24
[Ans: AB = 2.4 × 106 J, CD = -8 × 105 J, BC and DA zero, because constant volume change]
Answer:
Maharashtra Board Class 12 Physics Solutions Chapter 4 Thermodynamics 25
Data: PA = PB = 6 × 105 Pa, PC = PD = 2 × 105 Pa VA = VD = 2 L, VB = VC6 L, 1 L = 10-3m3
(i) The work done along the path A → B (isobaric process), WAB = PA (VB – VA) = (6 × 105 Pa)(6 – 2)(10-3 m3) = 2.4 × 103 J
(ii) WBC = zero as the process is isochoric (V = constant).
(iii) The work done along the path C → D (isobaric process), WCD = PC (VD – VC)
= (2 × 105 Pa) (2 – 6) (10-3m3) = -8 × 102J
(iv) WDA = zero as V = constant.

12th Physics Digest Chapter 4 Thermodynamics Intext Questions and Answers

Can you tell? (Textbook Page No. 76)

Question 1.
Why is it that different objects kept on a table at room temperature do not exchange heat with the table ?
Answer:
The objects do exchange heat with the table but there is no net transfer of energy (heat) as the objects and the table are at the same temperature.

Can you tell? (Textbook Page No. 77)

Question 1.
Why is it necessary to make a physical contact between a thermocouple and the object for measuring its temperature ?
Answer:
For heat transfer to develop thermoemf.

Can you tell? (Textbook Page No. 81)

Question 1.
Can you explain the thermodynamics involved in cooking food using a pressure cooker ?
Answer:
Basically, heat is supplied by the burning fuel causing increase in the internal energy of the food (system), and the system does some work on its surroundings. In the absence of any data about the components of food and their thermal and chemical properties, we cannot evaluate changes in internal energy and work done.

Maharashtra Board Class 12 Physics Solutions Chapter 4 Thermodynamics

Use your brain power (Textbook Page No. 85)

Question 1.
Verify that the area under the P-V curve has dimensions of work.
Answer:
Area under the P-V curve is \(\int_{V_{\mathrm{i}}}^{V_{\mathrm{f}}} P d V\), where P is the pressure and V is the volume.
Maharashtra Board Class 12 Physics Solutions Chapter 4 Thermodynamics 80

Use your brain power (Textbook Page No. 91)

Question 1.
Show that the isothermal work may also be expressed as W = nRT ln\(\left(\frac{\boldsymbol{P}_{\mathrm{i}}}{\boldsymbol{P}_{\mathrm{f}}}\right)\),
Answer:
In the usual notation,
W = nRT In \(\frac{V_{\mathrm{f}}}{V_{\mathrm{i}}}\) and Pi Vi = Pf Vf = nRT in an isothermal process
∴ \(\frac{V_{\mathrm{f}}}{V_{\mathrm{i}}}\) = \(\frac{P_{\mathrm{i}}}{P_{\mathrm{f}}}\) and W = nRT ln \(\frac{P_{\mathrm{i}}}{P_{\mathrm{f}}}\)

Use your brain power (Textbook Page No. 94)

Question 1.
Why is the P-V curve for an adiabatic process steeper than that for an isothermal process ?
Answer:
Adiabatic process : PVγ = constant
∴ VγdP + γPVγ-1 dV = 0
∴ \(\frac{d P}{d V}\) = – \(\frac{\gamma P}{V}\)
Isothermal process : PV = constant
∴ pdV + VdP = 0 ∴ \(\frac{d P}{d V}\) = \(\frac{P}{-V}\)
Now, γ > 1
Maharashtra Board Class 12 Physics Solutions Chapter 4 Thermodynamics 30
∴ \(\frac{d P}{d V}\) is the slope of the P – V curve.
∴The P – V curve for an adiabatic process is steeper than that for an isothermal process.

Question 2.
Explain formation of clouds at high altitude.
Answer:
As the temperature of the earth increases due to absorption of solar radiation, water from rivers, lakes, oceans, etc. evaporates and rises to high altitude. Water vapour forms clouds as water molecules come together under appropriate conditions. Clouds are condensed water vapour and are of various type, names as cumulus clouds, nimbostratus clouds, stratus clouds and high-flying cirrus clouds.

Can you tell? (Textbook Page No. 95)

Question 1.
How would you interpret Eq. 4.21 (Q = W) for a cyclic process ?
Answer:
It means ∆ u = 0 for a cyclic process as the system returns to its initial state.

Question 2.
An engine works at 5000 rpm, and it performs 1000 J of work in one cycle. If the engine runs for 10 min, how much total work is done by the engine ?
Answer:
The total work done by the engine = (1000 J/cycle) (5000 cycles/min) (10 min) = 5 × 107 J.

Do you know? (Textbook Page No. 101)

Question 1.
Capacity of an air conditioner is expressed in tonne. Do you know why?
Answer:
Before refrigerator and AC were invented, cooling was done by using blocks of ice. When cooling machines were invented, their capacity was expressed in terms of the equivalent amount of ice melted in a day (24 hours). The same term is used even today.
(Note : 1 tonne = 1000 kg = 2204.6 pounds, 1 ton (British) = 2240 pounds = 1016.046909 kg, 1 ton (US) = 2000 pounds = 907.184 kg.]

Maharashtra Board Class 12 Physics Solutions Chapter 4 Thermodynamics

Use your brain power (Textbook Page No. 105)

Question 1.
Suggest a practical way to increase the efficiency of a heat engine.
Answer:
The efficiency of a heat engine can be increased by choosing the hot reservoir at very high temperature and cold reservoir at very low temperature.

Maharashtra Board Class 12 Physics Solutions Chapter 14 Dual Nature of Radiation and Matter

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 14 Dual Nature of Radiation and Matter Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 14 Dual Nature of Radiation and Matter

1. Choose the correct answer.

i) A photocell is used to automatically switch on the street lights in the evening when the sunlight is low in intensity. Thus it has to work with visible light. The material of the cathode of the photocell is
(A) zinc
(B) aluminum
(C) nickel
(D) potassium
Answer:
(D) potassium

ii) Polychromatic (containing many different frequencies) radiation is used in an experiment on the photoelectric effect. The stopping potential
(A) will depend on the average wavelength
(B) will depend on the longest wavelength
(C) will depend on the shortest wavelength
(D) does not depend on the wavelength
Answer:
(C) will depend on the shortest wavelength

Maharashtra Board Class 12 Physics Solutions Chapter 14 Dual Nature of Radiation and Matter

iii) An electron, a proton, an α-particle and a hydrogen atom are moving with the same kinetic energy. The associated de Broglie wavelength will be longest for
(A) electron
(B) proton
(C) α-particle
(D) hydrogen atom
Answer:
(A) electron

iv) If NRed and NBlue are the number of photons emitted by the respective sources of equal power and equal dimensions in unit time, then
(A) NRed < NBlue
(B) NRed = NBlue
(C) NRed > NBlue
(D) NRed ≈ NBlue
Answer:
(C) NRed > NBlue

v) The equation E = pc is valid
(A) for all sub-atomic particles
(B) is valid for an electron but not for a photon
(C) is valid for a photon but not for an electron
(D) is valid for both an electron and a photon
Answer:
(C) is valid for a photon but not for an electron

2. Answer in brief.

i) What is photoelectric effect?
Answer:
The phenomenon of emission of electrons from a metal surface when electromagnetic radiation of appropriate frequency is incident on it is known as photoelectric effect.

ii) Can microwaves be used in the experiment on photoelectric effect?
Answer:
No

iii) Is it always possible to see photoelectric effect with red light?
Answer:
No

iv) Using the values of work function given in Table 14.1, tell which metal will require the highest frequency of incident radiation to generate photocurrent.
Maharashtra Board Class 12 Physics Solutions Chapter 14 Dual Nature of Radiation and Matter 2
Answer:
Gold.
[ Note : W0 = hv0, where h is Planck’s constant. The larger the work function (W0), the higher is the threshold frequency (v0). ]

v) What do you understand by the term wave-particle duality? Where does it apply?
Answer:
Depending upon experimental conditions or structure of matter, electromagnetic radiation and material particles exhibit wave nature or particle nature. This is known as wave-particle duality.

It applies to all phenomena. The wave nature and particle nature are liked by the de Broglie relation λ = h/p, where λ is the wavelength of matter waves, also called de Broglie waves / Schrodinger waves, p is the magnitude of the momentum of a particle or quantum of radiation and h is the universal constant called Planck’s constant.

[Note : It is the smallness of h (= 6.63 × 10-34 J∙s) that is very significant in wave-particle duality.]

Question 3.
Explain the inverse linear dependence of stopping potential on the incident wavelength in a photoelectric effect experiment.
Answer:
We have V0e = \(\frac{h c}{\lambda}\) – Φ, where V0 is the stopping potential, e is the magnitude of the charge on the electron, h is Planck’s constant, c is the speed of light in free space, λ is the wavelength of the electromagnetic radiation incident on a metal surface and Φ is the work function for the metal, h, c and e are constants. Φ is constant for a particular metal.
Hence, it follows that as \(\frac{1}{\lambda}\) increases, V0 increases.
The plot of V0 verses \(\frac{1}{\lambda}\) is linear. This is because the energy associated with a quantum of radiation (photon) is directly proportional to the frequency of radiation and hence inversely proportional to the wavelength of radiation.

Maharashtra Board Class 12 Physics Solutions Chapter 14 Dual Nature of Radiation and Matter

Question 4.
It is observed in an experiment on photoelectric effect that an increase in the intensity of the incident radiation does not change the maximum kinetic energy of the electrons. Where does the extra energy of the incident radiation go? Is it lost? State your answer with explanatory reasoning.
Answer:
When electromagnetic radiation with frequency greater than the threshold frequency is incident on a metal surface, there is emission of electrons. It is observed that not every incident photon is effective in liberating an electron. In fact, the number of electrons emitted per second is far less than the number of photons incident per second. The photons that are not effective in liberation of electrons are reflected (or scattered) or absorbed resulting in rise in the temperature of the metal surface. The maximum kinetic energy of a photoelectron depends on the frequency of the incident radiation and the threshold frequency for the metal. It has nothing to do with the intensity of the incident radiation. The increase in intensity results in increase in the number of electrons emitted per second.

Question 5.
Explain what do you understand by the de Broglie wavelength of an electron. Will an electron at rest have an associated de Broglie wavelength? Justify your answer.
Answer:
Under certain conditions an electron exhibits wave nature. Waves associated with a moving electron are called matter waves or de Broglie waves or- Schrodinger waves. The de Broglie wavelength of these matter waves is given by X = h/p, where h is Planck’s constant and p is the magnitude of the momentum of the electron.

If an electron is at rest, its momentum would be zero, and hence the corresponding de Broglie wavelength would be infinite indicating absence of a matter wave. However, according to quantum mechanics/wave mechanics, this is not possible.

Question 6.
State the importance of Davisson and Germer experiment.
Answer:
The Davisson and Germer experiment directly indicated the wave nature of material particles and quantitatively verified the de Broglie hypothesis for the existence of matter waves.

[Note : The aim of the experiment was not to verify wave like properties of electrons. The realisation came only later, an example of serendipity.]

[Note : Like X-rays, electrons exhibit wave nature under suitable conditions. When the wavelength of matter waves associated with moving electrons is comparable to the inter-atomic spacing in a crystal, electrons show diffraction effects. In 1927, Sir George Thomson (1892 – 1975), British physicist, with his student Alex Reid, observed electron diffraction with a metal foil. It is found that neutrons, atoms, molecules, Œ-particles, etc. show wave nature under suitable conditions.]

Question 7.
What will be the energy of each photon in monochromatic light of frequency 5 × 1014 Hz?
Answer:
Data: y = 5 × 1014 Hz, h = 6.63 × 10-34 Js,
1eV=1.6 × 10-19 J
The energy of each photon,
E = hv = (6.63 × 10-34 J.s)(5 × 1014 Hz)
= 3.315 × 10-19 J
= \(\frac{3.315 \times 10^{-19} \mathrm{~J}}{1.6 \times 10^{-19} \mathrm{~J} / \mathrm{eV}}\) = 2.072 eV

Maharashtra Board Class 12 Physics Solutions Chapter 14 Dual Nature of Radiation and Matter

Question 8.
Observations from an experiment on photoelectric effect for the stopping potential by varying the incident frequency were plotted. The slope of the linear curve was found to be approximately 4.1 × 10-15 V s. Given that the charge of an electron is 1.6 × 10-19 C, find the value of the Planck’s constant h.
Answer:
Data : Slope=4.1 × 10-15 V∙s, e = 1.6 ×10-19 C
V0e = hv – hv0
∴ V0 =\(\left(\frac{h}{e}\right) v-\left(\frac{h v_{0}}{e}\right)\)
∴ Slope = \(\frac{h}{e}\) ∴ Planck’s constant,
h = (slope) (e)=(4.1 × 10-15 V∙s)(1.6 × 10-19 C)
= 6.56 × 10 34J. (as 1 V = \(\frac{1 \mathrm{~J}}{1 \mathrm{C}}\))

Question 9.
The threshold wavelength of tungsten is 2.76 × 10-5 cm. (a) Explain why no photoelectrons are emitted when the wavelength is more than 2.76 × 10-5 cm.(b) What will be the maximum kinetic energy of electrons ejected in each of the following cases
(i) if ultraviolet radiation of wavelength λ = 1.80 × 10-5 cm and
(ii) radiation of frequency 4 × 1015 Hz is made incident on the tungsten surface.
Answer:
Data: λ0 = 2.76 × 10-5 cm = 2.76 × 10-7 m,
λ =1.80 × 10-5 cm = 1.80 × 10-7 m,
v = 4 × 1015 Hz, h = 6.63 × 10-34 J∙s,c = 3 × 108 m/s
(a) For λ > λ0, v < v0 (threshold frequency).
∴ hv < hv0. Hence, no photoelectrons are emitted.

(b) Maximum kinetic energy of electrons ejected
= hc\(\left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right)\)
=(6.63 × 10-34)(3 × 108)\(\left(\frac{10^{7}}{1.8}-\frac{10^{7}}{2.76}\right)\)J
= (6.63 × 10-19)(0.5555 – 0.3623)
= (6.63)(0.1932 × 10-19)J = 1.281 × 10-19 J
= \(\frac{1.281 \times 10^{-19} \mathrm{~J}}{1.6 \times 10^{-19} \mathrm{~J} / \mathrm{eV}}\) = 0.8006 eV

(c) Maximum kinetic energy of electrons ejected
= hv – \(\frac{h c}{\lambda_{0}}\)
=(6.63 × 10-34(4 × 1015) – \(\frac{\left(6.63 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{2.76 \times 10^{-7}}\)
= 26.52 × 10-19 – 7.207 × 10-19
= 19.313 × 10-19 J
= \(\frac{19.313 \times 10^{-19} \mathrm{~J}}{1.6 \times 10^{-19} \mathrm{~J} / \mathrm{eV}}\) = 12.07eV

Question 10.
Photocurrent recorded in the micro ammeter in an experimental set-up of photoelectric effect vanishes when the retarding potential is more than 0.8 V if the wavelength of incident radiation is 4950 Å. If the source of incident radiation is changed, the stopping potential turns out to be 1.2 V. Find the work function of the cathode material and the wavelength of the second source.
Answer:
Data: V0 = 0.8 V, λ = 4950 Å = 4.950 × 10-7 m,
V0‘ = 1.2V, h = 6.63 × 10-34 J∙s, c = 3 × 108 m/s.
(i) V0e = hv – Φ = \(\frac{h c}{\lambda}\) – Φ
∴ The work function of the cathode material,
Maharashtra Board Class 12 Physics Solutions Chapter 14 Dual Nature of Radiation and Matter 3

Maharashtra Board Class 12 Physics Solutions Chapter 14 Dual Nature of Radiation and Matter

Question 11.
Radiation of wavelength 4500 Å is incident on a metal having work function 2.0 eV. Due to the presence of a magnetic field B, the most energetic photoelectrons emitted in a direction perpendicular to the field move along a circular path of radius 20 cm. What is the value of the magnetic field B?
Answer:
Data: λ = 4500Å = 4.5 × 10-7 m,
Φ = 2.0eV = 2 × 1.6 × 10-19 J = 3.2 × 10-19 J,
h = 6.63 × 10-34 J∙s, c = 3 × 108 m/s,
r = 20 cm = 0.2 m, e= 1.6 × 10-19 C,
m = 9.1 × 10-31kg
Maharashtra Board Class 12 Physics Solutions Chapter 14 Dual Nature of Radiation and Matter 4
This is the value of the magnetic field.

Question 12.
Given the following data for incident wavelength and the stopping potential obtained from an experiment on photoelectric effect, estimate the value of Planck’s constant and the work function of the cathode material. What is the threshold frequency and corresponding wavelength? What is the most likely metal used for emitter?
Maharashtra Board Class 12 Physics Solutions Chapter 14 Dual Nature of Radiation and Matter 1
Answer:
Data: λ = 2536Å = 2.536 × 10-7 m,
λ’ = 3650Å = 3.650 ×10-7 m, V0 = 1.95V, V0‘ = 0.5V,
c = 3 × 108 mIs, e = 1.6 × 10-19 C

(i) V0e = \(\frac{h c}{\lambda}\) – Φ and V0‘e =\(\frac{h c}{\lambda^{\prime}}\) – Φ
∴ (V0 – V0‘)e = hc\(\left(\frac{1}{\lambda}-\frac{1}{\lambda^{\prime}}\right)\)
∴ (1.95 – 0.5(1.6 × 10-19)
= h (3 × 108\(\left(\frac{10^{7}}{2.536}-\frac{10^{7}}{3.650}\right)\)
∴ 2.32 × 10-19 = h(3 × 1015)(0.3943 – 0.2740)
∴ h = \(\frac{2.32 \times 10^{-34}}{0.3609}\) = 6.428 × 10-34 J∙s
This is the value of Planck’s constant.

(ii) Φ = \(\frac{h c}{\lambda}\) – V0e
Maharashtra Board Class 12 Physics Solutions Chapter 14 Dual Nature of Radiation and Matter 5
This is the work function of the cathode material.

(iii) Φ = hv0
∴ The threshold frequency, v0 = \(\frac{\phi}{h}\)
= \(\frac{4.484 \times 10^{-19} \mathrm{~J}}{6.428 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}}\) = 6.976 × 1014 Hz

(iv) v0 = \(\frac{c}{\lambda_{\mathrm{o}}}\) ∴ The threshold frequency, λ0 = \(\frac{c}{v_{\mathrm{o}}}\)
= \(\frac{3 \times 10^{8}}{6.976 \times 10^{14}}\) = 4.300 × 10-7 m = 4300 Å

(v) The most likely metal used for emitter : calcium

Question 13.
Calculate the wavelength associated with an electron, its momentum and speed
(a) when it is accelerated through a potential of 54 V
Answer:
Data : V = 54 V, m = 9.1 × 10-31 kg, e
e = 1.6 × 10-19 C, h = 6.63 × 10-34 J.s, KE = 150 eV
(a) We assume that the electron is initially at rest.
∴ Ve = \(\frac{1}{2}\)mv2
∴ v = \(\sqrt{\frac{2 V e}{m}}=\sqrt{\frac{2(54)\left(1.6 \times 10^{-19}\right)}{9.1 \times 10^{-31}}}\)
= \(\sqrt{19 \times 10^{12}}\) = 4.359 × 106 m/5
This is the speed of the electron.
p = mv = (9.1× 10-31)(4.359 × 106)
= 3.967 × 10-24 kg∙m/s
This is the momentum of the electron. The wavelength associated with the electron,
λ = \(\frac{h}{p}=\frac{6.63 \times 10^{-34}}{3.967 \times 10^{-24}}\) = 1.671 × 10-10 m
= 1.671 Å = 0.1671 nm

(b) when it is moving with kinetic energy of 150 eV.
Answer:
As KE ∝ \(\sqrt{V}\), we get
\(\frac{v^{\prime}}{v}=\sqrt{\frac{150}{54}}\) = 1.666
∴ v’ = 1.666v = (1.666)(4.356 × 106)
= 7.262 × 106 m/s
This is the speed of the electron.
p’ = mv’’=(9.1 × 10-31)(7.262 × 106)
= 6.608 × 10-24 kg∙m/s
This is the momentum of the electron. The
wavelength associated with the electron,
λ = \(\frac{h}{p^{\prime}}=\frac{6.63 \times 10^{-34}}{6.608 \times 10^{-24}} \) = 1.003 × 10-10 m
= 1.003 Å = 0.1003 nm

Question 14.
The de Broglie wavelengths associated with an electron and a proton are same. What will be the ratio of
(i) their momenta
(ii) their kinetic energies?
Answer:
Data : λ (electron) = λ (proton)
m (proton) = 1836 m (electron)
(i) λ = \(\frac{h}{p}\) As λ (electron) = λ (proton),
\(\frac{p(\text { electron })}{p \text { (proton) }}\) = 1, where p denotes the magnitude of momentum.

(ii) Assuming v «c,
KE = \(\frac{1}{2}\)mv2 = \(\frac{1}{2} \frac{m^{2} v^{2}}{m}=\frac{p^{2}}{2 m}\)
∴ \(\frac{\mathrm{KE} \text { (electron) }}{\mathrm{KE} \text { (proton) }}=\frac{m \text { (proton) }}{m \text { (electron) }}\) = 1836 as p is the same for the electron and the proton.

Maharashtra Board Class 12 Physics Solutions Chapter 14 Dual Nature of Radiation and Matter

Question 15.
Two particles have the same de Broglie wavelength and one is moving four times as fast as the other. If the slower particle
is an α-particle, what are the possibilities for the other particle?
Answer:
Data : λ1 = λ2, v1 = 4v2
λ = \(\frac{h}{p}=\frac{h}{m v}\) ∴ λ1 = \(\frac{h}{m_{1} v_{1}}\), λ2 = \(\frac{h}{m_{2} v_{2}}\)
∴ m1 = m2 \(\frac{v_{2}}{v_{1}}\) = m2\(\left(\frac{1}{4}\right)=\frac{m_{2}}{4}\)
As particle 2 is the a-particle, particle 1 (having the mass \(\frac{1}{4}\) times that of the a-particle) may be a proton or neutron.

Question 16.
What is the speed of a proton having de Broglie wavelength of 0.08 Å?
Answer:
Data : λ = 0.08 Å = 8 × 10-12m, h = 6.63 × 10-34 J∙s, m = 1.672 × 10-27 kg
λ = \(\frac{h}{m v}\) ∴ v = \(\frac{h}{\lambda m}=\frac{6.63 \times 10^{-34}}{\left(8 \times 10^{-12}\right)\left(1.672 \times 10^{-27}\right)}\)
∴ v = 4.957 × 104 m/s
This is the speed of the proton.

Question 17.
In nuclear reactors, neutrons travel with energies of 5 × 10-21 J. Find their speed and wavelength.
Answer:
Data : KE = 5 × 10-21 J, m = 1.675 × 10-27 kg, h = 6.63 × 10-34 J∙s
KE = \(\frac{1}{2}\) mv2 = 5 × 10-21 J
∴ v = \(\sqrt{\frac{2 \mathrm{KE}}{m}}=\sqrt{\frac{(2)\left(5 \times 10^{-21}\right)}{1.675 \times 10^{-27}}}\)
= 2.443 × 103 m/s
This is the speed of the neutrons. The de Broglie wavelength associated with the neutron,
λ = \(\frac{h}{m v}=\frac{6.63 \times 10^{-34}}{\left(1.675 \times 10^{-27}\right)\left(2.443 \times 10^{3}\right)}\)
= 1.620 × 10-10 m = 1.620 Å

Maharashtra Board Class 12 Physics Solutions Chapter 14 Dual Nature of Radiation and Matter

Question 18.
Find the ratio of the de Broglie wavelengths of an electron and a proton when both are moving with the (a) same speed, (b) same energy and (c) same momentum? State which of the two will have the longer wavelength in each case?
Answer:
Data: mp = 1836 me
Maharashtra Board Class 12 Physics Solutions Chapter 14 Dual Nature of Radiation and Matter 6

12th Physics Digest Chapter 14 Dual Nature of Radiation and Matter Intext Questions and Answers

Remember This (Textbook Page No. 316)

Question 1.
Is solar cell a photocell?
Answer:
Yes

Remember This (Textbook Page No. 317)

Question 1.
Can you estimate the de Broglie wavelength of the Earth?
Answer:
Taking the mass of the Earth as (about) 6 × 1024 kg, and the linear speed of the earth around the Sun as (about) 3 × 104 m/s, we have, the de Brogue wave length of the Earth as
λ = \(\frac{h}{p}=\frac{h}{M v}=\frac{6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}}{\left(6 \times 10^{24} \mathrm{~kg}\right)\left(3 \times 10^{4} \mathrm{~m} / \mathrm{s}\right)}\)
= 3.683 × 10-63 m (extremely small)

Question 2.
The expression p = E/c defines the momentum of a photon. Can this expression be used for momentum of an electron or proton?
Answer:
No

Remember This (Textbook Page No. 319)

Diffraction results described above can be produced in the laboratory using an electron diffraction tube as shown in figure. It has a filament which on heating produces electrons. This filament acts as a cathode. Electrons are accelerated to quite high speeds by creating large potential difference between the cathode and a positive electrode. On its way, the beam of electrons comes across a thin sheet of
Maharashtra Board Class 12 Physics Solutions Chapter 14 Dual Nature of Radiation and Matter 7
graphite. The electrons are diffracted by the atomic layers in the graphite and form diffraction rings on the phosphor screen. By changing the voltage between the cathode and anode, the energy, and therefore the speed, of the electrons can be changed. This will change the wavelength of the electrons and a change will be seen in the diffraction pattern. By increasing the voltage, the radius of the diffraction rings will decrease. Try to explain why?
Answer:
When the accelerating voltage is increased, the kinetic energy and hence the momentum of the electron increases. This decreases the de Brogue wavelength of the electron. Hence, the radius of the diffraction ring decreases.

Remember This (Textbook Page No. 320)

Question 1.
On what scale or under which circumstances are the wave nature of matter apparent?
Answer:
When the de Brogue wavelength of a particle such as an electron, atom, or molecule is comparable to the interatomic spacing in a crystal, the wave nature of matter is revealed in diffraction/interference.

Maharashtra Board Class 12 English Yuvakbharati Solutions Chapter 4.4 The Sign of Four

Balbharti Yuvakbharati English 12th Digest Chapter 4.4 The Sign of Four Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 12 English Yuvakbharati Solutions Chapter 4.4 The Sign of Four

12th English Digest Chapter 4.4 The Sign of Four Textbook Questions and Answers

CHARACTER:

(A1)

Question (i)
Read the extract again and complete the web by highlighting the qualities of the following characters:
Maharashtra Board Class 12 English Yuvakbharati Solutions Chapter 4.4 The Sign of Four 1
Maharashtra Board Class 12 English Yuvakbharati Solutions Chapter 4.4 The Sign of Four 2
Answer:
Maharashtra Board Class 12 English Yuvakbharati Solutions Chapter 4.4 The Sign of Four 3
Maharashtra Board Class 12 English Yuvakbharati Solutions Chapter 4.4 The Sign of Four 4

Maharashtra Board Class 12 English Yuvakbharati Solutions Chapter 4.4 The Sign of Four

Question (ii)
Describe the character of Mary Morstan from Dr. Watson’s point of view.
Answer:
From Dr. Watson’s point of view: When I first saw Mary, she was dressed simply but tastefully. I could see that she was a person of limited means. Her expression was sweet and pleasant, and I could make out that her nature was refined and sensitive. My calculations told me that she was about 27 years old. She was agitated by the mystery surrounding her life. I found her attractive, though her face did not have regular features or a beautiful complexion. Her eyes showed that she was a sympathetic person. I was much impressed by her and attracted to her.

Question (iii)
Sherlock Holmes is the leading character in the extract. Explain.
Answer:
It is Sherlock Holmes who is the detective and the leading character. Mary Morstan had come to ask his advice about a problem that she was facing. Holmes was the one who took the lead and found out about Major Sholto; it was Holmes who analysed the handwriting in the letter that Mary had received. Holmes was sharp, accurate, intelligent and methodical. He had an excellent record of solving cases, and his deductions were always correct. Watson was merely his friend who helped him and kept a record of his cases.

Question (iv)
Dr. Watson, the narrator, is one of the major characters in the novel. Illustrate.
Answer:
Dr. Watson is the narrator. He was present when the case was brought to Holmes by Mary Morstan. He is generally always with Holmes, helping him to solve cases. He accompanied Holmes whenever necessary.

He also kept a record of all the cases that Holmes was a part of. In this extract, he is present when Mary recounts her case, and he accompanies Holmes and Mary to meet the writer of the anonymous letter. (He marries Mary in the end.)

Question (v)
Holmes is always one step ahead of Dr. Watson in solving cases. Elucidate.
Answer:
Where Watson is emotional, simple and trustful, Holmes is sharp, objective and methodical. Holmes is also analytical and notices the little details which give him clues to solving a case. Watson does not, and hence is often on the wrong track. Holmes is the real detective, while Watson is merely his companion. Holmes is always ahead of Watson and solves cases which Watson is not even near to cracking.

PLOT:

(A2)

Question (i)
Arrange the sentences in correct sequence as per their occurrence in the extract.
Answer:

Jumbled Incidents Correct Sequence
1. Holmes put a revolver in his pocket. (a) Mary Morstan was a well-dressed young lady.
2. Holmes gave Winwood’s book ‘Martyrdom of Man’ to Dr. Watson. (b) Mary’s father was an officer in an Indian regiment.
3. Mary received a large and lustrous pearl through the post. (c) Mary received a large and lustrous pearl through the post.
4. Mary’s father was an officer in an Indian regiment. (d) Holmes gave Winwood’s book ‘Martyrdom of Man’ to Dr. Watson.
5. Mary Morstan was a well-dressed young lady. (e) Holmes put a revolver in his pocket.

Maharashtra Board Class 12 English Yuvakbharati Solutions Chapter 4.4 The Sign of Four

(ii) Discuss the importance of the following statements from the light of the extract.

Question (a)
The trio-Holmes, Dr. Watson and Mary decide to visit Lyceum Theatre.
Answer:
Mary had received an anonymous letter asking her to be outside the Lyceum Theatre on a particular night at seven o’clock. The letter said that it would be to Mary’s advantage if she came. The letter also mentioned that she could bring two friends with her. However, she did not have any friends who could accompany her, and so she asked Holmes and Watson if they could do so. They agreed. Hence, Holmes, Dr. Watson and Mary decide to visit Lyceum Theatre. This was the first step to solving the case.

Question (b)
Mary received pearls every year on the same day.
Answer:
Major Sholto, Mary’s father’s friend, had cheated Mary’s father of his share in the Agra treasure. When he died, Major Sholto informed his son Thaddeus of this. Though Thaddeus did not have the treasure, he tried to rectify the matter to a certain extent by sending Mary a rare and expensive pearl every year, on the same day as he sent the first one.

Question (c)
Holmes carefully examined the paper given by Mary.
Answer:
Mary had found a curious paper in her father’s desk which no one could understand. Holmes deduced from the colour of the paper that it was an important document. He felt it was related in some way to the mystery on hand. Hence, he examined it carefully to get some clues which would help to solve the mystery.

SETTING:

(A3)

Question (i)
Cite various references (lines) from the extract that tell us about the time and period of the events:
Answer:

Lines Time and period
1. He disappeared upon the 3rd of December, 1878. – nearly ten years ago. Mary’s father had disappeared about ten years before she met Holmes and Watson on a particular day.
2. About six years ago – to be exact, upon the 4th of May, 1882 – an advertisement appeared in the Times asking for the address of Miss Mary Morstan. The same day there arrived through the post a small card-board box addressed to me, which I found to contain a very large and lustrous pearl. Mary first received an expensive and rare pearl six years before she received an anonymous letter/before she came to meet Holmes.
3. This morning I received this letter, which you will perhaps read for yourself. Mary receives an anonymous letter on the morning of the day on which she consults Holmes.
4. Major Sholto, of Upper Norword, late of the 34th Bombay Infantry, died upon the 28th of April, 1882. Within a week of his death Captain Morstan’s daughter receives a valuable present, which is repeated from year to year. Mary begins to receive the pearls immediately after Major Sholto’s death.
5. At the Lyceum Theatre the crowds were already thick at the side-entrances. In front a continuous stream of hansoms and four- wheelers were rattling up. Holmes, Watson and Mary reach the Lyceum Theatre on the evening of the day Mary receives the anonymous letter, as instructed by the writer of the letter. This was in the year 1888.
6. We had hardly done so before the driver whipped up his horse, and we plunged away at a furious pace through the foggy streets. This happens when Holmes, Watson and Mary are taken by the driver to meet the writer of the anonymous letter, on the evening when Mary receives it.
7. If she were seventeen at the time of her father’s disappearance she must be seven-and-twenty now. Watson, who is attracted to Mary, calculates that Mary must be twenty -seven years old in 1888 when she meets him and Holmes.
8. In the year 1878 my father, who was senior captain of his regiment, obtained twelve months’ leave and came home. This was the time, ten years earlier, when Captain Morstan disappeared.

Maharashtra Board Class 12 English Yuvakbharati Solutions Chapter 4.4 The Sign of Four

Question (ii)
Explain by citting references from the extract the ways the series of actions moves from London to India.
Answer:
The extract begins when Mary Morstan meets Sherlock Holmes at his house in London. They then meet Thaddeus Sholto in a rundown neighbourhood of London. Thaddeus reveals that his father Major Sholto had mistakenly killed Captain Morstan in London. They then go to Bartholomew Sholto’s house to get the treasure; however, Bartholomew is found dead.

Holmes follows Jonathan Small and Tonga, who have escaped by a steam launch, over the river Thames in London. When Small is captured, he tells them about the time he spent in India, where he was an accomplice in stealing the Agra treasure. Thus, the narration goes to India. Major Sholto and Captain Morstan were also at one time stationed in India.

Question (iii)
The extract begins when Mary Morstan meets Sherlock Holmes at his house. After that Holmes, Dr.Watson and Mary visit some places in London. Explain in detail the various places mentioned in the extract.
Answer:
Holmes, Dr.Watson and Mary were taken down the Strand, which was crowded, badly lit and humid. All kinds of people-sad, happy, old and young could be seen moving about in the dim light. Watson found it eerie and ghostlike, and he felt nervous and depressed. They then reached the Lyceum Theatre, where the crowds were pouring in.

A continuous stream of horse carriages could be seen, with stylish people getting out of them. Near the Lyceum Theatre they were met by a coachman who took them in his coach through Rochester Row and Vincent Square onto Vauxhall Bridge Road. They were on the Surrey side, on the bridge from where they got glimpses of the river Thames with lamps shining on the silent water.

The cab then took them through a maze of streets. Holmes could identify Wordsworth Road, Priory Road, Lark Hall Lane, Stockwell Place, Robert Street and Cold Harbor Lane. They were all rundown places. The cab took them further to a rather grim and shady neighbourhood with dull brick houses and cheap and showy public houses at the corner.

Holmes mentions that this was not a very fashionable or rich neighbourhood. This was followed by rows of two-storied villas each with a small front garden, and then again there were never-ending lines of new brick buildings, which were an extension of the city. The houses in the area were all dark and appeared uninhabited.

At last the cab drew up at the third house in a new terrace, which was also dark except for a light in the kitchen. However, when they knocked the door was opened instantly, and an Oriental figure of a servant clad in a yellow turban, white loose-fitting clothes, and a yellow sash stood there. It was strange to find an Oriental figure framed in the doorway of a cheap suburban house.

Question (iv)
Basically the setting of the extract is in London but it has some references of India, too. Explain how the settings of the extract contribute to the theme of the novel.
Answer:
The setting of the extract is in London, where Mary meets Holmes and Watson to explain her problem. She talks about her father being an officer in an Indian regiment. When he returned to England on leave, he called Mary to meet him at a London hotel, but disappeared mysteriously before she could do so. His only friend in London was a Major Sholto. Holmes finds that Major Sholto was also from the 34th Bombay Infantry.

Mary shows Holmes a piece of paper belonging to her father. The paper was of Indian origin, and three of the names written on it were also Indian. Holmes, Watson and Mary go to meet the anonymous letter writer at a rundown suburban house in London. Later they chase Jonathan Small and Tonga, who were trying to escape by boat on the river Thames. When Jonathan Small was captured, he spoke of being an accomplice in stealing the Agra treasure.

He was sent to the Andaman Islands, where Major Sholto and Captain Morstan were prison guards. At the end of the extract, the door of the anonymous letter writer’s house was opened by an Indian servant. His master used an Indian name to call him. Thus, we have a mingling of incidents both in London as well as in India, where the case had its roots.

Maharashtra Board Class 12 English Yuvakbharati Solutions Chapter 4.4 The Sign of Four

Question (v)
Describe in brief the importance of the following places in the extract.
(a) London
(b) Lyceum Theatre
(c) Edinburgh
(d) Agra
(e) Andaman Islands
Answer:
(a) London: The case starts here with Mary Morstan meeting Holmes at his place in London. They go to meet Thaddeus Sholto in London. They also chase Jonathan Small and Tonga in London. Tonga is killed and Small captured. Small then narrates the entire story.

(b) Lyceum Theatre: This is the place near which the writer of the anonymous letter told Mary Morstan to reach if she wished to get justice.

(c) Edinburgh: Mary spent her childhood till she was seventeen at a boarding school in Edinburgh.

(d) Agra: When Jonathan Small was standing guard one night at the Agra fortress, he was overpowered by two Sikh troopers, who forced him to waylay a servant of a Rajah and steal a valuable fortune in pearls and jewels. This was called the ‘Agra treasure’.

(e) Andaman Islands: Jonathan Small was arrested and imprisoned on the Andaman Islands for the robbery of the Agra treasure. After 20 years, Small made a deal with John Sholto and Arthur Morstan, who were the prison guards. Sholto would recover the treasure and in return send a boat to pick up Small and the Sikhs. Sholto double-crossed both Morstan and Small and stole the treasure for himself. Small vowed vengeance and four years later escaped from the Andaman Islands with an islander named Tonga after they both killed a prison guard.

Question (vi)
Complete:
Name the places/cities in India and England which are mentioned/have appeared in the extract. Describe their importance.
Answer:

India Importance London Importance
Andaman Islands Major Sholto and Captain Morstan were stationed here and in charge of the troops; Jonathan Small was also imprisoned here. Baker Street The residence of Sherlock Holmes and Dr. Watson. This was the place which Mary Morstan came to, to consult Holmes.
Agra Jonathan Small was a gatekeeper at the Agra fortress when he was forced to be an accomplice in the theft of the Rajah’s jewels. Langham Hotel This was the place Mary’s father stayed at when he came to London. He invited Mary to the hotel to meet him; but disappeared before her arrival.
Bombay (Now Mumbai) Major Sholto, and Captain Morstan were both from the regiment ‘the 34th Bombay Infantry’. Lyceum Theatre Mary was supposed to meet the writer of the anonymous letter or his messenger at the third pillar from the left outside the Lyceum Theatre.
River Thames Jonathan Small, who tried to escape by boat along the river Thames, was captured. His accomplice Tonga was killed.

Maharashtra Board Class 12 English Yuvakbharati Solutions Chapter 4.4 The Sign of Four

THEME:

(A4)

Question (i)
Write in brief the theme of the extract.
Answer:
The theme of the extract revolves round the mystery of the disappearance of Mary Morstan’s father, the receipt of expensive pearls by Mary and the mysterious letter received by her. It also involves the journey of Holmes, Watson and Mary Morstan to a strange house to meet the writer of the mysterious letter. The theme of the novel revolves around the Agra treasure.

Question (ii)
Write 4-5 sentences about the meeting of Miss Morstan with Holmes.
Answer:
Miss Morstan met Holmes and Watson at their house in Baker Street. She then discussed with them the mysterious disappearance of her father a few years earlier, the receipt of an expensive pearl every year for the past six years, and the receipt of a mysterious letter that morning asking her to meet the writer of the letter. Miss Morstan was intensely agitated and confused and did not know what to do. She showed Holmes the pearls, the boxes in which they had come and the letter. Then they planned to follow the instructions and meet the writer of the letter.

Question (iii)
Write the central idea of the given extract of the novel, “The Sign of Four”.
Answer:
The central idea is the meeting of Mary Morstan with Holmes and Watson, and her explanation of her problems. It is also about the short trip made by the three to meet I the writer of the mysterious letter. This is Watson’s first meeting with Miss Morstan and his attraction towards her.

Question (iv)
Complete the following giving reasons:
Answer:
(a) Miss Morstan plans to meet Sherlock Holmes to ask his advice about the disappearance of her father, the receipt of expensive pearls and the mysterious letter received by her.

(b) Miss Morstan gives the reference of Mrs. Cecil Forrester because Mrs. Cecil Forrester was her employer, whom Holmes had once helped to solve a domestic complication. Mrs. Forrester had been impressed by his kindness and skill.

(c) It’s a singular case because Miss Morstan’s father had come back to England and contacted her, and had seemed happy. After fixing a meeting ; with her at his hotel, he had suddenly ; disappeared and was never seen again, Even his only friend in town, Major Sholto, had not known either of his ; arrival or disappearance.

(d) Holmes needed some references to find out details about Major Sholto, who was the only friend Mary’s father had in England, and who had said that he did not know about his arrival in England.

(e) Miss Morstan received a pearl every year, when she replied to an advertisement asking for her address, adding that it would be to her advantage.

(f) The coachman confirmed that neither of Miss Morstan’s companion was a police officer because this was the condition made by the writer of the mysterious letter, whom they were going to meet.

Maharashtra Board Class 12 English Yuvakbharati Solutions Chapter 4.4 The Sign of Four

LANGUAGE:

(A5)

(i) Elaborate the following lines in the light of the novel/extract, “The Sign of Four”:

Question (a)
“You really are an automaton – a calculating machine”.
Answer:
These words are said by Watson to Holmes when Mary Morstan had left after discussing her case. Watson is attracted to her and full of admiration for her. When he voices his admiration, Holmes says that he had not noticed if she is attractive or not. Watson is indignant and calls him a calculating machine.

Question (b)
“The letter speaks of giving her justice.”
Discuss.
Answer:
These are the words of Holmes to Watson, when they are discussing the letter that Mary Morstan has received from an unknown person. He wondered what was the ‘justice’ that the letter spoke of, and who had done ’ something wrong to Mary that she now needed justice.

Question (c)
“Our quest does not appear to take us to very fashionable regions.”
Answer:
These words are said by Holmes to Watson and Mary Morstan, when they are being driven by the coachman to some strange place. They were going through narrow streets in an unfriendly and grim neighbourhood, which had dull brick rows of houses and cheap and showy public houses at the comer. Holmes mentions that this was not a very fashionable or rich neighbourhood.

Maharashtra Board Class 12 English Yuvakbharati Solutions Chapter 4.4 The Sign of Four

Question (ii)
Following are some dialogues of the major characters in the extract. Find out who the speaker is, his/her tone, style, significance, etc. of the dialogue.
Answer:

Dialogue Speaker To whom it is said Tone, Style, Significance etc.
1.  “… you have once enabled my employer, Mrs. Cecil Forrester, to unravel a little domestic omplication. She was much impressed by your kindness and skill.” Miss Morstan Sherlock Holmes Polite, cultured. Mary proves her identity, and how she came to know about Sherlock Holmes.
2. “You will, I am sure, excuse me.” Watson Miss Morstan and Sherlock Holmes Polite and courteous; Watson wants to make a good impression on Miss Morstan, and doesn’t want to poke his nose if he is not wanted.
3. “Your statement is most interesting. Has anything else occurred to you?” Sherlock Holmes Miss Morstan Polite tone, acknowledging the story told by Miss Morstan, and trying to get further information.
4. “Are you the parties who come with Miss Morstan?” A coachman/  messenger sent by the letter-writer. Sherlock Holmes and Watson Firm but respectful; cautious and asking for affirmation; shows that the person who had invited Miss Morstan was being very cautious, and checking them out.
5. “The Sahib awaits you.” Khitmutgar (a male servant) Miss Morstan, Sherlock Holmes and Watson Respectful, formal. Shows some connection with the east, especially India.