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NCERT Exemplar Class 12 Chemistry Chapter 3 Electrochemistry are part of NCERT Exemplar Class 12 Chemistry. Here we have given NCERT Exemplar Class 12 Chemistry Chapter 3 Electrochemistry. https://www.cbselabs.com/ncert-exemplar-problems-class-12-chemistry-electrochemistry/

NCERT Exemplar Class 12 Chemistry Chapter 3 Electrochemistry

Multiple Choice Questions

Single Correct Answer Type
Question 1. Which cell will measure standard electrode potential of copper electrode?

Solution: (c) When copper electrode is connected to standard hydrogen electrode, it acts as cathode and its standard electrode potential can be measured.

To calculate the standard electrode potential of the given cell it is coupled with the standard hydrogen electrode in which pressure of hydrogen gas is one bar and the cone, of H+ ion in the solution is one molar and also the concentrations of the oxidized and the reduced forms of the species in the right hand half-cell are unity.

Question 2. Electrode potential for Mg electrode varies according to the equation


Solution:

Question 3. Which of the following statements is correct?

Solution: (c)E_cell is an intensive property as it does not depend upon mass of species (number of particles) but A_rG of the cell reaction is an extensive property because this depends upon mass of species (number of particles).

Question 4. The difference between the electrode potentials of two electrodes when no
current is drawn through the cell is called 
(a) cell potential (b) cell emf
(c) potential difference (d) cell voltage
Solution: (b) EMF is the difference between the electrode potentials of two electrodes • when no current is drawn through the cell.

Question 5. Which of the following statements is not correct about an inert electrode in a cell?
(a) It does not participate in the cell reaction.
(b) It provides surface either for oxidation or for reduction reaction.
(c) It provides surface for conduction of electrons.
(d) It provides surface for redox reaction.
Solution: (d) Inert electrode does not participate in redox reaction and acts only as source or sink for electrons. It provides surface either for oxidation or for reduction reaction.

Question 6. An electrochemical cell can behave like an electrolytic cell when

Solution: (c) If an external opposite potential is applied on the galvanic cell and increased reaction continues to take place till the opposing voltage reaches the value 1.1 V.
At this stage no current flow through the cell and if there is any further increase in the external potential then reaction starts functioning in opposite direction.
Hence, this works as an electrolytic cell.

Question 7. Which of the following statement about solutions of electrolytes is not correct?
(a) Conductivity of solution depends upon size of ions.
(b) Conductivity depends upon viscosity of solution.
(c) Conductivity does not depend upon salvation of ions present in solution.
(d) Conductivity of solution increases with temperature.
Solution: (c) Conductivity depends upon salvation of ions present in the solution. The greater the salvation of ions, the lesser is the conductivity.

Question 8.Using given below find strongest reduction agent.

Solution: (b) A negative value of standard reduction potential for Cr3+ to Cr means that the redox couple is a stronger reducing agent.

Question 9. Use the data given in Q. 8 and find out which of the following is the strongest oxidizing agent.

Solution: (c) The higher the positive value of standard reduction potential of metal ion, the higher will be its oxidizing capacity.
Since,

has value equal to 1.51 V hence it is the strongest oxidizing agent.

Question 10. Using the data given in Q. 8, find out in which option the order of reducing power is correct.

Solution:

Question 11. Use the data given in Q. 8 and find out the most stable ion in its reduced form.

Solution: (d) Mn⁺² is most stale in its reduced form due to highest E° value.

Question 12. Use the data given in Q.8 and find out the most stable oxidized species.

Solution:

Question 13. The quantity of charge required to obtain one mole of aluminium from

Al₂O₃ is (a) IF (b) 6F (c) 3F (d) 2F
Solution:

Question 14. The cell constant of a conductivity cell
(a) changes with change of electrolyte
(b) changes with change, of concentration of electrolyte
(c) changes with temperature of electrolyte
(d) remains constant for a cell
Solution: (d) The cell constant of a conductivity cell (a) remains constant for a cell.

Question 15. While charging the lead storage battery
(a)PbSO₄ anode is reduced to Pb
(b)PbSO₄ cathode is reduced to Pb
(c)PbSO₄ cathode is oxidized to Pb
(d)PbSO₄ anode is oxidized to Pb02
Solution: (a) While charging the lead storage battery the reaction occurring on cell is reversed and PbSO₄(s) on anode and cathode is converted into Pb and Pb02 respectively.
Hence, option (a) is the correct choice The electrode reactions are as follows:

Question 16.

Solution:

Question 17. In the electrolysis of aqueous sodium chloride solution, which of the half-cell reaction will occur at anode?

Solution: (d) During electrolysis of aqueous

(ii)The reaction at anode with lower value of E° should be preferred, but oxidation of 02 is kinetically slow process and needs high voltage thus reaction (i) takes place.
More than One Correct Answer Type

Question 18. The positive value of the standard electrode potential of Cu⁺²/Cu indicates
that
(a) this redox couple is a stronger reduction agent than the H^{/H₂ couple}
(b) this redox couple is a stronger oxidizing agent than H⁺/H₂
(c) Cu can displace H₂ from acid
(d) Cu cannot displace H₂  from acid
Solution: (b, d) The Lesser the E° value of redox couple, the higher the reducing power ’.

Since, 2 H⁺/H₂ has lesser SRP than Cu⁺²/Cu redox couple. Therefore,
(i) This redox couple is a stronger oxidizing agent than H⁺/H₂
(ii) Cu cannot displace H₂ from acid.
Hence, (b) and (d) are correct.

Question 19. E°ell for some half-cell reactions are given below. On the basis of these marks the correct answer will be

(a) In dilute sulphuric acid solution, hydrogen will be reduced at cathode.
(b) In concentrated sulphuric acid solution, water will be oxidized at anode.
(c) In dilute sulphuric acid solution, water will be oxidized at anode.
(d) In dilute sulphuric acid solution, SO^2-ion will be oxidized to tetrathionate ion at anode.
Solution: (a, c) In dilute sulphuric acid solution, hydrogen will be reduced at cathode.

while in concentrated solution of sulphuric acid, SO^-2 ions are oxidized to tetrathionate (SO2) ions.

Question 20. E°en = 1.1 V for Daniell cell. Which of the following expressions are correct description of state of equilibrium in this cell?

Solution:

Question 21. Conductivity of an electrolytic solution depends on
(b) concentration of electrolyte
(d) distance between the electrodes
Solution: (a, b) Conductivity of electrolyte solution is due to presence of mobile ions in the solution. This type of conductance is known as ionic conductance. Conductivity of these type of solutions depend upon
(i) the nature of electrolyte added
(ii) size of the ion produced and their solvation
(iii) concentration of electrolyte
(iv) nature of solvent and its viscosity
(v) temperature
While power of source or distance between electrodes has no effect on conductivity of electrolyte solution.
Hence, options (a) and (b) are the correct choices.

Question 22.

Solution:

However, the sum of molar conductivities of constituent ions gives the molar conductivity of water but here NH₄OH is a weak electrolyte of which complete decomposition is not possible.

Question 23. What will happen during the electrolysis of aqueous solution of CuSO₄ by using platinum electrodes?
(a) Copper will deposit at cathode.
(b) Copper will deposit at anode.
(c) Oxygen will be released at anode.
(d) Copper will dissolve at anode.
Solution:

Question 24. What will happen during the electrolysis of aqueous solution of CuSO₄ in the presence of Cu electrodes?
(a) Copper will deposit at cathode.
(b) Copper will dissolve at anode.
(c) Oxygen will be released at anode.
(d) Copper will deposit at anode.
Solution:

Question 25. Conductivity K, is equal to

Solution:

Question.26. Molar conductivity of ionic solution depends on .
(a) temperature
(b) distance between electrodes
(c) concentration of electrolytes in solution
(d) surface area of electrodes
Solution: (a, c) Molar conductivity of ionic solution depends on temperature and concentration of electrolytes in solution.

Question 27. For the given cell, Mg|Mg²⁺||Cu²⁺ || Cu
(a) Mg is cathode
(b) Cu is cathode
(c) The cell reaction Mg + Cu²⁺+ —»Mg²⁺ + Cu
(d) Cu is the oxidizing agent
Solution: (b, c) Left side of cell reaction represents oxidation half-cell i.e., oxidation of Mg and right side of cell represents reduction half-cell reactions i.e., reduction of copper.
(ii) Cu is reduced and reduction occurs at cathode.
(iii) Mg is oxidized and oxidation occurs at anode.
(iv) Whole cell reaction can be written as

Hence, options (b) and (c) both are correct choices.

Short Answer Type Questions

Question 28. Can absolute electrode potential of an electrode be measured?
Solution: No, absolute electrode potential of an electrode cannot be measured.

Question 29. Can

for cell reaction ever be equal to zero?
Solution:

Question 30. Under what condition is

Solution: At equilibrium i.e., when the cell is completely discharged,

Question 31. What does the negative sign in the expression

mean? Zn /Zn
Solution:

Question 32. Aqueous copper sulphate solution and aqueous silver nitrate solution are electrolysed by 1 ampere current for 10 minutes in separate electrolytic cells. Will the mass of copper and silver deposited on the cathode be same or different? Explain your answer.
Solution:

Mass of silver will be different because the equivalent mass of Ag is different.

Question 33. Depict the galvanic cell in which the cell reaction is Cu + 2Ag⁺ > 2Ag+Cu2
Solution: Cu | Cu²⁺ (aq, 1M) || Ag⁺ (aq, 1M) | Ag

Question 34. Value of standard electrode potential for the oxidation of Cl ions is more positive than that of water, even then in the electrolysis of aqueous sodium chloride, why is Cl- oxidized at anode instead of water?
Solution: Under the conditions of electrolysis of aqueous sodium chloride, oxidation of water at anode requires over potential and therefore, Cu is oxidised instead of water.

Question 35. What is electrode potential?
Solution: The electrical potential difference set up between the metal and its solution is called electrode potential.

Question 36. Consider the following diagram in which an electrochemical cell is coupled to an electrolytic cell. What will be the polarity of electrodes ‘A’ and ‘B’ in the electrolytic cell?

Solution: ‘A’ will have -ve polarity and ‘B’ will have +ve polarity.

Question 37. Why is alternating current used for measuring resistance of an electrolytic solution?
Solution: The alternating current is used to prevent electrolysis so that the concentration of ion in the solution remains constant.

Question 38. A galvanic cell has electrical potential of 1.1 V. If an opposing potential of 1.1 V is applied to this cell, what will happen to the cell reaction and current flowing through the cell?
Solution: When the opposing potential becomes equal to electrical potential, the cell reaction stops and no current flows through the cell. Thus, there is no chemical reaction.

Question 39. How will the pH of brine (aq. NaCl solution) be affected when it is electrolysed?
Solution: Aqueous solution of brine contains Na+, CP, H⁺ and H^–. Electrode process are given as follows:

Remaining solution will contain NaOH, which is base, therefore, pH will increase.

Question 40. Unlike dry cell, the mercury cell has a constant cell potential throughout its useful life. Why?
Solution: Electrolyte is not consumed in the cell process of mercury cell hence it will deliver the current at constant potential throughout its life.

Question 41. Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The Am of ‘B’ increases 1.5 times while that of A increases 25 times. Which of the two is a strong electrolyte? Justify your answer.
Solution: Electrolyte ‘B’ will be strong electrolyte because it is completely ionised and on dilution the molar conductance will increase to the small extent due to increase in speed of ions only.

Question 42. When acidulated water (dil. H2SO₄ solution) is electrolysed, will the pH of the solution be affected? Justify your answer.
Solution: The pH of solution will not be affected because there is no change in the concentration of  H⁺ions.

Question 43. In an aqueous solution, how does specific conductivity of electrolytes change with addition of water?
Solution: On the addition of water, number of ions per unit volume decreases and therefore conductivity decreases.

Question 44. Which reference electrode is used to measure the electrode potential of other electrodes?
Solution: The standard hydrogen electrode is used as a reference electrode whose electrode potential is taken to be. zero. The electrode potential of other electrodes is measured with respect to it.

Question 45. Consider a cell given below:
Cu |Cu²⁺ || Cl |Cl₂, Pt
Write the reactions that occur at anode and cathode.
Solution: The given cell is:

Question 46. Write the Nemst equation for the cell reaction in the Daniell cell. How will the Ecell be affected when concentration of Zn²⁺ ions is increased?
Solution: Daniell cell
Zn(s) | Zn²⁺ || Cu²⁺ | Cu(s)

Above equation shows that the cell potential will decrease with increase in the concentration of Zn2+ ion.

Question 47. What advantage do the fuel cells have over primary and secondary batteries?
Solution: Primary batteries contain a limited amount of reactants and are discharged when the reactants have been consumed. Secondary batteries can be recharged but take a long time to recharge. Fuel cell runs continuously as long as the reactants are supplied to it and products are removed continuously.

Question 48. Write the cell reaction of a lead storage battery when it is discharged. How does the density of the electrolyte change when the battery is discharged?
Solution:

Density of electrolyte decreases because water is formed and sulphuric acid consumed as the product during discharge of the battery.

Question 49. Why on dilution the Am of CH3COOH increases drastically, while that of
CH4COONa increases gradually? 
Solution: In the case of CH3COOH, which is a weak electrolyte, the number of ions increase on dilution due to an increase in degree of dissociation.

In the case of strong electrolyte such as CH3COONa, the number of ions remains the same but the interionic attraction decreases.

Matching Column Type Questions

Question 50. Match the terms given in Column 1 with the units given in Column II.

Solution: (i —» c), (ii —> d), (iii —> a), (iv —> b)

Question 51. Match the terms given in Column I with the items given in Column II.

Solution: (i -> d), (ii -> a), (iii -> b), (iv -> c)

Question 52. Match the items of Column I and Column II.

Solution: (i —> d), (ii -> c) (iii -> a), (iv -> b)

Question 53.Match the items of Column I and Column II.

Solution:(i -> d), (ii -> c), (iii -> b), (iv -> a)

Question 54.Match the items of Column I and Column II.

Solution: (i -> d), (ii -> c), (iii —> a, e), (iv -> b)

Question 55. Match the items of Column I and Column II on the basis of data given below:

Solution:(i->c),(ii->a),(iii->g),(iv->e),(v->d),(vi->b),(vii->f).

Assertion and Reason Type Questions

In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
(a) Both Assertion and Reason are true and the Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason are true and the Reason is not correct explanation for Assertion
(c) Assertion is true but the Reason is false.
(d) Both Assertion and Reason are false.
(e) Assertion is false but Reason is true.

Question 56. Assertion (A): Cu is less reactive than hydrogen.
Reason (R): E°u2+/Cu is negative.
Solution: (c) Cu is less reactive than hydrogen because E°u2+ Cu is positive.

Question 57. Assertion (A): Ecell should have a positive value for the cell to function. Reason (R). Ecadlode Eanode
Solution: (c) Ecell = Ecathode – Eanode. To have positive value of Ecell, Ecathode > Eanode.

Question 58. Assertion (A): Conductivity of all electrolytes decreases on dilution.
Reason (R): On dilution number of ions per unit volume decreases.
Solution: (a) Conductivity depends on number of ions per unit volume which decreases on dilution of electrolytes.

Question 59. Assertion (A): Lm for weak electrolytes shows a sharp increase when the electrolytic solution on dilution of solution.
Electrockemistrij 63
Reason (R): For weak electrolytes dissociate partially in concentrated solution. On dilution, their degree of dissociation increases hence their Am increases sharply.
Solution: (a) Weak electrolytes dissociate partially in concentrated solution. On dilution, their degree of dissociation increases hence their Am increases sharply.

Question 60. Assertion (A): Mercury cell does not give steady potential.
Reason (R): In the cell reaction, ions are not involved in solution.
Solution: (e) Mercury cell gives a steady potential because in the cell reaction ions are not involved in the solution.

Question 61. Assertion (A): Electrolysis of NaCl solution gives chlorine at anode instead of 02.
Reason (R): Formation of oxygen at anode requires overvoltage.
Solution: (a) Formation of oxygen has lower value of E° than formation of chlorine even then it is not formed because it requires overvoltage.

Question 62. Assertion (A): For measuring resistance of an ionic solution an AC source is used.
Reason (R): Concentration of ionic solution will change if DC source is used.
Solution: (a) Alternating current is used in the measurement of resistance of electrolyte solution because concentration changes with DC current due to electrolysis.

Question 63. Assertion (A): Current stops flowing when E_cell = 0.
Reason (R): Equilibrium of the cell reaction is attained.
Solution: (a) At equilibrium  E_cell = 0 and no current flows.

Question 64. Assertion (A): E , increases with increase in concentration of Ag⁺ ions.
Ag / Ag⁺
Reason (R): E + has a positive value.
Ag /Ag
Solution:

Question 65. Assertion (A): Copper sulphate can be stored in zinc vessel.
Reason (R): Zinc is less reactive than copper.
Solution: (d) Zinc will get dissolved in CuS04 solution since zinc is more reactive than copper.

Long Answer Type Questions

Question 66. Consider the following figure and answer the following questions.

(i) Cell ‘A’ has Ecell = 2 V and cell ‘B’ has Ecen = 1.1 V which of the two cells ‘A’ or ‘B’ will act as an electrolytic cell. Which electrode reactions will occur in this cell?
(ii) If cell ‘A’ has Ecen = 0.5 V and cell ‘B’ has Ecen = 1.1 V, what will be the reactions at anode and cathode?
Solution: (i) Cell ‘B’ will act as electrolytic cell because potential of ‘B’ is less than
that of ‘A’. Electrode process in the cell ‘B’ may be given as
<

Question 67. Consider the figure given below and answer the questions (i) to (vi) that follow.

(i) Redraw the diagram to show the direction of electron flow.
(ii) Is silver plate anode or cathode?
(iii)What will happen if salt bridge is removed?
(iv)When will the cell stop functioning?
(v)How will concentration of Zn⁺² cell functions?
(vi)How will the concentration of Zn⁺² ions and Ag⁺ ions be affected after the cell becomes ‘dead’?
Solution:(i)The cell is:
Zn(s) | Zn⁺² || Ag⁺ | Ag
(ii)Electron will flow from zinc anode to silver cathode in external circuit. Silver will act as cathode, since its standard reduction potential is greater than that of zinc.
(iii)Potential will drop to zero if salt bridge is suddenly removed.
(iv)Cell will stop functioning when it is discharged i.e., when cell potential is zero.
(v)Nemst equation for the cell is: 0.059,

Cell potential will decrease with increase in concentration of [Zn⁺²] while it will increase with the concentration of [Ag⁺].
(vi) When cell is dead or discharged, E will be zero and the cell will be at equilibrium. Then, concentration of Zn⁺² and Ag⁺ will not change.

Question 68. What is the relationship between Gibbs free energy of the cell reaction in a galvanic cell and the emf of the cell? When will the maximum work be obtained from a galvanic cell?
Solution: The required relations are:

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NCERT Exemplar Class 12 Chemistry Chapter 1 Solid State are part of NCERT Exemplar Class 12 Chemistry. Here we have given NCERT Exemplar Class 12 Chemistry Chapter 1 Solid State. https://www.cbselabs.com/ncert-exemplar-problems-class-12-chemistry-solid-state/

NCERT Exemplar Class 12 Chemistry Chapter 1 Solid State

Multiple Choice Questions
Single Correct Answer Type

Solid State Class 12 Mcq Learn Cbse Chapter 1

Question 1. Which of the following condition favours the existence of a substance in the solid state?
(a) High temperature (b) Low temperature
(c) High thermal energy (d) Weak cohesive forces
Solution: (b) At low temperature substance exists in solid state due to decrease in molecular motion which leads to strong cohesive forces i.e., forces which hold the constituent particles together.

Question 2. Which of the following is not a characteristic of a crystalline solid?
(a) Definite and characteristic heat of fusion
(b) Isotropic nature
(c) A regular periodically repeated pattern of arrangement of constituent particles in the entire crystal
(d) A true solid
Solution: (b) Anisotropy: Crystalline solids are anisotropic in nature, that is some of their physical properties like electrical resistance or refractive index show different values when measured along different directions in the same crystal. This arises from different arrangement of particles in different directions arrangement of particles along different directions

Isotropy: In case of amorphous substances, properties such as electrical conductivity, refractive index, thermal expansion, etc. are identical in all directions just as in case of gases or liquids. This property is called isotropy and the substances showing this property are called isotropic.

NCERT Exemplar Class 12 Chemistry Solid State

Question 3. Which of the following is an amorphous solid?
(a) Graphite (C) (b) Quartz glass (SiO₂)
(c) Chrome alum (d) Silicon carbide (SiC)
Solution: (b)

Solid State Exemplar Class 12 Chapter 1

Question 4. Which of the following arrangement shows schematic alignment of magnetic moments of antiferromagnetic substances?

Solution: (d)

NCERT Exemplar Solid State Class 12 Chapter 1 

Question 5. Which of the following is true about the value of refractive index of quartz glass?
(a) Same in all directions (b) Different in different directions
(c) Cannot be measured (d) Always zero
Solution: (a) Since quartz glass is an amorphous solid having short range order of constitutents. Hence, value of refractive index is same in all directions, can be measured and not be equal to zero always.

Question 6. Which of the following statement is not true about amorphous solids?
(a) On heating they may become crystalline at certain temperature
(b) They may become crystalline on keeping for a long time
(c) Amorphous solids can be moulded by heating
(d) They are anisotropic in nature
Solution: (d) Amorphous solids are isotropic because they show thermal and optical properties, same in all directions.

Solid State NCERT Exemplar Class 12 Chapter 1 

Question 7. The sharp melting point of crystalline solids is due to
(a) a regular arrangement of constituent particles observed over a short
distance in the crystal lattice .
(b) a regular arrangement of constituent particles observed over a long distance in the crystal lattice
(c) same arrangement of constituent particles in different directions
(d) different arrangement of constituent particles in different directions
Solution:(b) A solid is said to be crystalline if the various constituent structural units (atoms, ions or molecules) of which the solid is made, are arranged in a definite geometrical pattern within the solid.
The type of forces in crystalline solids are of long range order due to which they have sharp melting point. .

Question 8. Iodine molecules are held in the crystals lattice by
(a) London forces (b) dipole-dipole interactions
(c) covalent bonds (d) coulombic forces
Solution:(a) I₂ is a molecular solid

Solid State NCERT Exemplar Solutions Class 12 Chapter 1

Question 9. Which of the following is a network solid?
(a) S0₂ (solid) (b) I₂
(c) Diamond (d) H₂0 (ice)
Solution: (c) Diamond is a three-dimensional network solid in which each carbon atom is tetrahedrally bonded with four carbon atoms.

Question 10. Which of the following solids is not an electrical conductor?
Solution: Together by London force or dispersion force. This is soft and non-conductor of electricity.
Water is a hydrogen bonded molecular solid in which H and O are held together by polar covalent bond and each water molecular held together by hydrogen bonding. Due to non-ionic nature, they are not electrical conductor.

Solid State Exemplar Solutions Class 12 Chapter 1

Question 11. Which of the following is not the characteristic of ionic solids?
(a) Very low value of electrical conductivity in the molten state
(b) Brittle nature
(c) Very strong forces of interactions
(d) Anisotropic nature
Solution: (a)

NCERT Exemplar Class 12 Chemistry Chapter 1

Question 12. Graphite is a good conductor of electricity due to the presence of
(a) lone pair of electrons (b) free valence electrons
(c) cations (d) anions
Solution: (b) In graphite one carbon atom is attached to three other carbon atoms. One electron of carbon remains free. Due to this free valence electron graphite is an electrical conductor.

Question.13. Which of the following oxide behaves as conductor or insulator depending upon temperature?
(a) TiO (b) Si0₂ (c) TiO3 (d) MgO
Solution: (c) TiO3 behaves as conductor or insulator depending on temperature because of variation of energy gap between valence band and conduction band with the variation of temperature.

Question 14. Which of the following oxide shows electrical properties like metals?
(a)SO₂ (b) MgO
(c)SO₂(s) (d) CrO₂
Solution: (d)CrO₂, TiO and Re03 are some typical metal oxides which show electrical conductivity similar to metal. While SO₂, MgO and SO₂ are oxides of metal, semimetal and non-metal which do not show electrical properties.

Class 12 Chemistry Exemplar Solutions Chapter 1

Question 15. The lattice site in a pure crystal cannot be occupied by
(a) molecule (b) ion
(c) electron (d) atom
Solution: (c) Pure crystals have constituents i.e., atoms or molecules or ions as lattice points which are arranged in fixed stoichiometric ratio. Electron can occupy the lattice site only when there is imperfection in solid and not in a pure crystal.
Hence, existence of free electrons are not possible, it is possible on in case of imperfection in solid.

Question 16. Graphite cannot be classified as
(a) conducting solid (b) network solid
(c) covalent solid (d) ionic solid
Solution: (d) Constituent units of graphite are carbon atoms, held together by covalent bonding in 2D network structure. Thus, it is not an ionic solid.

NCERT Exemplar Class 12 Chemistry Solid State Solutions Chapter 1 

Question 17. Cations are present in the interstitial sites in
(a) Frenkel defect (b) Schottky defect
(c) vacancy defect (d) metal deficiency defect .
Solution: (a)

Question 18. Schottky defect is observed in crystals when
(a) some cations move from their lattice site to interstitial sites
(b) equal number of cations and anions are missing from the lattice
(c) some lattice sites are occupied by electrons
(d) some impurity is present in the lattice
Solution: (b)

Solid State Class 12 NCERT Exemplar Chapter 1

Question 19. Which of the following is true about the charge acquired by p-type semicon-ductors?
(a) Positive
(b) Neutral
(c) Negative
(d) Depends on concentration of p impurity
Solution: (b) p-Type semiconductors are neutral but they conduct electricity through positive holes.

Question 20. To get a n-type semiconductor from silicon, it should be doped with a
substance with valency
(a) 2 (b) 1
(c) 3 (d) 5
Solution: (d) Impurity of higher group is doped to get n-type semiconductor. Thus, silicon (valency = 4) should be doped with the element with valency equal to 5.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 1

Question 21. The total number of tetrahedral voids in the face centered unit cell is
(a) 6 (c) 10
(b) 8 (d) 12
Solution: (b) Fee unit cell contains 8 tetrahedral voids at centre of each 8 smaller cube of a unit cell as shown below

NCERT Exemplar Class 12 Solid State Chapter 1

Question 22. Which of the following point defects are shown by AgBr (s) crystals?
(A) Schottky defect (B) Frenkel defect
(C) Metal excess defect (D) Metal deficiency defect
(a) A and B (b) C and D
(c) A and C (d) B and D
Solution: (a) AgBr shows both Schottky and Frenkel defects. In AgBr, both Ag⁺ and Br ions are absent from the lattice causing Schottky defect. However, Ag⁺ ions are mobile so they have a tendency to move aside the lattice and trapped in interstitial site, hence cause Frenkel defect. ‘

Question 23. In which pair most efficient packing is present?
(a) hep and bcc (b) hep and ccp
(c) bcc and ccp (d) bcc and simple cubic cell
Solution: (b) Packing efficiency: It is the percentage of total filled space by particles

Since, packing efficiency for hep or ccp is calculated to be 74% which is maximum among all type of crystals.

Question 24. The percentage of empty space in a body centered cubic arrangement is
(a) 74 (b) 68 (c) 32 (d) 26
Solution: (c) Packing efficiency for bcc arrangement is 68% which represents total filled space in the unit cell. Hence, empty space in a body centered arrangement is 100 – 68 = 32%.

NCERT Exemplar Chemistry Class 12 Chapter 1

Question 25. Which of the following statement is not true about the hexagonal close packing?
(a) The coordination number is 12
(b) It has 74% packing efficiency
(c) Tetrahedral voids of the second layer are covered by the spheres of the third layer
(d) In this arrangement, spheres of the fourth layer are exactly aligned with those of the first layer.
Solution: (d) Hexagonal close packing can be arranged by two layers
A and B one over another which can be diagrammatically represented as

Here, we can see easily that 1st layer and 4th layer are not exactly aligned. Thus, statement (d) is not correct while other statements (a), (b) and (c) are true.

Question 26. In which of the following structure coordination number for cations and anions in the packed structure will be same?
(a) Cl^– ions form fee lattice and Na+ ions occupy all octahedral voids of the unit cell.
(b) Ca²⁺ ions form fee lattice and F- ions occupy all the eight tetrahedral voids of the unit cell
(c) O^2- ions form fee lattice and Na⁺ ions occupy all the eight tetrahedral voids of the unit cell
(d) S^2- ions form fee lattice and Zn²⁺ ions go into alternate tetrahedral voids of the unit cell.
Solution: (a) NaCl crystals have rock salt structure having fee lattice in which Cl^– ions are present at fee lattice points and face centre and Na⁺ occupies all the octahedral voids of given unit cell.
Where, coordination number of Na⁺ = 6
Coordination number of Cl^–= 6

Solid State Problems Class 12 Chapter 1 

Question 27. What is the coordination number in a square close packed structure in two dimensions? (a) 2 (b) 3 (c) 4 (d) 6
Solution: (c) Coordination number in a square closed packed structure in two dimensions is equal to 4 is shown as:

Question 28. Which kind of defect is introduced by doping?
(a) Dislocation defect (b) Schottky defect
(c) Frenkel defect (d) Electronic defect
Solution: (d) When electron rich or electron deficient impurity is added to a perfect crystal, it introduces electronic defect in them.

Question 29. Silicon doped with electron rich impurity forms
(a) p-type semiconductor (b) n-type semiconductor
(c) intrinsic semiconductor (d) insulator
Solution: (b) Silicon has four valence electrons. If it is doped with an electron rich impurity, the extra electron becomes delocalised and increases the conductivity. Since the increase in conductivity is due to negatively charged electron, hence it is called n-type semiconductor.

Solid State NCERT Exemplar Pdf Class 12 Chapter 1

Question 30. Which of the following statement is not true?
(a) Paramagnetic substances are weakly attracted by magnetic field
(b) Ferromagnetic substances cannot be magnetized permanently
(c) The domains in antiferromagnetic substances are oppositely oriented with respect to each other
(d) Pairing of electrons cancel their magnetic moment in the diamagnetic substances.
Solution: (b) Ferromagnetic species are strongly attracted in the magnetic field and can be permanently magnetised.
Hence, choice (b) is the correct answer while other three choices are correct.

Question 31. Which of the following is not true about the ionic solids?
(a) Bigger ions form the close packed structure
(b) Smaller ions occupy either the tetrahedral or the octahedral voids depending upon their size
(c) Occupation of all the voids is not necessary
(d) The fraction of octahedral or tetrahedral voids occupied depends upon the radii of the ions occupying the voids.
Solution: (d) The fraction of octahedral or tetrahedral voids occupied depends upon the radii of the ions present at the lattice points. As we know the radii of octahedral or tetrahedral void is related to radii of atoms (r) as Radius of octahedral void (R₀) = 0.414 r .
Radius of tetrahedral void (R₁ ) = 0.225 r  Where, r = radius of bigger atom involved.

Solid State Mcqs Learncbse Class 12 Chapter 1 

Question 32. A ferromagnetic substance becomes a permanent magnet when it is placed in a magnetic field because
(a) all the domains get oriented in the direction of magnetic field
(b) all the domains get oriented ill the direction opposite to the direction of magnetic field
(c) domains get oriented randomly
(d) domains are not affected by magnetic field.
Solution:(a) Ferromagnetic solids can be permanently magnetised and then all the domains get oriented in the direction of applied magnetic field.

Question 33. The correct order of the packing efficiency in different types of unit cells is…………
(a) fee < bee < simple cubic (b) fee > bee simple cubic
(c) fee < bee > simple cubic (d) bee < fee > simple cubic
Solution:(b) Packing efficiency in different types of unit cells can be tabulated as

Hence, correct order is fee (74%) > bee (68%) > simple cubic (52%).

Question 34. Which of the following defects is also known as dislocation defect?
(a) Frenkel defect (b) Schottky defect
(c) Non-stoichiometric defect (d) Simple interstitial defect
Solution: (a) In Frenkel defect, some cations occupy interstitial site and hence it is also called dislocation defect.

NCERT Exemplar Class 12 Chemistry Pdf Download With Solutions Chapter 1

Question 35. In the cubic close packing, the unit cell has
(a) 4 tetrahedral voids each of which is shared by four adjacent unit cells
(b) 4 tetrahedral voids within the unit cell
(c) 8 tetrahedral voids each of which is shared by four adjacent unit cells
(d) 8 tetrahedral voids within the unit cells.
Solution: (d) In the cubic close packing the unit cell has 8 tetrahedral voids within it and are located at each eight smaller cube of a unit cell.

Question 36. The edge lengths of the unit cells in terms of the radius of spheres constituting fee, bcc and simple cubic unit cells are respectively

Solution: (a) Note: Distance between two atoms is always measured from their centres
(i) If the crystal lattice consists of SCC, the atom which is present at the comers touch each other

(ii) In case of FCC, atom present at the comer and the centre of the face touch each other.

(iii)In case of BCC atom present at the corner and center of the body touch each other

Question 37. Which of the following represents correct order of conductivity in solids?

Solution: (a) Conductivity of metal, insulator and semiconductors can be represented in the term of k (Kappa) which depends upon energy gap between valence band and conduction band.

Question 38. Which of the following is not true about voids formed in three dimensional hexagonal close packed structure?
(a) A tetrahedral void is formed when a sphere of the second layer is present above triangular void in the first layer
(b) All the triangular voids are not covered by the spheres of the second layer
(c) Tetrahedral voids are formed when the triangular voids in the second layer lie above the triangular voids in the first layer and the triangular voids in the first layer and the triangular shapes of these voids do not overlap
(d) Octahedral voids are formed when the triangular voids in the second layer exactly overlap with similar voids in the first layer.
Solution: (c, d) Tetrahedral voids are formed when the triangular void in the second layer lie exactly above the triangular voids in the first layer and the triangular shape of these voids oppositely overlap.

Octahedral voids are formed when triangular void of second layer is not exactly overlap with similar void in first layer.

Question 39. The value of magnetic moment is zero in the case of antiferromagnetic substances because the domains …
(a) get oriented in the direction of the applied magnetic field
(b) get oriented opposite to the direction of the applied magnetic field
(c) are oppositely oriented with respect to each other without the application of magnetic field
(d) cancel out each other’s magnetic moment
Solution:  (c, d) In the case of antiferromagnetic substances, the magnetic moment becomes zero because the domains are oppositely oriented with respect to each other without the application of magnetic field which cancel out each other.

Question 40. Which of the following statements are not true?
(a) Vacancy defect results in a decrease in the density of the substance
(b) Interstitial defects results in an increase in the density of the substance
(c) Impurity defect has no effect on the density of the substance
(d) Frenkel defect results in an increase in the density of the substance
Solution: (c, d) Statements (c) and (d) can be correctly written as (c) Impurity defect
changes the density of substance as impurity has different than the ion present on perfect crystal e.g., When SrCl₂ is added to the NaCl crystal, it causes impurity defect, (d) Frenkel defect results neither decrease nor increase in density of substance.

Question 41. Which of the following statements are true about metals?
(a) Valence band overlap with conduction band
(b) The gap between valence band and conduction band is negligible
(c) The gap between valence band and conduction band cannot be determined
(d) Valence band may remain partially filled.
Solution:  (a, b, d) In metal, valence band overlap with conduction band. The gap between valence band and conduction band is negligible and valence band may remain partially filled.

Question 42. Under the influence of electric field, which of the following statements are true about the movement of electrons and holes in a p-type semiconductor?
(a) Electron will move towards the positively charged plate through electron holes
(b) Holes will appear to be moving towards the negatively charged plate
(c) Both electrons and holes appear to move towards the positively charged plate
(d) Movement of electrons is not related to the movement of holes
Solution:  (a, b) In p-type semiconductor, the conductivity is due to existence of hole. When electric field is applied to p-type semiconductor hole starts moving towards negatively charged plate and electron towards positively charged plate.
Flow of holes in p-type semiconductors Hole .

Question 43. Which of the following statements are true about semiconductors?
(a) Silicon doped with an electron rich impurity is a p-type semiconductor
(b) Silicon doped with an electron rich impurity is an n-type semiconductor
(c) Delocalised electrons increase the conductivity of doped silicon
(d) An electron vacancy increases the conductivity of type semiconductor
Solution: (b, c) Silicon (valence electron – 4) doped with electron rich impurity is an
n-type semiconductor due to extra electron and the delocalised electrons increase the conductivity of doped silicon.

Question 44. An excess of potassium ions makes KCl crystals appear violet or Lilac in
colour since
(a) some of the anionic sites are occupied by an unpaired electron
(b) some of the anionic sites are occupied by a pair of electrons
(c) there are vacancies at some’anionic sites
(d) F-centres are created which impart colour to the crystals
Solution: (a, d) .
When KC1 is heated in vapour of K, some of the Cl” leave their lattice site and create anion vacancies. This chloride ion wants to combine with K vapour to form potassium chloride. For doing so K atom loses electrons form K ions. This released electron diffuses into the crystal to get entrapped in the anion vacancy called F-centre. When visible light falls on the crystal, this entrapped electron gains energy, goes to the higher level when it comes back to the ground state, energy is released in the form of light.

Question 45. The number of tetrahedral voids per unit cell in NaCl crystal is
(c) twice the number of octahedral voids
(d) four times the number of octahedral voids
Solution: (b, c) NaCl has fee arrangement of CF ions. Thus,
Number of CF ions in packing per unit cell = 4
Number of tetrahedral voids = 2 x No. of particles present in close packing
=2×4=8
Number of tetrahedral voids = 2 x No. of octahedral voids

Question 46. Amorphous solids can also be called
(a) pseudo solids (b) true solids
(c) super cooled liquids (d) super cooled solids
Solution: (a, c) Amorphous solid has short range order which has a tendency to flow very slowly. Hence, it is also known as pseudo solids or super cooled liquids. Glass panes fixed to windows or doors of old buildings are invariably observed to be thicker at bottom than at the top. These are examples of amorphous solids.

Question 47. A perfect crystal of silicon (fig) is doped with some elements as given in the options. Which of these options show n-type semiconductors?


Solution: (a, c)
When group 15 elements are doped into a perfect crystal, it leads to the formation of n-type semiconductor.
Here, in (a) as (group 15, period 3) is doped to perfect Si-crystal and in (c) as (group 15, period 2) is doped to perfect Si-crystal.

Question 48. Which of the following statements are correct?
(a) Ferrimagnetic substances lose ferrimagnetism on heating and become paramagnetic
(b) Ferrimagnetic’substances do not lose ferrimagnetism on heating and remain ferrimagnetic
(c) Antiferromagnetic substances have domain structure similar to ferromagnetic substances and their magnetic moments are not cancelled by each other
(d) In ferromagnetic substances, all the domains get oriented in the direction of magnetic field and remain as such even after removing magnetic field.
Solution: (a, d) Ferrimagnetic substances lose ferrimagnetism on heating and become paramagnetic. In ferromagnetic substance, domains are aligned in parallel and antiparallel direction in unequal numbers.
In ferromagnetic substances, all the domains get oriented in the direction of magnetic field and remain as such even after removing magnetic field.

Question 49. Which of the following features are not shown by quartz glass?
(a) This is a crystalline solid
(b) Refractive index is same in all the directions
(c) This has definite heat of fusion
(d) This is also called super cooled liquid
Solution: (a, c) Quartz glass is an amorphous solid so it has not definite heat of fusion. This is due to short range order of molecule while quartz glass is also known as super cooled liquid and isotropic in nature.

Question 50. Which of the following cannot be regarded as molecular solid?
(a) SiC (b) AIN
(c) Diamond (d) I₂
Solution: (a, b, c) SiC, AIN and diamond are examples of network solid as they have three dimensional structure while, I₂ is a molecular solid, because such solid particles are held together by dipole-dipole interactions. SiC and AIN are interstitial solids. s

Question 51. In which of the following arrangements octahedral voids are formed?
(a) hep (b) bcc (c) simple cubic (d) fee
Solution: (a, d) In hep and fee arrangement, octahedral voids are formed. In fee, the octahedral voids are observed at edge and centre of cube while in bcc and simple cubic, no any octahedral voids are observed. In bcc, cubic voids formed.

Question 52. Frenkel defect is also known as
(a) stoichiometric defect (b) dislocation defect
(c) impurity defect (d) non-stoichiometric defect
Solution: (a, b) In Frenkel defect, dislocation of cations takes place and there is no change in stoichiometry of the crystal.

Question 53. Which of the following defects decrease the density?
(a) Interstitial defect (b) Vacancy defect
(c) Frenkel defect (d) Schottky defect
Solution: (b, d) Vacancy and Schottky defect which lead to decrease the density both are the types of a stoichiometric defect. In case of Frenkel defect and interstitial defect, there is no change in density of substance.

Short Answer Type Questions

Question 54. Why are liquids and gases categorized as fluids?
Solution: The liquids and gases have a property to flow i.e., the molecules of liquids and gases can easily move fast and tumble over one another freely. Because of their tendency to flow, these have been categorized as fluids.

Question 55. Why are solids incompressible?
Solution: The intemuclear distance between the constituent particles (atoms, molecules or ions) in solids are very less. On bringing them further closer, there will be large repulsive force between electron clouds of these particles. Therefore, solids cannot be compressed.

Question 56. In spite of long range order in the arrangement of particles why are the crystals usually not perfect?
Solution:Crystals have long range in the arrangement of particles but usually the crystals are not perfect this is because when crystallisation occurs at a fast rate or moderate rate, the constituent particles may not get sufficient time to arrange themselves in a perfect order.

Question 57. Why does table salt, NaCl sometimes appear yellow in colour?
Solution: The yellow colour of sodium chloride crystals is due to metal excess defect. In this defect, the unpaired electrons get trapped in anion vacancies. These sites are called F-centres. The‘yellow colour results by excitation of these electrons when they absorb energy from the visible light falling on the crystals.

Question 58. Why is FeO(s) not formed in stoichiometric composition?
Solution: Iron oxide (FeO) has rock salt structure.
In this case, O2 ions adopt on sites and Fe2+ ions should occupy non- stoichiometric. This is the ideal arrangement.
This oxide is always non-stoichiometric i.e., the composition of Fe2+ and O2 ions is not 1 : 1. It is 0.95 : 1 i.e. Fe0 95O(Wustite)
This composition can be obtained if a small number of Fe2+ ions are replaced by two-thirds of’Fe3+ ions in Oh sites.
Eventually there would be less amount of metal as compared to stoichiometric composition.

Question.59. Why does white ZnO (s) become yellow upon heating?
Solution: When ZnO is heated, it splits up to give Zn2+, electrons and colour because of the following reasons:
The excess Zn ions thus formed get entrapped in the interstitial site and electron in the neighborhood vacant interstitial sites. This electron is responsible for the colour and electrical conductivity in crystals.

Question 60. Why does the electrical conductivity of semiconductors increase with rise in temperature?
Solution: The energy gap between valence band and conduction band is small. At room temperature, they do not conduct electricity but when temperature is raised large number of electron from valence band get sufficient energy to jump to conduction band. This is known as thermodynamic conduction in intrinsic semiconductors. Thus, they become more conducting as the temperature increases.

Question 61. Explain why does conductivity of germanium crystals increase on doping with gallium?
Solution: p-type semiconductor:

•  When Ge is doped with group 13 elements, for example, gallium, the structure of crystal lattice does not change.
•  3 valence electrons of gallium are used up in the normal covalent bond.
• For one dopant atom, one hole is created because the place where fourth electron is missing is called vacancy or hole and is responsible for conduction of germanium doped with gallium.

Electron from neighbouring atom comes and fills the hole, thereby creating a hole in its original position.
Under the influence of electric field electrons move towards positively charged plates through these and conduct electricity. The holes appear to move towards negatively charged plates.

Question 62. In a compound, nitrogen atoms (N) make cubic close packed lattice and metal atoms (M) occupy one-third of the tetrahedral voids present. Determine the formula of the compound formed by M and N?
Solution: In ccp, no. of atoms per unit cell = 4
Thus, of tetrahedral voids = 2 x No. of atoms in ccp =2 x 4=8
Only one-third of tetrahedral voids are occupied by metal M. No. of

Question 63. Under which situations can an amorphous substance change to crystalline form?
Solution: An amorphous solid on heating at some temperature may become crystalline. Slow heating and cooling over a long period makes an amorphous solid acquires some crystalline character.

Matching Column Type Questions

Question 64. Match the defects given in Column I with the statements given in Column II.

Solution: (i) -> (c); (ii) -> (a); (iii) ->(d); (iv) ->(b)

Question 65. Match the type of unit cell given in Column I with the features given in Column II.

Solution: (i) —>(b, c); (ii) —>(c, d); (iii) —>(c, e); (iv) —> (a, d)
(i) For primitive unit cell, a = b = c
Total number of atoms per unit cell = 1/8 x 8 = 1
Here, 1/8 is due to contribution of each atom present at comer.
(ii) For body centered cubic unit cell, a = b = c.
This lattice contains atoms at comer as well as body centre. Contribution due to atoms at comer = 1/8 x 8 = 1 contribution due to atoms at body centre = 8

Question 66. Match the types of defect given in Column I with the statement given in Column II.

Solution: (i) —>(c); (ii) —>(a); (iii) —> (b)
(A) (i) Impurity defects: The defects introduced in the crystal lattice due to presence of the certain impurity are called impurity defects.
Example: Substitution of Na⁺ ions in NaCl by Sr²⁺ ions.
Structure with defect:

Impurity defect due to substitution of Na⁺ ions in NaCl by Sr²⁺ ions (Cation vacancy) ‘Schottky Defect’
(B) When NaCl is heated in vapour of sodium some of the Cf leave their lattice site and create anion vacancies. This chloride ion wants to combine with sodium vapour to form sodium chloride. For doing so sodium atom loses electrons form Na+ ions. This released electron diffuses into the crystal to get entrapped in the anion vacancy called F-centre.

(C) Metal deficiency is caused due to cation vacancy created by replacement of some lower valent ions by its higher valentions.
Note: Cation vacancies are found in crystals in which metals have different oxidation states.
Example: FeO, FeS, NiO

Question 67. Match the items given in Column I with the items given in Column II.

Solution:(i) -> (d); (ii)-> (c); (iii)->(b); (iv) ->(a)
(i) Mg in solid state show electronic conductivity due to presence of free electrons hence, they are known as electronic conductors.
(ii) MgCl₂ in molten state show electrolytic conductivity due to presence of electrolytes in molten state.
(iii) Silicon doped with phosphorus contain one extra electron due to which it shows conductivity under the influence of electric field and known as p-type semiconductor.
(iv) Germanium doped with boron contains one hole due to which it shows conductivity under the influence of electric field and known as n-type semiconductor.

Question 68. Match the type of-packing given in Column I with the items given in Column II.

Solution: (i) —> (c); (ii) —> (a); (iii) —>(d); (iv) —> (b)
(i) Square close packing in two dimensions each sphere have coordination number 4, as shown below.

(ii) Hexagonal close packing in two dimensions each sphere has coordina¬tion number 6 as shown below and creates a triangular void

(iii)Hexagonal close packing in 3 dimensions is a repeated pattern of sphere in alternate layers also known as ABAB pattern

(iv) Cubic close packing in a 3 dimensions is a repeating pattern of sphere in every fourth layer.

Assertion and Reason Type Questions:

In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices.
(a) Assertion and Reason both are correct statements and Reason is the correct explanation for Assertion.
(b) Assertion and Reason both are correct statements but Reason is not the correct explanation for Assertion.
(c) Assertion is correct but Reason is wrong.
(d) Assertion is wrong but Reason is correct.

Question 69. Assertion (A): The total number of atoms present in a simple cubic unit cell is one.
Reason (R): Simple cubic unit cell has atoms at its comers, each of which is shared between eight adjacent unit cells.
Solution: (a) In simple cubic unit cell, only comers are occupied by atoms. Thus, total number of atoms present in the unit cell will be one.

Question 70. Assertion (A): Graphite is a good conductor of electricity, however, diamond belongs to the category of insulators.
Reason (R): Graphite is soft in nature on the other hand diamond is very hard and brittle.
Solution: (b) Diamond is bad conductor of electricity because all valence e of carbon are involved in bonding. In graphite however 3 out of 4 valence electrons are involved in bonding, fourth electron remains free between adjacent layers which makes it a good conductor.
Graphite is soft because parallel layers are held together by week van der Waals force. However, diamond is hard due to compact three-dimensional network of bonding.

Question 71. Assertion (A): Total number of octahedral voids present in unit cell of cubic close packing including the one that is present at the body centre, is four. Reason (R): Besides the body centre, there is one octahedral void present at the centre of each ofthe six faces of the unit cell and each of which is shared between two adjacent unit cells.
Solution:(c) All edge centres and body centre represent octahedral void.
Total number of octahedral voids = 12 x 1/4 +1 = 4

Question 72. Assertion (A): The packing efficiency is maximum for the fee structure. Reason (R): The coordination number is 12 in fee structures.
Solution: (b) In fee unit cell, there is cep arrangement with packing efficiency of 74.01% which is maximum. In cep arrangement, coordination number is 12.

Question 73. Assertion (A): Semiconductors are solids with conductivities in the intermediate range from

Reason (R): Intermediate, conductivity in semiconductor is due to partially filled valence band.
Solution: (c) Conductance of semiconductors lies between metals and insulators, i.e., in the range of

Long Answer Type Questions

Question 74. With the help of a labelled diagram show that there are four octahedral voids per unit cell in a cubic close packed structure.
Solution: In cep, each cube consists of eight cubic components, number of atoms per unit cell in ccp is

Question.75. Show that in a cubic close packed structure, eight tetrahedral voids are present per unit cell.
Solution: In ccp, each cube consists of eight cubic components. Number of atoms per unit cell in ccp is

Position of tetrahedral Voids = At the centre of each cubic component Number of tetrahedral voids per unit cell in cubic close packing = 8×1=8 Number of tetrahedral Voids = 8.

Question 76. How does the doping increase the conductivity of semiconductors?
Solution:The conductivity of semiconductors is increased by adding an appropriate amount of suitable impurity or doping. Doping can be done with an impurity which is electron rich or electron deficient as compared to the intrinsic semiconductor, silicon or germanium. Such impurities introduce electronic defects in them. When silicon is doped with electron rich impurities the extra electron becomes delocalized. These delocalized electrons increase the conductivity of doped silicon due to the negatively charged electron, hence silicon doped with electron-rich impurity is called n-type semiconductor while electron-deficit impurities increase the conductivity through positive holes and this type of semiconductors are called /?-type semiconductors.

Question 77. A sample of ferrous oxide has actual formula Fe_0.93 O_1.00. In this sample, what fraction of metal ions are Fe²⁺ ions? What is the type of non-stoichiometric defect present in this sample? ’
Solution:

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NCERT Exemplar Class 12 Physics Chapter 15 Communication Systems are part of NCERT Exemplar Class 12 Physics. Here we have given NCERT Exemplar Class 12 Physics Chapter 15 Communication Systems.

NCERT Exemplar Class 12 Physics Chapter 15 Communication Systems

Multiple Choice Questions (MCQs)

Single Correct Answer Type

Question 1. Three waves A, B and C of frequencies 1600 kHz, 5 MHz and 60 MHz respectively are to be transmitted from one place to another. Which of the following is the most appropriate mode of communication?
(a) A is transmitted via space wave while B and C are transmitted via sky wave
(b) A is transmitted via ground wave, B via sky wave and C via space wave
(c) B and C are transmitted via ground wave while A is transmitted via sky wave
(d) B is transmitted via ground wave while A and C are transmitted via space wave
Solution: (b)
Key concept: The radio waves emitted from a transmitter antenna can reach the receiver antenna by the following mode of operation.
• Ground wave propagation
• Sky wave propagation
• Space wave propagation
Mode of communication frequency range:
• Ground wave propagation— 500 kHz to 1710 kHz
• Sky wave propagation — 2 MHz to 40 MHz
• Space wave propagation— 54 MHz to 42 GHz
So, A is transmitted via ground wave, B via sky wave and C via space wave.

Question 2. A loom long antenna is mounted on a 500 m tall building. The complex can become a transmission tower for waves with λ.
(a) ~400m    (b) -25 m    (c) -150 m     (d) -2400 m
Solution: (a) Length of the building (l) is
l = 500 m
and length of antenna = 100 m
and we know, wavelength of the wave which can be transmitted by
L =λ/4. So, λ~  4l= 4 x 100 = 400 m
Wavelength (λ) is nearly equal to 400 m.

Question 3. A1 kW signal is transmitted using a communication channel which provides attenuation at the rate of -2dB per km. If the communication channel has a total length of 5 km, the power of the signal received is [gain in  d_B =10 log₁₀(p₀/p_i)]
(a) 900 W    (b) 100 W    (c) 990 W    (d) 1010 W
Solution:

Question 4. A speech signal of 3 kHz is used to modulate a carrier signal of frequency 1 MHz, using amplitude modulation. The frequencies of the side bands will be
(a) 1.003 MHz and 0.997 MHz     (b) 3001 kHz and 2997 kHz
(c) 1003 kHz and 1000 kHz          (d) 1 MHz and 0.997 MHz
Solution: (a)
Key concept: The process of changing the amplitude of a carrier wave in accordance with the amplitude of the audio frequency (AF) signal is known as amplitude modulation (AM).
In AM, frequency of the carrier wave remains unchanged.
Side band frequencies: The AM wave contains three frequencies f_c, (f_c + fm) and (f_c -f_m),f_c is called carrier frequency, (f_c +f_m) and (f_c -f_m) are called side band frequencies.
(f_c +f_m): Upper side band (USB) frequency
(f_c -f_m): Lower side band (LSB) frequency
Side band frequencies are generally close to the carrier frequency.
According to the problem, frequency of carrier signal is f_c = 1 MHz and frequency of speech signal = 3 kHz
= 3 x 10^-3 MHz
= 0.003 MHz
We know that, Frequencies of side bands = (f_c ± f_m) = (1 + 0.003) and (1 – 0.003)
So, side band frequencies are 1.003 MHz and 0.997 MHz.

Question 5. A message signal of frequency ω_m is superposed on a carrier wave of frequency ω_c to get an Amplitude Modulated Wave (AM). The frequency of the AM wave will be

Solution: (b)
Key concept: The process of changing the amplitude of a carrier wave in accordance with the amplitude of the audio frequency (AF) signal is known as amplitude modulation (AM).
In AM, frequency of the carrier wave remains unchanged or we can say that the frequency of modulated wave is equal to the frequency of carrier wave. Now, according to the problem, frequency of carrier wave is f_c.
Thus the amplitude modulated wave also has frequency f_c.

Question 6. I-V Characteristics of 4 devices are shown in figure.

Solution: Key concept: A square law modulator is the device which can produce modulated waves by the application of the message signal and the carrier wave.
Square law modulator is used for modulation purpose. Characteristics shown by (i) and (iii) correspond to linear devices.
And by (ii) corresponds to square law device which shows non-linear relations. Some part of (iv) also follow square law.
Hence, (ii) and (iv) can be used for modulation.

Question 7. A male voice after modulation-transmission sounds like that of a female to the receiver. The problem is due to
(a) poor selection of modulation index (selected 0 < m <1)
(b) poor bandwidth selection of amplifiers
(c) poor selection of carrier frequency
(d) loss of energy in transmission.
Solution: (b) In this problem, the frequency of modulated signal received becomes more, due to improper selection of bandwidth.
This happens because bandwidth in amplitude modulation is equal to twice the frequency of modulating signal.
But, the frequency of male voice is less than that of a female.

Question 8. A basic communication system consists of
A. transmitter.
B. information source.
C. user of information.
D. channel.
E. receiver.
Choose the correct sequence in which these are arranged in a basic communication system.
(a) ABCDE   (b) BADEC   (c) BDACE    (d) BEADC
Solution: (b) A basic communication system consists of an information source, a transmitter, a link (channel) and a receiver or a communication system is the set-up used in the transmission and reception of information from one place to another.
The whole system consist of several elements in a sequence. It can be represented as the diagram given below:

Question 9. Identify the mathematical expression for amplitude modulated wave,

Solution:

One or More Than One Correct Answer Type

Question 10. An audio signal of 15 kHz frequency cannot be transmitted over long distances without modulation, because
(a) the size of the required antenna would be at least 5 km which is not convenient
(b) the audio signal cannot be transmitted through sky waves
(c) the size of the required antenna would be at least 20 km, which is not convenient
(d) effective power transmitted would be very low, if the size of the antenna is less than 5 km
Solution: (a, b, d)
Key concept: Size of the antenna or aerial. For transmitting a signal, we need an antenna or an aerial. This antenna should have a size comparable to the wavelength of the signal (at least 1/4 in dimension) so that the antenna properly senses the time variation of the signal. For an electromagnetic wave of frequency 20 kHz, the wavelength λ is 15 km. Obviously, such a long antenna is not possible to construct and operate. Hence direct transmission of such baseband signals is not practical. We can obtain transmission with reasonable antenna s if transmission frequency is high (for example,
if n is 1 MHz, then λ is 300 m). Therefore, there is a need of translating the information contained in our original low frequency baseband signal into high or radio frequencies before transmission.
Effective power radiated by an antenna: A theoretical study of radiation from a linear antenna (length l) shows that the power radiated is proportional to (1/λ)² . This implies that for the same antenna length, the power radiated increases with decreasing λ, i.e., increasing frequency. Hence, the effective power radiated by a long wavelength baseband signal would be small. For a good transmission,we need high powers and hence this also points out to the need of using high frequency transmission.

Question 11. Audio sine waves of 3 kHz frequency are used to amplitude modulate a carrier signal of 1.5 MHz. Which of the following statements are true?
(a) The side band frequencies are 1506 kHz and 1494 kHz
(b) The bandwidth required for amplitude modulation is 6 kHz
(c) The bandwidth required for amplitude modulation is 3 MHz
(d) The side band frequencies are 1503 kHz and 1497 kHz
Solution: (b, d)

Question 12. A TV transmission tower has a height of 240 m. Signals broadcast from this tower will be received by LOS communication at a distance of (assume the radius of earth to be (6.4 x 10⁶ m)
(a) 100 km     (b) 24 km        (c) 55 km        (d) 50 km
Solution: (b, c, d)

Question 13. The frequency response curve (figure) for the filter circuit used for production of AM wave should be

Solution:(a, b, c)
Key concept:
(i) Side band frequencies-. The AM wave contains three frequencies f_c ,(f_c +f_m) and (f_c-f_m),f_c is called carrier frequency, (f_c +f_m) and (f_c -f_m) are called side band frequencies.
(f_c + f_m)- Upper side band (USB) frequency
(f_c – f_m): Lower side band (LSB) frequency
Side band frequencies are generally close to the carrier frequency,
(ii) Bandwidth: The two side bands lie on either side of the carrier frequency at equal frequency interval ω_m.
So, bandwidth = {(ω_c + ω_m) – (ω_c – ω_m)} = 2ω_m

To produce an amplitude modulated wave, bandwidth is given by the difference between upper side band frequency and lower side band frequency. Bandwidth = ω_USB – ω_LSB = (ω_c + ω_m) – (ω_c – ω_m)

Question 14. In amplitude modulation, the modulation index m is kept less than or equal to 1 because
(a) m> 1, will result in interference between carrier frequency and message frequency, resulting into distortion.
(b) m > 1, will result in overlapping of both side bands resulting into loss of information
(c) m > 1, will result in change in phase between carrier signal and message signal.
(d) m > 1, indicates amplitude of message signal greater than amplitude of carrier signal resulting into distortion.
Solution: (b, d)
Key concept: Modulation index: The ratio of change of amplitude of carrier wave to the amplitude of original carrier wave Is called the modulation factor or degree of modulation or modulation index (m_a).

Very Short Answer Type Questions

Question 15. Which of the following would produce analog signals and which would produce digital signals?
(a) A vibrating tuning fork
(b) Musical sound due to a vibrating sitar string
(c) Light pulse
(d) Output of NAND gate
Solution: Analog and digital signals are the gateway of information or we can say that they are used to transmit information through electric signals. In both these signals, the information such as any audio or video is transformed into electric signals.
The difference between analog and digital technologies is that in analog technology, information is translated into electric pulses of varying amplitude. In digital technology, translation of information is into binary formal (zero or one) where each bit is representative of two distinct amplitudes. So, output of a NAND gate and a light pulse produces a digital signal.
Thus, (a) and (b) would produce analog signal and (c) and (d) would produce digital signals.

Question 16. Would sky waves be suitable for transmission of TV signals of 60 MHz frequency?
Solution: A signal to be transmitted through sky waves must have a frequency range of 1710 kHz to 40 MHz.
But, here the frequency of TV signals are 60 MHz which is beyond the required range (frequency range: there is a maximum frequency of EM waves called critical frequency, above which wave cannot reflect back).
So, sky waves will not be suitable for transmission of TV signals of 60 MHz frequency.
Important point: Sky wave propagation: These are the waves which are reflected back to the earth by ionosphere.
Ionosphere is a layer of atmosphere having charged particles, ions and electrons and extended above 80 km – 300 km from the earth’s surface.

Question 17. Two waves A and B of frequencies 2 MHz and 3 MHz, respectively are beamed in the same direction for communication via sky wave. Which one of these is likely to travel longer distance in the ionosphere before suffering total internal reflection?
Solution: We know that refractive index p of a layer is

The refractive index of wave B is more than refractive index of wave A because frequency of wave B is more than wave A (as refractive index increases with frequency increases).
Sin i / sin r = µ (lesser the value of r larger the value of µ )
For higher frequency wave (i.e., higher refractive index) the angle of refraction is less, i.e., bending is less. So, wave B travels longer distance in the ionosphere before suffering total internal reflection.
Importance point: Refractive index of a medium is that characteristic which decides speed of light in it.
Dependence of Refractive index:
(i) Nature of the media of incidence and refraction.
(ii) Colour of light or wavelength of light.
(iii) Temperature of the media: Refractive index decreases with the increase in temperature.
Total internal reflection: When a ray of light goes from denser to rarer medium it bends away from the normal and as the angle of incidence in denser medium increases, the angle of refraction in rarer medium also increases and at a certain angle, angle of refraction becomes 90°. This angle of incidence is called critical angle (C).
When angle of incidence exceeds the critical angle then light ray comes back into the same medium after reflection from interface. This phenomenon is called Total internal reflection (TIR).

Question 18. The maximum amplitude of an AM wave is found to be 15 V while its minimum amplitude is found to be 3 V. What is the modulation index?
Solution:

Question 19. Compute the LC product of a tuned amplifier circuit required to generate a carrier wave of 1 MHz for amplitude modulation.
Solution:

Question 20. Why is an AM signal likely to be more noisy than a FM signal upon transmission through a channel?
Solution: An AM signal likely to be more noisy than FM signal through a channel because in case of AM, the instantaneous voltage of carrier waver waves is varied by the modulating wave voltage So, during the transmission, noise signals can also be added and receiver assumes noise a part of the modulating signal. In case of FM, the frequency of carrier waves is changed as the change in the instantaneous voltage of modulating waves. This can be done by mixing and not while the signal transmitting in channel. So, noise does not affect FM signal or simply we can say that noise signals are difficult to filter out in AM reception whereas FM receivers easily filter out noise.
Important point: In frequency modulation m_f (frequency modulation index) is inversely proportional to modulating frequency f_m. While in PM it does not vary with modulating frequency. Moreover, FM is more noise immune.

Short Answer Type Questions

Question 21. Figure shows a communication system. What is the output power when input signal is of 1.01 mW? [Gain in d_B = 10 log₁₀ (P₀ / P₁)]

Solution:

Question 22. A TV transmission tower antenna is at a height of 20 m. How much service area can it cover if the receiving antenna is (i) at ground level, (ii) at a height of 25 m? Calculate the percentage increase in area covered in case (ii) relative to case (i).
Solution:

Question 23. If the whole earth is to be connected by LOS communication using space waves (no restriction of antenna size or tower height), what is the minimum number of antennas required? Calculate the tower height of these antennas in terms of earth’s radius.?
Solution:
Key concept: Distance or range of transmission tower, d_T =√2Rh_T
where, R is the radius of the earth (approximately 6400 km). h_T is the height of transmission tower, .
d_T is also called the radio horizon of the transmitting antenna.
Let us consider the figure given below to solve this problem.
Assume the height of transmitting antenna or receiving antenna in order to cover the entire surface of earth through communication is h₁, i.e. h_T = h_R and radius of earth is R. If d_M is the line-of-sight distance between the transmission and receiving antennas, then maximum distance

Question 24. The maximum frequency for reflection of sky waves from a certain layer of the ionosphere is found to be f_max= 9(N_max)^1/2, where N_max is the maximum electron density at that layer of the ionosphere.
On a certain day it is observed that signals of frequencies higher than 5 MHz are not received by reflection from the F₁ layer of the ionosphere while signals of frequencies higher than 8 MHz are not received by reflection from the F₂ layer of the ionosphere. Estimate the maximum electron densities of the F₁ and F₂ layers on that day.
Solution:

Question 25. On radiating (sending out) and AM modulated signal, the total radiated power is due to energy carried by ω_c, (ω_c – ω_m) and (ω_c + ω_m). Suggest ways to minimise cost of radiation without compromising on information.
Solution:
Key concept: Side band frequencies. The AM wave contains three frequencies ω_c, (ω_c + ω_m) and (ω_c – ω_m), ω_c is called carrier frequency, (ω_c + ω_m) and ( ω_c – ω_m) are called side band frequencies.
(ω_c + ω_m) = Upper side band (USB) frequency
(ω_c – ω_m) =Lower side band (LSB) frequency
Side band frequencies are generally close to the carrier frequency.
Only side band frequencies contain information in amplitude modulated signal, [only (ω_c+ ω_m) and (ω_c + ω_m)].
Here, the total radiated power is due to energy carried by ω_c, (ω_c – ω_m) and (ω_c+ω_m)
For reduction of cost of radiation without compromising on information ω_c can be left and transmitting the frequencies (ω_c + ω_m), (ω_c – ω_m) or both (ω_c + ω_m) and (ω_c – ω_m).

Long Answer Type Questions

Question 26. The intensity of a light pulse travelling along a communication channel decreases exponentially with distance x according to the relation I = I₀e^-ax, where I₀ is the intensity at x = 0 and α is the attenuation constant.
(a) Show that the intensity reduces by 75% after a distance of  (In4/α).
(b) Attenuation of a signal can be expressed in decibel (dB) according to the relation dB=10 log₁₀(I/I₀).What is the attenuation in dB/km for an optical fibre in which the intensity falls by 50% over a distance of 50 km?
Solution:

Question 27. A 50 MHz sky wave takes 4.04 ms to reach a receiver via re-transmission from a satellite 600 km above Earth’s surface. Assuming re-transmission time by satellite negligible, find the distance between source and receiver. If communication between the two was to be done by Line of Sight (LOS) method, what should size and placement of receiving and transmitting antenna be?
Solution: Let the receiver is at point A and source is at B.

Question 28. An amplitude modulated wave is as shown in figure. Calculate
(i) the percentage modulation,
(ii) peak carrier voltage and
(iii) peak value of information voltage

Solution:

Question 29. (i) Draw the plot of amplitude versus ω for an amplitude modulated wave whose carrier wave (ω > ω_c) is carrying two modulating signals, ω1 and ω₂ (ω₂> ω₁).
(ii) Is the plot symmetrical about ω_c? Comment especially about plot in region (ω < ω_c ).
(iii) Extrapolate and predict the problems one can expect if more waves are to be modulated.
(iv) Suggest solutions to the above problem. In the process can one understand another advantage of modulation in terms of bandwidth?
Solution:


(ii) In the plotted graph shown, we note that frequency spectrum is not symmetrical about ω_c. Crowding of spectrum is present for ω < ω_c.
(iii) If more modulating signals are present then there will be more crowding in the modulation signal in the region ω <ω_c. That will result more chances of mixing of signal.
(iv) To accommodate more signals, we should increase bandwidth and frequency carrier waves ω_c. This shows that large carrier frequency
enables to carry more information (i.e., more ω_m) and the same will in turn increase bandwidth.

Question 30. An audio signal is modulated by a carrier wave of 20 MHz such that the bandwidth required for modulation is 3 kHz. Could this wave be demodulated by a diode detector which has the values of R and C as
(i) R = 1 kΩ, C= 0.01 µF?
(ii) R= 10 kΩ, C=0.01 µF?
(iii) R = 10 kΩ, C = 0.1 µF?
Solution:

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NCERT Exemplar Class 12 Physics Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits are part of NCERT Exemplar Class 12 Physics. Here we have given NCERT Exemplar Class 12 Physics Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits.

NCERT Exemplar Class 12 Physics Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits

Multiple Choice Questions (MCQs)

Single Correct Answer Type

Question 1. The conductivity of a semiconductor increases with increase in temperature, because
(a) number density of free current carries increases
(b) relaxation time increases
(c) both number density of carries and relaxation time increase
(d) number density of carries increases, relaxation time decreases but effect of decrease in relaxation time is much less than increase in number density .
Solution: (d)

Question 2.

Solution: (b)
Key concept: P-N Junction Diode:
– When a P-type semiconductor is suitably joined to an N-type semiconductor, then resulting arrangement is called P-N junction or P-N junction diode.

(1) Depletion region: On account of difference in concentration of charge carrier in the two sections of P-N junction, the electrons from N-region diffuse through the junction into P-region and the hole from P region diffuse into N-region.
Due to diffusion, neutrality of both N and P-type semiconductor is disturbed, a layer of negative charged ions appear near the junction in the P-crystal and a layer of positive ions, appears near the junction in N-crystal. This layer is called depletion layer.

(2) Potential barrier: The potential difference created across the P-N junction due to the diffusion of electron and holes is called potential barrier.

Height of potential barrier is decreases when p-n junction is forward biased, it opposes the potential barrier junction, when p-n junction is reverse biased, it supports the potential barrier junction, resulting increase in potential barrier across the junction.

Question 3. In figure given on next page, assuming the diodes to be ideal
(a) D₁ is forward biased and D₂ is reverse biased and hence current flows from A to B
(b) D₂ is forward biased and D₁ is reverse biased and hence no current flows from B to A and vice-versa
(c) D₁ and D₂ are both forward biased and hence current flows from A to B
(d) D₁ and D₂ are both reverse biased and hence no current flows from A to B and vice-versa

Solution:

Question 4. A 220 V AC supply is connected between points A and B (figure). What will be the potential difference V across the capacitor?

Solution: (d)
Key concept: Half wave rectifier: When the P-N junction diode rectifies half of the ac wave, it is called half wave rectifier.

Question 5. Hole is
(a) an anti-particle of electron
(b) a vacancy created when an electron leaves a covalent bond
(c) absence of free electrons
(d) an artificially created particle
Solution: (b) Concept of holes in the semiconductor:

1. .When an electron is removed from a covalent bond, it leaves a vacancy behind. An electron from a neighbouring atom can move into this vacancy, leaving the neighbour with a vacancy. In this way the vacancy formed is called a hole (or cotter), and can travel through the material and serve as an additional current carriers.
2. A hole is considered as a seat of positive charge, having magnitude of charge equal to that of an electron.
3. Holes acts as a virtual charge, although there is no physical charge on it.
4. Effective mass of hole is more than an electron.
5. Mobility of hole is less than an electron.

Question 6. The output of the given circuit in figure is given below.

(a) would be zero at all times
(b) would be like a half wave rectifier with positive cycles in output
(c) would be like a half wave rectifier with negative cycles in output
(d) would be like that of a full wave rectifier
Solution: (c)

When the diode is forward biased during positive half cycle of input AC voltage, the resistance of p-n junction is low. The current in the circuit is maximum. In this situation, a maximum potential difference will appear across resistance connected in a series of circuit. This result into zero output voltage across p-n junction.
And when the diode is reverse biased during negative half cycle of AC voltage, the p-n junction is reverse biased. The resistance of p-n junction becomes high which will be more than resistance in series. That is why, there will be voltage across p-n junction with negative cycle in output, hence option (c) is correct.

Question 7. In the circuit shown in figure given below, if the diode forward voltage between A and B is

(a) 1.3 V    (b) 2.3 V    (c) 0    (d) 0.5 V
Solution:

Question 8. Truth table for the given circuit is

Solution:

One or More Than One Correct Answer Type

Question 9. When an electric field is applied across a semiconductor
(a) electrons move from lower energy level to higher energy level in the conduction band
(b) electrons move from higher energy level to lower energy level in the conduction band
(c) holes in the valence band move from higher energy level to lower energy level
(d) holes in the valence band move from lower energy level to higher energy level
Solution: (a, c)
In valence band electrons are not capable of gaining energy from external electric field. While in conduction band the electrons can gain energy from external electric field.
When electric field is applied across a semiconductor, the electrons in the conduction band (which is partially filled with electrons) get accelerated and acquire energy. They move from lower energy level to higher energy level. While the holes in valence band move from higher energy level to lower energy level, where they will be having more energy.

Question 10. Consider an n-p-n transistor with its base-emitter junction forward biased and collector base junction reverse biased. Which of the following statements are true?
(a) Electrons crossover from emitter to collector
(b) Holes move from base to collector
(c) Electrons move from emitter to base
(d) Electrons from emitter move out of base without going to the collector.
Solution: (a, c)

In normal operation base-emitter is forward biased, i.e., the positive pole of emitter base battery is connected to base and its negative pole is connected to the emitter. And collector base junction is reverse biased, i.e., the positive pole of the collector base battery is connected to collector and negative pole to base. Thus, electron moves from emitter to base and crossover from emitter to collector.

Question 11.

Solution: (b, c, d) According to above graph transfer characteristics of a base biased common emitter transistor, we note that .
(a) when V_i= 0.4 V, output voltage remain same,there is no collection current. So, transistor circuit is not in active state.
(b) when V_i = 1 V (This is in between 0.6 V to 2 V), the transistor circuit is in active state and when input is increasing output is decreasing because when CE is used as an amplifier input and output voltages are 180° out of phase. Then it is used as an amplifier.
(c) when V_i = 0.5 V, there is no collector current. The transistor is in cut off state. The transistor circuit can be used as a switch to be turned off.
(d) when V_i = 2.5 V, the collector current becomes maximum and transistor is in a saturation state and can used as switch turned on state.

Question 12. In a n-p-n transistor circuit, the collector current is 10 mA. If 95 per cent of the electrons emitted reach the collector, which of the following statements are true?
(a) The emitter current will be 8 mA
(b) The emitter current will be 10.53 mA
(c) The base current will be 0.53 mA
(d) The base current will be 2 mA
Solution: (b, c) According to the problem, the collector current is 95% of electrons reaching the collector after emission. And collector current, I_C = 10 mA

Question 13. In the depletion region of a diode 
(a) there are no mobile charges
(b) equal number of holes and elections exist, making the region neutral
(c) recombination of holes and electrons has taken place
(d) immobile charged ions exist
Solution: (a, b, d) On account of difference in concentration of charge carrier in the two sections of P-N junction, the electrons from N-rcgion diffuse through the junction into P-region and the hole from P-region diffuse into N-region.

Due to diffusion, neutrality of both N-and P-type semiconductor is disturbed, a layer of negative charged ions appear near the junction in the P-crystal and a layer of positive ions appears near the junction in N-crystal. This layer is called depletion layer.
The thickness of depletion layer is 1 micron = 10^-6 m.
Width of depletion layer ∞  1/Dopping
Depletion is directly proportional to temperature.
Important point: The P-N junction diode is equivalent to capacitor in which the depletion layer acts as a dielectric.

Question 14. What happens during regulation action of a Zener diode?
(a) The current and voltage across the Zener remains fixed
(b) The current through the series Resistance (R_s) changes
(c) The Zener resistance is constant
(d) The resistance offered by the Zener changes

Solution: (b, d) Symbolically zener diode represents like this:
In the forward bias, the zener diode acts as an ordinary diode. It can be used as a voltage regulator.

A zener diode when reverse biases offers constant voltage drop across in terminals as unregulated voltage is applied across circuit to regulate. Then during regulation action of a Zener diode, the current through the series resistance R_s changes and resistance offered by the Zener changes. The current through the Zener changes but the voltage across the Zener remains constant.

Question 15. To reduce the ripples in rectifier circuit with capacitor filter
(a) R_L should be increased
(b) input frequency should be decreased .
(c) input frequency should be increased
(d) capacitors with high capacitance should be used
Solution: (a, c, d)
Ripple factor may be defined as the ratio of r.m.s. value of the ripple voltage to the absolute value of the DC component of the output voltage, usually expressed as a percentage. However ripple voltage is also commonly expressed as the peak-to-peak value. Ripple factor (r) of a full wave rectifier using capacitor filter is given by

Ripple factor is inversely proportional to R_L, C and v.
Thus to reduce r, R_L should be increased, input frequency v should be increased and capacitance C should be increased.

Question 16. The breakdown in a reverse biased p-n junction is more likely to occur due to
(a) large velocity of the minority charge carriers if the doping concentration is small
(b) large velocity of the minority charge carriers if the doping concentration is large
(c) strongelectricfieldinadepletionregionifthedopingconcentrationissmall
(d) strong electric field in the depletion region if the doping concentration is large
Solution: (a, d)
Reverse biasing: Positive terminal of the battery is connected to the N-crystal and negative terminal of the battery is connected to P-crystal.
(i) In reverse biasing width of depletion layer increases
(ii) In reverse biasing resistance offered R_Reverse = 10⁵ Ω
(iii) Reverse bias supports the potential barrier and no current flows across the junction due to the diffusion of the majority carriers.
(A very small reverse current may exist in the circuit due to the drifting of minority carriers across the junction)
(iv) Break down voltage: Reverse voltage at which break down of semiconductor occurs. For Ge it is 25 V and for Si it is 35 V.

So, we conclude that in reverse biasing, ionization takes place because the minority charge carriers will be accelerated due to reverse biasing and striking with atoms which in turn cause secondary electrons and thus more number of charge carriers.
When doping concentration is large, there will be a large number of ions in the depletion region, which will give rise to a strong electric field.

Very Short Answer Type Questions

Question 17. Why are elemental dopants for Silicon or Germanium usually chosen from group XIII or group XV?
Solution: When pure semiconductor material is mixed with small amounts of certain specific impurities with valency different from that of the parent material, the number of mobile electrons/holes drastically changes. The process of addition of impurity is called doping. The size of the dopant atom should be compatible such that their presence in the pure semiconductor does not distort the semiconductor but easily contribute the charge carriers on forming covalent bonds with Silicon or Germanium atoms, which are provided by group XIII or group XV elements.

Question 18. Sn, C and Si, Ge are all group XIV elements. Yet, Sn is a conductor, C is an insulator while Si and Ge are semiconductors. Why?
Solution: The conduction level of any element depends on the energy gap between its conduction band and valence band.
In conductors, there is no energy gap between conduction band and valence band. For insulator, the energy gap is large and for semiconductor the energy gap is moderate.
The energy gap for Sn is 0 eV, for C is 5.4 eV, for Si is 1.1 eV and for Ge is 0.7 eV related to their atomic size. Therefore Sn is a conductor, C is an insulator, and Ge and Si are semiconductors

Question 19. Can the potential barrier across a p-n junction be measured by simply connecting a voltmeter across the junction?
Solution: We cannot measure the potential barrier across a p-n junction by a voltmeter because the resistance of voltmeter is very high as compared to the junction resistance. Potential of potential barrier for Ge is V_B = 0.3 V and for silicon is V_B = 0.7 V.
On the average the potential barrier in P-N junction is ~0.5 V.

Question 20. Draw the output waveform across the resistor in the given figure.

Solution: The diode act as a half wave rectifier, it offers low resistance when forward biased and high resistance when reverse biased. So the output is obtained only when positive input is given,so the output waveform is

Question 21. The amplifiers X, Y and Z are connected in series. If the voltage gains of X, Y and Z are 10,20 and 30, respectively and the input signal is 1 mV peak value, then what is the output signal voltage (peak value).
(i) if DC supply voltage is 10 V?
(ii) if DC supply voltage is 5 V?
Solution: Total voltage amplification is defined as the ratio of output signal voltage and input signal voltage.

Question 22. In a CE transistor amplifier, there is a current and voltage gain associated with the circuit. In other words there is a power gain. Considering power a measure of energy, does the circuit violate conservation of energy?
Solution:


The power gain is very high in CE transistor amplifier. In this circuit, the extra power required for amplified output is obtained from DC source. Thus, the circuit used does not violate the law of conservation.

Short Answer Type Questions

Question 23. (i) Name the type of a diode whose characteristics are shown in figure (a) and (b).
(ii) What does the point P in fig. (a) represent?
(iii) What does the points P and Q in fig. (b) represent?

Solution:
(i) Fig. (a) represents the characteristics of Zener diode and curve (b) is of solar cell.
(ii) In fig. (a), point P represents Zener breakdown voltage.
(iii) In fig. (b), the point Q represents zero voltage and negative current. Which means the light falling on solar cell with atleast minimum threshold frequency gives the current in opposite direction to that due to a battery connected to solar cell. But for the point Q the battery is short circuited. Hence it represents the short circuit current.
And the point Pin fig. (b) represents some open circuit, voltage on solar cell with zero current through solar cell.
It means, there is a battery connected to a solar cell which gives rise to the equal and opposite current to that in solar cell by virtue of light falling on it.

Question 24. Three photo diodes D₁, D₂ and D₃ are made of semiconductors having band gaps of 2.5 eV, 2 eV and 3 eV, respectively. Which ones will be able to detect light of wavelength 6000 Å ?
Solution:Key concept: In Photo diodes electron and hole pairs are created by junction photoelectric effect. That is the covalent bonds are broken by the EM radiations absorbed by the electron in the V.B. These are used for detecting light signals.

The incident radiation which is detected by the photodiode D₂ because energy of incident radiation is greater than the band-gap.

Question 25. If the resistance R₁ is increased (see figure), how will the readings of the ammeter and voltmeter change?

Solution: Let us redrawn the circuit diagram to find the change in reading of ammeter and voltmeter.


So, R₁ is increased, I_B is decreased.
Now, the current in ammeter is collector current I_C.
I_C =βI_B as I_B is decreased, I_C is also decreased and the reading of voltmeter and ammeter also decreased.

Question 26. Two car garages have a common gate which needs to open automatically when a car enters either of the garages or cars enter both. Draw a circuit that resembles this situation using diodes for this situation.
Solution: As car enters in either of the garages or both, the common gate opened automatically.
This means that if any one input is high, output will high otherwise low.
The device is shown like this:

Question 27. How would you set up a circuit to obtain NOT gate using a transistor?
Solution:
(1)It has only one input and only one output.
(2)Boolean expression is Y = Ᾱ and is read as “y equals not A” .
Logical symbol of NOT gate.

(3)Realization of NOT gate: The transistor is so biased that the collector voltage V_CC = V (Voltage corresponding to 1 state)
The resistors R and R_B are so chosen that if the input is low, i.e. 0, the transistor is in the cut off and hence the voltage appearing at the output will be the same as applied V = 5 V. Hence Y= V(or state I)
If the input is high, the transistor current is in saturation and the net voltage at the output Y is 0 (in state 0).

Question 28. Explain why elemental semiconductor cannot be used to make visible LEDs.
Solution:

Question 29. Write the truth table for the circuit shown in figure given below. Name the gate that the circuit resembles.

Solution:

Question 30.

Solution:

Long Answer Type Questions

Question 31. If each diode in figure has a forward bias resistance of 25 Ω and infinite resistance in reverse bias, what will be the values of the currents I₁, I₂, I₃ and I₄?

Solution:

Question 32. In the circuit shown in figure, when the input voltage of the base resistance is 10 V, V_BE is zero and V_CE is also zero. Find the values of I_B, I_C and β .

Solution:

Question 33. Draw the output signals C₁ and C₂ in the given combination of gates.

Solution:

Question 34. Consider the circuit arrangement shown in figure for studying input and output characteristics of n-p-n transistor in.CE configuration.
Select the values of R_B and R_C for a transistor whose V_BE = 0.7 V so that the transistor is operating at point Q as shown in the characteristics (see figure).

Solution:

Question 35. Assuming the ideal diode, draw the output waveform for the circuit given in figure, explain the waveform.

Solution:
Key concept: An ideal diode is a diode in which it has a very large resistance in reverse biased and very, low resistance in forward biased. So, it acts like a perfect conductor when voltage is applied forward biased and like a perfect insulator when voltage is applied reverse biased.
In reverse biased when the input voltage is equal to or less than 5 V diode,then it will offer high resistance in comparison to resistance (R) in series. Now, diode appears in open circuit. The input waveform is then passed to the output terminals. The result with sin wave input is to dip off all positive going portion above 5 V.
If input voltage is greater than +5 V, diode is in conducting state, then it will be conducting as if forward biased offering low resistance in comparison to R. But there will be no voltage in output beyond 5 V as the voltage beyond +5 V will appear across R.
When input voltage is negative, there will be opposition to 5 V battery in p-n junction input voltage becomes more than -5 V, the diode will be reverse biased. It will offer high resistance in comparison to resistance R in series. Now junction diode appears in open circuit. The input wave form is then passed on to the output terminals.
The output waveform will be like this (as shown below).

Question 36. Suppose a n-type wafer is‘created by doping Si crystal having 5 x 10²⁸ atoms/m³ with 1 ppm concentration of As. On the surface 200 ppm boron is added to create ‘p’ region in this wafer. Considering  n_i = 1.5 x 10¹⁶ m^-3,
(i) Calculate the densities of the charge carriers in the n and p regions,
(ii) Comment which charge carriers would contribute largely for the reverse saturation current when diode is reverse biased.
Solution: n-type wafer is created when As is implanted in Si crystal. The number of majority carriers electrons due to doping of As is

Question 37. An XOR gate has the following truth table.

It is represented by following logic relation Y = Ᾱ . B + A . B’ . Build this gate using AND, OR and NOT gates.
Solution: XOR gate can be realized by the combination of two NOT gates, two AND gates and one OR gate. According to the problem, the logic relation for the . given truth table is

Question 38. Consider a box with three terminals on top of it as shown in figure.

Three components namely, two germanium diodes and one resistor are connected across these three terminals in some arrangement.
A student performs an experiment in which any two of these three terminals are connected in the circuit as shown in figure.
The student obtains graphs of current-voltage characteristics for unknown combination of components between the two terminals connected in the circuit. The graphs are
(i) when A is positive and B is negative .



Solution: The V-I characteristics of these graph is discussed in points:
(a)In V-I graph of condition (i), a reverse characteristics is shown in fig. (c). Here A is connected to n-side of p-n junction I and B is connected top-side of p-n junction I with a resistance in series.
(b)In V-I graph of condition (ii), a forward characteristics is shown in fig. (d), where 0.7 V is the knee voltage of p-n junction I. 1/slope = (1/1000) Ω.
It means A is connected to n-side of p-n junction I and B is connected to p-side of p-n junction I and resistance R is in series of p-n junction I between A and B.
(c)In V-I graph of condition (iii), a forward characteristics is shown in figure (e) , where 0.7 V is the knee voltage. In this case p-side of p-n junction II is connected to C and n-side of p-n junction II to B.
(d)In V-I graphs of conditions (iv), (v), (vi) also concludes the above connection of p-n junctions I and II along with a resistance R.
Thus, the arrangement of p-n I, p-n II and resistance R between A, B and C will be as shown in the figure.

Question 39. For the transistor circuit shown in figure, evaluate V_E, R_B, R_E, given I_C= 1 mA,  V_CE= 3, V_BE = 0.5 V  and  V_CC= 12 V, β= 100.

Solution: Let us redraw the circuit diagram given here to solve this problem.

Question 40. In the circuit shown in figure, find the value of R_{c .}

.

Solution:

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NCERT Exemplar Class 8 Maths Chapter 4 Linear Equations in One Variable are part of NCERT Exemplar Class 8 Maths. Here we have given NCERT Exemplar Class 8 Maths Chapter 4 Linear Equations in One Variable.

NCERT Exemplar Class 8 Maths  Chapter 4 Linear Equations in One Variable

Multiple Choice Questions
Question. 1 The solution of which of the following equations is neither a fraction nor an integer?
(a) -3x + 2=5x + 2 (b)4x-18=2 (c)4x + 7 = x + 2 (d)5x-8 = x +4
Solution. For option (c)

Question. 2 The solution of the equation ax + b = 0 is
Solution.

Question. 3 If 8x – 3 = 25 + 17x, then x is
(a) a fraction (b) an integer
(c) a rational number (d) Cannot be solved
Solution. (c) Given, 8x-3 = 25+17x

Question. 4 The shifting of a number from one side of an equation to other is called
(a) transposition (b) distributivity
(c) commutativity (d) associativity .
Solution. (a) The shifting of a number from one side of an equation to other side is called transposition.
e.g. x +a = 0is the equation, x = -a
Here, number ‘a’ shifts from left hand side to right hand side.

Question. 5 If \(\frac { 5x }{ 3 }\)-4 =\(\frac { 2x }{ 5 }\) , then the numerical value of 2x – 7 is
(a)\(\frac { 19 }{ 13 }\) (b)\(\frac { -13 }{ 19 }\)
(c)0 (d)\(\frac { 13 }{ 19 }\)
Solution.(b)

Question. 6 The value of x, for which the expressions 3x – 4 and 2x + 1 become equal, is
(a) -3 (b) 0
(c) 5 x (d) 1
Solution. (c) Given expressions 3x – 4 and 2x + 1 are equal.
Then, 3x-4 = 2x + 1
3x- 2x = 1 + 4 [transposing 2x to LHS and -4 to RHS]
x = 5
Hence, the value of x is 5.

Question. 7 If a and b are positive integers, then the solution of the equation ax = b has to be always
(a) positive (b) negative (c) one (d) zero
Solution. (a) If ax = b, then x = \(\frac { b }{ a }\)
Since, a and b are positive integers. So,\(\frac { b }{ a }\) is also positive integer, Hence, the solution of the given equation has to be always positive.

Question. 8 Linear equation in one variable has
(a) only one variable with any power
(b) only one term with a variable
(c) only one variable with power 1
(d) only constant term
Solution. (c) Linear equation in one variable has only one variable with power 1.
e.g. 3x + 1 = 0,2y – 3 = 7 and z + 9 = – 2 are the linear equations in one variable.

Question. 9 Which of the following is a linear expression?
(a) \({ x }^{ 2 }\) +1 (b) y + \({ y }^{ 2 }\)
(c) 4 (d) 1 + z
Solution. (d) We know that, the algebraic expression in one variable having the highest power of the variable as 1, is known as the linear expression.
Here, 1 + z is the only linear expression, as the power of the variable z is 1.

Question.10 A linear equation in one variable has
(a) only one solution (b) two solutions
(c) more than two solutions (d) no solution
Solution. (a) A linear equation in one variable has only one solution.
e.g. Solution of the linear equation ax + b = 0 is unique, i.e. x = \(\frac { -b }{ a }\)

Question. 11 The value of S in \(\frac { 1 }{ 3 }\) + S = \(\frac { 2 }{ 5 }\) is
(a)\(\frac { 4 }{ 5 }\) (b)\(\frac { 1 }{ 15 }\)
(c)10 (d)0
Solution.(b) Given, \(\frac { 1 }{ 3 }\) + S = \(\frac { 2 }{ 5 }\)

Question.12 If –\(\frac { 4 }{ 3 }\) y = –\(\frac { 3 }{ 4 }\) then y is equal to
(a)\(-{ \left[ \frac { 3 }{ 4 } \right] }^{ 2 }\) (b)\(-{ \left[ \frac { 4 }{ 3 } \right] }^{ 2 }\)
(c)\({ \left[ \frac { 3 }{ 4 } \right] }^{ 2 }\) (d)\({ \left[ \frac { 4 }{ 3 } \right] }^{ 2 }\)
Solution.

Question. 13 The digit in the ten’s place of a two-digit number is 3 more than the digit in the unit’s place. Let the digit at unit’s place be b. Then, the number is
(a) 11b+30 (b) 10b+ 30
(c) 11 b + 3 (d) 10b + 3
Solution. (a) Let digit at unit’s place be b.
Then, digit at ten’s place = (3 + b)
Number = 10 (3 + b) + b – 30 + 10b + b = 11b + 30

Question. 14 Arpita’s present age is thrice of Shilpa. If Shilpa’s age three years ago was x, then Arpita’s present age is
(a) 3 (x – 3) (b)3x + 3
(c) 3x – 9 (d) 3(x + 3)
Solution. (d) Given, Shilpa’s age three years ago = x
Then, Shilpa’s present age = (x + 3)
Arpita’s present age = 3 x Shilpa’s present age = 3 (x + 3)

Question. 15 The sum of three consecutive multiples of 7 is 357. Find the smallest multiple.(a) 112 (b) 126 (c) 119 (d) 116
Solution.

Fill in the Blanks
In questions 16 to 32, fill in the blanks to make each statement true.
Question. 16 In a linear equation, the——— power of the variable appearing in the equation is one.
Solution. highest
e.g. x + 3 = O and x + 2 = 4 are the linear equations.

Question. 17 The solution of the equation 3x – 4 = 1 – 2x is————- .
Solution.

Question. 18 The solution of the equation 2y = 5y-\(\frac { 18 }{ 5 }\) is————.
Solution.

Question. 19 Any value of the variable, which makes both sides of an equation equal, is known as a———–of the equation.
Solution. e.g. x + 2 = 3 => x = 3-2 = 1 [transposing 2 to RHS]
Hence, x = 1 satisfies the equation and it is a solution of the equation.

Question. 20 9x – ……………….. = – 21 has the solution (- 2).
Solution. 3
Let 9x-m= -21 has the solution (-2).

Question. 21 Three consecutive numbers whose sum is 12 are——–,————-and———.
Solution.

Question. 22 The share of A when Rs 25 are divided between A and B, so that A gets Rs 8 more than B, is——–.
Solution.

Question. 23 A term of an equation can be transposed to the other side by changing its—-.
Solution. sign
e.g. x + a = 0 is a linear equation. .
=> x = -a
Hence, the term of an equation can be transposed to the other side by changing its sign.

Question. 24 On subtracting 8 from x, the result is 2. The value of x is——–.
Solution.

Question. 25 \( \frac { x }{ 5 }\) + 30 = 18 has the solution as——–.
Solution.

Question. 26 When a number is divided by 8, the result is -3. The number is——–.
Solution.

Question. 27 When 9 is subtracted from the product of p and 4, the result is 11. The value of p is—-.
Solution.

Question. 28 If \( \frac { 2 }{ 5 }\) x-2=5-\( \frac { 3 }{ 5 }\) x,then x=——-.
Solution.

Question. 29 After 18 years, Swarnim will be 4 times as old as he is now. His present age is——–.
Solution.

Question. 30 Convert the statement ‘adding 15 to 4 times x is 39’ into an equation.
Solution. 4x+ 15=39
To convert the given statement into an equation, first x is multiplied by 4 and then 15 is added to get the result 39. i.e. 4x + 15=39

Question. 31 The denominator of a rational number is greater than the numerator by 10. If the numerator is increased by 1 and the denominator is decreased by 1, then expression for new denominator is——.
Solution.

Question. 32 The sum of two consecutive multiples of 10 is 210. The smaller multiple is——-.
Solution.

True/False
In questions 33 to 48, state whether the statements are True or False.
Question. 33  3 years ago, the age of boy was y years. His age 2 years ago was (y — 2) years.
Solution. False
Given, 3 yr ago, age of boy = y yr
Then, present age of boy = (y + 3)yr
2 yr ago, age of boy = y + 3-2 = (y + 1)yr

Question. 34 Shikha’s present age is p years. Reemu’s present age is 4 times the present age of Shikha. After 5 years, Reemu’s age will be 15p years.
Solution. False
Given, Shikha’s present age = pyr
Then, Reemu’s present age = 4 x (Shikha’s present age) = 4pyr After 5 yr, Reemu’s age = (4p+5)yr

Question. 35 In a 2-digit number, the unit’s place digit is x. If the sum of digits be 9, then the number is (10x – 9).
Solution. False
Given, unit’s digit = x
and sum of digits = 9
Ten’s digit = 9 – x
Hence, the number = 10 (9 -x)+x = 90 -10x + x = 90 – 9x

Question. 36 Sum of the ages of Anju and her mother is 65 years. If Anju’s present age is y years, then her mother’s age before 5 years is (60 – y) years.
Solution. True
Given, Anju’s present age = y yr
Then, Anju’s mother age = (65 – y)yr
Before 5 yr, Anju’s mother age = 65 – y – 5 = (60 – y)yr

Question. 37 The number of boys and girls in a class are in the ratio 5 : 4. If the number of boys is 9 more than the number of girls, then number of boys is 9.
Solution. False
Let the number of boys be 5x and the number of girls be 4x.
According to the question, – 5x – 4x = 9 => x = 9
Hence, number of boys = 5 x 9 = 45

Question. 38 A and B are together 90 years old. Five years ago, A was thrice as old as B was. Hence, the ages of A and B five years back would be (x – 5) years and (85 – x) years, respectively.
Solution. True
Let the age of A be x yr.
Then, age of S = (90 – x) yr
Five years ago, the age of A = (x- 5) yr
The age of B= 90-x-5 = (85-x)yr
Hence, the ages of A and 8 five years back would be (x – 5) yr and (85 – x) yr, respectively.

Question. 39 Two different equations can never have the same answer.
Solution. False
Two different equations may have the same answer.
e.g.2x + 1 = 2 and 2x – 5 = – 4 are the two linear equations whose solution is \(\frac { 1 }{ 2 }\)

Question. 40 In the equation 3x – 3 = 9, transposing – 3 to RHS, we get 3x = 9.
Solution. False
Given, 3x – 3 = 9
=> 3x = 9 + 3 [transposing -3 to RHS]
=> 3x = 12

Question. 41 In the equation 2x = 4 – x, transposing – x to LHS, we get x = 4.
Solution. False
Given, 2x = 4-x
=> 2x + x = 4 [transposing -x to LHS]
=> 3x = 4

Question. 42

Solution.

Question. 43

Solution.

Question. 44 If 6x = 18, then 18x = 54.
Solution.

Question. 45 If \(\frac { x }{ 11 } \) , then x=\(\frac { 11 }{ 15 } \).
Solution.

Question. 46 If x is an even number, then the next even number is 2(x +1).
Solution. False
Given, x is an even number. Then, the next even number is (x + 2).

Question. 47 If the sum of two consecutive numbers is 93 and one of them is x, then the other number is 93 – x.
Solution. True
Given, one of the consecutive number = x
Then, the next consecutive number = x + 1

Question. 48 Two numbers differ by 40. When each number is increased by 8, the bigger becomes thrice the lesser number. If one number is x, then the other number is (40 – x).
Solution.

In Questions 49 to 78, solve the following.
Question. 49 \(\frac { 3x-8 }{ 2x } =1\).
Solution.

Question. 50 \(\frac { 5x }{ 2x-1 } =2\).
Solution.

Question. 51 \(\frac { 2x-3 }{ 4x+5 } =\frac { 1 }{ 3 }\) .
Solution.

Question. 52 \(\frac { 8}{ x } =\frac { 5 }{ x-1 }\) .
Solution.

Question. 53 \(\frac { 5(1-x)+3(1+x) }{ 1-2x } =8 \) .
Solution.

Question. 54 \(\frac { 0.2x+5 }{ 3.5x-3 } =\frac { 2 }{ 5 }\)
Solution.

Question. 55 \( \frac { y-(4-3y) }{ 2y-(3y+4y) } =\frac { 1 }{ 5 }\)
Solution.

Question. 56 \(\frac { x }{ 5 } =\frac { x-1 }{ 6 }\)
Solution.

Question. 57 0.4(3x-1)=0.5x +1
Solution.

Question. 58 8x-7-3x=6x-2x-3
Solution.

Question. 59 10x-5-7x=5x+15-8
Solution.

Question. 60 4t-3-(3t+1)=5t-4
Solution.

Question. 61 5(x-1)-2(x+8)=0
Solution.

Question. 62 \(\frac { x }{ 2 } -\frac { 1 }{ 4 } \left( x-\frac { 1 }{ 3 } \right) =\frac { 1 }{ 6 } \left( x+1 \right) +\frac { 1 }{ 12 }\)
Solution.

Question. 63 \(\frac { 1 }{ 2 } \left( x+1 \right) +\frac { 1 }{ 3 } \left( x-1 \right) =\frac { 5 }{ 12 } \left( x-2 \right)\)
Solution.

Question. 64 \(\frac { x+1 }{ 4 } =\frac { x-2 }{ 3 }\)
Solution.

Question. 65 \(\frac { 2x-1 }{ 5 } =\frac { 3x+1 }{ 3 }\)
Solution. Given \(\frac { 2x-1 }{ 5 } =\frac { 3x+1 }{ 3 }\)

Question. 66 1-(x-2)-[(x-3)-(x-1)]=0
Solution.

Question. 67 \(3x-\frac { x-2 }{ 3 } =4-\frac { x-1 }{ 4 }\)
Solution.

Question. 68 \( \frac { 3t+5 }{ 4 } -1=\frac { 4t-3 }{ 5 }\)
Solution.

Question. 69 \( \frac { 2y-3 }{ 4 } -\frac { 3y-5 }{ 2 } =y+\frac { 3 }{ 4 }\)
Solution.

Question. 70 0.25(4x-5)=0.75x +8
Solution.

Question. 71 \(\frac { 9-3y }{ 1-9y } =\frac { 8 }{ 5 }\)
Solution.

Question. 72 \( \frac { 3x+2 }{ 2x-3 } =-\frac { 3 }{ 4 }\)
Solution.

Question. 73 \( \frac { 5x+1 }{ 2x } =-\frac { 1 }{ 3 }\)
Solution.

Question. 74 \(\frac { 3t-2 }{ 3 } +\frac { 2t+3 }{ 2 } =t+\frac { 7 }{ 6 }\)
Solution.

Question. 75 \( m-\frac { m-1 }{ 2 } =1-\frac { m-2 }{ 3 }\)
Solution. Given \( m-\frac { m-1 }{ 2 } =1-\frac { m-2 }{ 3 }\)

Question. 76 4 (3p + 2) – 5 (6p – 1) = 2 (p – 8) – 6 (7p – 4)
Solution.

Question. 77 3(5x-2)+2(9x-11)=4(8x-7)-111
Solution.

Question. 78 0.16 (5x-2)=0.4x +7
Solution.

Question. 79 Radha takes some flowers in a basket and visits three temples one-by-one. At each temple, she offers one half of the flowers from the basket. If she is left with 3 flowers at the end, then find the number of flowers she had in the beginning.
Solution.

Question. 80 Rs 13500 are to be distributed among Salma, Kiran and Jenifer in such a way that Salma gets Rs 1000 more than Kiran and Jenifer gets Rs 500 more than Kiran. Find the money received by Jenifer.
Solution.

Question. 81 The volume of water in a tank is twice of that in the other. If we draw out 25 litres from the first and add it to the other, the volumes of the water in each tank will be the same. Find the volumes of water in each tank.
Solution.

Question. 82 Anushka and Aarushi are friends. They have equal amount of money in their pockets. Anushka gave 1/3 of her money to Aarushi as her birthday gift. Then, Aarushi gave a party at a restaurant and cleared the bill by paying half of the total money with her. If the remaining money in Aarushi’s pocket is Rs 1600, then find the sum gifted by Anushka.
Solution.

Question. 83 Kaustubh had 60 flowers. He offered some flowers in temple and found that the ratio of the number of remaining flowers to that of flowers in the beginning is 3 : 5. Find the number of flowers offered by him in the temple.
Solution.

Question. 84 The sum of three consecutive even natural numbers is 48. Find the greatest of these numbers.
Solution.

Question. 85 The sum of three consecutive odd natural numbers is 69. Find the prime number out of these numbers.
Solution.

Question. 86 The sum of three consecutive numbers is 156. Find the number which is a multiple of 13 out of these numbers.
Solution.

Question. 87 Find a number whose fifth part increased by 30 is equal to its fourth part decreased by 30.
Solution.

Question. 88 Divide 54 into two parts, such that one part is 2/7 of the other.
Solution.

Question. 89 Sum of the digits of a two-digit number is 11. The given number is less than the number obtained by interchanging the digits by 9. Find the number.
Solution.

Question. 90 Two equal sides of a triangle are each 4 m less than three times the third side. Find the dimensions of the triangle, if its perimeter is 55 m.
Solution.

Question. 91 After 12 years, Kanwar shall be 3 times as old as he was 4 years ago. Find his present age.
Solution.

Question. 92 Anima left one-half of her property to her daughter, one-third to her son and donated the rest to an educational institute. If the donation was worth Rs 100000, how much money did Anima have?
Solution.

Question. 93 If 1/2 is subtracted from a number and the difference is multiplied by 4, the result is 5. What is the number?
Solution.

Question. 94 The sum of four consecutive integers is 266. What are the integers?
Solution.

Question. 95 Hamid has three boxes of different fruits. Box A weighs \(2\frac { 1 }{ 2 }\) kg more than box B and Box C weighs \(10\frac { 1 }{ 4 }\)kg more than box B. The total weight of the three boxes is \(48\frac { 3 }{ 4 }\) kg. How many kilograms does box A weigh?
Solution.

Question. 96 The perimeter of a rectangle is 240 cm. If its length is increased by 10% and its breadth is decreased by 20%, then we get the same perimeter. Find the length and breadth of the rectangle.
Solution.

Question. 97 The age of A is five years more than that of B. 5 years ago, the ratio of their ages was 3 :2. Find their present ages.
Solution.

Question. 98 If numerator is 2 less than denominator of a rational number and when 1 is subtracted from numerator and denominator both, the rational number in its simplest form is 1/2. What is the rational number?
Solution.

Question. 99 In a two-digit number, digit in unit’s place is twice the digit in ten’s place. If 27 is added to it, digits are reversed. Find the number.
Solution.

Question. 100 A man was engaged as typist for the month of February in 2009. He was paid Rs 500 per day, but Rs 100 per day were deducted for the days he remained absent. He received Rs 9100 as salary for the month. For how many days did he work?
Solution.

Question. 101 A steamer goes downstream and covers the distance between two ports in 3 hours. It covers distance in 5 hours, when it goes upstream. If the stream flows at 3 km/h, then find what is the speed of the steamer upstream?
Solution.

Question. 102 A lady went to a bank with Rs 100000. She asked the cashier to give her Rs 500 and Rs 1000 currency notes in return. She got 175 currency notes in all. Find the number of each kind of currency notes.
Solution.

Question. 103 There are 40 passengers in a bus, some with Rs 3 tickets and remaining with Rs 10 tickets. The total collection from these passengers is Rs 295. Find how many passengers have tickets worth Rs 3?
Solution.

Question. 104 Denominator of a number is 4 less than its numerator. If 6 is added to the numerator, it becomes thrice the denominator. Find the fraction.
Solution.

Question. 105 An employee works in a company on a contract of 30 days on the condition that he will receive Rs 120 for each day he works and he will be fined Rs 10 for each day he is absent. If he receives Rs 2300 in all, for how many days did he remain absent?
Solution.

Question. 106 Kusum buys some chocolates at the rate of Rs 10 per chocolate. She also buys an equal number of candies at the rate of Rs 5 per candy. She makes a 20% profit on chocolates and 8% profit on candies. At the end of the day, all chocolates arid’ candies are sold out and her profit is Rs 240. Find the number of chocolates purchased.
Solution.

Question. 107 A steamer goes downstream and covers the distance between two ports in 5 hours, while it covers the same distance upstream in 6 hours. If the speed of the stream is 1 km/h, then find the speed of the steamer in still water.
Solution.

Question. 108 Distance between two places A and B is 210 km. Two cars start simultaneously from A and B in opposite directions and distance between them after 3 hours is 54 km. If speed of one car is less than that of other by 8 km/h, then find the speed of each.
Solution.

Question. 109 A carpenter charged Rs 2500 for making a bed. The cost of materials used is Rs 1100 and the labour charge is Rs 200 per hour. For how many hours did the carpenter work?
Solution.

Question. 110 For what value of x is the perimeter of shape 77 cm?

Solution.

Question. 111 For what value of x is the perimeter of shape 186 cm?

Solution.

Question. 112 On dividing Rs 200 between A and B, such that twice of A’s share is less than 3 times B’s share by 200, what is B’s share?
Solution.

Question. 113 Madhulika thought of a number, doubled it and added 20 to it. On dividing the resulting number by 25, she gets 4. What is the number?
Solution.

NCERT Exemplar Solutions

We hope the NCERT Exemplar Class 8 Maths Chapter 4 Linear Equations in One Variable help you. If you have any query regarding NCERT Exemplar Class 8 Maths Chapter 4 Linear Equations in One Variable, drop a comment below and we will get back to you at the earliest.