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Samacheer Kalvi 12th Chemistry Solutions Chapter 9 Electro Chemistry

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 9 Electro Chemistry

Samacheer Kalvi 12th Chemistry Chapter 9 Electro Chemistry Textual Evaluation Solved

Samacheer Kalvi 12th Chemistry Electro Chemistry Multiple Choice Questions
Question 1.
The number of electrons that have a total charge of 9650 coulombs is ………..
(a) 6.22 x 10²³
(b) 6.022 x 10²⁴
(c) 6.022 x 10²²
(d) 6.022 x 10^-34
Answer:
(c) 6.022 x 10²²
Hint: IF = 96500 C = 1 mole of e^– = 6.023 x 10²³ e^–
9650 C = $\frac{6.22 \times 10^{23}}{96500} \times 9650$ = 6.022 x 10²²

Question 2.
Consider the following half cell reactions:
Mn²⁺ + 2e^– → Mn E° = – 1.18V
Mn²⁺ → Mn³⁺ + e^– E = – 1.51V
The E for the reaction 3Mn²⁺ → Mn+2Mn³⁺, and the possibility of the forward reaction are respectively.
(a) 2.69V and spontaneous
(b) – 2.69 and non spontaneous
(c) 0.33V and Spontaneous
(d) 4.18V and non spontaneous
Answer:
(b) – 2.69 and non spontaneous
Hint: Mn⁺ + 2e^– → Mn(E⁰_red) = 1.18V
2[Mn²⁺ → Mn³⁺ + e^–] (E⁰_ox) = – 1.51V
3Mn²⁺+ → Mn³⁺ + 2Mn³⁺ + (E⁰_cell) = ?
E⁰_red = (E⁰_ox) + (E⁰_cell)
= – 1.51 – 1.18 and non spontaneous
= – 2.69 V
Since E° is – ve ∆G is +ve and the given forward cell reaction is non – spontaneous.

Question 3.
The button cell used in watches function as follows:
Zn_(s) + Ag₂O_(s) + H₂O₍₁₎ $\rightleftharpoons$ 2Ag_(s) + Zn²⁺_(aq) + 2OH^–_(aq) the half cell potentials are
Ag₂O_(s) + H₂O₍₁₎ + 2e^– →2Ag_(S) + 2OH^–_(aq) E° = 034V. The cell potential will be
(a) 0.84V
(b) 1.34V
(c) 1.10V
(d) 0.42V
Answer:
(c) 1.10V
Hint: Anodic oxidation: (Reverse the given reaction)
(E⁰_ox ) = 0.76V cathodic reduction
E⁰_cell = (E⁰_ox) + (E⁰_red) = 0.76 + 0.34 = 1.1V

Question 4.
The molar conductivity of a 0.5 mol dm^-3 solution of AgNO₃ with electrolytic conductivity of 5.76 x 10^-3S cm^-1at 298 K is ………….
(a) 2.88 S cm² mo1^-1
(b) 11.52 S cm² mol^-1
(c) 0.086 S cm² mol^-1
(d) 28.8 S cm² mol^-1
Answer:
(b) 11.52 S cm² mol^-1
Hint:
A = $\frac { k }{ M }$ x 10^-3 mol^-1 m³

= 11.52 S cm² mol^-1

Question 5.

Calculate A⁰_HOAC using appropriate molar conductances of the electrolytes listed above at infinite dilution in water at 25°C.
(a) 517.2
(b) 552.7
(c) 390.7
(d) 217.5
Answer:
(c) 390.7
Hint:

(426.2 + 91) – (126.5) = 390.7

Question 6.
Faradays constant is defined as
(a) charge carried by I electron
(b) charge carried by one mole of electrons
(c) charge required to deposit one mole of substance
(d) charge carried by 6.22 X 10¹⁰ electrons
Answer:
(b) charge carried by one mole of electrons
Hint:
IF = 96500 C = 1 charge of mole of e^– = charge of 6.022 x 10²³ e^–

Question 7.
How many faradays of electricity are required for the following reaction to occur
MnO^–₄ → Mn²⁺
(a) 5F
(b) 3F
(C) IF
(d) 7F
Answer:
(a) 5F
Hint:
7MnO^–₄ + 5e^– → Mn²⁺ + 4H₂O
5 moles of electrons i.e., 5F charge is required.

Question 8.
A current strength of 3.86 A was passed through molten Calcium oxide for 41 minutes and 40 seconds. The mass of Calcium in grams deposited at the cathode is (atomic mass of Ca is 40g / mol and IF = 96500C).
(a) 4
(b) 2
(c) 8
(d) 6
Answer:
(b) 2
Hint: m = ZIt
41mm 40sec = 2500 seconds
= $\frac { 40 x 3.86 x 2500 }{ 2 x 96500 }$
Z = $\frac { m }{ n x 96500 }$ = $\frac { 40 }{ 2 x 96500 }$
= 2g

Question 9.
During electrolysis of molten sodium chloride, the time required to produce 0.1 mol of chlorine gas using a current of 3A is ………..
(a) 55 minutes
(b) 107.2 minutes
(c) 220 minutes
(d) 330 minutes
Answer:
(b) 107.2 minutes
Hint: $\frac { m }{ ZI }$ (mass of 1 mole of Cl₂ gas = 71)
t = $\frac { m }{ ZI }$ mass of 0.1 mole of Cl₂ gas = 7.1 g mol^-1)

Question 10.
The number of electrons delivered at the cathode during electrolysis by a current of 1 A in 60 seconds is (charge of electron = 1.6 x 10^-19C)
(a) 6.22 x 10²³
(b) 6.022 x 10²⁰
(c) 3.75 x 10²⁰
(d) 7.48 x 10²³
Answer:
(c) 3.75 x 10²⁰
Hint: Q = It
= 1A x 60S
96500 C charge 6.022 x 10²³ electrons
60 C charge = $\frac{6.022 \times 10^{23}}{96500} \times 960$
= 3.744 x 10²⁰ electrons

Question 11.
Which of the following electrolytic solution has the least specific conductance?
(a) 2N
(b) 0.002N
(c) 0.02N
(d) 0.2N
Answer:
(b) 0.002N
Hint: In general, specific conductance of an electrolyte decreases with dilution. SO, 0.002N solution has least specific conductance.

Question 12.
While charging lead storage battery
(a) PbSO₄ on cathode is reduced to Pb
(b) PbSO₄ on anode is oxidised to PbO₄
(c) PbSO₄ on anode is reduced to Pb
(d) PbSO₄ on cathode is oxidised to Pb
Answer:
(c) PbSO₄ on anode is reduced to Pb.
Hint: Charging: anode: PbSO₄(s) + 2e^– → Pb (s) + SO₄^-2 (aq)
Cathode: PbSO₄(s) + 2H₂O (1) → PbO₂ (s) + SO₄^-2 (aq) + 2e_–

Question 13.
Among the following cells
I. Leclanche cell
II. Nickel – Cadmium cell
III. Lead storage battery
IV. Mercury cell
Primary cells are …………
(a) I and IV
(b) I and III
(c) III and IV
(d) II and III
Answer:
(a) I and IV

Question 14.
Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because
(a) Zinc is lighter than iron
(b) Zinc has lower melting point than iron
(c) Zinc has lower negative electrode potential than iron
(d) Zinc has higher negative electrode potential than iron
Answer:
(d) Zinc has higher negative electrode potential than iron
Hint: E⁰_Zn⁺|Zn = – 0.76V and E⁰_Fe²⁺|Fe = 0.44V. Zinc has higher negative electrode potential than iron, iron cannot be coated on zinc.

Question 15.
Assertion: pure iron when heated in dry air is converted with a layer Of rust.
Reason: Rust has the composition Fe₃O₄
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d) both assertion and reason are false.
Answer:
(d) both assertion and reason are false.
Hint: Both are false

Dry air has no reaction with iron
Rust has the composition Fe₂O₃ x H₂O

Question 16.
In H₂ – O₂ fuel cell the reaction occur at cathode is ……….
(a) O₂(g) + 2H₂O (l) + 4e^– → 4OH^–(aq)
(b) H⁺(aq) + OH^–(aq) → H₂O (l)
(c) 2H₂(g) + O₂(g) → 2H₂O (g)
(d) H⁺ + e^– → 1/2 H₂
Answer:
(a) O₂(g) + 2H₂O (l) + 4e^– → 4OH^–(aq)

Question 17.
The equivalent conductance of M/36 solution of a weak monobasic acid is 6mho cm² and at infinite dilution is 400 mho cm². The dissociation constant of this acid is ………….
(a) 1.25 x 10^-16
(b) 6.25 x 10 ^-6
(c) 1.25 x 10^-4
(d) 6.25 x 10^-5
Answer:
(b) 6.25 x 10 ^-6
Hint: α = $\frac { 6 }{ 400 }$
K_a = α²C = $\frac { 6 }{ 400 }$ x $\frac { 6 }{ 400 }$ x $\frac { 1 }{ 36 }$
= 6.25 x 10^-6

Question 18.
A conductivity cell has been calibrated with a 0.01M, 1:1 electrolytic solution (specific conductance (K = 1.25 x 10^-3 S cm^-1 ) in the cell and the measured resistance was 800Ω at 25⁰ C. The cell constant is,
(a) 10^-1 cm^-1
(b) 10^-1 cm^-1
(c) 1 cm^-1
(d) 5.7 x 10^-12
Answer:
(c) 1 cm^-1
Hint: R = p.$\frac { 1 }{ A }$
Cell constant = $\frac { R }{ ρ }$ = k.R $(\frac { 1 }{ ρ } =k)$ = 1.25 x 10^-3 f^-1cm^-1 x 800Ω = 1cm^-1

Question 19.
The conductivity of a saturated solution of a sparingly soluble salt AB (1:1 electrolyte) at 298K is 1.85 x 10^-5 S m^-1. Solubility product of the salt AB at 298K (Λ⁰_m)_AB = 14 x 10^-3 S m² mol^-1.
(a) 5.7 x 10^-2
(b) 1.32 x 10¹²
(c) 7.5 x 10^-12
(d) 1.74 x 10^-12
Answer:
(d) 1.74 x 10^-12

Question 20.
In the electrochemical cell: Zn|ZnSO₄ (0.01M)||CuSO₄ (1.0M)|Cu , the emf of this Daniel cell is E₁. When the concentration of ZnSO₄ is changed to 1.0 M and that CuSO₄ changed to 0.0 1M, the emf changes to E₂. From the followings, which one is the relationship between E₁ and E₂?
(a) E₁ < E₂
(b) E₁ > E₂
(c) E₂ = 0↑E₁
(d) E₁ = E₂
Answer:
(b) E₁ > E₂
Hint:

Question 21.
Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below:

Then the species undergoing disproportional is …………..
(a) Br₂
(b) BrO₄^–
(c) BrO₃^–
(d) HBrO
Answer:
(d) HBrO
Hint:

(E_cell)_A = – 1.82 + 1.5 = – 0.32V
(E_cell)_B = – 1.5 + 1.595 = + 0.095V
(E_cell)_C = 1.595 + 1.0652 = – 0.529V
The species undergoing disproportionation is HBrO

Question 22.
For the cell reaction
2Fe³⁺(aq) + ^2I(aq) → 2Fe²⁺ (aq) + I₂(aq)
EC⁰_cell = 0.24V at 298K. The standard Gibbs energy (∆, G⁰ ) of the cell reactions is …………
(a) – 46.32 KJ mol^-1
(b) – 23.16 KJ mol^-1
(c) 46.32 KJ mol^-1
(d) 23.16 KJ mor^-1
Answer:
(a) – 46.32 KJ mol^-1

Question 23.
A certain current liberated 0.504gm of hydrogen in 2 hours. How many grams of copper can be liberated by the same current flowing for the same time in a copper sulphate solution?
(a) 31.75
(b) 15.8
(c) 7.5
(d) 63.5
Answer:
(b) 15.8

Question 24.
A gas X at 1 atm is bubble through a solution containing a mixture of 1MY^– and 1MZ^-1 at 25°C . If the reduction potential of Z > Y> X, then
(a) Y will oxidize X and not Z
(b) Y will oxidize Z and not X
(c) Y will oxidize both X and Z
(d) Y will reduce both X and Z
Answer:
(a) Y will oxidize X and not Z

Question 25.
Cell equation: A²⁺ + 2B^– → A²⁺ + 2B
A²⁺ + 2e^– → AE° = + 0.34V and log₁₀ K = 15.6 at 300K for cell reactions find E° for
B¹ + e^– → B
(a) 0.80
(b) 1.26
(c) – 0.54
(d) – 10.94
Answer:
(a) 0.80

Samacheer Kalvi 12th Chemistry Electro Chemistry Short Answer

Question 1.
Define anode and cathode
Answer:

Anode: The electrode at which the oxidation occur is called anode.
Cathode: The electrode at which the reduction occur is called cathode.

Question 2.
Why does conductivity of a solution decrease on dilution of the solution?
Answer:
Conductivity always decreases with decrease in concentration (on dilution of the solution) both for weak as much as for strong electrolytes. ¡t is because the number of ions per unit volume that carry the current is a solution decreases on dilution.

Question 3.
State Kohlrausch Law. How is it useful to determine the molar conductivity of weak electrolyte at infinite dilution.
Answer:
Kohlrausch’s law:
It is defined as, at infinite dilution the limiting molar conductivity of an electrolyte is equal to the sum of the limiting molar conductivities of its constituent ions.

Determination of the molar conductivity of weak electrolyte at infinite dilution.
It is impossible to determine the molar conductance at infinite dilution for weak electrolytes experimentally. However, the same can be calculated using Kohlraushs Law. For example, the molar conductance of CH₃COOH, can be calculated using the experimentally determined molar conductivities of strong electrolytes HCI, NaCI and CH₃COONa.
Λ°_CH₃COONa = λ°_Na⁺ + λ°_CH₃COONa …….(1)
Λ°_HCl = λ°_H⁺ + λ°_Cl^– ………………(2)
Λ°_NaCl = λ°_Na⁺ + λ°_Cl^– …………….(3)
Equation (1) + Equation (2) – Equation (3) gives,
(Λ°_CH₃COONa) + (Λ°_HCl) – (Λ°_NaCl) = λ°_H⁺ + λ°_CH₃COONa = Λ°_CH₃COONa

Question 4.
Describe the electrolysis of molten NaCI using inert electrodes.
Answer:
1. The electrolytic cell consists of two iron electrodes dipped in molten sodium chloride and they are connected to an external DC power supply via a key.

2. The electrode which is attached to the negative end of the power supply is called the cathode and the one is which attached to the positive end is called the anode.

3. Once the key is closed, the external DC power supply drives the electrons to the cathode and at the same time
pull the electrons from the anode.

Cell reactions:
Na⁺ ions are attracted towards cathode, where they combines with the electrons and reduced to liquid sodium.
Cathode (reduction)
Na⁺_(I) + e^–Na₍₁₎
E⁰ = – 2.7 1V
Similarly, Cl^– ions are attracted towards anode where they losses their electrons and oxidised to chlorine gas. Anode (oxidation)
2Cl^–₍₁₎ Cl₂_(g) + 2e^–
E° = – 1 .36V
The overall reaction is,
2Na⁺_(l) + 2Cl^–_(l) → 2Na_(l) + Cl₂_(g) (g)
E⁰ = 4.07 V

The negative E° value shows that the above reaction is a non-spontaneous one. Hence, we have to supply a voltage greater than 4.07V to cause the electrolysis of molten NaCl. In an electrolytic cell, oxidation occurs at the anode and reduction occur at the cathode as in a galvanic cell, but the sign of the electrodes is the reverse i.e., in the electrolytic cell cathode is -ve and the anode is +ve.

Question 5.
State Faraday’s Laws of electrolysis.
Answer:
Faraday’s laws of electrolysis:
1. First law:
The mass of the substance (M) liberated at an electrode during electrolysis is directly proportional to the quantity of charge (Q) passed through the cell. M α Q

2. Second law:
When the same quantity of charge is passed through the solutions of different electrolytes, the amount of substances liberated at the respective electrodes are directly proportional to their electrochemical equivalents. M α Z

Question 6.
Describe the construction of Daniel’s cell. Write the cell reaction.
Answer:
The separation of half-reaction is the basis for the construction of Daniel’s cell. It consists of two half cells.

Oxidation half cell:
The metallic zinc strip dips into an aqueous solution of zinc sulfate taken in a beaker.

Reduction half cell:
A copper strip that dips into an aqueous solution of copper sulphate taken in a beaker.

Joining the half cell:
The zinc and copper strips are externally connected using a wire through a switch (k) and a load (example: voltmeter). The electrolytic solution present in the cathodic and anodic compartment are connected using an inverted U tube containing an agar-agar gel mixed with an inert electrolyte such as Kl1, Na₂SO₄ etc.,

The ions of inert electrolyte do not react with other ions present in the half cells and they are not either oxidized (or) reduced at the electrodes. The solution in the salt bridge Voltmeter cannot get poured out, but through which the ions can move into (or) out of the half cells. When the switch (k) closes the circuit, the electrons flow from zinc strip to copper strip. This is due to the following redox reactions which are taking place at the respective electrodes.

Anodic oxidation:
The electrode at which the oxidation occur is called the anode. In Daniel cell, the oxidation take place at zinc electrode, i.e., zinc is oxidised to Zn² ions and the electrons. The Zn² ions enters the solution and the electrons enter the zinc metal, then flow through the external wire and then enter the copper strip. Electrons are liberated at zinc electrode and hence it is negative ( – ve).
Zn(s) → Zn²⁺(aq) + 2e^– (loss of electron-oxidation)

Cathodic reduction:
As discussed earlier, the electrons flow through the circuit from zinc to copper, where the Cu²⁺ ions in the solution accept the electrons, get reduced to copper and the same get deposited on the electrode. Here, the electrons are consumed and hence it is positive (+ve).
Cu²⁺_(aq) + 2e^– → Cu_(s) (gain of electron-reduction)

Salt bridge:
The electrolytes present in two half cells are connected using a salt bridge. We have learnt that the anodic oxidation of zinc electrodes results in an increase in the concentration of Zn²⁺ in solution. i.e., the solution contains more number of Zn²⁺ ions as compared to SO₄^2- and hence the solution in the anodic compartment would become positively charged.

Similarly, the solution in the cathodic compartment would become negatively charged as the Cu²⁺ ions are reduced to copper i.e., the cathodic solution contain more number of SO^2-₄ ions compared to Cu²⁺.

Completion of circuit:
Electrons flow from the negatively charged zinc anode into the positively charged copper electrode through the external wire, at the same time, anions move towards anode and cations are move towards the cathode compartment. This completes the
circuit.

Consumption of Electrodes:
As the Daniel cell operates, the mass of zinc electrode gradually decreases while the mass of the copper electrode increases and hence the cell will function until the entire metallic zinc electrode is converted in to Zn²⁺ the entire Cu²⁺ ions are converted in to metallic copper.
Daniel cell is represented as
Zn_(s)|Zn²⁺_(aq)||Cu²⁺_(aq)|Cu_(s)

Question 7.
Why is anode in galvanic cell considered to be negative and cathode positive electrode?
Answer:
A galvanic cell works basically in reverse to an electrolytic cell. The anode is the electrode where oxidation takes place, in a galvanic cell, it is the negative electrode, as when oxidation occurs, electrons are left behind on the electrode.

The anode is also the electrode where metal atoms give up their electrons to the metal and go into solution. The electron left behind on it render ¡t effectively negative and the electron flow goes from it through the wire to the cathode.

Positive aqueous ions in the solution are reduced by the incoming electrons on the cathode. This why the cathode is a positive electrode, because positive ions are reduced to metal atoms there.

Question 8.
The conductivity of a 0.01%M solution of a 1:1 weak electrolyte at 298K is 1.5 x 10^-4 S cm^-1.

molar conductivity of the solution
degree of dissociation and the dissociation constant of the weak electrolyte

Given that
λ⁰cation = 248.2 S cm² mol^-1
λ⁰anion = 51.8 S cm² mol^-1
Answer:
1. Molar conductivity
C = 0.01M
k = 1.5 x 10^-4 S cm^-1
(or)
K = 1.5 x 10^-2 S m^-1
$\frac{\kappa \times 10^{-3}}{\mathrm{C}}$ S m^-1 mol^-1 m³ = $\frac{1.5 \times 10^{-2} \times 10^{-3}}{0.01}$ S m² mol^-1
Λ_m = 1.5 x 10^-3 s m^-1

2. Degree of dissociation
α = $\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\infty}^{\circ}}$ (or) α = $\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}$
= (248.2 + 51.8)S cm² mol^-1
= 300 S cm² mol^-1
K_a = $\frac{\alpha^{2} C}{1-\alpha}$
= $\frac{(0.05)^{2}(0.01)}{1-0.05}$
K_a = 2.63 x 10^-5

Question 9.
Which of 0.1M HCI and 0.1 M KCI do you expect to have greater molar conductance and why?
Answer:
Compare to 0.1M HCI and 0.1 M KCI, 0.1M HCl has greater molar conductance.

Molar conductance of 0.1M HCl = 39.132 x 10^-3 S m² mol^-1.
Molar conductance of 0.1 M KCl = 12.896 x 10^-3 S m² mol^-1

Because, H⁺ ion in aqueous solution being smaller size than K⁺ ion and H⁺ ion have greater mobility than K ion. When mobility of the ion increases, conductivity of that ions also increases. Hence, 0. 1M HCI solution has greater molar conductance than 0.1 M KCI solution.

Question 10.
Arrange the following solutions in the decreasing order of specific conductance.

0.01M KCI
0.005M KCI
0.1M KCI
0.25 M KCI
0.5 M KCI

Answer:
0.005M KCl > 0.01M KCI > 0.1M KCI > 0.25KCl > 0.5 KCI.
Specific conductance and concentration of the electrolyte. So if concentration decreases, specific conductance increases.

Question 11.
Why is AC current used instead of DC in measuring the electrolytic conductance?
Answer:
1. AC current to prevent electrolysis of the solution.

2. If we apply DC current to the cell the positive ions will be attracted to the negative plate and the negative ions to the positive plate. This will cause the composition of the electrolyte to change while measuring the equivalent conductance.

3. So DC current through the conductivity cell will lead to the electrolysis of the solution taken in the cell. To avoid such a electrolysis, we have to use AC current for measuring equivalent conductance.

Question 12.
0.1M NaCI solution is placed in two different cells having cell constant 0.5 and 0.25cm^-1 respectively. Which of the two will have greater value of specific conductance.
Answer:
The specific conductance values are same. Because the reaction (cation) of cell constant does not change.

Question 13.
A current of 1 .608A is passed through 250 mL of 0.5M solution of copper sulphate for 50 minutes. Calculate the strength of Cu²⁺ after electrolysis assuming volume to be constant and the current efficiency is 100%.
Answer:
Given, I = I .608A
t = 50 min (or) 50 x 60 = 3000 S
V = 250 mL
C = 0.5M
η = 100%
The number of Faraday’s of electricity passed through the CuSO₄ solution
Q = It
= Q = 1.608 x 3000
Q = 4824C
Number of Faraday’s of electricity = $\frac { 4824C }{ 96500C }$ = 0.5F
Electrolysis of CuSO₄
Cu²⁺_(aq) + 2e^– → Cu_(s)
The above equation shows that 2F electricity will deposit 1 mole of Cu²⁺
0.5F electicity will deposit $\frac { 1mol }{ 2F }$ x 0.5F = 0.025 mol
Initial number of molar of Cu²⁺ in 250 ml of solution = $\frac { 1mol }{ 250mL }$ x 250mL = 0.125 mol
Number of molar of Cu²⁺ after electrolysis 0.125 – 0.025 = 0.1 mol
Concentration of Cu²⁺ = $\frac { 0.1mol }{ 250mL }$ X 1000 mL = 0.4 M

Question 14.
Can Fe³⁺ oxidises Bromide to bromine under standard conditions?
Given:

Answer:
Required half cell reaction

E⁰_cell = (E⁰_ox) + (E⁰_red) = – 1.09 + 0.771 = – 0.319V
We know that ∆G° = – nFE⁰_cell
If E⁰_cell is -ve; ∆G is +ve and the cell reaction is non-spontaneous.
Hence, Fe³⁺ cannot oxidise Bromide to Bromine.

Question 15.
Is it possible to store copper sulphate in an iron vessel for a long time?
Given:

Answer:
E⁰_cell = (E⁰_ox) + (E⁰_red) = 0.44 V + 0.34V = 0.78V
These +ve E⁰_cell values shows that iron will oxidise and copper will get reduced i.e., the vessel
will dissolve. Hence it is not possible to store copper sulphate in an iron vessel.

Question 16.
Two metals M₁ and M₂ have reduction potential values of – xV and + yV respectively. Which will liberate H₂ in H₂SO₄?
Answer:
Metals having negative reduction potential acts as powerful reducing agent. Since M₁ has – xV, therefore M1 easily liberate H₂ in H₂SO₄. Metals having higher oxidation potential will liberate H₂ from H₂SO₄. Hence, the metal M₁ having + xV, oxidation potential will liberate H₂ from H₂SO₄.

Question 17.
Reduction potential of two metals M₁ and M₂ are

Predict which one Is better for coating the surface of iron.
Given:
Answer:
Oxidation potential of M₁ is more +ve than the oxidation potential of Fe which indicates that it will prevent iron from rusting.

Question 18.
Calculate the standard emf of the cell: Cd | Cd²⁺|| Cu²⁺ | Cu and determine the cell reaction. The standard reduction potentials of Cu²⁺ | Cu and Cd²⁺ | Cd are 0.34V and – 0.40 volts respectively. Predict the feasibility of the cell reaction.
Answer:
Cell reactions:

Cd(s) + 2e^– Cd²⁺ + Cu(s)
E⁰_cell = (E⁰_ox) + (E⁰_red) = 0.4 + 0.34
emf is +ve, so ∆G is (-)ve, the reaction is feasible.

Question 19.
In fuel cell H₂ and O₂ react to produce electricity. In the process, H₂ gas is oxidised at the anode and O₂ at cathode. If 44.8 litre of H₂ at 25°C and also pressure reacts in 10 minutes, what is average current produced? If the entire current is used for electro deposition of Cu from Cu²⁺, how many grams of Cu deposited?
Answer:
Oxidation at anode:
2H₂(g) + 4OH^– (aq) → 4H₂O (1) + 4e^–
1 mole of hydrogen gas produces 2 moles of electrons at 25°C and 1 atm pressure, 1 mole of hydrogen gas occupies = 22.4 litres
∴no. of moles of hydrogen gas produced = $\frac{1 \mathrm{mole}}{22.4 \text { litres }}$ x 44.8 litres = 2 moles of hydrogen
∴2 of moles of hydrogen produces 4 moles of electro i.e., 4F charge. We know that Q = It
I = $\frac { Q }{ t }$ = $\frac{4 \mathrm{F}}{10 \mathrm{mins}}$ = $\frac { 4×96500 }{ 10x60s }$
I = 643.33 A
Electro deposition of copper
Cu²⁺_(aq) + 2e^– → Cu_(s)
2F charge is required to deposit
1 mole of copper i.e., 63.5 g
If the entire current produced in the fuel cell i.e., 4 F is utilised for electrolysis, then 2 x 63.5 i.e., 127.0 g copper will be deposited at cathode.

Question 20.
The same amount of electricity was passed through two separate electrolytic cells containing solutions of nickel nitrate and chromium nitrate respectively. If 2.935g of Ni was deposited in the first cell. The amount of Cr deposited in the another cell? Given: molar mass of Nickel and chromium are 58.74 and 52gm^-1 respectively.
Answer:
Ni²⁺ (aq) + 2e^– → Ni (s)
Cr²⁺(aq) + 3e^– → Cr(s)
The above reaction indicates that 2F charge is required to deposit 58.7 g of Nickel form nickel nitrate and 3F charge is required to deposit 52g of chromium. Given that 2.935 gram of Nickel is deposited 2F
The amount of charge passed through the cell = $\frac { 2F }{ 58.7g }$ x 2.935g = 0.1F
If 0. IF charge is passed through chromium nitrate the amount of chromium deposited
= 52g x 0.IF = 1.733g

Question 21.
0.1M copper sulphate solution in which copper electrode is dipped at 25C. Calculate the electrode potential of copper.

Answer:
Given that
[Cu²⁺] = 0.1 M
E⁰_Cu²⁺|Cu = 0.34
E_cell = ?
Cell reaction is Cu²⁺(aq) + 2e^– → Cu (s)
E_cell = E⁰ – $\frac { 0.0591 }{ n }$ log $\frac{[\mathrm{Cu}]}{\left[\mathrm{Cu}^{2+}\right]}$ = 0.34 – $\frac { 0.0591 }{ 2 }$ log $\frac { 1 }{ 0.1 }$
= 0.34 – 0.0296 = 0.31 V

Question 22.
For the cell Mg(s) | Mg²⁺(aq) || Ag⁺ (aq) | Ag (s), calculate the equilibrium constant at 25°C and maximum work that can be obtained during operation of cell. Given:

Answer:

E⁰_cell = (E⁰_ox) + (E⁰_red) = 2.37 + 0.80 = 3.17 V
Overall reaction
Mg + 2Ag⁺ → Mg²⁺ + 2Ag
∆G° = -nFE°
= – 2 x 96500 x 3.17
= – 6.118 x 10⁵ J
We know that W_max = ∆G°
W_max = + 6.118 x 10₅ J
Relationship between ∆G° and K_eq is,
∆G = – 2.303 RT logK_eq
∆G = – 2.303 x 8.314 x 298 log K_eq [25°C = 298 K]
log K_eq = $\frac{6.118 \times 10^{5}}{2.303 \times 8.314 \times 298}$ = $\frac{6.118 \times 10^{5}}{5705.84}$
log K_eq = 107.223
K_eq = Antilog (107.223)

Question 23.
8.2 x 10¹² litres of water is available in a lake. A power reactor using the electrolysis of water in the lake produces electricity at the rate of 2 x 10⁶ Cs^-1 at an appropriate voltage. How many years would it like to completely electrolyse the water in the lake. Assume that there is no loss of water except due to electrolysis.
Answer:
Hydrolysis of water:
At anode: 2H₂O → 4H⁺ + O₂ + 4e^– …………..(1)
At cathode: 2H₂O + 2e^– → H₂ + 2OH^–
Overall reaction: 6H₂O → 4H^– + 4OH^– +2H₂+ O₂
(or)
Equation (1) + (2) x 2
= 2H₂O → 2H₂ + O₂
According to Faraday’s Law of electrolysis, to electrolyse two mole of Water
(36g ≃ 36 mL. of H₂O), 4F charge is required alternatively, when 36 mL of water is electrolysed,
the charge generated = 4 x 96500 C.
When the whole water which is available on the lake is completely electrolysed the amount of charge generated is equal to

Given that in 1 second, 2 x 10⁶ C is generated therefore, the time required to generate
96500 x 10¹⁵ C is = $\frac{1 \mathrm{S}}{2 \times 10^{6} \mathrm{C}}$ x 96500 x 10¹⁵C = 48250 x 10⁹ S
Number of year = $\frac{48250 \times 10^{9}}{365 \times 24 \times 60 \times 60}$ 1 year = 365 days
= 1.5299 x 10⁶
= 365 x 24 hours
= 365 x 24 x 60 min
= 365 x 24 x 60 x 60 sec

Question 24.
Derive an expression for Nernst equation.
Answer:
Nernst equation is the one which relates the cell potential and the concentration of the species involved in an electrochemical reaction.
Let us consider an electrochemical cell for which the overall redox reaction is,
Answer:
xA + yB = lC + mD
The reaction quotient Q is,
$\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}$
We know that,
∆G = ∆G⁰ + RT ln Q
∆G = – nFE_cell
∆G⁰ = -nFE⁰_cell
equation (1) becomes
– nFE_cell = -nFE⁰_cell + RT ln Q
Subsitute the Q value in equation (2)
– nFE_cell = – nFE⁰_cell + RT ln $\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}$. ………..(3)
Divide the whole equation (3) by – nF
E_cell = E°_cell – $\frac { RT }{ nF }$ ln $\left(\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\right)$
E_cell = E°_cell – $\frac { 2.303RT }{ nF }$ log $\left(\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\right)$ ……………(4)
This is called the Nernst equation.
At 25°C (298 K) equation (4) becomes,
E_cell = E°_cell – $\frac { 2.303×8.314×298 }{ nx96500 }$ log $\left(\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\right)$
E_cell = E°_cell – $\frac { 0.0591 }{ n }$ log $\left(\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\right)$

Question 25.
Write a note on sacrificial protection.
Answer:
In this method, the metallic structure to be protected is made cathode by connecting it with more active metal (anodic metal). So that all the corrosion will concentrate only on the active metal. The artificially made anode thus gradually gets corroded protecting the original metallic structure. Hence this process is otherwise known as sacrificial anodic protection. Al, Zn and Mg are used as sacrificial anodes.

Question 26.
Explain the function of H₂ – O₂ fuel cell.
Answer:
In this case, hydrogen act as a fuel and oxygen as an oxidant and the electrolyte is aqueous KOH maintained at 200°C and 20 – 40 atm. Porous graphite electrode containing Ni and NiO serves as the inert electrodes. Hydrogen and oxygen gases are bubbled through the anode and cathode, respectively.
Oxidation occurs at the anode:
2H_2(g)+ 4OH_-(aq) → 4H₂O₍₁₎ + 4e^–
Reduction occurs at the cathode O₂(g) + 2 H₂O(1) + 4e^– → 4 OH^– (aq)
The overall reaction is 2H₂(g) + O₂(g) → 2H₂O(1)
The above reaction is the same as the hydrogen combustion reaction, however, they do not react directly ie., the oxidation and reduction reactions take place separately at the anode and cathode respectively like H₂ – O₂ fuel cell. Other fuel cell like propane – O₂ and methane O₂ have also been developed.

Question 27.
Ionic conductance at infinite dilution of Al³⁺ and SO₄^2- are 189 and 160 mho cm² equiv^-1. Calculate the equivalent and molar conductance of the electrolyte Al₂(SO₄) at infinite dilution.
Answer:
1. Molar conductance

= (2 x 189) + (3 x 160)
= 378 + 480
= 858 mho cm² mol^-1

2. Equivalent conductnace

= $\frac { 189 }{ 3 }$ + $\frac { 160 }{ 2 }$
= 143 mho cm² (g equiv)^-1

Samacheer Kalvi 12th Chemistry Electro Chemistry Evaluate Yourself

Question 1.
Calculate the molar conductance of 0.01M aqueous KCI solution at 25°C . The specific conductance of KCl at 25°C is 14.114 x 10^-2 Sm^-1.
Answer:
Concentration of KCI solution = 0.01 M.
Specific conductance (K) = 14.114 x 10^-2 S m^-1
Molar conductance (Λ_m) = ?
Λ_m = $\frac{\kappa \times 10^{-3}}{M}$ = $\frac{14.114 \times 10^{-2} \times 10^{-3}}{0.01}$
S m^-1 mol^-1 m³
Λ_m = 14.114 x 10^-5 x 10² = 14.114 x 10^-3 Sm² mol^-1

Question 2.
The resistance of 0.15M solution of an electrolyte is 50. The specific conductance of the solution is 2.4 Sm^-1. The resistance of 0.5 N solution of the same electrolyte measured using the same conductivity cell is 480Ω. Find the equivalent conductivity of 0.5 N solution of the electrolyte.
Answer:
Given that R₁ = 50Ω
R₂ = 480Ω
K₁ =2.4Sm^-1
K₂ = ?
N₁ = 0.15N
N₂ = 0.5N

= 5 x 10^-4sm² gram equivalent^-1
We know that
$\frac{\text { Cell constant }}{\mathrm{R}}$
$\frac{\kappa_{2}}{\kappa_{1}}$ = $\frac{R_{1}}{R_{2}}$
k₂ = k₁ x $\frac{R_{1}}{R_{2}}$ = 2.4 Sm^-1 x $\frac{50 \Omega}{480 \Omega}$ = 0.25 Sm^-1

Question 3.
The emf of the following cell at 25°C is equal to 0.34v. Calculate the reduction potential of copper electrode.
Pt(s) | H₂(g,1atm) | H⁺ (aq,1M) || Cu²⁺(aq,1M) | Cu(s)
Answer:
SHE Value is zero
E°_cell = E°_R – E°_L
= 0.34 – 0 = 0.34V
The reduction potential of copper electrode = 0.34V

Question 4.
Using the calculated emf value of zinc and copper electrode, calculate the emf of the following cell at 25°C.
Zn (s) | Zn²⁺ (aq, 1M) || Cu²⁺(aq, 1M) | Cu (s)
Answer:
E°_cell = E°_R – E°_L
E_zn/zn²⁺ = 0.76V
E_Cu/Cu²⁺ = 0.76V
E°_cell = 0.76 – (- 0.34V)
E°_cell = 0.76 – (- 0.34)
E°_cell = + 1.1 V

Question 5.
Write the overall redox reaction which takes place in the galvanic cell,
Pt(s) | Fe²⁺(aq),Fe²⁺(aq) || MnO^–₄(aq), H⁺(aq), Mn²⁺(aq) || Pt(s)
Answer:
At Anode half cell – 5Fe²⁺_(aq) → 5Fe³⁺_(aq) + 5e^–
At cathode half cell – MnO^–_4(aq) + 8H⁺_(aq) + 5e^– → Mn²⁺_(aq) + 4H₂O₍₁₎
Overall redox reaction – 5Fe²⁺_(aq) + MnO^–₄(aq) + 8H⁺(aq) → 5Fe³⁺_(aq) + Mn²⁺_(aq) + 4H₂O₍₁₎

Question 6.
The electrochemical cell reaction of the Daniel cell is
Zn (s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu_(s)
What is the change in the cell voltage on increasing the ion concentration in the anode compartment by a factor 10?
Answer:
Zn (s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu_(s)
ln the case E°_cell = 1.1V
Reaction quotient Q for the above reaction is, Q = $\frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}$
E_cell = E°_cell – $\frac { 0.0591 }{ n }$ log $\frac{\left[Z n^{2+}\right]}{\left[C u^{2+}\right]}$
If suppose concentration of Cu²⁺ is 1 .OM then the concentration of Zn²⁺ is 10M (why because,
ion concentration in the anode compartment increased by a 10 factor)
E_cell = 1.1 – $\frac { 0.0591 }{ n }$ log $(\frac { 10 }{ 1 })$
= 1.1 – 0.02955 ……………….(1)
= 1.070 V (cell voltage decreased)
Thus, the initial voltage is greater than E° because Q < 1. As the reaction proceeds, [Zn²⁺] in the anode compartment increases as the zinc electrode dissolves, while [Cu²⁺] in the cathode compartment decreases as metallic copper is deposited on the electrode.

During this process, the Q = [Zn²⁺] [Cu²⁺] steadily increases and the cell voltage therefore steadily decreases. [Zn²⁺] will continue to increase in the anode compartment and [Cu²⁺] will continue to decrease in the cathode compartment. Thus the value of Q will increase further leading to a further
decrease in value.

Question 7.
A solution of a salt of metal was electrolysed for 150 minutes with a current of 0.15 amperes. The mass of the metal deposited at the cathode is 0.783g. calculate the equivalent mass of the metal.
Answer:
Given,
I = 0.15 amperes
t = 150 mins
= t = 15O x 6Osec
= t = 9000sec
Q = It
= Q = 0.15 x 9000 coulombs
= Q = 1350 coulombs
Hence, 135 coulombs of electricity deposit is equal to 0.783g of metal.
96500 coulombs of electricity, $\frac { 0.783 x 96500 }{ 1350 }$ = 55.97 gm of metal
Hence equivalent mass of the metal is 55.97

Samacheer Kalvi 12th Chemistry Electro Chemistry Example Problems

Question 1.
A conductivity cell has two platinum electrodes separated by a distance 1.5 cm and the cross sectional area of each electrode is 4.5 sq cm. Using this cell, the resistance of 0.5 N electrolytic solution was measured as 15?. Find the specific conductance of the solution.
Solution.
k = $\frac { 1 }{ 15Ω }$ x $\frac{1.5 \times 10^{-2} \mathrm{m}}{4.5 \times 10^{-4} \mathrm{m}^{2}}$ = 2.22S m^-1
l = 1.5cm = 1.5 x 10²m
A = 4.5 cm² x (10^-4)m²
R = 15Ω

Question 2.
Calculate the molar conductance of 0.025M aqueous solution of calcium chloride at 25°C. The specific conductance of calcium chloride is 12.04 x 10²S m³.
Answer:
Molar conductance
_Λm = $\frac{\left(\mathrm{Sm}^{-1}\right) \times 10^{-3}}{\mathrm{M}}$ mol^-1 m³
= $\frac{\left(12.04 \times 10^{-2} \mathrm{Sm}^{-1}\right) \times 10^{-3}}{0.025}$ mol^-1 m³ = 581 . 6 10^-2 S m² mol¹

Question 3.
The resistance of a conductivity cell is measured as 190Ω using 0.1M KCI solution (specific conductance of 0.1M KCI is 1.3 Sm^-1). When the same cell is filled with 0.003M sodium chloride solution, the measured resistance is 6.3K?. Both these measurements are made at a particular temperature. Calculate the specific and molar conductance of NaCl solution.
Answer:
Given that
K = 1.3 S m^-1 (for 0.1 M KCI solution)
R = 190Ω
$\left(\frac{l}{A}\right)$ = k.R = (1.3 S m^-1) (190?) = 247m^-1
k_(NaCl) = $\frac{1}{\mathrm{R}_{(\mathrm{NaCT})}}$$\left(\frac{l}{A}\right)$ = $\frac{1}{6.3 \mathrm{K} \Omega}$ (247m^-1)
= 39.2 x 10^-3Sm^-1
_Λm = $\frac{\kappa \times 10^{-3} \mathrm{mol}^{-1} \mathrm{m}^{3}}{\mathrm{M}}$ = $\frac{39.2 \times 10^{-3}\left(\mathrm{Sm}^{-1}\right) \times 10^{-3}\left(\mathrm{mol}^{-1} \mathrm{m}^{3}\right)}{0.003}$
_Λm = 13.04 x 10^-3Sm² mol^-1

Question 4.
The net redox reaction of a galvanic cell is given below
2 Cr _(s) + 3Cu²⁺_(aq) → 2Cr³⁺_(aq) + 3Cu_(s)
Write the half reactions and describe the cell using cell notation.
Answer:
Anodic oxidation: 2Cr_(s) → 2Cr³⁺_(aq) + 6e^– …………..(1)
Cathodic reduction: 3Cu²⁺ + 6e^– → 3Cu_(s) ………….(2)
Cell Notation is: Cr_(s) | CI³⁺_(aq) || Cu²⁺_(aq) | Cu_(s)

Question 5.
Let us calculate the emf of the following cell at 25°C using Nernst equation.
Cu (s) | Cu²⁺(0.25 aq, M) || Fe³⁺(0.05 aq M) | Fe²⁺(0.1 aq M) | pt(s)
Answer:
Given:
Half reactions are
Cu (s) → Cu²⁺(aq) + 2e^– ……………(1)
2 Fe³⁺(aq) + 2e^– → 2Fe²⁺(aq) ……………(2)
the overall reaction is Cu (s) + 2 Fe³⁺(aq) → Cu²⁺(aq) + 2Fe²⁺(aq), and n = 2
Apply Nernst equation at 25°C

Given standard reduction potetial of Cu²⁺ | Cu is o.34

Question 6.
A solution of silver nitrate Is electrolysed for 20 minutes with a current of 2 amperes. Calculate the mass of silver deposited at the cathode.
Answer:
Electrochemical reaction at cathode is Ag⁺ + e^– Ag (reduction)
m = ZIt
m = $\frac{108 \mathrm{gmol}^{-1}}{96500 \mathrm{Cmol}^{-1}} \times 2400 \mathrm{C}$
m = 2.68 g
z = $\frac{\text { molarmass of } \mathrm{Ag}}{(96500)}$ = $\frac{108}{1 \times 96500}$
I = 2Ag
t = 20 x 60s = 1200s
It = 2A x 1200S = 2400C

Samacheer Kalvi 12th Chemistry Electro Chemistry Additional Questions

Samacheer Kalvi 12th Chemistry Electro Chemistry 1 Marks Questions And Answers

I. Choose the best answer and write it.

Question 1.
Which one of the following is an example of the conductor?
(a) PVC
(b) Bakelite
(c) Aluminium
(d) Rubber
Answer:
(c) Aluminium

Question 2.
Which one of the following can act as an insulator?
(a) Bakelite
(b) Aluminium
(c) Copper
(d) NaCI Solution
Answer:
(a) Bakelite V

Question 3.
Which form of energy is converted to electrical energy in batteries?
(a) tidal energy
(b) Chemical energy
(c) mechanical energy
(d) atomic energy
Answer:
(b) Chemical energy

Question 4.
Electrochemical reactions are generally ………..
(a) Reduction reactions
(b) oxidation reactions
(c) Redox reactions
(d) condensation reactions
Answer:
(c) Redox reactions

Question 5.
Consider the following statements.
Answer:
(i) Energy can neither be created nor be destroyed but one form of energy can be converted to another form
(ii) In batteries, electrical energy is converted to chemical energy.
(iii) Electrochemjcal reactions are redox reactions.

Which of the above statement is/are not correct?
(a) i & ii only
(b) ii only
(c) i only
(d) iii only
Answer:
(b) ii only

Question 6.
Which one of the following represents Ohm’s law?
(a) V = IR
(b) R = $\frac { 1 }{ V }$
(c) I = $\frac { V }{ R }$
(d) R = VI
Answer:
(a) V = IR

Question 7.
The unit of resistivity is …………
(a) Ω m^-1
(b) Ω m
(c) m^-1Ohm²
(d) Ω^-1m^-1
Answer:
(b) Ω m

Question 8.
When cell constant is unit, the resistance is known as …………
(a) specific resistance
(b) conductance
(c) specific conductance
(d) equivalent conductance
Answer:
(a) specific resistance

Question 9.
The unit of specific resistance is equal to ………..
(a) Ohm metre
(b) Ohm^-1 metre
(c) Ohm^-1 metre^-1
(d) Ohm
Answer:
(a) Ohm metre

Question 10.
Which is the SI unit of conductance?
(a) Siemen^-1 (or) S^-1
(b) Siemen (or) S
(c) Sm^-1
(d) S^-1m^-1
Answer:
(b) Siemen (S)

Question 11.
Which one of the following represents specific conductance (kappa)?
(a) $\frac { I }{ C }$ . $\frac { l }{ a }$
(b) $\frac { I }{ P }$ . $\frac { a }{ I }$
(c) $\frac { 1 }{ 2 }$ . $\frac{a}{l^{2}}$
(d) $\frac { I }{ P }$ . $\frac { l }{ a }$
Answer:
(d) $\frac { I }{ P }$ . $\frac { l }{ a }$

Question 12.
Which one is the unit of specific conductance?
(a) Ohm m
(b) Ohm^-1 m
(c) Ohm m^-1
(d) Ohm^-1 m^-1.
Answer:
(d) Ohm^-1m^-1

Question 13.
Which one of the following formulas represents equivalent conductance?
(a) $\frac { I }{ P }$.$\frac { l }{ a }$
(b) $\frac { I }{ P }$.$\frac { A }{ l }$
(c) C x $\frac { l }{ a }$
(d) $\frac{\kappa \times 10^{-3}}{N}$
Answer:
(d) $\frac{\kappa \times 10^{-3}}{N}$

Question 14.
The unit of equivalent conductance is …………
(a) Sm²g equivalenr’
(b) Sm^-1
(c) Ohm^-1m^-1
(d) Ohm m
Answer:
(a) Sm²g equivalent^-1

Question 15.
Consider the following statements:
(i) Solvent of higher dielectric constant show very low conductance in solution.
(ii) Conductance is directly proportional to the viscosity of the medium.
(iii) Molar conductance of a solution increases with increase in dilution.
Which of the above statement is / are correct?
(a) (i) & (ii)
(b) (ii) and (iii)
(c) (iii) only
(d) (i) only
Answer:
(c) (iii) only

Question 16.
Consider the following statements:
(i) If the temperature of the electrolytic solution increases, conductance decreases.
(ii) Conductivity increases with the decrease in viscosity.
(iii) Molar conductance of a solution decreases with increase in dilution.
Which of the above statement is / are not correct?
(a) (i) & (iii)
(b) (i) and (ii)
(c) (iii) only
(d) (ii) only
Answer:
(a) (i) & (iii)

Question 17.
Which one of the following is used to measure conductivity of ionic solutions?
(a) metre scale
(b) wheat stone bridge
(c) Dynamo
(d) Ammeter
Answer:
(b) wheat stone bridge

Question 18.
Which of the following is used to calculate the conductivity of strong electrolytes?
(a) Kohlraush’s law
(b) Henderson equation
(c) Debye-Huckel and Onsagar equation
(d) Ostwald’s dilution law
Answer:
(c) Debye-Huckel and Onsagar equation

Question 19.
Which one of the following represents Debye-Huckel and Onsagar equation?

Answer:

Question 20.
The value of A in Debye – Huckel and Onsagar equation is ……….

Answer:

Question 21.
The value of B in Debye Huckel and onsagar equation is …………

Answer:

Question 22.
Kohlrausch’s law is applied to calculate
(a) molar conductance at infinite dilution of a weak electrolyte
(b) degree of dissociation of weak electrolyte
(c) solubility of a sparingly soluble salt
(d) all the above
Answer:
(d) all the above

Question 23.
In which of the following interconversion of electrical energy into chemical energy and vice versa take place?
(a) electrochemical cell
(b) electric cell
(c) Dynamo
(d) AC generator
Answer:
(a) electrochemical cell

Question 24.
Consider the following statements:
(i) In Galvanic cell, chemical energy is’converted into electrical energy.
(ii) In electrolytic cell, electrical energy is converted into chemical energy.
(iii) In voltaic cell, electrical energy is converted into chemical energy.

Which of the above statement is / are not correct?
(a) (i) & (ii)
(b) (iii) only
(c) (ii) only
(d) (i) only
Answer:
(b) (iii) only

Question 25.
In Galvanic cell, the Zinc metal strip placed gets ………….
(a) Oxidised
(b) reduced
(c) hydrolysed
(d) condensed
Answer:
(a) Oxidised

Question 26.
Consider the following statements:
(i) In Galvanic cell, Zinc is oxidised to Zn²⁺ ions and Cu²⁺ ions are reduced to copper
(ii) In Galvanic cell, Zn²⁺ ions are reduced to Zinc and copper is oxidised to Cu²⁺ ions
(iii) In Galvanic cell, Zn and copper both get oxidised.
Which of the above statement is / are correct?
(a) (i) only
(b) (ii) & (iii)
(c) (ii) only
(d) (iii) only
Answer:
(a) (i) only

Question 27.
The salt bridge used in Daniel cell contains
(a) Na₂SO₄ + NaCl
(b) Agar-Agar gel + Na₂SO₄
(c) Silica gel + CuSO₄
(d) ZnSO₄ + CuSO₄
Answer:
(b) Agar-Agar gel + Na₂SO₄

Question 28.
Consider the following statements.
(i) In Daniel cell, when the switch (k) closes the circuit, the electrons flow from Zinc strip to copper strip.
(ii) In Daniel cell, when the switch (k) closes the circuit, the electrons flow from copper strip to Zinc strip
(iii) In Daniel cell, when the Switch (k) opens the circuit, the electrons flow from Zinc to copper.
Which of the above statement is / are correct?
(a) (i) only
(b) (ii) & (iii)
(c) (ii) only
(d) (iii) only
Answer:
(a) (i) only

Question 29.
Which one of the following can act as an inert electrode?
(a) Graphite
(b) Copper
(c) Platinum
(d) either a (or) e
Answer:
(a) either a (or) c

Question 30.
The SI unit of cell potential is ………….
(a) Ampere
(b) Ohm
(c) Volt
(d) Ohm^-1
Answer:
(c) Volt

Question 31.
The emf of Daniel cell Zn_(s) + Zn²⁺_aq(1m) || Cu²⁺_aq(1m) | Cu_(S) iS equal to …………
(a) – 1.107 Volts
(b) 1.107 Volts
(c) 3.4 Volt
(d) 7.6 Volt
Answer:
(b) 1.107 Volts

Question 32.
Which instrument is used to measure potential difference?
(a) Ammeter
(b) Voltmeter
(c) Wheat stone bridge
(d) metre bridge
Answer:
(b) Voltmeter

Question 33.
The value of EMF of standard hydrogen electrode at 25°C is …………
(a) maximum
(b) zero
(c) negative
(d) positive
Answer:
(b) zero

Question 34.
The electrode used in SHE is made of ………….
(a) graphite
(b) copper
(c) platinum
(d) iron
Answer:
(c) Platinum

Question 35.
What is the charge of one electron?
(a) 1.602 x 10¹⁹C
(b) 1.6 x 10^-19C
(c) 9645C
(d) 96488C
Answer:
(b) 1.6 x 10^-19C

Question 36.
The maximum work that can be obtained from a galvanic cell is ………..
(a) + nFE
(b) – nFE
(c) 2F
(d) 96500 F
Answer:
(b) – nFE

Question 37.
For all spontaneous cell reactions, the value of ?G should be ………….
(a) constant
(b) zero
(c) negative
(d) positive
Answer:
(c) negative

Question 38.
The value of one Faraday is equal to …………..
(a) 96400 C
(b) 96500 C
(c) 1 .602 x 10^-19C
(d) 1 .602 x 10¹⁹C
Answer:
(b) 96500 C

Question 39.
The relationship between standard free energy change and equilibrium constant is expressed as ………..
(a) ΔG° = -RTInk_eq
(b) ΔG = RTlnk_eq
(c) ΔG° = $-\frac{1}{\mathrm{RT} \ln \mathrm{k}_{\mathrm{eq}}}$
(d) ΔG = RTlog_eq
Answer:
(a) ΔG° -RThilc

Question 40.
Which equation relates the cell potential and the concentration of the species involved in an electro chemical reaction?
(a) Henderson equation
(b) Arrhenius equation
(c) Debye Huckel Onsagar equation
(d) Nemst equation
Answer:
(a) Nernst equation

Question 41.
Which one of the following is Nernst equation.

Answer:

Question 42.
Gibbs free energy can be related to cell emf as follows.
(a) ΔG° = – nFE_cell
(b) ΔG° = – nFE°_cell
(c) ΔG = nFE_cell
(d) ΔG° = nFE°_cell
Answer:
(b) ΔG° = – nFE°_cell

Question 43.
Which one of the following represents Faraday’s first law?
(a) m = ZIt
(b) m = $\frac { Z }{ It }$
(c) m = $\frac { It }{ Z }$
(d) Z = mIt
Answer:
(a) m = ZIt

Question 44.
When 1 coulomb of electric current is passed the amount of substance deposited or liberated is known as ………..
(a) equivalent mass
(b) electro chemical equivalent
(c) molar mass
(d) 1 Faraday
Answer:
(b) electro chemical equivalent

Question 45.
The value of electro chemical equivalent is equal to ………..
(a) $\frac{96500}{\text { Equivalent mass }}$
(b) $\frac{96500}{\text { Molar mass }}$
(c) $\frac{\text { Molar mass }}{96500}$
(d) $\frac{\text { Equivalent mass }}{96500}$
Answer:
(d) $\frac{\text { Equivalent mass }}{96500}$

Question 46.
The mathematical expression of Faraday’s second law is …………
(a) m = ZIt
(b) $\frac{m_{1}}{E_{1}}=\frac{m_{2}}{E_{2}}=\frac{m_{3}}{E_{3}}$
(c) $\frac{m_{1}}{Z_{1}}=\frac{m_{2}}{Z_{2}}=\frac{m_{3}}{Z_{3}}$
(d) $Z=\frac{m}{I t}$
Answer:
(c) $\frac{m_{1}}{Z_{1}}=\frac{m_{2}}{Z_{2}}=\frac{m_{3}}{Z_{3}}$

Question 47.
Which one of the following is used in cell phone, dry cell in flashlight?
(a) Zn – Cu battery
(b) Li – ion battery
(c) Ag – Cu battery
(d) Na, NaCI battery
Answer:
(b) Li – ion battery

Question 48.
The primary batteries are ………..
(a) rechargeable
(b) non – rechargeable
(c) reversible
(d) renewable
Answer:
(b) non – rechargeable

Question 49.
Consider the following statements.
(i) The secondary batteries are rechargeable
(ii) Primary batteries are non – rechargeable
(iii) Batteries are used as a source of AC current at a constant voltage.
Which of the above statement is I are not correct?
(a) (i) & (ii)
(b) (iii) only
(c) (i) only
(d) (ii) only
Answer:
(b) (iii) only

Question 50.
The anode and cathode used in Leclanche cell are ………… respectively.
(a) Zinc, Graphite rod with MnO₂
(b) Graphite rod in MnO₂ and Zinc container
(c) Zn container and copper rod
(d) Copper container and Zinc rod
Answer:
(a) Zinc, Graphite rod with MnO₂

Question 51.
Which electrolyte is used in Leclanche cell?
(a) ZnSO₄ + CuSO₄
(b) NH₄CI + ZnCl₂
(c) NaCI + CuSO₄
(d) MnSO₄ + MnO₂
Answer:
(b) NH₄Cl + ZnCl₂

Question 52.
Which one of the following is used as cathode in Mercury button cell?
(a) Zinc
(b) Copper
(c) Zinc amalgamated with mercury
(d) HgO mixed with graphite
Answer:
(c) Zinc amalgamated with mercury

Question 53.
Which one of the following is used as anode in Mercury button cell?
(a) HgO mixed with graphite
(b) Zinc amalgamated with mercury
(c) Copper amalgamated with Mercury
(d) HgO mixed with Copper
Answer:
(a) HgO mixed with graphite.

Question 54.
The value of cell emf of Mercury button cell is ………..
(a) 1.35V
(b) – 076V
(c) 0.34V
(d) 100V
Answer:
(a) 1.35 V
Which one of the following is used in pacemakers, cameras and electronic watches?
(a) Li-ion battery
(b) Leclanche cell
(c) Galvanic cell
(d) Mercurry button cell
Answer:
(d) Mercury button cell

Question 56.
The electrolyte used in Mercury button cell is ………….
(a) Paste of kOH and ZnO
(b) CuSO₄ + ZnSO₄
(c) NaCl + MgCl₂
(d) NH₄CI + ZnCl₂
Answer:
(a) Paste of kOH and ZnO

Question 57.
Which of the following is an example of secondary batteries?
(a) Mercury button cell
(b) Leclanche cell
(c) Lead storage battery
(d) Daniel cell
Answer:
(c) Lead storage battery

Question 58.
Which of the following act as cathode and anode in Lead storage battery?
(a) Lead plate bearing PbO₂, spongy Lead
(b) Spongy lead, lead plate bearing PbO₂
(c) Lead Copper
(d) Mercury oxide, PbO
Answer:
(a) Lead plate bearing PbO₂, Spongy lead

Question 59.
Which one of the following is used as an electrolyte Lead storage battery?
(a) PbSO₄
(b) H₂SO₄
(c) CuSO₄
(d) HNO₃
Answer:
(b) H₂SO₄

Question 60.
The emf of lead storage battery is …………
(a) +1.1 V
(b) 2.4V
(c) 2V
(d) 11 . 2V
Answer:
(c) 2 V

Question 61.
The Lead storage battery is used in …………
(a) pacemakers
(b) automobiles
(c) electronic watches
(d) flash light
Answer:
(b) automobiles

Question 62.
Which one of the following is used in automobiles, trains and in inverters?
(a) Lithium ion battery
(b) Mercurry button cell
(c) Lead storage battery
(d) Leclanche cell
Answer:
(c) Lead storage battery

Question 63.
Which one of the following is used as an anode in Lithium ion battery?
(a) Porous graphite
(b) Lithium
(c) CoO₂
(d) Copper
Answer:
(a) Porous graphite

Question 64.
which one of the following is used as cathode in Lithium ion battery?
(a) Porous graphite
(b) Lithium
(c) CoO₂
(d) Chromium
Answer:
(c) CoO₂

Question 65.
Which one of the following is used in cellular phones, Laptop computers and in digital camera?
(a) Mercury button cell
(b) Lithium – ion battery
(c) H₂O₂ fuel cell
(d) Leclanche cell
Answer:
(b) Lithium – ion battery

Question 66.
Which one of the following is used as an electrolyte in H₂O₂ fuel cell?
(a) Aqueous CuSO₄
(b) Aqueous CoO₂
(c) Aqueous KOH
(d) NH₄CI + ZnCI₂
Answer:
(c) Aqueous KOH

Question 67.
Which one of the following is an example for electrochemical process?
(a) Chrome plating
(b) Rusting of iron
(c) Galvanisation
(d) All the above
Answer:
(a) All the above

Question 68.
The formula of rust is ………..
(a) Fe₂O₃
(b) Fe₂O₃.xH₂O
(c) FeO
(d) FeO.xH₂O
Answer:
(b) Fe₂O₃.xH₂O

Question 69.
Which one of the following is / are very important for rusting’?
(a) Oxygen
(b) Water
(c) Both a & b
(d) H₂O
Answer:
(c) Both a & b

Question 70.
The electro plating of Zinc over a metal is called …………..
(a) Electrolysis
(b) Redox reaction
(c) Galvanisation
(d) Passivation
Answer:
(c) Galvanisation

Question 71.
Consider the following statements.
(i) The standard reduction potential (E°) is a measure of oxidising tendency of the species.
(ii) The standard oxidation potential (E°) is a measure of oxidising tendency of the species.
(iii) The standard oxidation potential (E°) is a measure of redox tendency of the species.
Which of the above statement is / are not correct?
(a) (i) only
(b) (ii) only
(c) (ii) & (iii)
(d) (iii) only
Answer:
(c) (ii) & (iii)

Question 72.
On the basis of the electrochemical theory of aqueous corrosion, the reaction occuring at the cathode is …………

Answer:

Hint:
2H⁺_(aq) + 2e^– → 2H₂
2H⁺ + $\frac { 1 }{ 2 }$O₂ → H₂O
2H⁺ + $\frac { 1 }{ 2 }$O₂ +2e^– → H₂O
Balancing the above equation
4H⁺_(aq) + O₂ + 4e^– → 2H₂O

Question 73.
The standard reduction potential for the half reactions are as ………..
Zn → Zn²⁺ + 2e^– E° = +0.76V
Fe → Fe²⁺ + 2e^–E° = + 041 V.
So for cell reaction F²⁺ + Zn → Zn²⁺ + Fe is ………….
(a) – 0.35V
(b) +0.35V
(c) +1.17V
(d) – 1.117V
Answer:
(b) +0.35V
Hint:
In the reaction F²⁺ + Zn° → Zn²⁺ + Fe°
emf = E_cathode – E_anode
= – 0.41 – (- 0.76)
= – 0.41 + 0.76
emf = + 0.35V

Question 74.
The standard emf for the given cell reaction Zn + Cu²⁺ → Cu + Zn²⁺ is 1.10V at 25°C. The emf for the cell reaction when 0.1 M Cu²⁺ and 0.1 M Zn²⁺ solutions are used at 25°C is ……….
(a) 1.10V
(b) 0.110V
(c) – 1.10V
(d) – 110V
Answer:
(a) 1.10V
Hint:
E_cell = E°_cell – $\frac { 0.0592 }{ 2 }$ log$\frac{\left(Z n^{2+}\right)}{\left(C u^{2+}\right)}$
= 1.10 – $\frac { 0.0592 }{ 2 }$ log $\frac { 0.1 }{ 0.1 }$
= 1.10 – $\frac { 0.0592 }{ 2 }$ log 1
= 1.10 – $\frac { 0.0592 }{ 2 }$ x 0
= 1.10V

Question 75.
Which amount of chlorine gas liberated at anode, if 1 ampere current is passed for 30 minutes from NaCI solution?
(a) 0.66 moles
(b) 0.33 moles
(c) 0.66 g
(d) 0.33 g
Answer:
(c) 0.66 g
Hint:
2Cl^– → Cl₂ + 2e^–
Q = It.
Amount of current passed = 1 x 30 x 60 = 1800C
The amount of Cl₂ liberated by passing 1800 coulomb of electric charge
= $\frac{1 \times 1800 \times 71}{2 \times 96500}$
= 0.66g

Question 76.
When Zinc piece is kept in CuSO₄ solution the copper gets precipitated due to standard potential of Zinc is …………
(a) > copper
(b) < copper
(c) > Sulphate
(d) < Sulphate
Answer:
(b) < copper
Hint:
Standard potential of zinc < copper.

Question 77.
Which equation shows the relation between electrode potential (E) standard electrode potential (E°) and concentration of ions in solution is ………..
(a) Kohlrausch’s equation
(b) Nernst equation
(c) Ohm’s equation
(d) Faraday’s equation
Answer:
(b) Nernst equation

Question 78.
The standard electrode potential of SHE at 298K is ………
(a) 0.05 V
(b) 0.01 V
(c) 0.0 V
(d) 0.11 V
Answer:
(c) 0.0 V

Question 79.
The reaction Zn²⁺ + 2e^– → Zn has a standard potential of – 0.76 V. This means
(a) Zn cannot replace hydrogen from acids
(b) Zn is a reducing agent
(c) Zn is an oxidising agent
(d) Zn²⁺ is a reducing agent
Answer:
(b) Zn is a reducing agent.
Hint:
Since E⁰_Zn²⁺/Zn is negative, so Zn has a greater tendency to be oxidised than hydrogen. Hence it can act as reducing agent.

Question 80.
K, Ca and Li metals may be arranged in the decreasing order of their standard electrode potentials as ……..
(a) K, Ca, Li
(b) Ca, K, Li
(c) Li, Ca, K
(d) Ca, Li, K
Answer:
(b) Ca, K, Li

Question 81.
The correct order of chemical reactivity with water according to electrochemical series ………….
(a) K > Mg > Zn > Cu
(b) Mg > Zn > Cu > K
(c) K > Zn > Mg > Cu
(d) Cu > Zn > Mg > K
Answer:
(a) K > Mg > Zn > Cu
Hint:
The standard reduction potential of K⁺, Mg²⁺, Zn²⁺, Cu²⁺ increases in this order.

Question 82.
For a spontaneous reaction, the ΔG, the equilibrium constant (K) and E°_cell will be respectively
(a) ve, > 1, + ve
(b) + ve , > 1, – ve
(c) – ve, < 1, – ve (d) – ve, > 1, – ve
Answer:
(a) – ve, > 1, + ve.

Question 83.
E° values of mg²⁺/mg is – 2.37 V, Zn²⁺/ Zn is – 0.76V, and Fe²⁺ / Fe is – 0.44 V. Which of the following statement is correct?
(a) Zn will reduce Fe²⁺
(b) Zn will reduce mg²⁺
(c) mg oxidises Fe
(d) Zn oxidises Fe
Answer:
(a) Zn will reduce Fe²⁺
Hint:
E⁰_Zn²⁺/Zn<E⁰_Fe²⁺/Fe So Zn will reduce Fe²⁺. Zinc cannot reduce Mg²⁺ because
$\mathbf{E}_{\mathbf{Z n}^{2+} / \mathbf{z n}}^{\circ}>\mathbf{E}_{\mathrm{Mg}^{2+} / \mathrm{Mg}}^{\circ}$ On similar reason mg and Zn cannot oxidise Fe.

Question 84.
In which cell, the free energy of a chemical reaction is directly converted into electricity?
(a) Leclanche cell
(b) Fuel cell
(c) Lead storage battery
(d) Lithium ion battery
Answer:
(b) Fuel cell

Question 85.
Which of the following has the highest electrode potential?
(a) Li
(b) Cu
(c) Au
(d) Al
Answer:
(c)Au

Question 86.
Consider the following statements.
(i) A salt bridge is used to eliminate liquid junction potential
(ii) The Gibbs free energy change ∆G is related with electro motive force (E) as ∆G = – nFE.
(iii) Nernst equation for a single electrode potential is E = E° – $\frac { RT }{ nF }$ In $a_{m} n^{4}$
(iv) The efficiency of a hydrogen oxygen fuel cell is 23%.

Which of the above statement is / are not correct?
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (iv) only
(d) (i) only
Answer:
(c) (iv) only

Question 87.
The specific conductance of 0.1 N KCl solution at 23°C is 0.012 Ohm^-1 cm^-1. The resistance of the cell containing the solution at the same temperature was found to be 55 Ohm. The cell constant will be …………
(a) 0.142 cm^-1
(b) 0.66 cm^-1
(c) 0.9 18 cm^-1
(d) 1.12 cm^-1
Answer:
(c) 0.66 cm^-1
Hint:
k x $\frac { 1 }{ R }$ x cell constant
Cell constant = k x R
= 0.012 x 55
= 0.66 cm^-1

Question 88.
Which of the following reaction is used to make a fuel cell?
(a) Cd_(s) + 2Ni(OH)_3(s) → CdO_(s) + 2Ni (OH) + H₂O₍₁₎
(b) Pb_(s) + PbO_2(s) + 2H₂SO_4(aq) → 2 PbSO_4(s) + 2H₂O₍₁₎
(c) 2H_2(g) + O_(s) + 2H₂O₍₁₎
(d) 2Fe_(s) + O_2(g) + 4H⁺_(ag) + 2Fe_4(s) + 2H₂O₍₁₎
Answer:
(c) 2H_2(g) + O_(s) + 2H₂O₍₁₎

When lead storage battery is charged
(a) PbO₂ is dissolved
(b) PbSO₄ is deposited on lead electrode
(c) PbSO₄ is deposited on lead electrode

Question 89.
Which colourless gas evolves when NH₄CI reacts with Zinc in a dry cell battery?
(a) NH₃
(b) N₂
(c) H₂
(d) Cl₂
Answer:
(c) H₂
Hint: 2NH₄Cl + Zn → 2NH₃ + ZnCl₂ + H₂ ↑

Question 91.
A cell from the following which converts electrical energy into chemical energy?
(b) Electro chemical cell
(d) Lithium – ion battery
(a) dry cell
(c) Electrolytic cell
Answer:

Question 92.
When 9.65 Coulombs of electricity is passed through a solution of silver nitrate (Atomic weight of Ag = 107.85g), the amount of silver deposited is ……………
(a) 10.8 mg
(b) 5.4 mg
(c) 16.2 mg
(d) 21.2 mg
Answer:
(a) 10.8 mg
Hint:
W_Ag = $\frac{E_{A g} \times Q}{96500}$ = $\frac{108 \times 9.65}{96500}$
= 1.08 x 10^-2

Question 93.
What weight of copper will be deposited by passing 2 Faraday’s of electricity through a cupric salt (Atomic weight of Cu = 63.5)
(a) 2.0g
(b) 3.175g
(c) 63.5g
(d) 127.0g
Answer:
(c) 63.5 g
Hint:
Cu²⁺ + 2e^– → Cu
2 Faraday’s will deposit 1g atom of Cu = 63.5 g

Question 94.
In electrolysis of a fused salt, the weight of the deposit on an electrode will not depend on ….
(a) temperature of the bath
(b) current intensity
(c) electro chemical equivalent of ions
(d) time for electrolysis.
Answer:
(a) temperature of the bath

Question 95.
The mass deposited at an electrode is directly proportional to ………..
(a) atomic weight
(b) equivalent weight
(c) molecular weight
(d) atomic number
Answer:
(b) equivalent weight

Question 96.
Which solution will show the highest resistance during the passage of current?
(a) 0.05 N NaCl
(b) 2N NaCI
(c) 0.1N NaCI
(d) 1N NaCI
Answer:
(b) 2N NaCl

Question 97.
In a galvanic cell, the electrons flow from
(a) anode to cathode through the solution
(b) cathode to anode through the solution
(c) anode to cathode through the external circuit
(d) cathode to anode through the external circuit
Answer:
(c) anode to cathode through the external circuit

Question 98.
Rusting of iron is catalysed by which of the following?
(a) Fe
(b) O₂
(c) Zn
(d) H
Answer:
(d) H

Question 99.
The conductivity of strong electrolyte is ………..
(a) increase on dilution slightly
(b) decrease on dilution
(c) does not change with dilution
(d) depend upon density of electrolyte itself
Answer:
(a) increase on dilutions lightly

Question 100.
Which one is not a conductor of electricity?
(a) NaCl_(aqueous)
(b) NaCl_(solid)
(c) NaCl_(molten)
(d) Ag_(metal)
Answer:
(b) NaCl_(solid)
Hint:
In solid state, NaCl does not dissociate into ions so it does not conduct electricity.

Question 101.
The molar conductivity is maximum for the solution of concentration
(a) 0.00 1 m
(b) 0.005 m
(c) 0.002 m
(d) 0.004 m
Answer:
(a) 0.001 m
Hint:
molar conductance α $\frac{1}{\text { molarity }}$

Question 102.
Resistance of 0.2 m solution of an electrolyte is 50 Ohm^-1. The specific conductance of the solution is 1.3 Sm^-1. If resistance of 0.4 m solution of the same electrolyte is 260 Ohm^-1, its molar conductivity is ……..
(a) 62.5 Sm² mol^-1
(b) 6250 Sm² mol^-1
(c) 6.25 x 10^-4 Sm² mol^-1
(d) 625 x 10^-4 Sm² mol^-1
Answer:
(c) 6.25 x 10^-4 Sm² mol^-1

Question 103.
Saturated solution of KCI (or) Na₂SO₄ is used to make salt bridge because
(a) velocity of K⁺ is greater than that of Cl^–
(b) velocity of Cl^– is greater than that of K⁺
(c) velocity of both K⁺ and Cl^-1 are nearly the same
(d) KCI is highly soluble in water.
Answer:
(c) velocity of both K⁺ and Cl^– are nearly the same

Question 104.
Which of the following electrolytic solutions has the least specific conductance?
(a) 0.02 N
(b) 0.2 N
(c) 2 N
(d) 0.002 N
Answer:
(d) 0.002 N

Question 105.
An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to ……….
(a) increase in both the number of ions and ionic mobility of ions
(b) increase in number of ions
(c) increase in ionic mobility of ions
(d) 100% ionization of electrolyte at normal dilution
Answer:
(c) increase in ionic mobility of ions

Question 106.
Li occupies higher position in the electrochcmical series of metals as compared to Cu, since
(a) the standard reduction potential of Li⁺/Li is lower than that of Cu²⁺/Cu
(b) the standard reduction potential of Cu²⁺/Cu is lower than that of Li⁺/Li
(c) the standard oxidation potential of Li/Li⁺ is lower than that of Cu/Cu²⁺
(d) Li is smaller in size as compared to Cu.
Answer:
(a) the standard reduction potential of Li⁺/Li is lower than that of Cu²⁺/Cu

Question 107.
The one which decreases with dilution is …………
(a) conductance
(b) specific conductance
(c) equivalent conductance
(d) molar conductance
Answer:
(b) specific conductance

Question 108.
Corrosion of iron is essentially an electrochemical phenomenon where the cell reactions are …………..
(a) Fe is oxidised to Fe²⁺ and dissolved oxygen in water is reduced to OH^–
(b) Fe is oxidised to Fe²⁺ and H₂O is reduced to O₂^2-
(c) Fe is oxidised to Fe²⁺ and H₂O is reduced to O₂^–
(d) Fe is oxidised to Fe²⁺ and H₂O is reduced to O₂
Answer:
(a) Fe is oxidised to Fe²⁺ and dissolved oxygen in water is reduced to OH^-2

Question 109.
A button cell used in watches functions as following.
Zn_(s) + Ag₂O_(s) + H₂O₍₁₎ → 2 Ag_(s) + Zn²⁺_(aq)+ 2OHsup>-_(aq).
If half cell potentials are Zn²⁺_(aq) + 2e^– → Zn_(s)
E° = – O.76V
Ag₂O_(s) + H₂O₍₁₎ +2e^– 2Ag_(s) + 2OH^–_(aq)E⁰ = 0.34V
The cell potential will be ………..
(a) 1.10V
(b) 0.42V
(c) 0.84V
(d) 1.34V
Answer:
(a) 1.10V
Hint:
Cell potential = E_cathode – E_anode
= 0.34 – (- 0.76)
= 0.34 + 0.76
= 1.10V

Question 110.
Among the following cells Leclanche cell
(I) Nickel – cadmium cell
(II) Lead storage battery
(III) and Mercury Cell
(IV) primary cells are
(a) I & II
(b) I & III
(c) II & III
(d) I & IV
Answer:
(d) I & IV

II. Fill in the blanks.

……… is defined as the resistance of an electrolyte confined between two electrodes having unit cross sectional area and separated by a unit distance
The reciprocal of the specific resistance is called the and represented by the symbol ………
The SI unit of specific conductance is ………
The relation between equivalent conductance and the specific conductance is given as ………
Conductivity increases with the ……… in viscosity.
A°_m values of the weak electrolytes can be determined using ………
……… is a device in which a spontaneous chemical reaction generates an electric current.
……… is a device that converts electrical energy into chemical energy.
The salt bridge contains a agar-agar gel mixed with an inert electrolyte such as ………
The SI unit of cell potential is ………
The reference electrode SHE has emf of exactly ……… volt
The value of charge of one electron is equal to ………
For a spontaneous cell reaction, the should be ………
……… is a process in which electrical energy is used to cause a non-spontaneous chemical reaction.
The negative E° values shows that the reactions are ………
……… is defined as the amount of a substance deposited or liberated at the electrode by a charge of 1 Coulomb.
……… batteries are used in cell phones.
……… cell is used in pacemakers, electronic watches and cameras.
………battery is used in automobiles.
Rusting of iron is an ……… process.

Answer:

Specific resistance (or) Resistivity
Specific conductance, Kappa(k)
Sm^-1
Λ = $\frac{\kappa \times 10^{-3}}{N}$
decrease
Kohlraush’s law
Galvanic (or) Voltaic cell
Electrolytic cell
KCl (or) Na₂SO₄
Volt (V)
zero
1.6 x 10^-19C
negative
Electrolysis
non – spontaneous
electro chemical equivalent
Li-ion
Mercury button
Lead storage
electro chemical

III. Match the following

Question 1.

Answer:
i. c d a b

Question 2.

Answer:
ii. c a d b

Question 3.

Answer:
iii. d c a b

Question 4.

Answer:
iv. c a d b

Question 5.

Answer:
i. b c d a

Question 6.

Answer:
i. c a d b

IV. Assertion and reasons.

Question 1.
Assertion(A): If the temperature of the electrolytic solution increases, conductance also increases.
Reason (R): Increase in temperature increases the kinetic energy of the ions and decreases the attractive force between the oppositely charged ions and hence conductivity increases.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are wrong
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 2.
Assertion(A): Molar conductance of a solution increases with increase in dilution.
Reason (R): For a strong electrolyte, inter ionic forces of attraction decreases with dilution and so conductivity increases. For a weak electrolyte, degree of dissociation increases with dilution and conductivity increases.
(a) Both A and R are correct and R is the correct explanation of A .
(b) A and R are wrong
(c) A is correct but R is not the explanation of A
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 3.
Assertion(A): AC current is used in wheatstone bridge arrangement to measure conductivity of ionic solution.
Reason (R): If DC current is used in wheatstone bridge arrangement, it will lead to electrolys is of the solution taken in the cell. So AC current is used to prevent electrolysis.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are wrong
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 4.
Assertion(A): Strong electrolytes have low molar conductivity at high concentration.
Reason (R): For a strong electrolyte, at high concentration, the number of constituent ions of the electrolyte is high and hence the attractive force between the oppositely charged ions is also high
(a) Both A and R are correct and R is the correct explanation of A
(b) A is correct but R is wrong
(c) A is wrong but R is correct
(d) Both A and R are wrong
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 5.
Assertion(A): In Daniel cell, the salt bridge contains an agar-agar gel mixed with an inert electrolyte KCl (or) Na₂SO₄.
Reason (R): The ions of inert electrolyte do not react with other ions present in half cells and they are not either oxidised or reduced at electrodes.
(a) Both A and R are correct
(b) Both A and R are wrong
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct

Question 6.
Assertion(A): Current stops flowing when E_cell = 0
Reason (R): At E_cell = 0, Equilibrium of the cell reaction is attained.
(a) Both A and R are correct
(b) Both A and R are wrong
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct

Question 7.
Assertion(A): Copper Sulphate can be stored in a Zinc vessel.
Reason (R): Zinc is less reactive than Copper.
(a) Both A and R are correct
(b) Both A and R are wrong
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct

Question 8.
Assertion(A): As a lead storage battery gets discharged. density of electrolyte present in it decreases.
Reason (R): Lead and Lead dioxide both react with sulphuric acid to form lead sulphate.
(a) Both A and R are correct
(b) A is correct but R is wrong
(c) A is wrong but R is correct
(d) Both A and R are wrong.
Answer:
(a) Both A and R are correct

Question 9.
Assertion(A): The cell potential of mercury cell is 1.35V which remains constant.
Reason (R): In mercury cell, the electrolyte is a paste of KOH and ZnO.
(a) Both A and R are correct, but R is not the correct explanation of A
(b) Both A and R are correct, but R is the correct explanation of A
(c) A is wrong but R is correct
(d) A is correct but R is wrong
Answer:
(a) Both A and R are correct, but R is not the correct explanation of A

Question 10.
Assertion(A): If an iron rod is dipped in CuSO₄ solution, then blue colour of the solution turns red.
Reason (R): Iron is more reactive than copper and so iron displaces copper from CuSO₄ solution.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are wrong
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A

V. Find the odd one out and give the reasons.

Question 1.
(a) I α V
(b) I = $\frac { V }{ R }$
(c) V = IR
(d) R = $\frac { I }{ V }$
Answer:
(d) R is the odd one, and other three represents Ohm’s law

Question 2.
(a) m = Zit
(b) Z = $\frac { m }{ It }$
(c) m α It
(c) $\frac{\mathrm{m}_{1}}{\mathrm{Z}_{1}}$$ = $$\frac{\mathrm{m}_{2}}{\mathrm{Z}_{2}}$
Answer:
(d) $\frac{\mathrm{m}_{1}}{\mathrm{Z}_{1}}$$ = $$\frac{\mathrm{m}_{2}}{\mathrm{Z}_{2}}$ = is the odd one, all the others are Faraday’s I law but (ci) is Faraday’s ¡I law

Question 3.
(a) Pacemaker
(b) electronic watches
(c) trains
(d) cameras
Answer:
(c) train, Lead storage battery is used in trains, but in all others mercury button cell is used up.

Question 4.
(a) Automobiles
(b) Pacemaker
(c) Train
(d) Inverters
Answer:
(b) Pacemaker. In pacemaker mercury button cell is used whereas in other three, lead storage battery is used up.

Question 5.
(a) Cellular phone
(b) Laptop
(c) Digital Camera
(d) Electronic watch
Answer:
(d) Electronic watch. ¡n this mercury button cell is used whereas in others Li-ion battery is used up.

Samacheer Kalvi 12th Chemistry Electro Chemistry 2 Mark Questions and Answers

VI. Answer the following.

Question 1.
State Ohm’s law.
Answer:
At a constant temperature, the current flowing through the cell (I) is directly proportional to the voltage across the cell (V).
I α V(or) I = $\frac { V }{ R }$
V = IR

Question 2.
Define Resistivity. Give its unit.
p (rho) is called the specific resistance (or) resistivity and it is defined as the resistance of an electrolyte confined between two electrodes having unit cross sectional area and are separated by a unit distance. Unit of resistivity is Ohm metre (Ω m)

Question 3.
Define conductance and give its unit.
Answer:
The reciprocal of the resistance $(\frac { 1 }{ 2 })$ is known as the conductance of an electrolytic solution.
The SI unit of conductance is Ohm^-1 (or) Siemen (S)

Question 4.
Define specific conductance. Give its SI unit.
Answer:
The specific conductance is defined as the conductance of a cube of an electrolytic solution of unit dimensions. The SI unit of specific
k = C . $\frac { l }{ A }$
conductance is Sm^-1

Question 5.
What Is meant by resistance? Give its unit.
Answer:
Resistance is the opposition that a cell offers to the flow of electric current through it.
R α $\frac { l }{ A }$
The SI unit of R = Ohm (Ω)

Question 6.
What is molar conductance? Give its SI unit.
Answer:
The conductance of a conductivity cell in which the electrodes are separated by 1m and having V m³ of electrolytic solution that contains 1 moIe of an electrolyte is known as molar conductance.
Λ_m = k x V
The SI unit of Λ_m = S m² mol^-1

Question 7.
Define equivalent conductance. Give its SI unit.
Answer:
Equivalent conductance is defined as the conductance of ‘V’ m³ of electrolytic solution containing one gram equivalent of electrolyte in a conductivity cell in which the electrodes are one metre apart.
Λ = $\frac{\kappa \times 10^{-3}}{N}$
The SI unit of A = S m² gm equivalent’

Question 8.
What are electro chemical cells? Mention its types.
Answer:
Electro chemical cell is a device which inter converts chemical energy into electrical energy and vice versa. It consists of two separate electrodes which are in contact with an electrolyte solution. Electro chemical cells are classified into two types.

Galvanic cell and
Electrolytic cell.

Question 9.
Distinguish between galvanic cell and electrolytic cell.
Answer:
Galvanic Cell

It is a device in which a spontaneous chemical reaction generates an electric current.
It converts chemical energy into electrical energy. It is commonly known as Battery.
e.g., Daniel cell, Dry cell.
A salt bridge is used in this.

Electrolytic cell

It is a device in which an electric current from an external source drives a non spontaneous reaction
It converts electrical energy into chemical energy.
e.g., Electrolysis of molten NaCI.
Na salt bridge is used.

Question 10.
What is meant by Faraday? How is it calculated?
Answer:
One Faraday is defined as the charge of one mole of electron.
Charge of one electron = 1.6 x 10^-19C
Charge of 1 mole of electrons = 6.023 x 10²³ x 1.602 x 10^-19C
= 6.023 x 10²³ x 1.602 x 10^-19C
= 96488 C
i.e., IF ≃ 965O0C

Question 11.
Define corrosion. Give one example.
Answer:
The redox process which causes the deterioration of metal is called corrosion. Rusting of iron is an example of corrosion. It is an electro chemical process.

Question 12.
Can you store copper sulphate solution in a zinc pot?
Answer:
Zinc is more reactive than copper. Hence, it displaces copper from copper sulphate solution as follows
Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu (s) So, we cannot store copper sulphate solution in a zinc pot.

Question 13.
Why does the conductivity of a solution decrease with dilution?
Answer:
Conductivity of a solution is the conductance of ions present in a unit volume of solution. On dilution the number of ions per unit volume decreases. Hence, the conductivity decrease.

Question 14.
Suggest a way to determine 10m value of water.

We find out

Then

Question 15.
Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging? .
A lead storage battery consists of anode made up of lead, cathode made up of grid of lead packed with lead dioxide (PbO₂) and 38% solution of sulphuric acid is used as an electrolyte. When the battery is in use, the following reaction take place:
Anode: Pb(s) + SO₄^2-(aq) → PbSO₄(s) + 2e^–
Cathode: PbO₂(s) + SO₄^2-(aq) + 4H⁺(aq) + 2e^– → 2PbSO₄(s) + 2H₂O(1)
On charging the battery, the reverse reaction takes place, i.e. PbSO₄ deposited on the electrode is converted back into Pb and PbO₂ and H₂SO₄ regenerated.

Question 16.
What is meant by cell constant?
Answer:
Cell constant is the ratio of the distance between the electrodes (l) and the area of cross-section (A). It is denoted by 1. Its unit is cm^-1. Its SI unit is m^-1.

Question 17.
State two advantages of H₂ – O₂ fuel cell over ordinary cells.
Answer:

It is highly efficient.
It is pollution free

Question 18.
Why Om for CH₃COOH cannot be determined experimentally?
Answer:
Molar conductivity of weak electrolytes keeps on increasing with dilution and does not become constant even at very large dilution.

Question 19.
Which will have greater molar conductivity and why?
Answer:
Sol. (A) 1 mol KCI dissolved in 200 cc of the solution.
Sol. (B) 1 mol KCI dissolved in 500 cc of the solution.
Sol. (B) will have greater molar conductivity because
λ_m = k x V
with dilution K decreases but V increases, so that product will increase more.

Question 20.
Raju and his father were going in a boat in the river. Raju’s father was going to throw away the cell used in watches and hearing aids into the water. Raju prevented him doing so.

As a student of chemistry, why would you advise Raju’s father not to throw the cell in the water body?
What is the value associated with the above decision?

Answer:

The watch cells are made up of mercury. This mercury will pollute the water. Water contaminated with mercury leads to accumulation of mercury in the body of fishes and other aquatic life.
It helps in keeping the environment safe from pollution due to mercury.

Samacheer Kalvi 12th Chemistry Electro Chemistry 3 Mark Questions And Questions

VII.Answer the following questions.

Question 1.
Explain about conductivity cell with an example.
Answer:
1. Sodium chloride (or) potassium chloride is dissolved in a solvent like water, the electrolyte is completely dissociated to give its constituent ions (cations and anions).

2. When an electric field is applied to such an electrolytic solution, the ions present in the solution carry charge from one electrode to another. PLattnium electrode and thereby they conduct electricity. electrode

3. The conductivity of the electrolytic solution is measured using a conductivity cell, solution

Question 2.
Explain about the factors affecting electrolytic conductance. Conductivily Cell
Answer:

If the interionic attraction between the oppositely charged ions ofsolute increases, the conductance will decrease.
Solvent of high dielectric constant show high conductance in solution.
Conductance is inversely proportional to the viscosity of the medium. i.e., conductivity increases with the decrease in viscosity.
If the temperature of the electrolytic solution increases, conductance also increases.
Molar conductance of a solution increases with increase in dilution. This is because, for a strong electrolyte, inter ionic force of attraction decrease with dilution.
For a weak electrolyte, degree of dissociation increases with dilution.

Question 3.
Explain about the variation of molar conductivity with concentration by Kohiraush studies?
Answer:
Kohiraush observed that, increase of molar conductance of an electrolytic solution with the increase in dilution. He deduced the following empirical relationship between the molar conductance (Λ_m) and concentration of the
Λ_m = Λ⁰_m – k$\sqrt { c }$
For strong electrolytes such as KCl, NaCl the plot Λ_mV_s$\sqrt { c }$, gives a straight line. It is also observed that the plot is not a linear one for weak electrolyte.

Question 4.
For strong electrolytes the molar conductivity increases on dilution and reaches a maximum value at infinite dilution. Justify this statement.
Answer:
For a strong electrolyte, at high concentration, the number of constituent ions of the electrolyte in a given volume is high and hence the attractive force between the oppositely charged ions is also high. The ions also experienced a viscous drag due to greater solvation. These factors attribute for the low molar conductivity at high concentration.

When the dilution increases, the ions are far apart and the attractive forces decreases. At infinite dilution, the ions are so far apart, the interaction between them becomes insignificant and hence the molar conductivity increases and reaches a maximum value at infinite dilution.

Question 5.
For weak electrolyte, sudden increase In molar conductance with dilution. Prove this statement.
Answer:
For a weak electrolyte, at high concentration the plot is almost parallel to concentration axis with slight increase in conductivity as the dilution increases. When the concentration approaches zero, there is a sudden increase in the molar conductance and the curve is almost parallel to Am axis. This is due to the fact that the dissociation of the weak electrolyte increases with the increase in dilution (Ostwald’s dilution law).

Question 6.
Explain about Debye-Huckel and Onsagar equation.
Answer:
The influence of ion-ion interactions on the conductivity of strong electrolytes was studied by Debye and Huckel.
They considered that each ion is surrounded by an ionic atmosphere of opposite sign, and derived an expression relating the molar conductance of strong electrolytes with the concentration by It was further developed by Onsagar. For a uni – univalent electrolyte the Debye Huckel and Onsagar equation is given below.
Λ_m = Λ°_m (A + B Λ°m) $\sqrt { c }$
Where A and B are constants which depend only in the nature of the solvent and temperature.

Question 7.
What are the values of A and B in Debye Huckel and Onsagar equation?
Answer:
Debye Huckel and Onsagar equation is
Λ_m = Λ°_m (A + B Λ°m) $\sqrt { c }$
Where
$\mathrm{A}=\frac{82.4}{\sqrt{\mathrm{DT}} \mathrm{n}}$
$B=\frac{8.20 \times 10^{5}}{3 \sqrt{D T}}$
D = Dielectric constant of the medium
η = Viscosity of the medium
T = Temperature in Kelvin

Question 8.
How will you calculate degree of dissociation of weak electrolytes and dissociation constant using Kohlrausch’s law?
Answer:
1. The degree of dissociation of weak electrolyte can be calculated from the molar conductivity at a given concentration and the molar conductivity in infinite dilution using the formula

2. According to Ostwald’s dilution law K_a = $\frac{\alpha^{2} C}{1-\alpha}$
Substituting a value in the above equation

Question 9.
How would you calculate the solubility of sparingly soluble salt using Kohlrausch’s law?
Answer:
1. Substances like AgCl, PbSO₄ are sparingly soluble in water. The solubility product can be determined using conductivity experiments.

2. Let us consider AgCl as an example
AgCI_(s) → Ag⁺ + Cl^–
K_sp = [Ag⁺] [Cl^–]

3. Let the concentration of [Ag⁺] be ‘C’ mol L^-1.
If [Ag⁺] = C, then [Cl^–] is also equal to C mol L^-1.
K_sp = C.C
K_sp = C².

4. The relationship between molar conductance and equivalent conductance is

Substitute the concentration value in the relation K_sp = C²

Question 10.
What is the relationship between molar mass and electro chemical equivalent. Derive the equation.
Answer:
1. Consider the following general electro chemical redox reaction
Mⁿ⁺(aq) + ne^– → M_(s)

2. We can infer from the above equation that ‘n’ moles of electrons are required to precipitate

3. The quantity of charge required to precipitate 1 mole of Mⁿ⁺ = Charge of ‘n’ moles of electrons = nF

4. In other words, the mass of the substance deposited by one coulomb of charge. Molar mass (M)
Electro chemical equivalent Mⁿ⁺ = $\frac{\text { Molar mass }(\mathrm{M})}{\mathrm{n}(96500)}$
(or)
Z = $\frac{\text { Equivalent mass }}{96500}$

Question 11.
What is meant by standard reduction potential? What is its application?
Answer:
1. The standard reduction potential (E°) is a measure of the oxidising tendency of the species.
2. The greater the E° value means greater is the tendency shown by the species to accept electrons and undergo reduction.
3. So higher the E° values, lesser is the tendency to undergo corrosion.

Question 12.
What is meant by Electro chemical series? Mention the top most and the least placed element in that series.
Answer:
1. The standard aqueous electrode potential at 298 K for various metal-metal ion electrodes are arranged in the decreasing order of their standard reduction potential values is known as electro chemical series.

2. The strongest reducing agent is Li which has E° value as – 3.05 and it is in bottom of the series.

3. The strongest oxidising agent is F which has E° value as + 2.87 is the first element in that series.

Question 13.
Calculate the emf of the cell in which the following reaction takes place
Ni(s) + 2Ag⁺(0.002 M) —‘ Ni²⁺(0.160 M) + 2 Ag(s)
Given that E°_cell = 1.05 V
Answer:
Applying Nernst equation to the given cell reaction:

= 1.05 – 0.14 V = 0.91 V

Question 14.
If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?
Answer:
Q (coulomb) = 1 (ampere) x t(sec)
= 0.5 ampere x 2 x 60 x 60 = 3600C
A flow of IF, i.e., 96500 C is equivalent to the flow of 1 mole of electrons
i.e., 6.023 x 10²³electrons
3600 C is equivalent to flow of electrons
= $\frac{6.02 \times 10^{23}}{96500}$ x 3600 2.246 x 10²² electrons,

Question 15.
What are fuel cells? Write the electrode reactions of a fuel cell which uses the reaction of hydrogen with oxygen?
Answer:
A fuel cell is similar to a galvanic cell, it generates electricity directly by the electrochemical conversion of gaseous or liquid fuels fed into the cell as required.
At anode: H₂(g) + 2OH^–(aq) → 2H₂O(l) + 2e^–
At cathode: O₂(g) + 2H₂O(1) + 4e^– → 4OH_–(aq)
Overall reaction : 2H₂(g) + O₂(g) → 2H₂O(l)

Question 16.
Write the cell reaction which occur ¡n the lead storage battery

When the battery is in use,
When the battery is charging.

Answer:
1. Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(1)
At anode: Pb(s) + SO^2-₄ → PbSO₄(s) + 2e^–
At cathod: PbO₄(s) + SO^2-₄(aq) + 4H⁺(aq) + 2e^– → PbSO₄(s) + 2H₂O(1)

2. 2PbSO₄(s) + 2H₂O(s) → PbSO₄(s) + 2H₂SO₄(aq)

Question 17.
Describe the composition of anode and cathode in a mercury cell. Write the electrode reactions for this cell.
Answer:
Mercury cell: It consists of zinc mercury amalgam as anode, a paste of HgO and Carbon black is used as cathode. The electrolyte is a paste of KOll and ZnO.
At anode: Zn (amalgam) + 2OH^– → ZnO(s) + H₂O(l) + 2e^–
At cathode: HgO(s) + H₂O + 2e^– → Hg(1) + 2OH^–
The net reaction : Zn (amalgam) + HgO(s) → ZnO(s) + Hg(l)

Question 18.
How much copper is deposited on the cathode of an electrolytic cell If a current of 5 ampere is passed through a solution of copper sulphate for 45 minutes?
Answer:
[Molar mass of Cu = 63.5 g mol^-1, IF = 96,500 C mol^-1]
Cu²⁺(aq) + 2e^– → Cu(s)
m = Z xI x t
= $\frac{63.5}{2 \times 96500}$ x 5amp x 45 x 60 = $\frac { 857250 }{ 193000 }$ = 4.44g

Question 19.
How much time would it take in minutes to deposit 1.18 g of metallic copper on a metal object when a current of 2.0 A is passed through the electrolytic cell containing Cu²⁺ ions?
Answer:
[Molar mass of Cu = 63.5 g mol^-1, IF = 96,500 C mol^-1]
m = Z x I x t
1.18 = $\frac{63.5}{2 \times 96500}$
t = $\frac{1.18 \times 2 \times 96500}{2 \times 63.5}$
= 1793.23 sec
= $\frac { 1793.23 }{ 60 }$ = 29.88 min

Question 20.
What is a salt bridge? What is it used for?
Answer:
A salt bridge is a U – shaped tube containing concentrated solution of an inert electrolyte like KCl, KNO₃, K₂SO₄, etc., or a solidified solution of an electrolyte such as agar-agar and gelatine. It is used,

To complete the electrical circuit by allowing ions to flow from one solution to the other without mixing the two solutions.
To maintain the electrical neutrality of the solutions in the two half cells.

Question 21.
Calculate emf of the following cell of 25°C.
Answer:
Fe | Fe²⁺(0.001M) || H⁺(0.01M) | H₂(g) (1 bar) pt
E⁰_(Fe²⁺/Fe) = – 0.44V
E⁰_(H⁺/H₂) = 0.00V
Cell reaction
Fe + 2H⁺ Fe²⁺ + H₂
E⁰_cell = E⁰_cathode – E⁰_anode
E⁰_cell = 0.44 – $\frac{0.0591}{2}$ 10 g $\frac{\left[10^{-3}\right]}{\left[10^{-2}\right]^{2}}$
= 0.44 – $\frac{0.0591}{2}$ 10g 10
= 0.44 – $\frac{0.0591}{2}$
E⁰_cell = 0.4105V

Question 22.
What is corrosion? CO₂ is always present in natural water. Explain its effect (increases, stops or no effect) on rusting of Fe.
Answer:
Corrosion is a process of slowly eating away of the metal due to the attack of atmospheric gases on thc surface of metal resulting into the formation of compounds such as oxides, suiphides, carbonates, suiphates, etc.

Factors affecting corrosion are:

Reactivity of metal
Presence of impurities
Presence of air and moisture
Strains in metals
Presence of electrolytes

CO₂ increases the rusting of iron because greater the number of W ions, faster the rusting will take place.

Samacheer Kalvi 12th Chemistry Electro Chemistry 5 Marks Question

VIII.Answer the following questions.

Question 1.
How would you measure the conductivity of ionic solutions?
Answer:
1. The conductivity of an electrolytic solution is determined by using a wheatstone bridge arrangement in which one resistance is replaced by a conductivity cell filled with the electrolytic solution of unknown conductivity.

2. In the measurement of specific resistance of a metallic wire, a DC power supply is used. Here AC is used for this measurement to prevent electrolysis. Because DC current through the conductivity cell leads to the electrolysis of the solution taken in the cell.

3. A wheatstone bridge is constituted using known resistances P,Q, a variable resistance S and conductivity cell. An AC source (550 Hz to 5 KHz) is connected between the junctions A and C. A suitable detector is connected between the junctions B and D.

4. The variable resistance S is adjusted until the bridge is balanced and in this canditions, there is no current flow through the detector.

5. Under balanced condition
P/Q = R/S
R = P/Q x S ………..(1)
The resistance of the electrolytic solution (R) is calculated from the known resistance values P, Q and the measured S value using the equation (1).

6. Specific conductance (or) conductivity of an electrolyte can be calculated from the resistance value (R) using the following expression

7. The value of cell constant is usually provided by the cell manufacturer. Alternatively the cell constant may be determined using KCI solution whose concentration and specific conductance are known.

Question 2.
Explain about SHE (Standard Hydrogen Electrode).
Answer:
1. It is impossible to measure the emf of a single electrode, but we can H₂ out H₂ measure the potential difference between two electrodes (E_cell) using a voltmeter.

2. To calculate the emf of a single electrode, we need a reference electrode whose emf is known. For that purpose, Standard Hydrogen
Electrode (SHE) is used as the reference electrode.

3. SHE has been assigned an orbitary emf of exactly zero volt.

4. It consists of platinum electrode in contact with 1M HC1 solution and 1 atm hydrogen gas.

5. The hydrogen gas bubbled through the solution at 25°C. SHE can act as cathode as well as an anode.

6. The Half cell reactions are given below:
If SHE is used as a cathode, the reduction reaction is
2H² + 2e^– + H_{2(g,1 atm)} E°= 0 volt
If SHE is used as an anode, the oxidation reaction is
H_{2(g, 1 atm)} 2H⁺_{( aq, 1M)} + 2e^– E° = 0 volt.

Question 3.
How would you determine the reduction potential of Zn/Zn²⁺(aq)? (or) How will you calculate the reduction potential of Half cell?
Answer:
1. Consider the zinc electrode dipped in zinc sulphate solution using SHE

2. Step 1:
The following galvanic cell is constructed using SHE
Zn_(s) | Zn²⁺_(aq.1M) | H_2(g.1atm) | Pt_(s)

3. Step 2:
The emf of the above galvanic cell is measured using a voltmeter. In this case the measured emf of the above galvanic cell is 0.76 V

4. Calculation:
We know that.
E°_cell = (E°_ox)_zn|zn²⁺ + (E°_red)_SHE
E°_cell = 0.76 + 0V
= 0.76 V
This oxidation potential corresponds to the below mentioned half cell reaction which takes place at the cathode
Zn → Zn²⁺ + 2e^– (oxidation)
(y) The emf of the reverse reaction will give the reduction potential
Zn²⁺ + 2e^– → Zn
(E°_red)_zn²⁺|zn = – 0.76V

Question 4.
Derive the relationship between Gibb’s free energy and maximum work obtained from galvanic cell and equilibrium constant.
Answer:
1. In a galvanic cell, chemical energy is converted into electrical energy. The electrical energy produced by the cell is equal to the product of the total charge of the electrons and emf of the cell which drives these electrons between the electrodes.

2. If ‘n’ is the number of moles of electrons exchanged between the oxidising and reducing
agent in the overall cell reaction, then the electrical energy produced by the cell is given
as below.
Electrical energy = change of ‘n’ moles of electrons x E_cell ……………..(1)
Charge of I mole of electrons = one Faraday = 1F
Charge of n moles of electrons = nF ……………..(2)

3. Electrical energy = nFE_cell …………….(3)
This energy is used to do electric work. Therefore the maximum work that can be obtained from a galvanic cell is
Here the – ve sign is introduced to indicate that work is done by the system on the surroundings.

4. Second law of thermodynamics states that the maximum work done by the system is equal to the change in Gibbs free energy of the system
W_max = ΔG …………..(5)

5. ΔG = – nFE_cell (6)
For a spontaneous cell reactions, the AG should be negative. The above expression indicates that E_cell should be positive to get a negative AG value.

6. When all the cell components are in their standard state, the equation (6) becomes
ΔG° = – nFE°_cell …………….(7)

7. The standard free energy change is related to the equilibrium constant as per the following equation.
ΔG° = – RT In K_eq …………..(8)
Comparing equations (7) and (8)
nFE°_cell = RT In K_eq …………..(9)
nFE°_cell = $\frac { 2.303RT }{ nF }$ log K_eq ………………(10)

Question 5.
Describe about the working principle of Leclanche cell.
Answer:

1. Leclanche cell:
Anode: Zinc container
Cathode: Graphite rod in contact with MnO₂
Electrolyte: Ammonium chloride and Zinc chloride in water.
emf of the cell = 1.5 V

2. Cell reaction:
Oxidation at anode
Zn_(s) → Zn²⁺_(aq) + 2e_– ……………(1)
Reduction at cathode
2NH⁺_{4 (aq)} + 2e^– → 2NH_3(aq) + H_2(aq)

3. The hydrogen gas is oxidised to water by MnO₂
H_2(g)+ 2MnO_2(s) → Mn₂O_3(s) + H₂O₍₁₎ ……(3)
Adding equations 1,2,3 the overall redox reaction
Zn_(s) + 2NH⁺_4(aq) + 2MnO_2(s) → Zn²⁺_(aq) + Mn₂O_3(s) + H₂O₍₁₎ + 2NH₃ …………….(4)

4. The ammonia produced at the cathode combines with Zn²⁺ to form a complex ion [Zn (NH₃)₄]₂₊(aq). As the reaction proceeds, the concentration of NH₄⁺ will decrease and the aqueous NH₃ will increase which lead to the decrease in the emf of the cell.

Question 6.
Explain about the construction and uses of mercury button cell.
Answer:
1. Mercury button cell:
Anode: Zinc Amalgamated with mercury
Cathode: HgO mixed with graphite
Electrolyte: Paste of KOH and ZnO.

2. Oxidation occurs at anode:

3. Reduction occurs at cathode

Hg_(s) + 2OH^–_(aq)
Overall reaction is
Zn_(s) + HgO_(s) → ZnO_(s) + Hg₍₁₎

4. Cell emf: about 1.35 V

5. Uses:
It has higher capacity and longer life. It is used in pacemakers, electronic watches, cameras.

Question 7.
Describe about lead storage battery construction and its uses.
Answer:
Lead storage battery:
1. Anode – Spongy lead
Cathode – Lead plate bearing PbO₂
Electrolyte – 38% by mass of H₂SO₄ with density 1.2 g/ml

2. Oxidation occurs at the anode
Pb_(s) → Pb²⁺_(aq) + 2e^– ………..(1)
The Pb²⁺ ions combine with SO₄^2- to form PbSO₄ precipitate
Pb²⁺_(aq) + SO₄^2- → PbSO_4(s) …………(2)

3. Reduction occurs at the cathode
PbO_2(s) + 4H⁺_(aq) + 2e^– → + 2H₂O₍₁₎ ……….(3)
The Pb²⁺ ions also combines with SO₄^2- ions to form sulphuric acid to form PbSO₄ Precipitate
Pb²⁺_(aq) + SO^2-₂₊ → PbSO ………….(4)

4. The overall reaction is,
(1) + (2) + (3) + (4)
Pb_(s) + pbO_{2_(s)} + 4H⁺_(s) + 2SO^2-_4(s) → 2PbSO_{4_(s)} + 2H₂O₍₁₎

5. The emf of a single cell is about 2V. Usually six such cells are combined in series to produce 12 volts.

6. The emf of the cell depends on the concentration of H₂SO₄. As the cell reaction uses SO₄^2- ions, the concentration H₂SO₄ decreases. When the cell potential falls to about 1.8V, the cell has to be recharged.

7. Recharge of the cell – During recharge process, the role of anode and cathode is reversed and H₂SO₄ is regenerated. Oxidation occurs at cathode (now anode)

Reduction occurs at anode (now cathode)
PbSO₄(s) + 2e^– → Pb_(s) + SO^2-_4(sq)
Overall reaction
2 pbSO_4(s) + 2H₂O₍₁₎ → pb_(s) + pbO₂(s) + 4H⁺_(aq) + 2SO^2-_4(aq)
The above reaction is exactly the reverse of redox reaction which takes place while discharging.

8. Uses: Lead storage battery is used in automobiles, trains, inverters.

Question 8.
Describe about lithium-ion battery and its uses.
Answer:
1. Lithium-ion battery
Anode – Porous graphite
Cathode – Transition metal oxide as CoO₂
Electrolyte – Lithium salt in an organic solvent

2. At the anode oxidation occurs
Li_(s) → Li⁺_(aq) + e^–
At the cathode reduction Occurs.
Li⁺ + CoO_2(s) + e^– → Li CoO_2(s)

3. Overall reactions
Li_(s) + CoO₂ → Li CoO_2(s)

4. Both electrodes allow Li ions to move in and out of their structures. During discharge the Li ions produced at the anode moves towards cathode through the non-aquaeous electrolyte.

5. When a potential greater than the emf produced by the cell is applied across the electrode, the cell reaction is reversed and now the Li⁺ ions move from cathode to anode where they become embedded on the porous electrode. This is known as intercalation.

6. Uses: This Li-ion battery is used in cellular phones, Laptop computer and digital camera.

Question 9.
What is corrosion? Explain about the electrochemical mechanism of corrosion.
Answer:
1. The metal is oxidised by oxygen in the presence of moisture. The redox process which causes the deterioration of metal is called corrosion.

2. Corrosion of iron is known as Rusting and it is an electrochemical process.

3. Electrochernical mechanism of corrosion – The formation of rust requires both oxygen and water. Since it is an electrochemical redox process, it requires both an anode and cathode in different places on the iron. The iron surface and a droplet of water on the surface form a tiny galvanic cell.

The region enclosed by water is exposed to low amount of oxygen and it act as anode. The remaining area has high amount of oxygen and it act as cathode. So based on oxygen amount, an electrochem leal cell is formed.
Fe_(s) → Fe²⁺_(aq) + 2e^–
O_{2_(g)} + 4H⁺_(aq) + 4e^– → 2H₂O₍₁₎

4. At anode:
2Fe_(s) → 2Fe²⁺_(aq) + 4e^– E° = 1.23V
The electrons move through the iron metal from the anode to the cathode area where the oxygen dissolved in water is reduced to water.

5. At cathode:
The reaction of atmospheric carbondioxide with water gives carbonic acid which furnishes the H⁺ ions for reduction.
O_{2_(s)} + 4H⁺_(g) + 4H⁺_(aq) → 2H₂O₍₁₎
E° = 1.23 V
= + 1.67V

6. The electrical circuit is completed by the migration of ions through water droplet. The overall redox reaction is
2Fe_(s) + O_{2_(g)} + 4H⁺_(aq) → 2Fe³⁺_(aq) + 2H₂O(1).
E°= 0.44+ 1.23
= + 1.67 V

7. The positive emf value indicates that the reaction is spontaneous.

8. The Fe²⁺ ions are further oxidised to Fe³⁺ which on further reaction with oxygen to form rust.

Question 10.
Explain about the various protection methods to prevent corrosion.
1. Coating metal surface by paint

2. Galvanizing – By Coating with another metal such as zinc. Zinc is stronger oxidising agent than iron and hence it can be more easily corroded than iron. i.e., instead of iron, zinc is oxidised.

3. Cathodic protection: In this technique, unlike galvanizing, the entire surface of the metal to be protected need not be covered with a protecting metal instead, metals such as Mg (or) Zn which is corroded more easily than iron can be used as sacrificial anode and the iron material act as cathode. So iron protected but Mg or Zn gets corroded.

4. Passivation – The metal is treated with strong oxidising agent such as Conc. HNO₃. As a result, a protective layer is formed on the surface of the metal.

5. Alloy formation – The oxidising tendency of iron can be reduced by forming its alloy with other more anodic metals. Example – Stainless steel, an alloy of Fe and Cr.

Question 11.
(a) Give reasons for the following

Rusting of iron is quicker in saline water than in ordinary water.
Aluminium metal cannot be produced by the electrolysis of aqueous solution of aluminium salt.

(b) Resistance of a conductivity cell filled with 0.1 M KCI solution is 1oo ohms. If the resistance of the sanie cell when filled with 0.02 M KCI solution is 520 ohms, calculate the conductivity and molar conductivity of 0.02 M KCI solution. Conductivity of 0.01 M KCI solution is 1.29 S m^-1.
Answer:
(a)
1. It is because in saline water, there are inore H₄ ions. Greater the number of H₄ ions, quicker the rusting will take place.

2. It is because aluminium metal is more reactive than hydrogen and it will react with H₂O.

(b)
Cell constant = Conductivity (k) x Resistance (R)
= 1.29 S m^-1 x 100Ω
= 1.29 m^-1 = 1.29 cm^-1
λ_m = $\frac { 100k }{ M }$
k = $\frac { 1 }{ R }$ x $\frac { l }{ A }$
k = $\frac { 1 }{ 520Ω }$ x 1.29 cm^-1, where
k = 2.48 x 10^-3S cm^-1
λ_m = $\frac{100 \times 0.248 \times 10^{-2}}{0.02}$
= 124 S cm² mol¹

Question 12.
1. State two advantages of H₂ – O₂ fuel cell over ordinary cell.
2. Silver is electron deposited on a metallic vessel of total surface area 900 cm² by passing a current of 0.5 amp for two hours. Calculate the thickness of silver metal deposited. [Given: Density of silver = 10.5 g cm^-3 Atomic mass of silver = 108 u. IF = 96500 C mol^-1].
Answer:
1.

It is highly efficient and do not produce pollution.
The H₂O so produced can be used by astronauts for drinking purpose.

2.
m = Z x l x t
m = $\frac{108}{96500}$ x 0.5 x 2 x 60 x 60
= $\frac{108 \times 5}{965 \times 10}$ x 2 x 6 x 6 = 4.03g
4.03g = V x d
4.03 g = V x 10.5 g cm^-3
V = Area x thickness
v = 4.03
= 900 cm² x thickness
Thickness = $\frac{4.03}{10.5}$
$\frac{0.338 \mathrm{cm}^{3}}{900 \mathrm{cm}^{2}}$ = 4.26 x 10^-4cm

Question 13.
Distinguish between Leclanche cell and Lead storage battery.
Answer:
Leclanche Cell

It is a primary cell
It is a non-rechargeable cell
Anode : Zinc container
Cathode : Graphite rod in contact with MnO₂
Electrolytes: Ammonium chloride and zinc chloride in water
Emf of the cell = 1.5 V

Lead storage battery

It is a secondary cell
It is rechargeable cell
Anode : Spongy lead
Cathode : Lead plate bearing PbO₂
Electrolytes: 38% by mass of H₂SO₄ with density 1.2 g/ml
emf of the cell 2 V.

Question 14.
Account for the following

Aluminium undergo slow corrosion than iron.
H₂ – O₂ fuel cell ¡s more useful than other cells.

Answer:
1. Aluminium, copper, silver also undergo corrosion but at a slower rate than iron. For eg., let us consider the reduction of Aluminium
Al_(s) → Al³⁺_(aq) + 3e^–
Al³⁺ which reacts with oxygen in air to form a protective coating of AI₂O₃. This coating act as a protective film for the inner surface. So further corrosion is prevented.

2. H₂ – O₂ fuel cell is more advanced. Because it is highly efficient. it is pollution free. In this, H₂ – O₂ fuel cell, energy of combustion of fuel is directly converted to electrical energy.

Common Errors

Units of electrical quantities may get confused.
Writing cell notation may be difficult.

Rectifications

1.Resistance – R ohm (Ω)
Potential difference – V = Volt
Amount of current – 1 = ampere
Specific resistance (ohm m) (Ω m)
Conductivity = Siemen = S
Specific conductance – k = S m^-1
Molar conductance = S m² mol^-1
Equivalent conductance = S m² gram equivalent^-1
Cell constant = m^-1

2. Easy way is (AC)
solid I aqueous solution II aqueous solution I soLid

Tamil Nadu 12th Commerce Model Question Paper 4 English Medium

Leave a Comment / Class 12 / By Prasanna

Students can Download Tamil Nadu 12th Commerce Model Question Paper 4 English Medium Pdf, Tamil Nadu 12th Commerce Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 12th Commerce Model Question Paper 4 English Medium

Instructions:

The question paper comprises of four parts.
You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
All questions of Part I, II. III and IV are to be attempted separately.
Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about 50 words.
Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in about 150 words.
Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in about 250 words. Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 90

Part -I

Choose the correct answer. Answer all the questions: [20 x 1 = 20]

Question 1.
……………. system gives full scope to the Individual strength and Responsibility.
(a) MBO
(b) MBE
(c) MBM
(d) MBA
Answer:
(a) MBO

Question 2.
Capital market do not provide ……………..
(a) Short term funds
(b) Debenture funds
(c) Equity funds
(d) Long term funds
Answer:
(a) Short term funds

Question 3.
……………….. is the oldest stock exchange in the world.
a) London stock exchange
(b) Bombay stock exchange
(c) National stock exchange
(d) Amsterdam Stock Exchange
Answer:
(b) Bombay stock exchange

Question 4.
Human resource is a ……………. asset.
(a) Tangible
(b) Intangible
(c) Fixed
(d) Current
Answer:
(b) Intangible

Question 5.
Off the Job training is given ………….
(a) In the class room
(b) On off days
(c) Outside the factory
(d) In the playground
Answer:
(c) Outside the factory

Question 6.
…………………….. is the result of New Industrial Policy which abolished the ‘License system’.
(a) Globalisation
(b) Privatisation
(c) Liberalization
(d) None of these
Answer:
(c) Liberalization

Question 7.
The initiative was launched to modernize the Indian economy to make all government services available electronically.
(a) Stand up India
(b) Start up India
(c) Digital India
(d) Make in India
Answer:
(c) Digital India

Question 8.
A person shall hold office as a director …………….. in companies as per the Companies Act, 2013.
(a) 5
(b) 10
(c) 20
(d) 15
Answer:
(c) 20

Question 9.
Management is an …………….
(a) Art
(b) Science
(c) Art and science
(d) Art or science
Answer:
(c) Art and science

Question 10.
Which of the following is not a main function?
(a) Decision – making
(b) Planning
(c) Organising
(d) Staffing
Answer:
(a) Decision – making

Question 11.
Which is the first step in process of MBO?
(a) Fixing key result area
(b) Appraisal of activities
(c) Matching resources with activities
(d) Defining organizational objectives
Answer:
(d) Defining organizational objectives

Question 12.
Capital market is a market for ……………
(a) Short term finance
(b) Medium term finance
(c) Long term finance
(d) Both short term and medium term finance
Answer:
(c) Long term finance

Question 13.
When the NSEI was established …………..
(a) 1990
(b) 1992
(c) 1998
(d) 1997
Answer:
(b) 1992

Question 14.
The headquarters of SEBI is
(a) Calcutta
(b) Bombay
(c) Chennai
(d) Delhi
Answer:
(b) Bombay

Question 15.
Who is supreme in the market?
(a) Customer
(b) Seller
(c) Wholesaler
(d) Retailer
Answer:
(a) Customer

Question 16.
Who is the father of Consumer Movement?
(a) Mahatma Gandhi
(b) Mr. John F. Kennedy
(c) Ralph Nader
(d) Jawaharlal Nehru
Answer:
(c) Ralph Nader

Question 17.
Which of the below constitutes the essential element of contract of sale?
(a) Two parties
(b) Transfer of property
(c) Price
(d) All of the above
Answer:
(d) All of the above

Question 18.
Negotiable Instrument is freely transferable by delivery if it is a …………. instrument.
(a) order
(b) bearer
(c) Both A & B
(d) None of the above
Answer:
(b) bearer

Question 19.
How does a person who envisages the idea to form a company called…………. ?
Answer:
(a) Director
(b) Company secretary
(c) Registrar
(d) Promoter
Answer:
(d) Promoter

Question 20.
Who can become a secretary for a company?
(a) Individual person
(b) Partnership firm
(c) Co-operative societies
(d) Trade unions
Answer:
(a) Individual person

Part – II

Answer any seven in which question No. 30 is compulsory. [7 x 2 = 14]

Question 21.
List out the management tools.
Answer:
Tools of management have been developed such as, accounting, business law, psychology, statistics, econometrics, data processing, etc.

Question 22.
Define – MBO
Answer:
MBO is popularised in the USA by George Odiome. According to him, “MBO is a system wherein the superior and the sub-ordinate managers of an organisation jointly identify its common goals, define each individual’s major area of responsibility in terms of the result expected of him and use these measures as guides for operating the unit and assessing the contribution of each of its members”.

Question 23.
What are the components of organized sectors?
Answer:

Regulators
Financial Institutions
Financial Markets
Financial Services

Question 24.
Define the term “Money Market”.
Answer:
According to Crowther, “The money market is the collective name given to the various firms and institutions that deal in the various grades of near money”.

Question 25.
What is meant by Stock Exchange?
Answer:
Stock Exchange is an organized market for the purchase and sale of industrial and financial security, ft is also called stock market or share market.

Question 26.
What is Market?
Answer:
The word market is derived from the Latin word ‘Marcatus’ which means trade, commerce, merchandise, a place where business is transacted.

Question 27.
Who is a consumer?
Answer:
A consumer is one who consumes goods manufactured and sold by others or created (air, water, natural resources) by nature and sold by others.

Question 28.
Define Business environment.
Answer:
According to Bayard O Wheeler, Business environment is “The total of all things external to firms and industries which affect their organisation and operations”.

Question 29.
Mention any two features of entrepreneurs.
Answer:
Features of Entrepreneurs:

Spirit of Enterprise: Entrepreneur should be bold enough to encounter risk arising from the venture undertaken.
Self Confidence: Entrepreneur should have a self confidence in order to achieve high goals in the business.

Question 30.
Who is called as Promoters?
Answer:
Promotion stage begins when the idea to form a company comes in the mind of a person. The person who envisage the idea is called a ‘promoter’.

Part – III

Answer any seven in which question No. 31 is compulsory. (7 x 3 = 21]

Question 31.
Define the term management.
Answer:
“To manage is to forecast, to plan, to organise, to command, to co-ordinate and to control.” -Henry Fayol. It attempts to describe management in terms of what a manager does, and not what management is.

Question 32.
Write the features of MBO.
Answer:

An attempt is made by the management to integrate the goals of an organisation and individuals. This will lead to effective management.
MBO tries to combine the long run goals of organisation with short run goals.
Management tries to relate the organisation goals with society goals.

Question 33.
What are the limitations of Stock exchange?
Answer:

Lack of uniformity and control of stock exchanges.
Absence of restriction on the membership of stock exchanges.
Failure to control unhealthy speculation.

Question 34.
Define the term Recruitment.
Answer:
According to Edwin B. Flippo, “It is a process of searching for prospective employees and stimulating and encouraging them to apply for jobs in an organisation.”

Question 35.
What are the objectives of marketing?
Answer:
Baker and Anshen say,“The end of all the marketing activities is the satisfaction of human wants”.The following are the objectives of marketing:

Intelligent and capable application of modem marketing policies.
To develop the marketing field.
To develop guiding policies and their implementation for a good result.

Question 36.
What is B2B and B2C type of E-Commerce?
Answer:
B2B is an online business model that facilitates online sales transactions between two businesses. E-tailing is a business to consumer (B2C) transaction model. E-tailing is also called online retailing.

Question 37.
Explain the natural environment of business.
Answer:
Geo-physical environment – The natural, geographical and ecological factors have a bearing on the business. These are as follows:

The availability of natural resources like minerals oil .etc, since setting up of industries requires availability of raw materials.
The weather and climatic conditions and availability of water and other natural resources is essential for the agricultural sector.
Topographical factors like the terrain impacts type of business since the demand and consumption pattern may vary in these regions.
Location of certain industries is influenced by the geographical conditions.
Availability of natural harbours and port facilities for transporting goods.

Question 38.
Explain the meaning of Agreement to sell.
Answer:
The property (ownership or title) in the goods has to pass at a future time or after the fulfilment of certain conditions specified in the contract.

Question 39.
What is political environment?
Answer:
To commence business, various factors are needed. Apart from all other factors, political environment is also essential. It means that the concessions, incentives provided by the government drive them to enter into venture. The government also provides support in the form of loans, subsidies and other taxes.

Question 40.
What is Special Resolution?
Answer:
A special resolution is the one which is passed by not less than 75% of majority. The number of votes, cast in favour of the resolution should be three times the number of votes cast against it.

Part – IV

Answer all the following questions. [7 x 5 = 35]

Question 41(a).
Explain the features of Stock Exchange. (Any 5)
Answer:
There are various features of a stock exchange. They are given below:

Market for Securities: Stock exchange is a market, where securities of corporate bodies, government companies are bought and sold.
Deals in Second Hand Securities: It deals with shares, debentures bonds and securities already issued by the companies.
Regulates Trade in Securities: Stock exchange does not buy or sell any securities on its own account. It regulates the trade activities so as to ensure free and fair trade.
Allows Dealings only in Listed Securities: In the stock exchange only listed securities are purchased and sold. Unlisted securities cannot be traded in the stock exchange.
Association of Persons: A stock exchange is an association of persons or- body of individuals which may be registered or unregistered.

[OR]

(A) Explain the different methods of recruitment.
Answer:
Recruitment means selecting the right person for the right job. There are basically two ways by which an organisation can recruit its employees – Internal and External sources. External sources can further be classified into Direct and Indirect sources.
Internal Sources – Transfer, Upgrading, Promotion, Demotion, Recommendation by existing Employees, Job rotation, Retired employees, Dependants, Previous applicants, Acquisitions and Mergers.

External Sources:

Direct – Advertisements, Unsolicited applicants, Walk-ins, Campus Recruitment, Recruitment at Factory gate, Rival firms, e-Recruitment.
Indirect – Employee referral, Govemment/Public Employment Exchanges, Employment Agencies, Employment Consultancies, Professional Associations, Deputation, Word of mouth, Labour Contractors, Job Portals, Outsourcing, Poaching.

Question 42(a).
Explain the management process in detail.
Answer:
The substance of management should be identified as a process. A process is something that what a person does in the context of his individual duties and responsibilities assigned by his or her immediate higher authority.
There are twin purposes of the management process:

Maximum productivity or profitability
Maximum human welfare and satisfaction.

There are five parts of management as a process:

Co-ordination of resources: The manager of an enterprise must effectively coordinate all activities and resources of the organisation, namely, men, machines, materials and money, the four M’s of management.
Management is a Process: The manager achieves proper coordination of resources by means of the managerial functions of planning, organising, staffing, directing (or leading and motivating) and controlling.
Management is a Purposive Process: It is directed toward the achievement of predetermined goals or objectives. Without an objective, we have no destination to reach or a path to follow to arrive at our destination, i.e. a goal, both management and organisation must be purposive or goal-oriented.
Management is a Social Process: It is the art of getting things done through other people.
Management is a Cyclical Process: It represents planning-action-control-replanning cycle, i.e., an ongoing process to attain the planned goals.

[OR]

(b) Distinguish between new issue market and secondary market.

Basis For Comparison	New Issue Market	Secondary Market
Meaning	The market place for new shares is called primary market.(Initial Issues Market)	The place where formerly issued securities are traded is known as Secondary Market. (Resale Market)
Buying	Direct	Indirect
Financing	It supplies funds to budding enterprises and also to existing companies for expansion and diversification.	It does not provide funding to companies.
How can securities be sold?	Only once	Multiple times
Buying and Selling between	Company and Investors	Investors
Gained person	Company	Investors
Intermediary	Underwriters	Brokers
Price	Fixed price	Fluctuates, depends on the demand and supply force.
Organizational difference	Not rooted to any specific spot or geographical location.	It has physical existence.

Question 43(a).
Discuss about the evolution of marketing.
Answer:
Marketing is one of the business functions. The development of marketing is evolutionary rather than revolutionary.
Evolution of Marketing:

Barter System: The goods are exchanged against goods, without money.
Production Orientation: This was a stage where producers, instead of being concerned with the consumer preferences, concentrated on the mass production of goods.
Sales Orientation: The selling became the dominant factor, without any efforts for the satisfaction of the consumer needs.
Marketing Orientation: Customers’ importance was realised but only as a means of disposing of goods produced.
Consumer Orientation: Under this stage only such products are brought forward to the markets which are capable of satisfying the tastes and preferences of consumers.

[OR]

(b) Explain advantages and disadvantages of E-tailing.
E-tailing or electronic retailing refers to selling of goods and services through a shopping website:
Advantages:

Customer can buy the product at anytime from anywhere.
Direct contact of end consumer by the manufacturer cuts down the cost.
Customer can buy whatever they want by browsing the various sites.

Disadvantages:

E-tailing needs a strong advertisement and for which it has to spend large amount.
It is not suitable for small size business.

Question 44(a).
What are the rights of consumers?
Answer:
As a consumer, everyone should know the basic rights as well as about the counts and procedures to be followed.
The rights of consumers as per Consumer Protection Act are given below:
(1) Right to Protection of Health and Right of Safety: There may be products that cause physical danger to consumers’ health, lives and property. The health hazards which are likely to arise have to be eradicated or reduced altogether.

(ii) Right to be Informed: Consumers should be given all the relevant facts about the products. The manufacturer and the dealer should disclose all the material facts relating to the product.

(iii) Right to choose: Consumer satisfaction can be increased by giving the consumer the widest choice. From the widest range of products in quality and brand as well as price, the consumer can choose the goods.

(iv) Right to be Heard: Consumers have every right to ventilate and register the dissatisfaction, disagreements and get the complaint heard and aired.

(v) Right to Seek Redressal: The aggrieved party is to be granted compensation within a reasonable time.

(vi) Right to Consumer Education: The consumer has a right to acquire knowledge and stay well-informed all through his life.

[OR]

(b) Explain the advantages and disadvantages of liberalization.
Liberalization means relaxation of various government restrictions in the areas of social and economic policies of the country.
Advantages of Liberalization:

Increase in foreign investment: If a country liberalizes its trade, it will make the country more attractive for inward investment.
Increase the foreign exchange reserve: Relaxation in the regulations covering foreign investment and foreign exchange has paved way for easy access to foreign capital.
Increase in consumption: Liberalization increases the number of goods available for consumption within a country

Disadvantages of Liberalization:

Increase in unemployment: Due to liberalization some industries grow, some decline. Therefore, there may be unemployment from certain industries closing.
Increased dependence on foreign nations: Trade liberalization means firms will face greater competition from abroad.
Unbalanced development: Trade liberalization may be damaging for developing economies which cannot compete against free trade.

Question 45(a).
Mention the presumptions of Negotiable Instruments.
Answer:
Presumptions of Negotiable Instrument:

Every negotiable instrument is presumed to have been drawn and accepted for consideration.
Every negotiable instrument bearing, a date is presumed to have been made or drawn on such a date.
It is presumed to have been accepted within a reasonable time after the date and before its maturity.
The transfer of a negotiable instrument is presumed to have been made before maturity,
When a negotiable instrument has been lost, it is presumed to have been duly stamped
The holder of a negotiable instrument is presumed to be a holder in due course.

[OR]
(b) Explain in detail on classification according to the type of business.
Classification of Entrepreneur according to the type of business:

Business Entrepreneur: He is called solo entrepreneur. He is the one who finds out an idea for a new product or service and establish a business enterprise.
Trading Entrepreneur: Trading entrepreneurs are those who restrict themselves to buying and selling finished goods.
Industrial Entrepreneur: These are entrepreneurs who manufacture products to cater to the needs of the consumers.
Corporate Entrepreneur: He is called as promoter. He takes initiative necessary to start an entity under corporate format.
Agricultural Entrepreneur: These entrepreneurs are those who raise farm products and market them.

Question 46(a).
Explain any five Government Entrepreneurial schemes.
Answer:
To support and strengthen the Start-up culture in India, the Government has launched various schemes. They are as follows:
(i) The Modified Special Incentive Package Scheme (M-SIPS): The M-SIPS scheme provides capital subsidy of 20% in SEZ and 25% subsidy in non-SEZ for business units engaged in manufacturing of electronic goods.

(ii) New Gen Innovation and Entrepreneurship Development Centre (New Gen IEDC):
It provides a limited one time non-recurring financial assistance to entrepreneurship up to Rs.25 lakhs in the fields of chemicals, technology, health care, defence, etc.

(iii) Dairy Entrepreneurship Development Scheme: Dairy Entrepreneurship Development Scheme aims at helping entrepreneurs in the fields of agriculture, pets and animals and dairy farms.

(iv) Single Point Registration Scheme: A great scheme for micro and small enterprises which provides an exemption from payment of Earnest Money Deposit.

(v) Atal Incubation Centres (AIC): The Government of India has set up the Atal Innovation Mission (AIM) at NITI Aayog in 2016 with the overarching purpose of promoting a culture of innovation and entrepreneurship in the country.

[OR]
(b) Who are the KMP?
Answer:
Companies Act, 2013 has introduced many new concepts and Key Managerial Personnel (KMP) is one of them. KMP covers the traditional roles of managing director and whole time director and also includes some functional heads.

Key Managerial Personnel: The definition of the term Key Managerial Personnel is contained in Section 2(51) of the Companies Act, 2013. This Section states:

The Chief Executive Officer
The Managing Director or Manager
The Company Secretary
The Whole-time Director
The Chief Financial Officer
Such other officer as may be prescribed

Question 47(a).
Discuss the liabilities of Company Secretary.
Answer:
As an officer, a company secretary has extensive duties and liabilities. The success of the company depends upon his efficient functions and capacity to perform.
Liabilities:

It is duty of the secretary to arrange for Board meetings and shareholders annual general meeting.
The secretary controls and supervises the day-to-day activities of the company.
Also he should prepare details for issue of allotment letters, share certificates and dividend warrants.
To arrange for filing statement in lieu of prospectus.
The secretary should send notice of general meeting to every member of the company.
Being a principal officer, a company secretary can sign contracts and proceedings of company meetings.
He is liable to maintain share registers and register of directors and contracts.
To prepare minutes of every general meeting and Board meetings within 30 days.

[OR]
(b) Explain the role of business in consumer protection.
Business enterprises should do the following towards protecting consumers.

Avoidance of Price Hike: Business enterprises should desist from hiking the price in the context of acute shortage of goods.
Avoidance of Hoarding: Business enterprises should not indulge in hoarding and black marketing to earn maximum profit.
Guarantees for Good Quality: Business enterprises should not give false warranty for the products.
Product Information: Business enterprises should disclose correct, complete and accurate information about the product viz. size, quality, quantity, weight etc.
Truth in advertising: Business enterprises should not convey false, untrue, bogus information relating to the product through the advertisement.
Money Refund Guarantee: Where the product becomes defective, business enterprises should replace it with new one or refund the purchase price.

Samacheer Kalvi 12th Bio Botany Solutions Chapter 9 Plant Breeding

Tamilnadu Samacheer Kalvi 12th Bio Botany Solutions Chapter 9 Plant Breeding

Samacheer Kalvi 12th Bio Botany Plant Breeding Text Book Back Questions and Answers

Question 1.
Assertion: Genetic variation provides the raw material for selection.
Reason: Genetic variations are differences in genotypes of the individuals.
(a) Assertion is right and reason is wrong.
(b) Assertion is wrong and reason is right.
(c) Both reason and assertion is right.
(d) Both reason and assertion is wrong.
Answer:
(c) Both reason and assertion is right.

Question 2.
While studying the history of domestication of various cultivated plants recognized earlier.
(a) Centres of origin
(b) Centres of domestication
(c) Centres of hybrid
(d) Centres of variation
Answer:
(a) Centres of origin

Question 3.
Pick out the odd pair.
(a) Mass selection – Morphological characters
(b) Purline selection – Repeated self pollination
(c) Clonal selection – Sexually propagated
(d) Natural selection – Involves nature
Answer:
(a) Mass selection – Morphological characters

Question 4.

Answer:
(b) i – III, ii – I, iii – IV, iv – II

Question 5.
The quickest method of plant breeding is __________
(a) Introduction
(b) Selection
(c) Hybridization
(d) Mutation breeding
Answer:
(d) Mutation breeding

Question 6.
Desired improved variety of economically useful crops are raised by __________
(a) natural selection
(b) hybridization
(c) mutation
(d) biofertilisers
Answer:
(b) hybridization

Question 7.
Plants having similar genotypes produced by plant breeding are called __________
(a) clone
(b) haploid
(c) autopolyploid
(d) genome
Answer:
(a) clone

Question 8.
Importing better varieties and plants from outside and acclimatising them to local environment is called __________
(a) cloning
(b) heterosis
(c) selection
(d) introduction
Answer:
(d) introduction

Question 9.
Dwarfing gene of wheat is __________
(a) pall
(b) Atomita 1
(c) Norin 10
(d) pelita 2
Answer:
(c) Norin 10

Question 10.
Crosses between the plants of the same variety are called __________
(a) interspecific
(b) intervarietal
(c) intravarietal
(d) intergeneric
Answer:
(c) intravarietal

Question 11.
Progeny obtained as a result of repeat self pollination a cross pollinated crop to called __________
(a) pure line
(b) pedigree line
(c) inbreed line
(d) heterosis
Answer:
(a) pure line

Question 12.
Jaya and Ratna are the semi dwarf varieties of __________
(a) wheat
(b) rice
(c) cowpea
(d) mustard
Answer:
(b) rice

Question 13.
Which one of the following are the species that are crossed to give sugarcane varieties with high sugar, high yield, thick stems and ability to grow in the sugarcane belt of North India?
(a) Saccharum robustum and Saccharum officinarum
(b) Saccharum barberi and Saccharum officinarum
(c) Saccharum sinense and Saccharum officinarum
(d) Saccharum barberi and Saccharum robustum
Answer:
(b) Saccharum barberi and Saccharum officinarum

Question 14.
Match column I (crop) with column II (Corresponding disease resistant variety) and select the correct option from the given codes.

Answer:
(b) ii, i, iii, iv

Question 15.
A wheat variety, Atlas 66 which has been used as a donor for improving cultivated wheat, which is rich in __________
(a) iron
(b) carbohydrates
(c) proteins
(d) vitamins
Answer:
(c) proteins

Question 16.
Which one of the following crop varieties correct matches with its resistance to a disease?

Answer:
(a) Pusa Komal – Bacterial blight

Question 17.
Which of the following is incorrectly paired?
(a) Wheat – Himgiri
(b) Milch breed – Sahiwal
(c) Rice – Ratna
(d) Pusa Komal – Brassica
Answer:
(d) Pusa Komal – Brassica

Question 18.
Match list I with list II:

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

Question 19.
Differentiate primary introduction from secondary introduction
Answer:

Primary Introduction: Primary introduction – When the introduced variety is well adapted to the new environment without any alternation to the original genotype.
Secondary Introduction: Secondary introduction – When the introduced variety is subjected to selection to isolate a superior variety and hybridized with a local variety to transfer one or a few characters to them.

Question 20.
How are microbial innocnlants used to increase the soil fertility?
Answer:

Rhizobium biofertilizer culture will fix the atmospheric nitrogen by which the paddy field yield will increase by 15-40 %
Azolla will get quickly decomposed in soil and increase the yield of rice crop.
Arbuscular mycorrhizae have the abitity to dissolve phosphates found in the soil.
Sea weed Liquid fertilizer will provide carbohydrate for plants. Thus the microbial innoculants used to increase the soil fertitty.

Advantages :

They are efficient in fixing nitrogen solubilising phosphate and decomposing cellulose.
They are designed to improve the soil fertility plant growth and also the member and biological activity of beneficial microoganisms in the soil.
They are eco-friendly organic agro inputs and are more efficient and cost effective than chemical fertilizers.

Question 21.
What are the different types of hybridization?
Answer:
Types of Hybridization
According to the relationship between plants, the hybridization is divided into.

Intravarietal hybridization – The cross between the plants of same variety. Such crosses are useful only in the self-pollinated crops.
Intervarietal hybridization – The cross between the plants belonging to two different varieties of the same species and is also known as intraspecific hybridization. This technique has been the basis of improving self-pollinated as well as cross pollinated crops.
Interspecific hybridization – The cross between the plants belonging to different species belonging to the same genus is also called intragenic hybridization. It is commonly used for transferring the genes of disease, insect, pest and drought resistance from one species to another.

Question 22.
Explain the best-suited type followed by plant breeders at present?
Answer:

It develops new plant varieties by the process of selection and seeks to achieve the expression of genetic material which is already present within the species.
Conventional plant breeding methods resulting in hybrid varieties had a tremendous impact on agriculture productivity over the last decades.

Methods of plant breeding :

plant introductions method
Hybridization
Heterosis
mutation
ploidy
Tissue culture
Green revolution
Biotechnological Technique

Question 23.
Write a note on heterosis.
Answer:
The superiority of the F₁ hybrid in performance over its parents is called heterosis or hybrid vigour. Vigour refers to increase in growth, yield, greater adaptability of resistance to diseases, pest and drought. Vegetative propagation is the best suited measure for maintaining hybrid vigour, since the desired characters are not lost and can persist over a period of time. Many breeders believe that its magnitude of heterosis is directly related to the degree of genetic diversity between the two parents. Depending on the nature, origin, adaptability and reproducing ability heterosis can be classified as:

Euheterosis- This is the true heterosis which is inherited and is further classified as
Mutational Euheterosis – Simplest type of euheterosis and results from the sheltering or eliminating of the deleterious, unfavourable often lethal, recessive, mutant genes by their adaptively superior dominant alleles in cross pollinated crops.
Balanced Euheterosis – Well balanced gene combinations which is more adaptive to environmental conditions and agricultural usefulness.
Pseudoheterosis – Also termed as luxuriance. Progeny possess superiority over parents in vegetative growth but not in yield and adaptation, usually sterile or poorly fertile.

Question 24.
List out the new breeding techniques involved in developing new traits in plant breeding.
Answer:
In the milestones of plant breeding methods Genetic Engineering, Plant tissue culture, protoplasmic fusion or somatic hybridization molecular marking, and DNA fingerprinting are some of the modern plant breeding tools used to improve the crop varieties.

New plant engineering Techniques / New Breeding Techniques (NBT): NBT is a collection of methods that could increase and accelerate the development of new traits in plant breeding.

The various methods of achieving these changes in traits include the following.

Cutting and modifying the genome during the repair process by tools like CRISPR / Cas
Genome editing to introduce changes in few base pairs using a technique called oligonucleotide, directed mutagenesis (ODM)
Transferring a gene from an identical or closely related species (cisgenesis)
Organising processes that alter gene activity without altering the DNA itself (epigenetic methods)

Samacheer Kalvi 12th Bio Botany Plant Breeding Additional Questions and Answers

1 – Mark Questions

Question 1.
____________ is the process of bringing a plant species under human control.
(a) Emasculation
(b) Hybridization
(c) Domestication
(d) Acclimatization
Answer:
(c) Domestication

Question 2.
Which of the following scientist developed the world’s first cotton hybrid?
(a) Dr. B.P. Pal
(b) C.T. Patel
(c) Dr. K. Ramiah
(d) N.G.P. Rao
Answer:
(b) C.T. Patel

Question 3.
Identify the incorrect statement:
(a) Bio-inoculants are efficient in solubilizing phosphate
(A) Bio-inoculants are eco-friendly organic agro outputs
(c) Bio-inoculants are obtained from dead organic matters
(d) Bio-inoculants are designed to improve soil fertility
Answer:
(c) Bio-inoculants are obtained from dead organic matters

Question 4.
Which is not a free-living nitrogen-fixing species?
(a) Azotobacter
(b) Clostridium
(c) Nostop
(d) Anabaena
Answer:
(d) Anabaena

Question 5.
Arbuscular mycorrhizae is a symbiotic association between ________
(a) Algae and fungi
(b) Angiosperm roots and fungi
(c) Blue-green algae and Azolla fern
(d) Cyanobacteria and corolloid root
Answer:
(b) Angiosperm roots and fungi

Question 6.
Azolla is best suited biofertilizer for ____________
(a) Sugar cane cultivation
(b) Paddy cultivation
(c) Wheat cultivation
(d) Cotton cultivation
Answer:
(A) Paddy cultivation

Question 7.
Assertion (A): SLF promotes vigorous growth and provide resistance against diseases.
Reason (R): SLF is made from kelp containing more than 70 minerals.
(a) Both A and R are true. R explains A.
(A) A is true R is false
(c) A is false R is true
(d) Both A and R are false
Answer:
(a) Both A and R are true. R explains A.

Question 8.
Assertion (A): Pure line varieties show homozygosity.
Reason (R): Pure line species are obtained through cross-pollination.
(a) Both A and R are true. R explains A.
(b) A is true R is false
(c) A is false R is true
(d) Both A and R are false
Answer:
(b) A is true R is false

Question 9.
Assertion (A): Hybrids show increased growth and elevated yield.
Reason (R): F₁ hybrids show Heterosis.
(a) Both A and R are true. R explains A.
(b) A is true R is false
(c) A is false R is true
(d) Both A and R are false
Answer:
(a) Both A and R are true. R explains A.

Question 10.
Statement (1): Trichoderma species is a free-living bacteria.
Statement (2): It acts as a potent biocontrol agent
(a) Statement 1 is correct and Statement 2 is incorrect
(b) Statement 1 is incorrect and Statement 2 is correct
(c) Both the statements are correct
(d) Both the statements are incorrect
Answer:
(b) Statement 1 is incorrect and Statement 2 is correct

Question 11.
Statement (1): Clonal selection is carried out in asexually propagating plants.
Statement (2): Clones show similar genotypes.
(a) Statement 1 is correct and Statement 2 is incorrect
(b) Statement 1 is incorrect and Statement 2 is correct
(c) Both the statements are correct
(d) Both the statements are incorrect
Answer:
(c) Both statements are correct

Question 12.
Removal of anthers from a flower to overcome self-pollination and the phenomenon is.
Answer:
Emasculation

Question 13.
Identify the proper sequence of the hybridization techniques.
(a) Emasculation → Selection → Bagging → Crossing → Harvesting
(b) Harvesting → Selection → Crossing → Emasculation → Bagging
(c) Selection → Harvesting → Crossing → Emasculation → Bagging
(d) Selection → Emasculation → Bagging → Crossing → Harvesting
Answer:
(d) Selection → Emasculation → Bagging → Crossing → Harvesting

Question 14.
Intraspecific hybridization is also termed as
(a) Intravarietal hybridization
(b) Intervarietal hybridization
(c) Interspecific hybridization
(d) Intergeneric hybridization
Answer:
(b) Intervarietal hybridization

Question 15.
The period of opening of a flower is ________
Answer:
Anthesis

Question 16.
The superiority of hybrids over parents only in vegetative growth not in yield. This phenomenon is termed as ________
(a) Euheterosis
(b) Balanced euheterosis
(c) Luxuriance
(d) Mutational heterosis
Answer:
(c) Luxuriance

Question 17.
The term green revolution was coined by ________
(a) William S Gaud
(b) M.S. Swaminathan
(c) Dr. B.P. Pal
(d) Dr. N.E. Borlaug
Answer:
(a) William S Gaud

Question 18.
Which is the second gamma garden in India?
Answer:
Indian Agricultural Research Institute (IARI)

Question 19.
Match the following:

Answer:
(a) A – ii, B – i, C – iv, D – iii

Question 20.
Who is popularly called the “father of the green revolution in India”?
(a) Nel Jeyaraman
(b) Dr. M.S. Swaminathan
(c) Dr. Nammalvar
(d) N.G.P. Rao
Answer:
(a) Dr. M. S. Swaminathan

Question 21.
Pusa swamim variety of Brassica species show resistance to ________
(a) White rust
(b) Leaf curl
(c) Black rot
(d) Hill bunt
Answer:
(a) White rust

Question 22.
The first established Atomic Garden in India was ________
(a) Bhabha Atomic Research Institute
(b) Indira Gandhi Centre for Atomic Research
(c) Indian Agricultural Research Institute
(d) Bose Research Institute
Answer:
(d) Bose Research Institute

Question 23.
Triticale is polyploid breed of ________
(a) Triticum cereale x Secale sativus
(b) Triticum durum x Secale cereale
(c) Triticum cereale x Secale sativus
(d) Triticum sativus x Secale cereale
Answer:
(b) Triticum durum x Secale cereale

Question 24.
Raphanobrassica is an example for ________
(a) Autopolyploid
(b) Allopolyploid
(c) Polyploid
(d) Polysomy
Answer:
(b) Allopolyploid

Question 25.
Atlas 66 is a improved variety of ________
(a) Rice
(b) Maize
(c) Wheat
(d) Spinach
Answer:
(c) Wheat

Question 26.
Pusa Sawani variety of okra is resistant against ________
(a) Aphids
(b) Fruit borers
(c) Shoot and fruit borers
(d) Jassids
Answer:
(c) Shoot and fruit borers

Question 27.
________ was the first person to develop the world’s first hybrid of sorghum.
Answer:
N.G.P. Rao

Question 28.
Identify the incorrect statement:
(i) SLF is obtained from Kelps – a brown algae
(ii) Azolla is a fern
(iii) Rhizobium is found in association with root nodules
(iv) AM forms symbiotic relation with angiosperm roots
(a) i only
(b) ii, iii, and iv only
(c) none of the above
(d) all the above
Answer:
(c) none of the above

Question 29.
Damping-off of tomato is controlled by ________
(a) Beauveria species
(b) Trichoderma species
(c) Acacia species
(d) Pseudomonas species
Answer:
(a) Beauveria species

Question 30.
Match the following:

Answer:
(a) A – ii, B – iv, C – i, D – iii

Question 31.
Identify the correct symbiotic association
(a) Rhizobium × Corolloid roots
(b) Arbuscular Mycorrhizae × Angiospermic roots
(c) Azolla × Amantia
(d) Azospirillum × Azolla
(b) Arbuscular Mycorrhizae × Angiospermic roots
Answer:
(b) Arbuscular Mycorrhizae × Angiospermic roots

Question 32.
Atomita 2 – rice is a product by.
(a) Polyploid breeding
(b) Hybridization
(c) Mutation breeding
(d) Clonal selection
Answer:
(c) Mutation breeding

Question 33.
Luxuriance is the term used on par with ________
(a) Heterosis
(b) Anthesis
(c) Hybrids
(d) Mutant breeds
Answer:
(a) Heterosis

Question 34.
The term pure line was coined by.
Answer:
Johannsen

2 – Mark Questions

Question 1.
What is meant by the domestication of plants?
Answer:
Domestication is the process of bringing a plant species under the control of humans and gradually changing it through careful selection, genetic alteration, and handling so that it is more useful to people.

Question 2.
Mention any two free-living nitrogen fixing bacteria.
Answer:
Azotobacter and Clostridium.

Question 3.
What are bio-pesticides? Why they are considered better than synthetic pesticides?
Answer:
Bio-pesticides are biologically based agents used for the control of plant pests. They are in high use due to their non-toxic, cheaper and eco-friendly characteristics as compared to chemical or synthetic pesticides.

Question 4.
Name any four plants used in Green leaf manuring.
Answer:

Cassia fistula
Sesbania grandiflora
Azadirachta indica
Pongamia pinnata

Question 5.
What is plant breeding?
Answer:
Plant breeding is the science of improvement of crop varieties with a higher yield, better quality, resistance to diseases, and shorter durations which are suitable to a particular environment.

Question 6.
Define acclimatization.
Answer:
The newly introduced plant has to adapt itself to the new environment. This adjustment or adaptation of the introduced plant in the changing environment is called acclimatization.

Question 7.
Who coined the term “pure line”? Define it.
Answer:
Johannsen in 1903 coined the word pure line. It is a collection of plants obtained as a result of repeated self-pollination from a single homozygous individual.

Question 8.
Define the following terms:
Answer:
(a) Emasculation
(b) Bagging
Emasculation: It is a process of removal of anthers to prevent self-pollination before anthesis (period of opening of a flower).
Bagging: The stigma of the flower is protected against any undesirable pollen grains, by covering it with a bag.

Question 9.
What is Heterosis?
Answer:
The superiority of the F₁ hybrid in performance over its parents is called heterosis or hybrid vigour. Vigour refers to an increase in growth, yield, greater adaptability of resistance to diseases, pest and drought.

Question 10.
What does the term ‘luxuriance’ stand for in plant breeding? Explain.
Answer:
Pseudoheterosis – Also termed as luxuriance. Progeny possesses superiority over parents in vegetative growth but not in yield and adaptation, usually sterile or poorly fertile.

Question 11.
Mutagens are substances that induce mutation. Name any two physical and chemical mutagens.
Answer:

UV short waves, X-rays – Physical mutagens.
Nitromethyl, Urea – Chemical mutagens.

Question 12.
Write in brief about Atomic Garden.
Answer:
Atomic Garden: This is a form of mutation breeding where plants are exposed to radioactive sources typically cobalt-60 or cesium-137 in order to generate desirable mutation in crop plants.

Question 13.
State any one advantage and one disadvantage of polyploid breeding.
Answer:

Advantage: Polyploidy often exhibits hybrid vigour and increased tolerance to biotic and abiotic stresses.
Disadvantage: Polyploidy results in reduced fertility due to meiotic error resulting in seedless varieties.

Question 14.
What are polyploids? Mention its nature.
Answer:
The majority of flowering plants are diploid (2n). The plants which possess more than two sets of a chromosome are called polyploids. Polyploidy often exhibits increased hybrid vigour, increased heterozygosity, increase tolerance to both biotic and abiotic stresses and buffering of deleterious mutations.

Question 15.
Name any two allopolyploid plant species.
Answer:

Triticale (Triticum durum x Secale cereale).
Raphanobrassica (Brassica oleraceae x Raphanus sativus).

Question 16.
What is Biofortification?
Answer:
Biofortification is breeding crops with higher levels of vitamins and minerals or higher protein and healthier fats. It is the most practical means to improve public health.

Question 17.
Name the insect-resistant varieties developed in the following crops.

Okra
Rapeseed mustard.

Answer:

Okra: Pusa Sawani and Pusa A-4 varieties of Okra are resistant to shoot and fruit borers.
Rapeseed mustard: Pusa Gaurav variety of rapeseed mustard shown

Question 18.
In which plant, and by whom the first natural hybridization was performed?
Answer:
The first natural hybridization was done by Cotton Mather in maize.

3 – Mark Questions

Question 19.
In 1926, Vavilov initially proposed eight main geographic centres of crop origin. Mention any six of them.
Answer:
China, India, South America, Mediterranean, Mesoamerica and South East Asia.

Question 20.
Name any three eminent plant breeders of Indian origin.
Answer:

Dr. M.S. Swaminathan
C.T. Patel
Dr. B.P. Pal

Question 21.
How does Rhizobium act as an efficient bio-fertilizer?
Answer:
Bio-fertilizers containing rhizome bacteria are called rhizome bio-fertilizer culture. Symbiotic bacteria that reside inside the root nodules convert the atmospheric nitrogen into a bio-available form to the plants. This nitrogen-fixing bacterium when applied to the soil undergoes multiplication in billions and fixes the atmospheric nitrogen in the soil. Rhizobium is best suited for the paddy fields which increase the yield by 15 – 40%.

Question 22.
Azolla increases the yield of paddy crop – support your answer.
Answer:
Azolla is a free-floating water fern that fixes the atmospheric nitrogen in association with nitrogen-fixing blue-green alga Anabaena Azolla. It is used as a bio-fertilizer for wetland rice cultivation and is known to contribute 40 – 60 kg/ha/crop. The agronomic potential of Azolla is quite significant particularly for increasing the yield of rice crop, as it quickly decomposes in the soil.

Question 23.
Mention any three advantages of Arbuscular mycorrhizal association.
Answer:

Arbuscular mycorrhizae have the ability to dissolve the phosphate found in soil.
Provides resistance against diseases, germs, and unfavourable weather conditions.
It assures water availability.

Question 24.
What makes the Trichoderma and effective bio-pesticide?
Answer:
Trichoderma species are free-living fungi that are common in soil and root ecosystems. They have been recognized as bio-control agent for

the control of plant disease
ability to enhance root growth development
crop productivity
resistance to abiotic stress and
uptake and use of nutrients.

Question 25.
Write a note on Green manuring.
Answer:
Green manuring is defined as the growth of green manure crops and the use of these crops directly in the field by ploughing. One of the main objectives of green manuring is to increase the content of nitrogen in the soil. Also, it helps in improving the structure and physical properties of the soil. The most important green manure crops are Crotalaria juncea, Tephrosia purpurea and Indigofera tinctoria.

Question 26.
Point out the objectives of plant breeding.
Answer:

To increase yield, vigour and fertility of the crop.
To increase tolerance to environmental conditions, salinity, temperature, and drought.
To prevent the premature falling of buds and fruits, etc.
To improve synchronous maturity.
To develop resistance to pathogens and pests.
To develop photosensitive and thermos-sensitive varieties

Question 27.
Give an account on clonal selection.
Answer:
In an asexually propagated crop, progenies derived from a plant resemble in genetic constitution with the parent plant as they are mitotically divided. Based on their phenotypic appearance, the clonal selection is employed to select an improved variety from a mixed population (clones). The selected plants are multiplied through vegetative propagation to give rise to a clone. The genotype of a clone remains unchanged for a long period of time.

Question 28.
Write a short note on autopolyploidy with an example.
Answer:
When chromosome number is doubled by itself in the same plant, is called autopolyploidy.
Example: A triploid condition in sugarbeets, apples, and pear has resulted in an increase in vigour and fruit size, large root size, large leaves, flowers, more seeds, and sugar content in them. It also resulted in seedless tomatoes, apples, watermelon, and orange.

5-Mark Questions

Question 29.
Give a comparative account on Seaweed liquid fertilizer.
Answer:
Seaweed liquid fertilizer (SLF) contains cytokinin, gibberellins, and auxin apart from macro and micronutrients. Most seaweed-based fertilizers are made from kelp (brown algae) which grows to length of 150 meters. Liquid seaweed fertilizer is not only organic but also eco-friendly. The alginates in the seaweed reacts with metals in the soil and form long, cross-linked polymers in the soil. These polymers improve the crumbling in the soil, swell up when they get wet and retain moisture for a long time.

They are especially useful in organic gardening which provides carbohydrates for plants. Seaweed has more than 70 minerals, vitamins and enzymes. It promotes vigorous growth. Improves resistance of plants to frost and disease. Seeds soaked in seaweed extract germinate much rapidly and develop a better root system.

Question 30.
Explain the steps involved in hybridization.
Answer:
The steps involved in hybridization are as follows:

Selection of Parents: Male and female plants of the desired characters are selected. It should be tested for their homozygosity.
Emasculation: It is a process of removal of anthers to prevent self-pollination before anthesis (period of opening of a flower).
Bagging: The stigma of the flower is protected against any undesirable pollen grains, by covering it with a bag.
Crossing: Transfer of pollen grains from the selected male flower to the stigma of the female emasculated flower.
Harvesting seeds and raising plants: The pollination leads to fertilization and finally seed formation takes place.
The seeds are grown into a new generation which is called hybrid.

Higher Order Thinking Skills (HOTs) Questions

Question 1.
Given below are the examples of symbiotic association in which one partner was mentioned. Write the name of the mutual co-partners.
(a) Azollafem + ______________
(b) Root nodules of legume plants + ______________
(c) phycomycetous fungi + ______________
(a) Anabaena azolla (blue green alga)
(b) Rhizobium
Answer:
(c) Angiosperm roots

Question 2.
Provide an example for each of the following agricultural components.

Biopesticide
Green manure crop
Bio-Fertiliser

Answer:

Bio pesticide – E.g. Trichoderma species
Green manure – E.g. Crotalariajuncea
Biofertilizer – E.g. Azolla

Question 3.
State the objective of using green manuring.
Answer:
Green manuring helps to increase the content of nitrogen soil and also improves the structure and physical property of soil.

Question 4.
Why plant Breeding is carried out by farmers and scientists?
Answer:
Plant breeding is a proposal manipulation of plant species in order to develop improved crop varieties with a higher yield, better quality, resistance to disease and quick maturation.

Question 5.
Yesterday, Ramu visited his friend’s Orchard, wherein he noticed few flowers of certain guava trees are covered using thin paper bags.

Name the process carried out there.
Why it was done so?

Answer:

The process is referred to as bagging in plant hybridization.
It is done after emasculation to protect the contact of the stigma of the flower against any other pollen grain.

Question 6.
Who am I?

Father of the Green Revolution.
Father of Indian Green Revolution.

Answer:

Dr. Norman E. Borlaug.
Dr. M.S. Swaminathan.

Question 7.
A plant breeder developed hybrid sugarcane by grafting two different varieties with desirable characters. The resultant hybrid showed excellent growth and productivity with increased sucrose content compared to its parental forms.

What does this phenomenon refer to?
How this condition can be maintained through further generation?

Answer:

Heterosis or Hybrid vigour
Vegetative propagation is the best-suited measure to maintain the vigourisity of hybrid.

Samacheer Kalvi 10th Tamil Model Question Paper 5

Tamil Nadu Samacheer Kalvi 10th Tamil Model Question Paper 5

நேரம் : 3.00 மணி
மதிப்பெண்கள் : 100

(குறிப்புகள்:

இவ்வினாத்தாள் ஐந்து பகுதிகளைக் கொண்டது. அனைத்து பகுதிகளுக்கும் விடையளிக்க – வேண்டும். தேவையான இடங்களில் உள் தேர்வு வினாக்கள் கொடுக்கப்பட்டுள்ளது. காக
பகுதி I, II, III, IV மற்றும் Vல் உள்ள அனைத்து வினாக்களுக்குத் தனித்தனியே விடையளிக்க வேண்டும்.
வினா எண். 1 முதல் 15 வரை பகுதி-1ல் தேர்வு செய்யும் வினாக்கள் தரப்பட்டுள்ளன. ஒவ்வொரு வினாவிற்கும் ஒரு மதிப்பெண். சரியான விடையைத் தேர்ந்தெடுத்து குறியீட்டுடன் எழுதவும்.
வினா எண் 16 முதல் 28 வரை பகுதி-IIல் இரண்டு மதிப்பெண் வினாக்கள் தரப்பட்டுள்ளன: ஏதேனும் 9 வினாக்களுக்கு மட்டும் விடையளிக்கவும்.
வினா எண் 29 முதல் 37 வரை பகுதி-IIIல் மூன்று மதிப்பெண் வினாக்கள் தரப்பட்டுள்ளன. –
ஏதேனும் 6 வினாக்களுக்கு மட்டும் விடையளிக்கவும்.
வினா எண் 38 முதல் 42 வரை பகுதி-IVல் ஐந்து மதிப்பெண் வினாக்கள் தரப்பட்டுள்ளன. ஏதேனும் 5 வினாக்களுக்கு மட்டும் விடையளிக்கவும்.
வினா எண் 43 முதல் 45 வரை பகுதி-Vல் எட்டு மதிப்பெண் வினாக்கள் தரப்பட்டுள்ளன. அனைத்து வினாவிற்கும் விடையளிக்கவும்.

பகுதி – 1 (மதிப்பெண்கள்: 15)

(i) அனைத்து வினாக்களுக்கும் விடையளிக்கவும்.
(ii) கொடுக்கப்பட்ட நான்கு விடைகளில் சரியான விடையினைத் தேர்ந்தெடுத்துக் குறியீட்டுடன் விடையினையும் சேர்த்து எழுதுக. [15 x 1 = 15]

(குறிப்பு: விடைகள் தடித்த எழுத்தில் உள்ளன)

Question 1.
பரிபாடல் அடியில் ‘விசும்பும் இசையும்’ என்னும் தொடர் எதனைக் குறிக்கிறது?
(அ) வானத்தையும் பாட்டையும்
(ஆ) வானத்தையும் புகழையும்
(இ) வானத்தையும் பூமியையும்
(ஈ) வானத்தையும் பேரொலியையும்
Answer:
(ஈ) வானத்தையும் பேரொலியையும்

Question 2.
” மருந்தே ஆயினும் விருந்தொடு உண்’ என்னும் தொடர் இடம் பெற்ற நூல்……
(அ) கொன்றை வேந்தன்
(ஆ) குறுந்தொகை
(இ) திருக்குறள்
(ஈ) நற்றிணை
Answer:
(இ) திருக்குறள்

Question 3.
“இங்கு நகரப் பேருந்து நிற்குமா” என்று வழிப்போக்கர் கேட்டது………….. வினா ”அதோ அங்கே நிற்கும்” என்று மற்றொருவர் கூறியது ……….. விடை
(அ) ஐயவினா, வினா எதிர்வினாதல்
(ஆ) அறிவினா, மறைவிடை
(இ) அறியாவினா, சுட்டு விடை
(ஈ) கொளல்வினா, இனமொழி விடை
Answer:
(இ) அறியாவினா, சுட்டு விடை

Question 4.
தொல்காப்பியம் குறிப்பிடும் இசைக்கருவி………
(அ) ஜால்ரா
(ஆ) பறை
(இ) உறுமி
(ஈ) நாகசுரம்
Answer:
(இ) உறுமி

Question 5.
வீட்டைத் துடைத்துச் சாயம் அடித்தல்’ – இவ்வடி குறிப்பிடுவது …………
(அ) காலம் மாறுவதை
(ஆ) வீட்டைத் துடைப்பதை
(இ) இடையறாது அறப்பணி செய்தலை
(ஈ) வண்ணம் பூசுவதை
Answer:
(இ) இடையறாது அறப்பணி செய்தலை

Question 6.
காற்றின் மெல்லிய ……….. பூக்களைத் தலையாட்ட வைக்கிறது. கைகளின் நேர்த்தியான ………. பூக்களை மாலையாக்குகிறது. தொழிற்பெயர்களைப் பொருத்தித் தொடர்களை முழுமை செய்க.
(அ) தொடுத்தல், எடுத்தல்
(ஆ) தொடுதல், தொடுத்தல்
(இ) எடுத்தல், தொடுதல்
(ஈ) தொடுத்தல், தொடுதல்
Answer:
(ஆ) தொடுதல், தொடுத்தல்

Question 7.
வாய்மையே மழைநீராகி ‘ இத்தொடரில் வெளிப்படும் அணி.
(அ) உவமை
(ஆ) தற்குறிப்பேற்றம்
(இ) உருவகம்
(ஈ) தீவகம்
Answer:
(இ) உருவகம்

Question 8.
”சங்க இலக்கியங்கள் ஐந்திணைகளுக்குமான ஒழுக்கங்களை இருதிணைகளும் பயன்பெற எடுத்தியம்புகின்றன” – இத்தொடரில் அமைந்துள்ள தொகைச் சொற்களின் பொருத்தமான விரியைக் கண்டறிக.
(அ) குறிஞ்சி, முல்லை, மருதம், நெய்தல், பாலை – நல்வினை, தீவினை
(ஆ) குறிஞ்சி, முல்லை , மருதம், நெய்தல், பாலை – உயர்திணை, அஃறிணை
(இ) குறிஞ்சி, முல்லை , நெய்தல், பாலை, மருதம் – அறம், பொருள், இன்பம்
(ஈ) குறிஞ்சி, மலை, முல்லை , காடு, மருதம், வயல், நெய்தல், கடல் – பனை, திணை
Answer:
(ஆ) குறிஞ்சி, முல்லை , மருதம், நெய்தல், பாலை – உயர்திணை, அஃறிணை

Question 9.
‘கேள்வியினான்’ என்னும் சொல்லின் இலக்கணக் குறிப்பு ….
(அ) வினையாலணையும் பெயர்
(ஆ) தொழிற்பெயர்
(இ) வினைத்தொகை
(ஈ) பண்புத்தொகை
Answer:
(அ) வினையாலணையும் பெயர்

Question 10.
கலையரங்கத்தில் எனக்காகக் காத்திருக்கிறவரை அழைத்து வாருங்கள் – இத்தொடர்………….
(அ) வினாத்தொடர்
(ஆ) தனித்தொடர்
(இ) செய்தித்தொடர்
(ஈ) கலவைத் தொடர்
Answer:
(இ) செய்தித்தொடர்

Question 11.
”காய்ந்த இலையும் காய்ந்த தோகையும்” நிலத்துக்கு நல்ல உரங்கள் – இத்தொடரில் அடிக் கோடிட்ட பகுதி குறிப்பிடுவது …..
(அ) இலையும் சருகும்
ஆ) தொகையும் ஓலையும்
(இ) தாளும் ஓலையும்
(ஈ) சருகும், சண்டும்
Answer:
(ஈ) சருகும், சண்டும்

பாடலைப் படித்துப் பின்வரும் வினாக்களுக்கு (12, 13, 14. 15) விடை தருக.
”நனந்தலை உலகம் வளைஇ நேமியொடு
வலம்புரி பொறித்த மாதாங்கு தடக்கை
நீர் செல், நிமிர்ந்த மாஅல் போல,
பாடு இமிழ் பனிக்கடல் பருகி, வலன் ஏர்பு,
கோடு கொண்டு எழுந்த கொடுஞ் செலவு எழிலி
பெரும்பெயல் பொழிந்த சிறுபுன் மாலை”

Question 12.
இப்பாடலின் ஆசிரியர்…………
(அ) ஒளவையார்
(ஆ) நப்பூதனார்
(இ) நல்வேட்டனார்
(ஈ) பெருங்கௌசிகனார்
Answer:
(ஆ) நப்பூதனார்

Question 13.
இப்பாடல் இடம் பெற்றுள்ள நூல்……….
(அ) மலைபடுகடாம்
(ஆ) பரிபாடல்
(இ) முல்லைப்பாட்டு
(ஈ) நற்றிணை
Answer:
(இ) முல்லைப்பாட்டு

Question 14.
‘நனந்தலை உலகம்’ – இத்தொடரின் பொருள்?
(அ) சிறிய உலகம்
(ஆ) நனைந்த உலகம்
(இ) சுற்றும் உலகம்
(ஈ) அகன்ற உலகம்
Answer:
(ஈ) அகன்ற உலகம்

Question 15.
பாடலில் இடம்பெற்றுள்ள அடி எதுகைச் சொற்களைக் குறிப்பிடுக.
(அ) பாடு இமிழ், கோடுகொண்டு
(ஆ) கோடுகொண்டு, பெரும்பெயல்
(இ) நனந்தலை, வலம்புரி
(ஈ) பொறித்த, பொழிந்த
Answer:
(அ) பாடு இமிழ், கோடுகொண்டு

பகுதி – II (மதிப்பெண்கள்: 18)

பிரிவு – 1

எவையேனும் நான்கு வினாக்களுக்கு மட்டும் குறுகிய விடையளிக்க.
21 ஆவது வினாவிற்குக் கட்டாயமாக விடையளிக்க வேண்டும். [4x 2 = 8]

Question 16.
விடைகளுக்கேற்ற வினாக்கள் அமைக்க.
(அ) ஒரு நாட்டின் வளத்திற்குத் தக்கபடியே, அந்நாட்டு மக்களின் அறிவொழுக்கங்களும் அமைந்திருக்கும்.
(ஆ) 1956ஆம் ஆண்டு நவம்பர் 1ஆம் நாள் கன்னியாகுமரி மாவட்டம் தமிழ்நாட்டுடன் இணைந்தது. விடை
Answer:
(அ) ஒரு நாட்டின் அறவொழுக்கம் எப்படி அமைதல் வேண்டும்?
(ஆ) கன்னியாகுமரி மாவட்டம் தமிழ்நாட்டுடன் எப்போது இணைந்தது?

Question 17.
யாருக்கு இவ்வுலகமே உரிமை உடையதாகும் எனத் திருக்குறள் குறிப்பிடுகிறது?
Answer:
நடுநிலையாகக் கடமை தவறாமல் இரக்கம் காட்டுபவருக்கு இவ்வுலகமே உரிமை உடையதாகும்.

Question 18.
‘நமக்கு உயிர் காற்று. காற்றுக்கு வரம் மரம் – மரங்களை வெட்டி எறியாமல் நட்டு வளர்ப்போம்’ – இது போன்ற உலக காற்று நாள் விழிப்புணர்வுக்கான இரண்டு முழக்கத் தொடர்களை எழுதுக.
Answer:

உணவு மனிதனுக்கு ஆதாரம்
மரம் மண்ணுக்கு ஆதாரம்
மரம் வளர்ப்போம்! மழை பெறுவோம்.

Question 19.
“மன்னும் சிலம்பே! மணிமே கலைவடிவே!
முன்னும் நினைவால் முடிதாழ வாழ்த்துவமே” – ஐம்பெருங்காப்பியங்களுள் இவ்வடிகளில் இடம்பெற்றுள்ள இருகாப்பியங்களைத் தவிர எஞ்சியுள்ள காப்பியங்களின் பெயர்களை எழுதுக.
Answer:
சீவகசிந்தாமணி, வளையாபதி, குண்டலகேசி

Question 20.
மொழிபெயர்ப்புக் குறித்து மணவை முஸ்தபா குறிப்பிடுவது யாது?
Answer:
”ஒரு மொழியில் உணர்த்தப்பட்டதை வேறொரு மொழியில் வெளியிடுவது மொழிபெயர்ப்பு” என்கிறார் மணவை முஸ்தபா.

Question 21.
‘கண்’ என முடியும் திருக்குறளை எழுதுக.
Answer:
பண் என்னாம் பாடற் கியைபின்றேல் கண் என்னாம் கண்ணோட்டம் இல்லாத கண்

பிரிவு – 2

எவையேனும் ஐந்து வினாக்களுக்கு மட்டும் குறுகிய விடையளிக்க. [5 x 2 = 10]

Question 22.
அண்ணன் நேற்று வீட்டிற்கு வந்தது. அண்ணன் புறப்படும் போது அம்மா வழியனுப்பியது. வழுவை வழாநிலையாக மாற்றுக)
Answer:
அண்ணன் நேற்று வீட்டிற்கு வந்தான். அண்ணன் புறப்படும் போது அம்மா வழியனுப்பினார்கள்.

Question 23.
பொருத்தமானவற்றைச் சொற்பெட்டியில் கண்டு எழுதுக.
Answer:
தோற்பாவை, விருது, தோற்பவை, விருந்து
(அ) வாழ்க்கையில் தோற்பவை மீண்டும் வெல்லும் – இதைத் தத்துவமாய்த் தோற்பாவைக் கூத்து சொல்லும்.
(ஆ) தெருக்கூத்தில் நடிகருக்குக் கைத்தட்டலே விருந்து அதில் வரும் காசு குறைந்தாலும் அதுவேயவர் விருது.

Question 24.
கொடுக்கப்பட்டுள்ள இரு சொற்களைப் பயன்படுத்தி ஒரு தொடர் அமைக்க.
Answer:
சிறு – சீறு
வயலில் உள்ள சிறு பாம்பாயினும் சீறும்.

Question 25.
கலைச்சொற்கள் தருக
Answer:
(அ) Modern Literature – நவீன இலக்கியம்
(ஆ) Epic Literature – காப்பிய இலக்கியம்

Question 26.
தொழிற்பெயருக்கும், வினையாலணையும் பெயருக்கும் உள்ள வேறுபாடுகள் இரண்டினை எழுதுக.
Answer:

Question 27.
பொருத்தமான நிறுத்தக் குறிகளை இடுக.
Answer:
சேரர்களின் பட்டப் பெயர்களில் கொல்லி வெற்பன் மலையமான் போன்றவை குறிப்பிடத்தக்கவை கொல்லி மலையை வென்றவர்கள் கொல்லி வெற்பன் எனவும் பிற மலைப்பகுதிகளை வென்றவர்கள் மலையமான் எனவும் பெயர் சூட்டிக்கொண்டனர் இதற்குச் சங்க இலக்கியத்தில் சான்றுகள் உள்ளன.

விடை : சேரர்களின் பட்டப் பெயர்களில் கொல்லி வெற்பன், மலைய குறிப்பிடத்தக்கவை. கொல்லி மலையை வென்றவர்கள் கொல்லி வெற்பன்’ எனவும், பிற மலைப்பகுதிகளைவென்றவர்கள் ‘மலையமான்’ எனவும் பெயர் சூட்டிக் கொண்டனர். இதற்குச் சங்க இலக்கியத்தில் சான்றுகள் உள்ளன.

Question 28.
அறியேன் – பகுபத உறுப்பிலக்கணம் தருக.
அறியேன் = அறி + ய் + ஆ + ஏன்
அறி – பகுதி
ய் – உடம்படுமெய் சந்தி
ஆ – எதிர்மறை இடைநிலை புணர்ந்து கெட்டது
ஏன் – தன்மை ஒருமை வினைமுற்று விகுதி

பகுதி – III (மதிப்பெண்க ள்: 18)

பிரிவு – 1

எவையேனும் இரண்டு வினாக்களுக்கு மட்டும் சுருக்கமாக விடையளிக்க. [2 x 3 = 6]

Question 29.
நீவிர் அறிந்த செயற்கை நுண்ணறிவுப் பயன்பாடு மூன்றினை எழுதுக.
Answer:

செயற்கை நுண்ணறிவைக் கொண்ட இயந்திரம் மனிதர்களுடன் சதுரங்கம் முதலான விளையாட்டுகளை விளையாடுகிறது.
கண் அறுவை மருத்துவம் செய்கிறது.
சமைக்கிறது.

Question 30.
ஜெயகாந்தன் தம் கதைமாந்தர்களின் சிறந்த கூறுகளைக் குறிப்பிடத் தவறுவதில்லை என்று அசோகமித்திரன் கூறுகிறார். இக்கூற்றை மெய்ப்பிக்கும் கூறு ஒன்றைத் தர்க்கத்திற்கு அப்பால் கதை மாந்தர் வாயிலாகக் குறிப்பிடுக.
Answer:

தோல்வி நிச்சயம் என்ற மனப்பான்மையுடன் போன நான் வழக்கத்திற்கு மாறாக அன்று – தோற்றுப்போனேன்.
தோல்வி நிச்சயம் என்ற என் மனப்போக்கு தோற்றது. என் வாழ்க்கையே நிர்ணயிக்கப்பட்டது.
பிச்சைக்காரனுக்கு பிச்சை போட்டதில் நாலணாவில் அந்த நல்ல நாளைக் கொண்டாடிவிட்ட நிறைவு பிறந்தது.
காலணாதான் கடன் தரலாம் தருமத்தைத் தரமுடியுமா? தருமத்தை யாசித்துத் தந்தால்தான் பெற முடியும்.
ஒருவனுக்கு என்ன கிடைக்க வேண்டுமோ அதுதான் கிடைக்கும் நாம் எப்படி முயற்சி செய்தாலும் நமக்குக் கிடைப்பது தான் கிடைக்கும்.

Question 31.
உரைப்பத்தியைப் படித்து வினாக்களுக்கு விடை தருக.
தம் வீட்டிற்கு வரும் விருந்தினரை முகமலர்ச்சியோடு வரவேற்று உண்ண உணவும் இருக்க இடமும் கொடுத்து அன்பு பாராட்டுவதே விருந்தோம்பல். விருந்தினர் என்றால் உறவினர் என்று இக்காலத்தினர் கருதுகின்றனர். உறவினர் வேறு, விருந்தினர் வேறு. முன்பின் அறியாத புதியவர்களுக்கே விருந்தினர் என்று பெயர். அதனால் தான் ‘விருந்தே புதுமை’ என்று தொல்காப்பியர் அன்றே கூறியுள்ளார்.

(அ) விருந்தினர் என்போர் யாவர்?
Answer:
விருந்தினர் என்றால் உறவினர் என்று இக்காலத்தினர் கருதுகின்றனர். உறவினர் வேறு, விருந்தினர் வேறு. முன்பின் அறியாத புதியவர்களுக்கே விருந்தினர் என்று பெயர்.

(ஆ) விருந்து குறித்து தொல்காப்பியர் கூறியது யாது?
Answer:
‘விருந்தே புதுமை’ என்று தொல்காப்பியர் அன்றே கூறியுள்ளார்.

(இ) இவ்வுரைப்பத்திக்குப் பொருத்தமான தலைப்பு ஒன்று தருக.
Answer:
விருந்தோம்பல்

பிரிவு – 2

எவையேனும் இரண்டு வினாக்களுக்கு மட்டும் சுருக்கமாக விடையளிக்க.
34 ஆவது வினாவிற்குக் கட்டாயமாக விடையளிக்க வேண்டும். [2 x 3 = 6]

Question 32.
“உள்வாய் வார்த்தை உடம்பு தொடாது” இடஞ்சுட்டிப் பொருள் விளக்குக.
Answer:
இடம் :
காலக்கணிதம் என்னும் நூலின் ஆசிரியர் கண்ணதாசன் அவர்கள் அமைந்த பாடல் வரிகள். விளக்கம் :
நம் வாயிலிருந்து வரும் வார்த்தைகள் ஒருபோதும் நம் உடலைக் காயப்படுத்துவது கிடையாது.

பாடலின் பொருள் :
மாற்றம் எனது மானிடத் தத்துவம். மாறுகின்ற உலகின் மகத்துவம் அறிவேன். எவை நன்மை, எவை தீமை என்பதை நான் அறிவேன். தலைவர்கள் மாறலாம். அவைகள் மாறலாம். தத்துவம் மட்டும் அட்சய பாத்திரம் ஆகும். எடுத்துக் கொள்பவர் எடுத்துக் கொள்ளட்டும். குரைப்போர் குரைத்துக் கொள்ளட்டும். வாயிலிருந்து வரும் வார்த்தைகள் ஒருபோதும் நம் உடலைக் காயப்படுத்துவது கிடையாது. நானே தொடக்கமாகவும், நானே முடிவாகவும், நான் கூறுவதே நாட்டின் சட்டமாகும்.

Question 33.
வைத்தியநாதபுரி முருகன், அணிந்திருக்கும் அணிகலன்களுடன் குழந்தையாகச் செங்கீரை ஆடிய நயத்தினை விளக்குக.
Answer:

திருவடியில் அணிந்த சிறு செம்பொன்னால் ஆன கிண்கிணிகளோடு சிலம்புகளும் சேர்ந்து ஆடட்டும்.
இடையில் அரைஞாண் மணியோடு ஒளிவீசுகின்ற அரைவடங்கள் ஆடட்டும்.
பசும்பொன் என ஒளிரும் தொந்தியுடன் சிறுவயிறு சரிந்தாடட்டும். பட்டம் கட்டிய நெற்றியில் விளங்குகின்ற பொட்டின் வட்டி வடிவான சுட்டி பதிந்தாடட்டும்.
கம்பிகளால் உருவான குண்டலங்களும் காதின் குழைகளும் அசைந்தாடட்டும்.
உச்சிக் கொண்டையும் அதில் சுற்றிக் கட்டப்பட்டுள்ள ஒளியுள்ள முத்துகளோடு ஆடட்டும்.
தொன்மையான வைத்தியநாதபுரியில் எழுந்தருளிய முருகனே! செங்கீரை ஆடி அருள்க!
இவற்றுடன் அழகிய பவளம் போன்ற திருமேனியும் ஆட, செங்கீரை ஆடுக.

Question 34.
அடிபிறழாமல் எழுதுக.
”நவமணி வடக்க யில்போல் ” எனத் தொடங்கும் தேம்பாவணிப் பாடல்.
Answer:
நவமணி வடக்க யில்போல்
நல்லறப் படலைப் பூட்டும்
தவமணி மார்பன் சொன்ன
தன்னிசைக்கு இசைகள் பாடத்
துவமணி மரங்கள் தோறும்
துணர் அணிச் சுனைகள் தோறும்
உவமணி கானம் கொல் என்று
ஒலித்து அழுவ போன்றே – வீரமாமுனிவர்

(அல்ல து)

“வாளால் அறுத்துச் சுடினும் எனத் தொடங்கும் பெருமாள் திருமொழிப் பாடல்.
வாளால் அறுத்துச் சுடினும் மருத்துவன் பால்
மாளாத காதல் நோயாளன் போல் மாயத்தால்
மீளாத் துயர்தரினும் வித்துவக் கோட்டம்மா! நீ
ஆளா உனதருளே பார்ப்பன் அடியேனே. – குலசேகராழ்வார்

பிரிவு – 3

எவையேனும் இரண்டு வினாக்களுக்கு மட்டும் சுருக்கமாக விடையளிக்க. [2 x 3 = 6]

Question 35.
”கண்ணே கண்ணுறங்கு!
காலையில் நீயெழும்பு!
மாமழை பெய்கையிலே
மாம்பூவே கண்ணுறங்கு
பாடினேன் தாலாட்டு!
ஆடி ஆடி ஓய்ந்துறங்கு!” – இத்தாலாட்டுப் பாடலில் அமைந்துள்ள தொகாநிலைத் தொடர் வகைகளை எழுதுக.
Answer:
அடுக்குத் தொடர், எழுவாய் தொடர், உரிச்சொல் தொடர்

Question 36.
‘பொருளல் லவரைப் பொருளாகச் செய்யும் பொருளல்ல தில்லை பொருள் – இக்குறட்பாவினை அலகிட்டு வாய்பாடு தருக.
Answer:

Question 37.
தன்மை அணி குறித்து எழுதுக.
Answer:
தன்மையணி :
எவ்வகைப்பட்ட பொருளாக இருந்தாலும் இயற்கையில் அமைந்த அதன் உண்மையான இயல்புத் தன்மையினைக் கேட்பவர்களின் மனம் மகிழுமாறு உரிய சொற்களை அமைத்துப் பாடுவது தன்மையணியாகும். இதனைத் தன்மை நவிற்சி அணி என்றும் கூறுவர். இவ்வணி நான்கு வகைப்படும். பொருள் தன்மையணி, குணத் தன்மையணி, சாதித் தன்மையணி, தொழிற் தன்மையணி என்பதாகும்.

(எ.கா.) மெய்யிற் பொடியும் விரித்த கருங்குழலும்
கையில் தனிச்சிலம்பும் கண்ணீரும் – வையைக் கோன்
கண்டளவே தோற்றான், அக்காரிகைதன் சொற்செவியில்
உண்டளவே தோற்றான் உயிர்

சிலம்பு – வழக்குரை காதை வெண்பா

பாடலின் பொருள்: உடம்பு முழுக்கத் தூசியும் விரித்த கருமையான தலைமுடியும் கையில் ஒற்றைச் சிலம்போடு வந்த தோற்றமும் அவளது கண்ணீரும் கண்ட அளவிலேயே வையை நதி பாயும் கூடல் நகரத்து அரசனான பாண்டியன் தோற்றான். அவளது சொல், தன் செவியில் கேட்டவுடன் உயிரை நீத்தான். அணிப்பொருத்தம். கண்ணகியின் துயர் நிறைந்த தோற்றத்தினை இயல்பான உரிய சொற்களின் மூலம் கூறியமையால் இது தன்மை நவிற்சியணி எனப்படும்.

பகுதி – IV (மதிப்பெண்க ள் : 25)

அனைத்து வினாக்களுக்கும் விடையளிக்க. [5 x 5 = 25]

Question 38.
(அ) சிலப்பதிகார மருவூர்ப்பாக்க வணிக வீதிகளை இக்கால வணிக வளாகங்களோடும் அங்காடிகளோடும் ஒப்பிட்டு எழுதுக.
Answer:
சிலப்பதிகார மருவூர்ப்பாக்க வணிக வீதிகள் :

புகார் நகர மருவூர்ப்பாக்கத்தின் வணிக வீதிகளில் வண்ணக்குழம்பு, சுண்ணப்பொடி, குளிர்ந்த மணச்சாந்து.
பூ, நறுமணப் புகைப்பொருள்கள், அகில் முதலான மணப்பொருள்கள் விற்பவர்கள் வீதிகளில் வணிகம் செய்து கொண்டிருக்கின்றனர்.

கூலக்கடைத் தெருக்கள்:

இங்குப் பட்டு, பருத்தி நூல், முடி இவற்றினைக் கொண்டு அழகாகப் பின்னிக்கட்டும் கைத்தொழில் வல்லுநரான நெசவாளர் வாழும் வீதிகள் உள்ளன.
இங்குப் பட்டும் பவளமும், சந்தனமும் அகிலும், முத்தும் மணியும் பொன்னும் அளக்க முடியாத அளவிற்குக் குவிந்து கிடக்கும் வளம் நிறைந்த அகன்ற வணிக வீதிகளும் உள்ளன.
மேலும் இவ்வீதிகளில் வேறு பலப்பல பண்டங்களின் விற்பனையும் நடைபெறுகின்றது.
எட்டுவகைத் தானியங்களும் குவிந்து கிடக்கும் கூலக்கடைத் தெருக்களும் உள்ளன.

மருவூர்ப்பாக்கத்தின் வணிகம்:

மருவூர்ப்பாக்கத்தின் தெருக்களில், பிட்டு வணிகம் செய்பவரும் அப்பம் சுடுபவரும் கள் விற்கும் வலைச்சியரும் மீன் விற்கும் பரதவரும் உள்ளனர்.
வெண்மையான உப்பு விற்கும் உமணரும் வெற்றிலை விற்பவரும் ஏலம் முதலான ஐந்து நறுமணப்.
பொருள் விற்பவரும் பல வகையான இறைச்சிகள் விற்பவரும் எண்ணெய் வணிகரும் இங்கு வணிகம் செய்கின்றனர்.
இவற்றுடன் அத்தெருக்களில் பல்வகைப் பொருள்களை விற்கின்ற கடைகளும் உள்ளன.
வெண்கலம், செம்புப் பாத்திரங்கள் செய்வோர், மரத்தச்சர், இரும்புக்கொல்வர், ஓவியர், மண் பொம்மைகள் செய்பவர், சிற்பிகள் ஆகியோர் உள்ளனர்.
பொற்கொல்லர், இரத்தின் வேலை செய்பவர், தையற்காரர், தோல் பொருள் தைப்பவர், துணியாலும் கட்டைகளாலும் பொம்மைகள் செய்பவர் ஆகியோர் உள்ளனர்.
இவ்வாறாகப் பழுதின்றிக் கைத்தொழில் பல செய்யும் மக்கள் வாழும் பகுதிகள் இங்கு நிறைந்துள்ளன.
குழலிலும் யாழிலும் குரல், துத்தம், கைக்கிளை, உழை, இளி, விளரி, தாரம் என்னும் ஏழு இசைகளைக் (ஸ, ரி, க, ம, ப, த.
நி என்னும் ஏழு சுரங்களை குற்றமில்லாமல் இசைத்துச் சிறந்த திறமையைக் காட்டும் பெரும்பாணர்களின் இருப்பிடங்களும் உள்ளன.
இவர்களுடன் மருவூர்ப்பாக்கத்தின் தெருக்களில் சிறுசிறு கைத்தொழில் செய்வோர், பிறருக்கு ஏவல் செய்வோர் வாழும் இடங்களும் உள்ளன.
இவை அனைத்தும் குற்றமின்றிச் சிறப்புடன் அமைந்து விளங்கப் பரந்து கிடந்தன.

இக்கால வணிக வளாகங்கள் அங்காடிகள் :

இக்கால வணிக வளாகங்கள், அங்காடிகள் ஒப்பனை செய்யப்பட்டு – வணிகருக்கும் மக்களுக்குமான உறவைக் குறைத்து வருகின்றன. மனதை மயக்கும் வகையில் இயங்குகிறது.
இக்காலத்தில் வியாபாரத்திற்கும், பணம் விளம்பரத்திற்கும் மட்டுமே முக்கியத்துவம் அளிக்கப்படுகிறது.
மனித மாண்புக்கு முக்கியத்துவம் அளிக்கப்படுவதில்லை. இக்கால வணிக வளாகங்களில் ஆடை முதல் ஆபரணங்கள் வரை, குண்டூசி முதல் கடப்பாரை வரை.
தானியங்கள் முதல் உணவுப்பண்டங்கள், வரை விற்பனை செய்யப்படுகின்றன.
பாத்திரங்கள் மற்றும் பல்வேறு தொழில் செய்பவர்களின் கிடங்குகள், இசைக்கருவிகள்.
விற்பனையிடங்கள் போன்றவை எங்கும், பரந்து காணப்படுகின்றன.

(அல்லது)
(ஆ) ஆற்றுப்படுத்துதல் என்பது அன்றைக்குப் புலவர்களையும் கலைஞர்களையும் வள்ளல்களை நோக்கி நெறிப்படுத்துவதாக இருந்தது. இன்றைய நிலையில் ஆற்றுப்படுத்துதல் ஒரு வழிகாட்டுதலாக மாறியிருப்பதை விளக்குக.

ஆற்றுப்படுத்துதல் என்பது வள்ளலை நாடி எதிர்வருபவர்களை அழைத்து யாம் இவ்விடத்தைச் சென்று இன்னவெல்லாம் பெற்று வருகின்றோம்.
நீயும் அந்த வள்ளலிடம் சென்று வளம் பெற்று வாழ்வாயாக என்று கூறுதல் ஆற்றுப்படை ஆகும்.
ஆற்றுப்படுத்துதல் என்பது இன்றைய நிலையில் ஒரு வழிகாட்டுதலாக இருக்கிறது.
தன்னிடம் இல்லை என்றோ அல்லது தெரியாது என்றோ யார் வந்தாலும் அவர்களுக்கு வழிகாட்டுதலாகவும் இருக்கிறது.
அவர்களுக்கு அறிவுரை கூறி அவர்களை வழிகாட்டுகின்றனர். அன்றைய ஆற்றுப்படுத்துதல் இன்றைய வழிகாட்டுதலாக மாறியுள்ளது.
இது ஒவ்வொரு நிலையிலும் மாற்றம் அடைந்துள்ளது. உதவி தேவைப்படுபவர்களுக்கு பெரும் உதவியாக இருந்து வருகிறது.
இதுவே இன்றைய ஆற்றுப்படுத்துதல் ஆகும்.

Question 39.
(அ) உங்கள் தெருவில் மின்விளக்குகள் பழுதடைந்துள்ளன. அதனால் இரவில் சாலையில் நடந்து சொல்வோருக்கும் வாகன ஓட்டிகளுக்கும் ஏற்படும் இடையூறுகளைக் குறிப்பிட்டு, புதிய மின்விளக்குகள் பொருத்தும்படி மின்வாரிய அலுவலருக்குக் கடிதம்
எழுதுக.

அனுப்புநர்
பொது மக்கள்,
பூந்தோட்டம்,
விழுப்புரம் – 5.

பெறுநர்
மின்வாரிய இயக்குநர்,
மின்வாரிய அலுவலகம்,
பூந்தோட்டம்,
விழுப்புரம்-5.

ஐயா,
பொருள்: தெரு விளக்கு பழுதுநீக்கித் தருமாறு விண்ணப்பம் அளித்தல் – சார்பு. வணக்கம். எங்கள் பகுதியில் ஏறக்குறைய மூவாயிரம் பேர் வாழ்கிறார்கள். தெருக்களில் விளக்குகள் ஒளிவழங்குவது இல்லை. அதனால் தெருக்களில் நாய்கள் படுத்து உறங்குவது தெரியாமல் மிதித்து விடுகின்றனர். அதனால் நாய்கள் தெரு வழியே செல்வோரைக் கடித்துவிடுகின்றன. நாய் கடியினால் வருந்துவோர்களின் எண்ணிக்கை மிகுதியாக உள்ளது.

தெரு விளக்குகள் இயங்காமையால் தெருவில் நடந்து செல்வோர், விபத்துக்கும் ஆளாகின்றனர். தவிர முகமூடிக் கொள்ளையர் தொடர்ச்சியாக வீடுகளில் புகுந்து திருடிச் செல்கின்றனர். உயிர்க் கொலையும் செய்கின்றனர். தெரு விளக்குகளை விரைவாகச் சீர்செய்து எங்கள் துன்பத்தைப் போக்க ஆவன செய்யுமாறு அன்புடன் வேண்டுகிறோம்.

நன்றி.

இங்ஙனம்,
உங்கள் உண்மையுள்ள,
பொதுமக்கள்

பூந்தோட்டம்,
4.4.2019.
உறைமேல் முகவரி

பெறுநர்
மின்வாரிய இயக்குநர்,
மின்வாரிய அலுவலகம்,
பூந்தோட்டம்,
விழுப்புரம் – 5.

(அல்லது)

(ஆ) “மரம் இயற்கையின் வரம்” எனும் தலைப்பில், மாநில அளவில் நடத்தப்பெற்ற கட்டுரை போட்டியில் முதல் பரிசு பெற்ற தோழனை வாழ்த்தி மடல் எழுதுக.
எண், 20/3 மாடவீதி,
மதுரை,
5.5.2019

அன்புள்ள நவீன்குமார்,
நாங்கள் அனைவரும் இங்கு நலம். நீயும் உன் குடும்பத்தாரும் எப்படி இருக்கிறீர்கள்? சென்ற மாதம் தமிழக அரசின் சார்பில் நடைபெற்ற இலக்கிய மன்றக் கட்டுரைப் போட்டியில் கலந்து கொண்டு ”மரம் இயற்கையின் வரம்” என்ற தலைப்பில் நீ எழுதிய கட்டுரை அனைவரிடமும் இயற்கையைக் காப்பாற்ற வேண்டும், மரங்களை நட்டு வளர்க்க வேண்டும் என்ற அவசியத்தை உணர்த்தியுள்ளது. ஆகவே உன் ஈடுபாட்டைப் பார்த்து எனக்கும் கட்டுரைப் போட்டியில் கலந்துகொள்ள ஆர்வம் வருகிறது. பல போட்டிகளில் நீ பெற்ற பரிசுப்பொருள்கள் உன் வீட்டில் ஏராளமாகக் குவிந்து கிடக்கின்றன. அனைத்தையும் பார்த்து நான் பலமுறை வியந்துள்ளேன்.

உன்னை நண்பனாக அடைந்ததற்கு நான் மிகவும் பெருமைப்படுகிறேன். நீ இன்னும் பல போட்டிகளில் கலந்து கொண்டு வெற்றிகளைப் பெற வாழ்த்துகிறேன். நீ அடுத்தமுறை போட்டியில் கலந்து கொள்ளும் போது எனக்குத் தெரியப்படுத்து. நான், நீ எவ்வாறு போட்டிக்குத் தயாராகிறாய் என்பதை அறிந்து கொள்கிறேன். உன் பதிலுக்காகக் காத்திருக்கிறேன்.

இப்படிக்கு,
உன் அன்புள்ள,
ப. அன்பரசன்.

உறைமேல் முகவரி
பெறுநர்
க. நவீன்குமார்,
5. காளையார் கோவில்,
முத்தமிழ் நகர்,
ஈரோடு – 638001

Question 40.
படம் உணர்த்தும் கருத்தை நயமுற ஐந்து தொடர்களில் எழுதுக.
Answer:

வறியவர்க்கு ஒன்று ஈவது ஈகை
மறித்துக் கொடுக்காமல் தடுப்பது தீமை
முடிந்ததைக் கொடுப்பது மேதை
வாடி நிற்கும் வறியவர்க்கு கொடுப்பது புகழ்
கொடுப்பதை தடுத்து நிறுத்துவது இகழ்

Question 41.
தட்டச்சர் பணிவாய்ப்பு வேண்டித் தன்விவரப் பட்டியல் ஒன்றை நிரப்புக
Answer:
அனுப்புநர்
கவிமணி.
63, மறைமலை அடிகள் தெரு
மேலூர்
மதுரை – 600 008.

பெறுநர்
சாரதா கல்வி நிலையம்.
மேலூர்.
மதுரை – 600008

ஐயா,
பொருள்: தட்டச்சர் பணி வேண்டி – விண்ணப்பம். தங்கள் அலுவலகத்தில் தட்டச்சர் பணிக்கான விண்ணப்பங்கள் வேண்டி நீங்கள் அறிவித்துள்ள செய்தியை 10.03.2019 அன்று வெளியான நாளிதழ் மூலம் அறிந்தேன். இந்தப் பணிக்கு நானும் விண்ணப்பம் கோருகிறேன்.

நான் 2018ஆம் ஆண்டு மார்ச் மாதம் நடந்த 10ஆம் வகுப்பு பொதுத்தேர்வில் வெற்றி பெற்றுள்ளேன். மேலும், தட்டச்சு இயக்குவதற்கான அடிப்படைப் பயிற்சியும் பெற்றுள்மோன் அப்பணியில் பணியாற்ற எனக்கு வாய்ப்பளித்தால் என் பணியைச் சிறப்பாக செய்வதுடன் காத கல்வி நிலைய வளர்ச்சிக்கும் பாடுபடுவேன் என்று உறுதியளிக்கிறேன். எனது சான்றிதழ்களைத் தங்கள் பார்வைக்கு இத்துடன் அனுப்பியுள்ளேன்.

நன்றி

இங்ஙனம்.
தங்கள் உண்மையுள்ள
கவிமணி
மாதிரி வினாத்தாள் 1

இணைப்பு:

தன் விவரக் குறிப்பு.
கல்விச் சான்றிதழ்

தன் விவரப் பட்டியல்
பெயர் – கவிமணி
பிறந்த நாள் மற்றும் வயது – 10.06.2001; 18
பாலினம் – ஆண்
தந்தையின் பெயர் – மணி
முகவரி – 73, மறைமலை அடிகள் தெரு, மேலூர், மதுரை.
பத்தாம் வகுப்பில் பெற்ற மதிப்பெண் – 450 / 500.
தாய்மொழி – தமிழ்
பயின்ற மொழி – தமிழ், ஆங்கிலம்
தட்டச்சு – தமிழ், ஆங்கிலம் (இளநிலைத் தட்டச்சர்)
கணினி – வேர்டு, பவர்பாயிண்ட், சி+++

மேற்கண்ட விவரங்கள் அனைத்தும் உண்மையென உறுதி கூறுகிறேன். தங்கள் நிறுவனத்தில் தட்டச்சர் பணி தந்தால் என் பணியைச் சிறப்பாகச் செய்வேன் என உறுதி கூறுகிறேன்.

நன்றி,

இடம் : மதுரை
தேதி : 18.04.2019

இங்ஙனம்,
தங்கள் உண்மையுள்ள,
கவிமணி.

உறைமேல் முகவரி
பெறுநர்
சாரதா கல்வி நிலையம்,
மேலூர்,
மதுரை – 625 008.

Question 42.
(அ) தமிழகப் பாரம்பரிய கலைகளைப் பாதுகாக்கவும் வளர்க்கவும் மேலும் பரவலாக்கவும் நீங்கள் செய்யவிருப்பனவற்றை வரிசைப்படுத்தி எழுதுக.
Answer:
பிறந்தநாள் விழாக்களில் மயிலாட்டம் முதலான கலைகளை நிகழ்த்த முனைவேன் எங்கள் குடும்ப விழாக்களில் பொம்மலாட்டம் நிகழ்த்த ஏற்பாடு செய்வேன்.

கோவில் திருவிழாக்களில் ஒயிலாட்டம் முதலான கலைகளை நிகழ்த்தலாம்.
பள்ளி ஆண்டு விழாக்களில் கரகாட்டம் போன்றவற்றை நடத்த வேண்டும்.
திருமண விழாவில் தமிழகப் பாரம்பரியக் கலைகளை நிகழ்த்தலாம்.
தெருக்கூத்து, நாடகங்களில் தமிழர் பாரம்பரியக் கலைகளை அரங்கேற்றலாம்.
ஊர்த் திருவிழாக்களில் தாரை தப்பட்டை நிகழ்த்த ஏற்பாடு செய்யலாம்.

மொழிபெயர்க்க.

(அ) The Golden sun gets up early in the morning and starts its bright rays to fade away the dark. The milky clouds start their wandering. The colourful birds start twitting their morning melodies in percussion. The cute butterflies dance around the flowers. The flowers’ fragrance
fills the breeze. The breeze gently blows everywhere and makes everything pleasant.

விடை தங்க நிற கதிர்களைக் கொண்ட கதிரவன் காலை எழுந்து இருளை அகற்றுகிறான். பால் போன்ற மேகங்கள் தன் ஓட்டத்தைத் தொடர்கின்றன. பல வண்ண பறவைகள் தன் காலை இன்னிசை பாடலை தொடங்குகின்றன. அழகிய பட்டாம் பூச்சிகள் பூக்களைச் சுற்றி நடனம் புரிகின்றன. பூக்களின் மணம் காற்றில் தவழ்கின்றன. மெதுவாக வீசும் காற்று எங்கும் இனிமையை நிரப்புகின்றது.

பகுதி – V (மதிப்பெண்கள் : 24)

அனைத்து வினாக்களுக்கும் விரிவாக விடையளிக்க. [3 x 8 = 24]

Question 43.
(அ) சங்க இலக்கியங்களில் போற்றப்பட்ட அறங்களில் உம்மைக் கவர்ந்த இரண்டு அறங்களைக் குறிப்பிட்டு, அவை நடைமுறை வாழ்வில் பொருந்துவதை விளக்குக.
Answer:
அறத்தில் வணிக நோக்கம் கொள்ளாமை:
அறம் செய்வதில் வணிக நோக்கம் இருக்கக்கூடாது என்பது சங்ககால மக்களின் கருத்தாக இருந்தது. இப்பிறப்பில் அறம் செய்தால் அதன் பயனை மறுபிறப்பில் பெறலாம் என்ற வணிக நோக்குக் கூடாது எனக் கூறப்பட்டது.

“இம்மைச் செய்தது மறுமைக்கு ஆம் எனும்
அறவிலை வணிகன் ஆஅய் அல்லன்”

புறம் எனச் சங்ககால வள்ளல்களில் ஒருவரான ஆய்பற்றி ஏணிச்சேரி முடமோசியார் குறிப்பிட்டுள்ளார். நோக்கமின்றி அறம் செய்வதே மேன்மை தருவது என்பது இதில் உணர்த்தப்பட்டுள்ளது.

அரசியல் அறம் :
மன்னர்களுடைய செங்கோலும் வெண்கொற்றக்குடையும் அறத்தின் குறியீடுகளாகப் போற்றப்பட்டன. அரசன் செங்கோல் போன்று நேரிய ஆட்சியை மேற்கொள்ள வேண்டும் என்பது பல பாடல்களில் வலியுறுத்தப்பட்டுள்ளது. நீர்நிலை பெருக்கி நிலவளம் கண்டு உணவுப்பெருக்கம் காண்பதும் அதனை அனைவருக்கும் கிடைக்கச் செய்வதும் அரசனின் கடமையாகச் சொல்லப்பட்டது. குற்றங்களை.

அறத்தின் அடிப்படையில் ஆராய்ந்து தண்டனை வழங்க வேண்டும் என்கிறார் ஊன் பொதிப் பசுங்குடையார். அரசன் அறநெறியில் ஆட்சி செய்வதற்கு அமைச்சரும் உதவினர். நன்றும் தீதும் ஆய்தலும், அன்பும் அறனும் காத்தலும் அமைச்சர் கடமை என்கிறது மதுரைக்காஞ்சி செம்மை சான்ற காவிதி மாக்கள்’ என்று அமைச்சர்களை மாங்குடி மருதனார் போற்றுகிறார்.

(அல்லது)

(ஆ) மலர்ந்தும் மலராத பாதிமலர் போல
வளரும் விழி வண்ணமே – வந்து
விடிந்தும் விடியாத காலைப் பொழுதாக
விளைந்த கலை அன்னமே
நதியில் விளையாடி கொடியில் தலை சீவி
நடந்த இளந் தென்றலே – வளர்
பொதிகை மலை தோன்றி மதுரை நகர் கண்டு
பொலிந்த தமிழ் மன்றமே

– கவிஞர் கண்ணதாசனின் இப்பாடலில் தவழும் காற்றையும் கவிதை நயத்தையும் பாராட்டி உரை செய்க.

முன்னுரை :
இயற்கையின் கூறுகளில் காற்றின் பங்கு கூடுதலானது எங்கும் நிறைந்திருப்பது உயிர்களின் உயிர் மூச்சுக் காற்றைக் கண்களால் காண முடியாது. மெய்யால் மட்டுமே உணரக்கூடியது காற்று நம்மை மெல்லத் தொட்டுச் சென்றால் தென்றல் எனப்படுகிறது.

தென்றல் காற்று:
தெற்கிலிருந்து வீசுவதால் தென்றல் காற்று எனப்படுகிறது. மரம், செடி, கொடி, ஆறு, மலை, பள்ளத்தாக்கு எனப் பல தடைகளைத் தாண்டி வருவதால் வேகம் குறைந்து இதமான இயல்பு கொள்கிறது. இந்த மென்காற்றை இளந்தென்றல் என்பர்.

இலக்கியத்தில் தென்றல்:
தென்றல் காற்று பலவித மலர்களின் நறுமணத்தை அள்ளி வரும் பொழுது கூடவே வண்டுகளையும் அழைத்து வருவதால் இளங்கோவடிகள் ” வண்டொடு புக்க மணவாய்த் தென்றல் என நயம்பட உரைக்கிறார். பலபட்டடைச் சொக்கநாதப் புலவர் எழுதிய பத்மகிரிநாதர் தென்றல் விடு தூது என்னும் சிற்றிலக்கியத்தில்.

“நந்தமிழும் தண்பொருநை நன்னதியும் சேர் பொருப்பிற்
செந்தமிழின் பின்னுதித்த தென்றலே “

என தென்றலை பெண்ணொருத்தி அன்போடு அழைக்கிறாள்.

கண்ணதாசனின் கவிதை நயம்:
முழுவதும் மலராத மலர் மணத்தையும் அழகையும் கூட்டி வைத்திருக்கும். அம்மலரைப்போல வளரும் கண்ணின் வண்ணமே எனவும் விடிந்தும் விடியாத குளிர்ந்த காலை நேரத்தில் தோன்றிய கலை அன்னமே எனவும், நதியில் விளையாடி கொடிகளில் பாய்ந்து தலைசீவி தவழ்ந்து நடந்து வருகின்ற இளம் தென்றலே எனவும் பொதிகை மலையில் அகத்தியரால் வளர்க்கப்பட்டு மதுரை தமிழ் சங்கங்களில் அழகாய் வளர்ந்த தமிழே எனவும் குழந்தையைக் கண்ணதாசன் பாடுகிறார்.

முடிவுரை:
இவ்வாறாக இலக்கியப் படைப்புகளிலும் திரையிசைப் பாடல்களிலும் தென்றல் காற்று இன்றளவும் நீங்கா இடம் பெற்றுள்ளது. மென்துகிலாய் உடல் வருடி மாயங்கள் செய்வது தென்றல் காற்றேயாகும்.

Question 44.
(அ) கற்கை நன்றே கற்கை நன்றே
பிச்சை புகினும் கற்கை நன்றே’ என்கிறது வெற்றிவேற்கை. மேரியிடமிருந்து பறிக்கப்பட்ட புத்தகம், அச்சிறுமியின் வாழ்க்கையில் கல்விச் சுடரை ஏற்றிய
கதையைப் பற்றிய உங்களின் கருத்துகளைக் குறிப்பிடுக.
Answer:
மேரியின் ஆர்வம்:

ஒரு சிறு குழந்தையின் மனதில் உனக்குப் படிக்கத் தெரியாது உன்னால் படிக்க முடியாது என்று கூறி.
கையில் இருந்த புத்தகத்தை வெடுக்கென்று வாங்கிய நிகழ்வு அந்தக் குழந்தையைப் படிக்க வேண்டும் என்று தூண்டியது.
நாம் படிக்க வேண்டும். என்னாலும் படிக்க முடியும் யார் தடுத்தாலும் என்னால் படிக்க முடியும் என்று தனக்குத் தானே சொல்லிக் கொண்டு காத்திருந்தாள்.

அறிவுச் சுடர்:

தன் அப்பாவிடம் நான் படிக்க வேண்டும். என்னைப் படிக்க வையுங்கள் என்று கேட்டாள்.
அதற்கு நமக்கு என்று படிக்கத் தனியாக பள்ளிக் கூடம் இல்லை.
நாம் எல்லாம் படிக்க முடியாது படிக்க வேண்டும் என்ற எண்ணத்தைத் தூக்கிபோட்டு விட்டு வேலைக்குச் செல் என்று தந்தை கூறினார்.
ஆனால் அவள் நம்பிக்கையை விடவில்லை தினமும் நான் படிக்க வேண்டும்.
நான் படிப்பேன் என்று தனக்குத் தானே சொல்லிக் கொண்டுதான் வேலைக்குச் செல்வாள் இப்படியே காலங்கள் சென்றன.
அவளை நோக்கி வாய்ப்பு வந்தது. ஒருவர் படிக்க வைப்பதாகக் கூறினார்.
அதை அவள் பயன்படுத்திக் கொண்டாள் நன்றாகப் படித்துப் பட்டம் பெற்றாள்.
அந்தச் சமூகத்தில் அவள்தான் முதல் பட்டதாரிப் பெண்.
கிடைத்த வாய்ப்பை நன்றாக பயன்படுத்தி அந்தச் சமூகம் முன்னேற அவளே அறிவுச் சுடராக மாறினாள்.

(அல்லது)
(ஆ) ‘கு.அழகிரிசாமியின் ஒருவன் இருக்கிறான்’ சிறுகதையில் மனிதத்தை வெளிப்படுத்தும் கதை மாந்தர் குறித்து எழுதுக.
Answer:
காஞ்சிப்புரத்துக்காரன்:

சட்டைப்பையிலிருந்து கடிதத்தையும் மூன்று ரூபாயையும் எடுத்து. குப்புசாமிகிட்டே குடுத்துடுங்க.
இல்லே, தங்கவேலு கிட்ட வேணும்னாலும் குடுத்துடுங்க. இன்னொரு சமயம் பட்டணம் வந்தா ஆசுபத்ரிலே போயி பார்க்கிறேன்”
என்று சொல்லிவிட்டுக் கடிதத்தையும் ரூபாயையும் என்னிடம் கொடுத்தான். அப்புறம் ஒரு நிமிஷம் எதையோ யோசித்துப் பார்த்தேன்.
மனசுக்குள் கணக்கு போடுகிறவன் போல் அவனுடைய முகபாவனையும் தலையசைப்பும் இருந்தன.
மறு நிமிஷத்திலேயே, ” இந்தாருங்க. இதையும் குப்புசாமிக்குக் குடுக்கச் சொல்லுங்க” என்று சொல்லித் தன்.
இடது கையில் தொங்கிய துணிப் பையிலிருந்து இரண்டு சாத்துக்குடிப் பழங்களை எடுத்துக் கொடுத்தான்.
”என் பசங்களுக்கு நாலு பழம் வாங்கினேன்.
போகட்டும். இவரு ஆசுபத்திரிலே இருக்கிறாரு. நாம்ப வேறு என்னத்தைச் செய்யப் போறோம்?”
இத்துடனும் அவன் நிறுத்தவில்லை ! தன் உபயமாக ஒரு ரூபாய் நோட்டு ஒன்றை எடுத்து என்னிடம்.
கொடுத்து, குப்புசாமியிடமோ தங்கவேலுவிடமோ சேர்க்கச் சொன்னான்.
அவன் குப்புசாமிக்காகத்தான் கொடுத்தானோ, குப்புசாமிக்காகக் காஞ்சிபுரத்தில் இருந்து கொண்டு கண்ணீர் வடிக்கும் அந்த வீரப்பன்.
குப்புசாமியின் உயிருக்குக் கொடுக்கும் மதிப்பைக் கண்டுதான் கொடுத்தானோ?
என்று அழகிரிசாமியின் ஒருவன் இருக்கிறான் ‘ சிறுகதையில் மனிதத்தை வெளிப்படுத்தும்.
கதை மாந்தரான – காஞ்சிபுரத்துக்காரனை ஆசிரியர் காட்டுகின்றார்.

Question 45.
(அ) குமரிக் கடல் முனையையும் வேங்கட மலைமுகட்டையும் எல்லையாகக் கொண்ட தென்னவர் திருநாட்டிற்குப் புகழ் தேடித்தந்த பெருமை தகைசால் தமிழன்னையைச் சாரும். எழில்சேர் கன்னியாய் திகழும் அவ்வன்னைக்கு , பிள்ளைத் தமிழ் பேசி, சதகம் சமைத்து, பரணி பாடி, கலம்பகம் கண்டு, உலா தந்து. அந்தாதி கூறி, கோவை யாத்து இவற்றையெல்லாம் அணியாகப் பூட்டி, அழகினைக் கூட்டி அகம்மிக மகிழ்ந்தனர் செந்நாப் புலவர்கள். இக்கருத்துகளைக் கருவாகக் கொண்டு சான்றோர் வளர்த்த தமிழ்’ என்னும் தலைப்பில் கட்டுரை எழுதுக.
Answer:
சான்றோர் வளர்த்த தமிழ்
முன்னுரை:
‘கல் தோன்றி மண் தோன்றாக் காலத்தே முன்தோன்றி மூத்த தமிழ்’ என்னும் பழைமையுடைய செந்தமிழ் மொழியை உயர்தனிச் செம்மொழி’ என்று வரையறுத்தவர் பரிதிமாற் கலைஞர் என்று பலராலும் போற்றப்படும் வி.கோ. சூரிய நாராயண சாஸ்திரியார் ஆவார். அகத்தியர் வளர்த்த தமிழ்’ பொதியமலைத் தமிழ் போன்ற தொடர்கள் தமிழின் பழைமையை விளக்கும் சான்றுகளாகும். உயர், தனி, செம்மை என்ற மூன்று அடைமொழிகள் கொண்டு தமிழ் விளங்கக் காரணம் என்ன என்பதைக் காண்பதே இக்கட்டுரையின் நோக்கம்.

பொருளுரை தொன்மை:
குமரிக்கண்டம் எனப்பட்ட லெமூரியாக் கண்டத்திலுள்ள மக்கள் பேசிய மொழி தமிழ் என்பது எலியர் கருத்து. மனித இனம் எப்போது தோன்றியதோ அப்போது தோன்றியது தமிழ். உலக மொழிகளுள் பழைமையும் இலக்கிய இலக்கண வளமும் உடையவை கிரேக்கம், இலத்தீன், சீனம், அரபு, சமஸ்கிருதம், தமிழ் என்பன. பிற மொழிகள் காலவெள்ளத்தில் சிதைந்து மாறுபட்டு விளங்குகின்றன. பத்தாயிரம் ஆண்டு கட்டு முன்பே பேசப்பட்டும், இன்றும் அழியாமல் நிலைத்து நிற்கும் சிறப்பான தன்மை தமிழ் மொழிக்கு அமைந்த பண்பு எனலாம்.

உயர்மொழி :
தான் பேசப்படும் நாட்டிலுள்ள பலமொழிகளுக்கும் தலைமையும், அவற்றை விட மேன்மைத்தன்மையும் உள்ள மொழியே, உயர்மொழி எனப்படும் என்று கூறுவார் பரிதிமாற்மலைஞர். இதன்படி பார்த்தால் திராவிட மொழிகளாகிய தெலுங்கு, கன்னடம், மலையாளம், துளு ஆகிய மொழிகளுக்கு எல்லாம் தலைமையும் மேன்மையும் பெற்றிருப்பதால் தமிழ் உயர்மொழியே’ ஆகும்.

தமிழ் தனிமொழி:
தான் வழங்கும் நாட்டிலுள்ள மற்றைய மொழிகளின் உதவியில்லாமல் தனித்தியங்க வல்ல ஆற்றலுடைய மொழி ‘தனிமொழி’ எனப்படும். பிறமொழிகளுக்குச் செய்யும் உதவி மிகுந்தும், பிற மொழிகள் தனக்குச் செய்யும் உதவி குறைந்தும் காணப்படுவது நம் தமிழ்மொழியில் மட்டுமே. பிற மொழிகளின் உதவி இல்லாமல் தனித்தியங்கும் ஆற்றல் பெற்றிருப்பதால் தமிழ் மொழி தனி மொழி’ எனப்படும்.

தமிழ் – செம்மொழி;
‘திருந்திய பண்பும், சீர்த்த நாகரிகம் பெற்ற தாய்மொழி புகல் செம்மொழியாகும்’ என்பது செம்மொழியின் இலக்கணம். இவ்வரையறை தமிழ் மொழிக்கும் பொருந்துகிறது. தமிழ் மொழியினுள் இடர்ப்பட்ட சொல் முடிவுகளும், தெளிவற்ற பொருள் முடிவுகளும் இல்லை. சொல்லையும் சொல்லுபவன் கருதிய பொருளைக் கேட்பவன் தெளிவாக உணர முடியும் பழையன கழிதலும் புதியன புகுதலும் என்பதைத் தமிழ்மொழி இன்றளவும் ஏற்றிருப்பதால் ‘தமிழ் செம்மொழி ஆகும்.

தமிழின் பொதுப்பண்பு:
மேல்நாட்டு அறிஞர்களான போப், கால்டுவெல், வீரமாமுனிவர் போன்றவர் தமிழினைக்கற்று இலக்கியத்திற்கு வளம் சேர்த்தனர். மதம், மொழி, இனம், நிறம், கடந்து ‘யாதும் ஊரே யாவரும் கேளிர்’ என்று முதல் முழக்கமிட்டது தமிழ் மொழியே ஆகும். உண்பது அமிழ்தமே ஆயினும் தனியராய் உண்ணோம் என்று உணர்த்தினவர் தமிழர். “தீதும் நன்றும் பிறர்தர வாரா” பிறப்பொக்கும் எல்லா உயிர்க்கும் ; உண்பது நாழி உடுப்பவை இரண்டே போன்ற உயர்ந்த சிந்தனைகளை உலகுக்கு உணர்த்தியது தமிழ் இல்வாழ்வையும், புறவாழ்வையும் அகம், புறம் என்று பிரித்து குறிஞ்சி, முல்லை மருதம், நெய்தல் பாலை என்று ஐவகை நிலம் வகுத்து, முதல், கரு, உரிப்பொருள் வகுத்து இலக்கியம் கண்டு இலக்கணம் இயம்பியது நம் செந்தமிழ் மொழியாகும்.

அரசின் கடமை :
மூவாயிரம் ஆண்டுகள் பழைமை வாய்ந்த தமிழைச் செம்மொழியாக அறிவித்தால் பிற நாட்டவர் தமிழ் மொழியைப் பயில்வர். நம் செந்தமிழ் இலக்கியம், இலக்கணம் பிற மொழியாளர்களால் ஆராயப்படும். தமிழ் உலகம் முழுவதும் ஏற்றம் பெற்று புதிய நூல்கள் ஆக்கம் பெறும். உலக அளவில் உயரிய மதிப்பு கூடும். மொழிக் களஞ்சியங்களில் தமிழ்க்கலை வெளிப்படும்.

முடிவுரை:
தமிழைச் செம்மொழி ஆக்கியதோடு மட்டுமன்றி வணிகத் துறையை எட்டிப்பிடித்துச் செல்வ மொழியாகவும் மாற்றிட வழிவகை செய்ய வேண்டும்.

(அல்லது)
(ஆ) குறிப்புகளைப் பயன்படுத்திக் கட்டுரை ஒன்று தருக.
முன்னுரை – விண்வெளியில் தமிழரின் அறிவு – கல்பனா சாவ்லா – விண்ணியல் அறிவில் வருங்காலத்தில் மேற்கொள்ள வேண்டியவை – முடிவுரை

முன்னுரை:
விண்வெளிக்குப் பயணம் செய்த இந்தியாவின் முதல் பெண்மணி என்ற பெருமைக்குரியவர் கல்பனா சாவ்லா. நடுத்தரக் குடும்பத்தில் பிறந்து உலகமே வியந்த விண்வெளி வீராங்கனையாக வாழ்ந்த கல்பனா சாவ்லா அமெரிக்காவின் கொலம்பியா ஓடத்தில் இருந்து விண்வெளிக்குப் பறந்து ஆராய்ச்சிகளை மேற்கொண்டவர்.

இளமைப் பருவம் :
இந்தியாவின் ஹரியானா மாநிலத்தில் 1961 ஆம் ஆண்டு ஜூலை 1 ஆம் தேதி கர்னாஸ் என்ற ஊரில் பிறந்தார் கல்பனா. வீட்டின் நான்கு பிள்ளைகளில் அவர்தான் கடைக்குட்டி அவர் தந்தை ஒரு வியாபாரி, தாய் இல்லத்தரசி பொம்மை வைத்து விளையாடும் வயதில் கல்பனாவுக்குப் பிடித்த பொழுதுபோக்கு விமான ஓவியங்கள் தீட்டி அழகு பார்ப்பது விமானங்களின் சத்தம் கேட்டாலே வீட்டில் இருந்து தெருவுக்கு ஓடி வந்து அந்த அலுமினியப் பறவை, புள்ளியாக மறையும் வரை கண்கள் சுருக்கிப் பார்த்துக்கொண்டே நிற்கும் குழந்தைகளில் ஒருவர்தான் கல்பனாவும்.

கல்வி:
கர்னாவில் உள்ள அரசுப் பள்ளியில் ஆரம்பக் கல்வியை முடித்த கல்பனாவுக்கு அந்த வயதிலேயே விண்வெளி வீரராக வேண்டும் என்ற இலக்கு மனதில் பதிந்துவிட்டது. சண்டிகரில் உள்ள பஞ்சாப் பொறியியல் கல்லூரியில் விமானப் பொறியியல் பயில் விரும்பினார். ஆனால் அது அப்போது ஆண்களின் படிப்பாக இருந்ததால் பெற்றோர் அனுமதிக்கவில்லை என்றாலும் கல்பனாவின் பிடிவாதத்தை அவர்களால் மாற்ற முடியவில்லை அந்தக் கல்லூரியில் பொறியியல் பட்டம் பெற்ற கல்பனாவை 1982 ல் அமெரிக்கா வரவேற்றது. 1984 ஆம் ஆண்டு டெக்காஸ் விண்வெளி பொறியியல் துறையில் முதுகலைப் பட்டம் பெற்றார் கல்பனா. நான்கு ஆண்டுகள் கழித்துக் கொலோரடோ பல்கலைக்கழகத்தில் விமானப் பொறியியல் துறையில் முனைவர் பட்டம் பெற்றார்.

கனவு நனவானது :
1993 ஆம் ஆண்டு ஒரு தனியார் நிறுவனத்தில் ஆய்வு விஞ்ஞானியாகச் சேர்ந்தார். அதற்கு அடுத்த ஆண்டே அவரின் விண்வெளி வீரர் கனவு நனவாகத் தொடங்கியது. நாஸாவில் விண்வெளி வீரர் பயிற்சி பெறுவதற்கு விண்ணப்பித்திருந்த சுமார் 3000 பேரில் ஆறு பேர் மட்டும் தேர்வானார்கள். அவர்களுள் கல்பனாவும் ஒருவர். ஜான்ஸன் விண்வெளி தளத்தில் பல்வேறு உடல் மருத்துவ பரிசோதனைகள், கடுமையான நேர்காணல்கள் ஆகியவற்றைக் கடந்து வெற்றிகரமாகத் தேர்ச்சிப் பட்டியலில் இடம் பிடித்தார்.

முதல் பயணம் :
1995 ஆம் ஆண்டு பயிற்சிகள் முடிந்து விண்வெளி வீராங்கனையாகத் தகுதி பெற்ற கல்பனாவின் முதல் விண்வெளிப் பயணம் 1997 ஆம் ஆண்டு நவம்பர் 19 ஆம் தேதி நிகழ்ந்தது. ஆறு வீரர்களுடன் கொலம்பிய விண்வெளி ஊர்தியான எஸ்.டி.எஸ் 87-ல் பயணம் செய்வதற்குத் தேர்வு செய்யப்பட்ட கல்பனாவுக்கு அதில் ஆராய்ச்சி குறித்த முக்கியப் பொறுப்புகளும் தரப்பட்டன. பூமியை சுமார் 252 தடவை சுற்றிய அந்த விண்கலத்தில் சுமார் 10 மில்லியன் மைல் தொலைவு பயணித்த கல்பனா சகவிண்வெளி வீரர்களுடன் வெற்றிகரமாகப் பூமிக்குத் திருப்பினார்.

கல்பனாவின் ஆர்வம் :
விமானம் மற்றும் கிளைடர்களை ஓட்டக் கற்றுக் கொடுக்க தகுதிச் சான்றிதழ் பெற்றதோடு மட்டுமல்லாமல் ஓட்டவும் அனுமதி பெற்றிருந்தார் கல்பனா. கொலம்பியா விண்வெளிப் பயணம் மேற்கொள்ளும் முன் கல்பனா முழுமனதோடு ஒரு காரியத்தில் ஈடுபடுபவர்களைப் பார்த்தால் எனக்கும் ஊக்கம் ஏற்படும் என்றார். ஆராய்ச்சியாளர்களின் சுயசரிதைகளை விரும்பிப் படிக்கும் அவர், தன் ஆசிரியர்களுக்கு எப்போதும் தன் நன்றிகளைத் தெரிவித்தபடி இருப்பார்.

இரண்டாம் பயணம் :
முதல் பயணத்தை வெற்றிகரமாக முடித்த கல்பனா. பிறகு அதே கொலம்பியா விண்கலத்தில் 2003 ஜனவரி 16 ஆம் தேதி ஆறு விண்வெளி வீரர்களுடன் மீண்டும் விண்ணுக்குப் பயணித்தார். பல விண்வெளி ஆராய்ச்சிகளை மேற்கொண்ட அவர்கள் பயணத்தை முடித்துக் கொண்டு வெற்றிகரமாக திட்டமிட்டபடி பிப்ரவரி ஒன்றாம் தேதி தரையிறங்க ஆயத்தமானார்கள். பூமியைத் தொட 16 நிமிடங்களே இருந்த நிலையில் அந்த சோக சம்பவம் நிகழ்ந்தது. கொலம்பியா விண்கலம் வெடித்துச் சிதறியது. 41 வயது கல்பனா தேவதையாக விண்ணில் கலந்தார்.

முடிவுரை :
இந்திய அரசு கல்பனா சாவ்லாவின் சாதனைகளை நினைவு கூறும் வகையில் 2011 ஆம் ஆண்டு முதல் ஆண்டுதோறும் வீரதீரச் சாதனைகள் புரிந்த பெண்களுக்குக் கல்பனா சாவ்லா விருது வழங்கி கவுரவப்படுத்துகிறது. அந்த விண்வெளி தேவதை நம் வீட்டின் பல குட்டி தேவதைகளுக்குப் பிரியமான.

Tamil Nadu 12th Maths Model Question Paper 4 English Medium

Leave a Comment / Class 12 / By Prasanna

Students can Download Tamil Nadu 12th Maths Model Question Paper 4 English Medium Pdf, Tamil Nadu 12th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 12th Maths Model Question Paper 4 English Medium

Instructions:

The question paper comprises of four parts.
You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
questions of Part I, II. III and IV are to be attempted separately
Question numbers 1 to 20 in Part I are objective type questions of one -mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
Question numbers 21 to 30 in Part II are two-marks questions. These are to be answered in about one or two sentences.
Question numbers 31 to 40 in Parr III are three-marks questions, These are to be answered in about three to five short sentences.
Question numbers 41 to 47 in Part IV are five-marks questions. These are to be answered) in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 90

Part – I

I. Choose the correct answer. Answer all the questions. [20 × 1 = 20]

Question 1.
The rank of the matrix $\left[\begin{array}{rrrr}
1 & 2 & 3 & 4 \\
2 & 4 & 6 & 8 \\
-1 & -2 & -3 & -4
\end{array}\right]$ is ………….
(a) 1
(b) 2
(c) 4
(d) 3
Answer:
(a) 1

Question 2.
If $\left|z-\frac{3}{z}\right|$ , then the least value of | z | is
(a) 1
(b) 2
(c) 3
(d) 5
Answer:
(a) 1

Question 3.
The product of all four values of ‘$\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)^{\frac{3}{4}}$ is ………….
(a) -2
(b) -1
(c) 1
(d) 2
Answer:
(c) 1

Question 4.
The value of cos^-1(-1) + tan^-1(∞) + sin^-1 1 = ………….
(a) 3π/2
(b) -π
(c) 2π
(d) 3π
Answer:
(c) 2π

Question 5.
The value of sin ^-1(1) + sin^-1(0) is
(a) π/2
(b) 0
(c) 1
(d) π
Answer:
(a) π/2

Question 6.
The length of the diameter of the circle which touches the x -axis at the point (1,0) and passes through the point (2, 3)
(a) 6/5
(b) 5/3
(c) 10/3
(d) 3/5
Answer:
(c) 10/3

Question 7.
Area of the greatest rectangle inscribed in the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is ………….
(a) 2ab
(b) ab
(c) $\sqrt{a b}$
(d) a/b
Answer:
(a) 2ab

Question 8.
The angle between two vectors $\vec{a}$ and $\vec{b}$ if $|\vec{a} \times \vec{b}|=\vec{a} \cdot \vec{b}$ is ………….
(a) π/4
(b) π/3
(c) π/6
(d) π/2
Answer:
(a) π/4

Question 9.
The distance between the planes x + 2y + 3z + 7 = 0 and 2x + 4y + 6z + 7 = 0 is ………….

Answer:
(a) $\frac{\sqrt{7}}{2 \sqrt{2}}$

Question 10.
The tangent to the curve y² – xy + 9 = 0 is vertical when
(a) y = 0
(b) y = ± √3
(c) y = 1/2
(d) y = ± 3

Question 11.
The curve y² = (x – a) (x – b)², a, b> 0 and a > b does not exist for
(a) x ≥ a
(b) x = b
(c) b < x < a
(d) x = a

Question 12.
The percentage error of fifth root of 31 is approximately how many times the percentage error in 31 ?
(a) 1/31
(b) 1/5
(c) 5
(d) 31
Answer:
(b) 1/5

Question 13.
The differential of y if y =$\frac{x-2}{2 x+3}$ is ………….

Answer:
(c) $\frac{7}{(2 x+3)^{2}} d x$

Question 14.
If $\int_{0}^{a} \frac{1}{4+x^{2}} d x=\frac{\pi}{8}$ then a is ………….
(a) 4
(b) 1
(c) 3
(d) 2
Answer:
(d) 2

Question 15.
If n is off then $\int_{0}^{\frac{\pi}{2}} \cos ^{n} x d x$ is

Answer:

Question 16.
The solution of $\frac{d y}{d x}+p(x) y=0$ is
(a) y = ce^∫pdx
(b) y = ce^-∫pdx
(c) x = ce^-∫pdx
(d) x = ce^∫pdx
Answer:
(b) y = ce^-∫pdx

Question 17.
The order and degree of the differential equation are $\frac{d^{2} y}{d x^{2}}-y+\left(\frac{d y}{d x}+\frac{d^{3} y}{d x^{3}}\right)^{\frac{3}{2}}=0$ ………….
(a) 2, 3
(b) 3,3
(c) 3, 2
(d) 2, 2
Answer:
(b) 3,3

Question 18.
The random variable X has the probability density function $f(x)=\left\{\begin{array}{cc}
a x+b, & 0<x<1 \\
0, & \text { otherwise }
\end{array}\right.$ and E(X) = —, then a and b are respectively ………….
(a) 1 and 1/2
(b) 1/2 and 1
(c) 2 and 1
(d) 1 and 2
Answer:
(a) 1 and 1/2

Question 19.
If the function f(x) = 1/12 for a < x < b represents a probability density function of a continuous random variable X, then which of the following cannot be the value of a and b?
(a) 0 and 12
(b) 5 and 17
(c) 7 and 19
(d) 16 and 24
Answer:
(d) 16 and 24

Question 20.
Which one of the following is not true?
(a) Negation of a negation of a statement is the statement itself.
(b) If the last column of the truth table contains only T then it is a tautology.
(c) If the last column of its truth table contains only F then it is a contradiction
(d) If p and q are any two statements then p ↔ q is a tautology.
Answer:
(d) If p and q are any two statements then p ↔ q is a tautology.

Part – II

II. Answer any seven questions. Question No.30 is compulsory [7 x 2 = 14]

Question 21.
If adj [A] = $\left[\begin{array}{ccc}
0 & -2 & 0 \\
6 & 2 & -6 \\
-3 & 0 & 6
\end{array}\right]$ then find A^-1
Answer:

Question 22.
Obtain the condition that the roots of x³ + px² + qx + r = 0 are in A.P.
Answer:
Let the roots be in A.R Then, we can assume them in the form α – d, α, α + d. Applying the Vietas formula (α – d) + α + (α + d) = -y = $-\frac{p}{1}$ => 3α = -p => α = $-\frac{p}{3}$
But, we note that a is a root of the given equation. Therefore, we get

Question 23.
Show that $\cot ^{-1}\left(\frac{1}{\sqrt{x^{2}-1}}\right)=\sec ^{-1} x$ |x| > 1
Answer:

We construct a right triangle with the given data.
From the triangle, sec α = $\frac{x}{1}$ = x. Thus, α = sec^-1 x.

Question 24.
Evaluate $\lim _{x \rightarrow 1} x^{\frac{1}{x-1}}$
Answer:

Question 25.
The radius of a circular disc is given as 24 cm with a maximum error in measurement of 0.02 cm. (i) Use differentials to estimate the maximum error in the calculated area of the disc, (ii) Compute the relative error.
Answer:
r = 24 cm, dr = 0;02 cm

Question 26.
Evaluate $\int_{-\log 2}^{\log 2} e^{-|x|} d x$
Answer:

Question 27.
Show that the following expressions is a solution of the corresponding given differential equation.
y = ae^x + be^-x ; y” – y = 0
Answer:
y = ae^x + be^-x …(1) Differential equation: y”- y = 0
Differentiate with respect to ‘x’
y’ = ae^x – be^-x
Again differentiate with respect to ‘x’
y”= ae^x + be^-x
y” = y ⇒ y” – y = 0
∴ (1) is the solution of the given differential equation.

Question 28.
Suppose X is the number of tails occurred when three fair coins are tossed once simultaneously. Find the values of the random variable X and number of points in its inverse images.
Answer:
Let X is the random variable denotes the number of tails when three coins are tossed simultaneously.
Sample space S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
∴ ‘X’ takes the values 0, 1, 2, 3
i.e., X (HHH) = 0 ; X (HHT, HTH, THH) = 1 ; X (HTT, THT, TTH) = 2 ; X (TTT) = 3

Question 29.
Verify the (i) closure property (it) commutative property (iii) associative property ‘ (iv) existence of identity and (v) existence of inverse for the arithmetic operation + on Z_o = the set of all odd integers.
Answer:
Consider the set Z₀ of all odd integers Z₀ = {2k + 1 : k ∈ Z} = 5,—3,-1, 1 ,3 ,5, …}. + is not a binary operation on Z₀ because when x = 2m + 1, y = 2n + 1,
x + y = 2(m + n) + 2 is even for all m and n. For instance, consider the two odd numbers 3,7 ∈ Z₀. Their sum 3 + 7 = 10 is an even number. In general, if x, y ∈ Z 0, then (x + y)∉ Z ₀
Other properties need not be checked as it is not a binary operation.

Question 30.
Simplify i¹⁹⁴⁸ – i¹⁸⁶⁹
Answer:

Part – III

III. Answer any seven questions. Question No. 40 is compulsory. [7 x 3 = 21]

Question 31.
Find the inverse of the matrix $\left[\begin{array}{rrr}
2 & -1 & 3 \\
-5 & 3 & 1 \\
-3 & 2 & 3
\end{array}\right]$
Answer:

Question 32.
Find the value of $\left(\frac{1+\sin \frac{\pi}{10}+i \cos \frac{\pi}{10}}{1+\sin \frac{\pi}{10}-i \cos \frac{\pi}{10}}\right)^{10}$
Answer:

Question 33.
Foci (0, ± 4) and end points of major axis are(0, ± 5). Find the equation of the ellipse.
Answer:
From the diagram we see that major axis is along y-axis.
Also a = 5 and ae = 4
⇒ 5e = 4 ⇒e = 4/5
Now a = 5 ⇒ a² = 25
ae = 4 ⇒ ae² =16
So b²= a²(1 – e²) = a²– a²e² = 25 – 16 = 9

Question 34.
Show that the two lines $\vec{r}=(\vec{i}-\vec{j})+t(2 \vec{i}+\vec{k}) \text { and } \vec{r}=(2 \vec{i}-\vec{j})+s(\vec{i}+\vec{j}-\vec{k})$ lines and find the distance between them.
Answer:

Question 35.
Find the local maximum and minimum of the function x² y² on the line x+y = 10.
Answer:
Let the given function be written as f(x) = x²(10 – x)². Now,
f(x) = x² (100 – 20x + x²) = x⁴ – 20x³ + 100x²
Therefore, f’ (x) = 4x³ – 60x² + 200x = 4x (x² – 15x + 50)
f'(x) = 4x(x² – 15x + 50) = 0 ⇒ x = 0, 5, 10
and f ” (x) = 12x² – 120x + 200

The stationary points of /(x) are x = 0, 5, 10 at these points the values of f” (x) are respectively 200, -100 and 200. At x = 0, it has local minimum and its value is f(0) = 0. At x = 5, it has local maximum and its value is f(5) = 625. At x =10, it has local minimum and its value is f(10) = 0.

Question 36.
Let U(x, y, z) = x² – xy + 3sin z, x, y, z ∈ R. Find the linear approximation for U at (2,-1,0).
Answer:
Linear approximation is given by

Now U_x = 2x – y, U_y = – x and U_z = 3cos z.
Here (x₀,y₀,z₀) = (2, -1,0), hence U_x(2, -1,0) = 5, U_y(2, -1,0) = -2 and U_z (2, -1,0) = 3.
Thus L(x, y, z) = 6 + 5(x – 2) – 2(y + 1) + 3(z – 0) = 5x – 2y + 3z – 6 is the required linear approximation for U at (2,-1,0).

Question 37.
Find the volume of a sphere of radius a.
Answer:
By revolving the upper semicircular region enclosed between the circle x² + y² = a² and the x-axis, we get a sphere of radius a.
The boundaries of the region are y = $\sqrt{a^{2}-x^{2}}$ x-axis x = -a and x= a Hence, the volume of the sphere is given by

Question 38.
Solve the following differential equations: ydx + (1 + x²) tan^-1 x dy = 0
Answer:
ydx + (1 + x²) tan^-1 x dy =0
(1 + x²) tan^-1 x dy= -y dx
Separating the variables

log y = – log (tan^-1x) + log c
log y + log (tan^-1 x) = log c
log (y tan^-1 x) = log c
y tan^-1 x = c

Question 39.
Prove that $\boldsymbol{q} \rightarrow \boldsymbol{p} \equiv \neg \boldsymbol{p} \rightarrow \neg \boldsymbol{q}$
Answer:
Truth table for q → p

Truth table for $\neg p \rightarrow \neg q$

The entries in the column corresponding to q → p and $\neg p \rightarrow \neg q$ are identical and hence they are equivalent.

Question 40.
If on an average 1 ship out of 10 do not arrive safely to ports. Find the mean and the standard deviation of the ships returning safely out of a total of 500 ships.
Answer:
Probability of a ship arriving safely

Part – IV

IV. Answer all the questions. [7 x 5 = 35]

Question 41.
(a) If F(α) = $\left[\begin{array}{ccc}
\cos \alpha & 0 & \sin \alpha \\
0 & 1 & 0 \\
-\sin \alpha & 0 & \cos \alpha
\end{array}\right]$ show that F[(α)]^-1 = F(-α)
Answer:
Let A = F(α)
So F[(α)]^-1 = F(-α)

Expanding the determinant – along R2 We get
-0() + 1 [cos²α + sin²α] – 0() = 1 ≠ 0
So A^-1 exists

[OR]

(b) If ω ≠ 1 is a cube root of unity, show that the roots of the equation (z – 1)³ + 8 = 0 are -1,1- 2ω , 1- 2ω²
Answer:
(z- 1)³ + 8 = 0 ⇒ (z – 1)³ = -8

Question 42.
(a) Solve: $(\sqrt{3}+\sqrt{2})^{x}+(\sqrt{3}-\sqrt{2})^{x}=10$
Answer:

[OR]

(b) If a₁ a₂, a₃,… a_n is an arithmetic progression with common difference d. Prove that

Answer:

Question 43.
(a) Show that the linex-y + 4 = 0isa tangent to the ellipse x² + 3y² = 12. Also find the coordinates of the point of contact.
Answer:
The given ellipse is x² + 3y² =12

(ie.,) Here a² = 12 and b² = 4
The given line is x – y + 4 = 0
(ie.,) y = x + 4
Comparing this line with y = mx + c
We get m = 1 and c = 4
The condition for the line y = mx + c

LHS = c² = 4² = 16
RHS: a²m² + b² = 12(1)² + 4=16
LHS = RHS ⇒ The given line is a tangent to the ellipse. Also the point of contact is

(b) Using vector method, prove that cos(α – β) = cos α cos β + sin α sin β
Answer:
Take two points A and B on the unit circle with centre
as origin ‘O’. so $|\overrightarrow{\mathrm{OA}}|=|\overrightarrow{\mathrm{OB}}|=1$
$\mathrm{AO} x=\alpha ; \quad \mathrm{BO} x=\beta \Rightarrow \mathrm{LAOB}=\alpha-\beta$
Let $\vec{i}$ and $\vec{j}$ be the unit vectors along the x, y respectively.
The co-ordinates of A and B be (cos α, sin α) and (cos β, sin β) respectively.

Question 44.
(a) A farmer plans to fence a rectangular pasture adjacent to a river. The pasture must contain 1,80,000 sq. mtrs in order to provide enough grass for herds. No fencing is needed along the river. What is the length of the minimum needed fencing material?
Answer:
Given Area = 180000 sqm
Let length be x and breadth be y
Now xy = 180000
⇒ y = $ \frac{180000}{x}$
Now perimeter = 2x + y(∵one side is along ther river)

at x = 300 m, p is positive ⇒ x = 300 is a minimum point
180000
when x = 300 m, y = $ \frac{180000}{300}$ = 600 m
So minimum perimeter = 2x + y
= 2(300) + 600 = 1200 m

[OR]

(b) If u = log (tan x + tan y + tan z), prove that $\Sigma \sin 2 x \frac{\partial u}{\partial x}=2$
Answer:

Question 45.
(a) Evaluate as the limit of sums: $\int_{1}^{2}\left(x^{2}-1\right) d x$
Answer:
Let f(x) = x² – 1 for 1≤ x ≤ 2
We divide the interval [1,2] into n equal sub-intervals each of length h. We have a = 1, b = 2

Here,f(a) = f(1) = (1)² – 1 = 0
f(a + h) = f(1 + h) = (1 + h)² – 1 = 1 + h² + 2h – 1 = h² + 2h
f(a + 2h) = f(1 + 2h) = (1 +2h)² = 1 + 4h² + 4h – 1 = 4h² + 4h
f[a + (n -1) h] = f[1 + (n -1) A] = [1 + (n -1) h]² – 1
= 1 + (n – 1)² h² + 2(n – 1) h – 1 = (n – 1)² h² + 2(n – 1) h²

(b) Solve $\frac{d y}{d x}=\frac{x-y+5}{2(x-y)+7}$
Answer:

Integrating both sides, we get
2z + 3 log |z + 2| =x + c
That is, 2(x – y) + 3 log |x – y + 2| = x + c

Question 46.
(a) For the random variable X with the given probability mass function as below, find the mean and variance.
$f(x)=\left\{\begin{array}{ll}
2(x-1), & 1<x<<2 \\
0 & , \quad \text { otherwise }
\end{array}\right.$
Answer:

(b) Show that the set { [1], [3], [4], [5], [9]} under multiplication modulo 11 satisfies closure, associative, identity and inverse properties.
Answer:
G = {[1], [3], [4], [5], [9]}
* is defined by multiplication modulo 11.
To prove G is an abelian group with respect to *
Since we are given a finite number of elements i.e., since the given set is finite, we can frame the multiplication table called Cayley’s table.
The Cayley’s table is as follows:

G₁: The elements in the above table are [1], [3], [4], [5] and [9] which are elements of G.
.’. closure axiom is verified.
G₂ : Consider [3], [4], [5] which are elements of G.
{[3]*[4]}*[5] = [1]*[5] = [5] ……….(1)
[3] *{[4]*[5]} = [3]*[9] = [5] ……….(2)
(1) = (2) ⇒ (a * b) * c = a * (b * c) i.e., associative axiom is verified.
G₃ : The first row elements are the same as that of the given elements in the same order.
i.e., from the table, the identity element is [1] ∈ G. So identity axiom is verified.

inverse axiom is verified.
G₅ : From the table * is commutative i.e., the entries equidistant from the leading diagonal on either sides are equal ⇒ a * b = b * a

Question 47.
(a) For the ellipse 4x² + y² + 24x – 2y + 21 = 0, find the centre, vertices, and the foci. Also prove that the length of latus rectum is 2.
Answer:
Rearranging the terms, the equation of ellipse is ,
4x² + 24x + y² – 2y + 21 = 0
That is, 4(x² + 6x + 9 – 9) + (y² – 2y + 1 – 1) + 21 =0
4 (x +3)² – 36 + (y-1)² – 1 + 21 = 0
4 (0+ 3)² + (y – 1)² = 16

Centre is (-3,1) a = 4, b = 2, and the major axis is parallel to y -axis
c² = 16 – 4 = 12
c = ± 2 √3
Therefore, the foci are (-3, 2√ +1) and (-3, -2√3 + 1).
Vertices are (1,± 4 +1). That is the vertices are (1, 5) and (1, -3), and
The length of Latus rectum = $\frac{2 b^{2}}{a}$ = 2 units

[OR]

(b) If $\vec{a}=-2 \hat{i}+3 \hat{j}-2 \hat{k}, \vec{b}=3 \hat{i}-\hat{j}+3 \hat{k}, \vec{c}=2 \hat{i}-5 \hat{j}+\hat{k}$ find $(\vec{a} \times \vec{b}) \times \vec{c}$ and $\vec{a} \times(\vec{b} \times \vec{c})$
State whether they are equal.
Answer:

Tamil Nadu 12th Commerce Model Question Paper 3 English Medium

Leave a Comment / Class 12 / By Prasanna

Students can Download Tamil Nadu 12th Commerce Model Question Paper 3 English Medium Pdf, Tamil Nadu 12th Commerce Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 12th Commerce Model Question Paper 3 English Medium

Instructions:

The question paper comprises of four parts.
You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
All questions of Part I, II. III and IV are to be attempted separately.
Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about 50 words.
Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in about 150 words.
Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in about 250 words. Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 90

Part – I

Choose the correct answer. Answer all the questions: [20 x 1 = 20]

Question 1.
Scientific management is developed by ……………….
(a) Fayol
(b) Taylor
(c) Mayo
(d) Jacob
Answer:
(b) Taylor

Question 2
keeps Management Alert to Opportunities and Threats by Identifying Critical Problems.
(a) MBA
(b) MBE
(c) MBM
(d) MBO
Answer:
(b) MBE

Question 3.
Primary market is a Market where securities are traded in the …………..
(a) First time
(b) second time
(c) Three time
(d) Several time
Answer:
(a) First time

Question 4.
There are stock exchange in the country.
(a) 21
(b) 24
(c) 20
(d) 25
Answer:
(a) 21

Question 5.
Human Resource management is both and ……………
(a) science and art
(b) theory and practice
(c) history and geography
(d) none of the above
Answer:
(a) science and art

Question 6.
A poor quality of selection will mean extra cost on ………… and supervision.
(a) training
(b) recruitment
(c) work quality
(d) none of these
Answer:
(a) training

Question 7.
The marketer initially wants to know in the marketing is ………..
(a) qualification of the customers
(b) quality of the product
(c) background of the customers
(d) needs of the customers
Answer:
(d) needs of the customers

Question 8.
Which is gateway to internet?
(a) portal
(b) CPU
(c) modem
(d) webnaire
Answer:
(c) modem

Question 9.
Sale of Goods Act was passed in the year?
(a) 1962
(b) 1972
(c) 1982
(d) 1985
Answer:
(c) 1982

Question 10.
The chairman of the state consumer protection council is ……………
(a) Judge of a High Court
(b) Chief minister
(c) Finance minister
(d) None of the above
Answer:
(a) Judge of a High Court

Question 11
………… means permitting the private sector to set up industries which were previously
reserved for public sector.
(a) Liberalisation
(b) privatisation
(c) Globalisation
(d) Public enterprise
Answer:
(b) privatisation

Question 12.
Which of the below is not a good?
(a) stocks
(b) dividend due
(c) crops
(d) water
Answer:
(b) dividend due

Question 13.
The transferee of a Negotiable Instrument is the one
(a) Who transfer the instrument
(b) On whose name it is transferred
(c) Who enchases it
(d) None of the above
Answer:
(b) On whose name it is transferred

Question 14.
Entrepreneur is not classified as …………
(a) Risk bearer
(b) Innovator
(c) Employee
(d) Organizer
Answer:
(b) Innovator

Question 15.
Which of the following is the activity of a business entrepreneur?
(a) production
(b) marketing
(c) operation
(d) all of the above
Answer:
(d) all of the above

Question 16.
…………… is designed to transform India to a global design and manufacturing hub.
(a) Digital India
(b) Make in India
(c) Startup India
(d) Design India
Answer:
(b) Make in India

Question 17.
For which type of capital a company pays the prescribed fees at the time of registration?
(a) Subscribed capital
(b) Authorised capital
(c) Paid-up capital
(d) Issued capital
Answer:
(b) Authorised capital

Question 18.
Which ………….. Director is appointed by a Financial institution?
(a) Nominee
(b) Additional
(c) Women
(d) Shadow
Answer:
(a) Nominee

Question 19.
Which meeting will be held only once in the life time of the company?
(a) Statutory
(b) Annual general
(c) Extra ordinary
(d) Class general
Answer:
(a) Statutory

Question 20.
Distribution of work in groupwise or section-wise is called as
(a) co-ordinating
(b) controlling
(c) staffing
(d) organising
Answer:
(d) organising

Part – II

Answer any seven in which question No. 30 is compulsory. [7 x 2 = 14]

Question 21.
Who is a manager?
Answer:
A manager is a dynamic and life giving element in every business. Without efficient management it cannot be possible to secure the best allocation and utilisation of human, material and financial resources.

Question 22.
What is meant by Motivation?
Answer:
The goals are achieved with the help of motivation. Motivation includes increasing the speed of performance of a work and developing a willingness on the part of workers. This is done by a resourceful leader.

Question 23.
Write a note on financial market.
Answer:
A financial market is an institution or arrangement that facilitates the exchange of financial instruments such as equity shares, preference shares, debentures, deposits and loans, corporate stocks and bonds, government bonds, and more exotic instruments such as options and futures contracts.

Question 24.
What is commercial bill market?
Answer:
The Commercial Bill is an instrument drawn by a seller of goods on a buyer of goods. A bill of exchange issued by a commercial organization to raise money for short-term needs. These bills are of 30 days, 60 days and 90 days maturity.

Question 25.
Write any two objectives of SEBI.
Answer:
The first objective of SEBI is to regulate, stock exchanges so that efficient services may be provided to all the parties operating there. Another objective is to supervise/check the activities of the brokers and other middlemen in order to control the capital market.

Question 26.
What is an interview?
Answer:
According to Scott and others “an interview is a purpose full exchange of ideas, the answering of questions and communication between two or more persons.”

Question 27.
What is green marketing?
Answer:
Green marketing involves developing and promoting products and services which satisfy customers’ wants and needs for quality, performance, affordable pricing and convenience – all without causing a detrimental impact on the environment.

Question 28.
Define Consumerism.
Answer:
“Consumerism is an attempt to enhance the rights and powers by buyers in relation to sellers”.

Question 29.
What is internal environment?
Answer:
Internal environment refers to those factors within an organisation e.g. Policies and programmes, organisational structure, employees, financial and physical resources.

Question 30.
Define Bill of Exchange.
Answer:
According to section 5 of the Negotiable Instruments Act, “a bill of exchange is an instrument in writing containing an unconditional order, signed by the maker, directing a certain person to pay a certain sum of money only to, or to the order of a certain person or to the bearer of the instrument”.

Part – III

Answer any seven in which question No.40 is compulsory. [7 x 3 = 21]

Question 31.
Is management an Art or Science?
Answer:
Management is neither a science nor an art, but a combination of both requiring people holding managerial positions to apply the scientific management principles and displaying popular managerial skills to accomplish the organizational goals as efficiently and as quickly as possible so as to be competitive in the globalised environment of business.

Question 32.
Give the meaning and definition of financial market.
Answer:

Meaning: A market wherein financial instruments such as financial claims, assets and securities are traded is known as a ‘financial market’.
Definition: According to Brigham, Eugene F, “The place where people and organizations wanting to borrow money are brought together with those having surplus funds is called a financial market.”

Question 33.
Explain Bull and Bear.
Answer:
Speculators in a stock market are of different types based on animals behaviour. They are:

Bull: A Bull or Tejiwala is an operator who expects a rise in prices of securities in the future.He purchases the securities expecting price of rise in future.
Bear: A bear or Mandiwala speculator expects prices to fall in future and sells securities at present with a view to purchase them at lower price.

Question 34.
What is stress interview?
Answer:
This type of interview is conducted to test the temperament and emotional balance of the candidate interviewed. Interviewer deliberately creates stressful situation and tries to assess the suitability of the candidate by observing his reaction and response to the stressful situations.

Question 35.
Explain the importance of social marketing.
Answer:
The primary aim of social marketing is ‘social good’ such as anti-tobacco, anti-drug, anti¬pollution, anti-dowry, road safety, protection of girl child, against the use of plastic bags. Social marketing promotes the consumption of socially desirable products and develops health consciousness. It helps to eradicate social evils that affect the society and quality of life.

Question 36.
Is Consumer Protection necessary?
Answer:
Consumer is supposed to be the king in the business. Consumer satisfaction is compulsorily necessary. Because the business depends upon the consumer. If there is no consumer, no need of production or sales. So the consumer is to be fully satisfied and respected in the market.

Question 37.
Discuss in detail about existing goods.
Answer:
Existing goods are those owned or possessed by the seller at the time of contract of sale. Goods possessed even refer to sale by agents or by pledgers. Existing goods may be either:

Specific Goods
Ascertained Goods
Generic or Unascertained Goods

Question 38.
Explain the nature of a Negotiable Instrument.
Answer:
A negotiable instrument is transferable from one person to another without any formality, such as affixing stamp, registration, etc. When the instrument is held by holder in due course in the process of negotiation, it is cured of all defects in the instrument with respect to ownership. Though a bill, a promissory note or a cheque represents a debt, the transferee is entitled to sue on the instrument in his own name in case of dishonour, without giving notice to the debtor that he has become its holder.

Question 39.
What is ‘Startup India’?
Answer:
Through the Startup India initiative, Government of India promotes entrepreneurship by mentoring, nurturing and facilitating startups throughout their life cycle.

Question 40.
What do you mean by Statutory Meeting?
Answer:
According to Companies Act, every public company, should hold a meeting of the shareholders within 6 months but not earlier than one month from the date of commencement of business of the company. This is the first general meeting of the public company, which is called the Statutory Meeting. This meeting is conducted only once in the lifetime of the company.

Part – IV

Answer all the following questions. [7 x 5 = 35]

Question 41.
(a) Describe the principles of scientific management.
Principles of scientific management propounded by Taylor are:

(i) Science, Not Rule of Thumb: Rule of Thumb means decisions taken by manager as per their personal judgments. According to Taylor, even a small production activity like loading iron sheets into box cars can be scientifically planned. This will help in saving time as well as human energy. Decisions should be based on scientific enquiry with cause and effect relationships.

(ii) Harmony, Not Discord: Taylor emphasized that there should be complete harmony between the workers and the management since if there is any conflict between the two,it will not be beneficial either for the workers or the management. Both the management and the workers should realize the importance of each other.

(iii) Mental Revolution: The technique of Mental Revolution involves a change in the attitude of workers and management towards each other. Both should realize the importance of each other and should work with full cooperation. Management as well as the workers should aim to increase the profits of the organisation.

(iv) Cooperation, Not Individualism: This principle is an extension of principle of Harmony, not discord’ and lays stress on mutual cooperation between workers and the management. Cooperation, mutual confidence, sense of goodwill should prevail among both, managers as well as workers. The intention is to replace internal competition with cooperation. Both ‘Management’ and ‘Workers’ should realize the importance of each other.

(v) Development of each and every person to his or her greatest efficiency and prosperity: Efficiency of any organisation also depends on the skills and capabilities of its employees to a great extent. Thus, providing training to the workers was considered essential in order to learn the best method developed through the use of scientific approach.

[OR]

(b) What are the advantages of MBE?
Answer:
Management by exception provides the following advantages:

It saves the time of managers because they deal only with exceptional matters. Routine problems are left to subordinates.
It focuses managerial attention on major problems. As a result, there is better utilisation of managerial talents and energy.
It facilitates delegation of authority. Top management concentrates on strategic decisions and operational decisions are left to the lower levels. There is increase in span of control. This leads to motivation and development of subordinates.
It is a technique of separating important information from unimportant one. It forces managers to review past history and study related business data for identifying deviations. There is better use of knowledge of trends, history and available business data.MBE keeps management alert to opportunities and threats by identifying critical problems. It can avoid uninformed and impulsive action.
Management by exception provides better yardsticks for judging results. It is helpful in objective performance appraisal.

Question 42(a).
Define Money Market and Capital Market. Explain the difference between the Money Market and Capital Market.
Answer:

Money Market Definition: According to Crowther, “the money market is the collective name given to the various firms and institutions that deal in the various grades of near money.”
Capital Market Definition: According to Arun K. Datta, capital market may be defined as “a complex of institutions investment and practices with established links between the demand for and supply of different types of capital gains”.

Question 43.
Differences between the Money Market and Capital Market
Answer:

S.No.	Features	Money Market	Capital Market
1.	Duration of Funds	It is a market for short-term loanable funds for a period of not exceeding one year.	It is a market for long-term funds exceeding period of one year.
2.	Supply of Funds	This market supplies funds for financing current business operations working capital requirements of industries and short period requirements of the government.	This market supplies funds for financing the fixed capital requirements of trade and commerce as well as the long-term requirements of the government.
3.	Deals with Instruments	It deals with instruments like commercial bills (bill of exchange, treasury bill, commercial papers etc.).	It deals with instruments like shares, debentures, Government bonds, etc.,
4.	Money Value	Each single money market instrument is of large amount. A treasury bill is of minimum for one lakh. Each certificate of deposits or commercial paper is for minimum of Rs 25 lakh.	Each single capital market instrument is of small amount. Each share value is Rs 10. Each debenture value is Rs 100.
5.	Role of Major Institution	The central bank and commercial banks are the major institutions in the money market.	Development banks and Insurance companies play a dominant role in the capital market.

[OR]

(b) Describe the significance of Human Resource Management.
The role of human resource management is the process of acquiring, training, appraising, and compensating employees. The significance of human resource management is given below:

(i) To identify manpower needs: Determination of manpower needs in an organisation is very important as it is a form of investment.

(ii) To ensure the correct requirement of manpower: At any time the organisation should not suffer from shortage or surplus manpower which is made possible through human resource management.

(iii) To select right man for right job: Human resource management ensures the right talent to select the right employee for the right job.

(iv) To update the skill and knowledge: Human resource management enables employees to remain up-to-date through training and development programmes.

(v) To appraise the performance of employees: Periodical appraisal of performance of employees through human resource management activities boosts up good performers and motivates slow performers.

Question 43.
(a) Define training. Discuss various types of training.
Answer:
According to Edwin B. Flippo “Training is the act of increasing the Knowledge and skills of an employee for doing particular jobs”. Training may be mainly divided into

On the job training
Off the job training

On the Job Training: On the job training refers to the training which is given to the employee at the workplace. The following are the on the job training methods.

Coaching Method: In this method of training, the superior teaches or guides the new employee about the knowledge and skills.
Mentoring Method: Mentoring is the process of sharing knowledge and experience of an employee.
Internship Training Method: A superior gives training to a subordinates or understudy like an assistant to a manager.

(b) Off the Job Training: It is the training method wherein the workers leam the job role away from the actual work place. The following are types of off the job training:

Lecture Method: Under this method trainees are educated about concepts, theories, principles in any particular area.
E-learning Method: E-leaming is the use of technological process to access of a traditional classroom or office.

[OR]

(b) What are the needs for consumer protection?
Answer:
Consumer is to be protected from the cheating business people. Though the consumer is said to be the king, his interests are neglected. Consumer protection is applicable to public sector, financial and co-operative enterprises. Recently even medical services have been brought under consumer movement. Satisfaction and wellbeing of the consumer should be the main objective of business. But in real practice consumer is not protected and safeguarded. Thus there is a need for consumer protection or movement.

Question 44(a).
Describe the economic and socio-cultural environment of business.
Answer:
Economic Environment: The multiple variables in the. macro environment system which has a bearing on a business include:

The nature of economy is based on the stage of development. The countries across the globe can be categorised on the basis of growth and per capita income as developed nations, developing nations and under developed nations.
The economic systems can be classified as Capitalistic, Socialistic and Mixed economy.
The economic policies of a nation are: Monetary policy, Fiscal policy, Industrial policy, Foreign exchange policy and Export-import policy.

Socio-Cultural environment: Social environment refers to the sum total of factors of the society in which the business is located. It is dynamic and includes the behaviour of individuals, the role and importance of family, customs, religion and languages, the ethical values.
[OR]

b. Distinguish between sale and agreement to sell.

Basis for Comparison	Sale	Agreement to Sell
1. Ownership	The property (ownership or title) in the goods passes from the seller to the buyer immediately.	The property (ownership or title) in the goods has to pass at a future time or after the fulfilment of certain conditions.
2. Risk of Loss	Where the goods sold under the contract of sale are destroyed, the loss falls fully on the buyer as the ownership has already passed.	Where the goods under the agreement to sell are destroyed, the loss falls fully on the seller as the ownership is still vested with seller.
3. Consequences of violating the contract	Where the buyer fails to pay the price, the seller cannot seize the goods.	Where the buyer violates the contract, the seller can repossess the goods from the former.
4. Nature of contract	It is an executed contract.	It is an executory contract, i.e., contract yet to be performed.
5. Insolvency of the Buyer	In a sale, if a buyer becomes insolvent before he pays for the goods even though the goods sold are under the possession of the seller, the latter has to return them to the Official Receiver.	If the buyer becomes insolvent before the payment of the price, the seller can retain the goods if they are under his possession.

Question 45.
(a) Distinguish a cheque and a bill of exchange.
Answer:

SI. No.	Basic of Difference	Bill of Exchange	Cheque
1.	Drawn	A bill of exchange can be drawn on any person including a banker.	A cheque can be drawn only on a particular banker.
2.	Payability	It is payable on demand or on the expiry of a certain period.	It is payable on demand only.
3.	Acceptance	In case of time bill, acceptance by the drawee is necessary before he can be made liable on it.	A cheque does not require any acceptance.
4.	Grace period	Three days of grace are allowed while calculating the maturity date in the case of time bill.	No days of grace are allowed in the case of a cheque for the simple reason that is always. payable on demand.
5.	Notice	When a bill is dishonoured, notice of dishonour is necessary.	Notice is not necessary for a cheque.
6.	Discounting	A bill can be discounted with a bank.	A cheque cannot be discounted.
7.	Crossing	A bill cannot be crossed.	A cheque can be crossed either generally or specially so as to ensure payment to the rightful owner.

[OR]
(b) What are the characteristics of an entrepreneur?
Characteristics of an Entrepreneur:
(i) Spirit of Enterprise: Entrepreneur should be bold enough to encounter risk arising from the venture undertaken.

(ii) Self-confidence: Entrepreneur should have self-confidence in order to achieve high goals ‘ in the business.

(iii) Flexibility: Entrepreneur should not doggedly stick to decisions in a rigid fashion.

(iv) Innovation: Entrepreneur should contribute something new or something unique to meet the changing requirements of customers namely new product, new method of production or distribution, adding new features to the existing product, uncovering a new territory for business, innovating new raw material, etc.

(v) Resource Mobilisation: Entrepreneur should have the capability to mobilise both tangible inputs like manpower, money materials, technology, market, method etc., which are scattered over a wide area and certain intangible inputs like motivation, morale and innovativeness.

Question 46.
(a) Discuss the nature of functional entrepreneurs.
Answer:
Nature of functional entrepreneurs:

Innovative Entrepreneur: He is a person who introduces new project. They observe the environment regarding information to the new venture.
Imitative Entrepreneur: He refers to the person who simply imitates existing knowledge or technology already in advance countries. Redesigning of products suited to the local conditions.
Fabian Entrepreneur: These are said to be conservatives and sceptical about any changes in their organisation. They are of risk-averse.
Drone Entrepreneur: Drone entrepreneurs are those who are totally opposed to changes in the environment. They used to operate in the niche market.

[OR]
(b) Brief different stages in Formation of a Company.
Section 3(1) of the Act states that a company may be formed for any lawful purpose by – (a) seven or more persons, where the company to be formed is to be a public company;
(b) two or more persons, where the company to be formed is to be a private company; or
(c) one person, where the company to be formed is to be One Person Company.

The process of formation of company consists of different stages:

1. Promotion: Promotion stage begins when the idea to form a company comes in the mind of a person.

2. Registration: The second stage in the formation of the company is incorporation or registration. In this stage, the promoter has to fix name of the company, prepare the necessary documents (Memorandum and Articles of Association), fix the registered
office, and name of the directors. After this, certificate of incorporation is issued.

3. Capital Subscription: A public limited company having its share capital has to pass through two stages. One of them is capital subscription.

The steps for this is:

Formalities for raising capital.00 Issuing prospectus
Appointing official banker
Pass resolution to make allotment

4. Commencement of Business: As per section 11 of the Act, a company having share capital should file with the Registrar, declaration stating that

Every subscriber has paid the value of shares.
Paid-up capital is not less than Rs.5 lakhs for a public limited company and Rs.l lakh in case of a private limited company.
It has filed the Registrar, regarding the verification of registered office.
After fulfilling these details, the Registrar will issue certificate of commencement of business.

Question 47.
(a) Explain composition of the board of directors.
Answer:
Composition of the Board off Directors:
(i) General Optimum Combination: Board of Directors shall have an optimum combination of executive and non-executive directors with at least one woman director.

(ii) When the non-executive Director is the Chairperson: In this case, at least one-third of the board of directors shall comprise of independent directors.

(iii) When the non-executive chairperson is a promoter or is related to any promoter or person occupying management positions at the level of board of director or at one level below the Board of Directors: In this case, at least one half of the board of directors of the company shall consist of Independent Directors.

[OR]

(b) Briefly state different types of company meetings.
A Company can convene meetings to discuss the performance of the company and also to take decisions. Under the companies Act 2013, company meetings may be classified as follows:

(i) Meetings of Shareholders: The meeting held for the shareholders of the company is shareholders meeting. This may be divided as follows:

Statutory Meeting: Every public company should hold a meeting of the shareholders within 6 months but not earlier than one month from the commencement of the business.
Annual General Meeting: Every year a meeting is held to transact the ordinary business of the company. It is called annual general meeting.
Extra-Ordinary General Meeting: If any meeting conducted in between two annual general meetings to deal with some urgent or special or extraordinary nature of business is called as extra-ordinary general meeting.

(ii) Meeting of the Board of Directors: To decide policy matters of the company, the board of directors meet frequently, which is known as meeting of the board of directors.

Board Meetings: Meetings of the directors are called board meetings. It may be convened to discuss the business and take formal decisions.
Committee Meetings: Every listed company and every other public company having a capital of? 10 crore is required to have audit committee. The meeting held by this committee is known as committee meetings.

(iii) Special Meeting:

Class Meeting: Meetings held by a particular class of share or debenture holders is known as Special or Class meeting, e.g. preference shareholders or debenture holders meeting.
Meetings of the Creditors: These are not meetings of a company. Meetings held with the creditors to discuss any crisis about the financial matters.

Tamil Nadu 12th Biology Model Question Paper 5 English Medium

Leave a Comment / Class 12 / By Prasanna

Students can Download Tamil Nadu 12th Biology Model Question Paper 5 English Medium Pdf, Tamil Nadu 12th Biology Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 12th Biology Model Question Paper 5 English Medium

General Instructions:

The question paper comprises of four parts. Questions for Botany and Zoology are asked separately.
You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
All questions of Part I, II, III and IV are to be attempted separately.
Question numbers 1 to 8 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
Question numbers 9 to 14 in Part II are two-marks questions. These are to be answered in about one or two sentences.
Question numbers 15 to 19 in Part III are three-marks questions. These are to be answered in about three to five short sentences.
Question numbers 20 and 21 in Part IV are five-marks questions. These are to be answered in detail. Draw diagrams wherever necessary.

Time: 2.30 Hours
Maximum Marks: 70

Bio-Botany [Maximum Marks: 35]

Part – I

Answer all the questions. [8 × 1 = 8]
Choose the most suitable answer from the given four alternatives and write the option code with the corresponding answer.

Question 1.
Size of pollen grain in Myosotis is _______ micrometer.
(a) 10
(b) 20
(c) 200
(d) 2000
Answer:
(a) 10

Question 2.
How many different kinds of gametes will be produced by a plant having the genotype AABbCC?
(a) Three
(b) Four
(c) Nine
(d) Two
Answer:
(d) Two

Question 3.
EcoRI cleaves DNA at ________.
(a) AGGGTT
(b) GTATAC
(c) GAATTC
(d) TATAGC
Answer:
(c) GAATTC

Question 4.
Choose the correct match from the following.

Answer:
1 – C, 2 – A, 3 – D, 4 – B

Question 5.
Photo synthetically active radiation wave length between the range of ________.
(a) 200-700nm
(b) 300-700nm
(c) 400-700 nm
(d) 500-700nm
Answer:
(c) 400-700 nm

Question 6.
For the given statements with respect to Eichhomia select the suitable answer.
Statement A: It drains of oxygen from water and is seen growing in standing water.
Statement B: It is an indogeneous species of our country.
(a) Statement A is correct and statement B is wrong.
(b) Both the statement A and B are correct.
(c) Statement A is wrong and statement B is correct.
(d) Both the statement A and Bare wrong.
Answer:
(d) Both the statement A and Bare wrong.

Question 7.
Which of the following is incorrectly paired?
(a) Wheat – Himgiri
(b) Milch breed – Sahiwal
(c) Rice – Ratna
(d) Pusa Komal – Brassica
Answer:
(d) Pusa Komal – Brassica

Question 8.
Find out the correctly matched pair:
(a) Rubber – Shora robusta
(b) Dye – Lawsonia inermis
(c) Timber – Cypress papyrus
(d) Pulp – Hevea brasiliensis
Answer:
(b) Dye – Lawsonia inermis

Part – II

Answer any four from the following questions. [4 × 2 = 8]

Question 9.
The endosperm of Angiosperm is different from Gymnosperm. Do you agree? Justify your answer.
Answer:
Yes, the endosperm of Angiosperm is different from Gymnosperm because in Angiosperm, the endosperm develops from primary endosperm nucleus which is formed after double fertilization whereas in Gymnosperms the endosperm is developed before the fertilization process.

Question 10.
Who are rediscovered Mendel’s experimental works?
Answer:
Hugo de Vries of Holland, Carl Corren’s of Germany, Erich von Tschermak of Austria.

Question 11.
Multiple Alleles-Define.
Answer:
Two or more alternate forms of single gene occupying the same locus in a chromosome are called as Multiple Alleles.

Question 12.
Differentiate Somaclonal variation, Gametoclonal variation.
Answer:

Somaclonal variation	Gametoclonal variation
Somatic variations found in plants is regenerated in vitro (i.e. variations found in leaf, stem, root, tuber or propagule).	Gametophytic variations found in plants is regenerated in vitro gametic origin (i.e. variations found in gametes and gametophytes)

Question 13.
What is called Phytoremediation?
Answer:
The process in which the plants can be used to remove toxic substances like cadmium from contaminated soils is known as Phytoremediation.

Soya bean and tomato tolerate cadmium poisoning by isolating cadmium and storing in few group of cells and prevent from affecting other cells.

Question 14.
What is ozone hole?
Answer:
The decline in the thickness of the ozone layer over restricted area is called ozone hole.

Part – III

Answer three from the following questions. Question No. 19 is compulsory. [3 × 3 = 9]

Question 15.
Differentiate continuous variation with discontinuous variations.
Answer:
1. Discontinuous Variation: Within a population there are some characteristics which show a limited form of variation. Example: Style length in Primula, plant height of garden pea. In discontinuous variation, the characteristics are controlled by one or two major genes which may have two or more allelic forms. These variations are genetically determined by inheritance factors.

Individuals produced by this variation show differences without any intermediate form between them and there is no overlapping between the two phenotypes. The phenotypic expression is unaffected by environmental conditions. This is also called as qualitative inheritance.

2. Continuous Variation: This variation may be due to the combining effects of environmental and genetic factors. In a population most of the characteristics exhibit a complete gradation, from one extreme to the other without any break.

Inheritance of phenotype is determined by the combined effects of many genes, (polygenes) and environmental factors. This is also known as quantitative inheritance. Example: Human height and skin color.

Question 16.
What is gene mapping? Mention its uses.
Answer:
Genetic Map: The diagrammatic representation of the portion of genes and related distance between the adjacent genes in called Genetic mapping.
Use: Gene map provides clues about the where the genes lies on that chromosome.

Question 17.
Lichen is considered as a good example of an obligate mutualism. Explain.
Answer:
Lichen is considered as a good example of obligate mutualism, because in Lichen, the algae (cyano bacteria) and fungi lives in a mutual relationship i.e., the algae being a autotroph synthesizes its food and provides the fungus and in turn the fungus helps the algae in absorption of water and nutrients and also protects form bacterial or other fungal infections.

Question 18.
Draw a pyramid from following details and explain. The quantities of organisms are given: Hawks 25 and plants-500 rabbit and mouse-125+125 snake and lizard- 50+25 respectively.
Answer:

The pyramid represents pyramid of numbers in grassland ecosystem.
The pyramid of numbers in grassland ecosystem is always upright because there is a gradual decrease in the number of organisms in each trophic level form producers to consumers and then to secondary consumers and finally to tertiary consumers.

Question 19.
What do you know about the objectives of plant breeding?
Answer:

To increase yield, vigour and fertility of the crop.
To increase tolerance to environmental condition, salinity, temperature and drought.
To prevent the premature falling of buds and fruits, etc.
To improve synchronous maturity.
To develop resistance to pathogens and pests.
To develop photosensitive and thermos-sensitive varieties.

Part – IV

Answer all questions. [2 × 5 = 10]

Question 20.
(a) With a suitable diagram explain the structure of an ovule.
Answer:

Ovule is also called megasporangium and is protected by one or two covering called integuments.

A mature ovule consists of a stalk and a body.
The stalk or the funiculus is present at the base and it attaches the ovule to the placenta.
The point of attachment of fiinicle to the body of the ovule is known as hilum. It is a junction between ovule and fiinicle.
In an inverted ovule, the funicle is adnate to the body of the ovule forming a ridge called raphe.
The body of the ovule has a central mass of parenchymatous tissue called nucellus which has large reserve food materials.
The nucellus is covered by integuments completely except at the top and forms a pore called micropyle.
The ovule with one or two integuments are called unitegmic or bitegmic ovules respectively.
The basal of the ovule where the nucellus, the integument and the funicle meet is called chalaza.
Inside the nucellus, toward micropylar end there is a large, oval sack like structure called embryo sac or female gametophyte.
The embryo sac develops from the functional megaspore.
In some species, the inner integument perform the nutritive function for the embryo sac and is called as endothelium or integumentary tapetum (Example : Asteraceae).
Group of cells found at the base of the ovule between chalaza and embryo sac is called Hypostase.
The thick walled cells found above micropylar end and embryo sac is called epistase.
There are two types of ovules based on the position of sporogenous.
(a) Tenuinucellate ovule: Ovules with hypodermal sporogerous cell with single layer of nucellus around it is called tenuinucellate. They have very small nucellus
(b) Crassinucellate ovule: Ovule with subhypodermal sporogenous cell is called crassinucellate ovule. They have large nucellus.
The thick walled cells found above micropylar end and embryo sac is called epistase.

[OR]

(b) What is single cell protein. Mention its application.
Answer:
Single Cell Protein (SCP) are dried cells of microorganisms that are used as protein supplement in foods or animal feeds.

Organism used in SCP production:
Bacteria – Methylophilus methylotrophus, Cellulomonas, Alcaligenes
Fungi – Agaricus campestris , Saccharomyces cerevisiae, Candida utilis
Algae – Spirulina, Chlorella and Chlamydomonas

Substrates for SCP production: Straw, molasses, animal manure, sewage, waste water from potato processing plants with starch.

Uses:

SCP offers an unconventional but possible solution to protein deficiency faced buy human beings.
SCP forms an important food source because of their protein content, carbohydrates, fats, vitamins and minerals.
It is used by Astronauts and Antarctica expedition scientists.
Since SCP can be grown even in sewage water, it also reduces environmental pollution.
250 gms of Methylophilus methylotrophus as its high rate of biomass production and growth can yield 25 tonnes of proteins.

Disadvantage of SCP: Although SCP has high nutritive value due to high content of protein, vitamin, lipid and amino acids, there are doubts on whether it could replace conventional protein source due to high nucleic acid content and slower in digestibility.

Question 21.
(a) Suggest a solution to water crises and explain its advantages.
Answer:
Rainwater harvesting is the accumulation and storage of rain water for reuse in-site rather than allowing it to run off. Rainwater can be collected from rivers, roof tops and the water collected is directed to a deep pit. The water percolates and gets stored in the pit. RWH is a sustainable water management practice implemented not only in urban area but also in agricultural fields, which is an important economical cost effective method for the future.

Environmental benefits of Rain Water Harvesting:

Promotes adequacy of underground water and water conservation.
Mitigates the effect of drought.
Reduces soil erosion as surface run-off is reduced.
Reduces flood hazards.
Improves groundwater quality and water table / decreases salinity.
No land is wasted for storage purpose and no population displacement is involved.
Storing water underground is an eco-friendly measure and a part of sustainable water storage strategy for local communities.

[OR]

(b) What are millets? What are its types? Give example for each type.
Answer:
Millets: (Siru Thaniyangal)

Millets is the term applied to a variety of very small seeds originally cultivated by ancient Africans and Asians.
They are gulten free and have less glycemic index.

Types of Millets:

Major millets
Minor millets.

Example for Major millets:

Finger millets – Ragi (Eleusine coracana)
Sorghum – Sorghum vulgare

Example for Minor millets:

Foxtail millet – (Setaria italica)
Kodo millet – Paspalum scrobiculatum

Bio-Zoology [Maximum Marks: 35]

Part – I

Answer all the questions. [8 × 1 = 8]
Choose the most suitable answer from the given four alternatives and write the option code with the corresponding answer.

Question 1.
Which among the following animals exhibit ovoviviparity?
(a) frog
(b) shark
(c) sheep
(d) hen
Answer:
(b) shark

Question 2.
21^st trisomy is observed in ________.
(a) Down’s syndrome
(b) Patau’s syndrome
(c) Turner’s syndrome
(d) Klinefelter’s syndrome
Answer:
(a) Down’s syndrome

Question 3.
Which among the codon codes for methionine?
(a) AUG
(b) UAA
(c) UUU
(d) AUC
Answer:
(a) AUG

Question 4.
The disease caused by Wuchereria bancrofti is ________.
(a) Malaria
(b) Filariasis
(c) Kala-azar
(d) Sleeping sickness
Answer:
(b) Filariasis

Question 5.
Which of the following pair is correctly matched? [Wrong question]
(a) Trichoderma polysporum – clot buster
(b) Aspergillus niger – butric acid
(c) Clostridium butyricum – cyclosporine A
(d) Streptococcus – citric acid

Question 6.
The first clinical gene therapy was done for the treatment of _______.
(a) SCID
(b) cancer
(c) AIDS
(d) cystic fibrosis
Answer:
(a) SCID

Question 7.
Given below are some examples for population interactions. Which among the following shows mutualism?
(a) Lion and deer
(b) Man and tape worm
(c) Hermit crab and sea-anemone
(d) Sucker fish and shark
Answer:
(c) Hermit crab and sea-anemone

Question 8.
The most abundant green house gas is ______.
(a) Carbon-dioxide
(b) Methane
(c) Sulphur-dioxide
(d) Nitrous oxide
Answer:
(a) Carbon-dioxide

Part – II

Answer any four of the following questions. [4 × 2 = 8]

Question 9.
Describe the structure of the head of a human sperm.
Answer:
The head comprises of two parts namely acrosome and nucleus. Acrosome is a small cap like pointed structure present at the tip of the nucleus and made of Golgi body of the spermatid. It contains hyaluronidase (sperm cysin) a proteolytic enzyme which helps to penetrate the ovum during fertilisation. The nucleus is flat and oval.

Question 10.
Mention any four salient features of Human Genome Project.
Answer:
The main goals of Human Genome Project are as follows:

Identify all the genes (approximately 30000) in human DNA.
Determine the sequence of the three billion chemical base pairs that makeup the human DNA.
To store this information in databases.
Improve tools for data analysis.

Question 11.
Name an opioid drug and its plant source. How does the drug affect Human body?
Answer:
Morphine is an opioid drug obtained from flowers of the poppy plants.
It is used as a strong pain killer during surgery.

Question 12.
How does the transgenic cow Rosie’s milk differ from normal cow’s Milk?
Answer:
Rosie was the first transgenic cow. It produced human protein enriched milk, which contained the human alpha lactalbumin (2.4 gm/litre). This milk was a nutritionally balanced food for infants than the normal milk of cows.

Question 13.
Differentiate natality from mortality.
Answer:

Question 14.
What is ozone depletion? How is it caused?
Answer:
Thinning of the stratospheric ozone layer is known as ozone depletion. Such depletion causes the ‘ozone hole’, resulting in poor screening of the harmful UV rays and increase in incidences of skin cancer. Ozone depression is due to certain gases like CFC’s.

Part – III

Write any three of the following in which question No.19 is compulsory. [3 x 3 = 9]

Question 15.
Explain multiple fission in Plasmodium with a diagram.
Answer:
In Multiple fission of plasmodium, the oocyte or schizont divides into many similar daughter cells simultaneously. Nucleus undergoes repeated mitosis producing many nuclei without the division of cytoplasm. Later the cytoplasm divides and encircles each nucleus forming many daughter cells oocyte undergoes sporogony forming sporozoites. Schizont undergoes schizogony forming merozoites.

Question 16.
Write the causes and the differences between Haemophilia and Thalassemia.
Answer:
Causes for Haemophilia: Haemophilia is also called bleeder’s disease is caused by a recessive X-linked gene. A person with a recessive gene for haemophilia lacks a normal clotting substance in blood, hence minor injuries cause continuous bleeding, leading to death.

Causes for Thalassemia: Thalassemia is an autosomal recessive disorder caused by.gene mutation resulting in excessive destruction of RBC’s due to the formation of abnormal haemoglobin molecules.

Haemophilia	Thalassemia
1. It is a X-linked recessive gene inheritance	1. It is a autosomal recessive gene inheritance
2. It follows criss-cross pattern of inheritance	2. It follows Mendelian pattern of inheritance

Question 17.
Explain the secondary treatment of waste water.
Answer:
Secondary treatment or biological treatment: The primary effluent is passed into large aeration tanks where it is constantly agitated mechanically and air is pumped into it. This allows vigorous growth of useful aerobic microbes into floe (masses of bacteria associated with fungal filaments to form mesh like structures). While growing, these microbes consume the major part of the organic matter in the effluent. This significantly reduces the BOD (Biochemical oxygen demand or Biological oxygen demand).

BOD refers to the amount of the oxygen that would be consumed, if all the organic matter in one litre of water were oxidized by bacteria. The sewage water is treated till the BOD is reduced. The greater the BOD of the waste water more is its polluting potential. Once the BOD of sewage water is reduced significantly, the effluent is then passed into a settling tank where the bacterial “floes” are allowed to sediment. This sediment is called activated sludge. A small part of activated sludge is pumped back into the aeration tank to serve as the inoculum.

The remaining major part of the sludge is pumped into large tanks called anaerobic sludge digesters. Here, the bacteria which grow anaerobically, digest the bacteria and the fungi in the sludge. During this digestion, bacteria produce a mixture of gases such as methane, hydrogen sulphide and CO2.
These gases form biogas and can be used as a source of energy.

Question 18.
Explain the steps involved in the preparation of human insulin artificially.
Answer:
Production of insulin by recombinant DNA technology started in the late 1970s. This technique involved the insertion of human insulin gene on the plasmids of E.coli. The polypeptide chains are synthesized as a precursor called pre-pro insulin, which contains A and B segments linked by a third chain (C) and preceded by a leader sequence. The leader sequence is removed after translation and the C chain is excised, leaving the A and B polypeptide chains.

Question 19.
Differentiate in-situ and ex-situ conservation by sighting examples for each.
Answer:
In-situ conservation:
Conservation of plants and animals in theimatural habitat.
Eg., Sacred Groves
A sacred grove or sacred woods are any grove of trees that are of special religious importance to a particular culture. Sacred groves feature in various cultures throughout the world.

Ex-situ conservation:
Conservation of selected or rare plants or animals in place outside their natural habitat.
E.g., Gene banks
Gene banks are a type of biorepository which preserve genetic materials. Seeds of different genetic strains of commercially important plants can be stored for long periods in seed banks, gametes of threatened species can be preserved in viable and fertile condition for long periods using ervopreservation techniques.

Part – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
(a) Explain any five techniques of Assisted Reproductive Technology (ART).
Answer:
Intra-uterine insemination (IUI): This is a procedure to treat infertile men with low sperm count. The semen is collected either from the husband or from a healthy donor and is introduced into the uterus through the vagina by a catheter after stimulating the ovaries to produce more ova. The sperms swim towards the fallopian tubes to fertilize the egg, resulting in normal pregnancy.

Zygote intra-failopian transfer (ZIFT): As in IVF, the zygote upto 8 blastomere stage is transferred to the fallopian tube by laparoscopy. The zygote continues its natural divisions and migrates towards the uterus where it gets implanted.

Intra uterine transfer (IUT): Embryo with more than 8 blastomeres is inserted into uterus to complete its further development.

Gamete intra-failopian transfer (GIFT): Transfer of an ovum collected from a donor into the fallopian tube. In this the eggs are collected from the ovaries and placed with the sperms in one of the fallopian tubes. The zygote travels toward the uterus and gets implanted in the inner lining of the uterus.

Intra-cytoplasmic sperm injection (ICSI): In this method only one sperm is injected into the focal point of the egg to fertilize. The sperm is carefully injected into the cytoplasm of the egg. Fertilization occurs in 75 – 85% of eggs injected with the sperms. The zygote is allowed to divide to form an 8 celled blastomere and then transferred to the uterus to develop a protective pregnancy.

[OR]

(b) (i) Differentiate divergent and convergent evolution with examples.
(ii) What are the various objections to Darwinism?
Answer:
(i) Divergent Evolution:
Divergent evolution is a result of homology. E.g: The wings of bird and the forelimbs of human both are homologous structures modified according to functions. In birds, it is used for flight and in humans used for writing and other purposes.

Convergent Evolution:
Convergent evolution is a result of analogy. E.g: Root modification in sweet potato, and stem modification in potato are analogous structures both performing same function i.e., storage.

(ii) Some objections raised against Darwinism were,

Darwin failed to explain the mechanism of variation.
Darwinism explains the survival of the fittest but not the arrival of the fittest.
He focused on small fluctuating variations that are mostly non-heritable.
He did not distinguish between somatic and germinal variations.
He could not explain the occurrence of vestigial organs, over specialization of some organs like large tusks in extinct mammoths and oversized antlers in the extinct Irish deer, etc.

Question 21.
(a) Explain the life cycle of Plasmodium.
Answer:
Plasmodium vivax is a digenic parasite, involving two hosts, man as the secondary host and female Anopheles mosquito as the primary host. The life cycle of Plasmodium involves three phases namely schizogony, gamogony and sporogony.

The parasite first enters the human blood stream through the bite of an infected female Anopheles mosquito. As it feeds, the mosquito injects the saliva containing the sporozoites. The sporozoite within the blood stream immediately enters the hepatic cells of the liver. Further in the liver they undergo multiple asexual fission (schizogony) and produce merozoites. After being released from liver cells, the merozoites penetrate the RBC’s.

Inside the RBC, the merozoite begins to develop as unicellular trophozoites. The trophozoite grows in size and a central vacuole develops pushing them to one side of cytoplasm and becomes the signet ring stage. The trophozoite nucleus then divides asexually to produce the schizont. The large schizont shows yellowish – brown pigmented granules called Schuffners granules. The schizont divides and produces mononucleated merozoites.

Eventually the erythrocyte lyses, releasing the merozoites and haemozoin toxin into the blood stream to infect other erythrocytes. Lysis of red blood cells results in cycles of fever and other symptoms. This erythrocytic stage is cyclic and repeats itself approximately every 48 to 72 hours or longer depending on the species of Plasmodium involved. The sudden release of merozoites triggers an attack on the RBCs.

Occasionally, merozoites differentiate into macro gametocytes and micro gametocytes. When these are ingested by a mosquito, they develop into male and female gametes, respectively. In the mosquito’s gut, the infected erythrocytes lyse and male and female gametes fertilize to form a diploid zygote called ookinete. The ookinete migrates to the mosquito’s gut wall and develop into an oocyte.

The oocyte undergoes meiosis by a process called sporogony to form sporozoites. These sporozoites migrate to the salivary glands of the mosquito. The cycle is now completed and when the mosquito bites another human host, the sporozoites are injected and the cycle begins a new.

The pathological changes caused by malaria, affects not only the erythrocytes but also the spleen and other visceral organs. Incubation period of malaria is about 12 days. The early symptoms of malaria are headache, nausea and muscular pain. The classic symptoms first develop with the synchronized release of merozoites, haemozoin toxin and erythrocyte debris into the blood stream resulting in malarial paroxysms – shivering chills, high fever followed by sweating. Fever and chills are caused partly by malarial toxins that induce macrophages to release tumour-necrosis factor (TNF-α) and interleukin.

[OR]

(b) Mention the reasons for the richness of biodiversity in the tropics.
Answer:
The reasons for the richness of biodiversity in the Tropics are,

Warm tropical regions between the tropic of Cancer and Capricorn on either side of equator possess congenial habitats for living organisms.
Environmental conditions of the tropics are favourable not only for speciation but also for supporting both variety and number of organisms.
The temperatures vary between 25°C to 35°C, a range in which most metabolic activities of living organisms occur with ease and efficiency.
The average rainfall is often more than 200 mm per year.
Climate, seasons, temperature, humidity, photo periods are more or less stable and encourage both variety and numbers.
Rich resource and nutrient availability.

Samacheer Kalvi 12th Computer Science Solutions Chapter 16 Data Visualization Using Pyplot: Line Chart, Pie Chart and Bar Chart

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Students can Download Computer Science Chapter 16 Data Visualization Using Pyplot: Line Chart, Pie Chart and Bar Chart Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Computer Science Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Computer Science Solutions Chapter 16 Data Visualization Using Pyplot: Line Chart, Pie Chart and Bar Chart

Samacheer Kalvi 12th Computer Science Data Visualization Using Pyplot: Line Chart, Pie Chart and Bar ChartText Book Back Questions and Answers

PART – I
I. Choose The Best Answer

Question 1.
Which is a python package used for 2D graphics?
(a) matplotlib.pyplot
(b) matplotlib.pip
(c) matplotlib.numpy
(d) matplotlib.plt
Answer:
(a) matplotlib.pyplot

Question 2.
Identify the package manager for Python packages, or modules.
(a) Matplotlib
(b) PIP
(c) plt.show( )
(d) python package
Answer:
(b) PIP

Question 3.
Read the following code: Identify the purpose of this code and choose the right option from the following.
C:\Users\YourName\AppData\Local\Programs\Python\Python36-32\Scripts>pip -version
(a) Check if PIP is Installed
(b) Install PIP
(c) Download a Package
(d) Check PIP version
Answer:
(d) Check PIP version

Question 4.
Read the following code: Identify the purpose of this code and choose the right option from the following.
C:\Users\Your Name\AppData\Local\Programs\Python\Python36-32\Scripts>pip list
(a) List installed packages
(b) list command
(c) Install PIP
(d) packages installed
Answer:
(a) List installed packages

Question 5.
To install matplotlib, the following function will be typed in your command prompt. Python -m pip install -U pip What does “-U”represents?
(a) downloading pip to the latest version
(b) upgrading pip to the latest version
(c) removing pip
(d) upgrading matplotlib to the latest version
Answer:
(b) upgrading pip to the latest version

Question 6.
Observe the output figure. Identify the coding for obtaining this output.

(a) import matplotlib.pyplot as pit
plt.plot( [1,2,3], [4,5,1 ])
plt.show( )

(b) import matplotlib.pyplot as pit plt.plot([l,2],[4,5])
plt.show( )

(d) import matplotlib.pyplot as pit
plt.plot([l,3],[4,l])
plt.show( )
Answer:
(a) import matplotlib.pyplot as pit
plt.plot( [1,2,3], [4,5,1 ])
plt.show( )

Question 7.
Read the code:
(a) import matplotlib.pyplot as pit
(b) plt.plot(3,2)
(c) plt.show( )
Identify the output for the above coding
(a)

(b)

(c)

(d)

Answer:
(c)

Question 8.
Which key is used to run the module?
(a) F6
(b) F4
(c) F3
(d) F5
Answer:
(d) F5

Question 9.
Identify the right type of chart using the following hints.
Hint 1: This chart is often used to visualize a trend in data over intervals of time.
Hint 2: The line in this type of chart is often drawn chronologically.
(a) Line chart
(b) Bar chart
(c) Pie chart
(d) Scatter plot
Answer:
(a) Line chart

Question 10.
Read the statements given below. Identify the right option from the following for pie chart.
Statement A: To make a pie chart with Matplotlib, we can use the plt.pie( ) function.
Statement B: The autopct parameter allows us to display the percentage value using the Python string formatting.
(a) Statement A is correct
(b) Statement B is correct
(c) Both the statements are correct
(d) Both the statements are wrong
Answer:
(c) Both the statements are correct

PART – II
II. Answer The Following Questions

Question 1.
Define: Data Visualization?
Answer:
Data Visualization is the graphical representation of information and data. The objective of Data Visualization is to communicate information visually to users. For this, data visualization uses statistical graphics. Numerical data may be encoded using dots, lines, or bars, to visually communicate a quantitative message.

Question 2.
List the general types of data visualization?
Answer:

Charts
Tables
Graphs
Maps
Infographics
Dashboards

Question 3.
List the types of Visualizations in Matplotlib?
Answer:
There are many types of Visualizations under Matplotlib. Some of them are:

Line plot
Scatter plot
Histogram
Box plot
Bar chart and
Pie chart

Question 4.
How will you install Matplotlib?
Answer:
We can install matplotlib using pip. Pip is a management software for installing python packages.
After installing Matplotlib, we will begin coding by importing Matplotlib using the command: import matplotlib.pyplot as plt

Question 5.
Write the difference between the following functions: plt.plot([1,2,3,4]), plt.plot([1,2,3,4]), [1.4.9.16] ).
Draw line plt.plot(l,2,3,4] generate x values and given values are y values.
[1.4.9.16]) to plot x verses y

PART – III
III. Answer The Following Questions

Question 1.
Draw the output for the following data visualization plot?
Answer:
import matplotlib.pyplot as pit
plt.bar([l ,3,5,7,9],[5,2,7,8,2],
label=”Example one”)
plt.bar([2,4,6,8,10],[8,6,2,5,6], label=”Example two”, color=’g’)
pit. legend))
plt.xlabel(‘bar number’)
plt.ylabel(‘bar height’)
pit.title(‘Epic Graph\nAnother Line! Whoa’)
pit. show( )

Question 2.
Write any three uses of data visualization?
Answer:

Data Visualization help users to analyze and interpret the data easily.
It makes complex data understandable and usable.
Various Charts in Data Visualization helps to show relationship in the data for one or more variables.

Question 3.
Write the coding for the following:
a. To check if PIP is Installed in your PC.
b. To Check the version of PIP installed in your PC.
c. To list the packages in matplotlib.
Answer:
In command prompt type pip — version. If it is installed already, you will get version. Python -m pip install – U pip
C:\Users\YourName\AppData\Local\Programs\Python\Python36-32\Scripts>pip – version
C:\Users\YourName\AppData\Local\Programs\Python\Python36-32\Scripts>pip list

Question 4.
Write the plot for the following pie chart output?
Answer:

‘import matplotlib.pyplot as pit
slices = [7,2,2,13]
activities = [‘sleeping’, ‘eating’, ‘working’, ‘playing’]
plt.pie(slices, labels = activities, autopct = ‘y.1.1 f % % ‘)
plt.title(‘Interesting Graph check it out’)
plt.show( )

PART – IV
IV. Answer The Following Questions

Question 1.
Explain in detail the types of pyplots using Matplotlib?
Answer:
Line Chart:
A Line Chart or Line Graph is a type of chart which displays information as a series of data points called ‘markers’ connected by straight line segments. A Line Chart is often used to visualize a trend in data over intervals of time – a time series – thus the line is often drawn chronologically.
Example: Line plot
import matplotlib.pyplot as plt
years= [2014,2015,2016,2017,2018]
totaljopulations = [8939007, 8954518,8960387,8956741, 8943721]
plt.plot (years, total_populations)
plt.title (“Year vs Population in India”)
plt.xlabel (“Year”)
plt.ylabel (“Total Population”)
plt.legend( )
pit. show( )
In this program,
Plt.title( ) → specifies title to the graph
Plt.xlabel( ) → specifies label for X-axis
Plt.ylabel( ) → specifies label for Y-axis
Output 4

Bar Chart
A BarPlot (or BarChart) is one of the most common type of plot. It shows the relationship between a numerical variable and a categorical variable.
Bar chart represents categorical data with rectangular bars. Each bar has a height corresponds to the value it represents. The bars can be plotted vertically or horizontally. It’s useful when we want to compare a given numeric value on different categories. To make a bar chart with Matplotlib, we can use the plt.bar( ) function.
The above code represents the following:
Labels → Specifies labels for the bars.
Usage → Assign values to the labels specified.
Xticks → Display the tick marks along the x-axis at the values represented. Then specify the label for each tick mark.
Range → Create sequence of numbers.

Pie Chart:
Pie Chart is probably one of the most common type of chart. It is a circular graphic which is divided into slices to illustrate numerical proportion. The point of a pie chart is to show the relationship of parts out of a whole.
To make a Pie Chart with Matplotlib, we can use theplt.pie( ) function. The autopct parameter allows us to display the percentage value using the Python string formatting.

Question 2.
Explain the various buttons in a matplotlib window?
Answer:

Home Button → The Home Button will help once you have begun navigating your chart. If you ever want to return back to the original view, you can click on this.
Forward/Back buttons → These buttons can be used like the Forward and Back buttons in your browser. You can click these to move back to the previous point you were at, or forward again.
Pan Axis → This cross-looking button allows you to click it, and then click and drag your graph around.
Zoom → The Zoom button lets you click on it, then click and drag a square that you would like to zoom into specifically. Zooming in will require a left click and drag. You can alternatively zoom out with a right click and drag.
Configure Subplots → This button allows you to configure various spacing options with your figure and plot.
Save Figure → This button will allow you to save your figure in various forms.

Question 3.
Explain the purpose of the following functions:
Answer:
a. plt.xlabel
b. plt.ylabel
c. plt.title
d. plt.legend( )
e. plt.show( )
After installing Matplotlib, we will begin coding by importing Matplotlib using the command: import matplotlib.pyplot as pit
Now you have imported Matplotlib in your workspace. You need to display the plots. Using Matplotlib from within a Python script, you have to add plt.show( ) method inside the file to display your plot.
With plt.xlabel and plt.ylabel, you can assign labels to those respective axis. Next, you can assign the plot’s title with plt.title, and then you can invoke the default legend with plt
legend( ).
plt.plot (years, total_populations)
plt.title → (“Year vs Population in India”)
plt.xlabel → (“Year”)
plt.ylabel (“Total Population”)
plt.legend( )
plt.show( )
Plt.title( ) → specifies title to the graph
Plt.xlabel( ) → specifies label for X-axis
Plt.ylabel( ) → specifies label for Y-axis

Practice Programs

Question 1.
Create a plot. Set the title, the x and y labels for both axes?
Answer:
import matplotlib.pyplot as plt
x = [1, 2, 3]
y = [5, 7, 4]
plt.plot(x,y)
plt.xlabel(‘x-axis’)
plt.ylabel(‘y-axis’)
plt.title(‘LINE GRAPH’)
plt.show( )

Question 2.
Plot a pie chart for your marks in the recent examination?
Answer:
Example
import matplotlib.pyplot as plt
sizes= [89, 80, 90, 100, 75]
labels= [“Tamil”, “English”, “Maths”, “Science”, “Social”]
plt.pie (sizes, labels Mabels, autopct = “%.2f”)
plt.axes( ).set_aspect (“equal”) plt.sbow( )
Output:

Question 3.
Plot a line chart on the academic performance of Class 12 students in Computer Science for the past 10 years?
Answer:
import matplotlib.pyplot as plt
years = [2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017, 2018]
cs = [65, 70, 75, 76, 78, 80, 82, 85, 87, 92]
plt.plot(years,cs)
plt.title(“COMPUTER SCIENCE ACADEMIC PERFORMANCE”)
plt.xlabel(“cs”)
plt.ylabel(“years”)
plt.show( )

Question 4.
Plot a bar chart for the number of computer science periods in a week?
Answer:
import matplotlib.pyplot as pit
labels = [“MON”, “TUE”, “WED”, “THUR”, “FRI”, “SAT”, ]
usage = [3, 2, 1,3,2, 2]
y_positions = range(len(labels))
plt.bar(y_positions,usage)
plt.xticks(y_positions, labels)
plt.ylabel(“PERIODS”)
plt.ylabel(“years”)
plt.title(“NO. OF CS PERIODS”)
plt.show( )

Samacheer kalvi 12th Computer Science Data Manipulation Through SQL Additional Questions and Answers

PART – I
1. Choose The Correct Answer

Question 1.
……………………….. is the graphical representation of information and data.
Answer:
Data Visualization

Question 2.
Pick the odd one out.
Tables, databases, Maps, Infographics, Dashboards
Answer:
Databases

Question 3.
The representation of information in a graphic format is called …………………………
(a) chart
(b) graphics
(c) Infographics
(d) graphs
Answer:
Infographics

Question 4.
Read the statements given below. Identify the right option from the following.
Statement A: Dashboards are used to detect patterns and relationships easily.
Statement B: Dashboard is a type of game.
(a) A is correct
(b) B is correct
(c) Both are correct
(d) Both are wrong
Answer:
(a) A is correct

Question 5.
Various types of visualizations are available in
(a) stdlib
(b) graphics
(c) library
(d) matplotlib
Answer:
(d) matplotlib

Question 6.
Find the Incorrect match from the following.
(a) Scatter plot – collection of points
(b) line charts – markers
(c) Box plot – Boxes
Answer:
(c) Box plot – Boxes

Question 7.
The terms minimum, first quartile, median, third quartile maximum are related to ……………………… type of plot.
(a) Box
(b) scatter
(c) pie
(d) Bar
Answer:
(a) Box

Question 8.
The position of a point in scatter plot is ……………………. value
(a) ID
(b) 2D
(c) 3D
(d) U shaped
Answer:
(b) 2D

Question 9.
Which method is used to display the plot?
(a) disp( )
(b) display( )
(c) print( )
(d) show( )
Answer:
(d) show( )

Question 10.
Find the wrong statement
If single list is given to the plot( ) command, matplotlib assumes
(a) it is as a sequence of x values
(b) sequence of y values
Answer:
(a) it is as a sequence of x values

Question 11.
……………………….. is a versatile command, which takes an arbitrary number of arguments
Answer:
plot( )

Question 12.
The name to the x axis in the plot is given by ………………………….
(a) label
(b) x
(c) x label
(d) x axis
Answer:
(c) x label

Question 13.
Which one of the following is the cross-looking button that allows you to click it, and drag the . graph around?
(a) x axis
(b) y axis
(c) pan axis
(d) plot axis
Answer:
(c) pan axis

Question 14.
……………………. plot shows the relationship between a numerical variable and a categorical variable.
(a) line
(b) Bar
(c) Scatter
(d) Box
Answer:
(b) Bar

Question 15.
…………………….. assign values to the labels specified in the bar chart.
(a) usage
(b) label
(c) values
(d) =
Answer:
(a) usage

Question 16.
……………………… displays the tick marks along the x axis at the values represented in bar chart.
Answer:
x ticks

Question 17.
A ……………………….. represents the frequency distribution of continuous variables.
Answer:
histogram

Question 18.
Find the Incorrect statement
(a) Histogram is drawn in such a way that there is gap between the bars.
(b) Histogram represents numerical data
(c) Bar graph shows categorical data
Answer:
(a) Histogram is drawn in such a way that there is gap between the bars.

Question 19.
Which function shows the percentage value in pie chart?
(a) percent
(b) percentage
(c) slice
(d) auto pet
Answer:
(d) auto pet

Question 20.
Match the following
1. pie chart – (i) Series of data point
2. Histogram – (ii) Parts out of a whole
3. line chart – (iii) Quartile
4. Box plot – (iv) Freqency distribution
(a) 1-i, 2-ii, 3-iii, 4-iv
(b) 1-iv, 2-iii, 3-ii, 4-i
(c) 1 -ii, 2-iv, 3-i, 4-iii
(d) 1-iv, 2-i, 3-ii, 4-iii
Answer:
(c) 1 -ii, 2-iv, 3-i, 4-iii

PART – II
II. Answer The Following Questions

Question 1.
What is meant by Infographics?
Answer:
Infographics → An infographic (information graphic) is the representation of information in a graphic format.

Question 2.
Define Scatter plot?
Answer:
Scatter plot: A scatter plot is a type of plot that shows the data as a collection of points. The position of a point depends on its two-dimensional value, where each value is a position on either the horizontal or vertical dimension.

Question 3.
What is Box Plot?
Answer:
Box plot: The box plot is a standardized way of displaying the distribution of data based on the five number summary: minimum, first quartile, median, third quartile, and maximum.

PART – III
III. Answer The Following Questions

Question 1.
Define Dashboard?
Answer:
Dashboard → A dashboard is a collection of resources assembled to create a single unified visual display. Data visualizations and dashboards translate complex ideas and concepts into a simple visual format. Patterns and relationships that are undetectable in text are detectable at a glance using dashboard.

Question 2.
Write a program to create pie chart?
Answer:
Example
import matplotlib.pyplot as plt
sizes= [89, 80, 90, 100, 75]
labels= [“Tamil”, “English”, “Maths”, “Science”, “Social”]
plt.pie (sizes, labels Mabels, autopet = “%.2f’)
plt.axes( ).set_aspect (“equal”)
plt.show( )
Output:

PART – IV
IV. Answer The Following Questions

Question 1.
Write a program for Bar chart?
Answer:
Example
import matplotlib.pyplot as pit
#Our data
labels= [“TAMIL”, “ENGLISH”, “MATHS”, “PHYSICS”, “CHEMISTRY”, “CS”]
usage= [79.8, 67.3, 77.8, 68.4, 70.2, 88.5]
# Generating they positions. Later, we’ll use them to replace them with labels.
y_positions =range (len(labels))
#Creating our bar plot plt.bar (y_positions, usage)
plt.xticks (y_positions, labels)
plt.ylabel (“RANGE”)
pit.title (“MARKS”)

Question 2.
Explain the key differences between Histogram and Bar Graph?
Answer:
Key Differences Between Histogram and Bar Graph:
The differences between Histogram and bar graph are as follows:
1. Histogram refers to a graphical representation; that displays data by way of bars to show the frequency of numerical data. A bar graph is a pictorial representation of data that uses bars to compare different categories of data.

2. A histogram represents the frequency distribution of continuous variables. Conversely, a bar graph is a diagrammatic comparison of discrete variables.

3. Histogram presents numerical data whereas bar graph shows categorical data.

4. The histogram is drawn in such a way that there is no gap between the bars. On the other hand, there is proper spacing between bars in a bar graph that indicates discontinuity, (v) Items of the histogram are numbers, which are categorised together, to represent ranges of data. As opposed to the bar graph, items are considered as individual entities.

5. In the case of a bar graph, it is quite common to rearrange the blocks, from highest to lowest. But with histogram, this cannot be done, as they are shown in the sequence of classes.

6. The width of rectangular blocks in a histogram may or may not be same while the width of the bars in a bar graph is always same.

Tamil Nadu 12th Commerce Model Question Paper 2 English Medium

Leave a Comment / Class 12 / By Prasanna

Students can Download Tamil Nadu 12th Commerce Model Question Paper 2 English Medium Pdf, Tamil Nadu 12th Commerce Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 12th Commerce Model Question Paper 2 English Medium

Instructions:

The question paper comprises of four parts.
You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
All questions of Part I, II. III and IV are to be attempted separately.
Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about 50 words.
Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in about 150 words.
Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in about 250 words. Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 90

Part – I

Choose the correct answer. Answer all the questions: [20 x 1 = 20]

Question 1.
Dividing the work into small tasks is known as …………..
(a) discipline
(b) unity
(c) division of work
(d) equity
Answer:
(c) division of work

Question 2.
Delegation of Authority is easily done with the help of ………..
(a) MBM
(b) MBE
(c) MBO
(d) MBA
Answer:
(c) MBO

Question 3.
Participants is the capital market includes ……………
(a) Individuals
(b) Corporate
(c) Financial institutions
(d) All of the above
Answer:
(d) All of the above

Question 4.
Stock exchanges deal in ……………
(a) Goods
(b) Services
(c) Financial securities
(d) Country’s currency
Answer:
(c) Financial securities

Question 5.
Planning is a …………….. function.
(a) selective
(b) pervasive
(c) both a and b
(d) None of the above
Answer:
(b) pervasive

Question 6
…………… refers to the process of identifying and attracting job seekers so as to build a pool of qualified job applicants.
(a) selection
(b) training
(c) recruitment
(d) Induction
Answer:
(c) recruitment

Question 7.
The spot market is classified on the basis of …………….
(a) commodity
(b) transaction
(c) regulation
(d) time
Answer:
(b) transaction

Question 8.
Which one represents a cluster of manufacturers, content providers and online retailers organized around an activity?
(a) Virtual mall
(b) Association
(c) Metomediary
(d) Portal
Answer:

Question 9.
………….. is the king of modem marketing.
(a) consumer
(b) wholesales
(c) producer
(d) retailer
Answer:
(a) consumer

Question 10.
GST stands for ……………..
(a) Goods, Sales and Tax
(b) Goods and Services Tax
(c) Global Services and Tax
(d) Government Services and Tax
Answer:
(b) Goods and Services Tax

Question 11.
In case of the sale, the has the right to sell.
(a) buyer
(b) seller
(c) hirer
(d) consignee
Answer:
(b) seller

Question 12.
What are the characteristics of an entrepreneur?
(a) spirit of enterprise
(b) flexibility
(c) self confidence
(d) all of the above
Answer:
(d) all of the above

Question 13.
………… is the government of India’s endeavour to promote culture of innovation and entrepreneurship.
(a) AIM
(b) STEP
(c) SEED
(d) AIC
Answer:
(a) AIM

Question 14.
A private company shall have a minimum of ………………
(a) seven directors
(b) five directors
(c) three directors
(d) two directors
Answer:
(d) two directors

Question 15.
Which of the following is verification function?
(a) planning
(b) Organisation
(c) staffing
(d) Controlling
Answer:
(d) Controlling

Question 16.
Primary market is also called as ………………
(a) secondary market
(b) Money Market
(c) new issue market
(d) indirect market
Answer:
(c) new issue market

Question 17.
A major player in the money market is the
(a) Commercial Bank
(b) Reserve Bank of India
(c) State Bank of India
(d) Central Bank
Answer:
(a) Commercial Bank

Question 18.
In which year SEBI was constituted as the regulator of capital markets in India?
(a) 1988
(b) 1992
(c) 2014
(d) 2013
Answer:
(a) 1988

Question 19.
Recruitment bridges gap between and
(a) Job seeker and Job provider
(b) Job seeker and agent
(c) Job provider and owner
(d) Owner and servant
Answer:
(a) Job seeker and Job provider

Question 20.
Vestibule training is provided
(a) On the job
(b) In the class room
(c) In a situation similar to actual working environment
(d) By the committee
Answer:
(c) In a situation similar to actual working environment

Part – II

Answer any seven in which question No.30 is compulsory. [7 x 2 = 14]

Question 21.
What is meant by Motivation?
Answer:
The goals are achieved with the help of motivation. Motivation includes increasing the speed of performance of a work and developing a willingness on the part of workers. This is done by a resourceful leader.

Question 22.
What is equity market?
Answer:
Equity Market is the financial market for trading in Equity Shares of Companies.

Question 23.
What is a CD market?
Answer:
Certificate of Deposits are short-term deposit instruments issued by banks and financial institutions to raise large sums of money. The Certificate of Deposit is transferable from one party to another. Due to their negotiable feature, they are also known as negotiable certificate of deposit.

Question 24.
What is Demat account?
Answer:
A demat account holds all the shares that are purchased in electronic or dematerialized form.

Question 25.
Give the meaning of Recruitment.
Answer:
Recruitment is the process of finding suitable candidates for the various posts in an organisation

Question 26.
What is meant by training?
Answer:
Training is the act of increasing / enhancing the new skill of problem solving activity and technical knowledge of an employee for doing the jobs themselves.

Question 27.
Define Marketing Mix.
Answer:
“Marketing mix is a pack of four sets of variables namely product variable, price variable, promotion variable, and place variable”.

Question 28.
What do you know about National Commission?
Answer:

The National Consumer Disputes Redressal Commission (NCDRC), India is a quasi-judicial commission in India which was set up in 1988 under the Consumer Protection Act of 1986.
The National Consumer Disputes Redressal Commission (NCDRC) is also called as National Commission.

Question 29.
What is Share?
Answer:
The term Share is viewed by a layman as a fraction or portion of total capital of the company which have equal denomination.

Question 30.
Define Meeting.
Answer:
A company meeting must be convened and held in perfect compliance with the various provisions of the Act and the rules framed thereunder

Part – III

Answer any seven in which question No. 31 is compulsory. [7 x 3 = 21]

Question 31.
Differentiate Management from Administration.
Answer:

Basis for Comparison	Management	Administration
Meaning	An organised way of managing people and things of a business organisation is called the management.	The process of administrating an organisation by a group of people is known as the administration.
Authority	Middle and lower level.	Top level
Role	Executive	Decisive
Concerned with	Policy Implementation.	Policy formulation.
Area of operation	It works under administration.	It has full control over the activities of the organisation.
Applicable for	Profit making organisations, i.e. business organisations.	Government offices, military clubs, business enterprises,hospitals,religious and educational organisations.

Question 32.
State the importance of staffing.
Answer:
Staffing refers to placement of right persons in the right jobs. Staffing includes selection of right persons, training to those needy persons, promotion of best persons, retirement of old persons, performance appraisal of all the personnel, and adequate remuneration of personnel. The success of any enterprise depends upon the successful performance of staffing function.

Question 33.
What are the various kinds of Capital Market? Explain.
Answer:
The capital market is divided into two i.e., primary market and secondary market.

Primary Market: Primary market is a market for new issues or new financial claims.
Secondary Market: Secondary Market may be defined as the market for old securities, in the sense that securities which are previously issued in the primary market are traded here.

Question 34.
Explain Stag and Lame Duck.
Answer:

Stag: A stag is a cautious speculator in the stock exchange. He applies for shares in new companies and expects to sell them at a premium.
Lame Duck: When a bear finds it difficult to fulfill his commitment, he is said to be struggling like a lame duck.

Question 35.
What is structured interview?
Answer:
Under this method, a series of questions to be asked by the interviewer are pre-prepared by the interviewer and only these questions are asked in the interview.

Question 36.
What is need for market and explain the concept of marketing?
Answer:
Market is needed for the producers and consumers. For this there are various concepts:

The products produced are to be marketed in the market for sales.
First create a consumer and then create products.
Keep respect and love customers than the products.
Customer is the king of the business.

Question 37.
What are the important legislations related to consumerism in India?
Answer:
Consumer Legislation:

The Indian Contract Act, 1982
The Sale of Goods Act, 1982
The Essential Commodities Act, 1955
The Agricultural Products Grading and Marketing Act, 1937
The Prevention of Food Adulteration Act, 1954
Weights and Measures Act, 1958
The Trademark Act, 1999

Question 38.
What are the political environment factors?
Answer:
Political and Legal environment – The framework for running a business is given by the political and legal environment.

Political stability is reflected by the following parameters like the election system, the law and order situation.
The image of the leader and the country in the international arena.
The constitution of the nation.

Question 39.
Distinguish between Negotiability and Assignability.
Answer:

S.No.	Basis of Difference	Negotiability	Assignability
1.	Legal Ownership	It passes to the transferee by mere endorsement in the case of a bearer instrument and by endorsement and delivery in the case of an order instrument.	An assignment can be made by observing certain formalities. For instance, an instrument is to be made in writing, duly stamped and signed by the transferor or his agent.
2.	Notice	Notice is not necessary for the holder of negotiable instrument to claim the payment from the debtor.	In case of actionable claim, notice of the assignment by the transferee regarding the transfer of debt to the debtor is necessary.
3.	Nature of title	Holder of negotiable instrument in due course gets a better title than even the transferor. It means that the transferee gets the instrument free from any defect existing in the title of the transferor or any prior party.	The transferee’s title to the instrument is subject to the defects of the transferor’s title. In other words, defects in the title of the transferor pass on to the transferee too.
4.	Consideration	Consideration is presumed.	The assignee has to prove the consideration for the transfer.

Question 40.
What do you understand by Issue of Securities at Premium?
Answer:
When shares are issued at a price above the face or nominal value, they are said to be issued at a premium. For example, a share having the face value of Rs.10 is issued at Rs.12. Here, Rs.2 is the premium. The amount of share premium has to be transferred to an account called the ‘Securities Premium Account’.

Part – IV

Answer all the following questions. [7 x 5 = 35]

Question 41.
(a) Briefly explain different types of Directors.
Answer:
Types of Directors as per Companies Act 2013:

Residential Director: According to Section 149(3) ofCompaniesAct2013, every company should appoint a director who has stayed in India for a total period of 182 days.
Independent Director: An independent director is an alternate director other than a Managing Director who is known as Whole-time director or Nominee director.
Small shareholder’s Director: Small shareholders can appoint a single director in a listed company.
Nominee Director: A director nominated by any financial institution in pursuance of the provisions of law.
Additional Directors: Any Individual can be appointed as Additional Directors by a company.
Alternate Directors: Alternate director is appointed by the Board of Directors, as a substitute who may be absent from India, for a period of 3 months.

[OR]

(b) Describe the steps for promoting Entrepreneurial venture.
Steps for promoting entrepreneurial venture:

Selection of the product: An entrepreneur may select a product according to his capacity and motivation after a thorough scrutiny of micro and macro environment of business.
Selection of form of ownership: Entrepreneur has to choose the form of organisation suitable and appropriate for his venture namely family ownership, partnership and private company.
Selection of site: Entrepreneur has to choose suitable plot for starting his venture. The plot may be industrial site, land by the private people or housing board plot allotted.
Designing capital structure: The entrepreneurs has to determine the source of finance for funding the venture. It may be own savings, loan from friends, relatives or loan from banks.
Acquisition of Manufacturing know-how: Entrepreneur can acquire manufacturing know-how from government research laboratories and industrial consultants.
Preparation of project report: Project reports needs to be prepared according to the format prescribed in the loan application forms.

Question 42.
(a) Distinguish between an Entrepreneur and an Intrapreneur.
Answer:

Basis of difference	Entrepreneur	Intrapreneur
Thinking	Entrepreneur is a free thinker.	Intrapreneur is forced to think independently but within scope of business activities undertaken in the enterprise.
Dependency	Entrepreneur is an independent person.	Intrapreneur is dependent on the entrepreneur. He is an employee.
Fund Mobilization	Entrepreneur has to mobilize funds to finance the venture.	Intrapreneur does not engage in fund mobilization. But can access funds mobilized by the entrepreneur.
Reward	Entrepreneur is rewarded by profit for the risk bearing exercise.	Intrapreneur does not share in profits of venture. But gets perquisites, salary, incentives, etc., for the service.
Risk Bearing	Entrepreneur bears the risk involved in the venture undertaken.	Intrapreneur does not bear any risk in the venture and does not even share the risk inherent in the project or work assigned. However Intrapreneur is accountable for the task or project assigned.
	Entrepreneur is owner, and doesn’t report to anybody in the venture.	Intrapreneur is a salaried employee. Intrapreneur works within control put in place in the organization and is made accountable for the activities undertaken.

[OR]

(b) Explain the micro environmental factors of business.
Answer:
Micro environmental factors are those, which are in the immediate environment of a business affecting its performance. These include the following:

Suppliers: In any organisation the suppliers of raw materials and other inputs play a vital role. Timely procurement of materials from suppliers enables continuity in production and reduces the cost of production.
Customers: The aim of any business is to satisfy the needs of its customers. The customer is the king of the business.
Competitors: All organisations face competition at all levels local, national and global. Competitors maybe for the same product or for similar products.
Financiers: The financiers of a business includes the debenture holders and financial institutions.
Marketing Channel members: The marketing intermediaries serve as a connecting link between the business and its customers. The middlemen like dealers, wholesalers and retailers ensure transfer of product to customers.

Question 43.
(a) Describe the various strategies pursued in recent day’s marketers.
Answer
The market scenario in the world today is changing very rapidly.
Strategies of Recent Marketers:

Due to the development in information technology, transportation, liberalisation, their buying habits are varying.
In the globalised business environment, the marketer must move goods faster and quicker to satisfy the needs of the customer.
It is possible to carryout all the business transactions over an electronic network.
They use a variety of tools like computers, laptops, tablet or android phone devices to access different websites.

[OR]

(b) How the market can be classified on the basis of Economics?
The Market can be classified on the basis of Economics as follows:
(a) Perfect Market: A market is said to be a perfect market, if it satisfies the following conditions:

Large number of buyers and sellers are there.
Prices should be uniform throughout the market.
Buyers and sellers have a perfect knowledge of market.
Goods can be moved from one place to another without restrictions.

(b) Imperfect Market: A market is said to be imperfect when

Products are similar but not identical.
Prices are not uniform.
There is lack of communication.

Question 44.
(a) Briefly explain the various types of tests.
Answer:
(1) Prospectus:
Several tests are conducted in the selection process to ensure whether the candidate possesses the necessary qualifications.

(2) Ability Test: Ability test may be divided into:

Aptitude Test
Achievement Test
Intelligence Test
Judgment Test.

(3) Aptitude Test: Aptitude test is a test to measure suitability of the candidates for the post.

(4) Achievement Test: This test measures a candidate’s capacity to achieve in a particular field.

(5) Intelligence Test: Intelligence test is designed to measure a variety of mental ability, individual capacity of a candidate.

(6) Judgment Test: This test is conducted to test the presence of mind and reasoning capacity of the candidates.

(7) Personality Test: It refers to the test conducted to find out the non-intellectual traits of a candidate. It can be further divided into:

(8) Interest Test: Interest test measures a candidate’s extent of interest in a particular area. Projective

(9) Test: This test measures the candidate’s values, personality of the candidate.

(10) Attitude Test: measures candidate’s tendencies towards the people, situation, action and related things.

[OR]

(b) Elaborate on the Managerial functions of Human Resource Management.
Managerial Functions of Human Resource Management:

Planning: Planning is deciding in advance what to do, how to do and who is to do it.
Organising: It includes division of work among employees by assigning each employee their duties, delegation of authority required.
Directing: It involves issue of orders and instructions along with guidance and motivation to get the best employees.
Controlling: The control process includes fixing of standards, measuring actual
performance, comparing actual with standard laid down, measuring deviations and taking corrective actions.

Question 45.
(a) What are the functions of SEBI?
Answer:
SEBI performs three key functions: quasi-legislative, quasi-judicial and quasi-executive.

Safeguarding the interests of investors by means of adequate education and guidance.
Regulating and controlling the business on stock markets.
Conduct inspection and inquiries of stock exchanges, intermediaries and self-regulating organizations.
Barring insider trading in securities.
Prohibiting deceptive and unfair methods used by financial intermediaries operating in securities market.
Registering and controlling the functioning of stock brokers, sub-brokers, share transfer agents, bankers, trustees, underwriters who are linked to securities market.
SEBI issues Guidelines and Instructions to businesses orgnisations for issue of shares.
Registering and controlling the functioning of collective investment schemes such as mutual funds.

[OR]

(b) Explain the Benefits of Stock Exchange.

Economic Development: It increases the economic development by ensuring steady flow of savings for production.
Fund Raising Platform: It enables the well-managed companies to raise funds by issue of shares.

Benefits to the Company:

Enhances Goodwill or Reputation: Companies whose shares are quoted on a stock exchange enjoy greater goodwill and credit standing.
Raises huge funds: Stock exchange helps the companies to raise huge funds by issue of shares and debentures.

Benefits to Investors:

Liquidity: Stock exchange helps to convert his shares into cash quickly.
Investor protection: The stock exchange safeguards, investor’s interest by following strict rules and regulations.

Question 46.
(A) Explain the characteristics of Money Market.

1. Short-term Funds: It is a market purely for short-term funds or financial assets called near money.

2. Maturity Period: It deals with financial assets having a maturity period upto one year only.

3. Conversion of Cash: It deals with only those assets which can be converted into cash readily without loss and with minimum transaction cost.

4. No Formal Place: Generally, transactions take place through phone, i.e., oral communication. Relevant documents and written communications can be exchanged subsequently.

5. Sub-markets: It is not a single homogeneous market. It comprises of several sub-markets each specialising in a particular type of financing.

6. Role of Market: The components of a money market are the Central Bank, Commercial Banks. Commercial banks generally play a dominant role in this market.

7. Highly Organized Banking System: The Commercial Banks are the nerve centre of the whole money market. They are the principal suppliers of short-term funds.

8. Existence of Secondary Market: There should be an active secondary market for these instruments.

9. Demand and Supply of Funds: There should be a large demand and supply of short-term funds.

8. Wholesale Market: It is a wholesale market and the volume of funds or financial assets traded in the market is very large.

9. Flexibility: Due to greater flexibility in the regulatory framework, there are constant endeavours for introducing new instruments.

10. Presence of a Central Bank: The central bank keeps their cash reserves and provides them financial accommodation in difficulties by discounting their eligible securities.

[OR]
(b) Briefly explain the functions of capital market.
(i) Savings and Capital Formation: In capital market, various types of securities help to mobilize savings from various sectors of population (Individuals, Corporate, Govt. etc.) The twin features of reasonable return and liquidity in stock exchange are definite incentives to the people to invest in securities. This accelerates the capital formation in the country.

(ii) Permanent Capital: The existence of a capital market/stock exchange enables companies to raise permanent capital. The investors cannot commit their funds for a permanent period but companies require funds permanently.

(iii) Industrial Growth: The stock exchange is a central market through which resources are transferred to the industrial sector of the economy.

(iv) Ready and Continuous Market: The stock exchange provides a central convenient place where buyers and sellers can easily purchase and sell securities.

(v) Reliable Guide to Performance: The capital market serves as a reliable guide to the performance and financial position of corporate, and thereby promotes efficiency.

(vi) Proper Channelization of Funds: The prevailing market price of a security and relative yield are the guiding factors for the people to channelize their funds in a particular company.

(vii) Provision of Variety of Services: The financial institutions functioning in the capital market provide a variety of services such as grant of long term and medium term loans to entrepreneurs.

(viii) Development of Backward Areas: Capital Markets provide funds for projects in backward areas. This facilitates economic development of backward areas.

(ix) Foreign Capital: Capital markets make possible to generate foreign capital. Indian firms are able to generate capital funds from overseas markets by way of bonds and other securities.

(x) Easy Liquidity: With the help of secondary market investors can sell off their holdings and convert them into liquid cash.

Question 47.
(a) Enumerate the different kinds of financial markets.
Financial Markets can be classified in different ways.
On the Basis of Type of Financial Claim

Debt Market is the financial market for trading in Debt Instrument (i.e. Government Bonds or Securities, Corporate Debentures or Bonds).
Equity Market is the financial market for trading in Equity Shares of Companies.

On the Basis of Maturity of Financial Claim

Money Market is the market for short term financial claim (usually one year or less) E.g. Treasury Bills, Commercial Paper, Certificates of Deposit.
Capital Market is the market for long term financial claim more than a year; E.g. Shares, Debentures.

On the Basis of Time of Issue of Financial Claim

Primary Market is a term used to include all the institutions that are involved in the sale of securities for the first time by the issuers (companies). Here the money from investors goes directly to the issuers.
Secondary Market is the market for securities that are already issued. Stock Exchange is an important institution in the secondary market.

On the Basis of Timing of Delivery of Financial Claim

Cash/Spot Market is a market where the delivery of the financial instrument and payment of cash occurs immediately, i.e. settlement is completed immediately.
Forward or Futures Market is a market where the delivery of asset and payment of cash takes place at a pre-determined time frame in future.

On the Basis of the Organizational Structure of the Financial Market

Exchange-Traded Market is a centralized organization (stock exchange) with standardized procedures.
Over-the-Counter Market is a decentralized market (outside the stock exchange) with customized procedures.

[OR]

(b) Explain the various functions of management.
Main Functions:
(i) Planning:- Planning is the primary function of management. Planning is a constructive reviewing of future needs so that present actions can be adjusted in view of the established goal.

(ii) Organising:- Organising is the process of establishing harmonious relationship among the members of an organisation and the creation of network of relationship among them.

(iii) Staffing:- Staffing function comprises the activities of selection and placement of competent personnel.

(iv) Directing:- Directing denotes motivating, leading, guiding and communicating with subordinates on an ongoing basis in order to accomplish pre-set goals.

(v) Controlling:- Controlling is performed to evaluate the performance of employees and deciding increments and promotion decisions.

(vi) Co-ordination:- Co-ordination is the synchronization of the actions of all individuals, working in the enterprise in different capacities.

(vii) Motivating:- The goals are achieved with the help of motivation. Motivation includes increasing the speed of performance of a work and developing a willingness on the part of workers.

Subsidiary Functions:

Innovation:- Innovation includes developing new material, new products, new techniques in production, new package, new design of a product and cost reduction.
Representation:- A manager has to act as representative of a company. It is the duty of every manager to have good relation with others.
Decision-making:- Every employee of an organisation has to take a number of decisions every day. Decision-making helps in the smooth functioning of an organisation.
Communication:- Communication is the transmission of human thoughts, views or opinions from one person to another person. Communication helps the regulation of job and co-ordinates the activities.

Tamil Nadu 12th Physics Model Question Paper 5 English Medium

Leave a Comment / Class 12 / By Prasanna

Students can Download Tamil Nadu 12th Physics Model Question Paper 5 English Medium Pdf, Tamil Nadu 12th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 12th Physics Model Question Paper 5 English Medium

Instructions:

The question paper comprises of four parts
You are to attempt all the parts. An internal choice of questions is provided wherever: applicable
All questions of Part I, II, III and IV are to be attempted separately
Question numbers 1 to 15 in Part I are Multiple choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
Question numbers 16 to 24 in Part II are two-mark questions. These are lo be answered in about one or two sentences.
Question numbers 25 to 33 in Part III are three-mark questions. These are lo be answered in about three to five short sentences.
Question numbers 34 to 38 in Part IV are five-mark questions. These are lo be answered in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 70

PART – 1

Answer all the questions. Choose the correct answer. [15 × 1 = 15]

Question 1.
When the current changes from +2A to -2A in 0.05 s, an emf of 8 V is induced in a coil. The co-efficient of self-induction of the coil is
(a) 0.2 H
(b) 0.4 H
(c) 0.8 H
(d) 0.1 H
Answer:
(d) 0.1 H

Question 2.
If λ_v , λ_x, λ_m, represent the wavelength of visible light, X-rays and microwaves, respectively, then
(a) λ_m > λ_v > λ_x
(b) λ_m > λ_x > λ_v
(c) λ_v > λ_m > λ_x
(d) λ_v > λ_x > λ_m
Answer:
(a) λ_m > λ_v > λ_x

Question 3.
The materials used in Robotics are:
(a) Aluminium and silver
(b) silver and gold
(c) Copper and gold
(d) Steel and aluminium
Answer:
(d) Steel and aluminium

Question 4.
Two wires of A and B with circular cross-section are made up of the same material with equal lengths. If R_A = 3R_B, then what is the ratio of radius of wire A to that of B?
(a) 3
(b) √3
(c) $\frac{1}{\sqrt{3}}$
(d) $\frac{1}{3}$
Answer:
(c) $\frac{1}{\sqrt{3}}$

Question 5.
The frequency range of 3 MHz to 30 MHz is used for
(a) Ground wave propagation
(b) Space wave propagation
(c) Sky wave propagation
(d) Satellite communication
Answer:
(c) Sky wave propagation

Question 6.
A ray of light strikes a glass plate at an angle 60°. If the reflected and refracted rays are perpendicular to each other, the refractive index of the glass is,
(a) √3
(b) $\frac{3}{2}$
(c) $\sqrt{\frac{3}{2}}$
(d) 2
Hint. Angle of refraction r = 60° ; Angle of incident i = 30°
sin i = n × sin r
n = $\frac{\sin 30^{\circ}}{\sin 60^{\circ}}=\sqrt{3}$
Answer:
(a) √3

Question 7.
If voltage applied on a capacitor is increased from V to 2V choose the correct conclusion
(a) Q remains the same, C is doubled
(b) Q is doubled, C doubled
(c) C remains same, Q doubled
(d) Both Q and C remain same
Answer:
(c) C remains same, Q doubled

Question 8.
The nucleus is approximately spherical in shape. Then the surface area of the nucleus having mass number A varies as
(a) A^2/3
(b) A^4/3
(c) A^1/3
(d) A^5/3
Hint. Volume of nucleus is proportional to mass number 4
$\frac { 4 }{ 3 }$ π R³ ∝ A : R = R₀ A^1/3
So, 4 π R² = R₀ A^2/3 ⇒ 4 π R² ∝ A^2/3
Surface area is proportional to (mass number)23
Answer:
(a) A^2/3

Question 9.
The given electrical network is equivalent to

(a) AND gate
(b) OR gate
(c) NOR gate
(d) NOT gate
Answer:
(c) NOR gate

Question 10.
A wire of length / carries a current I along the Y direction and magnetic field is given by $\vec{B}$ = $\frac{\beta}{\sqrt{3}}$(i +j +k)T- The magnitude of Lorentz force acting on the wire is
(a) $\sqrt{\frac{2}{\sqrt{3}}} \beta I l$
(b) $\sqrt{\frac{1}{\sqrt{3}}} \beta \mathrm{Il}$
(c) $\sqrt{2} \beta \mathrm{Il}$
(d) $\sqrt{\frac{1}{2}} \beta \mathrm{Il}$
Answer:
(a) $\sqrt{\frac{2}{\sqrt{3}}} \beta I l$

Question 11.
When a point charge of 6 pC is moved between two points in an electric field, the work done is 1.8 × 10^-5 J. This potential difference between the two points is :
(a) 1.08 V
(b) 1.08 μV
(c) 3 V
(d) 30 V

$=\frac{1.8 \times 10^{-5}}{6 \times 10^{-6}}=\frac{18}{6}$
V = 3V
Answer:
(c) 3 V

Question 12.
The wavelength λ_e of an electron and λ_p of a photon of same energy E are related by
(a) λ_p ∝ λ_e
(b) λ_p ∝ $\sqrt{\lambda_{e}}$
(c) λ_p ∝ $\frac{1}{\sqrt{\lambda_{e}}}$
(d) λ_p ∝ λ_e²
Hint:
de-Breglie wavelength of proton λ_e = $\frac{h}{\sqrt{2 m \mathrm{E}}}$
∴ ie λ_e ∝ $\frac{1}{\sqrt{\mathrm{E}}}$ ⇒ λ_e² ∝ $\frac{1}{E}$ …….. (1)
de-Breglie wave length of proton
λ_p = $\frac{h c}{E}$
λ_p ∝ $\frac{1}{E}$ …….. (2)
From (1) and (2)
λ_e² ∝ λ_p i.e., λ_p ∝ λ_e²
Answer:
(d) λ_p ∝ λ_e²

Question 13.
For a myopic eye, the defect is cured by using a:
(a) convex lens
(b) concave lens
(c) cylindrical lens
(d) plane glass
Answer:
(b) concave lens

Question 14.
In a tangent galvanometer experiment, for two different values of current if the deflection are 45° and 30° respectively, then the ratio of the current is:
(a) 2 : 3
(b) 3 : 2
(c) √3 : 1
(d) 1 : √3
Hint: I₁ = tan θ₁
I₂ = tan θ₂
$\frac{I_{1}}{I_{2}}=\frac{\tan 45^{\circ}}{\tan 30^{\circ}}=\frac{1}{\left(\frac{1}{\sqrt{3}}\right)}=\frac{\sqrt{3}}{1}$
Answer:
(c) √3 : 1

Question 15.
If the current gain a of a transistor is 0.98, what is the value of p of the transistor?
(a) 0.49
(b) 49
(c) 4.9
(d) 5
Hint: $\beta=\frac{\alpha}{1-\alpha}=\frac{0.98}{1-0.98}=49$
Answer:
(b) 49

PART – II

Answer any six questions in which Q. No 24 is compulsory. [6 × 2 = 12]

Question 16.
What is meant by Fraunhofer lines?
Answer:
When the spectrum obtained from the Sun is examined, it consists of large number of dark lines (line absorption spectrum). These dark lines in the solar spectrum are known as Fraunhofer lines.

Question 17.
Why steel is preferred in making Robots?
Answer:
Steel is several time stronger. In any case, because of the inherent strength of metal, robot bodies are made using sheet, bar, rod, channel, and other shapes.

Question 18.
State Lenz’s law.
Answer:
Lenz’s law states that the direction of the induced current is such that it always opposes the cause responsible for its production.

Question 19.
Why do clouds appear white?
Answer:
Clouds have large particles like dust and water droplets which scatter light of all colours almost equally. Hence clouds generally appear white.

Question 20.
Calculate the radius of ¹⁹⁷₇₉Au nucleus.
Answer:
According to the equation R = R₀A^1/3
R = 1.2 × 10^-15 × (197)^1/3 = 6.97 × 10^-15 m (or) R = 6.97 F

Question 21.
What is the need for a feedback circuit in a transistor oscillator?
Answer:
The circuit used to feedback a portion of the output to the input is called the feedback network. If the portion of the output fed to the input is in phase with the input, then the magnitude of the input signal increases. It is necessary for sustained oscillations.

Question 22.
Show graphically the variation of electric held £ (y-axis) due to a charged infinite plane sheet with distance r (x-axis) from the plate.
Answer:
It independent of the distance. It straight line parallel to x-axis.

Question 23.
Give any two applications of internet.
(i) Search engine: The search engine is basically a web-based service tool used to search for information on World Wide Web.

(ii) Communication: It helps millions of people to connect with the use of social networking: emails, instant messaging services and social networking tools.

(iii) E-Commerce: Buying and selling of goods and services, transfer of funds are done over an electronic network.

Question 24.
Calculate the magnetic field inside a solenoid when the number of turns is halved and the length of the solenoid and the area remain the same.
Answer:
Magnetic field inside a solenoid is B = $\frac{\mu_{0} \mathrm{NI}}{\mathrm{L}}$
If the number of turns is havled B’ = $\mathrm{B}^{\prime}=\frac{\mu_{0}\left(\frac{\mathrm{N}}{2}\right) \mathrm{I}}{\mathrm{L}} \Rightarrow \mathrm{B}^{\prime}=\frac{1}{2}\left(\frac{\mu_{0} \mathrm{NI}}{\mathrm{L}}\right)$

PART – III

Answer any six questions in which Q.No. 33 is compulsory. [6 × 3 = 18]

Question 25.
Two cells each of 5V are connected in series across a 8 12 resistor and three parallel resistors of 4 12, 6 12 and 12 12. Draw a circuit diagram for the above arrangement calculate
(i) the current drawn from the cell
(ii) current through each resistor.
Answer:
V₁ = 5 V; V₂ = 5 v
R₁ = 8 Ω; R₂ = 4Ω; R₃ = 6Ω; R₄ = 12Ω
Three resistors R₂, R₃ and R₄ are connected parallel combination

$\frac{1}{R_{p}}=\frac{1}{R_{2}}+\frac{1}{R_{3}}+\frac{1}{R_{4}}$
$=\frac{1}{4}+\frac{1}{6}+\frac{1}{12}=\frac{3}{12}+\frac{2}{12}+\frac{1}{12}=\frac{6}{12}$
R_p = 2Ω
Resistors R₁ and R_p are connected in series combination
R_s = R₁ + R_p = 8 + 2 = 10
R₂ = 10Ω
Total voltage connected series to the circuit
V = V₁ + V₂
= 5 + 5 =10
V = 10V

(i) Current through the circuit, I = $\frac{V}{R_{s}}=\frac{10}{10}$
I = IA
Potential drop across the parallel combination,
V’ = I R_p = 1 × 2 .
V’ = 2 V

(ii) Current in 4Ω resistor, I = $\frac{V^{\prime}}{R_{2}}=\frac{2}{4}$ = 0.5 A
Current in 6Ω resistor, I = $\frac{V^{\prime}}{R_{3}}=\frac{2}{6}$ = 0.33 A.
Current in 12Ω resistor, I = $\frac{V^{\prime}}{R_{4}}=\frac{2}{12}$ = 0.17 A

Question 26.
Explain the various energy losses in a transformer.
Answer:
Energy losses in a’ transformer: Transformers do not have any moving parts so that its efficiency is much higher than that of rotating machines like generators and motors. But there are many factors which lead to energy loss in a transformer.

1. Core loss or Iron loss: This loss takes place in transformer core. Hysteresis loss and eddy current loss are known as core loss or Iron loss. When transformer core is magnetized and demagnetized repeatedly by the alternating voltage applied across primary coil, hysteresis takes place due to which some energy is lost in the form of heat. Hysteresis loss is minimized by using steel of high silicon content in making transformer core. Alternating magnetic flux in the core induces eddy currents in it. Therefore there is energy loss due to the flow of eddy current, called eddy current loss which is minimized by using very thin laminations of transformer core.

2. Copper loss: Transformer windings have electrical resistance. When an electric current flows through them, some amount of energy is dissipated due to Joule heating. This energy loss is called copper loss which is minimized by using wires of larger diameter.

3. Flux leakage: Flux leakage happens when the magnetic lines of primary coil are not completely linked with secondary coil. Energy loss due to this flux leakage is minimized by winding coils one over the other.

Question 27.
Discuss the alpha – decay process with example.
Answer:
When unstable nuclei decay by emitting an α-particle (⁴₂He nucleus), it loses two protons and two neutrons. As a result, its atomic number Z decreases by 2, the mass number decreases by 4. We write the alpha decay process symbolically in the following way

Here X is called the parent nucleus and Y is called the daughter nucleus.
Example: Decay of Uranium ²³⁸₉₂U to thorium ²³⁴₉₀Th with the emission of ⁴₂He nucleus (α-particle)
²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He
As already mentioned, the total mass of the daughter nucleus and 2 He nucleus is always less than that of the parent nucleus. The difference in mass ( Δm = m_x – m_y m_α) is released as energy called disintegration energy Q and is given by
Q = (m_x – m_y – m_α) c²
Note that for spontaneous decay (natural radioactivity) Q > 0. In alpha decay process, the disintegration energy is certainly positive (Q > 0). In fact, the disintegration energy Q is also the net kinetic energy gained in the decay process or if the parent nucleus is at rest, Q is the total kinetic energy of daughter nucleus and the ⁴₂He nucleus. Suppose Q < 0, then the decay process cannot occur spontaneously and energy must be supplied to induce the decay.

Question 28.
Obtain the expression for energy stored in the parallel plate capacitor.
Answer:
Energy stored in the capacitor: Capacitor not only stores the charge but also it stores energy. When a battery is connected to the capacitor, electrons of total charge -Q are transferred from one plate to the other plate. To transfer the charge, work is done by the battery. This work done is stored as electrostatic potential energy in the capacitor.
To transfer an infinitesimal charge dQ for a potential difference V, the work done is given by
dW = VdQ ….(1)
Where V = $\frac{Q}{C}$
The total work done to charge a capacitor is
W = $\int_{0}^{0} \frac{\mathrm{Q}}{\mathrm{C}} d \mathrm{Q}=\frac{\mathrm{Q}^{2}}{2 \mathrm{C}}$ ….. (2)
This work done is stored as electrostatic potential energy (U_E) in the capacitor.
U_E = $\frac{\mathrm{Q}^{2}}{2 \mathrm{C}}=\frac{1}{2} \mathrm{CV}^{2}$ (∴Q = CV) …… (3)
where Q = CV is used. This stored energy is thus directly proportional to the capacitance of the capacitor and the square of the voltage between the plates of the capacitor. But where is this energy stored in the capacitor? To understand this question, the equation (3) is rewritten as follows using the results
C = $\frac{\varepsilon_{0} A}{d}$ and V = Ed
U_E = $\frac{1}{2}\left(\frac{\varepsilon_{0} \mathrm{A}}{d}\right)(\mathrm{E} d)^{2}=\frac{1}{2} \varepsilon_{0}(\mathrm{A} d) \mathrm{E}^{2}$ …… (4)
where Ad = volume of the space between the capacitor plates. The energy stored per unit Volume of space is defined as energy density u_E = $\frac{\mathrm{U}}{\text { Volume }}$ From equation (4) we get
U_E = $\frac{1}{2} \varepsilon_{0} \mathrm{E}^{2}$ ……. (5)
From equation (5), we infer that the energy is stored in the electric field existing between the plates of the capacitor. Once the capacitor is allowed to discharge, the energy is retrieved.

Question 29.
Explain any three recent advancements in medical technology.
Answer:
1. Virtual reality:
Medical virtual reality is effectively used to stop the brain from processing pain and cure soreness in the hospitalized patients. Virtual reality has enhanced surgeries by the use of 3D models by surgeons to plan operations. It helps in the treatment of Autism, Memory loss, and Mental illness.

2. Precision medicine:
Precision medicine is an emerging approach for disease treatment and prevention that takes into account individual variability in genes, environment, and lifestyle for each person. In this medical model it is possible to customise healthcare, with medical decisions, treatments, practices, or products which are tailored to the individual patient.

3. Health wearables:
A health wearable is a device used for tracking a wearer’s vital signs or health and fitness related data, location, etc. Medical wearables with artificial intelligence and big data provide an added value to healthcare with a focus on diagnosis, treatment, patient monitoring and prevention.
Note: Big Data: Extremely large data sets that may be analysed computationally to reveal patterns, trends, and associations, especially relating to human behaviour and interactions.

Question 30.
Two light sources with amplitudes 5 units and 3 units respectively interfere with each other. Calculate the ratio of maximum and minimum intensities.
Answer:
Amplitudes, a₁ = 5, a₂ = 3
Resultant amplitude, A = $\sqrt{a_{1}^{2}+a_{2}^{2}+2 a_{1} a_{2} \cos \phi}$
Resultant amplitude is maximum when,
Φ = 0, cos 0 = 1, A_max = $\sqrt{a_{1}^{2}+a_{2}^{2}+2 a_{1} a_{2}}$
A_max = $\sqrt{\left(a_{1}+a_{2}\right)^{2}}=\sqrt{(5+3)^{2}}=\sqrt{(8)^{2}}$ = 8 units
Resultant amplitude is minimum when, Φ = π, cos π = -1, A_max = $\sqrt{a_{1}^{2}+a_{2}^{2}-2 a_{1} a_{2}}$
A_min = $\sqrt{\left(a_{1}-a_{2}\right)^{2}}=\sqrt{(5-3)^{2}}=\sqrt{(2)^{2}}$ = 2 units
I ∝ A²
$\frac{I_{\max }}{I_{\min }}=\frac{\left(A_{\max }\right)^{2}}{\left(A_{\min }\right)^{2}}$
Substituting, $\frac{I_{\max }}{I_{\min }}=\frac{(8)^{2}}{(2)^{2}}=\frac{64}{4}=16$ (or) I_max : I_min = 16 : 1

Question 31.
An electron moves in a circular orbit with a uniform speed v it produces a magnetic field B at the centre of the circle. Prove that the radius of the circle is proportional to $\sqrt{\frac{v}{\mathbf{B}}}$
Answer:
The magnetic field produce by moving electron in circular path.

Question 32.
Give the construction and working of photo emissive cell.
Answer:

Photo emissive cell:
Its working depends on the electron emission from a metal cathode due to irradiation of light or other radiations.
Construction:

It consists of an evacuated glass or quartz bulb in which two metallic electrodes – that is, a cathode and an anode are fixed.
The cathode C is semi-cylindrical in shape and is coated with a photo sensitive material. The anode A is a thin rod or wire kept along the axis of the semi-cylindrical cathode.
A potential difference is applied between the anode and the cathode through a galvanometer G.

Working:

When cathode is illuminated, electrons are emitted from it. These electrons are attracted by anode and hence a current is produced which is measured by the galvanometer.
For a given cathode, the magnitude of the current depends on
- the intensity to incident radiation and
- the potential difference between anode and cathode.

Question 33.
In the circuit shown in the figure the input voltage V₁ = +5 V, V_BE = + 0.8 V and V_CE = + 0.12 V_CE. Find the values of I_B, I_C and β.
Answer:

PART – IV

Answer all the questions. [5 × 5 = 25]

Question 34.
(a) Obtain the expression for electric field due to an uniformly charged spherical shell at a distance its centre.
Answer:
Electric field due to a uniformly charged spherical shell: Consider a uniformly charged spherical shell of radius R and total charge Q. The electric field at points outside and inside the sphere is found using Gauss law.

Case (a) At a point outside the shell (r > R): Let us choose a point P outside the shell at a distance r from the center as shown in figure (a). The charge is uniformly distributed on the surface of the sphere (spherical symmetry). Hence the electric field must point radially outward if Q > 0 and point radially inward if Q < 0. So we choose a spherical Gaussian surface of radius r and the total charge enclosed by this Gaussian surface is Q. Applying Gauss law,
$\oint \overrightarrow{\mathrm{E}} \cdot d \overrightarrow{\mathrm{A}}=\frac{\mathrm{Q}}{\varepsilon_{0}}$ …(1)

The electric field $\vec{E}$ and $d \vec{A}$ point in the same direction (outward normal) at all the points on the Gaussian surface. The magnitude of $\vec{E}$ is also the same at all points due to the spherical symmetry of the charge distribution.

Substituting this value in equation (2).
E.4πr² = $\frac{Q}{\varepsilon_{0}}$
E.4πr² = $\frac{Q}{\varepsilon_{0}}$ (or) E = $\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r^{2}}$
In vector form $\overrightarrow{\mathrm{E}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Q}}{r^{2}} \hat{r}$

The electric field is radially outward if Q > 0 and radially inward if Q < 0. From equation (3), we infer that the electric field at a point outside the shell will be same as if the entire charge Q is concentrated at the center of the spherical shell. (A similar result is observed in gravitation, for gravitational force due to a spherical shell with mass M)

Case (b): At a point on the surface of the spherical shell (r = R): The electrical field at points on the spherical shell (r = R) is given by
$\overrightarrow{\mathrm{E}}=\frac{Q}{4 \pi \varepsilon_{0} R^{2}} \hat{r}$ …… (4)

Case (c) At a point inside the spherical shell (r < R): Consider a point P inside the shell at a distance r from the center. A Gaussian sphere of radius r is constructed as shown in the figure (b). Applying Gauss law

E.4πr² = $\frac{Q}{\varepsilon_{0}}$ ….. (5)
Since Gaussian surface encloses no charge, So Q = 0. The equation (5) becomes
E = 0 (r < R) …….. (6)
The electric field due to the uniformly charged spherical shell is zero at all points inside the shell.

[OR]

Question 34.
(b) Write any five properties of electromagnetic waves.
Answer
Properties of electromagnetic waves:

Electromagnetic waves are produced by any accelerated charge.
Electromagnetic waves do not require any medium for propagation. So electromagnetic wave is a non-mechanical wave.
Electromagnetic waves are transverse in nature. This means that the oscillating electric field vector, oscillating magnetic field vector and propagation vector (gives direction of propagation) are mutually perpendicular to each other.
Electromagnetic waves travel with speed which is equal to the speed of light in vacuum, or freer space, c = $\frac{1}{\sqrt{\varepsilon_{0} \mu_{0}}}$ = 3 × 10⁸ ms^-1
The speed of electromagnetic wave is less than speed in free space or vacuum, that is, v < c. In a medium of refractive index,

Question 35.
(a) What is modulation? Explain the types of modulation with necessary diagrams.
Answer:
Modulation: For long distance transmission, the low frequency baseband signal (input signal) is superimposed onto a high frequency radio signal by a process called modulation.
There are 3 types of modulation based on which parameter is modified. They are
(i) Amplitude modulation,
(ii) Frequency modulation, and
(iii) Phase modulation.

(i) Amplitude Modulation (AM):
If the amplitude of the carrier signal is modified according to the instantaneous amplitude of the baseband signal, then it is called amplitude modulation. Here the frequency and the phase of the carrier signal remain constant. Amplitude modulation is used in radio and TV broadcasting.

The signal shown in figure (a) is the message signal or baseband signal that carries information, figure (b) shows the high frequency carrier signal and figure (c) gives the amplitude modulated signal. We can see clearly that the carrier wave is modified in proportion to the amplitude of the baseband signal.
Amplitude Modulation

(ii) Frequency Modulation (FM):
The frequency of the camer signal is modified according to the instantaneous amplitude of the baseband signal in frequency modulation. Here the amplitude and the phase of the carrier signal remain constant. Increase in the amplitude of the baseband signal increases the frequency of the carrier signal and vice versa. This leads to compressions and rarefactions in the frequency spectrum of the modulated wave. Louder signal leads to compressions and relatively weaker signals to rarefactions. When the amplitude of the baseband signal is zero in Figure (a), the frequency of the modulated signal is the same as the carrier signal. The frequency of the modulated wave increases when the amplitude of the baseband signal increases in the positive direction
(A, C). The increase in amplitude in the negative half cycle (B,D) reduces the frequency of the modulated wave (figure (c)).

(iii) Phase Modulation (PM):
The instantaneous amplitude of the baseband signal modifies the phase of the camer signal keeping the amplitude and frequency constant is called phase modulation. This modulation is used to generate frequency modulated signals. It is similar to frequency modulation except that the phase of the carrier is varied instead of varying frequency. The carrier phase changes according to increase or decrease in the amplitude of the baseband signal. When the modulating signal goes positive, the amount of phase
lead increases with the amplitude of the modulating signal. Due to this, the carrier signal is compressed or its frequency is increased.

On the other hand, the negative half cycle of the baseband signal produces a phase lag in the carrier signal. This appears to have stretched the frequency of the carrier wave. Hence similar to frequency modulated wave, phase modulated wave also comprises of
compressions and rarefactions. When the signal voltage is zero (A, C and E) the carrier frequency is unchanged.

[OR]

Question 35.
(b) Find the expression for the mutual inductance between a pair of coils and show that (M₁₂ = M₂₁).
Answer:
Mutual induction: When an electric current passing through a coil changes with time, an emf is induced in the neighbouring coil. This phenomenon is known as mutual induction and the emf is called mutually induced emf.

Consider two coils which are placed close to each other. If an electric current q is sent through coil i₁ the magnetic field produced by it is also linked with coil 2.
Let Φ₂₁ be the magnetic flux linked with each turn of the coil 2 of N₂ turns due to coil 1, then the total flux linked with coil 2 (N₂Φ₂₁) is proportional to the current i₁ in the coil 1.
N₂Φ₂₁ ∝ i₁
N₂Φ₂₁ = M₂₁i₁ or M₂₁ = $\frac{N_{2} \Phi_{21}}{i_{1}}$
The constant of proportionality M₂₁ is the mutual inductance of the coil 2 with respect to coil 1. It is also called as coefficient of mutual induction. If i₁ = 1 A, then M₂₁ = N₂Φ₂₁.
Therefore, the mutual inductance M₂₁ is defined as the flux linkage of the coil 2 when 1A current flows through coil 1.
When the current i₁ changes with time, an emf ξ₂ is induced in coil 2. From Faraday’s law of electromagnetic induction, this mutually induced emf ξ₂₁ is given by

The negative sign in the above equation shows that the mutually induced emf always opposes the change in current with respect to time. If $\frac{d i_{1}}{d t}$ = 1 As^-1, then M₂₁ = -ξ₂.
Mutual inductance M₂₁ is also defined as the opposing emf induced in the coil 2 when the rate of change of current through the coil I is l As^-1.
Similarly, if an electric current i₂ through coil 2 changes with time, then emf ξ₁ is induced in coil 1. Therefore, N

where M₁₂ is the mutual inductance of the coil I with respect to coil 2. It can be shown that for a given pair of coils, the mutual inductance is same. i.e., M₂₁ = M₁₂ = M.
In general, the mutual induction between two coils depends on size, shape, the number of turns of the coils, their relative orientation and permeability of the medium.

Question 36.
(a) Derive the expression for the radius of the orbit of the electron and its velocity using Bohr atom model.
Answer:
Radius of the orbit of the electron and velocity of the electron:
Consider an atom which contains the nucleus at rest and an electron revolving around the nucleus in a circular orbit of radius rn as shown in figure. Nucleus is made up of protons and neutrons. Since proton is positively charged and neutron is electrically neutral, the charge of a nucleus is purely the total charge of protons.

Let Z be the atomic number of the atom, then +Ze is the charge of the nucleus. Let -e be the charge of the electron. From Coulomb’s law, the force of attraction between the nucleus and the electron is

$\vec{F}$_coloumb = $\frac{1}{4 \pi \varepsilon_{0}} \frac{(+\mathrm{Ze})(\mathrm{e})}{\mathrm{r}_{\mathrm{n}}^{2}} \hat{\mathrm{r}}$
$=-\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e^{2}}{r_{n}^{2}} \hat{r}$
This force provides necessary centripetal force
$\vec{F}$_centripetal = $\frac{m v_{n}^{2}}{r_{n}} \hat{r}$
where m be the mass of the electron that moves with a velocity DW in a circular orbit. Therefore,
|$\vec{F}$_coloumb| = |$\vec{F}$_centripetal|
$\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e^{2}}{r_{n}^{2}}=\frac{m v_{n}^{2}}{r_{n}}$
r_n = $\frac{4 \pi \varepsilon_{0}\left(m \mathrm{v}_{n} r_{n}\right) 2}{Z m e^{2}}$ ……. (1)
From Bohr’s assumption, the angular momentum quantization condition mυ_nr_n = l_n = n_n

where n ∈ N . Since, ε₀, h, e and n are constants. Therefore, the radius of the orbit becomes
r_n = a₀$\frac{n^{2}}{Z}$
Where a₀ = $\frac{\varepsilon_{0} h^{2}}{\pi m e^{2}}$ = 0.529Å. This is known as Bohr radius which is the smallest radius of the orbit in an atom. Bohr radius is also used as unit of length called Bohr. 1 Bohr = 0.53 Å. For hydrogen atom (Z = 1), the radius of n^th orbit is
r_n = a₀n²
For the first orbit (ground state),
r₁ = a₀ = 0.529 Å
For the second orbit (first excited state),
r₂ = 4a₀ = 2.116 Å
For the third orbit (second excited state),
r₃ = 9a₀ = 4.761 Å and so on.
Thus the radius of the orbit from centre increases with n, that is, r_n ∝ c n² as shown in figure. Further, Bohr’s angular momentum quantization condition leads to
mυ_nr_n = mυ_na₀n² = n$\frac{h}{2 \pi}$
υ_n = $\frac{h}{2 \pi m a_{0}} \frac{Z}{n}$
υ_n ∝ $\frac{1}{n}$
Note that the velocity of electron decreases as the principal quantum number increases as shown in figure. This curve is the rectangular hyperbola. This implies that the velocity of electron in ground state is maximum when compared to excited states.

[OR]

Question 36.
(b) Discuss the working and theory of cyclotron in detail.
Answer:

Cyclotron: Cyclotron is a device used to accelerate the charged particles to gain large kinetic energy. It is also called as high energy accelerator. It was invented by Lawrence and Livingston in 1934.

Principle: When a charged particle moves normal to the magnetic field, it experiences magnetic Lorentz force.

Construction:The particles are allowed to move in between two semicircular metal containers called Dees (hollow D – shaped objects). Dees are enclosed in an evacuated chamber and it is kept in a region with uniform magnetic field controlled by an electromagnet. The direction of magnetic field is normal to the plane of the Dees. The two Dees are kept separated with a gap and the source S (which ejects the particle to be accelerated) is placed at the center in the gap between the Dees. Dees are connected to high frequency alternating potential difference.

Working: Let us assume that the ion ejected from source S is positively charged. As soon as ion is ejected, it is accelerated towards a Dee (say, Dee – 1) which has negative potential at that time. Since the magnetic field is normal to the plane of the Dees, the ion undergoes circular path. After one semi-circular path in Dee-1, the ion reaches the gap between Dees. At this time, the polarities of the Dees are reversed so that the ion is now accelerated towards Dee-2 with a greater velocity. For this circular motion, the centripetal force of the charged particle q is provided by Lorentz force.
$\frac{m v^{2}}{r}$ = qv B ⇒ r = $\frac{m}{q \mathrm{B}} v$ ⇒ r ∝ v
From the equation, the increase in velocity increases the radius of circular path. This process continues and hence the particle undergoes spiral path of increasing radius. Once it reaches near the edge, it is taken out with the help of deflector plate and allowed to hit the target T. Very important condition in cyclotron operation is the resonance condition. It happens when the frequency f at which the positive ion circulates in the magnetic field must be equal to the constant frequency of the electrical oscillator f_osc From equation
f_osc = $\frac{q \mathbf{B}}{2 \pi m} \Rightarrow \mathrm{T}=\frac{1}{f_{\mathrm{osc}}}$
The time period of oscillation is T = $\frac{2 \pi m}{q B}$
The kinetic energy of the charged particle is K E = $\frac{1}{2} m v^{2}=\frac{q^{2} \mathbf{B}^{2} r^{2}}{2 m}$
Limitations of cyclotron
(a) the speed of the ion is limited
(b) electron cannot be accelerated
(c) uncharged particles cannot be accelerated

Question 37.
(a) Obtain lens maker’s formula and mention its significance.
Answer:
Lens maker’s formula and lens equation: Let us consider a thin lens made up of a medium of refractive index n₂ is placed in a medium of refractive index n₁ . Let R₁ and R₂ be the radii of curvature of two spherical surfaces (1) and (2) respectively and P be the pole. Consider a point object O on the principal axis. The ray which falls very close to P, after refraction at the surface (1) forms image at I’. Before it does so, it is again refracted by the surface (2). Therefore the final image is formed at I. The general equation for the refraction at a spherical surface is given by

For the refracting surface (1), the light goes from to n₁ to n₂

If the object is at infinity, the image is formed at the focus of the lens. Thus, for u = oo, v =/ Then the equation becomes.

If the refractive index of the lens is n₂ and it is placed in air, then n₂ = n and n₁ = 1. So the equation (4) becomes,
$\frac{1}{f}=(n-1)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)$

The above equation is called the lens maker’s formula, because it tells the lens manufactures what curvature is needed to make a lens of desired focal length with a material of particular refractive index. This formula holds good also for a concave lens. By comparing the equations (3) and (4) we can write,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ ……… (6)
This equation is known as lens equation which relates the object distance it and image distance v with the focal length / of the lens. This formula holds good for a any type of lens.

Question 37.
(b) Explain the construction and working of a full wave rectifier.
Answer:
Full wave rectifier:
The positive and negative half cycles of the AC input signal pass through the full wave rectifier circuit and hence it is called the full wave rectifier. It consists of two p-n junction diodes, a center tapped transfonner, and a load resistor (RL). The centre is usually taken as the ground or zero voltage reference point. Due to the centre tap transformer, the output voltage rectified by each diode is only one half of the total secondary voltage.

During positive half cycle:
When the positive half cycle of the ac input signal passes through the circuit, terminal M is positive, G’ is at zero potential and N is at negative potential. This forward biases diode D₁ and reverse biases diode D₂ Hence, being forward biased, diode D₁ conducts and current flows along the path MD₁ AGC . As a result, positive half cycle of the voltage appears across R_L in the direction G to C.

During negative half cycle:
When the negative half cycle of the ac input signal passes through the circuit, terminal N is positive, G is at zero potential and M is at negative potential. This forward biases diode D₂ and reverse biases diode D₁. Hence, being forward biased, diode D₂ conducts and current flows along the path ND₂ BGC . As a result, negative half cycle of the voltage appears across R_L in the same direction from G to C

Hence in a full wave rectifier both positive and negative half cycles of the input signal pass through the circuit in the same direction as shown in figure (b). Though both positive and negative half cycles of ac input are rectified, the output is still pulsating in nature.

The efficiency ( η) of full wave rectifier is twice that of a half wave rectifier and is found to be 81.2 %. It is because both the positive and negative half cycles of the ac input source are rectified.

Question 38.
(a) (i) Derive the expression for the de Broglie wavelength of an electron.
Answer:
An electron of mass m is accelerated through a potential difference of V volt. The kinetic energy acquired by the electron is given by
$\frac{1}{2} m v^{2}$ = eV
Therefore, the speed v of the electron is v = $\sqrt{\frac{2 e \mathrm{V}}{m}}$
Hence, the de Broglie wavelength of the electron is λ = $\frac{h}{m v}=\frac{h}{\sqrt{2 e m \mathrm{V}}}$
Substituting the known values in the above equation, we get

For example, if an electron is accelerated through a potential difference of 100V, then its de Broglie wavelength is 1.227 Å . Since the kinetic energy of the electron, K = eV, then the de Broglie wavelength associated with electron can be also written as
λ = $\frac{h}{\sqrt{2 m \mathrm{K}}}$

(ii) An electron is accelerated through a potential difference of 81V. What is the de Broglie wavelength associated with it? To which part of the electromagnetic spectrum does this wavelength correspond?
de-Broglie wavelength of an electron beam accelerated through a potential difference of V volts is

X-ray is the part of electromagnetic spectrum does this wavelength corresponds. X-ray has the wavelengths ranging from about 10^-8 to 10^-12 m.

[OR]

Question 38.
(b) (i) How will you measure the internal resistance of a cell by potentiometer?
Answer:
Measurement of internal resistance of a cell by potentiometer:
To measure the internal resistance of a cell, the circuit connections are made as shown in figure. The end C of the potentiometer wire is connected to the positive terminal of the battery Bt and the negative terminal of the battery is connected to the end D through a key K₁ This forms the primary circuit.

The positive terminal of the cell ξ whose internal resistance is to be determined is also connected to the end C of the wire. The negative terminal of the cell ξ is connected to a jockey through a galvanometer and a high resistance. A resistance box R and key K₂ are connected across the cell ξ. With K₂ open, the balancing point J is obtained and the balancing length CJ = l₁ is measured. Since the cell is in open circuit, its emf is
ξ ∝ l₁ ….. (1)
A suitable resistance (say, 10 H) is included in the resistance box and key K₂ is closed.
Let r be the internal resistance of the cell. The current passing through the cell and the resistance R is given by
I = $\frac{\xi}{R+r}$
The potential difference across R is
V = $\frac{\xi R}{R+r}$
When this potential difference is balanced on the potentiometer wire, let /2 be the balancing length.
Then $\frac{\xi R}{R+r} \propto l_{2}$ ….. (2)
From equations (1) and (2)

Substituting the values of the R, l₁ and l₂, the internal resistance of the cell is determined. The experiment can be repeated for different values of R. It is found that the internal resistance of the cell is not constant but increases with increase of external resistance connected across its terminals.

(ii) A cell supplies a current of 0.9 A through a 1Ω resistor and a current of 0.3 A through a 2 Ω resistor. Calculate the internal resistance of the cell.
Current from the cell 1,I₁, = 0.9 A and Resistor, R₁ = 1Ω
Current from the cell 2,I₂ = 0.3 A and Resistor, R₂ = 2Ω
Current in the circuit, I₁ = $\frac{\xi}{r+R_{1}}$
ξ = I₁r + I₁R₁ ….. (1)
Current in the circuit, I₂ = $\frac{\xi}{r+R_{2}}$
ξ = I₂r + I₂R₂ ….. (2)
From equation (1) and (2),

|r| = |-0.5| Ω
Magnitude of internal resistance r = 0.5 Ω

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