## Samacheer Kalvi 12th Chemistry Guide Chapter 8 Ionic Equilibrium

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Chemistry Guide Pdf Chapter 8 Ionic Equilibrium Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

### 12th Chemistry Guide Ionic Equilibrium Text Book Questions and Answers

Part – I Text Book Evaluation

Question 1.
Concentration of the Ag+ ions in a saturated solution of Ag2C2O4 is 2.24 x 10-4 mol L-1 solubility product of Ag2C2O4 is ………………
(a) 2.42 x 10-8 mol3 L-3
(b) 2.66 x 10-12 12 mol3 L-3
(c) 45 x 10-11 mol3 L-3
(d) 5.619 x 10-12 mol3 L-3
(d) 5.619 x 10-12 mol3 L-3
Solution:
Ag2C2O4 2Ag+ + C2 O42-
[Ag+] = 2.24 x 10-4 mol L-1
[C2 O42-] = $$\frac{2.24 \times 10^{-4}}{2}$$ mol L-1
= 1.12 x 10-4 mol L-1
Ksp = [Ag+]2 [C2O42-]
=(2.24 x 10-4 mol L-1)2 (1.12 x 10-4 mol L-1)
=5.619 x 10-12 mol3 L-3

Question 2.
Following solutions were prepared by mixing different volumes of NaOH of HCl different concentrations.
(i) 60 mL $$\frac { M }{ 10 }$$ HCI + 40 mL $$\frac { M }{ 10 }$$ NaOH
(ii) 55 mL $$\frac { M }{ 10 }$$ HCl + 45 mL $$\frac { M }{ 10 }$$ NaOH
(iii) 75 mL $$\frac { M }{ 5 }$$ HCI +25 mL $$\frac { M }{ 5 }$$ MNaOH
(iv) 100 mL $$\frac { M }{ 10 }$$ HCI+ 100 mL $$\frac { M }{ 10 }$$ NaOH

pH of which one of them wilt be equal to 1?
(a) (iv)
(b) (i)
(c) (ii)
(d) (iii)
(d) (iii) 75 mL $$\frac { M }{ 5 }$$ HCI + 25 mL $$\frac { M }{ 5 }$$ NaOH
No of moles of HCl = 0.2 x 75 x 10-3 = 15 x 10-3
No of moles of NaOH = 0.2 x 25 x 10-3 = 5 x 1o-3
No of moles of HCl after mixing = 15 x 10-3 – 5 x 10-3
∴ Concentration of HCl
$$\frac{\text { No of moles of }}{\mathrm{HCl} \text { Vol inlitre }}$$
for (iii) solution, pH of 0.1 M HCI = – 1og10 (0.1) = 1.

Question 3.
The solubility of BaSO4 in water is 2.42 x 10-3 gL-1 at 298K. The value of its solubility product
(Ksp) will be …………………..
(Given molar mass of BaSO4 = 233g mol-1)
(a) 1.08 x 10-14 mol2L2
(b) 1.08 x 10-12 mol2L2
(c) 1.08 x 10-10 mol2 L2
(d) 1.08 x 10-8 mol2L-2
(c) 1.08 x 10-10 mol2 L2
Solution:
BaSO4 $$\rightleftharpoons$$ Ba2+ + SO42-
Ksp = (s) (s)
Ksp = (s)2
= (2.42 x 10-3g L-1)2
= $$\left(\frac{2.42 \times 10^{-3} \mathrm{~g} \mathrm{~L}^{-1}}{233 \mathrm{~g} \mathrm{~mol}^{-1}}\right)^{2}$$
= (0.01038 x 1o-3)2
= (1.038 x 10-5)2
= 1.077 x 10-10
= 1.08 x 10-10 mol2 L2

Question 4.
pH of a saturated solution of Ca(OH)2 is 9. The Solubility product (Ksp) of Ca(OH)2
(a) 0.5 x 10-15
(b) 0.25 x 10-10
(c) 0.125 x 10-15
(d) 0.5 x 10-10
(a) 0.5 x 10-15
Solution:
Ca(OH)2 $$\rightleftharpoons$$ Ca2+ + 2OH
Given that pH = 9
pOH = 14 – 9 = 5
[p0K = – 1og10[OH]]
[OH] = 10-pOH
[OH] =10-5M
Ksp = [Ca2+] [OH]2
= $$\frac{10^{-5}}{2} \times\left(10^{-5}\right)^{2}$$ = 0.5 x 10-15

Question 5.
Conjugate base for bronsted acids H2O and HF are ………………
(a) OH and H2FH+, respectively
(b) H3O+ and F, respectively
(c) OH and F, respectively
(d) H3O+ and H2F+, respectively
(c) OH and F, respectively
Solution:

∴ Conjugate bases are OH and F respectively

Question 6.
Which will make basic buffer?
(a) 50 mL of 0.1M NaOH + 25mL of 01M CH3COOH
(b) 100 mL of 0.1M CH3COOH + 100 mL of 0.1M NH4OH
(c) 100 mL of 0.1M HCI + 200 mL of 0.1M NH4OH
(d) 100 mL of 0.1M HCI + 100 mL of O.1 M NaOH
(c) 100 mL of 0.1M HCI + 200 mL of 0.1M NH4OH
Solution:
Basic buffer is the solution which has weak base and its salt

Question 7.
Which of the following fluro – compounds is most likely to behave as a Lewis base?
(a) BF3
(b) PF3
(c) CF4
(d) SiF4
(b) PF3
Solution:
BF3 → electron deficient → Lewis acid
PF3 → electron rich → Lewis base
CF4 → neutral → neither lewis acid nor base
SiF4 → neutral → neither lewis acid nor base

Question 8.
Which of these is not likely to act as lewis base?
(a) BF3
(b) PF3
(c) CO
(d) F
(a) BF3
Solution:
BF3 → electron deficient → Lewis acid
PF3 → electron rich → Lewis base
CO → having lone pair of electron → Lewis base
F → unshared pair of electron → lewis base

Question 9.
What is the decreasing order of strength of bases?
OH, NH2, H – C ≡ C and CH3 – CH2
(a) OH > NH2 > H – C ≡ C > CH3 – CH2
(b) NH2 > OH > CH3 – CH2 > H – C ≡ C
(c) CH3 – CH2, > NH2 > H – C ≡ C > OH
(d) OH > H – C ≡ C > CH3 – CH2 > NH2
(c) CH3 – CH2, > NH2 > H – C ≡ C > OH
Solution:
Acid strength decreases in the order
HOH > CH ≡ CH > NH3 > CH3CH3
Its conjucate bases arc in the reverse order
CH3 – CH2 > NH2 > H – C ≡ C > OH

Question 10.
The aqueous solutions of sodium formate, anilinium chloride and potassium cyanide are respectively
(a) acidic, acidic, basic
(b) basic, acidic, basic
(c) basic, neutral, basic
(d) none of these
(b) basic, acidic, basic
Solution:

Question 11.
The percentage of pyridine (C5H5N) that forms pyridinium ion (C5H5NH) in a 0.10M aqueous pyridine solution (Kb for C5H5N = 1.7 x 10-9) iS ……………..
(a) 0.006%
(b) 0.013%
(c) 0.77%
(d) 1.6%
(b) 0.013%

Percentage of dissociation
= $$\sqrt { 1.7 }$$ x 10-4 x 100 = 1.3 x 10-2 = 0.013%

Question 12.
Equal volumes of three acid solutions of pH 1,2 and 3 are mixed in a vessel. What will be the H+ ion concentration in the mixture?
(a) 37 x 10-2
(b) 10-6
(c) 0.111
(d) none of these
(a) 3.7 x 10-2
pH = – log10 [H+]
[H+] = 10-pH
Let the volume be x mL
V1M1 + V2M2 + V3M3 = VM
x mL of 10-1M + x mL of 10-2M + x mL of 10-3 M
= 3 x mL of [H+]
= 3 x mL of [H+]

= 0.037 = 3.7 x 10-2

Question 13.
The solubility of AgCl (s) with solubility product 1.6 x 10-10 in 0. 1 M NaCl solution would be ………….
(a) 1.26 x 10-5 M
(b) 1.6 x 10-9 M
(c) 1.6 x 10-11 M
(d) Zero
(b) 1.6 x 10-9 M
AgCl (s) $$\rightleftharpoons$$ Ag+(aq) + Cl(aq)

Question 14.
If the solubility product of lead iodide is 3.2 x 10-8, its solubility will be …………..
(a) 2 x 10-3M
(b) 4 x 10-4 M
(c) 1.6 x 10-5 M
(d) 1.8 x 10-5 M
(a) 2 x 10-3M
Solution:
PbI2 (s) $$\rightleftharpoons$$ Pb2+ (aq) + 2I (aq)
Ksp = (s) (2s)2
3.2 x 10-8 = 4s3
S = $$\left(\frac{3.2 \times 10^{-8}}{4}\right)^{1 / 3}$$
= (8 x 10-9)1/3 = 2 x 10-3MNaOH

Question 15.
Using Gibb’s free energy change, ∆G0 = 57.34 KJ mol-1, for the reaction, X2Y(s) $$\rightleftharpoons$$ 2X+ + Y2-(aq), calculate the solubility product of X2Y in water at 300K (R = 8.3 J K-1 Mol-1)
(a) 10-10
(b) 10-12
(c) 10-14
(d) can not be calculated from the given data
(a) 10-10
Solution:

Question 16.
MY and NY3, are insoluble salts and have the same Ksp values of 6.2 x 10-13 at room temperature. Which statement would be true with regard to MY and NY3?
(a) The salts MY and NY3 are more soluble in O.5 M KY than in pure water
(b) The addition of the salt of KY to the suspension of MY and NY3 will have no effect on
(c) The molar solubities of MY and NY3 in water are identical
(d) The molar solubility of MY in water is less than that of NY3
(d) The molar solubility of MY in water is less than that of NY3
Solution:
Addition of salt KY (having a common ion Y) decreases the solubility of MY and NY3 due to common ion effect. Option (a) and (b) are wrong.
For salt MY, MY $$\rightleftharpoons$$ M+ + Y
Ksp = (s) (s)
6.2 x 10-13 = s2
therefore $$\mathrm{s}=\sqrt{6.2 \times 10^{-13}} \simeq 10^{-7}$$
for salt NY3,
NY3 ⇌ N3+ + 3Y
Ksp = S (3S)3
Ksp = 27s4
$$S=\left(\frac{6.2 \times 10^{-13}}{27}\right)^{1 / 4} \mathrm{~s} \simeq 10^{-4}$$
The molar solubility of MY in water is less than of NY3

Question 17.
What is the pH of the resulting solution when equal volumes of 0.1M NaOH and 0.01M HCl are mixed?
(a) 2.0
(b) 3
(c) 7.0
(d) 12.65
(d) 12.65
Solution:
x ml of 0.1 m NaOH + x ml of 0.01 M HCI
No. of moles of NaOH = 0.1 x x x 10-3 = 0.l x x 10-3
No. of moles of HCl = 0.01 x x x 10-3 = 0.01 x x 10-3
No. of moles of NaOH after mixing = 0.1 x x 10-3 – 0.01x  x 10-3
= 0.09x x 10-3
Concentration of NaOH = $$\frac{0.09 x \times 10^{-3}}{2 x \times 10^{-3}}=0.045$$
[OH] = 0.045
pOH = – log (4.5 x 10-2)
= 2 – log 4.5
= 2 – 0.65 = 1.35
pH = 14 – 1.35 = 12.65

Question 18.
The dissociation constant of a weak acid is 1 x 10-3 . In order to prepare a buffer solution with a pH =4, the [Acid] / [Salt] ratio should be ………………..
(a) 4:3
(b) 3:4
(c) 10:1
(d) 1:10
(d) 1:10
Solution:
Ka = 1 x 10-3 ; pH = 4

Question 19.
The pH of 10-5 M KOH solution will be …………..
(a) 9
(b) 5
(c) 19
(d) none of these
(a) 9
Solution:

[OH] = 10-5M.
pH = 14 – pOH .
pH = 14 – ( – log [OH])
= 14 + log [OH]
= 14 + log 10-5
= 14 – 5 = 9

Question 20.
H2PO4 the conjugate base of …………….
(a) PO43-
(b) P2O5
(c) H3PO4
(d) HPO42-
(c) H3PO4
Solution:

Question 21.
Which of the following can act as lowery – Bronsted acid well as base?
(a) HCl
(b) SO42-
(c) HPO42-
(d) Br
(c) HPO42-
Solution:
HPO42- can have the ability to accept a proton to form HPO4
It can also have the ability to donate a proton to form PO4-3.

Question 22.
The pH of an aqueous solution is Zero. The solution is ……………..
(a) slightly acidic
(b) strongly acidic
(c) neutral
(d) basic
(b) strongly acidic
Solution:
pH = – log10[H+]
[H+] =10-pH
= 100 = 1
[H+] = 1 M
The, solution is strongly acidic

Question 23.
The hydrogen ion concentration of a buffer solution consisting of a weak acid and its salts is given by ………………

a
Solution:
According to Henderson equation

Question 24.
Which of the following relation is correct for degree of hydrolysis of ammonium acetate?

c) h = $$\sqrt{\frac{K_{h}}{K_{a} \cdot K_{b}}}$$
Solution:
h = $$\sqrt{\frac{K_{h}}{K_{a} \cdot K_{b}}}$$

Question 25.
Dissociation constant of NH4OH is 1.8 x 10-5 the hydrolysis constant of NH4Cl would be …………….
(a) 1.8 x 10-19
(b) 5.55 x 10-10
(c) 5.55 x 10-5
(d) 1.80 x 10-5
(b) 5.55 x 1010
Solution:
$$\mathrm{K}_{\mathrm{h}}=\frac{\mathrm{K}_{\mathrm{W}}}{\mathrm{K}_{\mathrm{b}}}=\frac{1 \times 10^{-14}}{1.8 \times 10^{-5}}$$
= 0.55 x 10-9
= 5.5 x 10-10

Question 1.
What are lewis acids and bases? Give two example for each.

 Lewis acids: Lewis bases: 1. Electron pair acceptors Electron pair donors 2. EX: BF3 , AlCl3 EX: NH3 ,H2O

Question 2.
Discuss the Lowry – Bronsted concept of acids and bases.

• Lowry Bronsted acids – proton donors
• Lowry Bronsted bases – proton acceptors
• An acid1 after donating a proton becomes a base1 (conjugate base)
• A base2 after accepting a proton becomes an acid2 (conjugate acid)
• In general Lowry – Bronsted acid — base reaction is
• In the above reaction
• A conjugate acid – base pair differs by a proton.

Limitation : BF3, AlCl3 etc., that do not donate protons are known to behave as acid s.

Question 3.
Indentify the conjugate acid base pair for the following reaction in aqueous solution.
i) HS (aq) + HF $$\rightleftharpoons$$ F (aq) + H2S (aq)
ii) HPO42- + SO32- $$\rightleftharpoons$$ PO43- + HSO3
iii) NH4+ + CO32- $$\rightleftharpoons$$ NH3 + HCO3

Question 4.
Account for the acidic nature of HClO4. In terms of Bronsted – Lowry theory, identify its conjugate base.
HClO4 $$\rightleftharpoons$$ H+ + ClO4
1. According to Lowry – Bronsted concept, a strong acid has weak conjugate base and a weak acid has a strong conjugate base.

2. Let us consider the stabilities of the conjugate bases ClO4 , ClO3, CIO2 and ClO formed from these acid HClO4, HClO3, HCIO2, HOCI respectively.

These anions are stabilized to greater extent, it has lesser attraction for proton and therefore, will behave as weak base. Consequently, the corresponding acid will be strongest because weak conjugate base has strong acid and strong conjugate base has weak acid.

3. The charge stabilization mercases in the order, ClO < ClO2 < ClO3 < ClO4 .

This means ClO4 will have maximum stability and therefore will have a minimum attraction for W. Thus CIO4 will be weakest base and its conjugate acid HCIO4 is the strongest acid.

4. CIO4 is the conjugate base of the acid HClO4.

Question 5.
When aqueous ammonia is added to CuSO4 solution, the solution turns deep blue due to the formation of tetrammine copper (II) complex, [Cu(H2O)6]2+(aq) + 4NH3 (aq) $$\rightleftharpoons$$ [Cu(NH3)4]2+ (aq), among H2O and NH3 which is stronger Lewis base.

• [Cu(H2O)6]2+(aq) + 4NH3 (aq) $$\rightleftharpoons$$ [Cu(NH3)4]2+ (aq) + H2
• Nitrogen less electronegative than oxygen and donates its lone pair of electrons readily. Hence NH3 is a stronger Lewis base.
• If a better Lewis base (ligand) is available, a Lewis acid (central metal ion) will react (Ligand exchange reaction)
• In this reaction, H2O is exchanged with NH3.
• The Lewis acid Cu2+ exchanges the Lewis base H with better Lewis base N-H3 to form [Cu(NH3)4]2+
• Hence NH3 is a stronger Lewis base than H2O in this reaction.

Question 6.
The concentration of hydroxide ion in a water sample is found to be 2.5 x 10-6 M. Identify the nature of the solution.
The concentration of OH ion in a water sample is found to be 2.5 x 10 M
pOH = – log10 [OH ]
pOH = – 1og10 [2.5 x 10-6]
= – log10 [2.5] – log10 [10-6]
= – 0.3979 – ( – 6)
= – 0.3979 + 6
pOH = 5.6

We know that,
pH + pOH = 14
pH + 5.6 = 14
pH = 14 – 5.6
pH = 8.4
pH = 8.4, shows the nature of the solution is basic.

Question 7.
A lab assistant prepared a solution by adding a calculated quantity of HCl gas 25°C to get a solution with [H3O+] = 4 x 105 M. Is the solution neutral (or) acidic (or) basic.
[H3O+] = 4 x 10-5M
pH = – log10 [H3O+]
pH = – 1og10[4 x 105]
pH = – log10 [4] – log10 [10-5]
pH = – 0.6020 – ( – 5) = – 0.6020 + 5
pH = 4.398
Therefore, the solution is acidic.
Since pH is less than 7, the solution is acidic

Question 8.
Calculate the pH of 0.04 M HNO3 solution.
Concentration of HNO3 = 0.04M
[H3O+] = 0.04 mol dm-3
pH = – 1og[H3O+]
= – log (0.04)
= – log(4 x 10-2)
= 2 – log4 = 2 – 0.6021
= 1.3979
= 1.40

Identify the degree and leading coefficient calculator of polynomial functions.

Question 9.
Define solubility product.
Solubility product:
It is defined as the product of the molar concentration of the constituent ions, each raised to the power of its stoichiometric coefficient in a balanced equilibrium equation.

Question 10.
Define ionic product of water. Give its value at room temperature.
1. The product of the concentration of H+ and OH ions in water at a particular temperature is known as ionic product.
2. The ionic product of water at room temperature (25°C) is,
Kw = [H+] [OH+] (or)
Kw= [H3O+] [OH+]
Kw =(1 x 10-7) (1 x 10-7)
Kw= 1 x 10-14 mol2 dm-6

Question 11.
Explain common ion effect with an example.
Common ion Effect:
When a salt of a weak acid is added to the acid itself, the dissociation of the weak acid is suppressed further. Acetic acid is a weak acid. It is not completely dissociated in aqueous solution and hence the following equilibrium exists.
CH3COOH (aq) $$\rightleftharpoons$$ H+(aq)+ CH3COO (aq)

However, the added salt, sodium acetate, completely dissociates to produce Na+ and CH3COO ion.
CH3COONa (aq) → Na+ (aq) + CH3COO (aq) Hence, the overall concentration ofCH3COO is increased, and the acid dissociation equilibrium is disturbed.

We know from Le chatelier’s priñciple that when a stress is applied to a system at equilibrium, the system adjusts itself to nullify the effect produced by that stress. So, in order to maintain the equilibrium, the excess CH3COO ions combines with H ions to produce much more unionized CH3COOH i.e.,

the equilibrium will shift towards the left. In other words, the dissociation of CH3COOH is suppressed. Thus, the dissociation of a weak acid (CH3COOH) is suppressed in the presence of a salt (CH3COONa) containing an ion common to the weak electrolyte. It is called the common ion effect.

Question 12.
Derive an expression for Ostwald’s dilution law.

• Ostwald’s dilution law relates the dissociation constant of the weak acid (Ka) with its degree of dissociation (a) and the concentration ( C)
• Ostwald’s dilution law states that, when dilution increases, the degree of dissociation of weak electrolyte also increases
• Degree of dissociation
$$\alpha=\frac{\text { Number of moles dissociated }}{\text { Total number of moles }}$$
• Consider the dissociation of a weak acid (CH3COOH)

Question 13.
Define pH.
pH of a solution is defined as the negative logarithm of base 10 of the molar concentration of the hydronium ions present in the solution.
pH = – log10 [H3O+] (or) pH = – log10 [H+]

Question 14.
Calculate the pH of 1.5 x 10-3 M solution of Ba(OH)2
Acidity of Ba(OH)2 in 2
∴ Normality = Molarity x acidity
= 15 x 10-3 x 2
= 3 x 10-3
So, [OH] = Normality = 3 x 10-3
pOH = – log [OH]
= -log3 x 10-3
= -log3 – log10-3
= log 3 +3 log 10
= 3 – log3
= 3 – 0.4771
pOH = 2.5229
pH+ pOH = 14
pH = 14 – pOH
=14 – 2.5229
= 11.4771
pH = 11.48

Question 15.
50 ml of 0.05 M HNO3 is added to 50 ml of 0.025 M KOH. Calculate the pH of the resultant solution.
Solution:
Number of millinioles VrnI x Molarity
Millimoles of HNO3 = 50 x 0.05 = 2.5
Millimoles of KOH = 50 x 0.025 = 1.25
After mixing millimoles of
HNO3 remaining = 2.5 – 1.25 = 1.25
Total volume = 50 + 50 = 100 ml
Molarity = $$\frac{\text { Number of millimoles }}{\mathrm{Vml}}$$
= 1.25/100
Molarity = 1.25 x 10-2
HNO3 is mono basic.
∴ Normality Molarity x basicity
= 1.25 x 10-2 x 1
= 1.25 x 10-2
∴[H3O+] = 1.25 x 10-2
pH = – log[H3O+]
= – log 1.25 x 10-2
= – log 1.25 – log 10-2
= – log 1.25 + 2 log 10
=2-log 1.25 = 2-0.0969
pH = 1.9031

Question 16.
The Ka value for HCN is 10-9. What is the pH of 0.4 M HCN solution?
Ka =10-9
c = O.4M
pH = – log10 [H3O+]

∴ pH = – log(2 x 10-5)
= – log 2 – log (10-5)
= – 0.3010 + 5
pH = 4.699

Question17.
Calculate the extent of hydrolysis and the pH of 0.1 M ammonium acetate Given that.
Ka = Kb = 1.8 x 10-5
Solution:
Ka = Kb = 1.8 x 10-5

Question 18.
Derive an expression for the hydrolysis constant and degree of hydrolysis of salt of strong acid and weak base.
Consider the reaction between the strong acid HCl and the weak base NH4OH to form the salt NH4Cl and water.
$$\mathrm{HCl}_{(\mathrm{aq})}+\mathrm{NH}_{4} \mathrm{OH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{NH}_{4} \mathrm{Cl}_{(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}$$

In aqueous solution NH4C1 is completely dissociated as follows:
$$\mathrm{NH}_{4} \mathrm{Cl}_{(\mathrm{aq})} \rightarrow \mathrm{NH}_{4}{ }^{+}{ }_{(\mathrm{aq})}+\mathrm{Cl}^{-}{ }_{(\mathrm{aq})}$$

NH4+ is the strong conjugate acid of the weak base NH4OH.
Hence it has a tendency to react with 0H from water to form unionised NH4OH.
$$\mathrm{NH}_{4(\mathrm{aq})}^{+}+\mathrm{H}_{2} \mathrm{O}_{(l)} \rightleftharpoons \mathrm{NH}_{4} \mathrm{OH}_{(\mathrm{aq})}+\mathrm{H}^{+}(\mathrm{aq})$$
No such tendency is shown by Cl the weak conjugate base of the strong acid HCI.

In the above reaction 11+ ion is formed.
∴ [H+] > [OH] and the solution is acidic and the pH is less than 7.
Equilibrium constant (hydrolysis constant) for the above equilibrium is

Question 19.
Solubility product of Ag2CrO4 is 1 x 10-12. What is the solubility of Ag2CrO4 in 0.01 M AgNO3 solution?

[Ag+] = 2 S + 0.01
0.01 > > 2 S, 2 S can be neglected.
∵ [Ag+] = 0.01 = 1 x 10-2
[CrO4] = S; Ksp= 1 x 10-12
∴ Ksp = [Ag+]2 [CrO42-]
1 x 10-12 = (1 10-2)2 x s
∴ S = $$\frac{1 \times 10^{-12}}{1 \times 10^{-4}}$$ ⇒ S = 1 x 10-8M

Question 20.
Write the expression for the solubility product of Ca3(PO4)2
Ca3(PO4)2 (s) $$\rightleftharpoons$$ 3Ca2+ (3s) + 2PO43- (2s)
Solubility of Ca3(PO4)2 is,
Ksp = [Ca2+]3 . [PO43-]2
Ksp = (3s)3 . (2s)2
Ksp= 27 s3 . 4s2
Ksp = 108s5.

Question 21.
A saturated solution, prepared by dissolving CaF2(s) in water, has [Ca2+] = 3.3 x 10-4 M. What is the Kspof CaF2?
CaF2 (s) $$\rightleftharpoons$$ Ca2+(aq) + 2F(aq)
[F] = 2 [Ca2+] = 2 x 33 x 10-4 M
= 6.6 x 10-4 M
= [Ca2+] [F]2
= (3.3 x 10-4) (6.6 x 10-4)2
= 1.44 x 10-10

Question 22.
Ksp of AgCl is 1.8 x 10-10. Calculate molar solubility in 1 M AgNO3

[Cl] = S
Ksp = [Ag+] [Cl]
1.8 x 10-10 = (1) (S)
S = 1.8 x 10-10M

Question 23.
A particular saturated solution of silver chromate Ag2CrO4 has [Ag+] = 5 x 10-5 and [CrO4]2- = 4.4 x 10 M. What is the value of for Ag2CrO4?
Ag2CrO4 (s) $$\rightleftharpoons$$ 2Ag+(aq) + CrO42-(aq)
Ksp = [Ag+]2 [CrO42-]
= (5 x 10-5)2 (4.4 x 10-4)
= 1.1 x 10-12

Question 24.
Write the expression for the solubility product of Hg2CI2.

Question 25.
Ksp of Ag2CrO4 is 1.1 x 10-12 What is the solubility of Ag2CrO4 in 0.1M K2CrO4.

Question 26.
Will a precipitate be formed when 0.150 L of 0.1 M Pb(NO3)2 and 0.100 L of 0.2 M NaCl are mixed? (PbCI2) = 1.2 x 10-5.
Total volume = 0.150 + 0.100 = 0.250L

Number of moles of Pb2 = V x M
= 0.150 x 0.1

Number of moles of Cl = V x M = 0.100 x 0.2

Ionic Product = [Pb2+] [Cl]2
= (0,06) (0.08)2
= 3.84 x 10-4
Given Ksp= 1.2 x 10-5
Ionic product > Ksp
∴ Precipitation of PbCl2 will occur

Question 27.
of Al(OH)3 is 1 x 10-15 M. At what pH does 1.0 x 10-13 M AI3+ precipitate on the addition of buffer of NH4CI and NH4OH solution?
Al(OH)3 Al3+ (aq) + 3OH (aq)
Ksp = [Al3+] [OH]3
Al(OH)3 precipitates when
[Al3+] [OH]3 > Ksp
(1 x 10-3)[OH]3 > Ksp
[OH]3 > 1 x 10-12
[OH] > 1 x 10-4M
[OH] = l x 10-4 M
pOH = – 1og10[OH] = – log (1 x 10-4) = 4
pH = 14 – 4 = 10
Thus, Al (OH)3 precipitates at a pH of 10

III. Evaluate Yourself

Question 1.
Classify the following as acid (or) base using Arrhenius concept

1. HNO3
2. Ba(OH)2
3. H3PO4
4. CH3COOH

1. HNO3:
Nitric acid, dissociates to give hydrogen ions in water.
HNO3 is acid.

2. Ba(OH)2:
Barium hydroxide dissociates to give hydroxyl ions in water.
Ba(OH)2 is base.

3. H3PO4:
Orthophosphoric acid dissociates to give hydrogen ions in water.
H3PO4 is acid.

4. CH3COOH:
Acetic acid dissociates to give hydrogen ions in water.
CH3COOH is acid.

Question 2.
Write a balanced equation for the dissociation of the following in water and identify the conjugate acid-base pairs.
i) NH4
ii) H2SO4
iii) CH3COOH.
Evaluate.

Question 3.
Identify the Lewis acid and the Lewis base in the following reactions.
(i). CaO + CO2 → CaCO3
ii) CH3 – O – CH3 – AlCl3

(i). CaO + CO2 → CaCO3
CaO – Lewis base – All metals oxides are Lewis bases
CO2 – Lewis acid – CO2 contains a polar double bond.

ii)

Question 4.
H3BO3 accepts hydroxide ion from water as shown below
H3BO3 (aq) + H2O(l) = B(OH)4 + H+
Predict the nature of H3BO3 using Lewis concept.

Question 5.
At a particular temperature, the Kw of a neutral solution was equal to 4 x 10-14. Calculate the concentration of [H3O+] and [OH].
Given solution is neutral
[H3O+] = [OH]
Let [H3O+] = x ; then [OH] = x
Kw = [H3O+] [OH]
4 x 10-14 = x . x
x2 = 4 x 10-14
X = $$\sqrt{4 \times 10^{-14}}=2 \times 10^{-7}$$

Question 6.
a) Calculate pH of 10-8 M H2SO4
b) Calculate the concentration of hydrogen ion in moles per litre of a solution whose pH is 5.4
c) Calculate the pH of an aqueous solution obtained by mixing 50 ml of 0.2 M HCI with 50 ml 0.1 M NaOH
a.

In this case the concentration of H2SO4 is very low and hence [H3O] from water cannot be neglected
∴ [H3O+] = 2 x 10-8 (from H2SO4) + 10-7 (from water)
= 10-8(2+ 10)
= 12 x 10-8 = 1.2 x 10-7
pH = – log10[H3O+]
= – log10( 1.2 x 10-7)
= 7 – log101.2
= 7 – 0.0791 = 6.9209

b. pH of the solution = 5.4
[H3O+] = antilog of (- pH)
= antilog of (- 5.4)
= antilog of (-6 + 0.6) = $$\overline{6} .6$$
= 3.981 x 10-6
i.e., 3.98 x 10-6 mol dm-3

c. No. of moles of HCl = 0.2 x 50 x 10-3 = 10 x 10-3
No. of moles of NaOH =0.1 x 50 x 10-3 = 5 x 10-3
No. of moles of HCl after mixing = 10 x 10-3 – 5 x 10-3
= 5 x 10-3
after mixing total volume = 100 mL
∴ Concentration of HCl in moles per litre
$$=\frac{5 \cdot 10^{-3} \mathrm{~mole}}{100 \cdot 10^{-3} \mathrm{~L}} ;\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=5 \times 10^{-2} \mathrm{M}$$
[H3O+] = 5 x 10-2 M
pH = – log ( 5 x 10-2)
= 2 – log 5
= 2 – 0.6990
= 1.30

Question 7.
Kb for NH4OH is 1.8 x 10-5 Calculate the percentage of ionisation of 0.06 M ammonium hydroxide solution.

Question 8.
1. Explain the buffer action in a basic buffer containing equimolar ammonium hydroxide and ammonium chloride.
2. Calculate the pH of a buffer solution consisting of 0.4M CH3COOH and 0.4 M CH3COONa. What is the change in the pH after adding 0.01 mol of HCI to 500m1 of the above buffer solution.
Assume that the addition of HCI causes negligible change In the volume. Given: (K = 1.8 x 105).
1. Dissociation of buffer components
NH4OH (aq) $$\rightleftharpoons$$ NH4+ (aq) + OH (aq)
NH4CI → NH4+ + Cl
The added H+ ions are neutralized by NH4OH and there is no appreciable decrease in pH.
NH4OH(aq) + H+ $$\rightleftharpoons$$ NH4+(aq) + H2O (1)
NH4 (aq) + OH (aq) → NH4OH (aq)
The added OH ions react with NH4 to produce unionized NH4OH. Since NH4OH is a weak base, there is no appreciable increase in pH.

2. pH of buffer

Addition of 0.01 mol HCI to 500 ml of buffer
= $$\frac{0.01 \mathrm{~mol}}{500 \mathrm{~mL}}=\frac{0.01 \mathrm{~mol}}{\frac{1}{2} \mathrm{~L}}$$
= 0.02

∴ pH = – log (1.99 x 10-5)
= 5 – log 1.99
= 5 – 0.30
= 4.70

Question 9.
1. How can you prepare a buffer solution of pH9. You are provided with 0.1 M NH4OH solution and ammonium chloride crystals. (Given: pKb for NH4OH is 4.7 at 25°C)

2. What volume of 0.6 M sodium formate solution is required to prepare a buffer solution of pH 4.0 by mixing it with 100 ml of 0.8 M formic acid. (Given: pKa for formic acid is 3.75.)
1. We know that pH + pOH = 14
9 + pOH = 14
= pOH = 14 – 9 = 5

[NH4Cl] = 0.1 M x 1.995
= 0. 1995 M
=0.2 M
Amount of NH4CI required to prepare 1 litre 0.2 M solution = Strength of NH4CI x molar
mass of NH4CI
= 0.2 x 535
= 10.70 g
10.70 g ammonium chloride is dissolved in water and the solution is made up to one litre to get 0.2 M solution. On mixing equal volume of the given NH4OH solution and the prepared NH4CI solution will give a buffer solution with the required pH value (pH = 9).

2.

[Sodium formate] = number of moles of HCOONa
= 0.6 x V x 10-3
[formic acid] = number of moles of HCOOH
= 0.8 x 100 x 10-3
[formic acid] = number of moles of HCOOH
= 0.8 x 100 x 10-3
= 80 x 10-3
4 = 3.75 + log $$\frac { 0.6V }{ 80 }$$
0.25 = log $$\frac { 0.6V }{ 80 }$$
antilog of 0.25 = $$\frac { 0.6V }{ 80 }$$
0.6V = 1.778 x 80
= 1.78 x 80
= 142.4
V = $$\frac { 142.4 mL }{ 0.6 }$$ = 237.33 mL

Question 10.
Calculate the
i) hydrolysis constant
ii) degree of hydrolysis and
iii) pH of 0.05M sodium carbonate solution pKa for HCO3 is 10.26.
Sodium carbonate is a salt of weak acid, H2CO3 and a strong base, NaOH and hence the solution is alkaline due to hydrolysis.

Given
Kw = 1 x 10-14
c = 0.05 M
PKa = 10.26
Ka = – log Ka
Ka = antilog of (- pKa)
Ka = antilog of (- 10.26)
Ka = 5.49 x 10-11

h = 6.034 x 10-2

Question 1.
The Latin word acidus means
a) bitter
b) sour
c) sweet
d) salty
b) sour

Question 2.
According to Arrhenius concept, an acid is
a) hydrogen ion donor
b) hydrogen ion acceptor
c) hydroxyl ion donor
d) electron donor
a) hydrogen ion donor

Question 3.
According to Arrhenius concept, a base is
a) hydrogen ion donor
b) hydroxyl ion acceptor
c) hydroxyl ion donor
d) electron acceptor
c) hydroxyl ion donor

Question 4.
Arrhenius theory does not explain the behaviour of acids and bases in
a) aqueous solvents
b) non-aqueous solvents
c) water
d) None of the above
b) non-aqueous solvents

Question 5.
According to Lowry-Bronsted theory, an acid is
a) proton donor
b) proton acceptor
c) hydroxyl ion donor
d) electron donor
a) proton donor

Question 6.
According to Lowry-Bronsted theory, a base is
a) proton donor
b) proton acceptor
c) hydroxyl ion acceptor
d) electron donor
b) proton acceptor

Question 7.
According to Lowry-Bronsted theory, an equilibrium exists between an acid and its
a) base
b) conjugate acid
c) conjugate base
d) water
c) conjugate base

Question 8.
A conjugate acid-base pair differs only by
a) an electron
b) a proton
c) a hydroxyl ion
d) none of the above
b) a proton

Question 9.
Lowry-Bronsted theory could not explain the acidic behaviour of
a) BF3
b) AlCl3
c) SO2
d) all the above
d) all the above

Question 10.
According to Lewis concept an acid is
a) proton donor
b) proton acceptor
c) electron pair acceptor
d) electron-pair donor
c) electron pair acceptor

Question 11.
According to Lewis concept a base is
a) hydroxyl ion donor
b) hydroxyl ion acceptor
c) electron pair acceptor
d) electron-pair donor
d) electron-pair donor

Question 12.
In a coordination complex the central metal ion acts as
a) Arrhenius acid
b) Lewis base
c) Lowry-Bronsted acid
d) Lewis acid
d) Lewis acid

Question 13.
Which of the following is true for acidic solutions?
a) [H3O+] > [OH]
b) [H3O+] < [OH]
c) [H3O+] = [OH]
d) [H3O+] < [OH]
a) [H3O+] > [OH]

Question 14.
Electron deficient molecules can act as
a) Arrhenius base
b) Lewis base
c) Lowry-Bronsted base
d) Lewis acid

Question 15.
Molecules with one or more lone pairs of electrons can act as
a) Arrhenius acid
b) Lewis base
c) Lowry-Bronsted acid
d) Lewis acid
b) Lewis base

Question 16.
Which among the following is a Lewis base?
a) BF3
b) SO3
C) SF4
d) CaO
d) CaO

Question 17.
Which among the following is not a Lewis base?
a) MgO
b) CO2
c) H2O
d) Na2O
b) CO2

Question 18.
Which among the following is a Lewis acid?
a) NH3
b) CaO
c) RNH2
d) FeCl3
d) FeCl3

Question 19.
Which among the following is not a Lewis acid?
a) SiF4
b) CH2=CH2
c) BeF2
d) Fe3+
b) CH2=CH2

Question 20.
The conjugate acid of H2O is
a) HCl
b) OH
c) H3O+
d) HSO4
c) H3O+

Question 21.
The conjugate base of H2O is
a) HCl
b) OH
c) H3O+
d) HSO4
b) OH

Question 22.
Which of the following can act both as Bronsted acid and Bronsted base?
a)Cl
b) H3O+
c) HCO3
d) CO32-
c) HCO3

Question 23.
In which of the following cases, the sparingly soluble salt solution is unsaturated?
a) Ionic product > solubility product (Ksp)
b) Ionic product < solubility product (Ksp)
c) Ionic product = solubility product (KcD)
d) Both (a) and (b)
b) Ionic product > solubility product (Ksp)

Question 24.
Which of the following is the strongest conjugate base?
a) Cl
b) SO42-
c) CH3COO
d) NO3
c) CH3COO
Reason: The conjugate base of a weak acid is a stronger base.

Question 25.
Among the following the Ka value for a strongest acid is
a) 1.8 x 10-4
b) 1.8 x 10-5
c) 1.8 x 105
d) 1.8 x 104
c) 1.8 x 105

Question 26.
Which among the following is the strongest base?
a) HSO4
b) H2O
C) F
d) H
d) H
H2 is the weakest acid and its conjugate base
H is the strongest base.

Question 27.
Which of the following salts do not undergo salt hydrolysis?
a) Sodium acetate
b) Ammonium acetate
c) Ammonium chloride
d) Sodium nitrate
d) Sodium nitrate

Question 28.
If the hydrogen ion concentration of the solution is 10-5M, its hydroxyl ion concentration is
a) 10-5 M
b) 10-9M
c) 10-14M
d) 10-7M
Ans:
b) 10-9M

Question 29.
If the hydrogen ion concentration of a solution is 10~SM, its pOH is
a) 5
b) 9
c) 14
d) 7
b) 9 Reason: pH = -log[H30+] = -log 10-5 = 5
pH + pOH = 14; pOH = 14 – pH = 14 – 5 = 9

Question 30.
If the pH of a solution is 9, the solution is
a) acidic
b) neutral
c) basic
d) strongly acidic
c) basic
Reason:.pH < 7; pOH > 7 = Acidic
pH > 7; pOH < 7 = Basic

Question 31.
If the pH of a solution is zero the solution is
a) acidic
b) neutral
c) basic
d) strongly acidic

Question 32.
The pOH of IN HCI is
a) 0
b) 1
c) 7
d) 14
d) 14
Reason: pH = -log[H3O+] = -log 1 = 0
pOH = 14 – pH = 14 – 0 = 14

Question 33.
As [H3O+] of a solution increases, its pH
a) increases
b) deposes
c) remains the same
d) becomes
b) decreases

Question 34.
The concentration of a solution of acetic acid changes from 10-4 M to 10-2 M, its degree of dissociation
a) increases
b) decreases
c) remains the same
d) becomes zero
b) decreases
Reason: As dilution increases (Concentration decreases) degree of dissociation increases. Here concentration increases. Hence degree of dissociation decreases.

Question 35.
Which of the following represents Ostwald’s dilution law for a binary electrolyte whose degree of disscociation is a and concentration C

c

Question 36.
NH4OH is a weak base because
a) It has a low vapour pressure
b) It is completely ionised
c) it is only partially ionised
d) It has low density
c) it is only partially ionised

Question 37.
The concentration of acetic acid changes from 10-2M to 10-4 M its degree of dissociation
a) increases
b) decreases
c) remains the same
d) becomes zero
a) increases

Question 38.
For a weak acid the hydrogen ion concentration is given as

d) both (a) & (b)

Question 39.
When ammonium chloride is added to ammonium hydroxide, the degree of dissociation of ammonium hydroxide
a) increases
b) decreases
c) remains the same
d) becomes zero
b) decreases
Reason: Common ion effect

Question 40.
The relationship between the solubility product (Ksp) and molar solubility (S) for Ag2 (OO4) is
a) Ksp = S3
b) Ksp = S2
c) Ksp = 4S3
d) Ksp = 3S2
c) Ksp = 4S3

Question 41.
The decrease in degree of dissociation of HF on addition of NaF is known as
a) buffer action
b) neutralization
c) common ion effect
d) hydrolysis
c) common ion effect

Question 42.
Which is not a buffer solution?
a) CH3COOH + CH3COONa
b) HCl + NaCl
c) NH4OH + NH4Cl
d) H2CO3+NaHCO3
b) HCl + NaCl
Reason: A strong acid and its salt is not a buffer solution. A weak acid or base and its salt is a buffer solution.

Question 43.
Which among the following does not undergo hydrolysis?
a) NH4Cl
b) CH3COONa
c) NaCl
d) CH3COONH4
c) NaCl
Reason: NaCt is a salt of strong acid and strong base.

Question 44.
Hydrolysis of CH3COONa gives
a) acidic solution
b) basic solution
c) neutral solution
d) No solution
b) basic solution
Reason: CH3COONa is a salt of weak acid and strong base. Hence CH3COO undergoes hydrolysis giving a basic solution.

Question 45.
The pH of the solution resulting from the hydrolysis of NH4Cl is
a) 7
b) greater than 7
c) less than 7
d) 14
c) less than 7
Reason: NH4Cl is a salt of strong acid and weak base. NH4C+ undergoes hydrolysis giving an acidic solution. Hence pH is less than 7.

Question 46.
For the hydrolysis of salt of weak acid and weak base which among the following is true?
a) Kh Ka = Kw
b) Kh Kb = kw
c) Ka Kb Kh = Kw
d) Kh= $$\frac{K_{1} K_{b}}{K_{w}}$$
c) Ka Kb Kh = Kw

Question 47.
The pH of the solution obtained from the hydrolysis of salt of strong acid and weak base

b

Question 48.
$$X_{m} Y_{n(s)} \stackrel{H_{2} O}{\rightleftharpoons} m X_{(a q)}^{n+}+n Y_{(a q)}^{m-}$$
The solubility product of the salt XmYn is given as
a) Ksp = [Xm+]n[Yn-]m
b) Ksp = m[Xn+] n[Ym-]
c) Ksp = [Xn+]m [Ym-]n
d) Ksp =[XmYn]m+n
c) Ksp = [Xn+]m [Ym-]n

Question 49.
The aqueous solution sdium acetate, ammonium chloflde, sodi maitrate are respectively. PTA -61
a) Neutral,acidic,basic
b) acidic, basic, neutral
c) basic, acidic, neutral
d) basic, aidk, basic
d) basic, acidic, basic

Question 50.
The relationship between the solubility product and molar solubility of Al2(SO4)3 is
a) S2
b) 4S3
c) 108S5
d) 27S5
c) 108S5
Reason:

II. Pick out the correct statements

Question 1.
i) Ka measures the strength of an acid
ii) Larger the pKa value, stronger is the acid.
iii) Weak acids undergo complete ionisation
iv) Acids with Ka value greater than 10 are considered as strong acids.
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (i) & (iv)
d) (i) & (iv)
Correct statement:
ii) Larger the pKa value, weaker is the acid, (or) Lower the pKa value, stronger is the acid.
iii) Weak acids undergo partial ionisation, (or) strong acids undergo complete ionisation.

Question 2.
i) Degree of dissociation
$$=\frac{\text { Number of moles dissociated }}{\text { Total number of moles }}$$
ii) dilution law is applicable to strong acids.
iii) Strong acids undergo complete dissociation
iv) In a weak acid when dilution increases by 100 times the dissociation increases by 10 times.
a) (i) & (ii) b) (ii) & (iii)
c) (ii) & (iv) d) (i) & (iv)
c) (ii) & (iv)
Correct statement:
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (iii) & (iv)
b) (i) & (iii)
Correct statement:
ii) Ostwald’s dilution law is applicable to strong acids.
iv) In a weak acid when dilution increases by 100 times the dissociation increases by 10 times.

Question 3.
i) A buffer solution is a mixture of a weak acid and its conjugate base
ii) Common ion effect is based on Le Chatelier’s principle.
iii) Blood contains a buffer solution of HCl and NaCl
iv) For a buffer solution
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
$$\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\mathrm{acid}]}{[\mathrm{salt}]}$$
a) (i) & (ii)
Correct statement:
iii) Blood contains a buffer solution of H2CO3 and NaHCO3
iv) For a buffer solution
pH = pKa + log$$\frac{[\text { acid }]}{[\text { salt }]}$$

Question 4.
i) The solution obtained from the hydrolysis of CH3COONa is acidic
ii) The solution obtained from the hydrolysis of NaCl is neutral
iii) The nature of the solution obtained from the hydrolysis of a salt of weak acid and weak base depends on the strength of acid or base.
iv) The solution obtained from the hydrolysis of NH4C/ is basic
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
b) (ii) & (iii)
Correct statement:
i) The solution obtained from the hydrolysis of CH3COONa is basic
iv) The solution obtained from the hydrolysis of NH4Cl is acidic

III. Pic out the correct statements

Question 1.
For the equilibrium
$$\mathrm{HClO}_{4}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{ClO}_{4}^{-}$$
Which are the incorrect statements?
i) HClO4 is the conjugate acid of H2O
ii) H2O is the conjugate base of H3O+
iii) H3O+ is the conjugate base of H2O
iv) $$\mathrm{ClO}_{4}^{-}$$ is the conjugate base of HClO4
a) (i) & (ii)
b) (i) & (iii)
c) (ii) & (iv)
d) (iii) & (iv)
b) (i) & (iii)
Correct statement:
i) HClO4 is the conjugate acid of $$\mathrm{ClO}_{4}^{-}$$
iii) H3O+ is the conjugate acid of H2O

Question 2.
i) All metal ions are Lewis acids.
ii) All metal oxides are Lewis acids.
iii) All anions are Lewis bases
iv) Molecules that contain polar double bond
are Lewis bases
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (ìv)
d) (i) & (iv)
c) (ii) & (iv)
Correct statement:
ii) All metal oxides are Lewis bases.
iv) Molecules that contain polar double bond are Lewis acids.

Question 3.
i) The conjugate base of OH is O2-
ii) The conjugate acid of $$\mathbf{N H}_{2}^{-}$$ is NH3
iii) The conjugate acid of H2SO4 is $$\mathbf{H S O}_{4}^{-}$$
iv) The conjugate base of H2O is H3O+
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
c) (iii) & (iv)
Correct statement:
iii) The conjugate acid of H2SO4 is $$\mathbf{H S O}_{4}^{-}$$
iv) The conjugate base of H2O is H3O+

Question 4.
i) If the hydrogen ion concentration of a solution is 10-9 M, it is basic
ii) 1f the pOH of a solution is 14, it is strongly basic.
iii)As pH of a solution increases, its pOH also increases.
iv) At 25°C the sum of pH and pOH of a solution
is 14.
a) (i) & (ii) b) (ii) & (iii) c) (iii) & (iv)
d) (j) & (iv) Ans: b) (ii) & (iii)
Correct statement:
ii) If the pOH of a solution is 14 it is strongly acidic. (or) 1f the pH of a solution is 14, it is strongly basic
iii) As pH of a solution increases, its pOH decreases. (or) As pH of a solution decreases, its pOH increases.

IV. Assertion and reason

Question 1.
Assertion (A): Ionic product of water Kw increases with increase in temperature
Reason (R): Dissociation of water is an exothermic reaction.
a) Both A & R are correct, R explains A
b) Both A & R are correct, R does not explain A.
c) A is correct, but R is wrong.
d) A is wrong, but R is correct.
c) A is correct, but R is wrong.
Correct R: Dissociation of water is an endothermic reaction.

Question 2.
Assertion (A): Carbanion is a Lewis base
Reason (R): Carbanion can donate a pair of electron.
a) Both A & R are correct, R explains A
b) Both A & R are correct, R does not explain A.
c) A is correct, but R is wrong.
d) A is wrong, but R is correct.
a) Both A& R are correct, R explains A Correct statement:

Question 3.
Assertion (A): An aqueous solution of IICI is acidic
Reason (R): An aqueous solution of HCl
contains less H30* than OH- ions.
a) Both A & R are correct, R explains A
b) Both A & R are correct, R does not explain A.
c) A is correct, but R is wrong.
d) A is wrong, but R is correct.
c) A is correct, but R is wrong.
Correct R: An aqueous solution of HCl contains more H3O+ than 0H ions.

Question 4.
Assertion (A): When sodium acetate is added to acetic acid, the dissociation of acetic acid increases.
Reason (R): The addition of CH3COO ion shifts the equilibrium of dissociation of acetic acid to the left.
a) Both A & R are correct, R explains A
b) Both A & R are correct, R does not explain A.
c) A is correct, but R is wrong.
d) A is wrong, but R is correct.
d) A is wrong, but R is correct.
Correct A: When sodium acetate is added to acetic acid, the dissociation of acetic acid decreases.

V. Match the following:

Question 1.

i) c
ii) d
iii) a
iv) b

Question 2.

i) c
ii) d
iii) b
iv) a

VII. Two Mark Questions

Question 1.
What are Arrhenius acids and base? Give examples.

 Arrhenius acids Arrhenius bases 1. Hydrogen ion donor in water Hydroxyl ion donor in water Ex. HCl, H2SO4 Ex. NaOH, Ca(OH)2

Question 2.
What are the limitations of Arrhenius concept of acids and bases?

• Does not explain the behaviour of acids and bases in non-aqueous solvents such as acetone, THF etc.,
• Does not account for the basic nature of the substances like ammonia which do not contain a hydroxyl group.

Question 3.
How does BF3 act as Lewis acid?

Boron has a vacant 2p-orbital to accept the lone pair of electrons donated by ammonia to form a new coordinate covalent bond. Hence BF3 acts as a Lewis acid.

Question 4.
Arrange HCl, HCOOH and CH3COOH in their increasing order of acid strength if their Ka values at 25°C are 2 x 106, 1.8 x 10-4 and 1.8 x 10-5 respectively.

• As Ka value increases, the acid strength increases.
• Increasing order of Ka value is 1.8 x 10-5 < 1.8 x 10-4 < 2 x 106
• The increasing order of acid strength is CH3COOH < HCOOH < HCl.

Question 5.
What is neutralization?
The reaction in which an acid reacts with a base to form salt and water is called neutralization.

Question 6.
What do yo mean by salt hydrolysis?
Salt hydrolysis is the reaction of the cation or the anion or both the ions of the salt with water to produce either acidic or basic solution.

Question 7.
What is Buffer index (β) ?
It is defined as the number of gram equivalents of acid or base added to 1 litre of the buffer solution to change its pFl by unity.
$$\beta=\frac{\mathrm{dB}}{\mathrm{d}(\mathrm{pH})}$$
dB = number of gram equivalents of acid/base added to one litre of buffer solution. d(pH) = The change in the pH after the addition of acid / base.

Question 8.
How solubility product is determined by molar solubility?
Solubility product can be calculated from the molar solubility i.e,, the maximum number of moles of solute that can be dissolved in one litre of the solution.

Question 9.
Define pOH
pOH of a solution is defined as the negative logarithm of base 10 of the molar concentration of hydroxyl ions present in the solution. pOH = -log10[OH]

Question 10.
Write the pH value of the following substances.
(a) Vinegar
(b) Black coffee
(c) Baking soda
(d) Soapy water

 Substance pH Vinegar 2 Black coffee 5 Baking soda 9 soapy water 12.

VII.Three Mark Questions

Question 1.
Derive the relation between pH and pOH

Question 2.
What are buffer solutions? Explain their types with examples.
Solutions which resist drastic changes in its pH upon the addition of small amount of acids or bases are called buffer solutions.
A buffer solution consists of a mixture of a weak acid and its conjugate base (salt) or a weak base and its conjugate acid (salt)

Types:
Acidic buffer: Solution of acetic acid and sodium, acetate
Basic buffer: Solution of ammonium hydroxide and ammonium chloride.

Question 3.
What are conjugate acid-base pairs? Give example.
Acid1 + Base2 ⇌ Acid2 + Base1
The species that remains after the donation of
a proton is a base (Base}) and is called the conjugate base of the Bronsted acid (Acid}). In other words, chemical species that differ only by a proton are called conjugate acid – base pairs.
Conjugate acid – base pair

VIII. Five Mark Questions.

Question 1.
Write a note on Lewis’s concepts of acids and bases ?

Question 2.
Derive an expression for the hydrolysis constant and degree of hydrolysis of salt of weak acid and strong base. Answer:
Consider the reaction between the weak acid CH3COOH and the strong base NaOH to form the salt CH3COONa and water.
$$\mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{aqq}}+\mathrm{NaOH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COONa}_{(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{I})}$$

In aqueous solution CH3COONa is completely dissociated as follows,
$$\mathrm{CH}_{3} \mathrm{COONa}_{(\mathrm{aq})} \longrightarrow \mathrm{CH}_{3} \mathrm{COO}_{(\mathrm{aq})}+\mathrm{Na}_{(l)}^{+}$$
CH3 COO is the strong conjugate base of the weak acid CH3COOH
$$\mathrm{CH}_{3} \mathrm{COO}_{\text {(aq) }}^{-}+\mathrm{H}_{2} \mathrm{O}_{(l)} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{aq})}+\mathrm{OH}_{\text {(ач) }}^{-}$$
No such tendency is shown by Na+, the weak conjugate acid of the strong base NaOH.
In the above reaction OH ion is formed.
[OH] > [H+] and the solution is basic and the pH is greater than 7.
Equilibrium constant (hydrolysis constant) for the above

Dissociation of the weak acid is CH3COOH(aq) $$\mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}{ }_{(\mathrm{aq})}+\mathrm{H}^{+}{ }_{(a q)}$$
Dissociation constant of the weak acid

Question 3.
Explain buffer action of acidic buffer?

• The chemical reaction which is responsible to resist the pH change of a buffer solution is called its buffer action.
• To resist changes in its pH on the addition of an acid or base, the buffer solution should contain both acidic as well as basic components.
• These components neutralise the effect of added acid or base.
• At the same time, these components should not consume each other.
• Consider an acidic buffer containing CH3COOH and CH3COONa.

When an acid is added to this buffer, the added H+ ions are consumed by the conjugate base CH3COO to form undissociated CH3COOH. This keeps the H+ concentration and the pH unchanged.
$$\mathrm{CH}_{3} \mathrm{COO}_{(a \mathrm{q})}^{-}+\mathrm{H}_{(\mathrm{aq})}^{+} \longrightarrow \mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{ay}}$$

When a base is added to this buffer, the added OH- ions are consumed by H30+. To maintain the equilibrium more CH3COOH is dissociated keeping the H+ Concentration and the pH unchanged

Question 4.
Derive Henderson-Hasselbalch equation.
Consider an acidic buffer

• The weak acid is dissociated only to a small extent.
• Due to the common ion effect, the dissociation is further suppressed.
• Hence the equilibrium concentration of the acid is nearly equal to the initial concentration of the unionised acid.
• Similarly, the equilibrium concentration of the conjugate base is nearly equal to the initial concentration of the added salt.

Equations (4) & (5) are known as
Henderson-Hasselbaich equations.

Question 5.
Derive an expression for the hydrolysis constant and degree of hydrolysis of salt of a weak acid and weak base.

• Consider the reaction between the weak acid CH3COOH and the weak base NH4OH to form the salt CH3COONH4 and water.
• $$\mathrm{NH}_{4}^{+}$$ is the strong conjugate acid of the weak base NH4OH.
• CH3COO is the strong conjugate base of the weak acid CH3COOH.
• Hence both $$\mathrm{NH}_{4}^{+}$$ and CH3COO have the tendency to react with water.
• The nature of the solution depends on the strength of acid or base.
• if Ka >Kb then the solution is acidic and pH > 7.
• if Ka < Kb then the solution is basic and pH > 7.1
• if Ka = Kb then the solution is neutral and pH = 7.

Consider the hydrolysis of both $$\mathrm{NH}_{4}^{+}$$ and CH3COO

Dissociation of a weak acid

Dissociation of weak base

Question 6.
0.1 M Solution of HF is a weak acid. But 5M solution of HF is a stronger acid. Why?

• Hydrofluoric acid is a much stronger acid when it is concentrated than when it is diluted. As the concentration of hydrofluoric acid approaches 100 percent.
• It’s acidity increases because of home association, where a base and conjugate acid form a bond.
• 3HF ⇌ H2F+ + HF2
• The FHF-bifluoride anion is stabilized by a strong hydrogen bond between hydrogen and fluorine.
• The stated ionization constant of hydrofluoric acid, 10 – 3.15, does not reflect the true acidity of concentrated HF solutions.
• Hydrogen bonding also accounts for the higher boiling point of HF compared to other hydrogen halides.

Problems based on pH.

Question 1.
What is the pH of 0.001M NaOH?
Solution:
[OH] = Normality = Molarity x acidity
= 0.001 x 1 = 1 x 10-3
pOH = -log 10 [OH] = -log 1 x 10-3
= -log 1 – log 10-3 [log 1 = 0]
= 0 + 3log10 [log10 = 1]
pOH = 3
pH + pOH = 14
pH = 14 – pOH
= 14-3
pH = 11

Question 2.
What is the pH of a solution with hydronium ion concentration 6.2 x 10-9 M.
Solution:
[H3O+] = 6.2 x 10-9
pH = -log10[H3O+]
= – log 6.2 x 10-9
= – log 6.2 – log 10-9
= – log 6.2 + 9 log 10
= 9 – log 6.2
= 9 -0 .7924
pH = 8.2076

Question 3.
What is the pH of 10-2 M Ca(OH)2?
Solution:
[OH] = Normality = Molarity x acidity
= 10-2 x 2
[OH] = 2 x 10-2
pOH = -log10[OH ]
= -log 2 x 10-2
= -log 2 – log 10-2
= -log 2 + 21ogl0
= 2 – log 2
= 2 – 0.3010
pOH = 1.6990
pH+pOH = 14
pH = 14 -pOH
= 14 -1.6990
pH = 12.3010

Question 4.
Calculate the hydroxyl ion concentration of a solution with pH = 5
Solution:
[H3O+] = 10pH
[H3O+] = 10-5
[H3O+][OH] = 10-14
[OH] = $$\frac{10^{-14}}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}$$
= $$\frac{10^{-14}}{10^{-5}}$$
[OH] = 10-9

Question 5.
Calculate the pH of 10-8M HNO3
[H3O+] = 10-8 which is less than 10 7
[H3O+] from water should be added.
[H3O+] = 10 “(from water) + 10-8(from HN03)
[H3O+] = 10-7 (1+10-1)
= 10-7 (1+0.1)
[H3O+] = 1.1 x 10-7
pH = – log10 [H3O+]
= – log 1.1 x 10-7
= – log 1.1 – log 10-7
= – log 1.1 + 7 log 10
= 7 – log 1.1
= 7 – 0.0414
pH = 6.9586

Question 6.
The dissociation constant of 0.1 M weak acid is 1 x 10-5. Calculate its pH.

Question 7.
If a solution with pH = 5 is diluted 100 times. What will be the pH of the resulting solution?
Solution:
pH = 5; [H3O+] = 10-pH
[H3O+] = 10-5
After dilution
[H3O+] = $$\frac{10^{-5}}{100}=10^{-7} \mathrm{M}$$
As [H3O+] = 10-7 M, the [H3O+] contributed
Final [H3O+] = 10-7 (from water) + 10-7 (from the solution
= 10-7(1+1)
[H3O+] = 2 x 10-7M
= – log [H3O+]
= – log 2 x 10-7
= – log 2 – log 10-7
= – log 2 + 7 log 10
= 7 – log 2
= 7 – 0.3010
= 6.6990

Question 8.
Calculate the hydrogen ion concentration in human blood. (pH of human blood is 7.4)
Solution:
pH = 7.4
pH = – logio [H3O+]
log [H3O+] = -pH
[H30+] = Antilog of (-pH)
= Antilog of (-7.4)
= Antilog of (-7-0.4)
= Antilog of (-7-0.4+1-1)
= Antilog of (-8+0.6)
= Antilog of (8.6)
[H3O+] = 3.98 x 10-8

Question 9.
Calculate the pH of a solution obtained by mixing 50ml of 0.4 N HC1 and 50ml of 0.2N NaOH.
Solution:
Milli equivalent of 50 ml, 0.4N HCl
= Vm1 x N = 50 x 0.4 = 20
Milli equivalent of 50 ml, 0.2N NaOH
= Vm1 x N = 50 x 0.2 = 10
Remaining milli equivalent of Hcl = 20-10 =10

Question 10.
Calculate the pH of 10-7M HCl.
If we do not consider [H3O+] from the ionisation of H2O, then [H3O+] = [HCl]= 10-7M)
In this case the concentration of the acid is very low (10-7M). Hence, the [H3O+] (10-7M) formed due to the auto ionisation of water cannot be neglected.
[H3O+] = 10-7 (from HCl) + 10-7 (from H2O)
= 10-7 (1+1) = 2 x 10-7
pH = -log10 [H3O+]
= log10[2 x 10-7] = – log 2 + log 10-7
= -log 2 – (-7) log1010
= 7 – log2 .
= 7 – 0.3010=6.6990
= 6.70

Problems based on Henderson – Hasselbaich equations

Question 1.
A buffer solution contains equal volumes of 0.2 M NH4OH and 0.02 M NH4Cl. Kb of NH4OH is 1 x 10-5. Calculate the pH of the buffer.
Solution:
[Base] = 0.2 M; [salt] = 0.02 M
Kb = 1 x 10-5
pKb = -log Kb = – log 1 x 10-5 = 5
For a basic buffer
$$\mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{[\mathrm{Salt}]}{[\text { Base }]}$$
= 5+log 0.02/0.2
= 5+1og10-1
= 5 – 1
pOH = 4
pH + pOH = 14
pH = 14 – pOH
= 14 -4
pH = 10

Question 2.
Find the pH of a buffer solution containing 0.20 mole per litre CH3COONa and 0.15 mole per litre CH3COOH, Ka for acetic acid is 1.8 x 10-5
Solution:
[acid] = 0.15 M; [salt] = 0.20 M
Ka = 1.8 x 10-5
PKa = -log Ka = – log L8 x 10-5
= – log 1.8 – log10-5
= – log 1.8 + 5 log 10
= 5 – log 1.8
= 5 – 0.2553 pKa
PKa = 4.7447
For an acidic buffer
$$\mathrm{pK}_{\mathrm{b}}=\mathrm{pOH}-\log \frac{[\text { Salt }]}{[\text { Base }]}$$
= 4.7447 + log 0.20/0.15
= 4.7447 + log 4/3
= 4.7447 + log 4 – log 3
= 4.7447+0.6021-0.4771
pH = 4.8697

Question 3.
Calculate the pKb of NH4OH, if the pH of a buffer solution containing 0.1N NH4OH and 0.1 N NH4Cl is 9.25.
Solution;)
pH of basic buffer = 9.25
∴ pOH = 14 – pH
= 14 – 9.25
pOH = 4.75
[Base] = 0.1N ; [Salt] = 0.1 N
$$\mathrm{pK}_{\mathrm{b}}=\mathrm{pOH}-\log \frac{[\text { Salt }]}{[\text { Base }]}$$
$$\mathrm{pK}_{\mathrm{b}}=\mathrm{pOH}-\log \frac{[\text { Salt }]}{[\text { Base }]}$$
PKb = 4.75 – log M = 4.75 – log 0.1/0.1 = 4.75 – 0 = 4.75

Question 4.
Calculate the pH of the solution by mixing 1.5 mole of HCN and 0.15 mole of KCN in water and making up the total volume to 0.5 litre.
Solution:
Ka = 5 x 10-10;
fAcid] = 1.5/0 5 – 3.0
[Salt] = 0.15/0.5 = 0.3
pKa = – log Ka
= – log 5 x 10-10
= – log 5 – log 10-10
= – log 5 + 10 log-10
= 10 – log 5
= 10 – 0.6990
pK-10 = 9.3010
$$\mathrm{pH}=\mathrm{pKa}+\log \frac{[\mathrm{Salt}]}{[\mathrm{Acid}]}$$
= 9.3010 + log 0.3/3.0
= 9.3010 + log 10-10
= 9.3010 -1 log 10
pH = 8.3010

Problems based on Ostwald’s dilution law

Question 1.
Calculate the degree of dissociation of 0.1 N CH3COOH if Ka for CH3COOH is 1 x 10-5.
Solution:

Question 2.
What is the dissociation constant of 0.1 M HCN which is 0.01% ionised?
Solution:
C = 0.1M ; α = 0.01% = $$\frac{0.01}{100}$$ = 1 x 10 -4
Ka = ?
Ka = C α2 = 0.1 x (1 x 10-4)2 = 1 x 10-9

Question 3.
A weak monobasic acid is 1% ionised in 0.1M solution at 25°C. What is the percentage of ionisation in its 0.025M Solution?
Solution:
C =0.1 M; α = 1% = 1/100 = 10-2 Ka = ?
C =0.025 M % a = ?
Ka = C α2 = 0.1 x (10-2)2 = 10-5
$$\alpha=\sqrt{\frac{K a}{C}}=\sqrt{\frac{10^{-5}}{0.025}}=0.02$$
% α =100 x α = 100 x 0.02 = 2%

Question 4.
A mono basic weak acid solution has a molarity of 0.005 and pH of 5. What is its percentage ionisation in this solution?
Solution:
C = 0.005 M
pH = 5
∴ [H3O+] = 10-ph = 10-5
HA → H+ + A
ka = $$\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}$$
[H+] = [A] = 10-5
[HA] = 0.005M

α = 2 x 10-3
% α = 100 x α
= 100 x 2 x 10-3 = 0.2%

Problems based on Solubility Product

Question 1.
Calculate the solubility product of AgCl if the solubility of AgCl is 1.3 x 10-5 moles per litre at 25°C.
Solution:
S = 1.3 x 10-5 M ; Ksp = ?

Question 2.
Calculate the solubility of Calcium fluoride in a saturated solution if its solubility product is 3.2 x 10-11 M3.
Solution:
Ksp = 3.2 x 10-11 M3; S = ?

Question 3.
The solubility of AgCl is 0.0014 g per litre at 18°C. Calculate its solubility product at 18°C. Molecular weight of AgCl is 143.5. Solubility in

Ksp = [Ag+][Cl]
Ksp = S x S
Ksp = S2
Ksp = (9.756 x 10-6M)2
Ksp = 95.1795 x 10-12
Ksp = 9.518 x 10-11 M2

Question 4.
The solubility product of a salt having general formula MX2 in water is 4 x 10-12. What is the concentration of M2+ ions in the aqueous solution of the salt?
Solution:

Question 5.
Solid Ba (N03)2 is gradually dissolved in 1 x 10-4 m Na2CO3 solution. At what concentration of Ba2+ will a precipitate begin to form?
(Ksp for BaCO3 = 5.1 x 10-9 )
Solution:

Ionic Product > Ksp, Precipitation will occur.
∴ After attaining 5.1 x 10-5 M [Ba2+] concentration, a precipitate will begin to form.

## Samacheer Kalvi 12th Computer Applications Guide Chapter 18 Electronic Data Interchange – EDI

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Computer Applications Guide Pdf Chapter 18 Electronic Data Interchange – EDI Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Computer Applications Solutions Chapter 18 Electronic Data Interchange – EDI

### 12th Computer Applications Guide Electronic Data Interchange – EDI Text Book Questions and Answers

Part I

Question 1.
EDI stands for
a) Electronic Details Information
b) Electronic Data Information
c) Electronic Data Interchange
d) Electronic Details Interchange
a) Electronic Details Information

Question 2.
Which of the following is an internationally recognized standard format for trade, transportation, insurance, banking and customs?
a) TSLFACT
b) SETFACT
c) FTPFACT
d) EDIFACT
d) EDIFACT

What is the Greatest Common Factor of 12 and 18? … Greatest common factor (GCF) of 12 and 18 is 6.

Question 3.
Which is the first industry-specific EDI standard?
a) TDCC
b) VISA
c) Master
d) ANSI
a) TDCC

Question 4.
UNSM stands for
a) Universal Natural Standard Message
b) Universal Notations for Simple Message
c) United Nations Standard Message
d) United Nations Service Message
c) United Nations Standard Message

Question 5.
Which of the following is a type of EDI?
a) Direct EDI
b) Indirect EDI
c) Collective EDI
d) Unique EDI
a) Direct EDI

Question 6.
Who is called the father of EDI?
a) Charles Babbage
b) Ed Guilbert
c) Pascal
d) None of the above
b) Ed Guilbert

Question 7.
a) UNA, UNZ
b) UNB, UNZ
c) UNA, UNT
d) UNB, UNT
b) UNB, UNZ

Question 8.
EDIFACT stands for
a) EDI for Admissible Commercial Transport
b) EDI for Advisory Committee and transport
c) EDI for Administration, Commerce, and Transport
c) EDI for Administration, Commerce, and Transport

Question 9.
The versions of EDIFACT are also called as
a) Message types
b) Subsets
c) Directories
d) Folders
c) Directories

Question 10.
Number of characters in a single EDIFACT messages
a) 5
b) 6
c) 4
d) 3
b) 6

Part II

Question 1.
Define EDI.
The Electronic Data Interchange (EDI) is the exchange of business documents between one trade partner and another electronically. It is transferred through a dedicated channel or – through the Internet in a predefined format without much human intervention.

Question 2.
List few types of business documents that are transmitted through EDI.

1. Delivery notes
2. Invoices
3. Purchase orders
5. Functional acknowledgments etc.

Question 3.
What are the 4 major components of EDI?
There are four major components of EDI. They are:

1. Standard document format
2. Translator and Mapper
3. Communication software
4. Communication network

Question 4.
What is meant by directories inEDIFACT?

• The versions of EDIFACT are also called as directories.
• These EDIFACT directories will he revised twice a year.

Question 5.
Write a note on EDIFACT subsets.
Due to the complexity, branch-specific subsets of EDIFACT have developed. These subsets of EDIFACT include only the functions relevant to specific user groups.
Example:

• CEFIC – Chemical industry
• EDIFURN – furniture industry

Part III

Question 1.
Write a short note on EDI.

• The Electronic Data Interchange (EDI)is the exchange of business documents between one trade partner and another electronically,
• It is transferred through a dedicated channel or through the Internet in a predefined format without much human intervention,
• It is used to transfer documents such as delivery notes, invoices, purchase orders, advance ship notices, functional acknowledgments, etc.

Question 2.
List the various layers of EDI.
Electronic data interchange architecture specifies four different layers namely

1. Semantic layer
2. Standa, us translation layer
3. Transport layer
4. Physical layer

These EDI layers describe how data flows from one computer to another.

Question 3.
Write a note on UN/EDIFACT.

• United Nations / Electronic Data Interchange for Administration, Commerce, and Transport
• (UN / EDIFACT) is an international EDI – a standard developed under the supervision of the United Nations.
• In 1987, the UN / EDIFACT syntax rules were approved as ISO: IS09735 standard by the International Organization for Standardization.
• EDIFACT includes a set of internationally agreed standards, catalogs, and guidelines for the electronic exchange of structured data between independent computer systems.

Question 4.
Write a note on the EDIFACT message.

• The basic standardization concept of EDIFACT is that there are uniform message types called United Nations Standard Message (UNSM).
• In so-called subsets, the message types can be specified deeper in their characteristics depending on the sector.
• The message types, all of which always have exactly one nickname consisting of six uppercase English alphabets.
• The message begins with UNH and ends with UNT.

Question 5.
EDIFACT has the following punctuation marks that are used as standard separators.
Character Uses

 Character Uses Apostrophe (‘) segment terminator Plus sign (+) segment tag and data element separator Colon (;) component data element separator Question mark (?) Release character Period (.) decimal point

Part IV

Explain In Detail

Question 1.
Briefly explain various types of EDI.
The types of EDI were constructed based on how EDI communication connections and the conversion were organized. Thus based on the medium used for transmitting EDI documents the following are the major EDI types.

1. Direct EDI
2. EDI via VAN
3. EDI via-FTP/VPN, SFTP, FTPS
4. Web EDI
5. Mobile EDI
6. Direct EDI/Point-to-Point

It is also called as Point-to-Point EDI. It establishes a direct connection between various business stakeholders and partners individually. This type of EDI suits to larger businesses with a lot of day to day business transactions.

EDI via VAN:
EDI via VAN (Value Added Network) is where EDI documents are transferred with the support of third-party network service providers. Many businesses prefer this network model to protect them from the updating ongoing complexities of network technologies.

EDI via FTP/VPN, SFTP, FTPS:
When protocols like FTP/VPN, SFTP, and FTPS are used for the exchange of EDI-based documents through the Internet or Intranet it is called EDI via FTP/VPN, SFTP, FTPS.

Web EDI:
Web-based EDI conducts EDI using a web browser via the Internet. Here the businesses are allowed to use any browser to transfer data to their business partners. Web-based EDI is easy and convenient for small and medium organizations.

Mobile EDI:
When smartphones or other such handheld devices are used to transfer EDI documents it is called mobile EDI. Mobile EDI applications considerably increase the speed of EDI transactions.

Question 2.
What are the advantages of EDI?

• EDI was developed to solve the problems inherent in paper-based transaction processing and in other forms of electronic communication.
• Implementing an EDI system offers a company greater control over its supply chain and allow it to trade more effectively. It also increases productivity and promotes operational efficiency.

The following are the other advantages of EDI.

• Improving service to end-users
• Increasing productivity
• Minimizing errors
• Slashing response times
• Automation of operations
• Cutting costs

Question 3.
Write about the structure of EDIFACT.

• EDIFACT is a hierarchical structure where the top level is referred to as an interchange, and lower levels contain multiple messages.
• The messages consist of segments, which in turn consist of composites.
• The final iteration is a data element.

Segment Tables

• The segment table lists the message tags.
• It contains the tags, tag names, requirements designator, and repetition field.
• The requirement designator may be mandatory (M) or conditional (C).
• The (M) denotes that the segment must appear at least once. The (C) denotes that the segment may be used if needed.
• Example: CIO indicates repetitions of a segment or group between 0 and 10.

EDI Interchange

• Interchange is also called an envelope.
• The top-level of the EDIFACT structure is Interchange.
• An interchange may contain multiple messages. It starts with UNB and ends with UNZ

EDIFACT message

• The basic standardization concept of EDIFACT is that there are uniform message types called United Nations Standard Message (UNSM).
• In so-called subsets, the message types can be specified deeper in their characteristics depending on the sector.
• The message types, all of which always have exactly one nickname consisting of six uppercase English alphabets.
• The message begins with UNH and ends with UNT

Service messages

• To confirm/reject a message, CONTRL and APERAK messages are sent.
• CONTRL- Syntax Check and Confirmation of Arrival of Message
• APERAK – Technical error messages and acknowledgment

Data exchange

• CREMUL – multiple credit advice
• DELFOR- Delivery forecast
• IFTMBC – Booking confirmation

EDIFACT Segment

• It is the subset of messages.
• A segment is a three-character alphanumeric code.
• These segments are listed in segment tables.
• Segments may contain one, or several related user data elements.

EDIFACT Elements

• The elements are the piece of actual data.
• These data elements may be either simple or composite.

EDI Separators
EDIFACT has the following punctuation marks that are used as standard separators.

### 12th Computer Applications Guide Electronic Data Interchange – EDI Additional Important Questions and Answers

Part A

Question 1.
……………………. is the exchange of business documents between one trade partner and another electronically.
(a) EDI
(b) UDI
(c) FDI
(d) DDI
(a) EDI

Question 2.
First EDI standards were released by ………..
a) EDI
b) EFT
c) EDIA
d) TDCC
d) TDCC

Question 3.
(a) EDI
(b) XML
(c) EDIF
(d) EFT
(a) EDI

Question 4.
………… establishes a direct connection between various business stakeholders
and partners individually.
a) Direct EDI
b) EDI via VAN
c) Web EDI
d) Mobile EDI
a) Direct EDI

Question 5.
Electronic data interchange architecture specifies ……………. different layers.
a) two
b) three
c) four
d) five
c) four

Question 6.
TDCC was formed in the year …………………….
(a) 1964
(b) 1966
(c) 1968
(d) 1970
(c) 1968

Question 7.
In ……………… UN created the EDIFACT to assist with the global reach of technology in E-Commerce.
a)1985
b)1978
c)1974
d)1975
a)1985

Question 8.
Expand EDIA
(a) Electronic Data Interchange Authority
(b) Electronic Data Information Association
(c) Electronic Data Interchange Association
(d) Electronic Device Interface Amplifier
(c) Electronic Data Interchange Association

Question 9.
Which of the following is for the exchange of EDI-based documents through the Internet?
a) FTP/VPN
b) SFTP
c) FTPS
d) All of the above
d) All of the above

Question 10.
EDIA has become …………………….. committee.
(a) ANSIXI2
(b) ANSIXI3
(c) ANSIXI4
(d) ANSIX15
(a) ANSIXI2

Fill In The Blanks:

1. ……….. was developed to solve the problems inherent in paper-based transaction processing.
EDT

2. ………….. is also called as Point-to-Point EDI.
Direct EDT

3. Interchange is also called…………..
Envelope

Paperless

5. EFT is …………….. Payment
Paperless

6. ………… is “the computer-to-computer interchange of strictly formatted messages.
EDI

7. …………….. EDI is easy and convenient for small and medium organizations.
Web-based

8. The …………. is the most critical part of the entire EDI.
standard

Abbreviations

1. EDI – Electronic Data Interchange
2. EFT – Electronic Transfer
3. TDCC – Transportation Data Coordinating Committee
4. EDIA – Electronic Data Interchange Association
5. ANSI – American National Standards Institute
6. VAN – Value Added Network
7. ANSI ASC – American National Standards Institute Accredited Standard Committee
8. GTDI – Guideline for Trade Data Interchange
9. UN/ECE/ – United -Nations Economic Commission for Europe
10. UN/EDIFACT -United Nations / Electronic Data Interchange for Administration, Commerce, and Transport
11. UNSM -United Nations Standard Message

Assertion And Reason
Question 1.
Assertion (A): According to the National Institute of Standards and Technology, EDI is the computer-to-computer interchange of strictly formatted messages that represent documents other than monetary instruments.
Reason(R): The Electronic Data Interchange (EDI) is the exchange of business documents between one trade partner and another electronically.
a) Both (A) and (R) are correct and (R) ¡s the correct explanation of (A)
b) Both (A) and (R) are correct, but (R) is not the correct explanation of (A)
c) (A) is true and (R) is false
d) (A) is false and (R) is true
b) Both (A) and (R) are correct, but (R) is not the correct explanation of (A)

Question 2.
Reason(R): The Electronic Data Interchange (EDI) is the exchange of business documents between one trade partner and another electronically.
a) Both (A) and (R) are correct and (R) is the correct explanation of (A)
b) Both (A) and (R) are correct, but (R) is not the correct explanation of (A)
c) (A) is true and (R) is false
d) (A) is false and (R) is true
d) (A) is false and (R) is true

Question 3,
Assertion (A): United Nations / Electronic Data Interchange for Administration, Commerce, and Transport (UN / EDIFACT) is an international EDI – a standard developed under the supervision of the United Nations.
Reason(R): In 1985, the UN / EDIFACT syntax rules were approved as ISO: IS09735 standard by the International Organization for Standardization.
a) Both (A) and (R) are correct and (R) is the correct explanation of (A)
b) Both (A) and (R) are correct, but (R) is not the correct explanation of (A)
c) (A) is true and (R) is false
d) (A) is false and (R) is true
c) (A) is true and (R) is false

Question 4.
Assertion (A): The segment table lists the message tags.
Reason(R): It contains the tags, tag names, requirements designator, and repatriation field.
a) Both (A) and (R) are correct and (R) is the correct explanation of (A)
b) Both (A) and (R) are correct, but (R) is not the correct explanation of (A)
c) (A) is true and (R) is false
d) (A) is false and (R) is true
a) Both (A) and (R) are correct and (R) is the correct explanation of (A)

Question 5.
Assertion (A): The top level of EDIFACT structure is Interchange.
Reason(R): Interchange is also called an envelope. An interchange may contain multiple messages. It starts with UNB and ends with UNZ
a) Both (A) and (R) are correct and (R) is the correct explanation of (A)
b) Both (A) and (R) are correct, but (R) is not the correct explanation of (A)
c) (A) is true and (R) is false
d) (A) is false and (R) is true
a) Both (A) and (R) are correct and (R) is the correct explanation of (A)

Question 1.
Who is the father of EDI?
Ed Guilbert is called the father of EDI

Question 2.

Question 3.
What is Paperless Payment?
Transfer of money from one bank account to another, via computer-based systems, is known as Paperless payment

Question 4.
What is another name of Direct EDI?
Another name of Direct EDI is Point-to-Point EDI.

Question 5.
How many alphabets require for EDI messages?
Every EDI message requires six uppercase English Alphabets

Match The Following:

1. EDI – Booking confirmation
3. EDIFACT – Envelope
4. Interchange – Delivery forecast
5. CEFIC – Directories
6. EDIFURN – Chemical industry
7. EDIGAS – Technical error
8. CONTRL – Multiple credit advice
9. APERAK – Furniture industry
10. CREMUL – Arrival of Message
12. IFTMBC – Paperless Payment

2. Paperless Payment
3. Directories
4. Envelope
5. Chemical industry
6. Furniture industry
8. Arrival of Message
9. Technical error
11. Delivery forecast
12. Booking confirmation

Find The Odd One On The Following

1. (a) Deliver/ Notes
(b) Invoices
(d) EDIFACT
(d) EDIFACT

2. (a) EDIFACT
(b) XML
(c) CDMA
(d) ANSI ASCX12
(c) CDMA

3. (a) Direct EDI
(b) InDirectEDI
(c) Web EDI
(d) Mobile EDI
(b) InDirectEDI

4. (a) FTP/VPN
(b) HTTP
(c) SFTPP
(d) FTPS
(b) HTTP

5. (a) Dial-Up Line
(b) I way
(c) point to point
(d) Internet
(c) point to point

6. (a) Email
(b) MIME
(c) HTTP
(d) ANSI X12
(d) ANSI X12

7. (a) Transport Layer
(b) Semantic Layer
(c) Application Layer
(d) physical Layer
(c) Application Layer

8. (a) Standards
(b) Catalogs
(c) TDCC
(d) guidelines
(c) TDCC

9. (a) CREMUL
(b) DELFOR
(c) APERAK
(d) IFTMBC
(c) APERAK

10. (a) Segment Terminator
(b) : – component data
(c) ? – data element separator
(d). – decimal point
(c) ? – data element separator

Important Years To Remember:

 1975 First EDI standards were released by TDCC 1977 Drafting and using an EDI project begin 1978 TDCC is renamed as Electronic Data Interchange Association (EDIA) 1979 ANSI ASC developed ANSI X12 1985 UN created the EDIFACT 1986 UN/EDIFACT is officially proposed 1987 UN / EDIFACT syntax rules were approved

Part B

Question 1.
What is VAN?
A value-added network is a company, that is based on its own network, offering EDI services to other businesses. A value-added network acts as an intermediary between trading partners. The principal operations of value-added networks are the allocation of access rights and providing high data security.

Question 2.
What are the types of EDI?

1. Direct EDI
2. EDI via VAN
3. EDI via FTP/VPN, SFTP, FTPS
4. Web EDI
5. Mobile EDI

Question 3.
Write a short note on the Segment Table?
Segment Tables:
The segment table lists the message tags. It contains the tags, tag names, requirements designator, and repetitation field. The requirement designator may be mandatory (M) or conditional (C). The (M) denotes that the segment must appear atleast once. The (C) denotes that the segment may be used if needed.

Question 4.
Mention some International accepted EDI Standards.

• EDIFACT
• XML
• ANSI
• ASC XI2,

Part C

Question 1.
Write a short note on EDIFACT Structure.

• EDIFACT is a hierarchical structure where the top level is referred to as an interchange, and lower levels contain multiple messages.
• The messages consist of segments, which in turn consist of composites.
• The final iteration is a data element.

Question 2.
What is EDI interchange?

• The top-level of the EDIFACT structure is Interchange.
• An interchange may contain multiple messages.
• It starts with UNB and ends with UNZ

Question 3.
What is the EDI segment?

• A segment is a three-character alphanumeric code.
• These segments are listed in segment tables.
• Segments may contain one, or several related user data elements.

Question 4.
Write a note on EDI Interchange?
EDI Interchange:
Interchange is also called an envelope. The top-level of the EDIFACT structure is Interchange. An interchange may contain multiple messages. It starts with UNB and ends with UNZ.

Part D

Question 1.
Explain EDI standards?
EDI Standards:

• The standard is the most critical part of the entire EDI. Since EDI is the data transmission and information exchange in the form of an agreed message format, it is important to develop a unified EDI standard.
• The EDI standard is mainly divided into the following aspects: basic standards, code-standards, message standards, document standards, management standards, application standards, communication standards, and security standards.
• The first industry-specific EDI standard was the TDCC published by the Transportation Data Coordinating Committee in 1975.
• Then other industries started developing unique standards based on their individual needs. E.g. WINS in the warehousing industry.
• Since the application of EDI has become more mature, the target of trading operations is often not limited to a single industry.
• In 1979, the American National Standards Institute Accredited Standard Committee (ANSI ASC) developed a wider range of EDI standards called ANSI XI2.
• On the other hand, the European region has also developed an integrated EDI standard. Known as GTDI (Guideline for Trade Data Interchange).
• ANSI X12 and GTDI have become the two regional EDI standards in North America and Europe respectively.
• After the development of the two major regional EDI standards and a few years after the trial, the two standards began to integrate and conduct research and development of common EDI standards.
• Subsequently, the United Nations Economic Commission for Europe (UN/ECE/WP.4) hosted the task of the development of international EDI standards. In 1986, UN/EDIFACT is officially proposed. The most widely used EDI message standards are the United Nations EDIFACT and the ANSI X12.

Question 2.
Draw the structure of the UN/EDIFACT message.

## Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium

Students can Download Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium Pdf, Tamil Nadu 12th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

## TN State Board 12th Chemistry Model Question Paper 4 English Medium

Time: 3 Hours
Maximum Marks: 70

Instructions:

1. The question paper comprises of four parts
2. You are to attempt all the parts An internal choice of questions is provided wherever applicable
3. All questions of Part I, II, III, and IV are to be attempted separately
4. 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
5. Question numbers 16 to 24 in Part II are two-mark questions. These are lo be answered in about one or two sentences
6. Question numbers 25 to 33 in Part III are three-marks questions. These are to be answered in about three to five short sentences
7. Question numbers 34 to 38 in Part IV are five-mark Questions These are to answered in detail. Draw diagrams wherever necessary

Part – 1

Answer all the questions. Choose the correct answer. [15 × 1 = 15]

Question 1.
Depressing agents used to separate ZnS from PbS is …………..
(a) NaCN
(b) NaCl
(c) NaNO3
(d) NaNO2
(a) NaCN

Question 2.
The basic structural unit of silicates is
(a)(SiO3)2-
(b) (SiO4)2-
(c) (SiO)
(d) (SiO4)4-
(d) (SiO4)4-

Question 3.
………………… is a pungent smelling gas.
(a) Ammonia
(b) Nitric acid
(c) Fluorite
(d) Sodium chloride
(a) Ammonia

Question 4.
Consider the following statement.
(i) All the actinoids are non radioactive.
(ii) Neptunium and other heavier elements are produced by artificial transformation of naturally occurring elements by nuclear reactions.
(iii) Most.of the actinoids have long half lives.
Which of the above statements is/are not correct.
(a) i only
(b) i and ii
(c) ii and iii
(d) i and iii
(d) i and iii

Question 5.
How many geometrical isomers are possible for [Pt (Py) (NH3) (Br) (Cl)]?
(a) 3
(b) 4
(c) 0 3
(d) 15
(a) 3

Question 6.
Metal excess defect is possible in……………….
(a) AgCl
(b) AgBr
(c) KCl
(d) Fes
(c) KCl

Question 7.
The correct difference between first and second order reactions is that ……………………
(a) A first order reaction can be catalysed; a second order reaction cannot be catalysed.
(b) The half life of a first order reaction does not depend on [A0]; the half-life of a second order reaction does depend on [A0].
(c) The rate of a first order reaction does not depend on reactant concentrations; the rate of a second order reaction does depend on reactant concentrations.
(d) The rate of a first order reaction does depend on reactant concentrations; the rate of a second order reaction does not depend on reactant concentrations.
(b) The half life of a first order reaction does not depend on [A0]; the half-life of a second order reaction does depend on [A0].
Solution:
For a first order reaction

Question 8.
Concentration of the Ag+ ions in a saturated solution of Ag2C2O4 is 2.24 × 10-4 mol L-1 solubility product of Ag2C2O4 is
(a) 2.42 × 10-8 mol3 L-3
(b) 2.66 × 10-12mol3 L-3
(c) 4.5 × 10-11mol3 L-3
(d) 5.619 × 10-12mol3 L-3
(d) 5.619 × 10-12mol3 L-3
Solution:

Question 9.
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
(b) Both A and R are wrong

Question 10.
Fog is colloidal solution of……………..
(a) solid in gas
(b) gas in gas
(c) liquid in gas
(d) gas in liquid
(c) liquid in gas

dispersion medium-gas
dispersed phase-liquid

Question 11.
Match the following Column-I with Column-II using the code given below.
c
(a) A – 2,B – 3, C – 4, D – 1

Question 12.

(a) anilinium chloride
(b) O – nitro aniline
(c) benzene diazonium chloride
(d) m – nitro benzoic acid

(c) benzene diazonium chloride

Question 13.
Which one of the following is the IUPAC name of CH3 – CH2 – CH2 CN?
(a) Propiono nitrile
(b) Butane cyanide
(c) Isobutyro nitrile
(d) Butane nitrile
(d) Butane nitrile

Question 14.
The correct corresponding order of names of four aldoses with configuration given below Respectively is ……………
(a) L-Erythrose, L-Threose, L-Eiythrose, D-Threose
(b) D-Threose, D-Erythrose, L-Threose, L-Erythrose
(c) L-Erythrose, L-Threose, D-Erythrose, D-Threose
(d) D-Erythrose, D-Threose, L-Erythrose, L-Threose
(d) D-Erythrose, D-Threose, L-Erythrose, L-Threose

Question 15.
Tranquilisers are substances used for the treatment of ………………
(a) cancer
(b) AIDS.
(c) mental diseases
(d) blood infection
(c) mental diseases

Part – II

Answer any six questions. Question No. 22 is compulsory. [6 × 2 = 12]

Question 16.
What are leaching process?
This method is based on the solubility of the ore in a suitable solvent and the reactions in aqueous solution. In this method, the crushed ore is allowed to dissolve in a suitable solvent, the metal present in the ore is converted to its soluble salt or complex while the gangue remains insoluble. This process is also called chemical method.

Question 17.
What happens when PCI5 is heated?
On heating phosphorous pentachloride, it decomposes into phosphorus trichloride and chlorine.

Question 18.
Write the biologically importance of coordination compounds.
Biological important of coordination compounds:

(i) Ared blood corpuscles (RBC) is composed of heme group, which is Fe2+– Porphyrin complex, it plays an important role in carrying oxygen from lungs to tissues and carbon dioxide from tissues to lungs.

(ii) Chlorophyll, a green pigment present in green plants and algae, is a coordination complex containing Mg2+ as central metal ion surrounded by a modified Porphyrin ligand called corrin ring. It plays an important role in photosynthesis, by which plants converts CO2 and water into carbohydrates and oxygen.

(iii) Vitamin B12(cyanocobalamine) is the only vitamin consist of metal ion. it is a coordination complex in which the central metal ion is CO+ surrounded by Porphyrin like ligand.

(iv) Many enzymes are known to be metal complexes, they regulate biological processes. For example, Carboxypeptidase is a protease enzyme that hydrolytic enzyme important in digestion, contains a zinc ion coordinated to the protein.

Question 19.
Mention the factors affecting the reaction rate.
The rate of the reaction is affected by the following factors.

• Nature and state of the reactant .
• Concentration of the reactant
• Surface area of the reactant
• Temperature of the reaction
• Presence of a catalyst

Question 20.
What is meant by standard reduction potential? What is its application?

• The standard reduction potential (E°) is a measure of the oxidising tendency of the species.
• The greater the E° value means greater is the tendency shown by the species to accept electrons and undergo reduction.
• So higher the E° values, lesser is the tendency to undergo corrosion.

Question 21.
Give two important characteristics of physisorption.
Important characteristics of physisorption:

• It is reversible
• It has low heat of adsorption
• It has weak van der Waals forces of attraction with adsorbent.
• It increases with increase in pressure.
• It forms multimolecular layer.

Question 22.
Draw the major product formed when 1-ethoxyprop-l-ene is heated with one equivalent

This reaction follows SN1 mechanism because in this reaction the more stable carbocation is formed that is double bonded carbocation. Therefore, the given molecule reacts with HI to form ethanol and 1-iodo prop-1-ene.

Question 23.
Define Tautomerism. Give example. Why tertiary nitro alkanes do not exhibit tautomerism?
(i) Tautomerism is an isomerism in which the isomers change into one another with great ease of shifting of proton so that they exist together in equilibrium.

(ii) Tertiary nitro alkanes do not exhibit tautomerism due to absence of a-H atom.

Question 24.
Why do soaps not work in hard water?
Soaps are sodium or potassium salts of long-chain falty acids. Hard water contains calcium and magnesium ions. When soaps are dissolved in hard water, these ions displace sodium or potassium from insoluble calcium or magnesium salts of fatty acids. These insoluble salts separate as scum.

This is the reason why soaps do not work in hard water.

Part – III

Answer any six questions. Question No. 31 is compulsory. [6 × 3 = 18]

Question 25.
Write the application of Iron (Fe).

• Iron is one of the most useful metals and its alloys are used everywhere including bridges, electricity pylons, bicycle chains, cutting tools and rifle barrels.
• Cast iron is used to make pipes, valves and pumps stoves etc.
• Magnets can be made of iron and its alloys and compounds.

An important alloy of iron is stainless steel, and it is very resistant to corrosion. It is used in architecture, bearings, cutlery, surgical instruments and jewellery. Nickel steel is used for making cables, automobiles and aeroplane parts. Chrome steels are used for manufacturing cutting tools and crushing machines.

Question 26.
Give a reason to support that sulphuric acid is a dehydrating agent.
Sulphuric acid is highly soluble in water and has strong affinity towards water and hence it can be used as a dehydrating agent. When dissolved in water it forms mono (H2SO4.H2O) and di (H2SO4.H2O) hydrates and the reaction is exothermic.
The dehydration property can also be illustrated by its reaction with organic compounds such as sugar, oxalic acid and formic acid.

Question 27.
Give evidence that [CO(NH3)gCl]SO4 and [CO(NH3)5SO4]Cl are ionisation isomers.
When they are dissolved in water, they will give different ions in the solution which can be
tested by adding AgNO3 solution and BaCl2 solution. When Cl ions are the counter ions, a white precipitate will be obtained with AgNO3 solution. If SO42- ions are the counter ions, a white precipitate will be obtained with BaCl2 solution.

Question 28.
For a reaction, X + Y → Product; quadrupling [x] , increases the rate by a factor of 8. Quadrupling both [x] and [y], increases the rate by a factor of 16. Find the order of the reaction with respect to x and y. what is the overall .order of the reaction?
z = k [x]m [y]n …… (1)
8z = k [4x]m [y]n…… (2)
I 6z = k [4x]m [4y]n…… (3)
Dividing Eq (2) by Eq (1) we get,

1.5 order with respect to x.
Dividing Eq (3) by Eq (1) we get,

16 = 4m . 4n
16 = 42.4n
16/16 = 4n
1 = 4n
∴ n = 0 [Zero order with respect toy]
Overall order of the reaction, ‘
k[x]m[y]n
k [x]1.5 [y]0
Order = (1.5 + 0) = 1.5

Question 29.
Identify the conjugate acid base pair for the following reaction in aqueous solution

• HF and F , HS and H2S are two conjugate acid – base pairs.
• F is the conjugate base of the acid HF (or) HF is the conjugate acid of F .
• H2S is the conjugate acid of HS (or) HS is the conjugate base of H2S.

(ii)

• HPO42- and PO43- , SO3-2 and HSO3 are two conjugate acid – base pairs.
• PO43- is the conjugate base of the acid HPO42- (or)
HPO42- is the conjugate acid of PO43-.
• HSO3 is the conjugate acid of SO3-2 (or) SO3-2 is the conjugate base of HSO3

(iii)

• NH4+ and NH3, CO32- and HCO3 are two conjugate acid – base pairs.
• HCO3 is the conjugate of acid CO32- (or) CO32- is the conjugate bases of HCO3
• NH3 is the conjugate base of NH4+ (or) NH4+ is the conjugate acid of NH3.

Question 30.
In the following fields, how adsorption is applied?
(i) Medicine (ii) Metallurgy (iii) Mordant & Dyes (iv) indicators
(i) Medicine: – Drugs cure diseases by adsorption on body tissues.

(ii) Metallurgy:- Sulphide ores are concentrated by a process called froth floation in which lighter ore particles are adsorbed by pine oil.

(iii) Mordants and Dyes:- Most of the dyes are adsorbed on the surface of the fabric. Mordants are the substances used for fixing dyes onto the fabric.

(iv) In the precipitation titrations, the end point is indicated by an external indicator which changes its colour after getting absorbed on precipitate. It is used to indicate the end point of filtration.

Question 31.
Identify A, B, C, and D

Question 32.
Draw the structural formula and write the IUPAC name of
(i) N, N-dimethyl aniline (ii) Benzyl amine (iii) N-methyl benzylamine

Question 33.
Differentiate soap and detergents?
Soap :

1. Soaps are sodium or potassium salt of long chain fatty acid.
2. Soaps are made from animal (or) plant fats and oils.
3. Soaps have lesser cleansing action.
5. Soaps are less effective in hard water.
6. They have a tendency to form a scum in hard water.
7. Example: Sodium palmitate.

Detergent :

1. Detergent is sodium salt of alkyl hydrogen sulphate or alkyl benzene sulphonic acid.
2. Detergents are made from petrochemicals.
3. Detergents have more cleansing action.
5. Detergents are more effective even in hard water.
6. They do not form scum with hard water.
7. Example: Sodium lauryl sulphate.

Part – IV

Answer all the questions. [5 × 5 = 25]

Question 34.
(a) (i) Explain the role of carbon monoxide in the purification of nickel? (2)
(ii) Describe the structure of diborane. (3)
[OR]
(b) (i) What type of hybridisation occur in 1. BrF5. 2. BrF3 (2)
(ii) Most of the transition metals act as catalyst. Justify this statement. (3)
(a) (i) During the purification of Nickel by Mond’s process, carbon monoxide (CO) is used to convert impure nickel to nickel carbonyl.
Nickel carbonyl is an unstable compound. Heating to higher temperature decomposes it to give pure nickel.

(ii) In diborane two BH2 units are linked by two bridged hydrogens. Therefore, it has eight B-H bonds. However, diborane has only 12 valance electrons and are not sufficient to
form normal covalent bonds. The four terminal B-H bonds are normal covalent bonds (two centre – two electron bond or 2c-2e bond). The remaining four electrons have to used for the bridged bonds.
1. e. two three centred B-H-B bonds utilise two electrons each. Hence, these bonds are three centre- two electron bonds. The bridging hydrogen atoms are in a plane as shown in the figure.

In dibome, the boron is sp3 hybridised. Three of the four sp3 hybridised orbitals contains single electron and the fourth orbital is empty. Two of the half filled hybridised orbitals of each boron overlap with the two hydrogens to form four terminal 2c-2e bonds, leaving one empty and one half filled hybridised orbitals on each boron. The Three centre – two electron bonds, B-H-B bond formation involves overlapping the half filled hybridised orbital of one boron, the empty hybridised orbital of the other boron and the half filled Is orbital of hydrogen.

[OR]

(b) (i) 1. BrF5. is a AX5 type. Therefore is has sp3 d2 hybridisation. Hence, BrF5 molecule has square pyramidal shape.
2. BrF3 is a AX3 type. Therefore it has sp3 d hybridisation. Hence, BrF3 molecule has T-shape.

(ii) Many industrial processes use transition metals or their compounds as catalysts. Transition metal has energetically available d orbitals that can accept electrons from reactant molecule or metal can form bond with reactant molecule using its ‘d’ electrons.

For example, in the catalytic hydrogenation of an alkene, the alkene bonds to an active site by using its n electrons with an empty d orbital of the catalyst.

Question 35.
(a) (i) Why tetrahedral complexes do not exhibit geometrical isomerism. (2)
(ii) Explain about the importance and application of coordination complexes. (3)
[OR]
(b) Explain AAAA and ABABA and ABCABC type of three dimensional packing with the help of neat diagram.
(a) (i) In tetrahedral geometry

• AH the four ligands are adjacent or equidistant to one another.
• The relative positions of donor atoms of ligands attached to the central metal atom are same with respect to each other.
• It has plane of symmetry

Therefore, tetrahedral complexes do not exhibit geometrical isomerism.

(ii) 1. Phthalo blue – a bright blue pigment is a complex of copper (II) ion and it is used in printing ink and packaging industry.

2. Purification of Nickel by Mond’s process involves formation of [Ni (CO)4 ] which yields 99.5% pure on decomposition.

3 . EDTA is used as a chelating ligand for the separation of lanthanides, in softening of hard water and also in removing lead poisoning.

4. Coordination complexes are used in the extraction of silver and gold from their ores
by forming soluble cyano complex. These cyano complexes are reduced by zinc to
yield metals. This process is called Mac – Arthur Forrest cyanide process.

5. Some metal ions are estimated more accurately by complex fonnation. For eg., Ni present in Nickel chloride solution is estimated accurately forming an insoluble, complex called [Ni (DMG)2].

6. Many of the complexes are used as catalyst in organic and inorganic reactions. For e.g.,
(i) Wilkinson’s Catalyst – [(PPh3)3, Rh Cl] is used for hydrogenation of alkenes.
(ii) Ziegler – Natta Catalyst [TiCl4 + Al (C2H5) ]3 is used in the polymerisation of ethene.

7. In photography, when the developed film is washed with sodium thio sulphate solution (hypo), the negative film gets fixed. Undecomposed AgBr forms a soluble complex called sodium dithio sulphate argentate (I) which can be removed easily by washing the film with water.

AgBr + 2 Na2 S2O3 → Na3 [Ag (S2O3)2] + 2NaBr

[OR]

(b) (i) AAAA type of three dimensional packing:
This type of three dimensional packing arrangements can be obtained by repeating the AAAA type two dimensional arrangements in three dimensions, i.e., spheres in one layer sitting directly on the top of those in the previous layer so that all layers are identical. All spheres of different layers of crystal are perfectly aligned horizontally and also vertically, so that any unit cell of such arrangement as simple cubic structure as shown in fig.

In simple cubic packing, each sphere is in contact with 6 neighbouring spheres – Four in its own layer, one above and one below and hence the coordination number of the sphere in simple cubic arrangement is 6.

(ii) ABABA type of three dimensional packing:
In this arrangement, the spheres in the first layer (A type) are slightly separated and the second layer is formed by arranging the spheres in the depressions between the spheres in layer A as shown in figure. The third layer is a repeat of the first. This pattern ABABAB is repeated throughout the crystal. In this arrangement, each sphere has a coordination number of 8, four neighbors in the layer above and four in the layer below.

(iii) ABCABC type of three dimensional packing:
In this arrangement (FCC) second layer spheres are arranged at the dips of first layer. Third layer spheres are arranged in a manner such that it cover the octahedral void. Then no longer third layer is similar to first or second layer. Third layer gives different arrangement. Fourth Layer spheres are similar to first layer. If the first, second and third layer are represented as A,B,C then this type of packing gives the arrangement of layers as ABCABC… (Le.,), the first three layers do not resemble first, second and third layers respectively and the sequence is repeated with the addition of more layers.

In this arrangement atoms occupy 74% of the available space and thus has 26% vacant space. The coordination number is 12.

Voids: The empty spaces between the three dimensional layers are known as voids. There are
two types of common voids possible. They are tetrahedral and octahedral voids.

Tetrahedral void:
A void formed by three spheres of a layer in contact with each other and also with a sphere on the top or bottom r Layer a layer is a hole between four spheres. The spheres are arranged at the vertices Layerc of a regular tetrahedron such a hole or void is called tetrahedral void.

Octahedral void:
A hole or void formed by three spheres of a hexagonal layer and another three spheres of the adjacent layer is a hole between six spheres. The spheres are arranged at the vertices of a regular octahedron. Such a hole or void is abc arrangement – ccp structure called octahedral void.

The buffer capacity or index is defined as the concentration of acid or base to add in order to modify the pH of an aqueous solution.

Question 36.
(a) (i) The rate constant for a first order reaction is 1.54 × 10-3s-1 (2)
Calculate its half life time.
(ii) Explain about protective action of colloid. (3)
[OR]
(b) (i) What is buffer solution? Give an example for an acidic buffer and a basic buffer. (2)
(ii The value of kspof two sparingly soluble salts Ni(OH)2 and AgCN are 2.0 × 10-15 and 6 × 10-17 respectively. Which salt is more soluble? Explain. (3)
(a) (i) We know that, t1/2 = 0.693/ k
t1/2= O.693/1.54 ×10-3s -1 = 450s

(ii) 1. Lyophobic sols are precipitated readily even with small amount of electrolytes. But they are stabilised by the addition of small amount of lyophillic colloid.

2. A smaLl amount of gelatine sol is added to gold sol to protect gold sol.

3. Gold number is a measure of protecting power of a colloid. Gold number is defined as the number of milligrams of hydrophillic colloid that will just prevent the precipitation of 10 ml of gold sol on the addition of ImI of 10% NaCI solution. Smaller the gold number, greater the protective power.

[OR]

(b) (i) 1. Buffer is a solution which consists of a mixture of weak acid and its conjugate base (or) a weak base and its conjugate acid.

2. This buffer solution resists drastic changes in its pH upon addition of a small quantities of acids (or) bases and this ability is called buffer action.

3. Acidic buffer solution : Solution containing acetic acid and sodium acetate.
Basic buffer solution : Solution containing NH4O and NH4Cl.

(ii)

Ni(OH)2 is more soluble than AgCN.

Question 37.
[OR]
(b) (i) What happens when m-cresol is treated with acidic solution of sodium dichromate? (3)
(ii) Formic acid is more stronger than acetic acid. Justify this statement. (2)
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 SO42- to form PbSO4 precipitate

Pb2+(aq) + SO42- → PbS4(s) …………….(2)

3. Reduction occurs at the cathode

PbO2(s) + 4H+(aq) + 2e – → Pb2+(aq) + 2H2O ………….(3)

The Pb2+ ions also combines with SO42- ions to form sulphuric acid to form PbSO4 Precipitate

Pb2+(aq) + SO42-(aq) → PbSO4 …………..(4)

4. The overall reaction is,
(1) + (2) + (3) + (4)
pb(s) + PbO2(s) + 4H+ + 2SO42-(aq) → 2PbSO4(s) + 2H2O(l)

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 SO42- 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.

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.

[OR]

(b) (i) When m-cresol is treated with acidic solution of sodium dichromate it gives 4-hydroxy benzoic acid.

(ii) The electron releasing groups (+1 groups) increase the relative charge on the carboxylate ion and destabilise it and hence the loss of proton becomes difficult.
+I groups are CH3, – C2H5, – C3H7

Question 38.
(i) An aromatic compound ‘A’ on treatment with aqueous ammonia and heating
forms compound ‘B’ which on heating with Br2 and KOH forms a compound ‘C’
of molecular formula C6H7N. Write the structures and IUPAC names of compound
A,BandC. (3)
(ii) Lactose act as reducing sugar. Justify this statement. (2)
[OR]
(b) (i) Explain the mechanism of enzyme action? (3)
(ii) Which chemical is responsible for the antiseptic properties of dettol? (2)
(a) (i) Step-1: To find out the structure of compounds‘B’and‘C’.
1. Since compound ‘C’ with molecular formula C6H7N is formed from compound ‘B’ on treatment with Br2 + KOH (i.e, ., Hoffmann bromamide reaction). Therefore, compound ‘B’ must be an amide and ‘C’ must be an amine. The only amine having the molecular formula C6H7N is C6H5NH2 (i.e., aniline or benzenamine).

2. Since ‘C’ is aniline, therefore, the amide from which it is formed must be benzamide (C6H5CONH2). Thus, compound ‘B’ is benzamide:
The chemical equation showing the conversion of ‘B’ to ‘C’ is

Step-2: To find out the structure of compound ‘A’. Since compound ‘B’ is formed from compound ‘A’ by treatment with aqueous ammonia and heating. Therefore, compound ‘A’ must be benzoic acid or benzenecarboxylic acid.

(ii) Lactose is a disaccharide and contains one galactose unit and one glucose unit.
In lactose, the β-D galactose and β-D glucose are linked by β-1, 4 – glycosidic bond.
The aldehyde carbon is not involved in the glycosidic bond hence it retains its reducing property and is called a reducing sugar.

[OR]

(b) (i) Enzymes are bio catalysts that catalyse a specific bio chemical reaction. They generally activate the reaction by reducing the activation energy by stabilising the transition state.

In a typical reaction, enzyme E binds with the substrate molecule leversity to produce an enzyme substrate complex. During this stage the substrate is converted into product and the enzyme becomes free and is ready to bind to another substrate molecule.

(ii) (a) Chloroxylenol and (b) Terpineol are the chemicals responsible for the antiseptic properties of dettol. But among these two, chloroxylenol plays more important role. Chloroxylenol is an antiseptic and disinfectant which is used for skin disinfection and cleaning surgical instruments.

## Tamil Nadu 12th Physics Model Question Paper 3 English Medium

Students can Download Tamil Nadu 12th Physics Model Question Paper 3 English Medium Pdf, Tamil Nadu 12th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

## TN State Board 12th Physics Model Question Paper 3 English Medium

General 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 to be answered
in about one or two sentences.
• Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered
in about three to five short sentences.
• Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered
in detaiL Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 70

Part – I

Answer all the questions. Choose the correct answer. [15 x 1 = 15]

Question 1.
A thin conducting spherical shell of radius R has a charge Q which is uniformly distributed on its surface. the correct plot for electrostatic potential due to this spherical shell is ………

(b)

Question 2.
The temperature coefficient of resistance of a wire is 0.00125 per °C. At 300 K, its resistance is 1Ω. The resistance of the wire will be 2Ω at …………….
(a) 1154 K
(b) 1100 K
(c) 1400K
(d) 1127 K

Question 3.
n resistances, each of r Ω, when connected in parallel give an equivalent resistance of R Ω. If these resistances were connected in series, the combination would have a resistance in horns equal to ………..
(a) n2R
(b) $$\frac{R}{n^{2}}$$
(c) $$\frac{R}{n}$$
(d) nR
(a) n2R
Hint :
Resistance in parallel combination, R= r/n = r = Rn
Resistance in series combination , R = nr = n2R

Question 4.
A thin insulated wire forms a plane spiral of N = 100 tight turns carrying a current 1 = 8 m A (milli ampere). The radii of inside and outside turns are a = 50 mm and b = 100 mm respectively. The magnetic induction at the center of the spiral is……………………….
(a) 5 μT
(b) 7 μT
(c) 8 μT
(d) 10 μT
(b) 7 μT

Question 5.
In an oscillating LC circuit, the maximum charge on the capacitor is Q. The charge on the capacitor, when the energy is stored equally between the electric and magnetic fields, is……………………….
(a) $$\frac{Q}{2}$$
(b) $$\frac{Q}{\sqrt{3}}$$
(c) $$\frac{Q}{\sqrt{2}}$$
(d) Q
(c) $$\frac{Q}{\sqrt{2}}$$

Question 6.
The direction of induced current during electro magnetic induction is given by …………………………………..
(b) Lenz’s law
(c) Maxwell’s law
(d)  Ampere’s law
(b) Lenz’s law

Question 7.
In an electromagnetic wave in free space the rms value of the electric field is 3 V m . The peak value of the magnetic field is………………………..
(a) 1.414 x 10-8 T
(b) 1.0 x 10-8 T
(c) 2.828 x 10-8 T
(d) 2.0 x 10-8 T
(a) 1.414 x 10-8 T

Question 8.
Two point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm approximately. The maximum distance at which these dots can be resolved by the eye is, [take wavelength of light, λ = 500 nm]
(a) 1m
(b) 5m
(c) 3m
(d) 6m
(b) 5m
Hint:
Resolution limit sin θ =

Question 9.
Two mirrors are kept at 60° to each other and a body is placed at middle. The total number of images formed is ……………..
(a) six
(b) four
(c) five
(d) three
(a) six
Hint:
Number of images formed $$n=\frac{360}{\theta}-1=\frac{360}{60}-1=5$$

Question 10.
The threshold wavelength for a metal surface whose photoelectric work function is 3.313 eV is
{a) 4125 Å
(b) 3750  Å
(c) 6000 Å
(d) 2062.5 Å
(b) 3750  Å

Question 11.
Suppose an alpha particle accelerated by a potential of V volt is allowed to collide with a nucleus whose atomic number is Z, then the distance of closest approach of alpha particle to the nucleus is…..

(c)

Question 12.
The speed of the particle, that can take discrete values is proportional to ……………
(a) n-3/2
(b) n-3
(c) n1/2
(d) n
(d) n
Hint:
$$P=m v=\frac{n h}{2 a} ; V \propto n$$

Question 13.
Doping semiconductor results in ………………………………
(a) The decrease in mobile charge carriers
(b) The change in chemical properties
(c) The change in the crystal structure
(d) The breaking of the covalent bond
(c) The change in the crystal structure

Question 14.
The signal is affected by noise in a communication system …………………….
(a) At the transmitter
(b) At the modulator
(c) In the channel
(c) In the channel

Question 15.
The technology used for stopping the brain from processing pain is…………………………..
(a) Precision medicine
(b) Wireless brain sensor
(c) Virtual reality
(c) Virtual reality

Part – II

Answer any six questions. Question No. 21 is compulsory. [6 x 2 = 12]

Question 16.
Define ‘Electric field’.
The electric field at the point P at a distance r from the point charge q is the force experienced by a unit charge and is given by
$$\overrightarrow{\mathrm{E}}=\frac{\overrightarrow{\mathrm{F}}}{q_{0}}$$
The electric field is a vector quantity and its Sl unit is Newton per Coulomb (NC-1).

Question 17.
State the principle of potentiometer.
The basic principle of a potentiometer is that when a constant current flows through a wire of uniform cross-sectional area and composition, the potential drop across any length of the wire is directly proportional to that length.

Question 18.
Compute the intensity of magnetisation of the bar magnet whose mass, magnetic moment and density are 200 g, 2 A m2 and 8 g cm-3, respectively.

Question 19.
What is displacement current?
The displacement current can be defined as the current which comes into play in the region in ‘ which the electric field and the electric flux are changing with time.

Question 20.
What is principle of reversibility?
The principle of reversibility states that light will follow exactly the same path if its direction of travel is reversed.

Question 21.
Calculate the de Broglie wavelength of a proton whose kinetic energy is equal to. 81.9 x 1015 (Given: mass of proton is 1836 times that of electron).

Question 22.
Give the symbolic representation of alpha decay, beta decay and gamma decay.
Alpha decay: The alpha decay process symbolically in the following way
Beta decay: β decay is represented by $$_{ z }^{ A }X\rightarrow _{ z,-1 }^{ A }Y+e^{ + }+v$$
Gamma decay: The gamma decay is given $$\mathrm{z} \mathrm{X}^{*} \rightarrow_{Z}^{\mathrm{A}} \mathrm{X}$$+ gamma(y)rays

Question 23.
Explain the need for a feedback circuit in a transistor oscillator.
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 24.
Give the factors that are responsible for transmission impairments.

• Attenuation
• Distortion (Harmonic)
• Noise

Part – III

Answer any six questions. Question No. 32 is compulsory. [6 x 3 = 18]

Question 25.
A water molecule has an electric dipole moment of 6.3 x 10 Cm. A sample contains 10 water molecules, with all the dipole moments aligned parallel to the external electric field of magnitude 3 x 10s NC-1. How much work is required to rotate all the water molecules from θ = 0° to 90°?
When the water molecules are aligned in the direction of the electric field, it has minimum potential energy. The work done to rotate the dipole from θ = 0° to 90° is equal to the potential energy difference between these two configurations.
W = ΔU = U(90°) – U(0°)
As we know, U = -pE cos θ, Next we calculate the work done to rotate one water molecule
from θ = 0° to 90°,
For one water molecule, W = – pE cos 90° + pE cos 0° = pE
W = 6.3 x10-30 x 3 x 105 = 18.9 x 10-25
For 1022 water molecules, the total work done is Wtot = 18.9x 10 25 x 1022 = 18.9 x 10 3 J

Question 26.
What are the properties of an equipotential surface?
Properties of equipotential surfaces
(i) The work done to move a charge q between any two points A and B,
W = q (VB – VA ). If the points A and B lie on the same equipotential surface, work done is zero because VB – VA

(ii) The electric field is normal to an equipotential surface. If it is not normal, then there is a component of the field parallel to the surface. Then work must be done to move a charge between two points on the same surface. This is a contradiction. Therefore the electric field must always be normal to an equipotential surface.

Question 27.
An electronics hobbyist is building a radio which requires 150 ft in her circuit, but she has only 220 Ω 79 Ω and 92 Ω resistors available. How can she connect the available resistors to get desired value of resistance?
Required effective resistance = 150 ft
Given resistors of resistance, R = 220 ft, R = 79 ft, R = 92 ft
Parallel combination of R1 and R2

Question 28.
What is magnetic permeability?
Magnetic permeability: The magnetic permeability can be defined as the measure of ability of the material to allow the passage of magnetic field lines through it or measure of the capacity of the substance to take magnetisation or the degree of penetration of magnetic field through the substance.

Question 29.
A circular metal of area 0.03 m2 rotates in a uniform magnetic field of 0.4 T. The axis of rotation passes through the centre and perpendicular to its plane and is also parallel to the field. If the disc completes 20 revolutions in one second and the resistance of the disc is 4 Ω calculate the induced emf between the axis and the rim and induced current flowing in the disc.
A = 0.03 m2; B = 0.4T; f = 20rps; R = 4Ω
Area covered in 1 sec = Area of the disc x frequency
= 0.03 x 20 = 0.6 m2

Question 30.
Give the characteristics of image formed by a plane mirror.

• The image formed by a plane mirror is virtual, erect, and laterally inverted.
• The size of the image is equal to the size of the object.
• The image distance far behind the mirror is equal to the object distance in front of it.
• If an object is placed between two plane mirrors inclined at an angle 0, then the number of images n formed is as,$$n=\left(\frac{360}{\theta}-1\right)$$

Question 31.
An electron and an alpha particle have same kinetic energy. How are the de Broglie wavelengths associated with them related?

Question 32.
Assuming VCEsat = 0.2 V and β = 50, find the minimum base current (IB) required to drive the transistor given in the figure to saturation.

VCEsat = 0.2 V and β = 50
VCE = VCC -IC RC
0.2= 3 – Ic(1 k)

Question 33.

• The robots are much cheaper than humans.
• Robots never get tired like humans. It can work for 24 x 7. Hence absenteeism in work place can be reduced.
• Robots are more precise and error free in performing the task.

• Robots have no sense of emotions or conscience.
• They lack empathy and hence create an emotionless workplace.
• If ultimately robots would do all the work, and the humans will just sit and monitor them* health hazards will increase rapidly.

Part – IV

Answer all the questions. [5 x 5 = 25]

Question 34.
(a) Derive an expression for electrostatic potential energy of the dipole in a uniform electric field.
Expression for electrostatic potential energy of the dipole in a uniform electric field:
Considera dipole placed in the uniform electric field $$\overrightarrow{\mathrm{E}}$$ placed in the uniform electric field $$\overrightarrow{\mathrm{E}}$$ . A dipole experiences a torque when kept in an uniform electric field E.

This torque rotates the dipole to align it with the direction of the electric field. To rotate the dipole (at constant angular velocity) from its initial angle θ’ to another angle θ against the torque exerted by the electric field, an equal and opposite external torque must be applied on the dipole. The work done by the external torque to rotate the dipole from angle θ’ to θ at constant angular velocity is

This work done is equal to the potential energy difference between the angular positions 0 and 0′. U(θ) – (Uθ’) = ΔU = -pE cos θ + pE cos θ’.

If the initial angle is = θ’ = 90° and is taken as reference point, then U(θ’) + pE cos 90° = 0. The potential energy stored in the system of dipole kept in the uniform electric field is given by.
U = -pE cos θ = -p-E  ……………. (3)

In addition to p and E, the potential energy also depends on the orientation θ of the electric dipole with respect to the external electric field.

The potential energy is maximum when the dipole is aligned anti-parallel (θ = Jt) to the external electric field and minimum when the dipole is aligned parallel (θ = 0) to the external electric field.

This online reference angle calculator finds the quadrant of a trigonometric angle in a standard position.

[OR]

(b) Deduce the relation for the magnetic induction at a point due to an infinitely long straight conductor carrying current.
Magnetic field due to long straight conductor carrying current: Consider a long straight wire NM with current I flowing from N to M. Let P be the point at a distance a from point O.
Consider an element of length dl of the wire at a distance l from point O and $$\vec{r}$$ be the vector joining the element dl with the point P. Let θ be the angle between $$d \vec{l} \text { and } \vec{r}$$. Then, the magnetic field at P due to the element is
$$d \overrightarrow{\mathrm{B}}=\frac{\mu_{0} \mathrm{Id} l}{4 \pi r^{2}} \sin \theta$$(unit vector perpendicular to $$d \vec{l} \text { and } \vec{r}$$

The direction of the field is perpendicular to the plane of the paper and going into it. This can be determined by taking the cross product between two vectors $$d \vec{l} \text { and } \vec{r}$$
(let it be $$\hat{n}$$). The net magnetic field can be determined by integrating equation with proper limits. $$\overrightarrow{\mathrm{B}}=\int d \overrightarrow{\mathrm{B}}$$ From the figure, in a right angle triangle PAO,

This is the magnetic field at a point P due to the current in small elemental length. Note that we have expressed the magnetic field OP in terms of angular coordinate i.e. θ. Therefore, the net magnetic field at the point P which can be obtained by integrating $$d \overrightarrow{\mathrm{B}}$$ by varying the angle from θ = φ1 to θ =φ2 is

For a an infinitely long straight wire, 1 =0 and 2 =, the magnetic field is $$\overrightarrow{\mathrm{B}}=\frac{\mu_{0} I}{2 \pi a} \hat{n}$$ ………… (3)
Note that here $$\hat{n}$$ represents the unit vector from the point O to P.

Question 35.
(a) Obtain an expression for motional emf from Lorentz force.
Motional emf from Lorentz force: Consider a straight conducting rod AB of length l in a uniform magnetic field $$\vec{B}$$ which is directed perpendicularly into the plane of the paper. The length of the rod is normal to the magnetic field. Let the rod move with a constant velocity $$\vec{v}$$ towards right side. When the rod moves, the free electrons present in it also move with same velocity $$\vec{v}$$ in B. As a result, the Lorentz force acts on free electrons in the direction from B to A and is given by the relation  $$\overrightarrow{\mathrm{F}}_{\mathrm{B}}$$=-
$$e(\vec{v} \times \overrightarrow{\mathrm{B}})$$ …………. (1)

The action of this Lorentz force is to accumulate the free electrons at the end A. This accumulation of free electrons produces a potential difference across the rod which in turn establishes an electric field E directed along BA. Due to the electric field $$\overrightarrow{\mathrm{E}}$$, the coulomb force starts acting on the free electrons along AB and is given by
$$\overrightarrow{\mathrm{F}}_{\mathrm{E}}=-e \overrightarrow{\mathrm{E}}$$ ……….. (2)

The magnitude of the electric field E keeps on increasing as long as accumulation of electrons at the end A continues. The force $$\overrightarrow{\mathrm{F}}_{\mathrm{B}}$$ also increases until equilibrium is reached. At equilibrium, the magnetic Lorentz force $$\overrightarrow{\mathrm{F}}_{\mathrm{B}}$$ and the coulomb force F balance each other and no further accumulation of free electrons at the end A takes place,
i.e

The potential difference between two ends of the rod is
V= El
V= vBl
Thus the Lorentz force on the free electrons is responsible to maintain this potential difference and hence produces an emf

ε = B lv    ………….. (4)
As this emf is produced due to the movement of the rod, it is often called as motional emf.

[OR]

(b) Explain the Maxwell’s modification of Ampere’s circuital law.
Maxwell argued that a changing electric field between the capacitor plates must induce a magnetic field. As currents are the usual sources of magnetic fields, a changing electric field must be associated with a current. Maxwell called this current as the displacement current.

If ‘A’ be the area of the capacitor plates and ‘cf be the charge on the plates at any instant ‘f during the charging process, then the electric field in the gap will be

But $$\frac{d q}{d t}$$ is the rate of change of charge on the capacitor plates. It is called displacement current and given by $$\mathrm{I}_{d}=\frac{d q}{d t}=\varepsilon_{0} \frac{d \Phi_{\mathrm{E}}}{d t}$$

This is the missing term in Ampere’s Circuital Law. The total current must be the sum of the
conduction current Ic Hence, the modified form of the ampere’s law.
$$\oint_{c} \overrightarrow{\mathrm{B}} \cdot \overrightarrow{d l}=\mu_{0}\left(\mathrm{I}_{c}+\varepsilon_{0} \frac{d \Phi_{\mathrm{E}}}{d t}\right)$$

Question 36.
(a) Obtain the equation for resolving power of microscope.
Resolving power of microscope: The diagram related to the calculation of resolution of microscope. A microscope is used to see the details of the object under observation. The ability of microscope depends not only in magnifying the object objective but also in resolving two points on the object separated by a small distance dmin  Smaller the value of dmin better will be the resolving power of the microscope.

The radius of central maxima is, $$r_{0}=\frac{1.22 \lambda v}{a}$$ . In the place of focal length f we have the image distance v. If the difference between the two points on the object to be resolved is dmin, then the magnification m is, $$m=\frac{r_{o}}{d_{\min }}$$

To further reduce the value of dmin the optical path of the light is increased by immersing the objective of the microscope into a bath containing oil of refractive index n.

Such an objective is called oil immersed objective. The term n sin β is called numerical aperture NA.

[OR]

(b) Derive an expression for de Broglie wavelength of electrons.
An electron of mass m is accelerated through a potential difference of V volt. The kinetic energy acquired by the electron is given by
$$\frac{1}{2} m v^{2}=e \mathrm{V}$$
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 $$\lambda=\frac{h}{m v}=\frac{h}{\sqrt{2 e m \mathrm{V}}}$$
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
$$\lambda=\frac{h}{\sqrt{2 m \mathrm{K}}}$$

Question 37.
(a) Explain the variation of average binding energy with the mass number by graph and discuss its features.
We can find the average binding energy per nucleon $$\overline{\mathrm{BE}}$$.
It is given by

The average binding energy per nucleon is the energy required to separate single nucleon from the particular nucleus. $$\overline{\mathrm{BE}}$$ is plotted against A of all known nuclei.

Important inferences from of the average binding energy curve:

• The value of BE rises as the mass number increases until it reaches a maximum value of 8.8 MeV for A= 56 (iron) and then it slowly decreases.
• The average binding energy per nucleon is about 8.5 MeV for nuclei having mass number between A = 40 and 120. These elements are comparatively more stable and not radioactive.
• For higher mass numbers, the curve reduces slowly and BE for uranium is about 7.6 MeV. They are unstable and radioactive.  If two light nuclei with A < 28 combine with a nucleus with A < 56, the binding energy per nucleon is more for final nucleus than the initial nuclei. Thus, if the lighter elements combine to produce a nucleus of medium value A, a large amount of energy will be released. This is the basis of nuclear fusion and is the principle of the hydrogen bomb.
• If a nucleus of heavy element is split (fission) into two or more nuclei of medium value A, the energy released would again be large. The atom bomb is based on this principle and huge energy of atom bombs comes from this fission when it is uncontrolled.

[OR]

(b) State Boolean laws. Elucidate how they are used to simplify Boolean expressions with suitable example.
Laws of Boolean Algebra: The NOT, OR and AND operations are $$\bar{A}$$, A + B, A.B are the Boolean operations. The results of these operations can be summarised as:

Complement law:

The complement law can be realised as $$\overline{\mathrm{A}}$$ = A

Commutative laws
A + B = B +A
A . B = B . A

Associative laws
A + (B + C) = (A + B) + C
A . (B . C) = (A .B) . C

Distributive laws
A( B + C) = AB + AC
A + BC = (A + B) (A + C)
The above laws are used to simplify complicated expressions and to simplify the logic circuitry.

Question 38.
(a) Modulation helps to reduce the antenna size in wireless communication – Explain. Antenna size:
Antenna is used at both transmitter and receiver end. Antenna height is an important parameter to be discussed. The height of the antenna must be a multiple of $$\frac{\lambda}{4}$$
$$h=\frac{\lambda}{4}$$ ……………….. (1)
where X is wavelength $$\left(\lambda=\frac{c}{v}\right), c$$ c is the velocity of light and v is the frequency of the signal v to be transmitted.

An example
Let us consider two baseband signals. One signal is modulated and the other is not modulated. The frequency of the original baseband signal is taken as v = 10 kHz while the modulated signal is v = 1 MHz. The height of the antenna required to transmit the original baseband signal of frequency
v = 10kHz is

The height of the antenna required to transmit the modulated signal of frequency v = 1 MHz is

Comparing equations (2) and (3), we can infer that it is practically feasible to construct an antenna of height 75 m while the one with 7.5 km is not possible. It clearly manifests that modulated signals reduce the antenna height and are required for long distance transmission.

[OR]

(b) What are the possible harmful effects of usage of Nanoparticles? Why?
Possible harmful effects of usage of Nanoparticles:

The research on the harmful impact of application of nanotechnology is also equally important and fast developing. The major concern here is that the nanoparticles have the dimensions same as that of the biological molecules such as proteins. They may easily get absorbed onto the surface of living organisms and they might enter the tissues and fluids of the body.

The adsorbing nature depends on the surface of the nanoparticle. Indeed, it is possible to deliver a drug directly to a specific cell in the body by designing the surface of a nanoparticle so that it adsorbs specifically onto the surface of the target cell.

The interaction with living systems is also affected by the dimensions of the nanoparticles. For instance, nanoparticles of a few nanometers size may reach well inside biomolecules, which is not possible for larger nanoparticles. Nanoparticles can also cross cell membranes.

It is also possible for the inhaled nanoparticles to reach the blood, to reach other sites such as the liver, heart or blood cells. Researchers are trying to understand the response of living organisms to the presence of nanoparticles of varying size, shape, chemical composition and surface characteristics.

## Samacheer Kalvi 12th Maths Guide Book Back Answers Solutions

Subject Matter Experts at SamacheerKalvi.Guide have created Tamilnadu State Board 12th Maths Solutions Book Pdf Free Download New Syllabus of Volume 1 and Volume 2 in English Medium and Tamil Medium are part of Samacheer Kalvi 12th Books Solutions.

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11th Maths Book Back Answers Solutions Guide

### 12th Maths Book Volume 1 Solutions Answers Guide Pdf

Maths Solutions Book for 12th State Board English Medium Free Download Chapter 1 Applications of Matrices and Determinants

12th State Board Maths Solutions Book Pdf Download Chapter 3 Theory of Equations

12th State Board Maths Solutions Book Pdf English Medium Chapter 4 Inverse Trigonometric Functions

12th Maths Solution Book Tamil Medium Free Download Chapter 5 Two Dimensional Analytical Geometry – II

12th Maths Book Volume 1 Solutions Answers Guide Pdf Chapter 6 Applications of Vector Algebra

### 12th Maths Book Volume 2 Answers Solutions Guide Pdf

12th Maths Book Volume 2 Answers Solutions Guide Pdf Chapter 7 Applications of Differential Calculus

12th Maths Solution Book English Medium Free Download Chapter 8 Differentials and Partial Derivatives

12th Maths Solution Book Tamil Medium Pdf Chapter 9 Applications of Integration

Samacheer Kalvi 12th Maths Important Questions Chapter 10 Ordinary Differential Equations

Tamilnadu Samacheer Kalvi 12th Std Maths Guide Chapter 11 Probability Distributions

Samacheer Kalvi 12th Maths Book Solutions English Medium Chapter 12 Discrete Mathematics

12th Maths Guide Pdf Free Download 2021 Chapter 1 அணிகள் மற்றும் அணிக்கோவைகளின் பயன்பாடுகள்

12th Maths Book Back Answers Chapter 3 சமன்பாட்டியல்

Samacheer Kalvi 12th Maths Guide Chapter 4 நேர்மாறு முக்கோணவியல் சார்புகள்

Maths Guide for Class 12 Chapter 5 இரு பரிமாண பகுமுறை வடிவியல் – II

12th Maths Samacheer Solutions Chapter 6 வெக்டர் இயற்கணிதத்தின் பயன்பாடுகள்

TN 12th Maths Solutions Chapter 7 வகை நுண்கணிதத்தின் பயன்பாடுகள்

12th Samacheer Maths Solutions Chapter 8 வகையீடுகள் மற்றும் பகுதி வகைக்கெழுக்கள்

12th Std Maths Guide Chapter 9 தொகை நுண்கணிதத்தின் பயன்பாடுகள்

Samacheer Kalvi Guru 12th Maths Chapter 10 சாதாரண வகைக்கெழுச் சமன்பாடுகள்

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## Samacheer Kalvi 12th Maths Guide Chapter 4 Inverse Trigonometric Functions Ex 4.6

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 4 Inverse Trigonometric Functions Ex 4.6 Textbook Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.6

Inverse Function Calculator is an online tool that helps find the inverse value for the given function.

Choose the most suitable answer from the given four alternatives:

Question 1.
The value of sin-1(cos x), 0 ≤ x ≤ π is
(a) π – x
(b) x – $$\frac {π}{2}$$
(c) $$\frac {π}{2}$$ – x
(d) x – π
Solution:
(c) $$\frac {π}{2}$$ – x
Hint:
sin-1(cos x) = sin-1(sin($$\frac {π}{2}$$ – x)) = $$\frac {π}{2}$$ – x

Question 2.
If sin-1 x + sin-1 y = $$\frac {2π}{3}$$; then cos-1 x + cos-1 y is equal to
(a) $$\frac {2π}{3}$$
(b) $$\frac {π}{3}$$
(c) $$\frac {π}{6}$$
(d) π
Solution:
(b) $$\frac {π}{3}$$
Hint:
sin-1x + cos-1x + cos-1y + sin-1y = $$\frac {π}{2}$$ + $$\frac {π}{2}$$ = π
$$\frac {2π}{3}$$ + cos-1x + cos-1y = π
cos-1x + cos-1y = π – $$\frac {2π}{3}$$ = $$\frac {3π-2π}{3}$$ = $$\frac {π}{3}$$

Question 3.
sin-1$$\frac {3}{5}$$ – cos-1$$\frac {12}{13}$$ + sec-1$$\frac {5}{3}$$ – cosec-1$$\frac {13}{12}$$ is equal to
(a) 2π
(b) π
(c) 0
(d) tan-1$$\frac {12}{65}$$
Solution:
(c) 0
Hint:

Question 4.
If sin-1 x = 2sin-1 α has a solution, then
(a) |α| ≤ $$\frac {1}{√2}$$
(b) |α| ≥ $$\frac {1}{√2}$$
(c) |α| < $$\frac {1}{√2}$$
(d) |α| > $$\frac {1}{√2}$$
Solution:
(a) |α| ≤ $$\frac {1}{√2}$$
Hint:
If sin-1 x = 2sin-1 α has a solution then
–$$\frac {π}{2}$$ ≤ 2sin-1α ≤ $$\frac {π}{2}$$
–$$\frac {π}{4}$$ ≤ sin-1α ≤ $$\frac {π}{4}$$
sin($$\frac {-π}{4}$$) ≤ α ≤ sin$$\frac {π}{4}$$
–$$\frac {1}{√2}$$ ≤ α ≤ $$\frac {1}{√2}$$
-|α| ≤ $$\frac {1}{√2}$$

Question 5.
sin-1(cos x) = $$\frac {π}{2}$$ – x is valid for
(a) -π ≤ x ≤ 0
(b) 0 ≤ x ≤ π
(c) –$$\frac {π}{2}$$ ≤ x ≤ $$\frac {π}{2}$$
(d) –$$\frac {π}{4}$$ ≤ x ≤ $$\frac {3π}{4}$$
Solution:
(b) 0 ≤ x ≤ π
Hint:
sin-1 (cosx) = $$\frac {π}{2}$$ – x is valid for
cos x = sin ($$\frac {π}{2}$$ – x)
cos x ∈ [0, π]
∴ 0 ≤ x ≤ π

Question 6.
If sin-1 x + sin-1 y + sin-1 z = $$\frac {3π}{2}$$, the value of show that x2017 + y2018 + z2019 – $$\frac {9}{x^{101}+y^{101}+z^{101}}$$ is
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(a) 0
Hint:

Question 7.
If cot-1 x = $$\frac {2π}{5}$$ for some x ∈ R, the value of tan-1 x is
(a) –$$\frac {π}{10}$$
(b) $$\frac {π}{5}$$
(c) $$\frac {π}{10}$$
(d) –$$\frac {π}{5}$$
Solution:
(c) $$\frac {π}{10}$$
Hint:
tan-1 x + cos-1 $$\frac {π}{2}$$
tan-1x = $$\frac {π}{2}$$ – cot-1 x = $$\frac {π}{2}$$ – $$\frac {2π}{5}$$
= $$\frac {5π-4π}{10}$$ = $$\frac {π}{10}$$

Question 8.
The domain of the function defined by f(x) = sin-1 $$\sqrt {x-1}$$ is
(a) [1, 2]
(b) [-1, 1]
(c) [0, 1]
(d) [-1, 0]
Solution:
(a) [1, 2]
Hint:
f(x) = sin-1 $$\sqrt {x-1}$$
$$\sqrt {x-1}$$ ≥ 0
-1 ≤ $$\sqrt {x-1}$$ ≤ 1
∴ 0 ≤ $$\sqrt {x-1}$$ ≤ 1
0 ≤ x – 1 ≤ 1
1 ≤ x ≤ 2
x ∈ [1, 2]

Question 9.
If x = $$\frac {1}{5}$$ the value of cos(cos-1x + 2sin-1x) is
(a) –$$\sqrt{\frac {24}{25}}$$
(b) $$\sqrt{\frac {24}{25}}$$
(c) $$\frac {1}{5}$$
(d) –$$\frac {1}{5}$$
Solution:
(d) –$$\frac {1}{5}$$
Hint:
cos[cos-1x + sin-1x + sin-1x] = cos($$\frac {π}{2}$$ + sin-1x)
= -sin(sin-1x)
[∵ cos(90+θ) = -sin θ]
= -x = –$$\frac {1}{5}$$

Question 10.
tan-1($$\frac {1}{4}$$) + tan-1($$\frac {2}{9}$$) is equal to
(a) $$\frac {1}{2}$$cos-1($$\frac {3}{5}$$)
(b) $$\frac {1}{2}$$sins-1($$\frac {3}{5}$$)
(c) $$\frac {1}{2}$$tan-1($$\frac {3}{5}$$)
(d) tan-1($$\frac {1}{2}$$)
Solution:
(d) tan-1($$\frac {1}{2}$$)
Hint:

Question 11.
If the function f(x) = sin-1(x² – 3), then x belongs to
(a) [-1, 1]
(b) [√2, 2]
(c) [-2, -√2]∪[√2, 2]
(d) [-2, -√2]
Solution:
(c) [-2, -√2]∪[√2, 2]
Hint:
-1 ≤ x² – 3 ≤ 1
-1 + 3 ≤ x² ≤ 1 + 3
⇒ 2 ≤ x² ≤ 4
±√2 ≤ x ≤ ± 2
[-2, -√2]∪[√2, 2]

Question 12.
If cot-1 2 and cot-1 3 are two angles of a triangle, then the third angle is
(a) $$\frac {π}{4}$$
(b) $$\frac {3π}{4}$$
(c) $$\frac {π}{6}$$
(d) $$\frac {π}{3}$$
Solution:
(b) $$\frac {3π}{4}$$
Hint:
A + B + C = π (triangle)
cot-1 2 + cot-1 3 + C = π

Question 13.
sin-1(tan$$\frac {π}{4}$$) – sin-1($$\sqrt{\frac {3}{x}}$$) = $$\frac {π}{6}$$. Then x is root of the equation
(a) x² – x – 6 = 0
(b) x² – x – 12 = 0
(c) x² + x – 12 = 0
(d) x² + x – 6 = 0
Solution:
(b) x² – x – 12 = 0
Hint:

Question 14.
sin-1(2 cos²x – 1) + cos-1(1 – 2 sin²x) =
(a) $$\frac {π}{2}$$
(b) $$\frac {π}{3}$$
(c) $$\frac {π}{4}$$
(d) $$\frac {π}{6}$$
Solution:
(a) $$\frac {π}{2}$$
Hint:
sin-1(2 cos² x – 1) + cos-1(1 – 2 sin²x)
= sin-1 (2 cos² x – 1) + cos-1 (1 – sin² x – sin² x)
= sin-1(2 cos² x – 1) + cos-1(cos² x – (1 – cos²x))
= sin-1(2 cos² x – 1) + cos-1(cos² x – 1 + cos²x)
= sin-1(2 cos² x – 1) + cos-1(2 cos² x – 1)
= $$\frac {π}{2}$$ [∵ sin-1 x + cos-1 x = $$\frac {π}{2}$$]

Question 15.
If cot-1($$\sqrt {sinα}$$) + tan-1($$\sqrt {sinα}$$) = u, then cos 2u is equal to
(a) tan²α
(b) 0
(c) -1
(d) tan 2α
Solution:
(c) -1
Hint:
cot-1 x + tan-1 x = $$\frac {π}{2}$$
∴ u = $$\frac {π}{2}$$
cos 2u = cos 2($$\frac {π}{2}$$) = cos π = -1

Question 16.
If |x| ≤ 1, then 2 tan-1 x – sin-1$$\frac {2x}{1+x²}$$ is equal to
(a) tan-1x
(b) sin-1x
(c) 0
(d) π
Solution:
(c) 0
Hint:
sin-1$$\frac {2x}{1+x²}$$ = 2 tan-1x
∴ 2 tan-1 x – 2 tan-1 x = 0

Question 17.
The equation tan-1 x – cot-1 x = tan-1($$\frac {1}{√3}$$) has
(a) no solution
(b) unique solution
(c) two solutions
(d) infinite number of solutions
Solution:
(b) unique solution
Hint:
tan-1 x – cot-1 x = tan-1($$\frac {1}{√3}$$) …….. (1)
tan-1 x – cot-1 x = $$\frac {π}{2}$$ ……… (2)
2 tan-1 x = $$\frac {π}{6}$$ + $$\frac {π}{2}$$ = $$\frac {2π}{3}$$
tan-1 x = $$\frac {π}{3}$$
x = √3 which is uniqe solution.

Question 18.
If sin-1 x + cot-1($$\frac {1}{2}$$) = $$\frac {π}{2}$$, then x is equal to
(a) $$\frac {1}{2}$$
(b) $$\frac {1}{√5}$$
(c) $$\frac {2}{√5}$$
(d) $$\frac {√3}{2}$$
Solution:
(b) $$\frac {1}{√5}$$
Hint:

Question 19.
If sin-1 $$\frac {x}{5}$$ + cosec-1$$\frac {5}{4}$$ = $$\frac {π}{2}$$, then the value of x is
(a) 4
(b) 5
(c) 2
(d) 3
Solution:
(d) 3
Hint:

Question 20.
sin(tan-1 x), |x| < 1 is equal to
(a) $$\frac {x}{\sqrt{1-x^2}}$$
(b) $$\frac {1}{\sqrt{1-x^2}}$$
(c) $$\frac {1}{\sqrt{1+x^2}}$$
(d) $$\frac {x}{\sqrt{1+x^2}}$$
Solution:
(d) $$\frac {x}{\sqrt{1+x^2}}$$
Hint:
tan a = x
W.K.T 1 + tan² a = sec² a
1 + x² = sec² a
sec a = $$\sqrt{1+x^2}$$
$$\frac {1}{cosa}$$ = $$\sqrt{1+x^2}$$
cos a= $$\frac {1}{\sqrt{1+x^2}}$$
sin a = $$\sqrt{1-cos^2a}$$ = $$\sqrt{1-\frac {1}{1+x^2}}$$
$$\sqrt{\frac{1+x^2 -1}{1+x^2}}$$ = $$\frac {x}{\sqrt{1+x^2}}$$

## Samacheer Kalvi 11th Accountancy Guide Chapter 9 Rectification of Errors

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Accountancy Guide Pdf Chapter 9 Rectification of Errors Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors

### 11th Accountancy Guide Bank Rectification of Errors Text Book Back Questions and Answers

I. Multiple Choice Questions

Question 1.
Error of principle arises when ________.
a) There is complete omission of a transaction
b) There is partial omission of a transaction
c) Distinction is not made between capital and revenue items
d) There are wrong postings and wrong castings
c) Distinction is not made between capital and revenue items

Question 2.
Errors not affecting the agreement of trial balance are ________.
a) Errors of principle
b) Errors of overcasting
c) Errors of undercasting
d) Errors of partial omission
a) Errors of principle

Question 3.
The difference in trial balance is taken to ________.
a) The capital account
c) The suspense account
d) The profit and loss account
c) The suspense account

Question 4.
A transaction not recorded at all is known as an error of ________.
a) Principle
b) Complete omission
c) Partial omission
d) Duplication
b) Complete omission

Question 5.
Wages paid for installation of machinery wrongly debited to wages account is ah error of ________.
a) Partial omission
b) Principle
c) Complete omission
d) Duplication
b) Principle

Question 6.
Which of the following errors will not affect the trial balance?
a) Wrong balancing of an account
b) Posting an amount in the wrong account but on the correct side
c) Wrong totaling of an account
d) Carried forward wrong amount in a ledger account
b) Posting an amount in the wrong account but on the correct side

Question 7.
Goods returned by Senguttuvan were taken into stock, but no entry was passed in the books. While rectifying this error, which of the following accounts should be debited?
a) Senguttuvan account
b) Sales returns account
c) Returns outward account
d) Purchases returns account
b) Sales returns account

Question 8.
A credit purchase of furniture from Athiyaman was debited tg purchases account. Which of the following accounts should be debited while rectifying this error?
a) Purchases account
b) Athiyaman account
c) Furniture account
d) None of these
c) Furniture account

Question 9.
The total of purchases book was overcast. Which of the following accounts should be debited in the rectifying journal entry?
a) Purchases account
b) Suspense account
c) Creditor account
d) None of the above
b) Suspense account

Question 10.
Which of the following errors will be rectified using suspense account?
a) Purchases returns book was undercast by ₹ 100
b) Goods returned by Narendran was not recorded in the books ;
c) Goods returned by Akila ₹ 900 was recorded in the sales returns book as ₹ 90
d) A credit sale of goods to Ravivarman was not entered in the sales book.
a) Purchases returns book was undercast by ₹ 100

II. Very Short Answer Type Questions

Question 1.
What is meant by the rectification of errors?
Correction of errors in the books of accounts is not done by erasing, rewriting, or striking the figures which are incorrect. Correcting the errors that have occured is called Rectification.

Question 2.
What is meant by the error of principle?
The error of principle means the mistake committed in the application of fundamental accounting principles in recording a transaction in the books of accounts.

Question 3.
What is meant by the error of partial omission?
When the accountant has failed to record a part of the transaction, it is known as an error of partial omission. This error usually occurs in posting. This error affects only one account.

Question 4.
What is meant by the error of complete omission?
It means the failure to record a transaction in the journal or subsidiary book or failure to post both the aspects in the ledger. This error’ affects two or more accounts.

Question 5.
What are compensating errors?
The errors that make up for each other or neutralize each other are known as compensating errors. These errors may occur in related or unrelated accounts. Thus, excess debit or credit in one account may be compensated by excess credit or debit in some other account. These are also known as offsetting errors.

Question 1.
Write a note on the error of principle by giving an example.
It means the mistake committed in the application of fundamental accounting principles in recording a transaction in the books of accounts.
Example:
Entering the purchase of an asset in the purchases book. Machinery purchased on credit for ₹ 10,000 by M/s. Anbarasi garments manufacturing company entered in the purchases book.

Question 2.
Write a note on the suspense account.
1. When the trial balance does not tally, the amount of difference is placed to the debit (when the total of the credit column is higher than the debit column) or credit (when the total of the debit column is higher than the credit column) to a temporary account known as ‘suspense account

2. Suspense account will remain in the books until the location and rectification of errors.

3. Afterrectifying the errors and posting the rectification entries to the respective ledger accounts, the suspense account appearing in the ledger is to be balanced.

4. If all the errors are located and rectified, the suspense account gets closed.

Question 3.
What are the errors not disclosed by a trial balance?
Certain errors will not affect the agreement of trial balance. Though such errors occur in the books of accounts, the total debit and credit balance will be the same. The trial balance will tally. Errors of complete omission, error of principle, compensating error, a wrong entry in the subsidiary books are not disclosed by the trial balance.

Question 4.
What are the errors disclosed by a trial balance?
Certain errors affect the agreement of trial balance. If such errors have occurred in the books of accounts, the total of debit and credit balances will not be the same. The trial balance will not tally. The error of partial omission and error of commission affect the agreement of trial balance.
Examples of such errors are as follows:

• Entered in the journal but posted to one account and omitted to be posted to the other.
• Posting an amount to the wrong side of a ledger account
• Posting twice in a ledger account.
• Over-casting or under-casting in a subsidiary book.
• Posting a wrong amount to the wrong side of an amount.
• Errors in carrying forward the page total from one page to the next page of an account or subsidiary book.
• Errors arising in the balancing of an account.
• Omission to post an entry from a subsidiary book.

Question 5.
Write a note on one-sided errors and two-sided errors.
(a) Rectification of one-sided errors before preparing trial balance:

• When a one-sided error is detected before preparing the trial balance, no journal entry is required to be passed in the books.
• In such cases, the error can be rectified by giving an explanatory note in the account affected as to whether the concerned account is to be debited or credited.
• Example: Sales book is undercast by ₹ 100.
• In this case, the sales book is undercast by ₹ 100.
• The total of sales book is posted to the credit side of the sales account in the ledger.
• The under casting has resulted in an under-crediting of sales accounts by ₹ 100.
• This is an error of commission.
• The error is only in the sales account. There is short credit in the sales account by ₹ 100. Hence, it is rectified by crediting sales account by ₹ 100.

(b) Rectification of two-sided errors before preparing the trial balance:
1. When a two-sided error is detected before preparing the trial balance, it must be rectified by passing a rectifying journal entry in the journal proper after analyzing the error.
Example: Goods sold to Anand for ₹ 1,000 on credit was not entered in the sales book. The entry will be

Method of deriving the rectifying entries

IV. Exercises

Question 1.
State the account/s affected in each of the following errors
a) Goods purchased on credit from Saranya for ₹ 150 was posted to the debit side of her account.
b) The total of purchases book ₹ 4,500 was posted twice.
Solution:
a) Saranya account should be credit with ₹ 300
b) Purchases account should be credit with ₹ 4,500

Question 2.
State the account/s affected in each of the following errors
a) Goods sold to Vasu on credit for ₹ 1,000 was not recorded in the sales book.
b) The total of the sales book ₹ 2,500 was posted twice.
Solution:
a) Sales account should be credit with ₹ 1,000
b) Sales account should be Debited with ₹ 2,500

Question 3.
Rectify the following errors discovered before the preparation of the trial balance
a) Sales book was undercast by ₹ 100
b) Purchases returns book was overcast by ₹ 200
Solution:
a) Sales account should be Credited with ₹ 100
b) Purchases returns account should be debited with ₹ 200

Question 4.
Rectify the following errors before the preparation of trial balance
a) Returns outward book was undercast by ₹ 2,000
b) Returns inward book total was taken as ₹ 15,000 instead of ₹ 14,000
c) The total of the purchases account was carried forward ₹ 100 less.
Solution:
a) Purchases return account should be Credited with ₹ 2,000
b) Sales returns account should be Debited with ₹ 1,000
c) Purchases accounts should be debited ₹ 100

Question 5.
Rectify the following errors assuming that the trial balance is yet to be prepared
a) Sales book was undercast by ₹ 400
b) Sales returns book was overcast by ₹ 500
c) Purchases book was undercast by ₹ 600
d) Purchases returns book was overcast by ₹ 700
e) Bills receivable book was undercast by ₹ 800
Solution:
a) Sales account should be credited with ₹ 400
b) Sales return accounts should be Credited with ₹ 500
c) Purchases account should be debited with ₹ 600
d) Purchases returns account should be debited with ₹ 700
e) Bills received account should be debited with ₹ 800

Question 6.
Rectify the following errors before preparing trial balance
a) The total of purchases book was carried forward ₹ 90 less.
b) The total of purchases book was carried forward ₹ 180 more.
c) The total of sales book was carried forward ₹ 270 less.
d) The total of sales returns book was carried forward ₹ 360 more.
e) The total of purchase returns book was carried forward ₹ 450 less.
Solution:
a) Purchases account is to be Debited with ₹ 90
b) Purchases account is to be Credited with ₹ 180
c) Sales account is to be Credited with ₹ 270
d) Sales returns account is to be Credited with ₹ 360
e) Purchases returns account is to be Credit with ₹ 450

Question 7.
The following errors were located by the accountant before the preparation of trial balance. Rectify them.
a) The total of the discount column of ₹ 1,100 on the debit side of the cash book was not yet osted.
b) The total of the discount column on the credit side of the cash book was undercast by ₹ 500.
c) Purchased goods from Anbuchelvan on credit for ₹ 700 was posted to the debit side of his account.
d) Sale of goods to Ponmukil on credit for ₹ 78 was posted to her account as ₹ 87.
e) The total sales returns book of ₹ 550 was posted twice.
Solution:
a) Discount account should be debited with ₹ 1,100
b) Discount account should be Credited with ₹ 500
c) Anbuchelvan account should be Credited with ₹ 1,400
d) Ponmuki account should be Credit with ₹ 9
e) Sales return account should be Credit with ₹ 550

Question 8.
The accountant of a firm located the following errors before preparing the trial balance. Rectify them.
a) Machinery purchased for ₹ 3,000 was debited to the purchases account.
b) Interest received ₹ 200 was credited to the commission account.
c) An amount of ₹ 1,000 paid to Tamil selvan as salary was debited to his personal account.
d) Old furniture sold for ₹ 300 was credited to sales account.
e) Goods worth ₹ 800 purchased from Soundarapandian on credit was not recorded in the books of accounts.
Solution:
Journal Entries

Question 9.
Rectify the following errors which were located before preparing the trial balance.
a) Wages paid ₹ 2,000 for the erection of machinery was debited to wages account.
b) Sales returns book was short totaled by ₹ 1,000.
c) Goods purchased for ₹ 200 were posted as ₹ 2,000 to the purchases account.
d) The sales book was overcast by ₹ 1,500.
e) Cash paid to Mukil ₹ 2,800 which was debited to Akhil’s account as ₹ 2,000.
Solution:
Rectification of Errors:

Question 10.
Rectify the following errors which were located at the time of preparing the trial balance
a) The total of the discount column on the debit side of the cash book of ₹ 225 was posted twice.
b) Goods of the value of ₹ 75 returned by Ponnarasan were not posted to his account.
c) Cash received from Yazhini ₹ 1,000 was not posted.
d) Interest received ₹ 300 has not been posted.
e) Rent paid ₹ 100 was posted to rent account as ₹ 10
Solution:
Rectification of Errors

Question 11.
The following errors were located at the time of preparing the trial balance. Rectify them.
a) A personal expense of the proprietor ₹ 200 was debited to the traveling expenses account.
b) Goods of ₹ 400 purchased from Ramesh on credit was wrongly credited to Ganesh’s account.
c) An amount of ₹ 500 paid as salaries to Mathi was debited to his personal account.
d) An amount of ₹ 2,700 paid for the extension of the building was debited to the repairs account.
e) A credit sale of goods of ₹ 700 on credit to Mekala was posted to Krishnan’s account.
Solution:
Rectification of Errors

Question 12.
Rectify the following journal entries.

Solution:
Rectification of Errors

Question 13.
Rectify the following errors discovered after the preparation of the trial balance
a) Rent paid was carried forward to the next page ₹ 500 short.
b) Wages paid were carried forward ₹ 250 excess.
Solution:
a) Rent account should be debited with ₹ 500
b) Wages account should be Credited with ₹ 250

Question 14.
Rectify the following errors after preparation of trial balance
a) Salary paid to Ram ₹ 1,000 was wrongly debited to his personal account.
b) A credit sale of goods to Balu for ₹ 450 was debited to Balan.
Solution:
Rectification of Errors

Question 15.
Pass necessary journal entries to rectify the following errors located after the preparation of trial balance
a) Sales book was undercast by ₹ 1,000.
b) A amount of ₹ 500 paid for wages was wrongly posted to the machinery Account.
Solution:
Rectification of Errors

Question 16.
Give journal entries to rectify the following errors discovered after the preparation of trial balance
a) Purchases book was overcast by ₹ 10,000.
b) Repairs to the furniture of ₹ 500 were debited to the furniture account.
c) A credit sale of goods to Akilnilavan for ₹ 456 was credited to his account as ₹ 654.
Solution:
Rectification of Errors

Question 17.
Rectify the following errors located after the preparation of trial balance
a) Purchases book was undercast by ₹ 900.
b) Sale of old furniture for ₹ 1,000 was credited to sales account.
c) Purchase of goods from Arul for ₹ 1,500 on credit was not recorded in the books.
Solution:
Rectification of Errors

Question 18.
The following errors were located after the preparation of the trial balance. Pass journal entries to rectify them. Assume that there exists a suspense account.
a) The total sales book was undercast by ₹ 350.
b) The total of the discount column on the debit side of the cash book ₹ 420 was not posted.
c) The total of one page of the purchases book of ₹ 5,353 was carried forward to the next page as ₹ 5,533.
d) Salaries ₹ 2,400 were posted as ₹ 24,000.
e) Purchase of goods from Sembiyanmadevi on credit for ₹ 180 was posted to her account as ₹ 1,800
Solution:
Rectification of Errors

Question 19.
Rectify the following errors assuming that the trial balance is already prepared and the difference was placed to suspense account
a) Sales book was undercast by ₹ 250
b) Purchases book was undercast by ₹ 120
c) Sales book was overcast by ₹ 130
d) Bills receivable book was undercast by ₹ 75
e) Purchases book was overcast by ₹ 35.’
Solution:
Rectification of Errors

Question 20.
The following errors were located after the preparation of the trial balance. The difference in trial balance has been taken to the suspense account. Rectify them.
a) The total of purchases book was carried forward ₹ 70 less.
b) The total sales book was carried forward ₹ 340 more.
c) The total of purchases book was carried forward ₹ 150 more.
d) The total of sales book was carried forward ₹ 200 less.
e) The total of purchase returns book was carried forward ₹ 350 less.
Solution:
Rectification of Errors

Question 21.
The following errors were located by the accountant after the preparation of the trial balance. There exists a suspense account. Rectify them.
a) The total of the discount column of ₹ 1,180 on the debit side of the cash book was not posted.
b) Purchase of goods from Arivuchelvan on credit for ₹ 600 was posted to the debit side of his account.
c) The total of the discount column on the credit side of the cash book was undercast by ₹ 400.
d) The total sales returns book of ₹ 570 was posted twice.
e) Sold goods to Mukil on credit for ₹ 87 was posted to her account as ₹ 78.
Solution:
Rectification of Errors

Question 22.
The accountant of a firm located the following errors after preparing the trial balance. Rectify them assuming that there is a suspense account.
a) Machinery purchased for ₹ 3,500 was debited to purchases account.
b) ₹ 1,800 paid to Raina as salary was debited to his personal account.
c) Interest received ₹ 200 was credited to the commission account.
d) Goods worth ₹ 1,800 purchased from Amudhanila on credit was not recorded in the books of accounts,
e) Used furniture sold for ₹ 350 was credited to the sales account.
Solution:
Rectification of Errors

Question 23.
The book-keeper of a firm found that the trial balance was cut by ₹ 922 (excess credit). He placed the amount in the suspense account and subsequently found the following errors
a) The total discount column on the credit side of the cash book ₹ 78 was not posted in the ledger.
b) The total of purchases book was short by ₹ 1,000.
c) A credit sale of goods to Natarajan for ₹ 375 was entered in the sales book as ₹ 735.
d) A credit sale of goods to Mekala for ₹ 700 was entered in the purchases book.
You are required to give rectification entries and prepare a suspense account.
Solution:
Rectification of Errors

Suspense Account

Question 24.
The books of Raman did not agree. The accountant placed the difference of ₹ 1,270 to the debit of the suspense account. Rectify the following errors and prepare the suspense account
a) Goods taken by the proprietor for his personal use ₹ 75 was not entered in the books.
b) A credit sale of goods to Shanmugam for ₹ 430 was credited to his account as ₹ 340.
c) A purchase of goods on credit for ₹ 400 from Vivek was entered in the sales book. However, Vivek’s account was correctly credited.
d) The total of the purchases returns book ₹ 300 was not posted.
Solution:
Rectification of Errors

Suspense Account

### 11th Accountancy Guide Rectification of Errors Additional Important Questions and Answers

Question 1.
The errors can be classified into ……………… types.
(a) One
(b) Two
(c) Three
(d) Four
(d) Four

Question 2.
When one or both aspects of a transaction are recorded in the wrong category of an account, this is called ________.
a) Error of Principle
b) Error of Omission
c) Error of Commission
d) Error of original entry
a) Error of Principle

Question 3.
The errors that make up for each other or neutralize each other are known as ………………
(a) Errors of commission
(b) Errors of principle
(c) Errors of omission
(d) Compensating errors
(d) Compensating errors

Question 4.
₹ 60,000 paid on the extension of the building wrongly debited to Repairs account. This is called the error of ________.
a) Commission
b) Omission
c) Principle
d) None of these
c) Principle

Question 5.
Sales book is undercast by ₹ 100, classify the error
(a) Errors of principle
(b) Errors of commission
(c) Errors in casting
(d) Errors of omission
(c) Errors in casting

Question 6.
________ are those which cancel themselves out.
a) Error of principle
b) Error of Commission
c) Error of Omission
d) Compensating errors
d) Compensating errors

Question 7.
________ errors can be located in the preparation of trial balance.
a) Error of principle
b) Error of Commission
c) Error of Omission
d) Compensating errors
b) Error of Commission

Question 8.
An entry of ₹ 75 has been debited to Rajesh’s A/c as ₹ 57 is an error of ________.
a) Error of principle
b) Error of Commission
c) Error of Omission
d) Compensating errors
b) Error of Commission

Question 9.
Casting errors are the result of ________.
a) Wrong totaling
b) Wrong Balancing
c) the wrong carry forward
d) None of the above
a) Wrong totaling

Question 10.
Suspense account is usually closed when ________.
a) Accounts are finalized
b) After the completion of auditing
c) All the errors are rectified
d) None of the above
c) All the errors are rectified

Question 1.
What is error or omission?
The failure of the accountant to record a transaction or an item in the books of accounts is known as an error of omission. It can be a complete omission or partial omission.

Question 2.
At what stages the errors can occur?
The following types of errors may occur in various stages:

• At the stage of journalizing
• At the stage of posting
• At the stage of balancing
• At the stage of preparing trial balance

Question 3.
What do you mean by errors?
Errors mean recording or classifying or summarising the accounting transactions wrongly or omissions to record them by a clerk or an accountant unintentionally.

Question 4.
What is an error or omission?
The failure of the accountant to record a transaction or an item in the books of accounts is known as an error of omission. It can be a complete omission or partial omission.

Question 5.
What is an error of commission?
When a transaction is incorrectly recorded, it is known as error of commission. It usually occurs due to lack of concentration or carelessness of the accountant.

Question 6.
Write a note on one-sided errors.
When a one-sided error is detected before preparing the trial balance, no journal entry is required to be passed in the books. In such cases, the error can be rectified by giving an explanatory note in the account affected as to whether the concerned account is to be debited or credited.

Question 7.
Write a note on two-sided errors.
When a two-sided error is detected before preparing the trial balance, it must be rectified by passing a rectifying journal entry in the journal proper after analyzing the error.

Question 1.
What are the steps to be followed to locate the errors after the preparation of the trial balance?
While preparing trial balance, if it does not tally, it is an indication of the presence of errors in the books of accounts. The difference in the trial balance is transferred to the suspense account and then errors are to be located and rectified.

The following are the steps to be followed to locate errors after preparing a trial balance

1. The totals of debit and credit columns of trial balance are to be checked
2. The balances of various ledger accounts shown in the trial balance are to be checked to ensure whether they are shown in the respective columns (debit or credit).
3. The difference in the trial balance must be halved and compared with the balances of the ledger to verify whether any ledger balance is recorded on the wrong side of the trial balance.
4. The totals of all the subsidiary books are to be checked, especially if the difference is ₹ 1, ₹ 100, etc.
5. If the difference is divisible by ‘9’, the difference may be due to the transposition of figures in the books (Writing ₹ 127 as ₹ 172). Hence, the possibilities of transposition of figures shall be checked.
6. The accounts of all the creditors and debtors are to be verified.
7. The correctness of the balances of various ledger accounts is to be ensured.
8. The correctness of the balances of various ledger accounts is to be ensured.
9. All the amounts carried forward from one page to the next are to be verified. .
10. If the difference still exists, as a final step all the entries in the journals should be verified.

Question 2.
What are the steps to be followed to locate the errors before preparation of trial balance?
Errors may be located before preparing the trial balance either spontaneously or by intentional scrutiny of books of accounts.

The following are the steps to be followed to locate errors before preparing trial balance:

1. Scrutiny of entries made in the journal proper
2. Scrutiny of entries made in the subsidiary books
3. Checking the totals of the subsidiary books
4. Scrutiny of postings made to the ledger accounts
5. Scrutiny of balancing of ledger accounts

Question 3.
Spot out the location of errors before preparation of trial balance.
Errors may be located before preparing the trial balance either spontaneously or by intentional scrutiny of books of accounts. The following are the steps to be followed to locate errors before preparing trial balance

1. Scrutiny of entries made in the journal proper.
2. Scrutiny of entries made in the subsidiary books.
3. Checking the totals of the subsidiary books
4. Scrutiny of balancing of ledger accounts.
5. Scrutiny of postings made to the ledger accounts

Question 4.
Rectify the following errors

1. Purchases Book is overcast by ₹ 3,500
2. Sales Book is undercast by ₹ 2,000
3. Purchases returns books have been overcast by ₹ 7,600
4. Sales returns book has been undercast by ₹ 500

Solution:
Rectification:

1. Credit Purchases Account with ₹ 3,500
2. Credit Sales Account with ₹ 2,000
3. Debit Purchases Returns account with ₹ 7,600
4. Debit Sales returns account with ₹ 500

Question 5.
Rectify the following errors
1. Sales book undercast by ₹ 5,000
2. Machinery purchased for ₹ 9000 passed through purchases Book.
3. Sales to Ram for ₹ 11,000 debited to his account as ₹ 10,100.
4. Repairs to building ₹ 3,640 debited to buildings account.
Solution:

Question 6.
Rectify the following errors
1. The purchase of machinery from A for ₹ 3,000 has been entered in the purchase daybook.
2. Received ₹ 1,000 from M but credited to N’s account.
3. ₹ 800 paid as wages for the erection of machine has been charged to Repairs account.
4. ₹ 250 received from Ganesh, previously written off, has been credited to Ganesh account
Solution:

Question 7.
Rectify the following Errors
1. Purchase book is overcast by ₹ 6,000
2. Sales book carried forward ₹ 630 instead of ₹ 360
3. Purchase from Sreesha ₹ 5, 000 has been posted to the debit side of her account.
4. Sale of old machinery for ₹ 50,000 has been entered in Sales book.
Solution:

Question 8.
Rectify the following error
1. Salary paid to manager ₹ 8,000 debited to his personal account.
2. Total the discount column in the debit side of the cash book is wrongly cast short. ₹ 540.
3. Total of sales book has been added ₹ 2,400 excess
4. ₹ 230 received in respect of a book debt was posted to the sales account
5. Goods sold for ₹ 3,873 to Raju were returned to us and recorded in the sales book.
Solution:

Question 9.
When a Trial Balance failed to agree, ₹ 37,900 was transferred to the credit of the suspense account. The following errors were discovered. Give journal entries and prepare a suspense account.
1. Sales daybook was undercast by ₹ 40,000
2. Purchase of machinery for ₹ 60,000 was passed through the purchase book.
3. Goods sold to Velu for ₹ 4,500 were posted to his account as ₹ 5,400.
4. Purchase returns book was overcast by ₹ 2,000
5. The total sales book from one page was carried forward to the next page as ₹ 12,000 instead of ₹ 11,000.
Solution:
Rectification of Entries

Suspense Account

Question 10.
In considering the trial balance, a book-keeper finds that the debit total exceeds the credit total by ₹ 3,520. The amount is placed to the credit of a newly opened suspense account. The following mistakes were discovered. Pass the necessary entries for rectifying the mistakes and show the suspense account.
a) Sales daybook was overcast by ₹ 1000
b) A sale of ₹ 500 to Rajesh was wrongly debited to Ramesh
c) General expenses ₹ 180 was posted as ₹ 800.
d) Cash received from Ganesh was debited to his account ₹ 1,500
e) While carrying forward the total of one page of the purchases day book to the next the amount of ₹ 12,350 was entered as ₹ 13,250.
Solution:
Rectification of Entries

Suspense Account

Question 11.
Pass journal entries to rectify the following entries
1. ₹ 4,500 received in respect of a book debt was posted to sales account
2. Defective goods worth ₹ 260 returned to Saran were recorded through the sales returns book.
3. Goods sold for ₹ 950 to Rakesh were returned to us and recorded in the sales book.
4. A purchase of ₹ 2,100 from Banu on the last day of the year was taken into stock, but the invoice was not passed through the purchase book.
Solution:
Rectification of Entries

Question 12.
Show how you will rectify the following entries.
1. A Credit sales of ₹ 650 to Raja were debited to Kaja.
2. A purchase of goods for ₹ 750 from Shaji was debited to his account.
3. An office typewriter purchased for ₹ 6,500 was debited to the Repairs account.
4. A sum of ₹ 3,900 received from a debtor was debited to his account.
5. Purchase of goods for the consumption of proprietor was debited to purchase account ₹ 1,000
Solution:
Rectification of Entries

Question 13.
Write down the rectifying journal entries for the following errors and show the suspense account.
1. The sales returns books have been undercast by ₹ 5,000
2. Goods worth ₹ 1,500 sold to Batiya have been credited to his account.
3. Purchase of furniture ₹ 7,000 has been entered in the purchases account.
4. Cash Rs.4,500 from Aadhira has been posted to his account as ₹ 5,400.
5. A bill received from X for ₹ 4,000 has been posted to the Bills payable account.
Solution:
Rectification of Entries

Suspense Account

Question 1.
Explain the steps to be followed to locate errors after preparing a trial balance
The following steps are to be followed to locate errors after preparing a trial balance

1. The totals of the debit and credit columns of the trial balance are to be checked.
2. The balances of various ledger accounts shown in the trial balance are to be checked to ensure whether they are shown in the respective columns (debit or credit).
3. The difference in the trial balance must be halved and compared with the balances of the ledger to verify whether any ledger balance is recorded on the wrong side of the trial balance.
4. The totals of all the subsidiary books are to be checked, especially if the difference is ₹1 to ₹ 100 etc.
5. If the difference is divisible by 9 the difference may be due to the transposition of figures in the books. (Writing ₹ 127 as ₹ 172). Hence, the possibilities of transposition of figures shall be checked.
6. The accounts of all the creditors and debtors are to be verified.
7. Postings from the subsidiary books to different accounts in the ledger are to be checked.
8. The correctness of the balances of various ledger accounts is to be ensured.
9. All the amounts carried forward from one page to the next area to be verified.
10. If the differences still exists, as a final step all the entries in the journals should be verified.

Question 2.
Briefly explain the Classification or errors

The failure of the accountant to record a transaction or an item in the books of accounts is known as an error of omission. It can be a complete omission or partial omission.
1. Error of complete omission – It means the failure to record a transaction in the journal or subsidiary book or failure to post both the aspects in the ledger. This error affects two or more accounts.

2. Error of partial omission – When the accountant has failed to record a part of the transaction, it is known as an error of partial omission. This error usually occurs in posting. This error affects only one account.

3. Error of commission – When a transaction is incorrectly recorded, it is known as an error of commission. It usually occurs due to lack of concentration or carelessness of the accountant.

Errors of Principle: It means the mistake committed in the application of fundamental accounting principles in recording a transaction in the books of accounts.

4. Compensating errors – The errors that make up for each other or neutralize each other are known as compensating errors. These errors may occur in related or unrelated accounts. Thus, excess debit or credit in one account may be compensated by excess credit or debit in some other account. These are also known as offsetting errors.

5. Errors disclosed by the trial balance and errors not disclosed by the trial balance – Generally, one-sided errors are revealed by a trial balance. They will cause disagreement of totals of debit balances and credit balances. Two-sided errors are not revealed by a trial balance.

Question 1.
Rectify the following errors
i) Purchases book overcast by ₹ 1,300
ii) Sales book under cast by ₹ 2,500
Solution:
i) Credit – Purchases A/c with ₹ 1,300
ii) Credit – Sales A/c with ₹ 2,500

Question 2.
Rectify the following errors
i) Purchases return book overcast by ₹ 750
ii) Sales return book under cast by ₹ 600
Solution:
i) Debit – Purchases return A/c with ₹ 750
ii) Debit – Sales return A/c with ₹ 600

Question 3.
Rectify the following errors
i) Purchases book is carried forward ₹ 850 Less
ii) Sales book total is carried toward ₹ 2,500 More
Solution:
i) Debit – Purchases A/c with ₹ 850
ii) Debit – Sales A/c with ₹ 2,500

Question 4.
Rectify the following errors
i) A total of ₹ 7,580 in the purchases book has been carried forward as ₹ 8,570
ii) The total of the sales book ₹ 7,500 or page 20 was carried forward to page 21 as ₹ 5,570
iii) Purchases return book was carried forward as ₹ 1,520 instead of ₹ 5,120
Solution:
i) Credit – Purchases A/c with ₹ 990
ii) Credit – Sales A/c with ₹ 1,980
iii) Credit – Purchases return A/c with ₹ 3,600

Question 5.
Rectify the following errors
i) Purchases from Bagavathi for ₹ 4,500 has been posted to the debit side of her account
ii) Sales to Vijay for ₹ 1,520 has been posted to his credit as ₹ 1,250
Solution:
i) Purchases from Bagavathi should have been posted to the credit of Bagavathi’s A/c, but it has been . debited. Hence, credit Bagavathi’s A/c with double the amount i.e, ₹ 9,000

ii) Sales to Vijay has to be debited in Vijay’s account but his account is credited with ₹ 1,250. Hence Debit Vijay’s A/c with ₹ 1,250 to ₹ 1,520 i.e, ₹ 2,770

Question 6.
Rectify the following errors
i) Purchases from should for ₹ 750 has been omitted to be posted to the personal A/c
ii) Sales to Khader for ₹ 780 has been posted to his account as ₹ 870
Solution:
i) This is an error of omission. Posting must be to the credit of Shakila’s A/c. Hence, post ₹ 750 to the credit of Skakila’s A/c.

ii) Here Khader’s has been debited with a wrong amount i.e., with an excess amount. To rectify this error, the excess amount must be credited to his account. Hence, Credit Khader’s A/c with ₹ 90.

Question 7.
The following errors were found in the book of pradhu. Give the necessary entries to correct them.
i) Salary of ₹ 10,000 paid to Murali has been debited to his personal account.
ii) ₹ 3,500 paid for a typewriter was charged to the office expenses account.
iii) ₹ 8,000 paid for furniture purchased has been charged to the purchases account.
iv) Repairs made were debited to the building account for ₹ 500
v) An amount of ₹ 5,000 withdrawn by the proprietor for his personal use has been debited to the trade expenses account.
iv) ₹ 2,000 received from Shanthi and Co. has been wrongly entered as from Shakila and Co.
Solution:
In the book of Prathu – Rectification of Errors

Question 8.
Give journal entries to rectify the following errors
i) Purchases of goods from Devi amounting to ₹ 25,000 have been wrongly passed through the sales Book.
ii) Credit sale of goods ₹ 30,000 to Rajan has been wrongly passed through the purchases Book
iii) Sold old furniture for ₹ 3,500 passed through the sales Book.
iv) Paid wages for the construction of Building debited to wages to account ₹ 1,00,000
v) Paid ₹ 10,000 for the installation of machinery debited to wages account.
vi) On 31st December 2003 goods worth ₹ 5,000 were returned by Manjia and were taken into stock on the same date, but no entry was passed in the books.
Solution:
Rectification of Errors

Question 9.
An accountant could not tally the Trial Balance. The difference of ₹ 5,180 was Temporality placed to the credit of suspense account for preparing the final account. The following errors were taken located.
i) Commission of ₹ 500 paid, was posted twice, once to, discount allowed account and once to commission account.
ii) The sales book was undercast by ₹ 1,000
iii) A Credit sales of ₹ 2,780 to Raja though correctly entered in the sale book, was posted wrongly to her account as Rs.3,860
iv) A Credit purchase from Nataraj of ₹ 1,500, though correctly entered in the purchases book, was wrongly debited to his personal account.
v) Discount column of the payments side of the book was wrongly added as ₹ 2,800 instead of ₹ 2,400
You are required to
i) Pass the necessary rectifying entries.
ii) Prepare suspense Account.
Solution:
Rectification of Errors

Suspense Account

Question 10.
Rectify the following Journal Entries
Rectification of Errors

Solution:
Rectification of Errors

Question 11.
Rectify the following Errors
i) ₹ 12,000 paid of salary to cashier Govind, stands debited to his personal account.
ii) An amount of ₹ 5,000 withdrawn by the proprietor for his personal use has been debited to trade Expenses A/c
iii) Cash received from Bala ₹ 300 was credited to Balu.
iv) A credit sale of ₹ 2,000 to Janakiram has been wrongly passed through the purchase book.
Solution:
Rectification of Errors

Question 12.
Rectify the following Errors
i) Repairs made were debited to building account ₹ 5,000
ii) Mahesh networked goods worth ₹ 2,000 no Entry was passed in the Book to this Effects
iii) Purchases of goods from Antony amounting to ₹ 1,500 have been debited to his accounts.
iv) ₹ 5,200 paid for the purchases of the typewriter was charged to the office expenses account.
Solution:
Rectification of Errors

Question 13.
Rectify the following Errors
i) Credit purchases of goods from Madhan of ₹ 300 have been wrongly Entered in the sales book.
ii) ₹ 500 received from Seivan has been credited to Selvi’s account
iii) ₹ 1,000 received as interest was credited to the commission account.
iv) Sales book total ₹ 878 was wrongly totaled as ₹ 788.
v) The total of this discount column, on the debit side of the cash book has been added shortly by ₹ 400.
Solution:
Rectification of Errors

Question 14.
Rectify the following Errors
A Book keeper found his trial balance not balanced, placed the difference amount in the suspense account, and subsequently found the following errors
a) Sales book was overcast by ₹ 1,500
b) ₹ 2900 received from Vani in full settlement of her account of ₹ 3,000 was posted in the cash book but omitted to be entered in her account.
c) The total of the sales book ₹ 12,000 was debited to sales return accounts.
d) ₹ 1,000 received as interest was credited to interest as ₹ 100 Give rectifying entries and show the suspense account.
Solution:
Rectification of Errors

Suspense Account

## Samacheer Kalvi 11th Accountancy Guide Chapter 8 Bank Reconciliation Statement

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Accountancy Guide Pdf Chapter 8 Bank Reconciliation Statement Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

### 11th Accountancy Guide Bank Reconciliation Statement Text Book Back Questions and Answers

I. Multiple Choice Questions

Question 1.
A bank reconciliation statement is prepared by ________.
a) Bank

Question 2.
A bank reconciliation statement is prepared with the help of ________.
a) Bank statement
b) Cash book
c) Bank statement and bank column of the cash book
d) Petty cash book
c) Bank statement and bank column of the cash book

Question 3.
Debit balance in the bank column of the cash book means ________.
a) Credit balance as per bank statement
b) Debit balance as per bank statement
c) Overdraft as per cash book
d) None of the above
a) Credit balance as per bank statement

Question 4.
A bank statement is a copy of ________.
a) Cash column of the cash book
b) Bank column of the cash book
c) A customer’s account in the bank’s book
d) Cheques issued by the business
c) A customer’s account in the bank’s book

Question 5.
A bank reconciliation statement is prepared to know the causes for the difference between:
a) The balance as per the cash column of the cash book and bank column of the cashbook
b) The balance as per the cash column of the cash book and bank statement
c) The balance as per the bank column of the cash book and the bank statement
d) The balance as per petty cash book and the cash book
c) The balance as per the bank column of the cash book and the bank statement

Question 6.
When money is withdrawn from bank, the bank ________.
a) Credits customer’s account
b) Debits customer’s account
c) Debits and credits customer’s account
d) None of these
b) Debits customer’s account

Question 7.
Which of the following is not the salient feature of bank reconciliation statement?
a) Any undue delay in the clearance of cheques will be shown up by the reconciliation
b) Reconciliation statement will discourage the accountant of the bank from embezzlement
c) It helps in finding the actual position of the bank balance
d) Reconciliation statement is prepared only at the end of the accounting period
d) Reconciliation statement is prepared only at the end of the accounting period

Question 8.
Balance as per cash book is ₹ 2,000. Bank charge of ₹ 50 debited by the bank is not yet shown in the cash book. What is the bank statement balance now?
a) 1,950 credit balance
b) 1,950 debit balance
c) 2,050 debit balance
d) 2,050 credit balance
a) 1,950 credit balance

Question 9.
Balance as per bank statement is 1, 000. Cheque deposited, but not yet credited by the bank is 2, 000. What is the balance as per bank column of the cash book?
a) 3,000 overdraft
b) 3,000 favourable
c) 1,000 overdraft
d) 1,000 favourable
b) 3,000 favourable

Question 10.
Which one of the following is not a timing difference?
a) Cheque deposited but not yet credited
b) Cheque issued but not yet presented for payment
c) Amount directly paid into the bank
d) Wrong debit in the cash book
d) Wrong debit in the cash book

Question 1.
What is meant by bank overdraft?
It is not possible to have unfavourable cash balance in the cash book. But, it is possible to have unfavourable balance in the bank account. When the business is not having sufficient money in its bank account, it can borrow money from the bank. As a result of this, the amount is overdrawn from the bank.

Question 2.
What is a bank reconciliation statement?
The bank reconciliation statement is a statement that reconciles the balance as per the bank column of the cash book with the balance as per the bank statement by giving the reasons for such difference along with the amount. The internal record of the business (bank column of cash) can be reconciled with the external record (bank statement).

Question 3.
State any two causes of disagreement between the balance as per bank column of cash book and bank statement.

1. Cheques issued but not yet presented for payment.
2. Cheques deposited into the bank but not yet credited.

Question 4.
Give any two expenses which may be paid by the banker as per standing instruction.

1. Bank Charges
2. Interest

Question 5.
Substitute the following statements with one word/phrase
(a) A copy of the customer’s account issued by the bank.
(b) Debit balance as per bank statement.
(c) Statement showing the causes of disagreement between the balance as per cash book and balance as per bank statement.
(a) Pass book
(b) Pass book favourable
(c) (1) Timing difference, (2) Errors in recording

Question 6.
Do you agree with the following statements? Write “yes” if you agree, and write “no” if you Disagree

1. The bank reconciliation statement is prepared by the banker. – Yes
2. Adjusting the cash book before preparing the bank reconciliation statement is compulsory. – No
3. Credit balance as per bank statement is an overdraft. – No
4. Bank charges debited by the bank increases the balance as per bank statement. – No
5. Bank reconciliation statement is prepared to identify the causes of differences between balance as per bank column of the cash book and balance as per cash column of the cash book. – Yes

Question 1.
Give any three reasons for preparing a bank reconciliation statement.

1. To identify the reasons for the difference between the bank balance as per the cash book and bank balance as per bank statement.
2. To identify the delay in the clearance of cheques.
3. To ascertain the correct balance of the bank column of the cash book.
4. To discourage the accountants of the business as well as the bank from misusing funds.

Question 2.
What is meant by the term “cheque not yet presented?”

1. When the cheques are issued by the business, it is immediately entered on the credit side of the cash book by the business.
2. This may not be entered in the bank statement on the same day.
3. It will be entered in the bank statement only after it is presented with the bank.

Question 3.
Explain why does money deposited into the bank appears on the debit side of the cash book, but on the credit side of the bank statement?
When the cheques are deposited into the bank, the amount is debited in the cash book on the same day. But, these may not be shown in the bank pass book on the same day because these will be entered in the bank statement only after the collection of the cheques.

Question 4.
What will be the effect of interest charged by the bank, if the balance is an overdraft?

1. If the business has taken any loan or overdrawn, interest has to be paid by the business.
2. The entries for bank charges and interest are made in the bank statement.
3. The cash book shows more balance than the bank statement.

Question 5.
State the timing differences in BRS with examples.
The timing differences in BRS are:

1. cheques issued but not yet presented for payment
2. cheques deposited into the bank but not yet credited
3. bank charges and interest on loan and overdraft
4. interest and dividends collected by the bank
5. dishonour of cheques and bills
6. amount paid by parties directly into the bank
7. payment made directly by the bank to others
8. bills collected by the bank on behalf of its customer

IV. Exercises

Question 1.
From the following particulars prepare a bank reconciliation statement of Jayakumar as of 31st December 2016.
a) Balance as per cash book ₹ 7,130
b) Cheque deposited but not cleared ₹ 1,000
c) A customer has deposited ₹ 800 into the bank directly
Solution:
Bank reconciliation statement of Jayakumar as of 31st December 2016.

Question 2.
From the following particulars of Kamakshi traders, prepare a bank reconciliation statement as of 31st March 2018.
a) Debit balance as per cash book ₹ 10,500
b) Cheque deposited into bank amounting to ₹ 5,500 credited by bank but entered twice in the cash book
c) Cheques issued and presented for payment amounting to ₹ 7,000 omitted in the cash book
d) Cheque book charges debited by the bank ₹ 200 not recorded in the cash book.
e) Cash of ₹ 1,000 deposited by a customer of the business in cash deposit machine not recorded in the cash book.
Solution:
Bank reconciliation statement of Kamakshi traders as of 31st March 2018.

Question 3.
From the following information, prepare a bank reconciliation statement to find out the bank statement balance as of 31st December 2017.

Solution:
Bank reconciliation statement of Kamakshi traders as on 31st March, 2018.

Question 4.
On 31st March 2017, Anand’s cash book showed a balance of ₹ 1,12,500, Prepare bank reconciliation statement,
a) He had issued cheques amounting to ₹ 23,000 on 28,3.2017, of which cheques amounting to ₹ 9,000 have so far been presented for payment.

b) A cheque for ₹ 6,300 deposited into bank on 27.3.2017, but the bank credited the same only on 5th April 2017.

c) He had also received a cheque for ₹ 12,000 which, although entered by him in the cash book, was not deposited in the bank.

d) Wrong credit given by the bank on 30th March 2017 for ₹ 2,000.

e) On 30th March 2017, a bill already discounted with the bank for ₹ 3,000 was dishonoured, but no entry was made in the cash book.

f) Interest on debentures of ₹ 700 was received by the bank directly.

g) Cash sales of ₹ 4,000 wrongly entered in the bank column of the cash book.
Solution:
Bank reconciliation statement of Mr, Anand as on 31st March, 2017

Question 5.
From the following particulars of Siva and Company, prepare a bank reconciliation statement as of 31st December 2017.
a) Credit balance as per cash book ₹ 12,000
b) A cheque of ₹ 1,200 Issued and presented for payment to the bank, wrongly credited in the cash book
c) Debit side of bank statement was undercast by ₹ 100
Solution:
Bank reconciliation statement as of 31st December 2017.

Question 6.
From the following particulars of Raheem traders, prepare a bank reconciliation statement as of 31st March 2018.
a) Overdraft as per cash book ₹ 2,500
b) Debit side of cash book was under cash by ₹ 700
c) Amount received by the bank through RTGS amounting to ₹ 2,00,000, omitted in the cash book.
d) Two cheques issued for ₹ 1,800 and ₹ 2,000 on 29th March 2018. Only the second cheque is presented for payment.
e) Insurance premium on the car for ₹ 1,000 paid by the bank as per standing instruction not recorded in the cash book
Solution:
Bank reconciliation statement as of 31st March 2018.

Question 7.
From the following information, prepare a bank reconciliation statement as of 31st December,

Solution:
Bank reconciliation statement as on 31st December, 2017

Question 8.
Prepare bank reconciliation statement from the following data.

Solution:
Bank reconciliation statement

Question 9.
From the following particulars of Veera traders, prepare a bank reconciliation statement as on 31st December, 2017.
a) Credit balance as per bank statement ₹ 6,000
b) Amount received by bank through NEFT for ₹ 3,500, entered twice in the cash book.
c) Cheque dishonoured amounting to ₹ 2,500, not entered in cash book.
Solution:
Bank reconciliation statement as on 31st December, 2017.

Question 10.
Prepare bank reconciliation statement from the following data and find out the balance as per cash book as on 31st March, 2018.

Solution:
Bank reconciliation statement as on 31st March, 2018.

Question 11.
Ascertain the cash book balance from the following particulars as of 31st December 2017
i) Credit balance as per bank statement ₹ 2,500
ii) Bank charges of ₹ 60 have not been entered in the cash book
iii) Cheque deposited on 28th December 2017 for ₹ 1,000 was not yet credited by the bank
iv) Cheque issued on 24th December 2017 for ₹ 700, not yet presented for payment
v) A dividend of ₹ 400 collected by the bank directly but not entered in the cash book
vi) A cheque of ₹ 600 had been dishonoured, but no entry was made in the cash book
vii) Interest on term loan ₹ 1,200 debited by the bank but not accounted in the cash book
viii) No entry had been made in the cash book for a trade subscription of ₹ 500 paid vide banker’s order on 23rd December 2014
Solution:
Bank reconciliation statement as of 31st December 2017.

Question 12.
From the following particulars of Raja traders, prepare a bank reconciliation statement as on 31st January, 2018.
a) Balance as per bank statement ₹ 5,000

b) Cheques amounting to ₹ 800 had been recorded in the cash book as having been deposited into the bank on 25th January 2018, but were entered in the bank statement on 2nd February 2018.

c) Amount received by bank through NEFT amounting to ₹ 3,000, omitted in the cash book.

d) Two cheques issued for ₹ 3,000 and ₹ 2,000 on 29th March 2018. Only the first cheque is presented for payment.

e) Insurance premium on motor vehicles for ₹ 1,000 paid by the bank as per standing instruction not recorded in the cash book.

f) Credit side of cash book was undercast by ₹ 700

g) Subsidy received directly by the bank from the state government amounting to ₹ 10,000, not entered in the cash book.
Solution:
Bank reconciliation statement of Mr. Raja traders as of 31st January 2018.

Question 13.
From the following particulars of Simon traders, prepare a bank reconciliation statement as on 31st March 2018.
a) Debit balance as per bank statement ₹ 2,500
b) Cheques deposited amounting to ₹ 10,000, not yet credited by the bank.
c) Payment through net banking for ₹ 2,000, omitted in the cash book.
Solution:
Bank reconciliation statement as on 31st March 2018.

Question 14.
From the following particulars, ascertain the cash book balance as on 31st December 2016.
i) Overdraft balance as per bank statement ₹ 1,26,640
ii) Interest on overdraft entered in the bank statement, but not yet recorded in cash book ₹ 3,200
iii) Bank charges entered in the bank statement, but not found in cash book ₹ 600
iv) Cheques issued, but not yet presented for payment ₹ 23,360
v) Cheques deposited into the bank but not yet credited ₹ 43,400
vi) Interest on investment collected by the bank ₹ 24,000
Solution:
Bank reconciliation statement as of 31st December 2016.

Question 15.
From the following particulars of John traders, prepare a bank reconciliation statement as on 31st March, 2018.
a) Bank overdraft as per bank statement ₹ 4,000

b) Cheques amounting to ₹ 2,000 had been recorded in the cash book as having been deposited into the bank on 26th March 2018 but were entered in the bank statement on 4th April 2018.

c) Amount received by the bank through cash deposit machine amounting to ₹ 5,000, omitted in the cash book.

d) Amount of ₹ 3,000 wrongly debited to John traders account by the bank, for which no details are available.

e) Bills for collection credited by the bank till 29th March 2017 amounting to ₹ 4,000, but no advice received by John traders.

f) Electricity charges made through net banking for ₹ 900 were wrongly entered in the cash column of the cash book instead of the bank column.
Solution:
Bank reconciliation statement as of 31st March 2018.

Question 16.
Prepare bank reconciliation statement from the following data.

Solution:
Prepare bank reconciliation statement.

Question 17.
Prepare bank reconciliation statement as of 31st March 2017 from the following extracts of cash book and bank statement.
Cashbook (Bank column only)

Bank statement

Solution:
Bank reconciliation statement as on 31st March, 2017

Question 18.
A trader received his bank statement on 31st December 2017 which showed an overdraft balance of ₹ 12,000. On the same day, his cash book showed a debit balance of ₹ 2,000.
Analyze the following transactions. Choose the possible causes and prepare a bank reconciliation statement to show the causes of differences.
a) Cheque deposited for ₹ 2,000 on 21st December 2017. The bank credited the same on 26th December 2017.

b) Cheque issued for payment on 26th December 2017 amounting to ₹ 2,500, not yet presented until 31st, December 2017.

c) Bank charges amounting to ₹ 200 not yet entered in the cash book.

d) Online payment for ₹ 1,500 entered twice in the cash book.

e) Cheque deposited amounting to ₹ 1,000, but omitted in the cash book. The same cheque was dishonoured by bank, but not yet entered in the cash book.

f) Cheque deposited, not yet credited by bank amounting to ₹ 17,800.
Solution:
Bank reconciliation statement as of 31st December 2017.

### 11th Accountancy Guide Bank Reconciliation Statement Additional Important Questions and Answers

Question 1.
A Bank Reconciliation Statement is prepared with the help of ________.
a) Bank statement and bank column of the cash book
b) Journal
c) Ledger
d) None of the above
a) Bank statement and bank column of the cash book

Question 2.
Bank reconciliation statement is ______
a) Part of bank statement
b) Part of the cash book
c) A separate statement
d) A sub-division of journal
a) Part of bank statement

Question 3.
Uncollected cheques are also known as ______.
a) Outstanding cheques
b) Uncleared cheques
c) Outstation cheques
d) Both a & c
d) Both a & c

Question 4.
When a cheque is not paid by the bank it is called as ______.
a) Honoured
b) Endorsed
c) Dishonoured
d) None of these
c) Dishonoured

Question 5.
A bank reconciliation statement is prepared by ______.
a) Banker
c) Auditors₹
d) None of the above

Question 6.
The cheque which is deposited into the bank but not cleared at the end of a particular year is called ______.
a) Uncredited cheque
b) Unpresented cheque
c) Omitted cheque
d) Dishonoured cheque
b) Unpresented cheque

Question 7.
In cash book bank charges recorded in the ______.
a) Credit side
b) Debit Side
c) Both a & b
d) None of the above
a) Credit side

Question 8.
An amount of Rs,2G0Q is debited twice in the bank statement. What will the reflect when overdraft as per the cash book is the starting point ______.
a) ₹ 2000 will be deducted
b) ₹ 2000 will be added
c) ₹ 4000 will be deducted
d) ₹ 4000 will be deducted
b) ₹ 2000 will be added

Question 9.
If any amount is directly deposited into the bank then ______.
a) Cashbook will show less balance & bank book will show more
b) Cashbook will show more balance & bank book will show less
c) Cashbook will show double balance
d) Bank book will show double
a) Cashbook will show less balance & bank book will show more

Question 10.
Which of the following error results in an unadjusted cash book balance?
a) Outstanding cheques
b) Unpresented Cheques
c) deposit in Transit
d) Omission of Bank charges
d) Omission of Bank charges

Question 11.
Credit balance in the bank column of the cash book means ______.
a) Credit balance as per bank statement
b) Debit balance as per bank statement
c) Overdraft as per cash book
d) None of the above
b) Debit balance as per bank statement

Question 12.
When balance as per Cash Book is the starting point, to ascertain balance as per bank statement interest allowed by Bank is ______.
a) Subtracted
d) None of the above

Question 13.
When balance as per Cash Book is the starting point, to ascertain the balance as per bank statement interest charged by Bank is:
b) subtracted
d) None of the above
b) subtracted

Question 14.
When the balance as per Cash Book is the starting point to ascertain balance as per bank statement, direct deposits by customers are:
b) subtracted
d) None of the above

Question 15.
When the balance as per Cash Book is the starting point to ascertain balance as per bank statement, direct payments by the bank are:
b) subtracted
d) None of the above
b) subtracted

Question 16.
______ is not possible to have unfavourable cash balance in the cash book.
a) Bank statement
b) Bank overdraft
c) Cash overdraft
d) Cashbook
b) Bank overdraft

Question 17.
Bank overdraft is available only to the ______ holders.
a) Saving Account
b) Fixed Account
c) Joint Account
d) Current Account
d) Current Account

Question 18.
______ is simply a copy of the customer’s account in the books of a bank.
a) Cashbook
b) Bank statement
c) Bank Account
d) None of these
b) Bank statement

Question 19.
A bank statement is a copy of ______.
a) the cash column of a customer’s cash book
b) the bank column of a customer’s cash book
c) the customer’s account in the bank’s ledger
d) None of these
c) the customer’s account in the bank’s ledger

Question 20.
Debit balance in the Cash Book means ______.
a) overdraft as per bank statement
b) credit balance as per bank statement
c) overdraft as per Cash Book
d) None of these
b) credit balance as per bank statement

Question 1.
Differences between bank column of cash book and bank statement.
Bank column of cash book:

• It is prepared by a business concern.
• Cash deposits are entered on the debit side.
• Cash withdrawals are entered on the credit side.
• Cheque deposits are debited on the day of the deposit.
• Cheques issued are credited on the day of the issue of the cheque.
• Collections and payments as per standing instructions of the business are entered only after checking with the bank statement.
• It is balanced at the end of a specific period.

Bank statement:

• It is prepared by a bank (banker).
• Cash deposits are entered in the credit column.
• Cash withdrawals are entered in the debit column.
• Cheque deposits are credited only at the time of realization of the cheque.
• Cheques issued by customers are debited by the bank on the date on which the payment is made.
• Collections and payments as per standing instructions of the business are entered in the banker’s book on the date of realization of payment.
• It is balanced after each transaction.

Question 2.
What are the items recorded on the debit side of the bank column of the cash book?

1. Cheques deposited but not credited.
2. Credits in the passbook only.
• Interest credited in the bank statement
• Dividend and other income
• • Direct deposit by a party
3. Any error in the cash book/ bank statement has the effect of increasing the balance as per the bank statement.

Question 3.
What are the items recorded on the credit side of the bank column of the Cash Book?

1. Cheques deposited but not credited
2. Cheques dishonoured but not entered in the cash book
3. Debits in bank statement only
• Interest debited
• Insurance premium, loan instalment, etc., paid as per standing instructions
• Direct payment by banker
4. Any error in cash book/ bank statement which has the effect of decreasing the balance as per bank statement

Question 4.
What is meant by the term “Cheques deposited into bank but not yet credited?”
When the cheques are deposited into bank, the amount is debited in the cash book on the same day. But, these may not be shown in the bank pass book on the same day because these will be entered in the bank statement only after the collection of the cheques.

For example, the balances as per cash book and bank statement are ₹ 20,000 for X & Co. X & Co. receives a cheque on 25th March 2016, from ABC Limited for ₹ 5,000. On the same day, X & Co, debits its cash book with ₹ 5,000.

But bank credits X & Co’s account only when the cheque is collected from ABC Limited’s bank. This shows that is a time gap between depositing the cheque by the customer (X & Co) and collection of cheque by the bank.

Question 5.
What will be the effect of Interest and dividends collected by the bank?
The bank may collect dividends on its customer’s investment in shares and also interest on any investment. The entry for this will be made in the bank statement on the date of collection. But the entry is made in the cash book only when the bank statement is received by the customer. Till then, the cash book shows less balance than the bank statement.

Question 6.
What will be the effect of Dishonour of cheques and bills?
When the cheque is received from outside parties, it is deposited with the bank and debited in the cash book. If the cheque is dishonoured, the bank cannot collect the amount of such cheque from outside parties’ bank. It is not credited in the bank statement. As a result of this, the two records would differ.

While discounting the bills receivables, in the cash book it is entered in the debit side and in the bank statement it is credited. When the bill is presented by the bank to the drawee of the bill and the payment is not received, the bank debits the same to cancel the credit.

But, credit is made in the cash book only when the customer gets the entries made in the bank statement is received. The bank may also charge some amount for such dishonour.

Question 7.
What will be the effect of Amount paid by parties directly into the bank?
Sometimes, debtors or the customers of the business may directly deposit the money into bank account of the business. It may be done by directly visiting the branch of the bank by paying cash (including NEFT, RTGS) or swiping debit or credit or business card or depositing the money in cash deposit machine or transfer through online banking facility.

This will be credited in the banker’s book. But the entry is made in the cash book only when the bank statement is received by the customer. Until then, the cash book shows less balance than bank statement.

Question 8.
What will be the effect of Amount paid directly by the bank to others?
Sometimes the bank may be instructed to make payments such as, insurance premium, instalment of loan, etc., as an agent of the customer on behalf of its customer. In all such cases, debit is made in bank statement. But, the entry is made in the cash book only when the bank statement is received by the customer. Till then, the cash book shows more balance than bank statement.

Question 9.
What will be the effect of Bills collected by the bank on behalf of its customers?
When goods are sold by the business, the documents may be sent through the bank. When the bank collects the amount, it is credited in bank records. But, the entry is made in the cash book only when the bank statement is received by the business. Till then, the bank statement shows more balance than cash book.

Question 10.
Explain the differences arising due to errors in recording the entries.
Errors committed in recording the transactions by the business in the cash book:
Sometimes, errors may be committed in the cash book. For example, omission or wrong recording of transaction relating to cheques deposited or issued, wrong balancing, etc. In these cases, obviously, there will be differences between bank balance as per bank statement and bank balance as per cash book.

Errors committed in recording the transactions by the bank:
Sometimes errors may be committed in the banker’s book. For example, omission or wrong recording of transaction relating to cheques deposited and wrong balancing. In these cases, obviously, there will be differences between bank balance as per bank statement and bank balance as per cash book.

Question 11.
What is the need for bank reconciliation statement?

1. To identify the reasons for the difference between the bank balance as per the cash book and bank balance as per bank statement.
2. To identify the delay in the clearance of cheques.
3. To ascertain the correct balance of bank column of cash book.
4. To discourage the accountants of the business as well as bank from misusing funds.

Question 1.
Prepare bank reconciliation statement of Mr. Bala as on 31.03.2013.
a) Balance as per cash book ₹ 15,000
b) Cheques deposited but not cleared ₹ 1,000
c) Cheques issued but not yet present for payments ₹ 1500 d. Interest allowed by bank ₹ 200
Solution:
Prepare bank reconciliation statement

Question 2.
Prepare bank Reconciliation statement to find out balance as per bank statement on 31st March 2018,
1. Cheques deposited but not yet collected by the bank ₹ 1,000
2. Cheques issued but not yet presented for payment ₹ 2,000
3. Bank Interest Charged ₹ 200
4. Rent paid by bank as per standing Instruction ₹ 400
5. Cash book balances ₹ 600
Solution:
Prepare bank reconciliation statement

Question 3.
Form the following particulars of Ashok and company; prepare a bank reconciliation statement as on 31st March 2018.
a) Credit balance as per cash book ₹ 10,000
b) Cheques issued but not yet presented for payment ₹ 10,000
c) Cheques Deposited but not credited ₹ 9000
d) Rent collected by the bank as per standing Instruction ₹ 500
Solution:
Bank Reconciliation statement Mr. Ashok as on 31st March 2018

Question 4.
From the following information, Prepare bank Reconciliationstatement of Mr. Mohan as on 31st Dec. 2017 to find out the balance as per bank statement.
i) Overdraft as per cash book – 20000
ii) Cheques deposited but not yet credited – 10000
iii) Amount wrongly deposited by bank – 600
iv) Interest on overdraft debited by bank – 2000
v) Cheque issued but not yet present for payment – 2000
vi) Payment received from the customer directly by the bank – 1000
Solution:
Bank Reconciliation statement Mr. Mohan as on 31st Dec. 2018.

Question 5.
Prepare bank Reconciliation statement as on 31st December 2017. From the following in information.
a) Balance as perbank statement (pass book) is ₹ 50,000
b) Cheques deposited into bank amount into ₹ 7,000 were not yet collected.
c) Bank charges of ₹ 600 have not been entered in the cash book.
d) Cheques issued amount to ₹ 18,000 have not been presented by for payment.
Solution:
Bank Reconciliation statement as on 31st Dec. 2017.

Question 6.
From the following information, prepare bank Reconciliationstatement as of Mr. Pugazh as on 31st Dec. 2017.
i) Credit balance as per bank statement ₹ 12,000
ii) Cheques deposited on 28th December 2017 but not yet credited ₹ 4000
iii) Cheques issued for 20000 on 20th December 2017 but not yet presented for payment 6000.
iv) Interest on debentures directly in cash book ₹ 8000
v) Insurance premium on building directly paid by the bank ₹ 2000
vi) Amount wrongly credit by bank ₹ 1,000
Solution:
Bank Reconciliation statement of Mr. Pugazh as on 31st Dec. 2017.

Question 7.
From the following data, as certain the cash book balance as on 31st Dec. 2017.
1) Overdraft balance as per bank statement ₹ 13,000
2) Cheques deposited into the bank but not yet credited ₹ 21,000
3) Wrongly credit by the bank ₹ 1,000
4) Cheques issued, but not yet presented for payment₹ 6,000
5) Bank charges debited by bank ₹ 360
6) Insurance Premium on building directly paid by bank ₹ 200
Solution:
Bank Reconciliation statement on 31st Dec. 2017.

Question 8.
Prepare bank Reconciliation statement as on 31st Dec. 2017, From the following balance of cash book, and bank statement.
Cash book (Bank column)

Bank Statement

Solution:
Bank Reconciliation Statement on 31st Dec, 2017.

Question 9.
On 31st March 2017, the pass book of Mr, A showed a credit balance of Rs.92,500. A comparison of pass book and cash book revealed the following:

Solution:
Bank Reconciliation Statement of Mr. A as on 31st March 2017

Question 10.
The bank overdraft of Rajini on 31.12.2017 as per cash book is 90,000. From the following particulars, prepare bank reconciliation statement:

Solution:
Bank Reconciliation Statement as on 31,12.2017 3,000

Question 11.
Prepare a bank reconciliation statement from the following data as on 31.12.2017.
a) Balance as per cash book ₹ 1,25,500
b) Cheques issued but not presented for payment ₹ 9,000
c) Cheques deposited in bank but not collected ₹ 12,000
d) Bank paid insurance premium ₹ 5,000
e) Direct deposit by a customer ₹ 8,000
f) Interest on investment collected by bank ₹ 2,000
g) Bank charges ₹ 1,000
Solution:
Bank Reconciliation Statement as on 31.12.2017

Question 12.
The pass book of X with his bank shows a debit balance of Rs. 500 on 31.10.2017. On comparison of the pass book with the cash book, it is observed that:
i. Cheques issued by X in October 2017 amounted to ₹ 4, 535 of which cheques amounting to ₹ 3,535 were paid by the bank by 31st October 2017.

ii. X deposited cheques amounting to ₹ 5, 000 on 31st October 2017. These cheques were realised by the bank on 1st November, 2017.

iii. Y a customer of X had directly deposited a sum of ₹ 3, 000 on 24th October 2017 to the credit of X account with the bank. X recorded this receipt on 4th November, 2017.

iv. The bank had debited X’s account with ₹ 1, 520 on 31.1.2017 on account of a dishonoured bill. No entry for the same has been made in the account books.

v. On 31.10.1995 X’s account was credited with ₹ 130 being dividend collected by the bank. On the same day, his account was debited with ₹ 10 being bank charges. Both these entries were recorded by X only on 5th November, 2017.
Prepare the Bank Reconciliation Statement as at 31.10.2017.
Solution:
Bank Reconciliation Statement as on 31.10.2017

Question 13.
From the following particulars ascertain the bank balance as would appear in the pass‘book as on 31st December, 2016.
i. The bank overdraft (Credit balance) as per cash book on 31st December 2016 was ₹ 60,000
ii. Interest on overdraft, six months ending 31st December, amounting to ₹ 2,000 is debited in the pass book.
i. Bank charges for the above period also debited in the pass book which amounted to ₹ 500.
ii. Cheques issued but not presented for payment before 31st December amounted to ₹ 15,000.
iii. Cheques paid into the bank, but not cleared and credited before 31st December were ₹ 25,000.
iv. Interest on government securities collected by the bank and credited in the pass book amounted to ₹ 18,000.
Solution:
Bank Reconciliation Statement as on 31st March 2016

Question 14.
From the following information available from the books and records of X & Co., prepare BRS:

Solution:
Bank Reconciliation Statement for Bank A/c No. I

Bank Reconciliation Statement for Bank A/c No. II

Question 15.
From the following particulars ascertain the bank balance that would appear in the cash book of Son 31.12.2016.
i) The bank overdraft as per pass book on 31.12.2016 ₹ 6,340
ii) Interest on on for the year ending 31.12.2016 ₹ 160 is debited in the pass book
iii) Bank charges of ₹130 for the above period are also debited in the pass book.
iv) Cheques issued but not cashed prior to 31.12.2016 amounted to ₹ 11,168
v) Cheques paid into bank but not cleared before 31.12.2016 were ₹ 2170
vi) Interest on investments collected by the bankers and credited in the pass book, ₹ 1,200
Solution:
Bank Reconciliation Statement as on 31st December 2016

## Samacheer Kalvi 12th Business Maths Guide Chapter 10 Operations Research Miscellaneous Problems

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 10 Operations Research Miscellaneous Problems Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems

Question 1.
The following table summarizes the supply, demand and cost information for four factors S1, S2, S3, S4 Shipping goods to three warehouses D1, D2, D3.

Find an initial solution by using north west corner rule. What is the total cost for this solution?
Solution:
The given transportation table is

Here total supply = 5 + 8 + 7+14 = 34
Total amount =7 + 9 + 18 = 34
(i.e) Total supply = Total demand.
∴ The given problem is balanced transformation problem.
We can find an initial basic feasible solution to the given problem.
First allocation:

Transportation schedule:
S1 → D1; S2 → D1; S2 → D2;
S3 → D2; S3 → D3; S4 → D3
The transportation cost:
= (5 × 2) + (2 × 3) + (6 × 3) + (3 × 4) + (4 × 7) + (14 × 2)
= 10 + 6 + 18 + 12 + 28 + 28
= 102

Question 2.
Consider the following transportation problem

Determine an initial basic feasible solution using (a) Least cost method (b) Vogel’s approximation method.
Solution:
(a) Least cost method
Given transportation table is

Total Availability = Total Requirement = 100
∴ The given problem is balanced transformation problem.
Hence there exists a feasible solution to the given problem
First allocation:

Here
x12 = 10; x13 = 20; x21 = 30;
x22 = 10; x24 = 10; x32 = 20
Transportation Scheme:
O1 → D2; O1 → D3; O2 → D1;
O2 → D2; O2 → D4; O3 → D2
The transportation cost:
=(10 × 8) + (20 × 3) + (30 × 4) + (10 × 5) + (10 × 4) + (20 × 2)
= 80 + 60 + 120 + 50 + 40 + 40
= 390

(ii) Vogel’s approximation method:
Here Σai = Σbj = 100
∴ The given problem is balanced transformation problem.
Hence there exists a feasible solution to the given problem
First allocation:

Here
x11 = 30; x13 = 20; x22 = 20;
x24 = 10; x32 = 20
Transportation Scheme:
O1 → D1; O2 → D3; O2 → D2;
O2 → D4; O3 → D2;
Total transportation cost:
= (30 × 5) + (20 × 3) + (20 × 5) + (10 × 4) + (20 × 2)
= 150 + 60 + 100 + 40 + 40
= 390

Question 3.
Determine an initital basic feasible solution to the following transportation problem by using (i) North West Corner rule (ii) least cost method.

Solution:
(i) North West Corner rule:
The given transportation table is

Total supply = 25 + 35 + 40 = 100
Total Requirement = 30 + 25 + 45 + 100
(i.e) Total supply = Total requirement.
∴ The given problem is balanced transformation problem.
Hence there exists a feasible solution to the given problem
First allocation:

Transformation schedule:
S1 → D1; S2 → D1; S2 → D2;
S3 → D3; S3 → D3;
The transformation cost:
= (25 × 9) + (5 × 6) + (25 × 8) + (5 × 4) + (40 × 9)
= 225 + 30 + 200 + 20 + 360 = 835

(ii) Least cost method
The given transportation table is

(i.e) Total supply = Total requirement = 100
∴ The given problem is balanced transformation problem.
Hence there exists a feasible solution to the given problem
First allocation:

Transportation schedule:
S1 → D3; S2 → D3; S3 → D1
S3 → D2;
The transportation cost:
= (10 × 5) + (35 × 4) + (15 × 7) + (25 × 6) + (40 × 9)
= 50 + 140 + 105 + 150 + 360 = 805

Question 4.
Explain Vogel’s approximation method by obtaining initial basic feasible solution of the following transportation problem.

Solution:
The given transportation table is

Here Σai = Σbj =17
(i.e) Total supply = Total Demand
∴ The given problem is balanced transformation problem.
Hence there exists a feasible solution to the given problem
First allocation:

Here
x11 = 1; x12 = 5; x24 = 1;
x31 = 6; x33 = 15;
Transportation schedule:
O1 → D1; O1 → D2; O2 → D4
O3 → D1; O3 → D3
The transportation cost:
=(1 × 2) + (5 × 3) + (1 × 1) + (6 × 5) + (3 × 15) + (1 × 9)
= 2 + 15 + 1 + 30 + 45 + 9
= 102

Question 5.
A car hire company has one car at each of five depots a,b,c,d and e. A customer in each of the fine towers A, B, C, D and E requires a car. The distance (in miles) between the depots (origins) and the towers (destinations) where the customers are given the following distance matrix.

How should the cars be assigned to the customers so as to minimize the distance travelled?
Solution:
Here the number of rows and columns are equal.
∴ The given assignment problem is balanced.
Step 1.
Select the smallest element in each row and subtract this from all the elements in its row.

Step 2.
Select the smallest element in each column and subtract this from all the elements in its column.

Step 3. (Assignment)
Examine the rows with exactly one zero, mark the zero by □ mark other zeros, in its column by X

Step 4.
Now Examine the rows with exactly one zero, mark the zero by □ mark other zeros, in its column by X

Step 5.
Cover all the zeros of table 4 with three lives. Since three assignments were made please note that check [✓] Row C and E which have no assignment.

Step 6.
Develop the new revised tableau. Examine those elements that are not covered by a line in Table 5. Take the smallest element in each row and subtract from the uncovered cells, depots

Step 7.
Go to step 3 and repeat the procedure until you arrive at an optimal assignments.
depots

Step 8.
Determine an assignment

Here all the five assignments have been made. The optimal assignment schedule and total distance is

∴ The optimum Distance (minimum) 575 kms

Question 6.
A natural truck – rental service has a surplus of one truck in each of the cities 1, 2, 3, 4, 5 and 6 and a deficit of one truck in each of the cities 7, 8, 9, 10, 11 and 12. The distance (in kilometers) between the cities with a surplus and the cities with

How should the truck be dispersed so as to minimize the total distance travelled?
Solution:
Here the number of rows and columns are equal.
∴ The given assignment problem is balanced.
Step 1.
Select the smallest element in each row and subtract this from all the elements in its row.

Step 2.
Select the smallest element in each column and subtract this from all the elements in its column.

Step 3.
Examine the rows with exactly one zero, mark the zero by □ mark other zeros, in its column by X

Step 4.
Examine the Columns with exactly one zero. If there is exactly one zero, mark that zero by □ mark other zeros in its rows by X

Step 5.
Cover all the zeros of table 4 with five lines. Since three assignments were made

Step 6.
Develop the new revised tableau. Examine those elements that are not covered by a line in Table 5. Take the smallest element. This is l(one) in our case. By subtracting 1 from the uncovered cells.

Step 7.
Go to step 3 and repeat the procedure until you arrive at an optimal assignments.

Step 8.
Determine an assignment

Here all the six assignments have been made. The optimal assignment schedule and total distance is

∴The optimum Distance (minimum) = 125 kms

Question 7.
A person wants to invest in one of three alternative investment plans: Stock, Bonds and Debentures. It is assumed that the person wishes to invest all of the funds in a plan. The pay – off matrix based on three potential economic conditions is given in the following table

Solution:

(i) Maximin
Max (3000, 1000, 6000) = 6000. Since the maximum pay of is 6000, the alternative ‘Debentures’, is selected.

(ii) Minimax
Min (10000, 8000, 6000) = 6000, Since the minimum pay-off is 6000. the alternative ‘Debentures’ is selected.

## Samacheer Kalvi 12th Business Maths Guide Chapter 10 Operations Research Ex 10.4

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 10 Operations Research Ex 10.4 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Ex 10.4

Question 1.
The transportation problem is said to be unbalanced if
(a) Total supply ≠ Total demand
(b) Total supply = Total demand
(c) m = n
(d) m + n – 1
Solution:
(a) Total supply ≠ Total demand

Question 2.
In a non – degenerate solution number of allocation is
(a) Equal to m + n – 1
(b) Equal to m + n + 1
(c) Not equal to m + n – 1
(d) Not equal to m + n + 1
Solution:
(a) Equal to m + n – 1

Question 3.
In a degenerate solution number of allocations is
(a) Equal to m + n – 1
(b) Not equal to m + n – 1
(c) Less then m + n – 1
(d) Greater then m + n – 1
Solution:
(c) Less then m + n – 1

Question 4.
The Penalty in VAM represents difference between the first
(a) Two largest costs
(b) Largest and Smallest costs
(c) smallest two costs
(d) None of these
Solution:
(c) smallest two costs

Question 5.
Number of basic allocation in any row or column in an assignment problem can be
(a) Exactly one
(b) At least one
(c) At most one
(d) None of these
Solution:
(a) Exactly one

Question 6.
North – West Corner refers to
(a) Top left corner
(b) Top right corner
(c) Bottom right corner
(d) Bottom left corner
Solution:
(a) Top left corner

Question 7.
Solution for transportation problem using method is nearer to an optimal
(a) NWCM
(b) LCM
(c) VAM
(d) Row Minima
Solution:
(c) VAM

Question 8.
In an assignment problem the value of iedsion variable xij is
(a) 1
(b) 0
(c) 1 or 0
(d) none of them
Solution:
(c) 1 or 0

Question 9.
If number of sources is not equal to number of destinations, the assignment problem is called
(a) balanced
(b) unsymmetric
(c) symmetric
(d) unbalanced
Solution:
(d) unbalanced

Question 10.
The purpose of a dummy row or column in an assignment problem is to
(a) prevent a solution from becoming degenerate
(b) balance between total activities and total resources
(c) provide a means of representing a dummy problem
(d) None of the above
Solution:
(b) balance between total activities and total resources

Question 11.
The solution for an assignment problem is optimal if
(a) each row and each column has no assignment
(b) each row and each column has atleast one assignment
(c) each row and each column has atmost one assignment
(d) each row and each column has exactly one assignment
Solution:
(d) each row and each column has exactly one assignment

Question 12.
In an assignment problem involving four workers and three jobs, total number of assignments possible are
(a) 4
(b) 3
(c) 7
(d) 12
Solution:
(b) 3

Question 13.
Decision theory is concerned with
(a) analysis of information that is available
(b) decision making under certainty
(c) selecting optimal decisions in sequential problem
(d) All of the above
Solution:
(d) All of the above

Question 14.
A type of decision – making environment is
(a) certainty
(b) uncertainty
(c) risk
(d) all of the above
Solution:
(d) all of the above