Category: CBSE

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5

• Class 7 Maths The Triangle and Its Properties Exercise 6.1
• Class 7 Maths The Triangle and Its Properties Exercise 6.2
• Class 7 Maths The Triangle and Its Properties Exercise 6.3
• Class 7 Maths The Triangle and Its Properties Exercise 6.4
• Class 7 Maths The Triangle and Its Properties Exercise 6.5

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Exercise 6.5
Ex 6.5 Class 7 Maths Question 1.
PQR is a triangle, right angled at P. If PQ = 10 cm and PR = 24 cm, find QR.

Solution:
In right angled triangle PQR, we have
QR² = PQ² + PR² From Pythagoras property)
= (10)² + (24)²
= 100 + 576 = 676
∴ QR = \(\sqrt{676}\) = 26 cm
The, the required length of QR = 26 cm.

Ex 6.5 Class 7 Maths Question 2.
ABC is a triangle, right angled at C. If AB = 25 cm and AC = 7 cm, find BC.

Solution:
In right angled∆ABC, we have
BC² + (7)² = (25)² (By Pythagoras property)
⇒ BC² + 49 = 625
⇒ BC² = 625 – 49
⇒ BC² = 576
∴ BC = \(\sqrt{576}\) = 24 cm
Thus, the required length of BC = 24 cm.

Ex 6.5 Class 7 Maths Question 3.
A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.

Solution:
Here, the ladder forms a right angled triangle.
∴ a² + (12)² = (15)² (By Pythagoras property)
⇒ a²+ 144 = 225
⇒ a2 = 225 – 144
⇒ a² = 81
∴ a = \(\sqrt{81}\) = 9 m
Thus, the distance of the foot from the ladder = 9m

Ex 6.5 Class 7 Maths Question 4.
Which of the following can be the sides of a right triangle?
(i) 2.5 cm, 6.5 cm, 6 cm.
(ii) 2 cm, 2 cm, 5 cm.
(iii) 1.5 cm, 2 cm, 2.5 cm
Solution:
(i) Given sides are 2.5 cm, 6.5 cm, 6 cm.
Square of the longer side = (6.5)² = 42.25 cm.
Sum of the square of other two sides
= (2.5)² + (6)² = 6.25 + 36
= 42.25 cm.
Since, the square of the longer side in a triangle is equal to the sum of the squares of other two sides.
∴ The given sides form a right triangle.

(ii) Given sides are 2 cm, 2 cm, 5 cm .
Square of the longer side = (5)² = 25 cm Sum of the square of other two sides
= (2)² + (2)² =4 + 4 = 8 cm
Since 25 cm ≠ 8 cm
∴ The given sides do not form a right triangle.

(iii) Given sides are 1.5 cm, 2 cm, 2.5 cm
Square of the longer side = (2.5)² = 6.25 cm Sum of the square of other two sides
= (1.5)² + (2)² = 2.25 + 4
Since 6.25 cm = 6.25 cm = 6.25 cm
Since the square of longer side in a triangle is equal to the sum of square of other two sides.
∴ The given sides form a right triangle.

Ex 6.5 Class 7 Maths Question 5.
A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree . Find the original height of the tree.
Solution:
Let AB be the original height of the tree and broken at C touching the ground at D such that

AC = 5 m
and AD = 12 m
In right triangle ∆CAD,
AD² + AC² = CD² (By Pythagoras property)
⇒ (12)² + (5)² = CD²
⇒ 144 + 25 = CD²
⇒ 169 = CD²
∴ CD = \(\sqrt{169}\) = 13 m
But CD = BC
AC + CB = AB
5 m + 13 m = AB
∴ AB = 18 m .
Thus, the original height of the tree = 18 m.

Ex 6.5 Class 7 Maths Question 6.
Angles Q and R of a APQR are 25° and 65°. Write which of the following is true.
(i) PQ² + QR² = RP²
(ii) PQ² + RP² = QR²
(iii) RP² + QR² = PQ²

Solution:
We know that
∠P + ∠Q + ∠R = 180° (Angle sum property)
∠P + 25° + 65° = 180°
∠P + 90° = 180°
∠P = 180° – 90° – 90°
∆PQR is a right triangle, right angled at P
(i) Not True
∴ PQ² + QR² ≠ RP² (By Pythagoras property)
(ii) True
∴ PQ² + RP² = QP² (By Pythagoras property)
(iii) Not True
∴ RP² + QR² ≠ PQ² (By Pythagoras property)

Ex 6.5 Class 7 Maths Question 7.
Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.

Solution:
Given: Length AB = 40 cm
Diagonal AC = 41 cm
In right triangle ABC, we have
AB² + BC²= AC² (By Pythagoras property)
⇒ (40)² + BC² = (41)²
⇒ 1600 + BC² = 1681
⇒ BC² = 1681 – 1600
⇒ BC² = 81
∴ BC = \(\sqrt{81}\) = 9 cm
∴ AB = DC = 40 cm and BC = AD = 9 cm (Property of rectangle)
∴ The required perimeter
= AB + BC + CD + DA
= (40 + 9 + 40 + 9) cm
= 98 cm

Ex 6.5 Class 7 Maths Question 8.
The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.

Solution:
Let ABCD be a rhombus whose diagonals intersect each other at O such that AC = 16 cm and BD = 30 cm
Since, the diagonals of a rhombus bisect each other at 90°.
∴ OA = OC = 8 cm and OB = OD = 15 cm
In right ∆OAB,
AB² = OA² + OB² (By Pythagoras property)
= (8)²+ (15)² = 64 + 225
= 289
∴ AB =\(\sqrt{289}\)= 17 cm
Since AB = BC = CD = DA (Property of rhombus)
∴ Required perimeter of rhombus
= 4 × side = 4 × 17 = 68 cm.

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Chemistry Qualitative Analysis

Chemistry Lab ManualNCERT Solutions Class 12 Chemistry Sample Papers

Analytical chemistry deals with qualitative and quantitative analysis of the substances. In qualitative analysis, the given compound is analyzed for the radicals, i.e., cation and the anion, that it contains. Physical procedures like noting the colour, smell or taste of the substance have very limited scope because of the corrosive, poisonous nature of the chemical compounds. Therefore, what one has to resort to is the chemical analysis of the substance that has to be carried out along with the physical examination of the compound under consideration.
The common procedure for testing any unknown sample is to make its solution and then test this solution for the ions present in it. There are separate procedures for detecting cations and anions, therefore qualitative analysis is studied under cation analysis and anion analysis. The systematic procedure for qualitative analysis of an inorganic salt involves the following steps:
(а) Preliminary tests

1. Physical appearance (colour and smell).
2. Dry heating test.
3. Charcoal cavity test.
4. Charcoal cavity and cobalt nitrate test.
5. Flame test.
6. Borax bead test.
7. Dilute acid test.
8. Potassium permanganate test.
9. Concentrated sulphuric acid test.
10. Tests for sulphate, phosphate and borate.

(b) Wet tests for acid radical.
(c) Wet tests (group analysis) for basic radical.

Physical examination of the salt
The physical examination of the unknown salt involves the study of colour, smell and density. The test is not much reliable, but is certainly helpful in identifying some coloured cations. Characteristic smell helps to identify some ions such as ammonium, acetate and sulphide. (See Table 12.1 on next page)
Note:

1. If you have touched any salt, wash your hands at once. It may be corrosive to skin.
2. Never taste any salt, it may be poisonous. Salts of arsenic and mercury are highly poisonous.
3. Salts like sodium sulphide, sodium nitrite, potassium nitrite, develop a yellow colour.
   

Dry heating test 
This test is performed by heating a small amount of salt in a dry test tube. Quite valuable information can be gathered by carefully performing and noting the observations here. On heating, some salts undergo decomposition, thus, evolving the gases or may undergo characteristic changes in the colour of residue. These observations are tabulated in Table 12.2 along with the inferences that you can draw.
                                                       Table 12.2. Dry Heating Test




Note:

1. Use a perfectly dry test-tube for performing this test. While drying a test-tube, keep it in slanting position with its mouth slightly downwards so that the drops of water which condense on the upper cooler parts, do not fall back on the hot bottom, as this may break the tube.
2. For testing a gas, a filter paper strip dipped in the appropriate reagent is brought near the mouth of the test tube or alternatively the reagent is taken in a gas-detector and the gas is passed through it.
3. Do not heat the tube strongly at one point as it may break.
   

Charcoal cavity test
This test is based on the fact that metallic carbonates when heated in a charcoal cavity decom¬pose to give corresponding oxides. The oxides appear as coloured incrustation or residue in the cavity. In certain cases, the oxides formed partially undergo reduction to the metallic state producing metallic beads or scales.
Examples:


Procedure
While performing charcoal cavity test, make a small cavity on a charcoal block with the help of borer as shown in Fig. 12.2. Mix small amount of salt with double its quantity of sodium carbonate. Place it in the cavity made on the block of charcoal. Moisten with a drop of water and direct the reducing flame of the bunsen burner on the cavity by means of a mouth blowpipe as shown in Fig. 12.3. Heat strongly for sometime and draw inference according to the Table 12.3.



To obtain a reducing flame with the help of a mouth blow pipe, make the bunsen burner flame luminous by closing the air holes of the burner. Keep the nozzle of the blow pipe just outside the flame (Fig. 12.4) and blow gently on to the cavity.

Cobalt nitrate test
This test is applied to those salts which leave white residue in charcoal cavity test. The test is based on the fact that cobalt nitrate decomposes on heating to give cobalt oxide, CoO. This combines with the metallic-oxides, present as white residue in the charcoal cavity forming coloured compounds. For example, when a magnesium salt undergoes charcoal cavity test, a white residue of MgO is left behind. This on treatment with cobalt nitrate and subsequent heating forms a double salt of the formula MgO.CoO which is pink in colour. In addition to metallic oxides, phosphates and borates also react with cobalt oxide to form CO₃(PO₄)₂ and CO₃(BO₃)₂ which are blue in colour.
Some of the reactions involved are given below:

Procedure
Put one or two drops of cobalt nitrate solution on the white residue left after charcoal cavity test. Heat for one or two minutes by means of a blow pipe in oxidising flame. Observe the colour of the residue and draw inferences
Note:

1. Perform this test only if the residue in the charcoal cavity test is white.
2. Do not put more than 2 drops of cobalt nitrate on the white residue. Excess cobalt nitrate may decompose to give cobalt oxide which is black in colour.
3. Use dilute solution of cobalt nitrate.

Flame test
Certain salts on reacting with cone. HCl from their chlorides, that are volatile in non-luminous flame. Their vapours impart characteristic colour to the flame. This colour cfan give reliable information of the presence of certain basic radicals.
For proceeding to this test, the paste of the mixture with cone. HCl is introduced into the flame with the help of platinum wire (Fig. 12.5).

Procedure
Clean the platinum wire by dipping it in some cone. HCl taken on a watch glass and then heating strongly in the flame. This process is repeated till the wire imparts no colour to the flame. Now prepare a paste of the mixture with cone. HCl on a deem watch glass. Place small amount of this paste on platinum wire loop and introduce it into the flame. Note the colour imparted to the flame.

Borax bead test
This test is performed only for coloured salts.
Borax, Na₂B₄O₇.10H₂O, on heating gets fused and loses water of crystallisation. It swells up into a fluffy white porous mass which then melts into a colourless liquid which later forms , a clear transparent glassy bead consisting of boric anhydride and sodium metaborate.

Procedure
Borax, Na₂B₄O₇.10H₂O is heated in the loop of platinum wire, it swells and forms transparent colourless glassy bead. When this hot bead is touched with small amount of coloured salt and is heated again, it acquires a characteristic colour. The colour of bead gives indication of the type of the cation present. The colour of the bead is noted separately in oxidising and in reducing flame (Fig. 12.6).

                                                                           Table 12.6. Borax Bead Test

To remove the head from platinum wire, heat the head to redness. Tap the rod with finger stroke, till the bead jumps off (Fig. 12.7).

The identification of the acid radicals is first done on the basis of preliminary tests. Dry heating test is one of the preliminary tests performed earlier which may give some important information about the acid radical present. The other preliminary tests are based upon the fact that:


Take a small quantity of the salt in a test-tube and add 1-2 ml of dilute sulphuric add. Identify the gas and draw
Note:

1. Do not treat the salt with a large quantity of dilute acid.
2. Do not heat the salt with dilute acid.

Chemical Reactions Involved in Dil. H₂SO₄ Test
Dilute H₂SO₄ (or dilute HCl) decomposes carbonates, sulphides and nitrites in cold to give gases. These gases on identification indicate the nature of the add radical present in the salt.

Potassium permanganate test
To a pinch of salt in test tube add about 2 ml of dilute sulphuric acid. Boil off any gas evolved, add little more of dilute acid and then potassium permanganate solution dropwise. Note the changes as given in Table 12.8. This test helps in detection of Cl^–, Br^–, I^–, C₂O₄^2- and Fe²⁺ radicals.

Note:

1. As sulphides are oxidised by KMnO₄ so they have to be completely decomposed by heating with dilute sulphuric acid before this test is performed.
2. Potassium permanganate oxidises Fe2+ salts in cold. Oil H₂SO₄ acid is added to the salt and heated till sulphides, sulphites and nitrites are completely decomposed. Then KMnO₄ is added dropwise to cold solution.

Chemical Reactions Involved

Concentrated sulphuric acid test
This test is performed by treating small quantity of salt with cone, sulphuric acid (2-3 ml) in a test tube. Identify the gas evolved in cold and then on heating. Draw inferences from Table 12.9

Note:

1. Do not boil the salt with cone, sulphuric acid. On boiling, the acid may decompose to give SO₂ gas.
2. Nitrates give vapours of nitric acid (colourless) when heated with cone, sulphuric acid. When a paper pellet or copper chips is added, dense brown fumes evolve. Paper pellet acts as a reducing agent and reduces nitric acid to NO₂ (Reddish brown gas).
   

Chemical Reactions Involved in conc. H₂SO₄ Test

Tests for independent radicals (SO₄^2- and PO₄^3-)
As already diseussed these radicals are not detected by dilute or concentrated H₂SO₄ .They are tested individually.

1. Sulphate (SO₄^2- )
   Boil a small amount of salt with dilute HCl in a test tube. Filter the contents, and to the filtrate add few drops of BaCl2 solution. A white ppt. insoluble in cone. HCl indicates presence of sulphate.
2. Phosphate (PO₄^3-)
   Add cone. HNOs to the salt in a test tube. Boil the contents and add excess of ammonium molybdate solution. A yellow precipitate indicates presence of phosphate.

Confirmation of acid radicals by wet tests
The acid radical indicated by dil. H₂SO₄ or cone. H₂SO₄ tests is further confirmed by wet tests.

Preparation of solution for wet tests of acid radicals
The confirmatory tests for acid radicals are performed with salt solutions. The solution used for the purpose is any one of the following:

1. Aqueous solution or ‘water extract: Shake a little of the salt with water. If the salt dissolves, this aqueous solution obtained is used for the wet tests of acid radical and is called ‘water extract’ or ‘W.E.’. If the salt is not completely soluble in water, the salt is shaken with water and is filtered. The filtrate is treated as water extract.
2. Sodium carbonate extract: This is prepared only if the salt is insoluble in water.
   
   Preparation of Sodium Carbonate Extract. Mix about 1 g of the salt with about 2 g of pure sodium carbonate and boil it for 10-15 minutes with 20-25 ml of distilled water in a small conical flask having a funnel in its mouth (Fig. 12.8). The funnel acts as a condenser. This arrangement prevents the loss of water due to evaporation. Filter the solution, cool it and label it as sodium carbonate extract or S.E.
   Alternatively, sodium carbonate extract can be prepared in a test tube. A pinch of salt is mixed with double the amount of sodium carbonate and is boiled with distilled water for sometime. The suspension obtained is filtered.The filtrate is sodium carbonate extract. p.g 12 8. Preparation of sodium carbonate extract.
   Theory of Preparation of Sodium Carbon-ate Extract.
   When the salts are boiled with strong solution of sodium carbonate, double decompo-sition takes place resulting in the formation of the carbonates of heavy metallic radicals and sodium salts of the acid radicals. The sodium salts of corresponding add radicals being soluble in water pass into the solution and carbonates of heavy metals are predpitated
   

How to Use Sodium Carbonate Extract
Sodium carbonate extract always contains unreacted sodium carbonate in solution which has to be destroyed before using the extract for various tests. To do this, the extract is addified with some suitable acid and is boiled to expel carbon dioxide. The selection of add used for destroying excess Na₂CO₃depends upon the radical to be identified.
Now we describe in detail the confirmatory tests for various add radicals discussed so far.
Confirmation of Carbonate, CO₃^2-
(Indicated in dilute acid test by occurrence of brisk effervescence and evolution of carbon dioxide).

Note:

1. Do not use sodium carbonate extract for performing the tests of carbonates because it contains sodium carbonate.
2. Perform magnesium sulphate test only in case of soluble carbonates.

Confirmation of Sulphide, S^2-
(Indicated in dilute acid test by the evolution of hydrogen sulphide).

Confirmation of Sulphide, S^2-
(Indicated in dilute acid test by the evolution of hydrogen sulphide).

Confirmation of Nitrite, NO_2^–
(Indicated in dilute acid test by the evolution of brown vapours of nitrogen peroxide)

Chemical Reactions Involved in the Confirmation of Carbonate, Sulphide and Nitrite
Carbonate (CO₃^2- )

1. Reaction with dil. HCl
Carbonates on reaction with dil. HCl give CO₂ gas which turns lime water milky. In case of soluble carbonates this test is performed with water extract and in case of insoluble carbonates this test is performed with the solid salt.



Confirmation of Chloride,Cl^–
(No action with dilute H₂SO₄ but decomposed by cone. H₂SO₄ with the evolution of HCl gas).

Confirmation of Bromide, Br^–
(No action with dilute H₂SO₄ but decomposed by cone. H₂SO₄ with the evolution of bromine vapours).

Note. Chlorine water is prepared by adding dropwise cone. HCl to a small volume of KMnO₄ solution till the pink colour is just discharged, the resulting solutio/i is chlorine water.
Confirmation of Iodide, I^–
(No action with dilute H₂SO₄ but decomposed by cone. H₂SO₄ with the evolution of vapours
of iodine).

Confirmation of Nitrate, NO₃^–
(No action with dilute acids but decomposed by cone. H₂SO₄ with the evolution of brown vapours of nitrogen peroxide).

Confirmation of Acetate, CH₃COO^–
(No action with dilute acids but decomposed by cone. H₂SO₄ with the evolution of CH₃COOH vapours

Confirmation of Oxalate, C₂O_4^–
(No action with dilute acids but decomposed by cone. H₂SO₄with the evolution of CO₂ and CO gas)

Chemical Reactions Involved in the Confirmation of Chloride, Bromide, Iodide, Nitrate Acetate and Oxalate





Confirmation of Sulphate, SO₄^2-
(Not indicated in dilute and concentrated H₂SO₄ acid tests).

Confirmation of Phosphate, PO₄^3-
(Not indicated in dilute and concentrated H₂SO₄  acid test).

Chemical reactions involved in the confirmation of SO₄^2- and PO₄^3- 


Wet tests for basic (cations)
Preliminary tests such as dry heating test, charcoal cavity test, flame test and borax bead test may give us some indication about the cation present in the salt. However, the cation is finally detected and confirmed through a systematic analysis involving wet tests. For the sake of qualitative analysis the cations are classified into the following groups (Table 12.10).


Before carrying out the wet tests for the analysis of cation, the salt has to be dissolved in some suitable solvent to prepare its solution.

Preparation of solution for wet tests of basic radicals
The very first essential step is to prepare a clear and transparent solution of the salt under investigation. For this purpose, the under noted solvents are tried one after another in a systematic order. In case the salt does not dissolve in a particular solvent even on heating, try the next solvent. The following solvents are tried:

1. Distilled water (cold or hot).
2. Dilute HCl (cold or hot).
3. Cone. HCl (cold or hot).

Procedure for the preparation of solution
Take a small quantity of the given salt in a test tube. Add some suitable solvent into it and shake. If it does not dissolve even after heating for sometime, take the fresh quantity of the salt again and treat it in a similar manner with next solvent. The clear solution thus obtained is labelled as Original Solution (O.S.).

Important Notes:

1. In case some gas is evolved during the preparation of solution, let the reaction cease. Gas must be completely expelled by heating.
2. In case solution is prepared in dilute HCl, group I is absent. Proceed with group II.
3. If the salt is soluble in hot water, and on cooling white precipitates appear, lead chloride is indicated.
4. It is necessary to dilute the solution if it is made in concentrated acid before proceeding with the analysis.

The following table will help the students in the choice of a suitable solvent:

The separation of cations into various groups by making use of suitable reagents (known as group reagents) is based on the differences in chemical properties of cations. For example, if hydrochloric acid is added to a solution containing all cations, only the chlorides of lead, silver and mercury (ous) will precipitate, since all other chlorides are soluble. Thus, these cations form a group of ions which may be precipitated from solution by addition of group reagent HCl. Similarly, H₂S is a group reagent for group II. The following Table 12.11 clearly shows the group reagents for different groups and the form in which cations of the particular group are precipitated out.

Theory of precipitation of different groups
The classification of cations into different groups in the inorganic qualitative analysis is based upon the knowledge of solubility products of salts of these basic radicals. For example, chlorides
of Hg₂²⁺, Pb²⁺ and Ag⁺ have very low solubility products. On the basis of this knowledge these radicals are grouped together in group-I and are precipitated as their chlorides by adding dilute HCl to their solutions. For adjusting the conditions for precipitation, another concept called common ion effect plays very important role. Before we consider the precipitation of radicals of other groups, let us discuss in brief the concept of common ion effect.

Common ion effect
Weak acids and weak bases are ionised only to small extent in their aqueous solutions. In their solutions, unionised molecules are in dynamic equilibrium with ions. The degree of ionisation of a weak electrolyte (weak acid or weak base) is further suppressed if some strong electrolyte which can furnish some ion common with the ions furnished by weak electrolyte, is added to its solution. This effect is called common ion effect. For example, degree of ionisation of NH₄OH (a weak base) is suppressed by the addition of NH₄Cl (a strong electrolyte). The ionisation of NH₄OH and NH₄Cl in solution is represented as follows:

Due to the addition of NH₄Cl, which is strongly ionised in the solution, concentration of NH4+ ions increases in the solution. Therefore, according to Le-Chatelier’s principle equilibrium in equation (12.1) shifts in the backward direction in favour of unionised NH₄OH. In this way, addi¬tion of NH₄Cl suppresses the degree of ionisation of NH₄OH. Thus, the concentration of OH^– ions in the solution is considerably reduced and the weak base NH₄OH becomes a still weaker base.
The suppression of the degree of ionisation of a weak electrolyte (weak acid or weak base) by the addition of some strong electrolyte having a common ion, is called the common ion effect.
Application of concept of common ion effect in the qualitative analysis is illustrated as follows:
The cations of group II (Pb²⁺, Cu²⁺, AS³⁺) are precipitated as their sulphides. Solubility products of sulphides of group II radicals are very low. Therefore, even with low concentration of S^2- ions, the ionic products (Qs_p) exceed the value of their solubility products (KS_p) and the radicals of group II get precipitated. The low concentration of S^2- ions is obtained by passing H₂S gas through the solution of the salts in the presence of dil. HCl which suppresses degree of ionisation of H₂S by common ion effect.

It is necessary to suppress the concentration of S^2- ions, otherwise radicals of group IV will also get precipitated along with group II radicals.
Radicals of group IV (Ni²⁺, CO²⁺, Mn²⁺, Zn²⁺) are also precipitated as their sulphides. But solubility products of their sulphides are quite high. In order that ionic products exceed solubility products, concentration of S2- ions should be high in this case. High concentration of sulphide ions is achieved by passing H₂S gas through the solutions of the salts in the presence of NH₄OH. Hydroxyl ions from NH₄OH combine with H⁺ ions from H₂S. Due to the removal of H⁺ions the equilibrium of H₂S shifts in favour of ionised form.

Hence, concentration of S^2- ions increases. With this increased concentration of S^2- ions ionic products exceed solubility products and radicals of group IV get precipitated.
Radicals of group III (Fe³⁺, Al³⁺) are precipitated as their hydroxides by NH4OH in the presence of NH₄Cl. The purpose of NH₄Cl is to suppress the degree of ionisation of NH₄OH by common ion effect in order to decrease the concentration of OH^– ions.

The solubility products of hydroxides of group III radicals are quite low. Therefore, even with this suppressed concentration of OH^– ions their ionic products exceed solubility products and hence they get precipitated. If the concentration of OH^– ions is not suppressed, the radicals of groups IV, V and Mg2⁺ will also be precipitated along with radicals of group III.
Radicals of group V (Ba2⁺, Sa2⁺, Ca2⁺) are precipitated as their carbonates by the addition of (NH₄)₂ CO₃ in the presence of NH₄Cl and NH₄OH. NH₄Cl suppresses the degree of ionisation of (NH₄)₂ CO₃ by common ion effect and hence decreases the concentration of CO₃2- ions.

But solubility products of carbonates of group V radicals are quite low and hence even with the suppressed concentration of CO_3^2- ions their ionic products exceed solubility products and they get precipitated whereas Mg²⁺ and other radicals of group VI having relatively high solubility products are not precipitated.

Analysis of group Zero(NH₄⁺)
This group includes NH₄⁺ cation. During the analysis of cations NH₄Cl and NH₄OH are added in many steps. Therefore, H₄⁺ ion is detected in the beginning using solid salt.

Procedure
The solid salt is heated with concentrated solution of sodium hydroxide. In case, ammonia gas is evolved, NH₄⁺ is present. Evolution of NH₃ gas is confirmed by the following tests:

1. Characteristic ammoniacal smell.
2. The gas gives white fumes when a glass rod dipped in dil. HCl is brought near the mouth of the test tube.
3. When the gas is passed through Nessler’s reagent, it would give brown ppt. in case of NH₃.

Chemical Reactions Involved in Group-Zero Analysis

Analysis of group I (Silver Group)
This group includes Pb²⁺, Ag⁺ and Hg₂²⁺. But in the present context, we shall study only Pb²⁺. Group reagent for this group is dil. hydrochloric acid.

Procedure

1. To the original solution add dil. hydrochloric acid. If a white precipitate is formed, first group (Pb²⁺) is present.
2. Filter and wash the ppt. with cold water and examine as in Table 12.12.

Note:

1. If the original solution is prepared in cold dilute hydrochloric acid, first group is absent.
2. If the original solution is prepared in cone, hydrochloric acid, simply add water. White ppt. shows the presence of first group.

Chemical Reactions Involved in Group I Analysis

Analysis of group II (copper group)
This group includes Pb²⁺ and Cu²⁺ in IIA group and As³⁺ in IIB Group. These are precipitated as their sulphides. If group I is absent, the tests for radicals of group II are carried out. Group reagent for this group is H₂Sgas in the presence of dil. HCl.

Procedure
Take about 2 ml of the original solution in a test tube’. Make it acidic with dil. HCl and warm the contents. Through this solution pass H₂S gas from the Kipp’s apparatus by turning
the stop cock as shown in Fig. 12.10, Formation of the black or yellow precipitates indicates the presence of group II radical. If this is observed, pass more of H₂S gas to ensure complete precipitation of the radical sulphide. Centrifuge and separate the precipitates.

Identification of IIA and IIB Groups. Note the colour of the precipitate. If the precipitate is black in colour, it indicates Pb²⁺ or Cu²⁺. If the colour of precipitate is yellow this indicates As³⁺.

Chemical Reactions Involved in the Analysis of Group II

Analysis of group III (iron group)
The cations present in this group are Fe²⁺, Fe³⁺, Cr³⁺ and Al³⁺. Only Fe²⁺/Fe³⁺ and Al³⁺ are included  in the syllabus of this class. These cations are precipitated as hydroxides by adding ammonium hydroxide in presence of ammonium chloride. Thus, group reagent for this group is  NH₄OH in the presence of  NH₄Cl.
Procedure
In case, first and second groups are absent proceed for group III with the original solution. Take about 5 ml of the original solution and add 4-5 drops of cone, nitric acid. Boil the solution for sometime. Add to it about 2 g of solid  NH₄Cl and boil again. Cool the solution under tap water. Add excess of ammonium hydroxide to it and shake. A ppt. shows the presence of some cation of group III. Filter the ppt. and wash with water. Note the colour of the ppt. If the ppt. is reddish brown in colour, it indicates the presence of Fe³⁺ and if the colour is white, it indicates the presence of Al³⁺. Analyse the ppt. and draw inferences as in Table 12.14.

Note:

1. Test of Fe²⁺. The addition of cone, nitric acid in the analysis of group III serves to oxidise Fe²⁺ions to Fe³⁺ ions. Add cone, nitric acid only if the cation is Fe²⁺ otherwise the addition of nitric acid may be avoided. To test this, add a few drops of potassium ferricyanide solution to the original salt solution. A deep blue colouration shows Fe²⁺.
2. Use sufficient quantity of ammonium chloride, otherwise the hydroxides of higher group may be precipitated along with the radicals of third group.
3. Add NH₄OH until the solution gives the smell of ammonia.

Chemical Reactions Involved in the Analysis of Group III

Analysis of group IV (Zinc group)
The radicals present in this group are CO²⁺, Ni²⁺, Mn²⁺ and Zn²⁺. These are precipitated as sulphides by passing H₂S gas through the ammonical solution of the salt.
The group reagent for this group is H₂S  gas in the presence of NH₄Cl and NH₄OH.

Procedure
If there is no ppt. in the third group, then use the same ammonical solution for the fourth group. Pass H₂S gas through the solution. If some ppt. is formed, presence of some radical of group IV is indicated. Filter the ppt. and wash it with water. Note the colour of the ppt. and analyse the ppt. according to the Table 12.15.

Chemical Reactions Involved in the Analysis of Group IV
Passing of H₂S gas through the group III solution will precipitate the radicals CO²⁺, Ni²⁺, Mn²⁺ and Zn²⁺ as their sulphides. Formation of black ppt. (CoS or NiS) indicates cobalt or nickel. Formation of buff-coloured ppt. (MnS) indicates manganese and dirty white ppt. (ZnS) indictes zinc


2.Ammonium thiocyanate ether test
On addition of ether and a crystal of ammonium thiocyanate (shaking and allowing to stand), a blue colour due to the formation of ammonium cobhlti thiocyanate, is obtained in the ethereal layer.



Analysis of group V (calcium group)
Group V consists of three radicals: Ba²⁺, Sr²+ and Ca²⁺. These cations are precipitated as their carbonates.
Group reagent for this group is (NH₄)₂CO₃ in the presence ofNH₄Cl and NH₄OH.

Procedure
If the fourth group is absent, then proceed for radicals of group V.
To the O.S. add 2-3 gins of solid NH₄Cl, boil, cool and add NH₄OH till the solution smells of ammonia. Then add (NH₄)₂CO₃ solution. Appearance of white ppt. indicates the presence of group V cation. Filter and wash the ppt. with water. Dissolve the ppt. in hot dil. acetic acid. Divide the solution into three parts and proceed as in Table 12.16.

Note:

1. Proceed to test for group V cations in the order, Ba²⁺, Sr²⁺ and Ca²⁺. If Ba²⁺ is confirmed, do not test for Sr²⁺ or Ca²⁺ Similarly if Sr²⁺ is confirmed, do not test for Ca²⁺.
2. Original solution can be preferably used for testing Sr²⁺ and Ca²⁺.

Chemical Reactions Involved in the Analysis of Group V Radicals



When (NH₄)₂CO₃ is added to a salt solution containing NH₄Cl and NH₄OH, the carbonates of Ba²⁺, Sr²⁺ and Ca²⁺ are precipitated.

Analysis of group VI (Magnesium group)

Chemical Reactions Involved in Confirmation of Mg²⁺ 

Extra Questions for Class 6 Maths revision becomes an easy task during the annual exam. Students who find it challenging to understand the basics of the chapter must practice all the essential questions to understand better. These crucial questions and answers on Mensuration Class 6 Maths can aid the students’ preparation through the concepts of this chapter 10. https://meritbatch.com/mensuration-class-6-extra-questions/

Mensuration Class 6 Extra Questions Maths Chapter 10

Extra Questions for Class 6 Maths Chapter 10 Mensuration

Mensuration Class 6 Extra Questions Very Short Answer Type

Mensuration Class 6 Extra Questions Question 1.
The perimeter of a square is 64 cm. Find the length of each side.
Solution:
Perimeter of the square = 64 cm

Mensuration Questions For Class 6 Question 2.
Length and breadth of a rectangular table-top are 36 cm and 24 cm respectively. Find its perimeter.
Solution:
Length of the rectangular table-top = 36 cm
and its breadth = 24 cm.
∴ Perimeter of the table-top = 2 [length + breadth]
= 2 [36 cm + 24 cm]
= 2 x 60 cm = 120 cm.

Class 6 Maths Chapter 10 Extra Questions Question 3.
Which of the following figure has greater perimeter?

Solution:
Fig. (i) Perimeter of the square = 4 x side
= 4 x 4 cm = 16 cm
Fig. (ii) Perimeter of the rectangle
= 2 [length + breadth]
= 2[8 cm + 3 cm]
= 2 x 11 cm = 22 cm
Since 22 cm > 16 cm
∴ Rectangle has greater perimeter than the square.

Mensuration Class 6 Questions Question 4.
How much distance will you have to travel in going around each of the following figures?

Solution:
Distance travelled in going around Fig. (i)
= 12 cm + 3 cm + 12 cm + 3 cm = 30 cm
Distance travelled in going around Fig. (ii)
= 6 cm + 4 cm + 4 cm + 4 cm = 18 cm

Class 6 Maths Mensuration Extra Questions Question 5.
Find the perimeter of a square whose side is 15 cm.
Solution:
Side of the square = 15 cm
∴ Perimeter of the square = 15 cm x 4 = 60 cm

Mensuration Class 6 Extra Questions With Answers Question 6.
Find the cost of fencing a rectangular park 300 m long and 200 m wide at the rate of ₹4 per metre.
Solution:
Length of the park = 300 m
Breadth = 200 m
∴ Perimeter of the park = 2 [length + breadth]
= 2 [300 m + 200 m]
= 2 x 500 m = 1000 m.
Cost of fencing the rectangular park = 1000 x 4 = ₹4000

Class 6 Maths Ch 10 Extra Questions Question 7.
Find the area of a square field whose each side is 150 m.
Solution:
Side of the square field = 150 m
∴ Area of the square field = Side x Side
= 150 m x 150 m
= 22500 sq m.

Mensuration Class 6 Worksheet Question 8.
Length and breadth of a rectangular paper are 22 cm and 10 cm respectively. Find the area of the paper.
Solution:
Length of the rectangular paper = 22 cm
Breadth = 10 cm
∴ Area of the rectangular paper = length x breadth
= 22 cm x 10 cm
= 220 sq cm.

Mensuration Class 6 Extra Questions Short Answer Type

Questions On Mensuration For Class 6 Question 9.
Find the length of a rectangle given that its perimeter is 880 m and breadth is 88 m.
Solution:
Perimeter of the rectangle = 2 [length + breadth]
∴ 2 [length + breadth] = 880
length + breadth = 880 ÷ 2 = 440
∵ Breadth = 88 m
∴ Length = 440 m – 88 m = 352 m
Hence, the required length = 352 m.

Class 6 Mensuration Questions Question 10.
How many trees can be planted at a distance of 6 metres each around a rectangular plot whose length is 120 m and breadth is 90 m?
Solution:
Length of the rectangular plot = 120 m
Breadth = 90 m
∴ Perimeter of the rectangular plot
= 2 [length + breadth]
= 2 [120 m + 90 m]
= 2 x 210 m = 420 m
Now distance between two trees = 6 m
∴ Number of trees around the rectangular plot = 420 m ÷ 6 m = 70

Extra Questions Of Mensuration Class 6 Question 11.
A rectangular park is 30 metres long and 20 metres broad. A steel wire fence is put up all around it. Find the cost of putting the fence at the rate of ₹15 per metre.
Solution:
Length of the rectangular park = 30 m
Breadth = 20 m
∴ Perimeter of the rectangular park = 2(length + breadth)
= 2 [30 + 20] = 2 x 50 m = 100 m
∴ Cost of fencing all around the park = ₹15 x 100 = ₹1500

Class 6 Mensuration Questions Pdf Question 12.
Find the area of the figures A, B, C and D drawn on a squared paper in the following figure by counting squares.

Solution:
(A) Counting the squares, we have 8 squares
∴ Area = 8 sq units
(B) Counting the squares, we have 4 squares
∴ Area = 4 sq units
(C) Counting the squares, we have 5 squares
∴ Area = 5 sq units
(D) Counting the squares, we have 7 squares
∴ Area = 7 sq units

Mensuration Class 6 Extra Questions Higher Order Thinking Skills (HOTS)

Extra Questions On Mensuration For Class 6 Question 13.
A rectangle and a square have the same perimeter 100 cm. Find the side of the square. If the rectangle has a breadth 2 cm less than that of the square. Find the breadth, length and area of the rectangle.
Solution:
Perimeter of the square = 100 cm
Perimeter 100

= 25 cm.
∴ Breadth of the rectangle = 25 cm – 2 cm = 23 cm
Now perimeter of the rectangle = 100 cm
∴ 2 [length + breadth] = 100
length + breadth = 100 ÷ 2 = 50 cm
But breadth = 23 cm
∴ Length = 50 cm – 23 cm = 27 cm
Now, Area of the rectangle
= length x breadth = 27 cm x 23 cm
= 621 sq cm.

Mensuration Class 6 Question Bank Question 14.
Fencing the compound of a house costs ₹5452. If the rate is ₹94 per metre, find the perimeter of the compound. If the breadth is 10 m, find its length.
Solution:
Cost of fencing the compound = ₹5452
and the rate of fencing = ₹94 per metre
∴ Perimeter of the compound = 5452 ÷ 94 = 58 metres
Now breadth of the compound = 10 m.
2 [length + breadth] = 58 m
∴ length + breadth = 58 + 2 m = 29 m
∴ Length of the compound = 29 m – 10 m = 19 m.

NCERT Solutions for Class 6 Maths

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5

Get Free NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Continuity and Differentiability Exercise 5.5 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 5 Class 12 Continuity and Differentiability Ex 5.5 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

The topics and sub-topics included in the Continuity and Differentiability chapter are the following:

• Continuity and Differentiability
• Introduction
• Algebra of continuous functions
• Differentiability
• Derivatives of composite functions
• Derivatives of implicit functions
• Derivatives of inverse trigonometric functions
• Exponential and Logarithmic Functions
• Logarithmic Differentiation
• Derivatives of Functions in Parametric Forms
• Second Order Derivative
• Mean Value Theorem
• Summary

There are total eight exercises and one misc exercise(144 Questions fully solved) in the class 12th maths chapter 5 Continuity and Differentiability.

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5

Differentiate the functions given in Questions 1 to 11 w.r.to x

Ex 5.5 Class 12 Maths Question 1.
cos x. cos 2x. cos 3x
Solution:
Let y = cos x. cos 2x . cos 3x,
Taking log on both sides,
log y = log (cos x. cos 2x. cos 3x)
log y = log cos x + log cos 2x + log cos 3x,
Differentiating w.r.t. x, we get

Ex 5.5 Class 12 Maths Question 2.
\(\sqrt { \frac { (x-1)(x-2) }{ (x-3)(x-4)(x-5) } } \)
Solution:
\(y=\sqrt { \frac { (x-1)(x-2) }{ (x-3)(x-4)(x-5) } } \)
taking log on both sides
log y = log \(\sqrt { \frac { (x-1)(x-2) }{ (x-3)(x-4)(x-5) } } \)

Ex 5.5 Class 12 Maths Question 3.
(log x)^cosx
Solution:
let y = (log x)^cosx
Taking log on both sides,
log y = log (log x)^cosx
log y = cos x log (log x),
Differentiating w.r.t. x,

Ex 5.5 Class 12 Maths Question 4.
x – 2^sinx
Solution:
let y = x – 2^sinx,
y = u – v

Ex 5.5 Class 12 Maths Question 5.
(x+3)².(x + 4)³.(x + 5)⁴
Solution:
let y = (x + 3)².(x + 4)³.(x + 5)⁴
Taking log on both side,
logy = log [(x + 3)² • (x + 4)³ • (x + 5)⁴]
= log (x + 3)² + log (x + 4)³ + log (x + 5)⁴
log y = 2 log (x + 3) + 3 log (x + 4) + 4 log (x + 5)
Differentiating w.r.t. x, we get

Ex 5.5 Class 12 Maths Question 6.
\({ \left( x+\frac { 1 }{ x } \right) }^{ x }+{ x }^{ \left( 1+\frac { 1 }{ x } \right) }\)
Solution:
let \(y={ \left( x+\frac { 1 }{ x } \right) }^{ x }+{ x }^{ \left( 1+\frac { 1 }{ x } \right) }\)
let \(u={ \left( x+\frac { 1 }{ x } \right) }^{ x }and\quad v={ x }^{ \left( 1+\frac { 1 }{ x } \right) }\)

Ex 5.5 Class 12 Maths Question 7.
(log x)^x + x^logx
Solution:
let y = (log x)^x + x^logx = u+v
where u = (log x)^x
∴ log u = x log(log x)

Ex 5.5 Class 12 Maths Question 8.
(sin x)^x+sin^-1 √x
Solution:
Let y = (sin x)^x + sin^-1 √x
let u = (sin x)x, v = sin^-1 √x

Ex 5.5 Class 12 Maths Question 9.
x^sinx + (sin x)^cosx
Solution:
let y = x^sinx + (sin x)^cosx = u+v
where u = x^sinx
log u = sin x log x

Ex 5.5 Class 12 Maths Question 10.
\({ x }^{ x\quad cosx }+\frac { { x }^{ 2 }+1 }{ { x }^{ 2 }-1 } \)
Solution:
\(y={ x }^{ x\quad cosx }+\frac { { x }^{ 2 }+1 }{ { x }^{ 2 }-1 } \)
y = u + v

Ex 5.5 Class 12 Maths Question 11.
\({ (x\quad cosx) }^{ x }+{ (x\quad sinx) }^{ \frac { 1 }{ x } }\)
Solution:
\(y={ (x\quad cosx) }^{ x }+{ (x\quad sinx) }^{ \frac { 1 }{ x } }\)
Let u = (x cosx)^x
logu = x log(x cosx)

Find \(\\ \frac { dy }{ dx } \) of the functions given in Questions 12 to 15.

Ex 5.5 Class 12 Maths Question 12.
x^y + y^x = 1
Solution:
x^y + y^x = 1
let u = x^y and v = y^x
∴ u + v = 1,
\(\frac { du }{ dx } +\frac { dv }{ dx }=0\)
Now u = x

Ex 5.5 Class 12 Maths Question 13.
y^x = x^y
Solution:
y = x
x logy = y logx

Ex 5.5 Class 12 Maths Question 14.
(cos x)^y = (cos y)^x
Solution:
We have
(cos x)^y = (cos y)^x
=> y log (cosx) = x log (cosy)

Ex 5.5 Class 12 Maths Question 15.
xy = e^(x-y)
Solution:
log(xy) = log e^(x-y)
=> log(xy) = x – y
=> logx + logy = x – y
\(=>\frac { 1 }{ x } +\frac { 1 }{ y } \frac { dy }{ dx } =1-\frac { dy }{ dx } =>\frac { dy }{ dx } =\frac { y(x-1) }{ x(y+1) } \)

Ex 5.5 Class 12 Maths Question 16.
Find the derivative of the function given by f (x) = (1 + x) (1 + x²) (1 + x⁴) (1 + x⁸) and hence find f'(1).
Solution:
Let f(x) = y = (1 + x)(1 + x²)(1 + x⁴)(1 + x⁸)
Taking log both sides, we get
logy = log [(1 + x)(1 + x²)(1 + x⁴)(1 + x⁸)]
logy = log(1 + x) + log (1 + x²) + log(1 + x⁴) + log(1 + x⁸)

Ex 5.5 Class 12 Maths Question 17.
Differentiate (x² – 5x + 8) (x³ + 7x + 9) in three ways mentioned below:
(i) by using product rule
(ii) by expanding the product to obtain a single polynomial.
(iii) by logarithmic differentiation.
Do they all give the same answer?
Solution:
(i) By using product rule
f’ = (x² – 5x + 8) (3x² + 7) + (x³ + 7x + 9) (2x – 5)
f = 5x⁴ – 20x³ + 45x² – 52x + 11.
(ii) By expanding the product to obtain a single polynomial, we get

Ex 5.5 Class 12 Maths Question 18.
If u, v and w are functions of w then show that
\(\frac { d }{ dx } (u.v.w)=\frac { du }{ dx } v.w+u.\frac { dv }{ dx } .w+u.v\frac { dw }{ dx } \)
in two ways-first by repeated application of product rule, second by logarithmic differentiation.
Solution:
Let y = u.v.w
=> y = u. (vw)

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Hindi Medium Ex 5.5

NCERT Class 12 Maths Solutions

• Chapter 1 Relations and Functions
• Chapter 2 Inverse Trigonometric Functions
• Chapter 3 Matrices
• Chapter 4 Determinants
• Chapter 5 Continuity and Differentiability
• Chapter 6 Application of Derivatives
• Chapter 7 Integrals Ex 7.1
• Chapter 8 Application of Integrals
• Chapter 9 Differential Equations
• Chapter 10 Vector Algebra
• Chapter 11 Three Dimensional Geometry
• Chapter 12 Linear Programming
• Chapter 13 Probability Ex 13.1

CBSE Class 6 English Grammar Articles

Three words—a, an and the are called articles. They are divided in two groups.

A. The Definite Article (The)
1. Use of Definite Article
(a) Put the before the words which mean the inhabitants of a country.
The English live in England and the Indians in India.
But do not put the before the names of the language.
English is spoken in England and Hindi in India.
Exception
Plural nouns of a country’s inhabitants do not take the if they are thought of individually.
Indians are generally vegetarians.
Russians drink vodka.

(b) Put the before the names of mountain ranges.
The Alps, the Himalayas
But do not put the before the names of single mountains or hills.
Everest, Mont Blanc (not the Everest, the Mont Blanc)

(c) Use the before the names of rivers, canals, seas, oceans, valleys, deserts and forests :
Lucknow is on the Gomati. (not on GomatQ
Similarly we say—
The North Sea, The Indian Ocean, The Ganges, The Sahara

(d) Use the before the names of ships and trains :
The Rajdhani Express runs fast.
The Titanic was a very big ship.
But do not use the before the vehicles when they indicate a means of transport :
I will go to Delhi by bus.
We are going to Mumbai by train.

2. Omission of the Definite Article
(a) Do not put the before the names of substances if they are used in a general sense.
Gold is a precious metal. (not the golf)
Bread is made from flour. (not the bread … the flour)
Lead is very heavy. (not the lead)
But the must be used if the reference is to a particular kind or type.
The thieves stole the gold from that ship.
The bread in this hotel is of poor quality.

(b) Do not put the before the names of meals if they refer to the meals generally.
When do you have dinner ?
Have you had breakfast ?
Lunch is at 2.00 p.m.
But use the when the meal is a particular one.
The dinner will be taken at Ashoka.
We enjoyed the lunch given by the school.

(c) Do not use the before plural nouns when they are used in a general sense.
Books are necessary for students.
Apples grow in Simla.
Cars can run fast.

(d) Do not use the before the names of games.
Sania Mirza plays Tennis.
Chess is a game which requires skill.

(e) Do not put the before the names of the countries unless the name denotes that it is made of parts.
India, Italy, France and China are all republics.
But
The United States is very rich.
The U.S.S.R. has broken into smaller units.

(f) Do not use the before the names of the offices if these follow the names of the officers :
Prime Minister Dr. Manmohan Singh
President Dr. Abdul Kalam
D.C. Miss. Rajni Kothari
But if the names are not mentioned the must be used.
The Prime Minister, the President etc.

Exercise 1
(Solved)

Fill in the blanks with the where necessary.
1. The table was made of ………. wood.
2. ………. water in that pot is not fit for drinking.
3. ………. sweets we took after meals were tasty.
4. When do you have ………. breakfast ?
5. Are you attending ………. dinner tonight ?
6. Why do we wear ………. clothes ?
Answers:
1. × (No article)
2. The
3. The
4. × (No article)
5. the
6. × (No article)

B. The Indefinite Article (A and An)
(a) A is used before words beginning with a consonant, and an before words beginning with a vowel or with a letter h which is not sounded. The following are the main words which are spelt with unsounded h.

heir	heiress	heirloom	honest
honesty	honorarium	honorary	honour
honourable	honoured	hour	hour

(b) If a word begins with a vowel but is pronounced in the way as ‘y’ in yet, write a and not an before it.
a university, a European.

(c) A/an are to be used only before singular countable nouns.
a book, a city, an animal.
A/an should be used before the adjective if these singular countable nouns have an adjective before them.
a big city, a fine book, an ignorant person.

(d) Use a after the word such when it is applied to things which are countable.
I have never known such a cold winter.
Such a thing has never happened before.

(e) If instead of such, there is the word so, then put a after the adjective, just before the noun.
I have never known so cold a winter.   (not a so cold winter)

(f) When a is placed before the word few, it changes the meaning. Few means a small number when more might have been expected ; a few means a small number when none were expected.
Few boys were present in the class.    (ie., I had expected more)
A few boys were present in the class.
(i..e., I had expected none, still there was a small number of them present)
There is a similar difference between little and a little.
We have little time to spare.
(It means that we do not have as much time as we should like).
We have a little time to spare.
(It means that we are not so short of time that we cannot spare any).

Exercise 2
(Solved)

Fill in the blanks with a or an.
1. ………. man and ………. woman were sitting opposite me.
2. During our holiday, we stayed at ………. hotel.
3. For our dinner, we went to ………. restaurant.
4. As it was raining. I took ………. umbrella.
5. On the way, I saw ………. elephant.
6. I am ………. Indian whereas you are ………. Chinese.
Answers:
1. a, a
2. a
3. a
4. an
5. an
6. an, a

Exercise 3
(For Practice)

Fill in the blanks with ‘a’ or ‘an’.
1. Have you ever seen ………. lion ?
2. Does Sohan have ………. car ?
3. Is there ………. bank nearby ?
4. There isn’t ………. airport nearby.
5. Ram is ………. honest man.
6. He will return in ………. hour.

Exercise 4
(For Practice)

Insert ‘a’, ‘an’ or ‘the’.
Once there was ………. mouse. The mouse was always afraid of ………. cat. A magician took pity on ………. mouse. He turned it into ………. cat. Now cat was afraid of dogs. So, ………. magician turned ………. cat into ………. dog and finally into ………. tiger ………. tiger began to fear hunters in forest.
Then ………. magician said, “Be ………. mouse again. You are no better than mouse at heart.

Exercise 5
(For Practice)

Fill in the blanks with a few, a little, the few, the little.
1. Just ………. persons were present there.
2. I cannot prepare tea ; there is ………. milk in the house.
3. I have consumed ………. sugar you gave me.
4. ………. water of the pond will not last the year.
5. The class was not held as ………. students were present.
6. Don’t worry. I have ………. money in the bank.

Exercise 6
(For Practice)

Fill in the blanks with the where necessary.
1. Some soaps are made of ………. oils and some of only ………. chemicals.
2. ………. iron is found in India.
3. In India more people drink ………. milk than ………. wine.
4. ………. mangoes are grown in India. Pakistan and Srilanka.
5. Will you play ………. cricket ?
6. She plays ………. basketball very well.

Multiple Choice Questions

Read the statements given below and tick the correct option :
I. Identify as directed :
Question 1.
Identify the definite article
(i) a
(ii) an
(iii) the
(iv) it

Question 2.
Identify the indefinite articles
(i) a, the
(ii) an, the
(iii) a, an
(iv) both (i) and (ii)

II. Fill the correct option :
Question 3.
………. English is spoken in England
(i) A
(ii) The
(iii) An
(iv) None of these

Question 4.
Such ………. thing has never happened
(i) a
(ii) the
(iii) an
(iv) none of these

Question 5.
The tourist is ………. European
(i) a
(ii) the
(iii) an
(iv) none of these

Question 6.
………. ignorant person needs to learn
(i) A
(ii) The
(iii) An
(iv) None of these

Question 7.
………. money was in my purse
(i) A
(ii) The
(iii) An
(iv) None of these

III. Choose the correct option :
Question 8.
(i) A water in the pot is unfit
(ii) The water in the pot is unfit
(iii) An water in the pot is unfit
(iv) Its water in the pot is unfit

Question 9.
(i) The table is made of the wood
(ii) The table is made of wood
(iii) The table is made of a wood
(iv) That table is made of a wood

Question 10.
(i) It is great honour for me
(ii) It is a great honour for me
(iii) It is some great honour for me
(iv) It is more great honour for me

Question 11.
(i) I went to its university
(ii) I went to an university
(iii) I went to the university
(iv) I went to this university

Question 12.
(i) We have little time to spare
(ii) We have least time to spare
(iii) We have a little time to spare
(iv) We have the little time to spare
Answers
1. (iii)
2. (iii)
3. (iv)
4. (i)
5. (i)
6. (iii)
7. (ii)
8. (ii)
9. (ii)
10. (ii)
11. (iii)
12. (iii)

NCERT Solutions for Class 6 English

NCERT Solutions for Class 12 Accountancy Chapter 4 Reconstitution of a Partnership Firm – Retirement/Death of a Partner

NCERT Solutions CBSE Sample Papers AccountancyClass 12 Accountancy

DO IT YOURSELF

Question 1. Anita, Jaya and Nisha are partners sharing profits and losses in the ratio of 1:1:1 Jaya retires from the firm. Anita and Nisha decided to share the profit in future in the ratio 4 :3. Calculate the gaining ratio.

Question 2. Azad, Vijay and Amit are partners sharing profits and losses in the proportion of 1/4,1/8,10/16 and Calculate the new profit sharing ratio between continuing partners if (a) Azad retires; (b) Vijay retires; (c) Amit retires.

Question 3. Calculate the gaining ratio in each of the above situations.

Question 4.Anu, Prabha and Milli are partners. Anu retires. Calculate the future profit sharing ratio of continuing partners and gaining ratio if they agree to acquire her share : (a) in the ratio of 5:3; (b) equally.

Question 5. Rahul, Robin and Rajesh are partners sharing profits in the ratio of 3 : 2 : 1. Calculate the new profit sharing ratio of the remaining partners if (i) Rahul retires; (ii) Robin retires; (iii) Rajesh retires.

Question 6.Puja, Priya, Pratistha are partners sharing profits and losses in the ratio of 5 : 3 : 2. Priya retires. Her share is taken by Priya and Pratistha in the ratio of 2 : 1. Calculate the new profit sharing ratio.

Question 7.Ashok, Anil and Ajay are partners sharing profits and losses in the ratio of 1/2, 3/10 and 1/5. . Anil retires from the firm. Ashok and Ajay decide to share future profits and losses in the ratio of 3 : 2. Calculate the gaining ratio.

TEST YOUR UNDERSTANDING I
Choose the correct option in the following questions:
Question1. Abhishek, Rajat and Vivek are partners sharing profits in the ratio of 5:3:2. If Vivek retires, the New Profit Sharing Ratio between Abhishek and Rajat will be–
(a) 3:2 (b) 5:3 (c) 5:2 (d) None of these
Answer (b) 5 :3

Question 2. The old profit sharing ratio among Rajender, Satish and Tejpal were 2:2:1. The new profit sharing ratio after Satish’s retirement is 3:2. The gaining ratio is
(a) 3 : 2 (b) 2 : 1
(c) 1 : 1 (d) 2 : 2
Answer (c)

Question 3. Anand, Bahadur and Chander are partners. Sharing Profit equally on Chander’s retirement, his share is acquired by Anand and Bahadur in the ratio of 3 : 2. The new profit sharing ratio between Anand and Bahadur will be
(a) 8 : 7 (b) 4 : 5 (c) 3 : 2 (d) 2:3
Answer (a)

Question 4. In the absence of any information regarding the acquisition of share in profit of the retiring/deceased partner by the remaining partners, it is assumed that they acquire his/her share
(a) old profit sharing ratio (b) new profit sharing ratio (c) equal ratio (d) None of these
Answer (a) Old profit sharing ratio

TEST YOUR UNDERSTANDING II

• Choose the correct option in the following questions
Question 1. On retirement/death of a partner, the retiring/deceased partner’s capital account will be credited with
(a) his/her share of goodwill
(b) goodwill of the firm
(c) shares of goodwill of remaining partners
(d) None of the above
Answer (a) His/Her share of goodwill

Question 2. Gobind, Hari and Pratap are partners. On retirement of Gobind, the goodwill already appears in the Balance Sheet at T 24,000. The goodwill will be written off
(a) by debiting all partners’ capital accounts in their old profit sharing ratio
(b) by debiting remaining partners’ capital accounts in their new profit sharing ratio
(c) by debiting retiring partners’ capital accounts from his share of goodwill
(d) None of the above
Answer (a) By debiting all partners’ capital accounts in their old profit sharing ratio

Question 3.Chaman, Raman and Suman are partners sharing profits in the ratio of 5:3:2. Raman retires, the new profit sharing ratio between Chaman and Suman will be 1:1. The goodwill of the firm is valued at Rs. 1,00,000 Raman’s share of goodwill will be adjusted
(a) by debiting Chaman’s Capital account and Suman’s Capital Account with Rs 15,000 each.
(b) by debiting Chaman’s Capital account and Suman’s Capital Account with Rs. 21,429 and 8,571 respectively.
(c) by debiting only Suman’s Capital Account with Rs. 30,000.
(d) by debiting Raman’s Capital account with Rs. 30,000.
Answer (c)

Question 4. On retirement/death of a partner, the remaining partner(s) who have gained
due to change in profit sharing ratio should compensate the
(a) retiring partners only.
(b) remaining partners (who have sacrificed) as well as retiring partners.
(c) remaining partners only (who have sacrificed).
(d) none of these.

Answer (b) Remaining partners (who have sacrificed) as well as partners

DO IT YOURSELF II
Question 1. Vijay, Ajay and Mohan are friends. They passed B (Hons) from Delhi University in June, 2013. They decided to business of computer hardware.
On 1st of August, 2013, they introduced the capital of Rs. 50,000, Rs.30,000 and Rs. 20,000 respectively and started the business in partnership at Delhi. The profit sharing ratio decided between there was 4:2:1. The business was running successfully. But on 1st February, 2019, due to certain unavoidable circumstances and family circumstances, Ajay decided to settle in Pune and decided to retire from the partnership on 31st March, 2020; with the consent of partners, Ajay retires as on 31st March, 2020, the position of assets and liabilities are as follows

On the date of retirement, the following adjustments were to be made
(1) Firm’s goodwill was valued at Rs. 1,48,000.
(2) Assets and Liabilities are to be valued as under; Stock Rs. 72,000; Land and Buildings Rs. 1,35,600; Debtors Rs. 63,000; Machinery Rs.1,50,000; Creditors Rs. 84,000.
(3) Vijay to bring Rs. 1,20,000 and Mohan Rs. 30,000 as additional capital.
(4) Ajay was to be paid Rs. 97,200 in cash and the balance of his Capital Account to be transferred to his Loan Account Work out the amount

Do it yourself III

Question 1. The Balance Sheet of A, B and C who were sharing the profits in proportion to their capitals stood as on March 31, 2007.

B retired on the date of Balance Sheet and the following adjustments were to be made
(a) Stock was depreciated by 10%.
(b) Factory building was appreciated by 12%.
(c) Provision for doubtful debts to be created up to 5%.
(d) Provision for legal charges to be made at Rs. 265.
(e) The goodwill of the firm to be fixed at Rs. 10,000.
(f) The capital of the new firm to be fixed at Rs. 30,000. The continuing partners decide to keep their capitals in the new profit sharing ratio of 3:2.
Work out the final balances in capital accounts of the firm and the amount to bh brought in and/or withdrawn by A and C to make their capitals proportionate to then new profit sharing ratio.

Question 2. R, S and M were carrying on business in partnership
sharing profits in the ratio of 3 : 2 : 1 respectively. On March 31,
2011, Balance Sheet of the firm stood as follows

Shyam retired on the above mentioned date on the following terms
(a) Buildings to be appreciated by ? 8,800.
(b) Provision for doubtful debts to be made @ 5% on debtors.
(c) Goodwill of the firm to be valued at ? 9,000.
(d) Rs.5,000 to be paid to S immediately and the balance due to him to be treated as a loan carrying interest @ 6% per annum.
Prepare the balance sheet of the reconstituted firm.

Do it Yourself IV
Question 1. On December 31, 2007, the Balance Sheet of Pinki, Qureshi and Rakesh showed as under

The partnership deed provides that the profit be shared in the ratio of 2:1:1 and that in the event of death of a partner, his executors be entitled to be paid out
(a) The capital of his credit at the date of last Balance Sheet.
(b) His proportion of reserves at the date of last Balance Sheet.
(c) His proportion of profits to the date of death based on the average profits of the last three completed years, plus 10%.
(d) By way of goodwill, his proportion of the total profits for the three preceding years. The net profit for the last three years were
                           Rs.
2005              16,000
2006              16,000
2007              15,400

Rakesh died on April 1, 2007. He had withdrawn Rs. 5,000 to the date of his death. The investment were sold at par and R’s Executors were paid off. Prepare Rakesh’s Capital Account that of his executors.

Short Answer Type Questions

Question 1. What are the different ways in which a partner can retire from the firm?
Answer A partner can retire from the firm in three different ways. They are as follow
(i) Retirement Through Mutual Consent A partnership firm take its shape through mutual consent of partners in the same way. A partner may retire if all the partners agree on the decision of his/her retirement.
(ii) When There is a Provision in Partnership Deed When there is a provision for the retirement of a partner in the partnership deed in that case partner may retire from the firm by expressing his/her intention of leaving the firm though a notice to the other partners of the firm.
(iii) Retirement Through Written Notice In case when partnership among the partners is at will, then a partner may retire by giving notice in writing to all the other partners informing them about his/her intention to retire.

Question 2. Write the various matters that need adjustments at the time of retirement of partners.
Answer The following are the various matters that need to be adjusted at the time of retirement of partners/partner
(i) Revaluation of assets and liabilities of the new firm.
(ii) Calculation of goodwill of the new firm and its accounting treatment.
(iii) Calculation of new ratio of the remaining partners of the firm.
(iv) Computation of new gaining ratio among rest of the partners.
(v) Distribution of accumulated profits and losses and reserves among all the partners (including the retiring partner).
(vi) Treatment of Joint Life Policy.
(vii) Disposal of the amount due to the retiring partner
(viii) Adjustment of capital accounts of the remaining partners in their new profit sharing ratio.

Question 3. Distinguish between sacrificing ratio and gaining ratio.
Answer Difference between Sacrificing and Gaining Ratio

Question 4. Why do firm revaluate assets and reassess their liabilities on retirement or on the event of death of a partner?
Answer At the time of retirement or death of a partner, it becomes inevitable to revalue the assets and liabilities of the firm for ascertaining their true and fair values. The revaluation is necessary as the value of assets and liabilities may increase or decrease with the passage of time.
Further, it may be possible that there are certain assets and liabilities that remained unrecorded in the books of accounts. The retiring or the deceased partner may be benefitted or may bear loss due to change in the values of assets and liabilities. Therefore, the revaluation of the assets and liabilities is necessary in order to ascertain the true profit or loss that is to be divided among all the partners in their old profit sharing ratio.

Question 5. Why a retiring/deceased partner is entitled to a share of goodwill of the firm?
Answer Goodwill is an intangible asset of a firm that is earned by the efforts of all the partners of the firm. After the retirement or death of a partner, the fruits of the past performance and reputation will be shared only by the remaining partners. Thus, the remaining partners should compensate the retiring orthe deceased partner by entitling him/her a share of firm’s goodwill.

Long Answer Type Questions

Question 1. Explain the modes of payment to a retiring partner.
Answer Payment to a retiring partner can be made in the following ways
(i) Lump Sum Payment : A lump sum payment can be made to the retiring partner in full settlement. In that case, the following Journal entry will be passed

(ii) Opening the Loan Account Sometimes the amount due to the retiring partner is paid in instalments then the balancing figure of his/her capital account is transferred to his/her loan account, in this case, the retiring partner receives equal instalments along with the interest on the amount outstanding. In that case the following journal entries will be passed for transferring the amount paid to him/her in retiring partner’s loan account.

(iii)Some Payment in Cash and Some in Instalment Sometimes the amount due to the retiring partner is paid partly in cash and partly in equal instalments in that case a certain amount is paid in cash to the retiring partner and the rest amount due to him/her is transferred to his/her loan account. The following necessary journal entry is to be passed.

Question 2. How will you compute the amount payable to a deceased partner?
Answer In case of a death, the legal executor of the deceased partner is entitled for a claim which includes his share of profit or loss, interest on capital, interest on drawings In that case for computing the amount payable is calculated by preparing the deceased partner’s capital account as follows

Note: In the above capital account, the legal executor will be entitled for the balancing
figure that is the excess of the credit side over the debit side of the deceased partner’s capital account.

Question 3. Explain the treatment of goodwill at the time of retirement or on the event of death of a partner.
Answer At the time of retirement or is the event of death of a partner, the goodwill of the firm is adjusted among the partners in their gaining ratio with the share of goodwill of the retiring or the deceased partner. At the time of retirement or on the event of death of a partner, goodwill account is not opened hence only two situations are left for treating the goodwill first when
goodwill account is already there in the book or it appear in the books and second when the amount of goodwill is not appearing in the books.
The treatment of goodwill will be as follows in the above two situations
First Situation When Goodwill Already Appears in the Books of the Firm
Step 1 Write-off the Existing Goodwill When goodwill account already exist in the book of the firm or mentioned in the book first of all, it will be written oft and should be distributed among all the partners of the firm including the retiring or the deceased partner in their old profit sharing ratio. In that case, the journal entry will be as follows

Step 2 Adjusting Goodwill Through Partners’ Capital Account
After writing off the old goodwill, the amount of goodwill now needs to be adjusted through the partner’s capital account with the share of the goodwill of the retiring orthe deceased partner. The following journal entry is passed.

Second Situation When No Goodwill Appears in the Books of the Firm
In second case, when no goodwill appears in the books of the firm, the amount of goodwill will be adjusted through the partner’s capital account with the share of the goodwill of the retiring or the deceased partner. The following journal entry’is passed

Question 4. Discuss the various methods of computing the share in profits in the event of death of a partner.
Answer Computation of profit will be different in case of death of a partner as compare to the retirement. The reason is that in case of retirement everything is pre-planned but in case of death nothing is planned. In case of death, the share of profit.can be calculated by one of the two methods.
(i) On the Basis of Time
In this method, profit upto the date of the death of the partner is calculated on the basis of time passed till the death of the partner from the beginning of the year on the bases of the last year’s/years’ profit or average profit of last few years.
The assumption in this method is that the profit will be uniform throughout the current year. The share of the deceased partner profit will be calculated as follows

Example A, B, C and D are equal partners. The profit of the firm for the years 2009, 2010 and 2011 are ? 5,00,000, ? 7,00.000 and <9,00,000 respectively. C dies on June 30, 2012. The share of C in the firm’s profit will be calculated on the basis of average profit of last three years. Firm closes its books every year on December 31.
In this case, C’s share in the profits will be calculated for four months. i e., from January 1. 2012

(ii) On the Basis of Sale
In this method, profit up to the date of the death of the partner is calculated on the basis of sales affected till the date of the’ death of the partner from the beginning of the year. The assumption in this method is that the net profit margin for current year will be same as the previous year. The share of the deceased partner profit will be calculated as follows

Example A, B and C are equal partners. The last year’s sales and profit were ? 40,00,000 and ? 4,00,000. C died on June, 2012. Sales of the current year till the date of C’s death amounts to ? 15,00,000. Firm closes its books on December 31 every year.

Numerical Questions

Question 1. Aparna, Manisha and Sonia are partners sharing profits in the ratio of 3 : 2 :1. Manisha retires and goodwill of the firm is valued at Rs. 1,80,000. Aparna and Sonia decided to share future profits in the ratio of 3 : 2. Pass necessary journal entries.

Question 2. Sangeeta, Saroj and Shanti are partners sharing profits in the ratio of 2 : 3 : 5. Goodwill is appearing in the books at a value of Rs. 60,000. Sangeeta retires and goodwill is valued at Rs. 90,000. Saroj and Shanti decided to share future profits equally. Record necessary journal entries.

Question 3. Himanshu, Gagan and Naman are partners sharing profits and losses in the ratio of 3 : 2 :1 On March 31, 2007, Naman retires. The various assets and liabilities of the firm on the date were as follows Cash Rs.10,000, Building Rs. 1,00,000, Plant and Machinery Rs. 40,000, Stock Rs. 20,000, Debtors Rs. 20,000 and Investments Rs. 30,000.
The following was agreed upon between the partners on Naman’s retirement
(i) Building to be appreciated by 20%.
(ii) Plant and Machinery to be depreciated by 10%.
(iii) A provision of 5% on debtors to be created for bad and doubtful debts.
(iv) Stock was to be valued at Rs. 18,000 and Investment at Rs. 35,000.
Record the necessary journal entries to the above effect and prepare the revaluation account.

Question 4. Naresh, Raj Kumar and Bishwajeet are equal partners. Raj Kumar decides to retire. On the date of his retirement, the Balance Sheet of the firm showed the following : General Reserves Rs.36,000 and Profit and Loss Account (Dr) Rs. 15,000.
Pass the necessary journal entries to the above effect.

Question 5. Digvijay, Brijesh and Parakaram were partners in a firm sharing profits in the ratio of 2 : 2 : L Their Balance Sheet as on March 31, 2007 was as follows

Brijesh retired on March 31, 2007 on the following terms
(i) Goodwill of the firm was valued at ? 70,000 and was not to appear in the books.
(ii) Bad debts amounting to ? 2,000 were to be written off.
(iii) Patents were considered as valueless.
Prepare revaluation account, partners’ capital accounts and the balance sheet of Digvijay and Parakaram after Brijesh’s retirement

Note: If no information is given, the amount due to Brijesh (the balance of his capital account) is transferred to his loan account.

6. Radha, Sheela and Meena were in partnership sharing profits and losses in
the proportion of 3:2:1. On April 1, 2007, Sheela retires from the firm. On that
date, their Balance Sheet was as follows:

The terms were
(a) Goodwill of the Firm was valued at ? 13,000.
(b) Expenses owing to be brought down to ? 3,750.
(c) Machinery and Loose Tools are to be valued at 10% less than their book value.
(d) Factory premises are to be revalued at ? 24,300.
Prepare
1. Revaluation account
2. Partners’capital accounts
3. Balance sheet of the firm after retirement of Sheela

7. Pankaj, Naresh and Saurabh are partners sharing profits in the ratio of 3 : 2 :
1. Naresh retired from the firm due to his illness. On that date the Balance
Sheet of the firm was as follows:

Additional Information
(i) Premises have appreciated by 20%, stock depreciated by 10% and provision for doubtful debts was to be made 5% on debtors. Further, provision for legal damages is to be made for Rs. 1,200 and furniture to be brought up to Rs. 45,000.
(ii) Goodwill of the firm be valued at Rs. 42,000.
(iii) Rs. 26,000 from Naresh’s Capital account be transferred to his loan account and balance be paid through bank; if required, necessary loan may be obtained from Bank.
(iv) New profit sharing ratio of Pankaj and Saurabh is decided to be 5 :1.
Give the necessary ledger accounts the balance sheet of the firm after Naresh’s retirement.

8. Puneet, Pankaj and Pammy are partners in a business sharing profits and
losses in the ratio of 2 : 2 : 1 respectively. Their balance sheet as on March 31,
2007 was as follows:

Mr. Pammy died on September 30, 2007. The partnership deed provided the
following:
(i) The deceased partner will be entitled to his share of profit up to the date of
death calculated on the basis of previous year’s profit.
(ii) He will be entitled to his share of goodwill of the firm calculated on the
basis of 3 years’ purchase of average of last 4 years’ profit. The profits for
the last four financial years are given below:
for 2003–04; Rs. 80,000; for 2004–05, Rs. 50,000; for 2005–06, Rs. 40,000;
for 2006–07, Rs. 30,000.
The drawings of the deceased partner up to the date of death amounted to
Rs. 10,000. Interest on capital is to be allowed at 12% per annum.
Surviving partners agreed that Rs. 15,400 should be paid to the executors
immediately and the balance in four equal yearly instalments with interest
at 12% p.a. on outstanding balance.
Show Mr. Pammy’s Capital account, his Executor’s account till the settlement
of the amount due

(v) Calculation of Instalments Amount Due to Pammy’s Executor after paying some amount in cash was Rs.60,000 (= 75,400 -15,400) divided in 4 equal instalments (i.e., 60,000/4 = Rs. 15,000 each).

Question 9. Following is the Balance Sheet of Prateek, Rockey and Kushal as on March 31, 2007.

Rockey died on June 30, 2007. Under the terms of the partnership deed, the executors of a deceased partner were entitled to
(a) Amount standing to the credit of the Partner’s capital account.
(b) Interest on capital at 5% per annum.
(c) Share of goodwill on the basis of twice the average of the past three years’ profit.
(d) Share of profit from the closing date of the last financial year to
the date of death on the basis of last year’s profit.
Profits for the year ending on March 31, 2005, March 31, 2006 and March 31, 2007 were Rs. 12,000, Rs. 16,000 and Rs.14,000 respectively. Profits were shared in the ratio of capitals.
Pass the necessary journal entries and draw up Rockey’s capital account to be rendered to his executor.

10. Narang, Suri and Bajaj are partners in a firm sharing profits and losses in
proportion of 1 2 , 1 6 and 1 3 respectively. The Balance Sheet on April 1, 2007
was as follows:

Bajaj retires from the business and the partners agree to the following
(a) Freehold premises and stock are to be appreciated by 20% and 15% respectively.
(b) Machinery and furniture are to be depreciated by 10% and 7% respectively.
(c) Bad Debts reserve is to be increased to Rs. 1,500.
(d) Goodwill is valued at Rs. 21,000 on Bajaj’s retirement.
(e) The continuing partners have decided to adjust their capitals in their new profit sharing ratio after retirement of Bajaj. Surplus/deficit, if any, in their capital accounts will be adjusted through current accounts.

11. The Balance Sheet of Rajesh, Pramod and Nishant who were sharing profits in
proportion to their capitals stood as on March 31, 2007:

Pramod retired on the date of Balance Sheet and the following adjustments were made
(a) Stock was valued at 10% less than the book value.
(b) Factory buildings were appreciated by 12%.
(c) Reserve for doubtful debts be created up to 5%.
(d) Reserve for legal charges to be made at Rs. 265.
(e) The goodwill of the firm be fixed at Rs. 10,000.
(f) The capital of the new firm be fixed at Rs. 30,000. The continuing partners decide to keep their capitals in the new profit sharing ratio of 3 : 2.
Pass journal entries and prepare the balance sheet of the reconstituted firm after transferring the balance in Pramod’s capital account to his loan account.

Note (i) According to point (f) of the question the capital of remaining partner’s should beRs. 18.000and Rs. 12,000 respectively as the firms capital is fixed at?
Rs.30,000 and their share of profit is 3 : 2.
(ii) To match the answer of total of Balance sheet given in the question the difference of remaining partner’s capital (surplus or deficit) is to be adjusted with cash deposit or withdraw. In this situation the Balance sheet will change as given below

12. Following is the Balance Sheet of Jain, Gupta and Malik as on March 31, 2002.

The partners have been sharing profits in the ratio of 5:3:2. Malik decides to
retire from business on April 1, 2002 and his share in the business is to be
calculated as per the following terms of revaluation of assets and liabilities :
Stock, Rs.20,000; Office furniture, Rs.14,250; Plant and Machinery Rs.23,530;
Land and Building Rs.20,000.
A provision of Rs.1,700 to be created for doubtful debts. The goodwill of the firm
is valued at Rs.9,000.
The continuing partners agreed to pay Rs.16,500 as cash on retirement of
Malik, to be contributed by continuing partners in the ratio of 3:2. The balance
in the capital account of Malik will be treated as loan.
Prepare Revaluation account, capital accounts, and Balance Sheet of the
reconstituted firm.

13. Arti, Bharti and Seema are partners sharing profits in the proportion of 3:2:1
and their Balance Sheet as on March 31, 2003 stood as follows :

Bharti died on June 12, 2003 and according to the deed of the said partnership,
her executors are entitled to be paid as under :
(a) The capital to her credit at the time of her death and interest thereon @
10% per annum.
(b) Her proportionate share of reserve fund.
(c) Her share of profits for the intervening period will be based on the sales
during that period, which were calculated as Rs.1,00,000. The rate of profit
during past three years had been 10% on sales.
(d) Goodwill according to her share of profit to be calculated by taking twice
the amount of the average profit of the last three years less 20%. The profits
of the previous years were :
2001 – Rs.8,200
2002 – Rs.9,000
2003 – Rs.9,800
The investments were sold for Rs.16,200 and her executors were paid out. Pass
the necessary journal entries and write the account of the executors of Bharti.

14. Nithya, Sathya and Mithya were partners sharing profits and losses in the
ratio of 5:3:2. Their Balance Sheet as on December 31, 2002 was as follows :

Mithya dies on May 1, 2002. The agreement between the executors of Mithya
and the partners stated that :
(a) Goodwill of the firm be valued at 2.1/2 times the average profits of last four
years. The profits of four years were : in 1998, Rs.13,000; in 1999, Rs.12,000; in 2000, Rs.16,000; and in 2001, Rs.15,000.
(b) The patents are to be valued at Rs.8,000, Machinery at Rs.25,000 and
Premises at Rs.25,000.
(c) The share of profit of Mithya should be calculated on the basis of the profit
of 2002.
(d) Rs.4,200 should be paid immediately and the balance should be paid in 4
equal half-yearly instalments carrying interest @ 10%.
Record the necessary journal entries to give effect to the above and write the
executor’s account till the amount is fully paid. Also prepare the Balance Sheet
of Nithya and Sathya as it would appear on May 1, 2002 after giving effect to
the adjustments

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‘ज्ञापन’ शब्द में ‘वि’ उपसर्ग लगाने से ‘विज्ञापन’ बना है। इसका अर्थ है-जानकारी देना। वर्तमान समय में उत्पादकों द्वारा अपनी वस्तुओं को बेचने के लिए विज्ञापनों का जमकर प्रयोग किया जाता है। इन विज्ञापनों की भाषा आकर्षक और अतिशयोक्तिपूर्ण होती है जो कि लोगों पर जादू-सा असर करती है। किशोर और बच्चे उन्हीं वस्तुओं का प्रयोग करना चाहते हैं जिनका वे विज्ञापन देखते हैं। अब तो विज्ञापन हमारे खान-पान और रहन-सहन को बुरी तरह प्रभावित करता है।

उद्देश्य – विज्ञापन का उद्देश्य है-जानकारी पहुँचाना। इससे लोगों का ज्ञान बढ़ता है। उनके सामने चुनाव के विकल्प, गुण, मूल्य परखने की सुविधा सरलता से उपलब्ध हो जाती है। इससे विक्रेता भी लाभ कमाते हैं। विज्ञापन की सहायता से कम से कम खर्च में अधिकाधिक लोगों तक अपनी बात पहुँचाई जाती है। यही कारण है कि आज समाचार-पत्र, पत्रिकाएँ, दूरदर्शन के विभिन्न कार्यक्रम पोस्टर, बनैर यहाँ तक कि दीवारें भी रंगी नज़र आती हैं।

विज्ञापन लेखन में ध्यान देने योग्य बातें –

• विज्ञापन की भाषा आकर्षक और तुकबंदी युक्त होनी चाहिए।
• शब्द ऐसे होने चाहिए जो कम से कम होने पर अधिकाधिक अर्थ की अभिव्यक्ति करें।
• विज्ञापित वस्तु का चित्र साफ़ और स्पष्ट होना चाहिए।
• विज्ञापन में छूट, स्टॉक सीमित, जल्दी करें जैसे शब्द अवश्य होने चाहिए।
• विज्ञापन बड़े शब्दों में लिखा जाना चाहिए ताकि दूर से पढ़ा जा सके।
• विज्ञापन में चित्र रंगीन होने चाहिए।

विज्ञापन-लेखन के कुछ उदाहरण

1. आधुनिक सुविधाओं से युक्त किसी मोबाइल फ़ोन का विज्ञापन तैयार कीजिए।

2. रचना पेंसिलों के लिए विज्ञापन तैयार कीजिए।

3. सपना कलमों की बिक्री बढ़ाने हेतु विज्ञापन तैयार कीजिए।

4. मध्य प्रदेश राज्य पर्यटन निगम की ओर से पर्यटन को बढ़ावा देने के लिए एक विज्ञापन तैयार कीजिए।

5. राजस्थान पर्यटन निगम की ओर से पर्यटकों को आकर्षित करने हेतु विज्ञापन तैयार कीजिए।

6. उत्तर प्रदेश पर्यटन विभाग की ओर से यात्रियों को आकर्षित करने हेतु एक विज्ञापन तैयार कीजिए।

7. आधुनिक सुविधाओं वाले किसी होटल का विज्ञापन तैयार कीजिए।

8. आधुनिक सुविधाओं से युक्त लक्जरी कार का विज्ञापन तैयार कीजिए।

9. समीर ए०सी०. बनाने वाली कंपनी के लिए विज्ञापन तैयार कीजिए।

10. शिक्षार्थी कोचिंग सेंटर के लिए एक विज्ञापन तैयार कीजिए।

11. उत्तम बस्ते बनाने वाली कंपनी के लिए विज्ञापन तैयार कीजिए।

12. सफल बीज उत्पादन बनाने वाली कंपनी के लिए विज्ञापन तैयार कीजिए।

13. ‘हरियाली’ पौधशाला (नर्सरी) के लिए एक विज्ञापन तैयार कीजिए।

14. ‘खुशहाली’ मोमबत्तियाँ बनाने वाली कंपनी के लिए एक विज्ञापन तैयार कीजिए।

15. ‘निर्मल’ साबुन बनाने वाली कंपनी के लिए एक विज्ञापन तैयार कीजिए।

16. ‘पूजा’ धूप एवं अगरबत्तियाँ बनाने वाली कंपनी के लिए विज्ञापन तैयार कीजिए।

17. ममता क्रेच के लिए विज्ञापन तैयार कीजिए।

18. प्रकाश एल०ई०डी० बनाने वाली कंपनी बिक्री बढ़ाना चाहती है। उसके लिए विज्ञापन तैयार कीजिए।

19. ‘केश पुकार’ तेल बनाने वाली कंपनी के लिए विज्ञापन तैयार कीजिए।

20. ‘सुमन’ सूट विक्रेता कंपनी को बिक्री बढ़ाने हेतु विज्ञापन तैयार कीजिए।

21. ‘विकास’ छाता विक्रेता कंपनी को बिक्री बढ़ाने हेतु विज्ञापन तैयार कीजिए।

22. ‘एक्सलेंट’ लैपटाप बनाने वाली कंपनी के लिए विज्ञापन तैयार कीजिए।

NCERT Solutions for Class 10 Hindi 

NCERT Solutions Class 7 Science Chapter 9 Soil

Topics and Sub Topics in Class 7 Science Chapter 9 Soil:

Section Name	Topic Name
9	Soil
9.1	Soil teeming with life
9.2	Soil profile
9.3	Soil types
9.4	Properties of soil
9.5	Moisture in soil
9.6	Absorptions of water by soil
9.7	Soil and crops

Q1. Tick the most suitable answer in question 1 and 2.
In addition to the rock particles, the soil contains
(i) Air and water
(ii) Water and plants
(iii) Minerals, organic matter, air and water
(iv) Water, air and plants
Answer:
In addition to the rock particles, the soil contains
(iii) Minerals, organic matter, air and water.

Q2. The water holding capacity is the highest in
(i) Sandy soil
(ii) Clayey soil
(iii) Loamy soil
(iv) Mixture of sand and loam
Answer:
The water holding capacity is the highest in
(ii) Clayey soil

Q3. Match the items in Column I with those in Column II:

Column I	Column II
(i) A home for living organisms	(a) Large particles
(ii) Upper layer of the soil	(b) All kinds of soil
(iii) Sandy soil	(c) Dark in colour
(iv) Middle layer of the soil	(d) Small particles and packed tight
(v) Clayey soil	(e) Lesser amount of humus

Answer:

Column I	Column II
(i) A home for living organisms	(b) All kinds of soil
(ii) Upper layer of the soil	(c) Dark in colour
(iii) Sandy soil	(a) Large particles
(iv) Middle layer of the soil	(e) Lesser amount of humus
(v) Clayey soil	(d) Small particles and packed tight

Q4. Explain how soil is formed.
Answer:
Soil is formed through the process of weathering. Weathering is a process of physical breakdown and chemical decomposition of rocks and minerals near or at the surface of the earth. This physical and chemical decomposition is primarily done by wind, water, and climate. As a result of these processes, large rock pieces are converted into smaller pieces and eventually to soil.

Q5. How is clayey soil useful for crops?
Answer:
Following are the properties of clayey soil:

1. It has very good water holding capacity.
2. It is rich in organic matter.

For growing crops such as wheat, gram, and paddy, the soil that is good at retaining water and rich in organic matter is suitable. Therefore, clayey soils having these characteristics are useful for such kind of crops.

Q6. List the differences between clayey soil and sandy soil.
Answer:

Clayey Soil	Loamy Soil
(i) It has much smaller particles.	(i) It has much larger particles.
(ii) It can hold good amount of water.	(ii) It cannot hold water.
(iii) It is fertile.	(iii) It is not fertile.
(iv) Air content is low.	(iv) Air get trapped between the particles.
(iv) Particles are tightly packed	(iv) Particles are loosely packed
(iv) Good for growing various crops.	(iv) Not suitable for growing crops.

Q7. Sketch the cross section of soil and label the various layers.
Answer:

Q8. Razia conducted an experiment in the field related to the rate of percolation. She observed that it took 40 min for 200 mL of water to percolate through the soil sample. Calculate the rate of percolation.
Answer:

9. Explain how soil pollution and soil erosion could be prevented.
Answer:
Prevention of soil pollution:
The persistent build-up of toxic compounds in the soil is defined as soil pollution. To prevent soil pollution, its causes must be controlled.

1. Reduce the use of plastics: Plastics and polythene bags destroy the fertility of soil. Hence, these should be disposed off properly and if possible, their use should be avoided.
2. Industrial pollutants: Some waste products from industries and homes pollute soil. These pollutants should be treated chemically to make them harmless before they are disposed off.
3. Insecticides: Other pollutants of soil include pesticides and insecticides. Therefore, excessive use of these substances should be avoided.

Prevention of soil erosion:
Removal of the upper fertile layer of the soil (top soil) by strong wind and flowing water is known as soil erosion. Following steps can be taken to reduce soil erosion:

1. Mass awareness to reduce deforestation for industrial purposes.
2. Helping local people to regenerate degrading forest.
3. Planting trees.

10. Solve the following crossword puzzle with the clues given:

Across
2. Plantation prevents it.
5. Use should be banned to avoid soil pollution.
6. Type of soil used for making pottery.
7. Living organism in the soil.

Down
1. In desert soil erosion occurs through.
3. Clay and loam are suitable for cereals like.
4. This type of soil can hold very little water.
5. Collective name for layers of soil.

Answer:

Across
2. Plantation prevents it. → Erosion
5. Use should be banned to avoid soil pollution. → Polythene
6. Type of soil used for making pottery. → Clay
7. Living organism in the soil. → Earthworm

Down
1. In desert soil erosion occurs through. → Wind
3. Clay and loam are suitable for cereals like. → Wheat
4. This type of soil can hold very little water. → Sandy
5. Collective name for layers of soil. → Profile

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NCERT Solutions For Class 6 Maths Practical Geometry Exercise 14.2

NCERT Solutions For Class 6 Maths Chapter 14 Practical Geometry Ex 14.2

Exercise 14.2

Ex 14.2 Class 6 Maths Question 1.
Draw a line segment of length 7.3 cm using ruler.
Solution:
Step I: Mark at point P.
Step II : Place the O mark of the ruler against the point P.
Step III : Mark a point Q at a distance of 7.3 cm from P.
Step IV : Join P and Q.

Thus \(\overline { PQ }\) is the line segment of length 7.3 cm.

Ex 14.2 Class 6 Maths Question 2.
Construct a line segment of length 5.6 cm using ruler and compass.
Solution:
Step I: Draw any line L of suitable lengths.
Step II : Place the needle of the compass on the zero mark of the ruler and open it upto 5.6 mark.
Step III : Place the needle at any point A at the line and draw an arc to cut l at B.

Thus, \(\overline { AB }\) is the required line segment of length 5.6 cm.

Ex 14.2 Class 6 Maths Question 3.
Construct \(\overline { AB }\) of length 7.8 cm. From this, cut off \(\overline { AC }\) of length 4.7 cm. Measure \(\overline { BC }\) .
Solution:
Given that \(\overline { AB }\) = 7.8 cm and \(\overline { AC }\) = 4.7 cm.
Step I : Place zero mark of the ruler at A.
Step II : Mark a point B at a distance of 7.8 cm from A.
Step III : Mark another point C at a distance of 4.7 cm from A such that AC = 4.7 cm.
Step IV : On measuring the length of BC, we find that \(\overline { BC }\) = 3.1 cm.

Ex 14.2 Class 6 Maths Question 4.
Given \(\overline { AB }\) of length 3.9 cm. Construct \(\overline { PQ }\) such that the length of \(\overline { PQ }\) is twice that of \(\overline { AB }\). Verify by measurement.

(Hint : Construct \(\overline { PX }\) such that the length of \(\overline { PX }\) = length of \(\overline { AB }\) then cut off \(\overline { XQ }\) such that \(\overline { XQ }\) also has the length of \(\overline { AB }\).
Solution:
Step I: Draw a line l of suitable length.
Step II: Draw \(\overline { AB }\) = 3.9 cm
Step III: From the line, construct \(\overline { PX }\) = \(\overline { AB }\) = 3.9 cm.
Step IV: Again construct \(\overline { XQ }\)= \(\overline { AB }\) =3.9 cm
Verification: \(\overline { PX }\) + \(\overline { XQ }\) = \(\overline { AB }\) + \(\overline { AB }\)

Thus twice of \(\overline { AB }\) is equal to \(\overline { PQ }\)

Ex 14.2 Class 6 Maths Question 5.
Given \(\overline { AB }\) of length 7.3 cm and \(\overline { CD }\) of length 3.4 cm, construct a line segment \(\overline { XY }\) such that the length of XY is equal to the difference between the length of \(\overline { AB }\) and \(\overline { CD }\) . Verify the measurement.
Solution:
Step I : Construct \(\overline { AB }\) = 7.3 cm and \(\overline { CD }\) = 3.4 cm.

Step II: Take a point P on the given line l.
Step III: Construct \(\overline { PR }\) such that \(\overline { PR }\) = \(\overline { AB }\) = \(\overline { AB }\) = 7.3 cm.
Step IV:Construct \(\overline { RQ }\) = \(\overline { CD }\) = 3.4 cm such that PQ = \(\overline { AB }\) – \(\overline { CD }\) .
Verification : On measuring, we observe that \(\overline { PQ }\) = 3.9 cm = 7.3 cm 3.4 cm.
= \(\overline { AB }\) – \(\overline { CD }\)
Thus, \(\overline { PQ }\) = \(\overline { AB }\) – \(\overline { CD }\).

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