ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations MCQS

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations MCQS

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations MCQS

More Exercises

Choose the correct answer from the given four options (1 to 5) :

Question 1.
If x ∈ { – 3, – 1, 0, 1, 3, 5}, then the solution set of the inequation 3x – 2 ≤ 8 is
(a) { – 3, – 1, 1, 3}
(b) { – 3, – 1, 0, 1, 3}
(c) { – 3, – 2, – 1, 0, 1, 2, 3}
(d) { – 3, – 2, – 1, 0, 1, 2}
Solution:
x ∈ { -3, -1, 0, 1, 3, 5}
3x – 2 ≤ 8
⇒ 3x ≤ 8 + 2
⇒ 3x ≤ 10
⇒ x ≤ \(\\ \frac { 10 }{ 3 } \)
⇒ x < \(3 \frac { 1 }{ 3 } \)
Solution set = { -3, -1, 0, 1, 3} (b)

Question 2.
If x ∈ W, then the solution set of the inequation 3x + 11 ≥ x + 8 is
(a) { – 2, – 1, 0, 1, 2, …}
(b) { – 1, 0, 1, 2, …}
(c) {0, 1, 2, 3, …}
(d) {x : x∈R,x≥\(– \frac { 3 }{ 2 } \)}
Solution:
x ∈ W
3x + 11 ≥ x + 8
⇒ 3x – x ≥ 8 – 11
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations MCQS Q2.1

Question 3.
If x ∈ W, then the solution set of the inequation 5 – 4x ≤ 2 – 3x is
(a) {…, – 2, – 1, 0, 1, 2, 3}
(b) {1, 2, 3}
(c) {0, 1, 2, 3}
(d) {x : x ∈ R, x ≤ 3}
Solution:
x ∈ W
5 – 4x < 2 – 3x
⇒ 5 – 2 ≤ 3x + 4x
⇒ 3 ≤ x
Solution set = {0, 1, 2, 3,} (c)

Question 4.
If x ∈ I, then the solution set of the inequation 1 < 3x + 5 ≤ 11 is
(a) { – 1, 0, 1, 2}
(b) { – 2, – 1, 0, 1}
(c) { – 1, 0, 1}
(d) {x : x ∈ R, \(– \frac { 4 }{ 3 } \) < x ≤ 2}
Solution:
x ∈ I
1 < 3x + 5 ≤ 11
⇒ 1 < 3x + 5
⇒ 1 – 5 < 3x
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations MCQS Q4.1

Question 5.
If x ∈ R, the solution set of 6 ≤ – 3 (2x – 4) < 12 is
(a) {x : x ∈ R, 0 < x ≤ 1}
(b) {x : x ∈ R, 0 ≤ x < 1}
(c) {0, 1}
(d) none of these
Solution:
x ∈ R
6 ≤ – 3(2x – 4) < 12
⇒ 6 ≤ – 3(2x – 4)
⇒ 6 ≤ – 6x + 12
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations MCQS Q5.1

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HOTS Questions for Class 10 Science Chapter 5 Periodic Classification of Elements

HOTS Questions for Class 10 Science Chapter 5 Periodic Classification of Elements

Question 1.
Atoms of eight elements A, B, C, D, E, F, G and H have the same number of electronic shells but different number of electrons in their outermost shell. It was found that elements A and G combine to form an ionic compound. This compound is added in a small amount to almost all vegetable dishes during cooking Oxides of elements A and B are basic in nature while those of E and F are acidic. The oxide of D is almost neutral. Based on the above information answer the following questions :

  1. To which group or period of the periodic table do the listed elements belong ?
  2. What would be the nature of compound formed by a combination of elements B and F ?
  3. Which two elements could definitely be metals ?
  4. Which one of the eight elements is most likely to be found in gaseous state at room temperature ?
  5. If the number of electrons in the outermost shell of elements C and G be 3 and 7 respectively, write the formula of the compound formed by the combination of C and G.

Answer:
The clue for the answer is given by the compound which is added to almost all vegetables during cooking. It is sodium chloride (NaCl). Both these elements belong to third period :

  1. All the elements belong to third period. They have three shells (K, L, M) and the number of electrons vary from 1 to 8. These belong to different groups which are listed.
    HOTS Questions for Class 10 Science Chapter 5 Periodic Classification of Elements image - 1
  2. The element ‘B’ is Ca and ‘F’ is S. They combine to form CaS. It is an ionic compound.
  3. The elements A, B are definite metals because the elements present in group 1 and 2 are all metals.
  4. The element H is a noble gas element. It is most likely to be found in the gaseous state at room temperature.
  5. The electronic configuration of the element C(Z = 13) is 2, 8, 3 while that of G(Z = 17) is 2, 8, 7. The formula of the compound formed by their combination is CG3. It is actually AlCl3 and is formed as a result of electron sharing.
    HOTS Questions for Class 10 Science Chapter 5 Periodic Classification of Elements image - 2

More Resources

Question 2.
The following table shows the position of six elements A, B, C, D, E and F in the period table.

Groups

Periods

I 2 3 to 12 13 14 15 16 17

18

2.

A B C
3. D E

F

Using the above table answer the following questions :
(a) Which element will form only covalent compounds ?
(b) Which element is a metal with valency 2 ?
(c) Which element is a non-metal with valency of 3 ?
(d) Out of D and E, which one has more atomic radius and why ?
(e) Write a common name for the family of elements C and F.
Answer:
(a) The element ‘E’ present in group 14 is a non-metal. Its name is silicon (Si) and the compounds of the element are only covalent.
(b) The element ‘D’ present in group 2 is a metal known as magnesium (Mg). It exhibits valency 2 in its compounds.
(c) The elements ‘B’ present in group 15 is a non-metal. It is nitrogen (N) and exhibits valency 3 in its compounds.
(d) The element ‘D’ has more atomic radius than the element ‘E’ as the atomic size decreases along a period.
(e) The elements ‘C’ and ‘F’ present in group 18 belong to a family known as noble gases.

Question 3.
Two elements X and Y belong to group 1 and 2 respectively in the same period. Compare them with respect to :
(a) the number of valence electrons
(b) valency
(c) metallic character
(d) size of the atoms
(e) formulae of their oxides and chlorides.
Answer:
(a) The valence electrons present in element X (group 1) and element Y (group 2) are 1 and 2 respectively.
(b) The valency of the element X is one while that of the element Y is two.
(c) Metallic character decreases along a period. This means that the element X is more metallic as compared to element Y.
(d) Atomic size decreases along a period. As a result, the element Y has a smaller size than the element X.
(e) For element X : oxide (X2O), chloride (XCl).
For element Y : oxide (YO) and chloride (YCl2).

Question 4.
Atoms of seven elements A, B, C, D, E, F and G have a different number of electronic shells but have the same number of electrons in their outermost shells. The elements A and C combine with chlorine to form an acid and common salt respectively. The oxide of element A is liquid at room temperature and is a neutral substance while the oxides of the remaining six elements are basic in nature. Based on the above information, answer the following questions given ahead :

  1. What could the element A be ?
  2. Will elements A to G belong to the same period or same group of the periodic table ?
  3. Write the formula of the compound formed by the reaction of the element A with oxygen.
  4. Show the formation of the compound by a combination of element C with chlorine with the help of electronic structure.
  5. What would be the ratio of number of combining atoms in a compound formed by the combination element A with carbon ?
  6. Which one of the given elements is likely to have the smallest atomic radius ?

Answer:
Since all the seven elements have same number of electrons in their outer most shells, this means that they belong to the same group. From the available information, it becomes clear that the group is of alkali metals (group 1). Hydrogen (A) is also included. The elements present are;
HOTS Questions for Class 10 Science Chapter 5 Periodic Classification of Elements image - 3
Let us answer the questions asked

  1. The element A’ is hydrogen (H)
  2. The elements A’ to ‘G’ belong to the same group and not same period
  3. The formula of compound is H2O
  4. Compound between ‘C’ and chlorine (Cl) is sodium chloride (NaCl) formed by the transference of one electron.
    HOTS Questions for Class 10 Science Chapter 5 Periodic Classification of Elements image - 4
  5. The compound formed between A (H) and carbon is methane (CH4). In this the two elements are present in the ratio of 1 : 4
  6. The element A (hydrogen) has the smallest atomic radius since it is the first element of the group.

 

Hope given HOTS Questions for Class 10 Science Chapter 5 Periodic Classification of Elements helpful to you.

If you have any doubts, please comment below. We try to provide online math tutoring for you.

NCERT Exemplar Solutions for Class 10 Science Chapter 2 Acids, Bases and Salts

NCERT Exemplar Solutions for Class 10 Science Chapter 2 Acids, Bases and Salts

These Solutions are part of NCERT Exemplar Solutions for Class 10 Science. Here we have given NCERT Exemplar Solutions for Class 10 Science Chapter 2 Acids, Bases and Salts

NCERT Exemplar Solutions for Class 10 Science Chapter 2 Multiple Choice Questions

Question 1.
What happens when a solution of an acid is mixed with a solution of a base in a test tube ?
(i) The temperature of the solution increases
(ii) The temperature of the solution decreases
(iii) The temperature of the solution remains the same
(iv) Salt formation takes place
(a) (i) only
(b) (i) and (iii)
(c) (ii) and (iii)
(d) (i) and (iv)
Answer:
(d). Salt formation takes place in the neutralisation reaction. It is always exothermic and temperature increases.

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Question 2.
An aqueous solution turns red litmus solution blue. Excess addition of which of the following solutions would reverse the change ?
(a) Baking powder
(b) Lime
(c) Ammonium hydroxide solution
(d) Hydrochloric acid
Answer:
(d). The aqueous solution is of basic nature since red litmus changes to blue. The reaction can be reversed (acidic solution) by adding excess of hydrochloric acid.

Question 3.
During the preparation of hydrogen chloride gas on a humid day, the gas is usually passed through the guard tube containing anhydrous calcium chloride. The role of anhydrous calcium chloride taken in the guard tube is to
(a) absorb the evolved gas
(b) moisten the gas
(c) absorb moisture from the gas
(d) absorb Cl ions from the evolved gas
Answer:
(c). Anhydrous CaCl2 absorbs moisture to keep the gas in dry state. Otherwise, it will dissolve in moisture to form hydrochloric acid.

Question 4.
Which of the following salts does not contain any water of crystallisation ?
(a) Blue vitriol
(b) Baking soda
(c) Washing soda
(d) Gypsum
Answer:
(b). Baking soda (NaHCO3) does not contain any water of crystallisation.

Question 5.
Sodium carbonate is a basic salt because it is a salt of
(a) strong acid and strong base
(b) weak acid and weak base
(c) strong acid and weak base
(d) weak acid and strong base
Answer:
(d). Na2CO3 is a salt of weak acid (H2CO3) and strong base (NaOH).

Question 6.
Calcium phosphate is present in tooth enamel. Its nature is
(a) basic
(b) acidic
(c) neutral
(d) amphoteric
Answer:
(a). Ca3(PO4)2 is a salt of strong base Ca(OH)2 and weak acid H3PO4.

Question 7.
A sample of soil is mixed with water and allowed to settle. The clear supernatant solution turns the pH paper yellowish-orange. Which of the following would change the colour of this pH paper to greenish-blue ?
(a) Lemon juice
(b) Vinegar
(c) Common salt
(d) An antacid
Answer:
(d). The colour of the pH paper signifies that the solution is somewhat acidic. In order to change it to greenish-blue, we need an antacid.

Question 8.
Which of the following gives the correct increasing order of acidic strength ?
(a) Water
(b) Water
(c) Acetic acid
(d) Hydrochloric acid Water Acetic acid
Answer:
(a).

Question 9.
If a few drops of a concentrated acid accidentally spill over the hand of a student, what should be done ?
(a) Wash the hand with saline solution
(b) Wash the hand immediately with plenty of water and apply a paste of sodium hydrogen carbonate
(c) After washing hand with plenty of water, apply solution of sodium hydroxide on the hand
(d) Neutralise the acid with a strong alkali
Answer:
(b). Washing the hand initially with plenty of water gives partial relief from burning sensation. The paste of sodium hydrogen carbonate completely neutralises the effect of the acid.

Question 10.
Sodium hydrogen carbonate when added to acetic acid evolves a gas. Which of the following statements are true about the gas evolved ?
(i) It turns lime water milky
(ii) It extinguishes a burning splinter
(iii) It dissolves in a solution of sodium hydroxide
(iv) It has a pungent odour
(a) (i) and (ii)
(b) (i), (ii) and (iii)
(c) (ii), (iii) and (iv)
(d) (i) and (iv)
Answer:
(b). The gas evolved is carbon dioxide (CO2). The statements (i), (ii) and (iii) are true about the gas.

Question 11.
Common salt besides being used in kitchen can also be used as the raw material for making
(i) washing soda
(ii) bleaching powder
(iii) baking soda
(iv) slaked lime
(a) (i) and (ii)
(b) (i), (ii) and (iv)
(c) (i) and (iii)
(d) (i), (iii) and (iv)
Answer:
(c).

Question 12.
One of the constituents of baking powder is sodium hydrogen carbonate. The other constituent is :
(a) hydrochloric acid
(b) tartaric acid
(c) acetic acid
(d) sulphuric acid
Answer:
(b). Tartaric acid is the other consument.

Question 13.
To protect tooth decay, we are advised to brush our teeth regularly. The nature of the tooth paste commonly used is
(a) acidic
(b) neutral
(c) basic
(d) corrosive
Answer:
(c). The basic ingredient in the paste will neutralise any acid released from the sugary’ items which we eat.

Question 14.
Which of the following statements is correct about an aqueous solution of an acid and of a base ?
(i) Higher the pH, stronger the acid
(ii) Higher the pH, weaker the acid
(iii) Lower the pH, stronger the base
(iv) Lower the pH, weaker the base
(a) (i) and (iii)
(b) (ii) and (iii)
(c) (i) and (iv)
(d) (ii) and (iv)
Answer:
(d). Statements (ii) and (iv) are correct.

Question 15.
The pH of the gastric juices released during digestion is
(a) less than 7
(b) more than 7
(c) equal to 7
(d) equal to 0
Answer:
(a). Gastric juices generally release hydrochloric acid during digestion. Therefore the pH is less than 7.

Question 16.
Which of the following phenomena occur when a small amount of acid is added to water ?
(i) Ionisation
(ii) Neutralisation
(iii) Dilution
(iv) Salt formation
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iii)
(d) (ii) and (iv)
Answer:
(b). Water helps in the ionisation of acid and also in its dilution.

Question 17.
Which one of the following can be used as an acid- base indicator by a visually impared student ?
(a) Litmus
(b) Turmeric
(c) Vanilla essence
(d) Petunia leaves
Answer:
(c). Vanilla essence is an olfactory inidicator.

Question 18.
Which of the following substances will not give carbon dioxide on treatment with dilute acid ?
(a) Marble
(b) Lime stone
(c) Baking soda
(d) Lime
Answer:
(d). Since lime (CaO) does not contain carbon, it will not give any carbon dioxide gas.

Question 19.
Which of the following is acidic in nature ?
(a) Lime juice
(b) Human blood
(c) Lime water
(d) Antacid
Answer:
(a). Lime juice is of acidic nature. It contains citric acid.

Question 20.
In an attempt to demonstrate electrical conductivity through an electrolyte, the apparatus set up is given. Which among the following statement(s) is(are) correct ?
NCERT Exemplar Solutions for Class 10 Science Chapter 2 Acids, Bases and Salts image - 1
(i) Bulb will not glow because electrolyte is not acidic
(ii) Bulb will glow because HCl is a strong acid and furnishes ions for conduction.
(iii) Bulb will not glow because circuit is incomplete
(iv) Bulb will not glow because it depends upon the type of electrolytic solution
(a) (i) and (iii)
(b) (ii) and (iv)
(c) (ii) only
(d) (iv) only
Answer:
(c). Bulb will glow because the acid will furnish ions which conduct electricity.

Question 21.
Which of the following is used for dissolution of gold ?
(a) Hydrochloric acid
(b) Sulphuric acid
(c) Nitric acid
(d) Aqua regia.
Answer:
(d). For the details of aqua regia.

Question 22.
Which of the following is not a mineral acid ?
(a) Hydrochloric acid
(b) Citric acid
(c) Sulphuric acid
(d) Nitric acid.
Answer:
(b). Citric acid is an organic acid present in citrus fruits. It is not a mineral acid.

Question 23.
Which of the following is not a base ?
(a) NaOH
(b) KOH
(c) NH4OH
(d) C2H5OH.
Answer:
(d). Ethyl alcohol (C2H5OH) is not a base. It is an alcohol and is very weakly acidic in nature.

Question 24.
Which of the following statements is not correct ?
(a) All metal carbonates react with acid to give a salt, water and carbon dioxide
(b) All metal oxides react with water to give salt and acid
(c) Some metals react with acids to give salt and hydrogen
(d) Some non metal oxides react with water to form an acid
Answer:
(b). The statement is not correct as the metal oxide reacts with water to form metal hydroxide.

Question 25.
Match the chemical substances given in Column (A) with their appropriate application given in Column (B)

Column (A)   

Column (B)

(1) Bleaching powder

(i) Constituent of glass

(2) Baking soda

(ii) Production of H2 and Cl?
(3) Borax

(iii) Decolourisation

(4) Sodium chloride

(iv) Antacid

(a) 1—(ii), 2—(i), 3—(iv), 4—(iii)
(b) 1—(iii), 2—(ii), 3—(iv), 4—(i)
(c) 1—(iii), 2—(iv), 3—(i), 4—(ii)
(d) 1—(ii), 2—(iv), 3—(i), 4—(iii)
Answer:
(c).

Question 26.
Equal volumes of hydrochloric acid and sodium hydroxide solutions of same concentration are mixed and the pH of the resulting solution is checked with a pH paper. What would be the colour obtained ?
NCERT Exemplar Solutions for Class 10 Science Chapter 2 Acids, Bases and Salts image - 2
(a) Red
(b) Yellow
(c) Green
(d) Blue
Answer:
(c). pH paper will acqure green colour which indicates that the solution is of neutral nature.

Question 27.
Which of the following is(are) true when HCl (g) is passed through water ?
(i) It does not ionise in the solution as it is a covalent compound.
(ii) It ionises in the solution
(iii) It gives both hydrogen and hydroxyl ions in the solution
(iv) It forms hydronium ion in the solution due to the combination of hydrogen ion with water molecule
(a) (i) only
(b) (iii) only
(c) (ii) and (iv)
(d) (iii) and (iv)
Answer:
(c). Both are true.

Question 28.
Which of the following statements is true for acids ?
(a) Bitter and change red litmus to blue
(b) Sour and change red litmus to blue
(c) Sour and change blue litmus to red
(d) Bitter and change blue litmus to red
Answer:
(c). Both are the characteristics of acids.

Question 29.
Which of the following are present in a dilute aqueous solution of hydrochloric acid ?
(a) H3O+ + Cl
(b) H3O+ + OH
(c) Cl + OH
(d) unionised HCl
Answer:
(a). HCl will react with water as follows :
HCl (l) + H2O(l) ———> H3O+(aq) + Cl(aq)

Question 30.
Identify the correct representation of reaction occurring during chloralkali process
(а) 2NaCl(l) + 2H2O(l) ———–>2NaOH(s) + Cl2(g) + H2(g)
(b) 2NaCI(aq) + 2H2O (aq) ———–> 2NaOH(aq) + Cl2(g) + H2 (aq)
(c) 2NaCl(aq) + 2H2O(l) ———–> 2NaOH(aq) + Cl2(aq) + H2(g)
(d) 2NaCl (aq) + 2H2O (l) ———–> 2NaOH (aq) + Cl2(g) + H2(g)
Answer:
(d). It is the correct answer since the physical states of all the species involved are correct.

NCERT Exemplar Solutions for Class 10 Science Chapter 2 Short Answer Questions

Question 31.
Match the acids given in column (A) with their correct sources given in column (B)

Column (A)

Column (B)

 (a) Lactic acid

(i)          Tamarind

(b) Acetic acid

(ii)         Lemon
(c) Citric acid

(iii)       Vinegar

(d) Tartaric acid

(iv)        Curd

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

Question 32.
Match the important chemicals given in Column (A) with the chemical formulae given in Column (B)

Column (A)

Column (B)

(a) Plaster of Paris

(i) Ca(OH)2

(b) Gypsum

(ii)         CaSO4.½ H2O
(c) Bleaching Powder

(iii)       CaS04.2H7O

(d) Tartaric acid

(iv)        CaOCl2

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

Question 33.
What will be the action of the following substances on blue litmus paper ?
Answer:
Dry HCl gas, Moistened NH3 gas, Lemon juice, Carbonated soft drink, Curd, Soap solution.
It will become red, will remain blue, will become red, will become red, will become red, will remain blue.

Question 34.
Name the acid present in ant sting and give its chemical formula. Also give the common method to get relief from the discomfort caused by the ant sting.
Answer:
The acid present in ant sting is methanoic acid or formic acid. Its chemical formula is HCOOH. Being poisonous in nature, it causes pain and irritation. Even blisters may appear on the stung area. One should immediately rub the stung area with a mild base like baking soda (NaHCO3). It will react with formic acid to form salt and water. Its poisonous effect will be completely neutralised.
NCERT Exemplar Solutions for Class 10 Science Chapter 2 Acids, Bases and Salts image - 3
Remember : We should never apply sodium hydroxide pellets or solution directly or the affected portion. Being strongly corrosive in nature, it will aggrievate pain by forming more blisters. However, sodium hydrogen carbonate can be applied.

Question 35.
What happens when egg shell is placed in concentrated nitric acid taken in a beaker ?
Answer:
Egg shell contains calcium carbonate as its main constituent. When dipped in concentrated nitric acid, calcium carbonate reacts to evolve carbon dioxide. As a result, egg shell slowly dissolves.
NCERT Exemplar Solutions for Class 10 Science Chapter 2 Acids, Bases and Salts image - 4

Question 36.
A student prepares solutions of (i) an acid and (ii) a base in two separate beakers. She forgets to label the solutions and litmus paper is not available in the laboratory. Since both the solutions are colourless, how will she distinguish between the two ?
Answer:
Phenolphthalein and methyl orange are also acid-base indicators. They can be used in place of litmus.
Procedure : Transfer a portion of the colourless solutions to two glass tubes. Add one or two drops of henolphthalein indicator to these. An acid solution will remain colourless while the solution of base will ecome pink. Now, repeat the experiment with methyl orange indicator. In acid solution, the indicator will become reddish while in base, it will be yellowish.

Question 37.
How would you distinguish between baking soda and washing soda upon heating ?
Answer:
Baking soda is sodium hydrogen carbonate (NaHCO3). Upon heating, it will evolve CO2 gas which upon passing through lime water, will make it milky.
NCERT Exemplar Solutions for Class 10 Science Chapter 2 Acids, Bases and Salts image - 5
Washing soda is sodium carbonate decahydrate (Na2CO3.10H2O). Upon heating, it will not evolve any gas.
NCERT Exemplar Solutions for Class 10 Science Chapter 2 Acids, Bases and Salts image - 6

Question 38.
Salt ‘A’ commonly used in bakery products on heating gets converted into another salt ‘B which itself is used for the removal of hardness of water and a gas ‘C’ is evolved. The gas ‘C’ when passed through lime water, turns it milky. Identify A, B and C.
Answer:
The salt ‘A’ is sodium hydrogen carbonate (baking soda) and is commonly used in bakeries as a constitutent of baking power. Upon heating, it changes to sodium carbonate ‘B’ and evolves carbon dioxide gas ‘C’.
NCERT Exemplar Solutions for Class 10 Science Chapter 2 Acids, Bases and Salts image - 7
Sodium carbonate removes hardness from water while CO2 gas turns lime water milky.

Question 39.
In one of the industrial processes used for the manufacture of sodium hydroxide, a gas ‘X’ is formed as by-product. The gas ‘X’ reacts with lime water to give a compound ‘Y’ which is used as a bleaching agent in chemical industry. Identify ‘X’ and ‘Y’ giving the chemical equation of the reactions involved.
Answer:
Sodium hydroxide is manufactured by the electrolysis of a strong solution of sodium chloride (called brine). As a result, chlorine (X) is evolved at anode while hydrogen at cathode. Chlorine reacts with lime water containing slaked lime to form bleaching power (Y)
NCERT Exemplar Solutions for Class 10 Science Chapter 2 Acids, Bases and Salts image - 8

Question 40.
Fill in the missing data in the following table
NCERT Exemplar Solutions for Class 10 Science Chapter 2 Acids, Bases and Salts image - 9
Answer:
NCERT Exemplar Solutions for Class 10 Science Chapter 2 Acids, Bases and Salts image - 10

Question 41.
What are strong and weak acids ? In the following list of acids, separate strong acids from weak acids.
Hydrochloric acid, citric acid, acetic acid, nitric acid, formic acid, sulphuric acid.
Answer:
For the definitions of strong and weak acids and bases,
An acid may be defined as a substance which releases one or more H+ ions in aqueous solution. These ions exist as hydronium (H2O+) ions.
A base may be defined as a substance capable of releasing one or more OH ions in aqueous solution.
In general, mineral acids are strong acids while organic acids are weak. From the available list :
Strong acids : Hydrochloric acid, nitric acid, sulphuric acid.
Weak acids : Citric acid, acetic acid, formic acid.

Question 42.
When zinc metal is treated with a dilute solution of a strong acid, a gas is evolved which is utilised in the hydrogenation of oils. Name the gas evolved. Write the chemical equation of the reaction involved and also write a test to detect the gas formed.
Answer:
The gas evoloved is hydrogen. The gas burns brightly with a pop sound when a burning splinter is brought in its contact. This causes the hydrogenation of edible liquid oils to form solid fats also called Vanaspati ghee.
NCERT Exemplar Solutions for Class 10 Science Chapter 2 Acids, Bases and Salts image - 11

NCERT Exemplar Solutions for Class 10 Science Chapter 2 Long Answer Questions

Question 43.
In the following schematic diagram for the preparation of hydrogen gas as shown in the figure, what would happen if following changes are made ?
NCERT Exemplar Solutions for Class 10 Science Chapter 2 Acids, Bases and Salts image - 12
(a) In place of zinc granules, same amount of zinc dust is taken in the test tube
(b) Instead of dilute sulphuric acid, dilute hydrochloric acid is taken
(c) In place of zinc, copper turnings are taken
(d) Sodium hydroxide is taken in place of dilute sulphuric acid and the tube is heated.
Answer:
(a) Hydrogen gas would evolve at a greater speed because zinc dust provides more surface area as compared to zinc granules.
(b) Both would react in the same way. There would be no effect on the volume of the gas evolved.
(c) Copper does not react with either dilute HCl or dilute H2SO4. No gas would evolve in both the cases.
(d) Hydrogen gas would evolve in this case also.
NCERT Exemplar Solutions for Class 10 Science Chapter 2 Acids, Bases and Salts image - 13

Question 44.
For making cake, baking powder is taken. If at home your mother uses baking soda instead of baking powder in cake,
(a) how will it affect the taste of the cake and why ?
(b) how can baking soda be converted into baking powder ?
(c) what is the role of tartaric acid added to baking soda ?
Answer:
Baking powder is a mixture of baking soda and tartaric acid. Out of the two, only baking soda is actually used for making bread or cake fluffy. The role of tartaric acid is to neutralise sodium carbonate formed in the reaction.
(a) If baking soda is used for making cake in place of baking powder, then cake will taste bitter since there is no tartaric acid available to neutralise sodium carbonate formed.
NCERT Exemplar Solutions for Class 10 Science Chapter 2 Acids, Bases and Salts image - 14
(b) Baking soda can be converted to baking powder by adding appropriate amount of tartaric acid.
(c) Tartaric acid as pointed above will react with sodium carbonate which makes cake bitter. This means that the cake will not taste bitter.

Question 45.
A metal carbonate (X) on reacting with an acid gives a gas which when passed through a solution (Y) gives the carbonate back. On the other hand, a gas (G) that is obtained at anode during electrolysis of brine is passed on dry substance (Y). It gives a compound (Z), used for disinfecting drinking water. Identity X, Y, G and Z.
Answer:
The gas (G) obtained at anode during the electrolysis of brine is chlorine. The compound (Z) used for disinfecting drinking water is bleaching powder. It is formed on reacting chlorine with dry slaked lime i.e., Ca(OH)2. It is denoted as ‘Y’ This means that the metal carbonate ‘X’ is calcium carbonate. Upon heating, it evolves CO2 gas which gives back ‘X’ on reacting with calcium hydroxide. The chemical reactions involved are listed :
NCERT Exemplar Solutions for Class 10 Science Chapter 2 Acids, Bases and Salts image - 15

Question 46.
A dry pellet of a common base ‘B’, when kept in open absorbs moisture and turns sticky. The compound is also formed by chloralkali process. Identify ‘B’. What type of reaction occurs when B is treated with an acidic oxide ? Write a balanced chemical equation for one such solution.
Answer:
The available information suggests that the base ‘B’ is sodium hydroxide (NaOH). It is a deliquescent substance and becomes sticky on absorbing moisture from atmosphere. It is commercially formed by the electrolysis of a strong solution of sodium chloride (brine).
It reacts with an acidic oxide such as CO2 or SO, gas to form corresponding salt and water. For example,
NCERT Exemplar Solutions for Class 10 Science Chapter 2 Acids, Bases and Salts image - 16

Question 47.
A sulphate salt of Group 2 element of the Periodic Table is a white, soft substance which can be moulded into different shapes by making its dough. When this compound is left in open for some time, it becomes a solid mass and cannot be used for moulding purposes. Identify the sulphate salt. Why does it show such a behaviour ? Give the reaction involved.
Answer:
The available information suggests that the element present in group 2 of the Periodic Table* is calcium (Ca) and the sulphate salt (white in colour) is Plaster of Paris. It can be moulded into different shapes by making its dough with water. When left in the open, Plaster of Paris changes into Gypsum which is a solid mass quite hard in nature and can no longer be used for moulding.
NCERT Exemplar Solutions for Class 10 Science Chapter 2 Acids, Bases and Salts image - 17

Question 48.
Identify the compound ‘Y’ on the basis of the reactions given below. Also write the name and chemical formulae of A, B and C.
NCERT Exemplar Solutions for Class 10 Science Chapter 2 Acids, Bases and Salts image - 18
Answer:
The compound ‘Y’ is sodium hydroxide. It forms A, B and C as follows :
NCERT Exemplar Solutions for Class 10 Science Chapter 2 Acids, Bases and Salts image - 19

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NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals

NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals

These Solutions are part of NCERT Exemplar Solutions for Class 10 Science. Here we have given NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals

NCERT Exemplar Solutions for Class 10 Science Chapter 3 Multiple Choice Questions

Question 1.
Which of the following properties is generally not shown by metals ?
(a) Electrical conduction
(b) Sonorous nature
(c) Dullness
(d) Ductility.
Answer:
(c). Metals have shining lustre and they mostly do not show any dullness.

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Question 2.
The ability of metals to be drawn into thin sheets is known as
(a) ductility
(b) malleability
(c) sonorosity
(d) conductivity.
Answer:
(b).

Question 3.
Aluminium is used for making cooking utensils. Which of the following properties of the metal are responsible for the same ?
(i) Good thermal conductivity
(ii) Good electrical conductivity
(iii) Ductility
(iv) High melting point
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iii)
(d) (i) and (iv)
Answer:
(d). In order that a metal may be used for making cooking utensils, it must be good conductor of heat and also must have high melting point.

Question 4.
Which one of the following metals does not react with cold as well as hot water ?
(a) K
(b) Ca
(c) Mg
(d) Fe.
Answer:
(d).

Question 5.
Which of the following oxide(s) of iron would be obtained on prolonged reaction of iron with steam ?
(a) FeO
(b) Fe2O3
(c) Fe3O4
(d) Fe2O3 and Fe3O4
Answer:
(c).
3Fe (s) + 4H2O (g) ———> Fe3O4 (s) + 4H2 (g)

Question 6.
What happens when magnesium is treated with water ?
(i) It does not react with water
(ii) It reacts violently with water
(iii) It reacts less violently with water
(iv) Bubbles of hydrogen gas formed stick to the surface of the metal
(a) (i) and (iv)
(b) (ii) and (iii)
(c) (i) and (ii)
(d) (iii) and (iv)
Answer:
(d).
(iii) Magnesium reacts with water slowly to form magnesium hydroxide along with hydrogen gas
Mg(s) + 2H2O(l) ———-> Mg(OH)2(s) + H2(g)
(iv) Bubbles of hydrogen gas stick to the surface of the metal.

Question 7.
Generally metals react with acids to give salt and hydrogen gas. Which of the following acids does not give hydrogen gas on reacting with metals (except Mn and Mg) ?
(a) H2SO4
(b) HCl
(c) HNO3
(d) All of these.
Answer:
(c). HNO3 is a strong oxidising agent. It oxidises metals to metal oxides and then to metal nitrates and itself is reduced to either nitric oxide (NO), nitruos oxide (N2O) or nitrogen dioxide (NO2). However, both Mg and Mn evolve hydrogen on reacting with the acid.

Question 8.
Composition of aqua-regia by volume is :
(a) Dil HCl (3) : Cone HNO3 (1)
(b) Cone HCl (3) : Dil HNO3 (1)
(c) Cone HCl (3) : Cone HNO3 (1)
(d) Dil HC1 (3) : Dil HNO3 (1)
Answer:
(c). is the correct answer.

Question 9.
Which of the following are not ionic compounds ?
(i) CaCl2
(ii) HCl
(iii) CCl4
(iv) NaCl
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iii)
Answer:
(b). Both HCl and CCl4 are covalent compounds in nature.

Question 10.
Which one of the following properties is not generally exhibited by ionic compounds ?
(a) Solubility in water
(b) Electrical conductivity in solid state
(c) High melting and boiling points
(d) Electrical conductivity in molten state.
Answer:
(b). Ionic compounds are conducting due to the movement of ions. Since the ions do not move in the solid state, these compounds are therefore, not conducting in the solid state.

Question 11.
Which of the following metals exist in their native state in nature ?
(i) Cu
(ii) Au
(iii) Zn
(iv) Ag
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (ii) and (iv)
(d) (iii) and (iv)
Answer:
(c). Both Au and Ag are very little reactive. They exist in native state. However, the metal silver may also exist in some combined states (as minerals).

Question 12.
Metals are refined by using different methods. Which of the following metals are refined by electrolytic refining ?
(i) Ag
(ii) Cu
(iii) Na
(iv) Mg
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iii)
(d) (iii) and (iv)
Answer:
(a). Both Ag and Cu are very hard. These are called coinage metals and are purified by electrolytic refining.

Question 13.
Silver articles become black on prolonged exposure to air. This is due to the formation of
(a) AgCN
(b) Ag2O
(c) Ag2S
(d) Ag2S and AgCN
Answer:
(c). Air contains traces of hydrogen sulphide (H2S) gas which reacts with silver (Ag) to form Ag2S black in colour.

Question 14.
Galvanisation is a method of protecting iron from rusting by coating with a thin layer of
(a) Chromium
(b) Copper
(c) Zinc
(d) Tin
Answer:
(c). Coating the surface of iron with zinc is called galvanization.

Question 15.
Stainless steel is a very useful material for our life. In stainless steel, iron is mixed with
(a) Ni and Cr
(b) Cu and Cr
(c) Ni and Cu
(d) Cu and Au.
Answer:
(a). is the correct answer.

Question 16.
If copper is kept open in air, it slowly loses its shining brown surface and gains a green coating. It is due to the formation of :
(a) CuSO4
(b) CUCO3CO(OH)2
(c) CU(NO3)2
(d) CuO.
Answer:
(b). The green coating is due to the formation of basic copper carbonate which is CuCO3.Cu(OH)2.

Question 17.
Generally, metals are solid in nature. Which one of the following metals is found in liquid state at room temperature ?
(a) Na
(b) Fe
(c) Al
(d) Hg
Answer:
(d). Mercury (Hg) is a liquid at room temperature.

Question 18.
Which of the following metals are obtained by electrolysis of their chlorides in molten state ?
(i) Na
(ii) Ca
(iii) Fe
(iv) Cu
(a) (i) and (iv)
(b) (iii) and (iv)
(c) (i) and (iii)
(d) (i) and (ii)
Answer:
(d). Both Na and Ca are obtained by the electrolysis of their molten chlorides.

Question 19.
Generally non-metals are not lustrous. Which of the following non-metals is lustrous ?
(a) Sulphur
(b) Phosphorus
(c) Nitrogen
(d) Iodine.
Answer:
(d). Iodine is lustrous.

Question 20.
Which one of the following four metals would be displaced from the solution of its salts by other three metals ?
(a) Mg
(b) Cu
(c) Zn
(d) Fe
Answer:
(b). Cu is the least reactive and would be displaced from its salts by other metals.

Question 21.
5 mL each of concentrated HCl, HNO3 and a mixture of concentrated HCl (15 mL) and concentrated HNO3 (5 mL) were taken in test tubes labelled as A, B and C. A small piece of metal was put in each test tube. No change occurred in test tubes A and B but the metal got dissolved in test tube C. The metal could be
(a) Al
(b) Au
(c) Cu
(d) Ag
Answer:
(b). Gold (Au) dissolves in aqua regia.

Question 22.
An alloy is :
(a) an element
(b) a compound
(c) a homogeneous mixture
(d) a heterogeneous mixture
Answer:
(c). An alloy is a homogeneous mixture of two or more elements (metals or non-metals.)

Question 23.
An electrolytic cell consists of :
(i) positively charged cathode
(ii) negatively charged anode
(iii) positively charged anode
(iv) negatively charged cathode
(a) (i) and (ii)
(b) (iii) and (iv)
(c) (i) and (iii)
(d) (ii) and (iv)
Answer:
(b).

Question 24.
During electrolytic refining of copper, ir gets
(a) deposited on cathode
(b) deposited on anode
(c) deposited on cathode as well as anode
(d) remains in the solution.
Answer:
(a). Pure metal is always deposited on cathode.

Question 25.
An element ‘X’ is soft and can be cut with a knife. This is very reactive to air and cannot be kept in open air. It reacts vigorously with water. Identify the element from the following
(a) Mg
(b) Na
(c) S
(d) Mg
Answer:
(b). Sodium (Na) gives all the characteristics of the element X.

Question 26.
Alloys are homogeneous mixtures of a metal with a metal or non-metal. Which among the following alloys contain non-metal as one of its constituents ?
(a) Brass
(b) Gun metal
(c) Amalgam
(d) Steel.
Answer:
(d). Steel (stainless) contains very small percentage of carbon.

Question 27.
Which of the following statements is not correct for magnesium metal ?
(a) It burns in oxygen with dazzling flame
(b) It reacts with cold water to form magnesium oxide and evolves hydrogen gas
(c) It reacts with hot water to form magnesium hydroxide and evolves hydrogen gas
(d) It reacts with steam to form magnesium hydroxide and evolves hydrogen gas.
Answer:
(b). Magnesium (Mg) does not react with cold water since its position in the reactivity series of metals is not at the top.

Question 28.
Which among the following alloys contain mercury as one of its constituents ?
(a) Stainless steel
(b) German silver
(c) Solder
(d) Zinc amalgam.
Answer:
(d). All amalgams contain mercury as one of their constituents.

Question 29.
Reaction between the elements X and Y results in the compound Z. Whereas X loses electron, Y gains the same. Which of the following properties is not shown by Z ?
(a) Has high melting point
(b) Has low melting point
(c) Conducts electricity in molten state
(d) Occurs as solid
Answer:
(b). Z is an ionic compound. Therefore, it is not expected to have low melting point.

Question 30.
The electronic configurations of three elements X, Y and Z are X — 2, 8; Y — 2, 8, 6 and Z — 2, 8, 1. Which of the following is correct ?
(a) X is a metal
(b) Y is a metal
(c) Z is a non-metal
(d) Y is a non-metal and Z is a metal.
Answer:
(d). Y is sulphur (S) while Z is sodium (Na)

Question 31.
Although metals form basic oxides, which of the following metals form an amphoteric oxide ?
(a) Na
(b) Ca
(c) Zn
(d) Cu.
Answer:
(c). Zn forms amphoteric oxide ZnO which reacts with both the acids and the alkalies.

Question 32.
Generally, non-metals are not conductors of electricity. Which of the following is a good conductor of electricity ?
(a) Diamond
(b) Graphite
(c) Phosphorus
(d) Iodine
Answer:
(b). Graphite is a good conductor of electricity.

Question 33.
Electrical wires have a coating of an insulating material. The material, generally used is
(a) Lead
(b) Graphite
(c) PVC
(d) All can be used.
Answer:
(c). PVC (Poly Vinyl Chloride) coatings are generally used for insulating electrical wires.

Question 34.
Which of the following non-metals is a liquid ?
(a) Carbon
(b) Bromine
(c) Iodine
(d) Sulphur
Answer:
(b). Non-metal bromine is a liquid but in the molecular form (Br2).

Question 35.
Which of the following can undergo a chemical reaction ?
(a) MgSO4 + Zn
(b) ZnSO4 + Fe
(c) CaSO4 + Pb
(d) CuSO4 + Al
Answer:
(d). Only Al can displace Cu from CuSO4 solution since it is placed above Cu in the activity series.

Question 36.
Which one of the following figures correctly describes the process of electrolytic refining ?
NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals image - 1
Answer:
(c). Cu2+ ions are released from anode by the oxidation of copper (Cu). These migrate towards cathode and are reduced to copper (Cu).

NCERT Exemplar Solutions for Class 10 Science Chapter 3 Short Answer Questions

Question 37.
Iqbal treated a lustrous, divalent element M with sodium hydroxide. He observed the formation of bubbles in reaction mixture. He made the same observations when this element was treated with hydrochloric acid. Suggest how can he identify the produced gas. Write chemical equations for both the reactions.
Answer:
The divalent element M is a metal. It reacts with both sodium hydroxide (alkali) and dilute hydrochloric acid to evolve hydrogen gas
NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals image - 2
The gas burns with a pop sound when a burning match stick is brought near it.

Question 38.
During extraction of metals, electrolytic refining is used to obtain pure metals,
(a) Which material will be used as anode and cathode for refining of copper metal by this process
(b) Suggest a suitable electrolyte also,
(c) In this electrolytic cell, where do we get pure copper after passing electric current ?
Answer:
In the electrolytic refining of impure copper metal :
(a) A rod of impure copper is used as anode while that of pure copper as cathode.
(b) The electrolyte is a water soluble salt of copper. It is normally an aqueous solution of copper sulphate.
(c) On passing current, pure copper gets deposited at the cathode. It can be scrapped off later on. For more details.

Question 39.
Why should the metal sulphides and carbonates be converted to metal oxides in the process of extraction of metal from them ?
Answer:
It is quite easy to obtain a metal from metal oxide by carrying its reduction with a suitable reducing agent. However, metal sulphides and carbonates are converted to the oxide form and then reduced.

Question 40.
Generally, when metals are treated with mineral acids, hydrogen gas is liberated but when metals (except Mn and Mg), are treated with HNCfy, hydrogen is not liberated, why ?
Answer:
HNO3 acts as a powerful oxidising agent along with acting as acid. In this case, hydrogen gas which is intially involved is oxidised to water.

Question 41.
Compound A’ and aluminium are used to join railway tracks,
(a) Identify the compound A’
(b) Name the reaction
(c) Write down its reaction with aluminium.
Answer:
Railway trakcs as we all know, are made up of iron. This means that the compound A is an oxide of iron (Fe2CO3). It is reduced by aluminium by thermit reaction with aluminium.
NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals image - 3
The reaction is highly exothermic and as a result, iron is in molten state and can weld the broken railway tracks. The process is also called aluminothermie reduction or thermit process.

Question 42.
When a metal ‘X’ is treated with cold water, it gives a basic salt ‘Y’ with molecular formula XOH (Molecular mass = 56) and liberates a gas ‘Z’ which easily catches fire. Identify X, Y and Z and also write the reaction involved.
Answer:
The atomic mass of metal X = 56 – Mass of OH group = 56 — 17 = 39u. This shows that the metal ‘X’ is potassium (K) and the basic salt ‘Y’ is potassium hydroxide (KOH). It is formed by reacting potassium with cold water. Hydrogen gas ‘Z’ evolved in the reaction catches fire.
NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals image - 4

Question 43.
A non-metal X-exists in two different forms Y and Z. Y is the hardest natural substance, whereas Z is a good conductor of electricity. Identify X, Y and Z.
Answer:
Since “Y” is the hardest occuring non-metal, it is diamond. The non-metal “X” is carbon. Y is the crystalline allotropie form of carbon. The other allotropie form Z which is a good conductor of electricity is graphite.

Question 44.
The following reaction takes place when aluminium powder is heated with Mn02.
NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals image - 5
(a) Is aluminium getting reduced ?
(b) Is Mn02 getting oxidised ?
Answer:
(a) Aluminium is getting oxidised to Al2O3.
(b) Manganese dioxide is getting reduced to Mn.

Question 45.
What are the constituents of solder alloy ? Which property of solder makes it suitable for welding electrical wires ?
Answer:
The constituents of solder alloy are ; lead (Pb) and tin (Sn). Being a low melting solid and also a good conductor of electricity, solder is used for welding electrical wires and cables.

Question 46.
A metal ‘M’, which is used in thermite process, when heated with oxygen gives an oxide, which is amphoteric in nature. Identify ‘M’ and its oxide. Write down the reactions of oxide with HCl and NaOH.
Answer:
The information suggests that the metal ‘M’ is aluminium (Al). The formula of its oxide is Al2O3 which is amphoteric in nature. It reacts with both acid and base.
NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals image - 6

Question 47.
A metal that exists as a liquid at room temperature is obtained by heating its sulphide in the presence of air. Identify the metal and its ore and give the reaction involved.
Answer:
Mercury is the metal which exists as a liquid at room temper nature. It can be obtained by heating the sulphide ore in the presence of air. The process is known as roasting. The chemical reactions involved are :
NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals image - 7

Question 48.
Give the formulae of the stable binary compounds that would be formed by the combination of following pairs of elements.
(a) Ca and N2
(b) Li and O2
(c) Ca and Cl2
(d) K and
Answer:
(a) Calcium nitride (Ca3N2)
(b) Lithium oxide (Li2O)
(c) Calcium chloride (CaCl2)
(d) Potassium oxide (K2O).

Question 49.
What happens when
(a) ZnCO3 is heated in the absence of oxygen ?
(b) a mixture of Cu2O and Cu2S is heated ?
Answer:
(a) A mixture of ZnO(s) and CO2 is formed
NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals image - 8
(b) A mixture of copper and SO2 is formed.
NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals image - 9

Question 50.
A non-metal A is an important constituent of our food and forms two oxides B and C. Oxide B is toxic whereas C causes global warming
(a) Identify A, B and C
(b) To which Group of Periodic Table does A belong ?
Answer:
The non-metal A is carbon. It is an important constituent of our food in different forms. For example, glucose (C6H12O6) contains carbon. In fact, all food materials are organic compounds and these contain carbon as essential constituent. The two oxides of carbon are, carbon monoxide (B) and carbon dioxide (C). Carbon dioxide causes global warming.
(a) A = Carbon (C) ;
B = Carbon monoxide (CO) ;
C = Carbon dioxide (CO2)
(b) Carbon is the first member of group 14 in the Long form of Periodic Table.

Question 51.
Give two examples each of the metals that are good conductors and poor conductors of heat respectively.
Answer:
Good conductors of heat : Ag and Cu Poor conductors of heat : Pb and Hg.

Question 52.
Name one metal and one non-metal that exist in liquid state at room temperature. Also name two metals having melting point less than 310 K (37°C).
Answer:
The metal mercury (Hg) and non-metal (Br2) exist in liquid state at room temperature.
The two metals with melting points less than 310 K are ; cesium (Cs) with melting point 301 K and gallium (Ga) with melting point 303 K.

Question 53.
An element ‘A’ reacts with water to form a compound ‘B’ which is used in white washing. The compound ‘B’ on heating forms an oxide ‘C’ which on treatment with water gives back ‘B’. Identify ‘A’, ‘B’ and ‘C’ and give the reactions involved.
Answer:
Since the compound ‘B’ is used for white wash, it is calcium hydroxide and the element ‘A’ is calcium. Upon heating ‘B’ forms calcium oxide ‘C’. It reacts with water to give calcium hydroxide ‘B’ again.
NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals image - 10

Question 54.
An alkali metal A gives a compound B (molecular mass = 56) on reacting with water. The compound ‘B’ gives a soluble compound ‘C’ on treatment with aluminium oxide. Identify A, B and C and give the reaction involved.
Answer:
As explained under Question 42, the alkali metal A is potassium (K) and on reacting with water, it forms a compound ‘B’ which is potassium hydroxide. This upon reacting with aluminium oxide (Al2O3) forms potassium metaluminate ‘C’.
NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals image - 11

Question 55.
Give the reaction involved during extraction of zinc from its ore by
(a) roasting of zinc ore
(b) calcination of zinc ore
Answer:
NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals image - 12

Question 56.
A metal M does not liberate hydrogen from acids but reacts with oxygen to give a black coloured product. Identify M and black coloured product and also explain the reaction of M with oxygen.
Answer:
The avilable information suggests that the metal M is copper (Cu). It does not liberate hydrogen since it is placed below hydrogen in the reactivity series. Copper reacts with oxygen upon heating to form cupric oxide (CuO) which is black in colour.
NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals image - 13

Question 57.
An element forms an oxide A2O3 which is acidic in nature. Identify A as a metal or non-metal.
Answer:
Since the oxide A2O3 is acidic in nature, the element A is non-metal.

Question 58.
A solution of CuSO4 was kept in an iron pot. After few days, the iron pot was found to have a number of holes in it. Explain the reason in terms of reactivity. Write the equation of the reaction involved.
Answer:
Iron (Fe) is placed above copper (Cu) in the reactivity series. This means that a chemical reaction had occurred between iron (iron pot) and aqueous CuSO4 solution.
NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals image - 14
Since iron was consumed in the reaction, a number of holes appeared in the pot.

NCERT Exemplar Solutions for Class 10 Science Chapter 3 Long Answer Questions

Question 59.
A non-metal A which is the largest constituent of air, when heated with H2 in 1 : 3 ratio in the presence of catalyst (Fe) gives a gas B. On heating with O2 it gives an oxide C. If this oxide is passed into water in the presence of air, it gives an acid D which acts as a strong oxidising agent.
(a) Identify A, B, C and D.
(b) To which group of periodic table does this non-metal belong ?
Answer:
(a) The available information suggests that the non-metal A is nitrogen (N) and its molecular form is N2. It is the major constituent of air (about 79 per cent by volume). It reacts with H2 upon heating m the presence of iron catalyst (Fe) to form ammonia (NH3) gas(g)
NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals image - 15
Upon heating with oxygen, nitrogen forms intially nitric oxide (NO) and then nitrogen dioxide (C). The latter reacts with water in the presence of oxygen to form nitric acid (D). The acid acts as strong oxidising agent.
NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals image - 16
(b) The element nitrogen is the first member of group 15 of Modern periodic table.

Question 60.
Give the steps involved in the extraction of metals of low and medium reactivity from their respective sulphide ores.
Answer:
Extraction of Metals present low in the Activity series
Silver (Ag), gold (Au) and platinum (Pt) generally occur in the free or native state. This means that they can be isolated rather easily. Metals like copper (Cu) and mercury (Hg) are comparitively more reactive and occur in combined states. For example, the ore of mercury is cinnabar (HgS) while that of copper is copper glance (Cu2S). Both are converted into metallic form upon heating in air or oxygen.
NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals image - 17
Extraction of Metals present in the middle of the Activity series
The metals present in the middle of the series are zinc, iron, chromium, nickel, cobalt, lead etc. These are usually present as sulphides or carbonares in nature. However, it is quite easy to obtain a metal from its oxide form which is then reduced to the metallic state. The various steps involved in the process of extraction are briefly discussed.
(a) Calcination :
Calcination may be defined as the process of heating the ore below its melting point in the absence of air.
As a result of calcination, the following changes take place

  • Moisture is driven out and the ore becomes dry.
    NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals image - 18
  • Some hydrated ores decompose and become anhydrous by losing molecules of water of crystallisation. For example, heat
    NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals image - 19

(b) Roasting :
Roasting may be defined as the process of heating the ore below its melting point with excess of air.
As a result of roasting, the following changes occur :

  • Any organic matter if present, gets destroyed.
  • Impurities of non-metals such as sulphur, arsenic or phosphorus are converted into their volatile oxides which are removed. For example,
    NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals image - 20
  • Sulphides of the metals are converted into their oxides. For example,
    NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals image - 21
  • Small amounts of sulphides may also be converted into sulphates as as result of roasting. For example,

NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals image - 22

Question 61.
Explain the following
(a) Reactivity of Al decreases if it is dipped in cone. HNO3
(b) Carbon cannot reduce the oxides of Na or Mg.
(c) NaCl is not a conductor of electricity in solid state whereas it does conduct electricity in aqueous solution as in molten state
(d) Iron articles are galvanised.
(e) Metals like Na, K, Ca and Mg are never found in their free state in nature.
Answer:
(a) When Al metal is dipped in cone. HNO3 for sometime, it is oxidised initially to aluminium oxide (Al2O3). The oxide gets deposited on the surface of the metal and forms a protective coating an the surface. The metal is said to become passive towards air, acids and alkalies. Its reactivity therefore, decreases.
(b) Both Na and Mg are more reactive than carbon. Therefore, carbon is not in a position to reduce the oxides of these metals.
(c) NaCl is an ionic compound. Its electrical conductivity is due to the mobility of Na+ and Cl ions. These ions cannot move in the solid state. However, they can do so either in molten state of the salt or when it forms and aqueous solution in water.
(d) Iron has a tendency to get rusted in atmosphere by reacting with oxygen and water vapours present in air. In order to check rusting, iron articles are generally coated with zinc. This process is known as galvanization.
(e) All these metals are placed high in reactivity series of metals. They are quite reactive and do not exist in free states.

Question 62.
(i) Given below are the steps lor extraction of copper from its ore. Write the reaction involved.
(a) Roasting of copper (1) sulphide
(b) Reduction of copper (1) oxide with copper (1) sulphide.
(c) Electrolytic refining
(ii) Draw a neat and well labelled diagram for electrolytic refining of copper.
Answer:
NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals image - 23
This reaction in which one of the reactants (Cu2S) carries the reduction of the product (Cu2O) is known as auto-reduction.
(c) Reactions taking place in electrorefensing are :
NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals image - 24
(ii) For the diagram of electro-refining,
NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals image - 25

Question 63.
Of the three metals X, Y and Z, X reacts with cold water, Y with hot water and Z with steam only. Identify X, Y and Z and also arrange them in order of increasing reactivity.
Answer:
The answer is based on the relative positions of the metals in the reactivity series. The reactivity with water decreases down the series.

  • Metal X is Na or K
  • Metal Y is Mg or Ca
  • Metal Z is Al or Fe.

Order of increasing reactivity : Z < Y < X.

Question 64.
Two ores ‘X’ and ‘Y’ were taken. On heating, ore ‘X’ gives CO2 whereas ore ‘Y’ gives SO2. What steps will you take to convert them into respective metals ?
Answer:
Since the ore ‘A’ of the metal gives CO2 upon heating, it is some metal carbonate (MCO3). It can be converted to the metallic form as follows :
NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals image - 26
Since the ore ‘Y’ of the metal gives SO2 upon heating, it can be some metal sulphide (MS). It can be converted to the metallic form as follows :
NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals image - 27

Hope given NCERT Exemplar Solutions for Class 10 Science Chapter 3 Metals and Non-metals are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4

More Exercises

Question 1.
Solve the inequation 3x -11 < 3 where x ∈ {1, 2, 3,……, 10}. Also represent its solution on a number line
Solution:
3x – 11 < 3 => 3x < 3 + 11 => 3x < 14 x < \(\\ \frac { 14 }{ 3 } \)
But x ∈ 6 {1, 2, 3, ……., 10}
Solution set is (1, 2, 3, 4}
Ans. Solution set on number line
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q1.1

Question 2.
Solve 2(x – 3)< 1, x ∈ {1, 2, 3, …. 10}
Solution:
2(x – 3) < 1 => x – 3 < \(\\ \frac { 1 }{ 2 } \) => x < \(\\ \frac { 1 }{ 2 } \) + 3 => x < \(3 \frac { 1 }{ 2 } \)
But x ∈ {1, 2, 3 …..10}
Solution set = {1, 2, 3} Ans.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q2.1

Question 3.
Solve : 5 – 4x > 2 – 3x, x ∈ W. Also represent its solution on the number line.
Solution:
5 – 4x > 2 – 3x
– 4x + 3x > 2 – 5
=> – x > – 3
=> x < 3
x ∈ w,
solution set {0, 1, 2}
Solution set on Number Line :
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q3.1

Question 4.
List the solution set of 30 – 4 (2.x – 1) < 30, given that x is a positive integer.
Solution:
30 – 4 (2x – 1) < 30
30 – 8x + 4 < 30
– 8x < 30 – 30 – 4
– 8x < – 4 x > \(\\ \frac { -4 }{ -8 } \)
=> x > \(\\ \frac { 1 }{ 2 } \)
x is a positive integer
x = {1, 2, 3, 4…..} Ans.

Question 5.
Solve : 2 (x – 2) < 3x – 2, x ∈ { – 3, – 2, – 1, 0, 1, 2, 3} .
Solution:
2(x – 2) < 3x – 2
=> 2x – 4 < 3x – 2
=> 2x – 3x < – 2 + 4
=> – x < 2
=> x > – 2
Solution set = { – 1, 0, 1, 2, 3} Ans.

Question 6.
If x is a negative integer, find the solution set of \(\\ \frac { 2 }{ 3 } \)+\(\\ \frac { 1 }{ 3 } \) (x + 1) > 0.
Solution:
\(\\ \frac { 2 }{ 3 } \)+\(\\ \frac { 1 }{ 3 } \) x + \(\\ \frac { 1 }{ 3 } \) > 0
=> \(\\ \frac { 1 }{ 3 } \) x + 1 > 0
=> \(\\ \frac { 1 }{ 3 } \) x > – 1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q6.1
x is a negative integer
Solution set = {- 2, – 1} Ans.

Question 7.
Solve: \(\\ \frac { 2x-3 }{ 4 } \)≥\(\\ \frac { 1 }{ 2 } \), x ∈ {0, 1, 2,…,8}
Solution:
\(\\ \frac { 2x-3 }{ 4 } \)≥\(\\ \frac { 1 }{ 2 } \)
=> 2x – 3 ≥ \(\\ \frac { 4 }{ 2 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q7.1

Question 8.
Solve x – 3 (2 + x) > 2 (3x – 1), x ∈ { – 3, – 2, – 1, 0, 1, 2, 3}. Also represent its solution on the number line.
Solution:
x – 3 (2 + x) > 2 (3x – 1)
=> x – 6 – 3x > 6x – 2
=> x – 3x – 6x > – 2 + 6
=> – 8x > 4
=> x < \(\\ \frac { -4 }{ 8 } \) => x < \(– \frac { 1 }{ 2 } \)
x ∈ { – 3, – 2, – 1, 0, 1, 2}
.’. Solution set = { – 3, – 2, – 1}
Solution set on Number Line :
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q8.1

Question 9.
Given x ∈ {1, 2, 3, 4, 5, 6, 7, 9} solve x – 3 < 2x – 1.
Solution:
x – 3 < 2x – 1
x – 2x < – 1 + 3 => – x < 2 x > – 2
But x ∈ {1, 2, 3, 4, 5, 6, 7, 9}
Solution set = {1, 2, 3, 4, 5, 6, 7, 9} Ans.

Question 10.
Given A = {x : x ∈ I, – 4 ≤ x ≤ 4}, solve 2x – 3 < 3 where x has the domain A Graph the solution set on the number line.
Solution:
2x – 3 < 3 => 2x < 3 + 3 => 2x < 6 => x < 3
But x has the domain A = {x : x ∈ I – 4 ≤ x ≤ 4}
Solution set = { – 4, – 3, – 2, – 1, 0, 1, 2}
Solution set on Number line :
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q10.1

Question 11.
List the solution set of the inequation
\(\\ \frac { 1 }{ 2 } \) + 8x > 5x \(– \frac { 3 }{ 2 } \), x ∈ Z
Solution:
\(\\ \frac { 1 }{ 2 } \) +8x > 5x \(– \frac { 3 }{ 2 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q11.1

Question 12.
List the solution set of \(\\ \frac { 11-2x }{ 5 } \) ≥ \(\\ \frac { 9-3x }{ 8 } \) + \(\\ \frac { 3 }{ 4 } \),
x ∈ N
Solution:
\(\\ \frac { 11-2x }{ 5 } \) ≥ \(\\ \frac { 9-3x }{ 8 } \) + \(\\ \frac { 3 }{ 4 } \)
=> 88 – 16x ≥ 45 – 15x + 30
(L.C.M. of 8, 5, 4 = 40}
=> – 16x + 15x ≥ 45 + 30 – 88
=> – x ≥ – 13
=>x ≤ 13
x ≤ N.
Solution set = {1, 2, 3, 4, 5, .. , 13} Ans.

Question 13.
Find the values of x, which satisfy the inequation : \(-2\le \frac { 1 }{ 2 } -\frac { 2x }{ 3 } \le 1\frac { 5 }{ 6 } \), x ∈ N.
Graph the solution set on the number line. (2001)
Solution:
\(-2\le \frac { 1 }{ 2 } -\frac { 2x }{ 3 } \le 1\frac { 5 }{ 6 } \), x ∈ N
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q13.1

Question 14.
If x ∈ W, find the solution set of
\(\frac { 3 }{ 5 } x-\frac { 2x-1 }{ 3 } >1\)
Also graph the solution set on the number line, if possible.
Solution:
\(\frac { 3 }{ 5 } x-\frac { 2x-1 }{ 3 } >1\)
9x – (10x – 5) > 15 (L.C.M. of 5, 3 = 15)
=> 9x – 10x + 5 > 15
=> – x > 15 – 5
=> – x > 10
=> x < – 10
But x ∈ W
Solution set = Φ
Hence it can’t be represented on number line.

Question 15.
Solve:
(i)\(\frac { x }{ 2 } +5\le \frac { x }{ 3 } +6\) where x is a positive odd integer.
(ii)\(\frac { 2x+3 }{ 3 } \ge \frac { 3x-1 }{ 4 } \) where x is positive even integer.
Solution:
(i) \(\frac { x }{ 2 } +5\le \frac { x }{ 3 } +6\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q15.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q15.2

Question 16.
Given that x ∈ I, solve the inequation and graph the solution on the number line :
\(3\ge \frac { x-4 }{ 2 } +\frac { x }{ 3 } \ge 2 \) (2004)
Solution:
\(3\ge \frac { x-4 }{ 2 } +\frac { x }{ 3 } \) and \(3\ge \frac { x-4 }{ 2 } +\frac { x }{ 3 } \ge 2 \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q16.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q16.2

Question 17.
Given x ∈ {1, 2, 3, 4, 5, 6, 7, 9}, find the values of x for which -3 < 2x – 1 < x + 4.
Solution:
-3 < 2x – 1 < x + 4.
=> – 3 < 2x – 1 and 2x – 1 < x + 4
=> – 2x < – 1 + 3 and 2x – x < 4 + 1
=> – 2x < 2 and x < 5
=> – x < 1
=> x > – 1
– 1 < x < 5
x ∈ {1, 2, 3, 4, 5, 6, 7, 9}
Solution set = {1, 2, 3, 4} Ans.

Question 18.
Solve : 1 ≥ 15 – 7x > 2x – 27, x ∈ N
Solution:
1 ≥ 15 – 7x > 2x – 27
1 ≥ 15 – 7x and 15 – 7x > 2x – 27
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q18.1

Question 19.
If x ∈ Z, solve 2 + 4x < 2x – 5 ≤ 3x. Also represent its solution on the number line.
Solution:
2 + 4x < 2x – 5 ≤ 3x
2 + 4x < 2x – 5 and 2x – 5 ≤ 3x => 4x – 2x < – 5 – 2 ,and 2x – 3x ≤ 5
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q19.1

Question 20.
Solve the inequation = 12 + \(1 \frac { 5 }{ 6 } x\) ≤ 5 + 3x, x ∈ R. Represent the solution on a number line. (1999)
Solution:
12 + \(1 \frac { 5 }{ 6 } x\) ≤ 5 + 3x
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q20.1

Question 21.
Solve : \(\\ \frac { 4x-10 }{ 3 } \)≤\(\\ \frac { 5x-7 }{ 2 } \) x ∈ R and represent the solution set on the number line.
Solution:
\(\\ \frac { 4x-10 }{ 3 } \)≤\(\\ \frac { 5x-7 }{ 2 } \)
=> 8x – 20 ≤ 15x – 21
(L.C.M. of 3, 2 = 6)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q21.1

Question 22.
Solve \(\frac { 3x }{ 5 } -\frac { 2x-1 }{ 3 } \) > 1, x ∈ R and represent the solution set on the number line.
Solution:
\(\frac { 3x }{ 5 } -\frac { 2x-1 }{ 3 } \) > 1
=> 9x – (10x – 5) > 15
=> 9x – 10x + 5 > 15
=> – x > 15 – 5
=> – x > 10
=> x < – 10
x ∈ R.
.’. Solution set = {x : x ∈R, x < – 10}
Solution set on the number line
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q22.1

Question 23.
Solve the inequation – 3 ≤ 3 – 2x < 9, x ∈ R. Represent your solution on a number line. (2000)
Solution:
– 3 ≤ 3 – 2x < 9
– 3 ≤ 3 – 2x and 3 – 2x < 9
2x ≤ 3 + 3 and – 2x < 9 – 3
2x ≤ 6 and – 2x < 6 => x ≤ 3 and – x < 3 => x ≤ – 3 and – 3 < x
– 3 < x ≤ 3.
Solution set= {x : x ∈ R, – 3 < x ≤ 3)
Solution on number line
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q23.1

Question 24.
Solve 2 ≤ 2x – 3 ≤ 5, x ∈ R and mark it on number line. (2003)
Solution:
2 ≤ 2x – 3 ≤ 5 .
2 ≤ 2x – 3 and 2x – 3 ≤ 5
2 + 3 ≤ 2x and 2x ≤ 5 + 3
5 ≤ 2x and 2x ≤ 8.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q24.1

Question 25.
Given that x ∈ R, solve the following inequation and graph the solution on the number line: – 1 ≤ 3 + 4x < 23. (2006)
Solution:
We have
– 1 ≤ 3 + 4x < 23 => – 1 – 3 ≤ 4x < 23 – 3 => – 4 ≤ 4x < 20 => – 1 ≤ x < 5, x ∈ R
Solution Set = { – 1 ≤ x < 5; x ∈ R}
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q25.1

Question 26.
Solve tlie following inequation and graph the solution on the number line. (2007)
\(-2\frac { 2 }{ 3 } \le x+\frac { 1 }{ 3 } <3+\frac { 1 }{ 3 } \) x∈R
Solution:
Given \(-2\frac { 2 }{ 3 } \le x+\frac { 1 }{ 3 } <3+\frac { 1 }{ 3 } \) x∈R
\(-\frac { 8 }{ 3 } \le x+\frac { 1 }{ 3 } <\frac { 10 }{ 3 } \)
Multiplying by 3, L.C.M. of fractions, we get
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q26.1

Question 27.
Solve the following inequation and represent the solution set on the number line :
\(-3<-\frac { 1 }{ 2 } -\frac { 2x }{ 3 } \le \frac { 5 }{ 6 } ,x\in R\)
Solution:
\(-3<-\frac { 1 }{ 2 } -\frac { 2x }{ 3 } \le \frac { 5 }{ 6 } ,x\in R\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q27.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q27.2

Question 28.
Solve \(\frac { 2x+1 }{ 2 } +2(3-x)\ge 7,x\in R\). Also graph the solution set on the number line
Solution:
\(\frac { 2x+1 }{ 2 } +2(3-x)\ge 7,x\in R\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q28.1

Question 29.
Solving the following inequation, write the solution set and represent it on the number line. – 3(x – 7)≥15 – 7x > \(\\ \frac { x+1 }{ 3 } \), n ∈R
Solution:
– 3(x – 7)≥15 – 7x > \(\\ \frac { x+1 }{ 3 } \), n ∈R
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q29.1

Question 30.
Solve the inequation :
\(-2\frac { 1 }{ 2 } +2x\le \frac { 4x }{ 3 } \le \frac { 4 }{ 3 } +2x,\quad x\in W\). Graph the solution set on the number line.
Solution:
\(-2\frac { 1 }{ 2 } +2x\le \frac { 4x }{ 3 } \le \frac { 4 }{ 3 } +2x,\quad x\in W\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q30.1

Question 31.
Solve the inequation 2x – 5 ≤ 5x + 4 < 11, where x ∈ I. Also represent the solution set on the number line. (2011)
Solution:
2x – 5 ≤ 5x + 4 < 11 2x – 5 ≤ 5x + 4
=> 2x – 5 – 4 ≤ 5x and 5x + 4 < 11
=> 2x – 9 ≤ 5x and 5x < 11 – 4
and 5x < 7
=> 2x – 5x ≤ 9 and x < \(\\ \frac { 7 }{ 5 } \)
=> 3x > – 9 and x< 1.4
=> x > – 3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q31.1

Question 32.
If x ∈ I, A is the solution set of 2 (x – 1) < 3 x – 1 and B is the solution set of 4x – 3 ≤ 8 + x, find A ∩B.
Solution:
2 (x – 1) < 3 x – 1
2x – 2 < 3x – 1
2x – 3x < – 1 + 2 => – x < 1 x > – 1
Solution set A = {0, 1, 2, 3, ..,.}
4x – 3 ≤ 8 + x
4x – x ≤ 8 + 3
=> 3x ≤ 11
=> x ≤ \(\\ \frac { 11 }{ 3 } \)
Solution set B = {3, 2, 1, 0, – 1…}
A ∩ B = {0, 1, 2, 3} Ans.

Question 33.
If P is the solution set of – 3x + 4 < 2x – 3, x ∈ N and Q is the solution set of 4x – 5 < 12, x ∈ W, find
(i) P ∩ Q
(ii) Q – P.
Solution:
(i) – 3 x + 4 < 2 x – 3
– 3x – 2x < – 3 – 4 => – 5x < – 7
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q33.1

Question 34.
A = {x : 11x – 5 > 7x + 3, x ∈R} and B = {x : 18x – 9 ≥ 15 + 12x, x ∈R}
Find the range of set A ∩ B and represent it on a number line
Solution:
A = {x : 11x – 5 > 7x + 3, x ∈R}
B = {x : 18x – 9 ≥ 15 + 12x, x ∈R}
Now, A = 11x – 5 > 7x + 3
=> 11x – 7x > 3 + 5
=> 4x > 8
=>x > 2, x ∈ R
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q34.1

Question 35.
Given: P {x : 5 < 2x – 1 ≤ 11, x∈R)
Q{x : – 1 ≤ 3 + 4x < 23, x∈I) where
R = (real numbers), I = (integers)
Represent P and Q on number line. Write down the elements of P ∩ Q. (1996)
Solution:
P= {x : 5 < 2x – 1 ≤ 11}
5 < 2x – 1 ≤ 11
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q35.1

Question 36.
If x ∈ I, find the smallest value of x which satisfies the inequation \(2x+\frac { 5 }{ 2 } >\frac { 5x }{ 3 } +2\)
Solution:
\(2x+\frac { 5 }{ 2 } >\frac { 5x }{ 3 } +2\)
=>\(2x-\frac { 5x }{ 3 } >2-\frac { 5 }{ 2 } \)
=>12x – 10x > 12 – 15
=> 2x > – 3
=>\(x>-\frac { 3 }{ 2 } \)
Smallest value of x = – 1 Ans.

Question 37.
Given 20 – 5 x < 5 (x + 8), find the smallest value of x, when
(i) x ∈ I
(ii) x ∈ W
(iii) x ∈ N.
Solution:
20 – 5 x < 5 (x + 8)
⇒ 20 – 5x < 5x + 40
⇒ – 5x – 5x < 40 – 20
⇒ – 10x < 20
⇒ – x < 2
⇒ x > – 2
(i) When x ∈ I, then smallest value = – 1.
(ii) When x ∈ W, then smallest value = 0.
(iii) When x ∈ N, then smallest value = 1. Ans.

Question 38.
Solve the following inequation and represent the solution set on the number line :
\(4x-19<\frac { 3x }{ 5 } -2\le -\frac { 2 }{ 5 } +x,x\in R\)
Solution:
We have
\(4x-19<\frac { 3x }{ 5 } -2\le -\frac { 2 }{ 5 } +x,x\in R\)
Hence, solution set is {x : -4 < x < 5, x ∈ R}
The solution set is represented on the number line as below.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q38.1

Question 39.
Solve the given inequation and graph the solution on the number line :
2y – 3 < y + 1 ≤ 4y + 7; y ∈ R.
Solution:
2y – 3 < y + 1 ≤ 4y + 7; y ∈ R.
(a) 2y – 3 < y + 1
⇒ 2y – y < 1 + 3
⇒ y < 4
⇒ 4 > y ….(i)
(b) y + 1 ≤ 4y + 7
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q39.1

Question 40.
Solve the inequation and represent the solution set on the number line.
\(-3+x\le \frac { 8x }{ 3 } +2\le \frac { 14 }{ 3 } +2x,Where\quad x\in I\)
Solution:
Given : \(-3+x\le \frac { 8x }{ 3 } +2\le \frac { 14 }{ 3 } +2x,Where\quad x\in I\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q40.1

Question 41.
Find the greatest integer which is such that if 7 is added to its double, the resulting number becomes greater than three times the integer.
Solution:
Let the greatest integer = x
According to the condition,
2x + 7 > 3x
⇒ 2x – 3x > – 7
⇒ – x > – 7
⇒ x < 7
Value of x which is greatest = 6 Ans.

Question 42.
One-third of a bamboo pole is buried in mud, one-sixth of it is in water and the part above the water is greater than or equal to 3 metres. Find the length of the shortest pole.
Solution:
Let the length of the shortest pole = x metre
Length of pole which is burried in mud = \(\\ \frac { x }{ 3 } \)
Length of pole which is in the water = \(\\ \frac { x }{ 6 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 Q42.1

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Ex 4 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds

NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds

NCERT Exemplar Solutions for Class 10 Science Chapter 4 Multiple Choice Questions

Question 1.
Carbon exists in the atmosphere in the form of
(a) carbon monoxide only
(b) carbon monoxide and carbon dioxide in traces
(c) carbon dioxide only
(d) coal gas.
Answer:
(b). Both CO and CO2 gases are present in traces in the atmosphere.

More Resources

Question 2.
Which of the following statements are usually correct for carbon compounds ? These
(i) are good conductors of heat and electricity
(ii) are poor conductors of electricity
(iii) have strong forces of attraction between their molecules
(iv) do not have strong forces of attraction between their molecules
(a) (i) and (iii)
(b) (ii) and (iii)
(c) (i) and (iv)
(d) (ii) and (iv)
Answer:
(d). Carbon compounds are of covalent nature. That is why these are poor conductors of electricity and do not have strong forces of attraction.

Question 3.
A molecule of ammonia (NH3) has
(a) only single bonds
(b) only double bonds
(c) only triple bonds
(d) two double bonds and two single bonds.
Answer:
(a). All bonds are single covalent in nature.

Question 4.
Buckminster fullerene is an allotropie form of
(a) nitrogen
(b) sulphur
(c) carbon
(d) tin
Answer:
(c).

Question 5.
Which of the following are correct chain isomers of butane ?
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 1
(a) (i) and (iii)
(b) (ii) and (iv)
(c) (i) and (ii)
(d) (iii) and (iv)
Answer:
(c). Isomer (i) has a straight chain while (ii) is a branched chain isomer.

Question 6.
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 2
In the above reaction, alkaline KMnO4 acts as
(a) reducing agent
(b) oxidising agent
(c) catalyst
(d) dehydrating agent.
Answer:
(b). Alkaline KMnO4 is also known as Baeyer’s reagent. It acts as an oxidising agent.

Question 7.
Oils on treating with hydrogen in the presence of palladium or nickel catalyst form fats. This is an example of
(a) Addition reaction
(b) Substitution reaction
(c) Displacement reaction
(d) Oxidation reaction
Answer:
(a). Oils are unsaturated in nature. By the addition of hydrogen, they become saturated.

Question 8.
In which of the following compounds, —OH is the functional group ?
(a) Butanone
(b) Butanol
(c) Butanoic acid
(d) Butanal
Answer:
(b). —OH is the functional group in butanol which is an alcohol.

Question 9.
The soap molecule has a
(a) hydrophilic head and a hydrophobic tail
(b) hydrophobic head and a hydrophilic tail
(c) hydrophobic head and a hydrophobic tail
(d) hydrophilic head and a hydrophilic tail
Answer:
(a). ‘Head’ is a polar group and is attacked towards H2O molecules of water. ‘Tail’ is a long chain of hydrocarbons and is water repellent.

Question 10.
Which of the following is the correct representation of electron dot structure of nitrogen ?
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 3
Answer:
(d). Both the N atoms have a complete octet after electron sharing.

Question 11.
Structural formula of ethyne is
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 4
Answer:
(a).

Question 12.
Identify the unsaturated compounds from the following
(i) Propane
(ii) Propene
(iii) Propyne
(iv) Chloropropane
(a) (i) and (ii)
(b) (ii) and (iv)
(c) (iii) and (iv)
(d) (ii) and (iii)
Answer:
(d). Both propene and propyne are unsaturated hydrocarbons.

Question 13.
Chlorine reacts with saturated hydrocarbons at room temperature in the
(a) absence of sunlight
(b) presence of sunlight
(c) presence of water
(d) presence of hydrochloric acid
Answer:
(b). These reactions are called photochemical reactions.

Question 14.
In the soap micelles
(a) the ionic end of soap is on the surface of the cluster while the carbon chain is in the interior of the cluster.
(b) ionic end of soap is in the interior of the cluster and the carbon chain is out of the cluster,
(c) both ionic end and carbon chain are in the interior of the cluster
(d) both ionic end and carbon chain are on the exterior of the cluster
Answer:
(a).

Question 15.
Pentane has the molecular formula C5H12. h has
(a) 8 covalent bonds
(b) 10 covalent bonds
(c) 16 covalent bonds
(d) 14 covalent bonds
Answer:
(c).
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 5

Question 16.
Structural formula of benzene is
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 6
Answer:
(c). The carbon atom skeleton in benzene has alternate single and double bonds

Question 17.
Ethanol reacts with sodium and forms two products. These are :
(a) sodium ethanoate and hydrogen
(b) sodium ethanoate and oxygen
(c) sodium ethoxide and hydrogen
(d) sodium ethoxide and oxygen
Answer:
(c).
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 7

Question 18.
The correct structural formula of butanoic acid is
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 8
Answer:
(d).

Question 19.
Vinegar is a solution of
(a) 30% — 40% acetic acid in alcohol
(b) 5% — 8% acetic acid in alcohol
(c) 5% – 8% acetic acid in water
(d) 15% – 20% acetic acid in water
Answer:
(c).

Question 20.
Mineral acids are stronger acids than carboxylic acids because
(i) mineral acids are completely ionised
(ii) carboxylic acids are completely ionised
(iii) mineral acids are partially ionised
(iv) carboxylic acids are partially ionised
(a) (i) and (iv)
(b) (ii) and (iii)
(c) (i) and (ii)
(d) (iii) and (iv)
Answer:
(a). Mineral acids like HCl are completely ionised in solvent like water whereas carboxylic acids such as CH3COOH are only partially ionised.

Question 21.
Carbon forms four covalent bonds by sharing its four valence electrons with four univalent atoms e.g. hydrogen. After the formation of four bonds, carbon attains the electronic configuration of
(a) helium
(b) neon
(c) argon
(d) krypton
Answer:
(b). The compound formed is methane (CH4). In this, carbon atom has a complete octet and configuration of neon which is a noble gas element.

Question 22.
The correct electron dot-structure of a water molecule is
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 9
Answer:
(c).

Question 23.
Which of the following is not a straight chain hydrocarbon ?
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 10
Answer:
(d). Please note that the continuous chains of carbon atoms whether straight or bent are not branched in nature.

Question 24.
Which among the following are unsaturated hydrocarbons ?
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 11
(a) (i) and (iii)
(b) (ii) and (iii)
(c) (ii) and (iv)
(d) (iii) and (iv).
Answer:
(c). Both (ii) and (iv) are unsaturated in nature.

Question 25.
The IUPAC name of the compound
CH3—CH2—CHO is
(a) Propanal
(b) Propanone
(c) Ethanol
(d) Ethanal
Answer:
(a).

Question 26.
The heteroatoms present in CH3—O—CH2—CH2 (Br) are
(i) oxygen
(ii) carbon
(iii) hydrogen
(iv) bromine
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv)
Answer:
(d). Both oxygen (O) and bromine (Br) are heteroatoms. Please remember that apart from C and H atoms, all other atoms present in an organic compound are hetero atoms.

Question 27.
Which of the following does not belong to the same homologous series ?
(a) CH4
(b) C2H6
(c) C3H8
(d) C3H6.
Answer:
(d). It is an alkene while the rest are all alkane molecules in nature.

Question 28.
The first member of alkene family is
(a) ethyne
(b) ethene
(c) propyne
(d) ethane
Answer:
(b).

Question 29.
Which of the following represents saponification
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 12
Answer:
(d). This is an example of saponification reaction. All other options are not correct.

NCERT Exemplar Solutions for Class 10 Science Chapter 4 Short Answer Questions

Question 30.
Draw the electron dot structure of ethyne and also draw its structural formula
Answer:
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 13

Question 31.
Write the names of the following compounds :
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 14
Answer:
(a) Pentanoic acid
(b) But-l-yne
(c) Heptanal
(d) Pentan-l-ol

Question 32.
Identify and name the functional groups present in the following compounds.
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 15
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 16
Answer:
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 17

Question 33.
A compound X is formed by the reaction of a carboxylic acid C2H4O2 and an alcohol in presence of a few drops of H2SO4. The alcohol on oxidation with alkaline KMnO4 followed by acidification gives the same carboxylic acid as used in this reaction. Give the names and structures of (a) carboxylic acid,
(b) alcohol and
(c) the compound X. Also write the reaction. (CBSE Sample Paper 2017)
Answer:
The available information suggests that the alcohol which gives the same carboxylic acid upon oxidation has two carbon atoms. It is, therefore ethanol (C2H5OH). The structures of the different compounds are :
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 18

Question 34.
Why are detergents better cleansing agents than soaps ? Explain.
Answer:
A substance capable of removing grease and dirt from any fabric or body is called detergent. The detergents are of two types Le. soapy and non-soapy detergents. The soapy detergents are soaps whereas the non-soapy detergents are synthetic detergents or simply detergents. Although both are cleansing agents, they differ in chemical composition. In the present chapter, we shall briefly discuss the composition and cleansing action of soaps and synthetic detergents.
Soaps are the sodium and potassium salts of long chain fatty acids with general formula RCOONa or RCOOK. The acids present have the formula RCOOH where R may have following values.
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 19
These fatty acids exist as triesters of glycerol which is a trihydric alcohol. The triesters are also called triglycerides or simply glycerides and are the constituents of edible oils and fats. These are of animal and vegetable origin e.g. castor oil, linseed oil or soyabean oil. Chemically the triglycerides are formed as a result of esterification reaction.
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 20

Question 35.
Identify the functional groups present in the following compounds
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 21
Answer:
(a) >C = O
(b) —COOH
(c) —CHO
(d) —OH.

Question 36.
How is ethene prepared from ethanol ? Give the reaction involved in it.
Answer:
From ethanol: This method involves the slow oxidation of a dilute solution of ethanol (10-15 per cent) by oxygen present in air in the presence of an enzyme acetobactor.
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 22
The acid obtained is in the form of dilute solution called vinegar.
We have also studied under ethanol that it gets oxidised to ethanoic acid in the presence of dilute solution of alkaline KMnO4 or acidified K2Cr2O7.

Question 37.
Intake of small quantity of methanol can be lethal. Comment.
Answer:
From methanol: These days ethanoic acid is manufactured by the reaction between methanol and carbon monoxide in the presence of iodine-rhodium (I2 — Rh) catalyst mixture.
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 23

Question 38.
A gas is evolved when ethanol reacts with sodium. Name the gas evolved and also write the balanced chemical equation for the reaction involved.
Answer:
The gas evolved is hydrogen. The balanced chemical equation for the reactibn is :
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 24

Question 39.
Ethene is formed when ethanol at 443 K is heated with excess of concentrated sulphuric acid. What is the role of sulphuric acid in this reaction ? Write the balanced chemical equation of this reaction.
Answer:
Concentrated sulphuric acid acts as a dehydrating agent in the reaction. Ethene is formed as the
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 25

Question 40.
Carbon, Group (14) element in the Periodic Table, is known to form compounds with many elements. Write an example of a compound formed with
(a) chlorine (Group 17 of Periodic Table)
(b) oxygen (Group 16 of Periodic Table)
Answer:
(a) The compound is carbon tetrachloride (CCl4)
(b) The compound is carbon dioxide (CO2)

Question 41.
In electron dot structure, the valence shell electrons are represented by crosses or dots.
(a) The atomic number of chlorine is 17. Write its electronic configuration
(b) Draw the electron dot structure of chlorine molecule.
Answer:
(a) 2, 8, 7
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 26

Question 42.
Catenation is the ability of an atom to form bonds with other atoms of the same element. It is exhibited by both carbon and silicon. Compare the ability of catenation of the two elements. Give reasons.
Answer:
The tendency to show catenation is very small in case of Si as compared to C although both the elements belong to the same group (14). This is because of greater atomic size of Si atom (118 pm) than that of carbon (77 pm). As a result, the strength of Si—Si bond is less as compared to that of C—C bond. This means that lesser number of Si atoms can be linked to each other by covalent bonds as compared to carbon atoms.

Question 43.
Unsaturated hydrocarbons contain multiple bonds between the two C-atoms and show addition reactions. Give the test to distinguish ethane from ethene.
Answer:
Distinction between ethane and ethene can be done with the help of bromine water test. Whereas ethene dicolourises the yellow colour of bromine water, ethane does not.
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 27

Question 44.
Match the reactions given in Column (A) with the names given in column (B).
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 28
Answer:
(a)—(iv) ;
(b)—(i) ;
(c)—(ii) ;
(d)—(iii)

Question 45.
Write the structural formulae of all the isomers of hexane.
Answer:
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 29

Question 46.
What is the role of metal or reagents written on arrows in the given chemical reactions?
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 30
Answer:
(a) Nickel (Ni) acts as a hydrogenation catalyst for the reaction
(b) Cone. H2SO4 removes a molecule of H2O from the reaction mixture and acts as a dehydrating agent.
(c) Alkaline KMnO4 acts as an oxidising agent and oxidises ethanol to ethanoic acid.

NCERT Exemplar Solutions for Class 10 Science Chapter 4Long Answer Questions

Question 47.
A salt X is formed and a gas is evolved when ethanoic acid reacts with sodium hydrogen carbonate. Name the salt X and the gas evolved. Describe an activity and draw the diagram of the apparatus to prove that the evolved gas is the one which you have named. Also, write chemical equation of the reaction involved.
Answer:
The salt X formed in the reaction is sodium ethanoate. The gas evolved is carbon dioxide gas. For the activity and chemical reaction,
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 31

Question 48.
(a) What are hydrocarbons ? Give examples.
(b) Give the structural differences between saturated and unsaturated hydrocarbons with two examples of each.
(c) What is a functional group ? Give examples of four different functional groups.
Answer:
(a) Hydrocarbons are the organic compounds containing only carbon and hydrogen atoms as their constituents. These may be alkanes, alkenes and alkynes.
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 32
(b) Saturated hydrocarbons or alkanes contain either C—C or C—H bonds in their molecules. These are represented by the general formula CnH2n+2 For example,
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 33
Unsaturated hydrocarbons contain either atleast one >C = C< bond or triple —C ≡ C— bond in their molecules. These may be either alkenes or alkynes in nature. The general formula of alkenes is CnH2n while that of alkynes is CnH2n-2 for example,
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 34
(c) For the definition of functional group and example,
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 35

Question 49.
Name the reaction which is commonly used in the conversion of vegetable oils to fats. Explain the reaction in detail.
Answer:
The reaction is known as catalytic hydrogenation. For details, constult section 4.15.

Question 50.
(a) Write the formula and draw electron dot structure of carbon tetrachloride.
(b) What is saponification ? Write the reaction involved in this process.
Answer:
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 36
(b) For saponification reaction,
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 37
The reaction is known as saponification reaction because it is the basis for the formation of soap.

Question 51.
Esters are sweet-smelling substances and are used in making perfumes. Suggest some activity and the reaction involved for the preparation of an ester with well labelled diagram.
Answer:
Esters as pointed, are pleasant smelling compounds. These are therefore, commonly used as flavouring agents and also in perfumes. When an ester is reacted with water in the presence of a dilute acid like dilute HCl, acid and alcohol are formed as the product. The reaction is called ester hydrolysis.
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 38
Ester hydrolysis is the reverse of esterification reaction.
When an ester is reacted with an aqueous solution of base like NaOH or KOH, the product is an alcohol and salt of the acid. For example,NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 39

The reaction is known as saponification reaction because it is the basis for the formation of soap.

Question 52.
A compound C (molecular formula, C2H4O2) reacts with Na-metal to form a compound R and evolves a gas which burns with a pop sound. Compound C on treatment with an alcohol A in presence of an acid forms a sweet smelling compound S (molecular formula, C3H6O2). On addition of NaOH to C, it also gives R and water. S on treatment with NaOH solution gives back R and A.
Identify C, R, A, S and write down the reactions involved.
Answer:
From the available information, it is evident that

  1. Compound C with molecular formula C2H4O2 is ethanoic acid (CH3COOH)
  2. The compound R is sodium ethanoate and has formula CH3COONa
  3. Since the compound S has only three carbon atoms (C3H6O2) and has been formed by the action of an alcohol on compound C (C2H4O2), this means that the alcohol A has only one carbon atom. It is methanol (CH3OH).
  4. The compound S with a sweet smell is methyl ethanoate with formula CH3COOCH3.

The chemical reactions involved are as follows :
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 40

Question 53.
Look at given figure and answer the following questions :
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 41
(a) What change would you observe in the calcium hydroxide solution taken in tube B ?
(b) Write the reaction involved in test tubes A and B respectively.
(c) If ethanol is used instead of ethanoic acid, would you expect the same change ?
(d) How can a solution of lime water be prepared in the laboratory ?
Answer:
(a) It would become milky.
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 42
In test tube B, calcium hydroxide reacts with carbon dioxide to form calcium carbonate which is milky in colour.
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 43
(c) No, it would be different. No chemical reaction is possible between ethanol and sodium carbonate.
(d) Lime w’ater is prepared by keeping suspension of calcium hydroxide overnight in a beaker. The solution is decanted and is transferred to another beaker. It contains traces of calcium hydroxide and is called lime water.

Question 54.
How would you bring about the following conversions ? Name the process and write the reaction involved.
(a) ethanol to ethene.
(b) Methanol to Ethanoic acid. Write the reactions.
Answer:
(a) From ethanol: This method involves the slow oxidation of a dilute solution of ethanol (10-15 per cent) by oxygen present in air in the presence of an enzyme acetobactor.
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 44
The acid obtained is in the form of dilute solution called vinegar.
We have also studied under ethanol that it gets oxidised to ethanoic acid in the presence of dilute solution of alkaline KMnO4 or acidified K2Cr2O7.
(b) From methanol: These days ethanoic acid is manufactured by the reaction between methanol and carbon monoxide in the presence of iodine-rhodium (I2 — Rh) catalyst mixture.
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 45

Question 55.
Draw the possible isomers of the compound with molecular formula C3H6O and also give their electron dot structures.
Answer:
Two isomers are possible for the molecular formula C3H6O. These are known as functional isomers.
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 46

Question 56.
Explain the given reactions with the examples :
(a) Hydrogenation reaction
(b) Oxidation reaction
(c) Substitution reaction
(d) Saponification reaction
(e) Combustion reaction
Answer:
Hydrogenation of oils : The reaction is extremely useful in the hydrogenation of vegetable oils also called edible oils e.g. ground nut oil, cotton seed oil etc. These are also called cooking oils and are unsaturated in the sense that their molecules contain atleast one C=C bond in their structures. Upon passing hydrogen gas through oil in the presence of nickel catalyst, the double bond changes to single bond. As a result, the unsaturated oil changes to solid fat which is of saturated nature. Vegetable ghees such as Dalda, are of saturated nature and are formed by catalytic hydrogenation reaction.
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 47
Oxidation reaction :

  • Loss of hydrogen is known as oxidation.
  • Gain of oxygen is known as oxidation.

Therefore, it is an oxidation reaction.
Substitution Reactions : Substitution reactions are also called replacement reactions and you are quite familiar with these. In organic compounds, particularly the saturated hydrocarbons (or alkanes), these reactions are very common. One or more hydrogen atoms in the molecule of alkane such methane get substituted by chlorine atoms when the reaction is carried with chlorine in the presence of ultraviolet sun light.
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 48
Combustion reaction :
We have seen that all combustion reactions are basically oxidation reactions carried in the presence of air or oxygen. It is not necessary that the reactants may burn during combustion.

  • Blue flame signifies complete combustion of the fuel.
  • Yellow sooty flame signifies incomplete combustion of the fuel.

Question 57.
An organic compound A on heating with concentrated H2SOforms a compound B which on addition of one mole of hydrogen in presence of Ni forms a compound C. One mole of compound C on combustion forms two moles of CO2 and three moles of H2O. Identify the compounds A, B and C and write the chemical equations for the reactions involved.
Answer:
Since one mole of compound C on combustion forms two moles of CO2 and three moles of H2O the compound C is a hydrocarbon with formula C2H6. It is ethane. The compound B which forms C2H6 upon addition of hydrogen is ethene (C2H4). The organic compound A which forms ethene upon acidic dehydration is ethanol. The chemical equations for the reactions involved are :
NCERT Exemplar Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 49

 

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Value Based Questions in Science for Class 10 Chapter 2 Acids, Bases and Salts

Value Based Questions in Science for Class 10 Chapter 2 Acids, Bases and Salts

These Solutions are part of Value Based Questions in Science for Class 10. Here we have given Value Based Questions in Science for Class 10 Chapter 2 Acids, Bases and Salts

Question 1.
Ritesh was asked to determine the melting point of a given organic solid. For this, he used a bath containing cone. H2SO4. When he was looking at the thermometer, he lost his concentration and became a little casual. The beaker containing boiling sulphuric acid fell on his clothes. His clothes were burnt and he got severe burns on hands.

  1. Why did Ritesh meet with an accident ?
  2. Why did he get severe burns on the hands ?
  3. What lessons can a student learn from the above episode ?

Answer:

  1. The bath or the beaker containing cone. H2SO4 was not properly placed on the tripod stand. It lost balance and the acid fell over him.
  2. Sulphuric acid is a very powerful dehydrating agent. It removed water contents from the skin which got charred and burns appeared on hands.
  3. While working in the chemistry laboratory, a student must observe the following precautions :
    1. Must always wear an appron in the laboratory.
    2. Must always keep some distance from the table where the experiment is performed.
    3. Must always remain alert in the laboratory and must not lose concentration.

More Resources

Question 2.
Ram was suffering from a stomach pain for a number of days. He consulted a doctor who advised him to take two antacid tablets after each meal for about a week and avoid spicy food. Ram followed the advice strictly and was cured.

  1. What was the problem faced by Ram ?
  2. How did doctor help him ?
  3. Write the chemical equation if any
  4. What is the value associated with this ?

Answer:

  1. Ram had developed acidity in the stomach resulting in the formation of small ulcers which caused pain.
  2. The antacid tablets contain base like NaHCO3 or Mg(OH)2 which neutralise the affect of HCl released in the stomach.
  3. NaHCO3 + HCl ———–> NaCl + H2O + CO2 (Antacid)
  4. As far as possible, one should always avoid spicy and fried food stuff. They create acidity in the stomach. This leads in the formation of small ulcers which give pain. Apart from that one must always keep some antacid tablets or liquid gels.

Question 3.
Kamla was playing in the garden. She was stung by a wasp and started crying. Her mother immediately applied a coating of tooth paste on the affected area and then took her to the doctor.

  1. Why did Kamla cry ?
  2. What does wasp sting contain ?
  3. Why did her mother apply tooth paste on the affected area ?
  4. What values are displayed by this episode ?

Answer:

  1. Kamla cried because the sting by the wasp is very painful.
  2. Wasp sting contains in it formic acid (HCOOH)
  3. Tooth paste contains in it some basic ingredients which neutralise the effect of formic acid (HCOOH) and give relief.
  4. Kamla’s mother had a knowledge of chemistry. Tooth paste is readily available and she gave a first aid to her daughter.

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ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Chapter Test

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Chapter Test

More Exercises

Question 1.
If a man received ₹1080 as dividend from 9% ₹20 shares, find the number of shares purchased by him.
Solution:
Income on one share = \(\\ \frac { 9 }{ 100 } \) x 20
= Rs \(\\ \frac { 9 }{ 5 } \)
.’. No. of shares = 1080 x \(\\ \frac { 5 }{ 9 } \)
= 120 x 5 = 600 Ans.

Question 2.
Find the percentage interest on capital invested in 18% shares when a Rs 10 share costs Rs 12.
Solution:
Dividend on one share = 18% of Rs 10
= \(\\ \frac { 18\times 10 }{ 100 } \)
= Rs \(\\ \frac { 9 }{ 5 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Chapter Test Q2.1

Question 3.
Rohit Kulkami invests Rs 10000 in 10% Rs 100 shares of a company. If his annual dividend is Rs 800, find :
(i) The market value of each share.
(ii) The rate percent which he earns on his investment.
Solution:
Investment = Rs 10000
Face value of each share = Rs 100
Rate of dividend = 10%
Annual dividend = Rs 800
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Chapter Test Q3.1

Question 4.
At what price should a 9% Rs 100 share be quoted when the money is worth 6% ?
Solution:
If interest is 6 then investment = Rs 100
and if interest is 9, then investment
= Rs \(\\ \frac { 100\times 9 }{ 6 } \)
= Rs 150
Market value of each share = Rs 150 Ans

Question 5.
By selling at Rs 92, some 2.5% Rs 100 shares and investing the proceeds in 5% Rs 100 shares at Rs 115, a person increased his annual income by Rs 90. Find:
(i) the number of shares sold.
(ii) the number of shares purchased.
(iii) the new income.
(iv) the rate percent which he earns on his investment.
Solution:
Rate of dividend = 2.5% and market price = Rs 92
Let number of shares purchased = x.
Selling price of x shares = 92 x
Income from investing
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Chapter Test Q5.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Chapter Test Q5.2

Question 6.
A man has some shares of Rs. 100 par value paying 6% dividend. He sells half of these at a discount of 10% and invests the proceeds in 7% Rs. 50 shares at a premium of Rs. 10. This transaction decreases his income from dividends by Rs. 120. Calculate:
(i) the number of shares before the transaction.
(ii) the number of shares he sold.
(iii) his initial annual income from shares.
Solution:
Let no. of shares = x
Value of x shares = x × 100 = 100 x
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Chapter Test Q6.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Chapter Test Q6.2

Question 7.
Divide Rs. 101520 into two parts such that if one part is invested in 8% Rs. 100 shares at 8% discount and the other in 9% Rs. 50 shares at 8% premium, the annual incomes are equal.
Solution:
Total investment = Rs. 101520
Let investment in first part = x
and in second part = (101520 – x)
Market value of first kind of shares = Rs. 100 – Rs. 8
= Rs. 92
and rate of dividend = 8%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Chapter Test Q7.1

Question 8.
A man buys Rs. 40 shares of a company which pays 10% dividend. He buys the shares at such a price that his profit is 16% on his investment. At what price did he buy each share ?
Solution:
Face value of each share = Rs. 40
Dividend = 10%
Gain on investment = 10%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Chapter Test Q8.1

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ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends MCQS

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends MCQS

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends MCQS

More Exercises

Question 1.
If Jagbeer invest ₹10320 on ₹100 shares at a discount of ₹ 14, then the number of shares he buys is
(a) 110
(b) 120
(c) 130
(d) 150
Solution:
Investment = ₹ 10320
Face value of each share = ₹100
M.V. of each share = ₹100 – 14 = ₹86
No. of shares = \(\\ \frac { 10320 }{ 86 } \) = 120 (b)

Question 2.
If Nisha invests ₹19200 on ₹50 shares at a premium of 20%, then the number of shares she buys is
(a) 640
(b) 384
(c) 320
(d) 160
Solution:
Investment = ₹19200
Face value of each share = ₹50
M.V. = ₹50 x \(\\ \frac { 120 }{ 100 } \) = ₹60
Number of shares = \(\\ \frac { 19200 }{ 60 } \)
= 320 (c)

Question 3.
₹40 shares of a company are selling at 25% premium. If Mr. Jacob wants to buy 280 shares of the company, then the investment required by him is
(a) ₹11200
(b) ₹14000
(c) ₹16800
(d) ₹8400
Solution:
Face value of each share = ₹40
M.V. = 40 x \(\\ \frac { 125 }{ 100 } \)= ₹50
Number of shares = 280
Total investment = ₹280 x 50 = ₹ 14000 (d)

Question 4.
Arun possesses 600 shares of ₹25 of a company. If the company announces a dividend of 8%, then Arun’s annual income is
(a) ₹48
(b) ₹480
(c) ₹600
(d) ₹1200
Solution:
Number of shares = 600
F.V. of each share = ₹25
Rate of dividend = 8%
Annual income = 600 x 25 x \(\\ \frac { 8 }{ 100 } \)
= ₹1200 (d)

Question 5.
A man invests ₹24000 on ₹60 shares at a discount of 20%. if the dividend declared by the company is 10%, then his annual income is
(a) ₹3000
(b) ₹2880
(c) ₹ 1500
(d) 1440
Solution:
Investment = ₹24000
F.V. of each share = ₹60
M.V. at discount of 20% = 60 x \(\\ \frac { 80 }{ 100 } \)= ₹48
Rate of dividend = 10%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends MCQS Q5.1

Question 6.
Salman has some shares of ₹50 of a company paying 15% dividend. If his annual income is ₹3000, then the number of shares he possesses is
(a) 80
(b) 400
(c) 600
(d) 800
Solution:
F.V. of each share = ₹50
Dividend = 15%
Annual income = ₹3000
Let x be the share, then
F.V. of shares = x × 50 = ₹50x
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends MCQS Q6.1

Question 7.
₹25 shares of a company are selling at ₹20. If the company is paying a dividend of 12%, then the rate of return is
(a) 10%
(b) 12%
(c) 15%
(d) 18%
Solution:
F.V. of each share = ₹25 ,
M.V. = ₹20
Rate of dividend = 12%
Dividend on each share = \(\\ \frac { 12 }{ 100 } \) x 25 = ₹3
Return on ₹20 = ₹3
and on ₹100 = ₹ \(\\ \frac { 3 }{ 20 } \) x \(\\ \frac { 5 }{ 100 } \) = 15% (c)

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions

Other Exercises

Question 1.
Find the sum of n terms of the series :
(i) 4 + 44 + 444 + ….
(ii) 0.8 + 0.88 + 0.888 + ….
Solution:
(i) 4 + 44 + 444 + ….
= \(=\frac { 4 }{ 9 } \left[ 9+99+999+…. \right] \)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q1.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q1.2
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 2 Estimation 1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q1.4

Question 2.
Find the sum of infinite terms of each of the following geometric progression:
(i)\(1+\frac { 1 }{ 3 } +\frac { 1 }{ 9 } +\frac { 1 }{ 27 } +…\)
(ii)\(1-\frac { 1 }{ 2 } +\frac { 1 }{ 4 } -\frac { 1 }{ 8 } +…\)
(iii)\(\frac { 1 }{ 3 } +\frac { 1 }{ { 3 }^{ 2 } } -\frac { 1 }{ { 3 }^{ 3 } } +…\)
(iv)\(\sqrt { 2 } -\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ 2\sqrt { 2 } } -\frac { 1 }{ 4\sqrt { 2 } } +…\)
(v)\(\sqrt { 3 } +\frac { 1 }{ \sqrt { 3 } } +\frac { 1 }{ 3\sqrt { 3 } } +\frac { 1 }{ 9\sqrt { 3 } } +… \)
Solution:
(i)\(1+\frac { 1 }{ 3 } +\frac { 1 }{ 9 } +\frac { 1 }{ 27 } +…\) upto infinity
Sn = \(\\ \frac { a }{ 1-r } \)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q2.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q2.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q2.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q2.4

Question 3.
The second term of a G.P. is 9 and sum of its infinite terms is 48. Find its first three terms.
Solution:
In a G.P.
T2 = 9, sum of infinite terms = 48
Let a be the first term and r be the common ratio, therefore,
S = \(\\ \frac { a }{ 1-r } \)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q3.1

Question 4.
Find three geometric means between \(\\ \frac { 1 }{ 3 } \) and 432.
Solution:
Let G1, G2 and G3 be three means between
\(\\ \frac { 1 }{ 3 } \) and 432, then
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q4.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q4.2

Question 5.
Find :
(i) two geometric means between 2 and 16
(ii) four geometric means between 3 and 96.
(iii) five geometric means between \(3 \frac { 5 }{ 9 } \) and \(40 \frac { 1 }{ 2 } \).
Solution:
(i) Two G.M. between 2 and 16
Let G1 , and G1 be the G.M.,
then 2, G1, G2, 16
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q5.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q5.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q5.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q5.4

Question 6.
The sum of three numbers in G.P. is \(\\ \frac { 39 }{ 10 } \) and their product is 1. Find numbers
Solution:
Sum of three numbers in G.P = \(\\ \frac { 39 }{ 10 } \)
and their product = 1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q6.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q6.2

Question 7.
Find the numbers in G.P. whose sum is 52 and the sum of whose product in pairs is 624.
Solution:
Sum of 3 numbers in G.P. = 52
and their product in pairs = 624
Let numbers be a, ar, ar²
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q7.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q7.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q7.3

Question 8.
The sum of three numbers in G.P. is 21 and the sum of their squares is 189. Find the numbers.
Solution:
Sum of three numbers in G.P. = 21
Sum of their squares = 189
Let three numbers be a, ar, ar², then
a + ar + ar² = 21
=> a( 1 + r + r²) = 21….(i)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q8.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q8.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q8.3

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NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction

NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction

These Solutions are part of NCERT Exemplar Solutions for Class 10 Science. Here we have given NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction

Question 1.
Which of the following can make a parallel beam of light when light from a point source is incident on it ?
(a) Concave mirror as well as convex lens
(b) Convex mirror as well as concave lens
(c) Two plane mirrors placed at 90° to each other.
(d) Concave mirror as well as concave lens.
Answer:
(a).
Explanation : When a point source is placed at the focus of a concave mirror or a convex lens, light after reflecting from the concave mirror and light after refracting from a convex lens goes as a parallel beam.

More Resources

Question 2.
A 10 mm long owl pin is placed vertically in front of a concave mirror. A 5 mm long image of the owl pin is formed at 30 cm in front of the mirror. The focal length of this mirror is
(a) – 30 cm
(b) – 20 cm
(c) – 40 cm
(d) – 60 cm
Answer:
(b).
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 1

Question 3.
Under which of the following conditions a concave mirror can form a real image larger than the actual object ?
(a) When object is kept at a distance equal to its radius of curvature.
(b) When object is placed between the focus and centre of curvature.
(c) When object is kept at a distance less than its focal length.
(d) When object is kept at a distance greater than its radius of curvature.
Answer:
(b).
Explanation :
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 2

Question 4.
Figure shows a ray of light as it travels from medium A to medium B. Refractive index of the medium B relative to medium A is
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 3
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 4
Answer:
(a).
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 5

Question 5.
A light ray enters from medium A to medium B as shown in figure. The refractive index of medium B relative to medium A will be
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 6
(a) greater than unity
(b) less than unity
(c) equal to unity
(d) zero
Answer:
(a).
Explanation :
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 7
where v1 = speed of light in medium A
V2 = speed of light in medium B.
Since ray of light bends towards the normal, when it goes from medium A to rrtedium B, therefore, medium A is rarer and medium B is denser medium. Speed of light in rarer medium (v1) is greater than the speed of light in denser medium (v2). Hence, nBA > 1.

Question 6.
Beams of light are incident through the holes A and B and emerge out of a box through the holes C and D respectively as shown in figure. Which of the following could be inside the box ?
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 8
(a) A rectangular glass slab
(b) A convex lens
(c) A concave lens
(d) A prism.
Answer:
(a).
Explanation : Rectangular glass slab causes the lateral dis¬placement of a ray of light following on it. However, incident ray and emergent ray are paralled to each other.

Question 7.
A beam of light is incident through the holes on side A and emerges out of the holes on the other side of the box as shown in figure. Which of the following could be inside the box ?
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 9
(a) Concave lens
(b) Rectangular slab
(c) prism
(d) Convex lens
Answer:
(d).
Explanation :
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 10

Question 8.
Which of the following statements is true ?
(a) A convex lens has 4 dioptre power having focal length 0.25 m.
(b) A convex lens has – 4 dioptre power having focal length 0.25 m.
(c) A concave lens has 4 dioptre power having focal length 0.25 m.
(d) A concave lens has – 4 dioptre power having focal length 0.25 m.
Answer:
(a).
Explanation : Convex lens has positive power and positive focal length.
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 11

Question 9.
Magnification produced by a rear view mirror fitted in vehicles.
(a) is less than one
(b) is more than one
(c) is equal to one
(d) can be more than or less than one depending upon the position of the object in front of it.
Answer:
(a).
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 12
Rear view mirror is a convex mirror, which always forms an image whose size is less than the size of the object.

Question 10.
Rays from sun converge at a point 15 cm in front of a concave mirror. Where should an object be placed so that size of its image is equal to the size of the object ?
(a) 15 cm in front of the mirror
(b) 30 cm in front of the mirror
(c) between 15 cm and 30 cm in front of the mirror.
(d) more than 30 cm in front of the mirror.
Answer:
(b).
Explanation : Here, Focal length of concave mirror, f = -15 cm Radius of curvature of the mirror, R = 2f = -30 cm.
In case of concave mirror, size of image is equal to the size of the object if the object is placed at the centre of curvature (see figure 20). The distance of centre of curvature from the mirror = radius of curvature of the mirror.

Question 11.
A full length image of a distance tall building can definitely be seen by using
(a) concave mirror
(b) convex mirror
(c) plane mirror
(d) both concave as well as plane mirror.
Answer:
(b).
Explanation : Convex mirror forms full length of a distant tall object irrespective of the position of the object. However, plane mirror forms full size image of the object if the size of the plane mirror is half the size of the object. Concave mirror forms full size image of the object if the object is far away from it.

Question 12.
In torches, search lights and head lights of vehicles, the bulb is placed
(a) between the pole and the focus of the reflector
(b) very near to the focus of the reflector
(c) between the focus and centre of curvature of the reflector
(d) at the centre of curvature of the reflector.
Answer:
(b).
Explanation :
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 13

Question 13.
The laws of reflection hold good for
(a) plane mirror only
(b) concave mirror only
(c) convex mirror only
(d) all mirrors irrespective of their shape
Answer:
(d).

Question 14.
The path of a ray of light coming from air passing through a rectangular glass slab traced by four students are shown as A, B, C and D in the figure. Which one of them is correct ?
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 14
Answer:
(B)
Explanation : Glass slab causes the lateral displacement of a ray of light falling on it. However, incident ray and emergent ray are parallel to each other.

Question 15.
You are given water, mustard oil, glycerine and kerosene. In which of these media, a ray of light incident obliquely at same angle would bend the most ?
(a) Kerosene
(b) Water
(c) Mustard Oil
(d) Glycerine
Answer:
(d).
Explanation : The ray would bend the most, when it goes from rarer medium (say air) to the most denser medium. Since refrac¬tive index of glycerine is the highest among all these medium, so glycerine is the most denser medium.

Question 16.
Which of the following ray diagrams is correct for the ray of light incident on a concave mirror as shown in figure ?
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 15
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 16
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 17
Answer:
(d).
Explanation : Any ray of light parallel to the principal axis passes through the focus (F) after reflecting from the concave mirror.

Question 17.
Which of the following ray diagrams is correct for the ray of light incident on a lens shown in figure ?
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 18
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 19
Answer:
(a).
Explanation : A ray of light passing through the focus of a lens travels parallel to the principal axis after refracting through the lens.

Question 18.
A girl is standing in front of a magic mirror. She finds the image of her head bigger, the middle portion of her body of the same size and that of the legs smaller. The order of combinations for the magic mirror from the top is :
(a) Convex, plane and concave
(b) Plane, convex and concave
(c) Concave, plane and convex
(d) Convex, concave and plane.
Answer:
(c).
Explanation : Concave mirror forms a magnified (enlarged) image of the object if the object is placed close to the concave mirror (i.e. a distance less than its focal length). Plane mirror always forms the mage of the same size as that of the object.

Question 19.
Convex mirror always forms a diminished image of the object irrespective of the position of the object from the mirror. In which of the following, the image of an object placed at infinity will be highly diminished and point sized ?
(a) Concave mirror only
(b) Convex mirror only
(c) Convex lens only
(d) Concave mirror, convex mirror, concave lens and convex lens.
Answer:
(d).
Explanation :
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 20
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 21
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 22
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 23

Question 20.
Identify the device used as a spherical mirror or lens in the following cases, when the image formed is virtual and erect in each case.
(a) Object is placed between device and its focus, image formed is enlarged and behind it.
(b) Object is placed between the focus and device, image formed is enlarged and on the same side as that of an object.
(c) Object is placed between infinity and device, image formed is diminished and between focus and optical centre on the same side as that of the object.
(d) Obj ect is placed between infinity and device, image formed is diminished and between pole and focus, behind it.
Answer:
(a) Concave mirror.
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 24
(b) Convex lens.
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 25
(c) Concave lens.
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 26
(d) Convex mirror.
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 27

Question 21.
Why does a light ray incident on a rectangular glass slab immersed in any medium emerges parallel to itself? Explain using a diagram.
Answer:
Perform an experiment to demonstrate that light bends from its path, when it falls obliquely on the surface of a glass slab. Also show that angle of incidence is about equal to the emergent angle.

  1. Take a glass slab and place it on a white sheet of paper fixed on a drawing board.
  2. Mark the boundary ABCD of the glass slab.
  3. Fix two pins P1 and P2, vertically on the drawing board such that line joining the pins is inclined to the edge AB of the glass slab (Figure 37).
    NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 28
  4. Now, look through the glass slab from opposite side so that the images of pins P1 and P2 are seen exactly in line. Fix two pins P3 and P4, vertically on the drawing board such that pins P3, P4 and image of P1 and P2 are exactly in one line.
  5. Remove the glass slab and all pins. Join the points P1 and P2 with a line and extend this line to touch the edge AB at P. Similarly join the points P3 and P4 with a line and extend the line to touch the edge DC at Q.
  6. Join points P and Q with a straight line.
  7. Draw normal NN’ at P on the edge AB and a normal N1N1‘ at Q on the edge DC.
  8. Measure ∠P2 PN and ∠P3 QN1‘ using a protractor or dee.

This experiment shows that when light falls obliquely on a glass slab, it bends along PQ from its original path along PL.
Also ∠P2PN ≈ ∠P3QN1‘. That is, angle of incidence is about equal to the emergent angle.
Note : Distance between pins P1 and P2, and between P3 and P4 must be large.

Question 22.
A pencil when dipped in water in a glass tumbler appears to be bent at the interface of air and water. Will the pencil appear to be bent to the same extent, if instead of water, we use liquids like, kerosene or turpentine. Support your answer with reason.
Answer:
A pencil dipped in water appears to be bent at the interface of air and water due to the refraction of light. The refraction of light occurs because the speed of light changes when light travels from one medium to another. The pencil will not appear to be bent to the same extent, when it is dipped in kerosene or turpentine. This is because refractive index of kerosene or turpentine is greater than the index of water and hence speed of light (v = c/n) is less in kerosene or turpentine as compared to in water.

Question 23.
How is the refractive index of a medium related to the speed of light ?
Obtain an expression for refractive index of a medium with respect to another in terms of speed of light in these two media.
Answer:
Refractive index of a medium

NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 29

Question 24.
Refractive index of diamond with respect to glass is 1.6 and absolute refractive index of glass is 1.5.
Find out the absolute refractive index of diamond.
Answer:
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 30

Question 25.
A convex lens of focal length 20 cm can produce a magnified virtual as well as real image.
Is this a correct statement ? If yes, where shall the object be placed in each case for obtaining these images ?
Answer:
The statement is correct.
A convex lens of focal length 20 cm will produce a magnified Virtual image if object is placed at a distance less than 20 cm from the lens.
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 31
A convex lens of focal length 20 cm will produce a magnified real image if object is placed at a distance greater than 20 cm and less than 40 cm from the lens.
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 32
Sudha finds out that the sharp image of the window pane of her science laboratory is formed at a distance of 15 cm from the lens. She now tries to focus the building visible to her outside the window instead of the window pane without disturbing the lens.

Question 26.
In which direction will she move the screen to obtain a sharp image of the building ? What is the approximate focal length of this lens ?
Answer:
A real image can be obtained on the screen. Therefore, the lens used is convex lens as it forms real as well as virtual image. The distance of the real image formed by a convex lens from the lens decreases as the object distance from the lens increases. Hence, the screen has to be moved towards the lens to obtain the sharp image of the building.
Approximate focal length of the lens = 15 cm. The rays of light from the window pane are considered to come from infinity. These rays of light are focused by the convex lens at its focus (i.e. on the screen).

Question 27.
How are power and focal length of a lens related ? You are provided with two lenses of focal length 20 cm and 40 cm respectively. Which lens will you use to obtain more convergent light ? (CBSE 2012)
Answer:
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 33
The lens of focal length 20 cm or power 5.0 D will be used to have more convergent light. This is because lens of small focal length or large power strongly converges the parallel beam of light.

Question 28.
Under what condition in an arrangement of two plane mirrors, incident ray and reflected ray will always be parallel to each other, whatever may be the angle of incidence. Show the same with the help of diagram.
Answer:
Incident ray and reflected ray will always be parallel to each other if two plane mirrors are placed perpendicular to each other as shown in figure, i = r.
NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 34

Question 29.
Draw a ray diagram showing the path of rays of light when it enters with oblique incidence

  1. from air into water,
  2. from water into air.

Answer:

  1. When ray of light enters from air into water, it bends towards the normal as shown in figure I
  2. When ray of light enters from water into air, it bends away from the normal as shown in figure II.
    NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction image - 35

 

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