SOLUTIONS
INDEX
OF SELF-PRACTICE Q’S
FOR EXAMPLE
If you want to go to the answer of any Self-practice Q’s and you
know it comes under Objective Q's, then just click on Objective Q's link
within the chapter to navigate quickly.
Pg.
CHEMISTRY
Pg.
4 : Carbon and its Compounds
1 : Chemical Reactions
and Equations
Objective Q's
17
Objective Q's
01
CBQ
19
CBQ
02
VSA
20
VSA
03
SA-I
21
SA-I
03
SA-II
22
SA-II
04
LA
23
LA
04
BIOLOGY
5 : Life Processes
2 : Acids, Bases and Salts
Objective Q's
07
CBQ
07
VSA
08
SA-I
08
SA-II
09
LA
09
3 : Metals and Non-metals
Objective Q's
25
CBQ
25
VSA
26
SA-I
26
SA-II
27
LA
28
6 : Control and Coordination
Objective Q's
11
Objective Q's
30
CBQ
12
CBQ
30
VSA
13
VSA
32
SA-II
13
SA-I
32
LA
14
SA-II
33
LA
33
7 : How do Organisms
Reproduce?
Objective Q's
35
CBQ
36
VSA
39
SA-I
39
SA-II
41
LA
42
8 : Heredity
Objective Q's
46
CBQ
47
SA-I
48
SA-II
48
LA
50
PHYSICS
9 : Light – Reflection
and Refraction
SA-I
63
SA-II
64
LA
66
11 : Electricity
Objective Q's
68
CBQ
69
VSA
71
SA-I
71
SA-II
72
LA
74
12 : Magnetic Effects of
Electric Current
Objective Q's
79
CBQ
79
VSA
81
SA-II
81
LA
82
Objective Q's
52
CBQ
53
SA-I
55
BIOLOGY
SA-II
55
13 : Our Environment
LA
56
10 : The Human Eye and the
Colourful World
Objective Q's
60
CBQ
61
VSA
63
Objective Q's
83
CBQ
84
VSA
87
SA-I
87
SA-II
87
LA
88
ANSWER SHEET
SELF PRACTICE
18. (d) Helium or nitrogen
3. (d) (II) and (IV)
Explanation : An endothermic process involves
absorption of heat, such as sublimation and
evaporation need heat energy.
Related Theory

Dilution of sulphuric acid is an exothermic reaction
as a lot of heat is given out during this reaction.

Condensation of water vapours is the process by
which water vapour turns into liquid water. As water
vapour condenses into liquid, it loses energy in the
form of heat.
6. (a) Substance Oxidized : Fe; Reducing Agent : Fe
[CBSE Marking Scheme Term-1 SQP 2021]
Explanation: In a chemical reaction, the
substance to which oxygen is added or
hydrogen is removed is said to be oxidized and
the substance oxidized is the reducing agent as
it either removes oxygen atom from the other
substance or provides hydrogen to it.
In the reaction of iron with copper sulphate
solution:
CuSO4 + Fe → Cu + FeSO4
we observe that oxygen is added to Fe and it
becomes FeSO4 whereas oxygen is removed
from CuSO4 and it becomes Cu.
Therefore, Fe is getting oxidized and as it is
removing oxygen from CuSO4, it is also the
reducing agent.
Caution

Students get confused with reactants and products
while identifying the reduced and oxidised substances.
Only reactant substances are either oxidized or
reduced, not the products.
11. (c) (I) and (II) only
Explanation: In this reaction,
Explanation: Inert gases like helium or nitrogen
both can be used for the storage of fresh
sample of oil for a long time, as they do not
react with most elements, including oxygen.
Thus, these gases create an inert environment
for the oil and prevent its reaction (oxidation)
with any element in the environment. Hence,
stopping it from becoming rancid.
19. (a) X: Silver nitrate; Y: Silver chloride
Explanation: When silver nitrate (AgNO3)
solution is added to sodium chloride (NaCl)
solution, a white precipitate of silver chloride
(AgCl) is formed along with sodium nitrate
(NaNO3) solution.
23. (c) (A) is true, but (R) is false.
Explanation: Burning of magnesium ribbon
in air is a highly exothermic reaction in which
a white powder magnesium oxide (MgO) is
formed, along with the evolution of a large
amount of heat. It is due to the heat evolved
that it burns with a dazzling white flame.
25. (d) (A) is false, but (R) is true.
Explanation: A chemical equation is balanced
by making the number of atoms of each
element equal on both sides of the arrow
without altering the formula of the compounds
or elements involved in the reaction. It is
balanced by putting coefficients in front of the
compounds or elements.
28. (d) (A) is false, but (R) is true.
Explanation: The decomposition reaction of
silver chloride into silver and chlorine is an
endothermic process as it requires energy in
the form of sunlight for breaking down the
reactants.
29. (a) (A) and (R) are true and (R) is the correct
MnO2 + 4HCl → MnCl2 + Cl2 + H2O
MnO2 is losing oxygen to form MnCl2, so MnO2
is being reduced to MnCl2.
HCl is losing hydrogen to form Cl2, so HCl is
being oxidised to Cl2
MnO2 is the oxidising agent whereas HCl is the
reducing agent.
explanation of (A).
Explanation: A more reactive metal displaces
a less reactive metal from its compound. As
zinc and lead are both more active than copper
and placed above copper in the reactivity
series of metals, they displace copper from
its compound. Such reactions are called
displacement reactions.
Chemical Reactions and Equations
1
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33. (b) Both (A) and (R) are true but (R) is not the
correct explanation of (A).
Explanation: Iron (Fe) metal does not react
with cold or hot water but it reacts with steam
to form metal oxide and hydrogen.
The law of conservation of mass holds good
for a chemical reaction. The statement is true
but it does not explain about the equation
given in Assertion.
34. (C) (i) In the reaction, hydrogen is oxidized
to water by addition of oxygen whereas
Copper oxide is reduced to copper by
loss of oxygen.
(ii) Lead sulphide is oxidized and hydrogen
peroxide is reduced.
In the reaction given, lead sulphide gains
oxygen and therefore oxidizes to form
lead sulphate, whereas hydrogen
peroxide loses oxygen and reduces to
form water.
35. (D) (c) Hydrogen gas will be evolved
Explanation: If we add dilute hydrochloric
acid instead of dilute sulphuric acid to the
second test tube containing zinc granules,
the following reaction will take place:
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) + Heat
Bubbles will be observed as hydrogen gas
will be evolved. It is also an exothermic
reaction as heat is evolved during this
reaction. However, as zinc chloride formed
is soluble in the solution, no precipitate will
be formed.
37. (A) X is Zinc and Y is Copper
(B)
YSO4 - CuSO4 (Copper Sulphate)
The reaction between ‘X’ and YSO4 is
displacement reaction. As ‘X’ is more
reactive metal than ‘Y’ metal so it
displaces metal ‘Y’ from its salt solution.
(C) X(s) + YSO4(aq) → XSO4(aq) + Y(s)
Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)
Observations:
(i) The blue colour of CuSO4(YSO4)
solution fades due to the formation of
ZnSO4(XSO4) which is colourless.
(ii) A red-brown deposit of copper is
formed on the zinc.
38. (C) (d) Both (II) and (IV)
Explanation: Reactions in which two
compounds react by an exchange of ions
to form two new compounds are called
double displacement reactions.
For example, when sodium sulphate
solution is mixed with barium chloride
solution, a white precipitate of BaSO4 is
formed by the reaction of SO4 2– and Ba2+.
Na2SO4(aq) + BaCl2(aq) → BaSO4(s)
+ 2NaCl(aq)
2
Caution

Students get confused between displacement
reaction and double displacement reaction. A
double displacement reaction is also a displacement
reaction where two compounds react, and the
positive ions (cations) and the negative ions (anion)
of the two reactants switch places, forming two new
compounds.
(D) (a) All double displacement reactions are
precipitation reactions.
Explanation:
Double
displacement
reactions can be precipitation reactions
or neutralization reactions. A precipitation
reaction is a double displacement reaction
taking place between two aqueous ionic
compounds which forms a precipitate of
a new ionic compound. A neutralization
reaction is a double displacement reaction
taking place between an acid and a base to
form salt and water.
Neutralization reactions are generally not
precipitation reactions as the salt is soluble
in the solution.
So, all double displacement reactions need
not be precipitation reactions. An example of
neutralization reaction is:
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
The reaction between vinegar and baking
soda is an example of a neutralization
reaction, which is a double displacement
reaction.
NaHCO3 + CH3COOH(aq) → H2CO3
+ CH3COONa
When silver nitrate solution is added
to sodium chloride solution, a white
precipitate of silver chloride is formed
alongwith sodium nitrate solution:
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
39. (A) (b) Both (I) and (III)
Explanation: The reaction at (I) is a
combination reaction as copper combines
with the oxygen and forms copper oxide,
which is black in colour. The equation of the
reaction taking place is given below:
2Cu + O2 → 2CuO
However, the reaction at (II) is a
displacement reaction as iron displaces
hydrogen from dil. HCl to form hydrogen
gas.
Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g)
The reaction at (III) is a double displacement
reaction in which Cu2+ and H+ ions
exchange their positions and form two new
compounds, namely copper sulphide and
sulphuric acid.
Science Class X
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CuSO4(aq) + H2S(g) → CuS(s) + H2SO4(aq)
The reaction at (IV) is a decomposition
reaction as molten aluminium chloride
decomposes to form aluminium metal and
chlorine gas on passing electric current.
2AlCl3(l) → 2Al(l) + 3Cl2(g)
47. The balanced chemical equation for the
reaction taking place between carbon
monoxide and hydrogen gas at 340 atm to
form methyl alcohol or methanol is given
below:
(B) (d) Metal X: Zinc; Metal Y: Copper; Type of
Reaction: Displacement.
Explanation: The metal X is zinc, since
zinc is used for coating iron articles by
galvanization to protect it from rusting.
As copper sulphate solution is blue in
colour and it is given that it is a solution
of reddish brown metal, hence metal Y
is copper, since copper metal is reddish
brown in colour.
When zinc is added to copper sulphate
solution, it displaces copper from its
solution and forms zinc sulphate which
is colourless and hence we say that the
colour fades away.
The type of reaction is displacement
reaction as zinc displaces copper from
copper sulphate solution as it is more
reactive than copper.
The equation for the reaction taking place
is:
49. When iron (III) oxide is heated with aluminium
powder, then aluminium oxide and iron metal
are formed. A more reactive metal aluminium
displaces a less reactive metal iron from its
oxide.
Fe2O3(s) + 2Al(s) → Al2O3(s) + 2Fe(l)
53. (A) The
substance which gains oxygen
get oxidised and the one which loses
oxygen gets reduced.
Here in equations:
(i)
40. (A) Swapnil observed that the milk has
become sour due to oxidation of milk.
(B) The type of reaction that has taken place
is oxidation as lactose (milk sugar) present
in the milk changes to lactic acid.
(C) (i) Anti-oxidants are added to fat
containing foods to prevent the
development of rancidity due to
oxidation.
(ii) Corrosion is harmful as it:
 weakens the iron and steel objects
and structures.
 contamination of fluids in pipes or
vessels.
 a lot of money is spent every year
to replace the damaged iron and steel
structures.
(Any one)
42. (B) When magnesium oxide is dissolved in
water, it produces an aqueous solution of
magnesium hydroxide.
MgO(s) + H2O(l) → Mg(OH)2(aq)
Magnesium hydroxide solution is basic in
nature as it turns red litmus to blue.
45. When copper metal is heated over a flame,
a black coating of copper oxide (CuO) is formed.
The equation of the reaction taking place is:
Reduction
CuO(s) + H2(g) → Cu(s) + H2O(g)
So CuO is reduced and H2 is oxidised.
(ii)
Oxidation
Reduction
CuO(s) + Zn(s) → ZnO(s) + Cu(s)
Zn + CuSO4 → ZnSO4 + Cu
2Cu + O2 → 2CuO
340 atm
CO(g) + 2H2(g) → CH3OH(l)
Oxidation
CuO is getting reduced whereas Zn is being
oxidised.
(B) One industrial application of reduction:
In industries, metallic ores are reduced to
obtain the metal from them. For example,
calcium carbonate is reduced in industries
to get CaO and CO2.
56. When magnesium is kept in the open, the
oxygen from the atmosphere reacts with it.
This results in the formation of a protective
oxide layer on top of magnesium to form
magnesium oxide. When the magnesium
ribbon is rubbed using sandpaper, the layer
of oxide is removed, resulting in exposing the
magnesium underneath.
When magnesium ribbon is burnt in air, it
results in the formation of magnesium oxide.
This is a white coloured substance that is
obtained as residue (combination reaction).
2Mg + O2 → 2MgO
57. Element Q displaces P and R from their
compounds in reaction (I) and reaction (II)
respectively.
So element Q is more reactive than P and R.
Also element P displaces element R in reaction
(III). So we can say that
Chemical Reactions and Equations
3
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(A) Q is the most reactive element and R is the
least reactive element.
(B) The above reactions are displacement
reactions.
59. (A) Here, X reacts with barium chloride
to form a white ppt. (Y) and sodium
chloride. Therefore, X should be a sodium
compound and Y should be a barium salt.
When barium chloride solution is added
to sodium sulphate solution, a white
precipitate of barium sulphate is formed
alongwith sodium chloride solution.
BaCl2(aq) + Na2SO4(aq) → BaSO4(s) +
2NaCl(aq)
Therefore, X is Sodium sulphate and Y is
Barium sulphate.
(B) This is a double displacement reaction in
which there is an exchange of ions between
the reactants BaCl2 and Na2SO4.
63. (A) The temperature of the reaction mixture
rises when aluminium (Al) is added to HCl
because following exothermic reaction
takes place:
2Al + 6HCl → 2AlCl3 + 3H2 + Heat
(B) The bubbles are observed due to the
formation of Hydrogen gas. The reaction
gradually becomes slow due to the
formation of a coating of lead chloride on
lead, which prevents further reaction.
Pb + 2HCl → PbCl2 + H2(g)
64. The element is calcium and the substance X
is calcium oxide (CaO). Calcium oxide is used
extensively in the cement industry and calcium
is present in our bones in the form of calcium
phosphate Ca3(PO4)2. Calcium oxide (CaO) is
also known as quick lime.
Calcium oxide dissolves in water to form a
basic solution which turns red litmus blue.
CaO(s) + H2O(l) → Ca(OH)2
Calcium
Water
oxide
Calcium hydroxide
(slaked lime)
68. Precautions to be observed for studying
decomposition reaction:
(1) Use only hard boiling test tube.
(2) Hold the test tube in an inclined position
away from your body and do not point
the mouth of the boiling tube at your
neighbours or yourself.
(3) Use a pair of tongs for holding the boiling
tube while heating and don’t touch the
boiling tube with your bare hands.
(4) Do not inhale the gases emitted directly,
it should be inhaled by wafting gently
towards your nose.
(Write any 2 points to get full marks)
4
69. (A) NH3 is the reducing agent because it
gets oxidized to NO by the removal of
hydrogen and addition of oxygen. O2 has
been reduced to H2O by the addition of
hydrogen.
(B) H2O is the reducing agent. Here, F2 gets
reduced to HF (addition of hydrogen)
and H2O gets oxidized to O2 (removal of
hydrogen).
(C) CO is the reducing agent. Here, CO has
been oxidized to CO2 by the addition of
oxygen. Fe2O3 has been reduced to Fe by
the removal of oxygen.
(D) H2 is the reducing agent as it gets oxidized
to H2O by the addition of oxygen. O2 has
been reduced to H2O by the addition of
hydrogen.
76. A reduction reaction is a reaction in which
hydrogen is added to a substance or oxygen is
removed from a substance.
(A) In this reaction, Fe2O3 is losing oxygen and
forming Fe, whereas Al is gaining oxygen
and forming Al2O3.
Therefore, Fe2O3 is getting reduced and Al
is getting oxidized.
(B) In this reaction, PbO is losing oxygen and
forming Pb whereas C is gaining oxygen
and forming CO. Therefore, PbO is getting
reduced and C is getting oxidized.
78. (A) The ferrous sulphate crystals are hydrated
crystals (green color). When heated in a
boiling tube, the hydrated ferrous crystals
turn anhydrous and change their colour to
white.
(B) When heated more strongly, the
anhydrous crystals decompose and result
in the formation of ferric oxide as residue.
The gases obtained are sulphur dioxide and
sulphur trioxide. X — SO2, SO3, Y — Fe2O3
(C) The balanced chemical reaction is:
∆
2FeSO4(s) → Fe2O3(s) + SO2(g)
Ferrous
sulphate
Ferric
oxide
Sulphur
dioxide
+ SO3(g)
Sulphur
trioxide
79. (B) When potassium iodide is added to lead
nitrate solution, a yellow precipitate of lead
iodide is formed along with potassium
nitrate solution.
Pb(NO3)2(aq) + 2KI(aq) → PbI2(s)
+ 2KNO3(aq)
This is a double displacement reaction
in which two compounds react by the
exchange of ions to form two new
compounds. This is also called precipitation
reaction as an insoluble solid (precipitate)
is formed.
Science Class X
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80. (A) Type of reactions are:
(i) Double Displacement reaction.
(ii) Combination reaction
(iii) Decomposition reaction
(iv) Displacement reaction
flame increases because oxygen supports
burning. Oxygen gas burns brightly with
a wooden splinter, which proves the
combustible nature of this gas.
C + O2 → CO2 + Heat + Light
Wood Oxygen
(B) The balanced chemical equation for the
reaction between barium chloride and
aluminium sulphate to give aluminium
chloride and barium sulphate.
3BaCl2(aq) + Al2(SO4)3(aq) → 3BaSO4(s)
+ 2AlCl3(aq)
(D) H2 gas: Hydrogen (H2) gas burns with
a pop sound when a burning candle is
brought near it.
83. (D) Water is not a good conductor of electricity
sulphuric acid is added in the water to
make it a good conductor of electricity.
81. (A)CO2 gas: When CO2 gas is passed through
lime water, it forms insoluble calcium
carbonate which turns the solution milky.
This is known as lime water test.
Related Theory

The decomposition reaction which takes place
when electricity is passed through the compound
in the molten state or in aqueous solution, then
this reaction is called electrolytic decomposition
reactions or electrolysis.

Hydrogen gas collected is twice the volume of
Ca(OH)2(aq) + CO2 → CaCO3(s) + H2O
Lime water
Carbon
(Colourless)
dioxide
Calcium
carbonate
(White)
Water
The solution becomes clear in the excess
of CO2 because of the formation of soluble
calcium bicarbonate.
CO2 + H2O + CaCO3 → Ca(HCO3)2
Cabon Water Calcium
dioxide
carbonate
Calcium
bicarbonate
(Soluble)
(B) SO2 gas: Due to its acidic nature, sulphur
dioxide gas turns moist litmus paper from
blue to red.
Sulphur dioxide gas when passed through
acidic dichromate solution (orange in
colour) turns it to green because sulphur
dioxide is a strong reducing agent.
K2Cr2O7 + 3SO2 + H2SO4 → Cr2(SO4)3
Potassium
dichromate
Chromium
sulphate
+ K2SO4 + H2O
Sulphur dioxide (SO2) gas when passed
through acidic potassium permanganate
solution (purple in colour) turns it colourless,
because SO2 is a strong reducing agent
2KMnO4 + 2H2O + 5SO2 →
Potasssium
permanganate
(Purple)
K
Sulphur
dioxide
2SO4
Potassium
sulphate
(Colourless)
84. (A) Oxidizing agent: It is a substance which
loses oxygen or gains hydrogen.
For example in the given reaction:
CuO + H2 → Cu + H2O
CuO has given oxygen, hence it is oxidizing
agent.
(B) (i) Hydrogen gas combines with nitrogen
to form ammonia.
H2 + N2 → NH3
Balanced chemical equation:
3H2(g) + N2(g) → 2NH3(g)
(ii) Hydrogen sulphide gas burns in air to
give water and sulpur dioxide.
H2S + O2 → H2O + SO2
Balanced chemical equation:
2H2S(g) + 3O2(g) → 2H2O(l) + 2SO2(g)
(iii) Potassium
metal
reacts
with
water to give potassium hydroxide
and hydrogen gas.
K + H2O → KOH + H2
Balanced chemical equation:
2K(s) + 2H2O(l) →2KOH(s) + H2(g)
Caution

oxygen gas collected. It is 2 : 1 (by volume).
+ 2MnSO4 + 2H2SO4
Manganese
sulphate
(Colourless)
Carbon
dioxide
Students need to understand that an acid turns blue
litmus red and a base changes red litmus blue. This
is why SO2 gas turns blue litmus red as it is acidic in
nature.
(C) O2 gas: The evolution of oxygen (O2)
gas during a reaction can be confirmed
by bringing a burning candle near the
mouth of the test tube containing the
reaction mixture. The intensity of the
85. (A) Iron pipe reacts with oxygen in presence of
moisture in the air to form a red coloured
layer of rust. This process is called corrosion
(rusting).
By colouring the iron pipe, the contact
between the iron and oxygen can be
prevented. Hence, to avoid further corrosion
of the iron pipe, Sumit advised Samarth to
color the pipe.
Chemical Reactions and Equations
5
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86. When food materials containing fats and
Dilute HCl: Zn reacts with HCl and form zinc
chloride with the evolution of hydrogen gas.
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
oils are left for a long time, they start giving
unpleasant smell and taste.
When fats and oils are oxidised by the air, they
become rancid.
The condition produced by oxidation of fats
and oils in foods is marked by unpleasant smell
and taste called rancidity. Rancidity spoils the
food material and make it unfit for use.
Prevention:
(1) When food is stored in air-tight containers,
then there is a little exposure to oxygen
of air and the oxidation of fats and oils
present in food is slowed down.
(2) Anti-oxidants prevent the oxidation as the
fats and oils present in the foods do not
get oxidised easily and hence do not get
rancid.
(3) Packaged food bags are flushed with
nitrogen gas which prevent the food from
getting oxidised.
Zinc
Hydrochloric
acid
Zinc
6
Sulphuric
acid
Zinc
sulphate
Hydrogen
Hydrogen
Dilute HNO3: Reaction with dilute HNO3 is
different as compared to other acids because
nitric acid is an oxidizing agent and it oxidizes
H2 gas, evolved to H2O.
4Zn(s) + 10HNO3(aq) → 4Zn(NO3)2(aq)
+ 5H2O(l) + N2O(g)
Zinc reacts with cold and dilute nitric acid and
forms zinc nitrate, water and nitrous oxide.
NaCI solution: No reaction will take place as
sodium is more reactive than Zn.
Zn(s) + NaCl(aq) → No reaction
NaOH solution: Zinc reacts with NaOH
solution and forms sodium zincate and
hydrogen gas.
Zn(s) + 2NaOH(aq) → Na2ZnO2(aq) + H2(g)
Sodium zincate
88. Dilute H2SO4: Zn reacts with dilute sulphuric
acid to form zinc sulphate and hydrogen gas.
Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)↑
Zinc
chloride
Related Theory

In the case of the reaction of zinc granules with
sulphuric acid or hydrochloric acid, a salt is formed.
Science Class X
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ANSWER SHEET
SELF PRACTICE
9. (a) Water < acetic acid < hydrochloric acid
Explanation: A solution of hydrochloric acid will
have a much higher concentration of H+ ions
and hence, lower the pH level. That’s why it is a
strong acid.
Acetic acid is an organic acid and ionises only
partially in water. That is why it is a weak acid.
Water is almost neutral in nature. In pure water,
the number of positive hydrogen ions is equal
to the negative hydroxide ions.
11. (c) Red litmus paper remains red and blue
litmus paper turns red.
Explanation: When litmus paper is dipped in
acid, it does not change colour; instead, it turns
red. When the red litmus paper is dipped in the
base, the colour changes to blue.
Related Theory

Indicator
Blue
Red
Phenol- Methyl
Acid/Base Litmus Litmus phthalein Orange
Acid
Red
None
Colourless
Red
Base
None
Blue
Pink
Yellow
16. (c) (A): (iii); (B): (iv); (C): (i); (D): (ii)
Explanation:
(1) Bleaching powder (CaOCl2): Used for
bleaching or decolourisation of clothes.
(2) Baking soda (NaHCO3): Used as an
antacid for relieving stomach acidity as it
neutralises the acid present in the stomach.
(3) Washing soda (Na2CO3): Used in
preparation of glass.
(4) Sodium chloride (NaCl): Used for
production of NaOH, H2 and Cl2 gases
during chlor-alkali process.
17. (b) (I) and (III)
Explanation: When a small amount of acid
such as hydrochloric acid is added to water,
individual ions of the acid are dissociated. This
process is known as ionisation. Here, hydrochloric acid separates into positively charged
hydrogen ions and negatively charged chloride
ions.
HCl + H2O → H+ + Cl– + H2O
H2O + H+ → H3O+ (Hydronium ion)
The mixing of an acid with water is also
called dilution. Addition of water results in the
decrease in the concentration of ions, (H3O+)
per unit volume.
34. (c) (I), (II) and (III)
Explanation: Salts having the same positive or
negative radicals are said to belong to a same
family. Sodium chloride and sodium acetate
belong to the same family of salts as they both
have the same positive radical and belong to
the family of sodium salts.
Similarly, calcium sulphate, magnesium sulphate
also belong to the same family of salts as they
have the same negative radical and belong to the
family of sulphate salts.
Sodium carbonate and sodium hydrogen
carbonate belong to the same family of salts as
both have same positive radical (Na+) and belong
to the family of sodium salts.
42. (c) (A) is true but (R) is false.
Explanation: When zinc reacts with sodium
hydroxide solution, hydrogen gas is evolved.
However, such reactions are not possible with
all metals.
43. (d) (A) is false but (R) is true.
Explanation: The process of dissolving an acid
or a base in water is a highly exothermic one.
The acid must always be added slowly to
water with constant stirring. If water is added
to a concentrated acid, the heat generated
may cause the mixture to splash out and cause
burns.
50. (B) (a) Dry Blue Litmus paper: No change; Wet
Blue Litmus paper: Turns red.
Explanation: The gas evolved is hydrogen
chloride which has no effect on dry blue
litmus paper but changes the colour of
wet litmus paper to red. This is because
hydrogen chloride dissociates into hydrogen
ions only in the presence of water and the
H+ ions are responsible for the acidic nature,
due to which it turns wet blue litmus to red.
(E) (a) hydrogen ions combine with water to
form hydronium ions.
Explanation: Hydrogen ions cannot exist
alone, but they exist after combining with
water molecules.
Hydrogen ions must always be shown as
H+(aq) or hydronium ion (H3O+)
H+ + H2O → H3O+
Acids, Bases and Salts
7
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52. (A) (a) Strongly acidic
Explanation: A solution having pH 1 is
strongly acidic. The more acidic solution is,
the lower will be its pH.
(B) (b) (II) and (IV)
Explanation: A solution of sodium hydrogencarbonate is a basic in nature. The parent
acid is H2CO3 and the parent base is NaOH.
NaOH + H2CO3 → NaHCO3 + H2O
The basic salts have strong base and weak
acid. So, NaHCO3 is basic in nature. As we
already know the indicator red litmus paper
turns blue on adding base and colourless
phenolphthalein becomes pink in colour.
(C) (b) (III) and (IV)
Explanation: Aqueous solution of an
acid (Dil. HCl) and base (NaOH) conduct
electricity due to presence of charged
particles called ions in them.
HCl(aq) → H+(aq) + Cl–
NaOH(aq) → Na+ + OH–(aq)
Due to the presence of H+(aq) and OH–(aq) in
acids and bases respectively, these solution
conduct electricity. Though glucose contains
hydrogen but it does not produce hydrogen
ions or any other ions in it, So it does not
conduct electricity. Distilled water also does
not conduct electricity because it does not
produce ions.
53. (D) (d) (II) and (III)
Explanation: Washing soda is obtained
by heating baking soda and then
recrystallizing the sodium carbonate.
Sodium carbonate (Na2CO3) is its chemical
name. It's a mixture of a strong base
(sodium hydroxide) and a strong acid
(carbonic acid). Because a strong base with
a pH greater than 7 is present, its aqueous
solution is in the basic range.
Uses of washing soda:
(1) It has use in the glass, soap, and paper
industries.
(2) It is used to wash garments in the
laundry.
(3) It is used to remove the water's
permanent hardness.
Related Theory

Permanent hardness is defined as the presence
of magnesium and calcium salts in the form of
chlorides and sulphides in water. Washing soda can
be used to get rid of it.
54. (B) (b) High
concentration of hydroxide
ion (OH–) and low concentration of
hydronium on (H3O+)
8

Explanation: As detergents are basic in
nature, the concentration of hydroxide
ions will be high and concentration of
hydronium ions will be less.
Related Theory
hen bases are dissolved in water, they dissociate
W
into hydroxide ions.
For example,
NaOH(aq) → Na+(aq) + OH−(aq)
KOH(aq) → K+(aq) + OH−(aq)
(C) (c) S < Q < R < P
Explanation: The pH of a substance is
related to the concentration of hydrogen
or hydronium ions. An acid has a high
concentration of hydronium ions and a
low pH value whereas a base has a high
pH value and a low concentration of
hydronium ions.
So, we can say, lower the pH value, more
is the concentration of hydronium ions.
Therefore, the hydronium ion concentration
is least in S, followed by Q, R and then P
which has the maximum concentration of
hydronium ions.
55. (A)The substance will be an acid as it turns blue
litmus red.
56. (B)Chlorine gas is given off at the anode,
hydrogen gas at the cathode and sodium
hydroxide solution is formed near the
cathode.
57. Washing soda (sodium carbonate) and Baking
soda (sodium hydrogencarbonate)
61. (A) When an acid is dissolved in water, it forms
hydrogen or H+ ions are formed.
(B) Hydroxide or OH– ions are formed when a
base is dissolved in water.
64. The weak acid and dilute acid are not the
same. A dilute acid has water added to it and
its strength can still be more than that of a
weak acid. A weak acid in its concentrated
form is weak in strength and does not ionise
completely.
66. Comparing with the pH chart, we find that
orange colour corresponds to pH value of about
4 and blue colour to pH value of about 10.
(1) X is therefore acidic and pH is around 4
(2) Y is basic having a pH of about 10.
67. (A) Sodium hydrogen carbonate is used as an
antacid because it is alkaline in nature and
neutralises excess acid in stomach and
provides relief.
(B)Blue coloured copper sulphate crystals on
strong heating loses 5 molecules of water
of crystallisation and changes to anhydrous
copper sulphate which is white in colour.
Chemistry Class X
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Heat
CuSO4·5H2O(s) → CuSO4(s) + 5H2O(l)
Pentahydrate copper
Anhydrous
sulphate
(Blue)
copper
sulphate
(White)
Blue copper sulphate crystals is a hydrated
salt which on heating changes to white
anhydrous copper sulphate and 5 molecules
of water of crystallisation appear on the
upper cooler parts inside the test tube. If we
put 2-3 drops of water on white crystals,
they again turn blue.
74. Sodium chloride or NaCl is obtained by the
reaction between sodium hydroxide solution
and hydrochloric acid.
NaOH + HCl → NaCl + H2O
Sodium chloride is a neutral salt as it is formed
by the reaction between a strong acid and a
strong base.
It is called rock salt when found in the form of
large crystals which are often brown due to
impurities.
Rock salt is formed by evaporation of seawater,
as seawater contains many dissolved salts,
including sodium chloride.
Related Theory

Reactions between an acid and a base to produce a
salt and water are called Neutralization reactions.
76. Onion juice is an olfactory indicator. Those
substances whose smell or odour changes in
acidic or basic solution are called olfactory
indicators. The liquid ‘X‘ turns blue litmus red. It
is an acidic liquid.
Observations when liquid ‘X‘ reacts with
(A) Zinc granules: When zinc granules are
added in an acid in a test tube a vigorous
reaction takes place with evolution of
hydrogen gas.
The test tube becomes hot.
When a burning match stick is brought
near a gas filled bubble, the gas present in
the bubble burns with a pop sound.
(B) Solid sodium carbonate: When liquid ‘X‘
reacts with solid sodium carbonate, a salt,
carbon dioxide and water are formed.
Brisk effervescence of carbon dioxide gas is
produced.
When CO2 is passed through lime water,
lime water turns milky.
78. Equal lengths of magnesium ribbons are
taken in test tubes A and B. Hydrochloric acid
(HCl) is added to test tube A, while acetic acid
(CH3COOH) is added to test tube B.
In both cases, hydrogen gas is evolved. When
metal reacts with acid it forms salt and
hydrogen gas.
Metal + Acid → Salt + Hydrogen gas
Fizzing will occur more vigorously in test tube
(A) containing hydrochloric acid. This is because
hydrochloric acid is stronger acid than acetic
acid and reaction between magnesium ribbon
and HCl is faster in test tube (A) than the
reaction between Magnesium and acetic acid in
test tube B.
79. Preparation of washing soda from sodium
carbonate: Anyhydrous sodium carbonate is
dissolved in water i.e. recrystallization to form
sodium carbonate decahydrate.
Na2CO3 + 10H2O → Na2CO3.10H2O
Sodium carbonate
Washing soda
It is a basic salt.
Permanent hardness can be removed by
washing soda.
Related Theory

Washing soda is white crystalline solid which is
soluble in water and its solution is alkaline in nature.
(pH more than 7).

Hardness of water is due to hydrogen carbonates/
sulphates/chlorides of calcium or magnesium in
water.

Temporatory hardness of water can be removed by
boiling it.
86. (A) The aqueous solution of an acid conducts
electricity due to the presence of charged
particles called ions in it. For example,
when hydrochloric acid (HCl) is dissolved in
water, its solution contains hydrogen ions,
H+(aq) and chloride ions, Cl–(aq).
These ions can carry electric current. So,
due to the presence of H+­(aq) ions and Cl–
(aq) ions, a solution of hydrochloric acid
conducts electricity.
(B) When a base is dissolved in water, it
splits up into ions. Due to the presence of
ions, the solutions of bases also conduct
electricity.
(C) No, separation of H+ ions do not takes
place when HCl is added to a non-aqueous
solution. An acid always ionizes on
dissolving in water to produce hydrogen
ions.
(D) A concentrated acid is always diluted by
adding water to it. The process of mixing
water to a concentrated acid is a highly
exothermic process. In this process, a large
amount of heat is evolved.
(1) When concentrated acid is added
slowly to excess water, the heat is
evolved gradually and easily absorbed
by the large amount of water.
Acids, Bases and Salts
9
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(2) If water is added to excess concentrated acid, a large amount of heat is
evolved suddenly. This heat uses some
of the water to steam explosively. This
results in a splash of acid on our body
and causes acid burns.
87. (A)Baking soda is sodium hydrogen carbonate
(NaHCO3). It decomposes to sodium
carbonate, water and carbon dioxide on
heating.
2NaHCO3 → Na2CO3 + H2O + CO2
Baking powder is a mixture of sodium
hydrogen carbonate (NaHCO3) with tartaric
acid. It readily reacts with sodium carbonate
and neutralizes it.
Therefore use of baking soda will give a
bitter taste to cake due to the presence of
sodium carbonate as sodium carbonate is
basic in nature.
(B)Baking powder is formed by addition of
tartaric acid to baking soda.
10
Caution
 Students usually get confused with baking powder
and baking soda. While both products appear to
be identical, they are not. Baking soda is sodium
bicarbonate, which must be activated with an
acid and a liquid in order to help baked goods rise.
Baking powder, on the other hand, contains sodium
bicarbonate as well as an acid. To activate it, all it
takes is a liquid.
(C)Presence of tartaric acid in baking powder
neutralizes the effect of sodium carbonate
formed during decomposition of baking
soda. Tartaric acid is added to neutralize the
bitterness produced by the baking powder.
Also when baking powder mixes with water,
then the sodium hydrogen carbonate reacts
with tartaric acid to evolve carbon dioxide
gas which gets trapped in the wet dough
and bubbles out slowly making the cake
to rise and hence ‘soft and spongy’ thus
endowing them with a light, fluffy texture.
The equation which takes place can be
shown as:
NaHCO3 + H+ → Na+ + CO2 + H2O
Chemistry Class X
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ANSWER SHEET
SELF PRACTICE
[CBSE Marking Scheme Term-1 SQP 2021]
Explanation: Most metals react with the dilute
hydrochloric acid to produce metal salt and
hydrogen gas. However, hydrogen gas is not
evolved when less reactive metals such as
copper, mercury, silver or gold react with dilute
acid.
Metal + dil. Acid → Salt + Hydrogen
Calcium ion
2,8,8
××
The atomic number of chlorine is 17 and its
electronic configuration is 2, 8, 7. Chlorine atom
has 7 valence electrons, so it requires only one
more electron to complete its octet. Since one
calcium atom donates 2 electrons, so two
chlorine atoms take those two electrons and
form two chloride ions:
. .
2 Cl
. .
+
2e
. .
–
Two Chlorine atom
2 (2, 8, 7)
2 Cl
. .
.×
7. (d) (II) and (IV)
Ca2+ + 2e–
→
–
. .
3FeSO4 + 2Al → Al2(SO4)3 + 3Fe
Ca
Calcium atom
(2,8,8,2)
. .
Explanation: As aluminium is more reactive
than iron, it displaces iron from ferrous
sulphate solution. The green colour of ferrous
sulphate solution slowly becomes colourless.
donates its 2 valence electrons and forms a
stable calcium ion, Ca2+
.
4. (a) green solution slowly turns colourless
Two Chlorine ions
2 (2, 8, 8)
So, iron, magnesium and zinc react with dil. HCl
to form salt and hydrogen gas
Mg(s) + 2HCl(g) → MgCl2(aq) + H2(g)
Caution

Students should remember that all metals do not
react with dilute acids to evolve hydrogen gas.
Only those elements which lie above hydrogen in
the reactivity series displace hydrogen from dilute
acid. Also, except Mg and Mn, no other metal forms
hydrogen gas with dil. nitric acid.
10. (b)
It reacts with cold water to form
magnesium oxide and evolves hydrogen
gas.
Explanation: Magnesium metal never reacts
with cold water but reacts with both hot water
and steam to form magnesium hydroxide and
hydrogen gas is evolved.
Mg(s) + H2O(l) → Mg(OH)2 + H2(g)↑
Magnesium Water
Magnesium
hydroxide
Hydrogen
gas
Related Theory

Reaction of metals and water generally produces
hydrogen gas and their respective hydroxides. For
example, sodium metal reacts with water to produce
sodium hydroxide and liberates hydrogen gas.
..
–
..
××
12. (c) Ca2+ Cl
..
2
[CBSE Marking Scheme Term-1 SQP 2021]
Explanation: The atomic number of calcium
is 20, so its electronic configuration is 2, 8, 8,
2. It has 2 valence electrons. A calcium atom
In calcium chloride compound, the electron
arragement of calcium ion (2, 8, 8) and of
chloride ion (2, 8, 8) resemble that of an argon
atom which makes the calcium chloride very
stable compound. Calcium and chloride ions
have opposite charges, they attract each other.
This type of bond is ionic bond.
14. (d)
Beaker 1: KOH and H2 : Beaker 2: No
reaction takes place
Explanation: All the metals do not react in
the water. The intensity of reaction of a metal
with water depends on its chemical reactivity.
When a metal reacts with cold or hot water,
then the products formed are metal hydroxide
and hydrogen gas.
For example:
2K(s) + 2H2O(l) → 2KOH(aq) + H2(g) + Heat +
energy
The reaction is so violent and exothermic that
the evolved hydrogen gas catches fire. Silver
is very less reactive metal so it does not react
with water at all.
When a metal reacts with steam then the
products formed are metal oxide and hydrogen
gas.
Metals and Non-Metals
11
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29. (c) are generally brittle
Related Theory

How metals displace hydrogen from water. Water
is slightly ionised to give hydrogen ions (H+) and
hydroxide ions (OH–). When a reactive metal
combines with water, it gives its electrons to reduce
the hydrogen ions of water to hydrogen atoms, Two
hydrogen atoms combine to form hydrogen gas. The
unreactive metals like silver do not give electrons
easily, so they are not able to reduce the hydrogen
ions of water to hydrogen gas. Hence unreactive
metals do not displace hydrogen from water.
16. (c) PVC
Explanation: An insulating substance is
required to coat the electrical wire such as
PVC. PVC is a polymer and a bad conductor
of electricity. It is the most common insulating
material used to insulate electrical conductors
from electric charge, thus preventing direct
human contact with electricity.
Graphite is a good conductor of electricity.
Hence, it cannot be used as insulating material.
Sulphur, although a bad conductor of
electricity, is brittle in nature. So, it cannot be
used as an insulating material.
Explanation: Ionic compounds are generally
crystalline solids and hard due to the strong
force of attraction between the positive and
negative ions. They are generally brittle. They
have high melting and boiling points as a
large amount of energy is required to break
the strong inter-ionic attraction. These are
generally soluble in water but insoluble in
organic solvents like ether, kerosene, petrol, etc.
These conduct electricity in the molten state as
the electrostatic forces of attraction between
the oppositely charged ions are overcome due
to heat. Moreover, they also conduct electricity
when dissolved in water as its solution in water
contain ions. However, these do not conduct
electricity in the solid state due to their rigid
structure.
33. (a) Both (A) and (R) are true and (R) is the
correct explanation of (A).
Explanation: The reaction of sodium with
oxygen is highly exothermic. Moreover, as
the reaction of magnesium with oxygen is
less vigorous than sodium, magnesium is less
reactive than sodium.
24. (a) Converting metal sulphides into metallic
oxides and then using carbon to reduce it to
obtain pure metal
Explanation: The metal sulphide is converted
into metal oxide by heating strongly in the
presence of excess air. This process is called
roasting. The metal oxide is reduced to metal
by heating with carbon.
Heat
2ZnS(s) + 3O2(g)
ZnO(s) + C(s)
Heat
2ZnO(s) + 2SO2(g)
Caution

35. (d) (A) is false but (R) is true.
Explanation: Hydrogen gas is not evolved
when a metal reacts with nitric acid because
nitric acid (HNO3) is a strong oxidizing agent
which oxidizes the hydrogen produced to
water and itself gets reduced to any of the
oxides of nitrogen.
Zn(s) + CO(g)
26. (b) Covering the object with a layer of zinc
Explanation: Galvanization is the process of
coating a metal with a protective layer of zinc.
It is a standard way of protecting iron from
rusting.
41. (B) Copper reacts with moist carbon dioxide
in the air and forms green coloured basic
copper carbonate.
Related Theory

To prevent corrosion, many different types of
coatings can be applied to the exposed metal’s
surface. Paints, wax tapes, and varnish are all
examples of corrosion-resistant coatings. Adding an
electric charge to metals can help prevent corrosion
by inhibiting electrochemical processes.
28. (a) gallium and caesium
Explanation: Both gallium and caesium have
such low melting points that they will melt if
we keep them on our palms.
Sodium and potassium are alkali metals that
are so soft that they can be cut with a knife.
12
Students should be aware that exothermic reactions
can be dangerous as they release a lot of heat.
The extra energy is released as heat, causing
temperature rise in the reaction’s immediate vicinity.
Cu(OH)2. (CuCO3)
Related Theory

Basic copper carbonate is undesired layer formed on
the surface of the copper and the process is known
as corrosion.
43. (B)Yes, an aqueous solution of MgO will
conduct electricity as MgO is an ionic
compound and ions are free to move in
aqueous solution.
Science Class X
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44. (A)(d) Both (a) and (c)
Explanation: 24 carat gold is very soft, little
malleable and ductile. It is quite difficult to
work and the ornaments made of it breaks
very easily.
51. As the formula of oxide of E is E2O and E is a
good conductor of electricity, E is a metal with
valency 1. The formula of its compound with
chlorine will be ECl.
56. A is a non-metal.
Explanation: As the element ‘A’ forms two
oxides AO (neutral) and AO2 (acidic), ‘A’ is a
non-metal as oxides of non-metals are either
acidic or neutral. Carbon forms two oxides,
namely, carbon monoxide (CO) which is neutral
and carbon dioxide (CO2) which is acidic.
Al
Element
Electronic
Atomic
Valence
configuration
number
electrons
KLM
Na
11
2, 8, 1
1
Al
13
2, 8, 3
3
If sodium (Na) loses 1 electron (the electron)
from M shell, then its outermost shell will be
L having 8 electrons and has a stable octet.
The nucleus of this atom has 11 protons but
number of electrons become 10 so there is a
net positive charge and becomes Na+ cation.
Similarly in Al, 3 electrons are lost to become
Al3+.
17
2, 8, 7
2, 8, 7
69. (A) M—Magnesium
N—Magnesium oxide.
(B) 2Mg(s) + O2(g) →
Magnesium
7
Oxygen
2, 8, 8
2MgO(s)
Magnesium oxide
Or
2M + O2 → 2MO2
(C) ‘M‘ metal will undergo oxidation reaction
as oxygen is added to metal ‘M‘ and MO2
(metal oxide) is formed.
When a piece of shining metal ‘M‘ is burnt
in air, a white powder of metal oxide is
formed. Shining metal is magnesium
ribbon which burns with a dazzling white
flame and white powder is formed which is
magnesium oxide.
Z is Fe.
67. Elements which will form cation are (A) Na; (B)
Cl
Y is alkaline Earth metal, Mg or Ca.
Z (Fe) < Y (Mg) < X (Na)
Electronic
Atomic
Valence
Configuration
Number
Electrons
KLM
Cl atom has 7 valence electrons and it needs
1 electron to acquire stable octet in M shell.
After it gains one electron, the number of
protons will remain 17 but number of electrons
become 18 so then chlorine atom gets a unit
negative charge and becomes Cl– anion
Cl + e– → Cl–
66. X is alkali metal, Na or K.
X reacts with cold water, so it must be very
reactive like alkali metals, like sodium. Sodium
reacts with water to form sodium hydroxide
and hydrogen gas.
2Na + 2H2O → 2NaOH + H2
Y metal can react with hot water, so it must be
a little less reactive than X i.e., alkaline Earth
metal. So, Y can be magnesium (Mg) which
reacts with hot water to form magnesium
hydroxide.
Mg + 2H2O → Mg(OH)2 + H2
Z metal which reacts with steam must be iron
that forms iron (III) oxide with steam.
3Fe + 4H2O → Fe3O4 + 4H2
Hence, the increasing order of reactivity of the
given metals is:
Element
Related Theory

This reaction is combination and redox reaction also.
Here two elements magnesium and oxygen have
combined and a single compound magnesium oxide
is formed so its a combination reaction.
Addition of oxygen to magnesium is an oxidation
reaction but in terms of loss or gain of electrons
(electronic concept), it is a redox reaction.
According to electronic concept :
Oxidation is a process which involves loss of
electrons. Reduction is a process which involves gain
of electrons.
Mg has changed to Mg2+ ion by loss of electrons.
Mg → Mg2+ + 2e–
Hence, Mg is oxidised.
On the other hand, oxygen has changed into oxide,
O2– (in MgO) by gain of electrons.
O + 2e– → O–2
Hence, Oxygen has been reduced.
72. Alloy: An alloy is a homogenous mixture of
two or more metals, or a metal and a nonmetal.
Preparation of an alloy: It is first prepared by
melting the primary metal and then dissolving
the other elements in it in definite proportions.
It is then cooled at room temperature.
Metals and Non-Metals
13
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Advantages of making alloys: Alloys do not
get corroded/rusted by surrounding gases and
moisture present in the atmosphere. So, they
show resistance to corrosion.
(1) They have high melting and boiling points.
(2) They have high resistivity.
(3) They have high tensile strength.
Steel is preferred over iron because iron is very
soft and streches very easily when hot. When it
is mixed with a small amount of carbon (0.05),
it becomes hard and strong. It is used for the
construction of ships, bridges vehicles, etc.
Composition of stainless steel: Iron, Nickel,
Chromium.
73. (A) (i)
Aluminium
oxide
Hydrochloric
acid
2AlCl3(aq)(l) + 3H2O(l)
Aluminium chloride water
(ii) K2O
Potassium
Water
oxide
Iron
(steam)
Potassium
hydroxide
Iron (II, III) Hydrogen
oxide
Most of the metal oxides are basic in nature i.e., they
dissolve in water to give an alkaline solution.
But in some metal oxides, for example Al2O3 and ZnO
show both acidic as well as basic characters i.e., they
react with the acids as well as bases. Such metal
oxides are called amphoteric oxides.


Most of metal oxides are insoluble in water. But
some metal oxides like of sodium and potassium
dissolve in water to form an alkali.
Some metals react with hot water, some with cold
water and some do not react at all. But metals like
Al, Zn and Fe react only with steam to form metal
oxide and hydrogen.
(B) An element ‘X‘ displaces iron from the
aqueous solution of iron sulphate. A more
reactive metal displaces a less reactive
metal from its salt solution. It means
element ‘X‘ is more reactive than iron.
X + FeSO4(aq) → XSO4(aq) + Fe
The pale green colour of FeSO4 will fade
away slowly when element ‘X‘ is treated
with the aqueous solutions of FeSO4.
Copper is less reactive than iron, iron is less
reactive than X. It means copper is also less
reactive than X.
X + CuSO4(aq) → XSO4(aq) + Cu
The blue colour of CuSO4 will fade away
slowly.
The deposits of metal ‘X‘ will be seen on
the copper strip.
14
(Blue)
(Colourless)
(Colourless)
(Colourless)
76. Chemical properties used to differentiate
between metal and non-metals:
Metals
(1) Metals are electropositive in nature as
they lose electrons
and form positive ions.
K → K+ + e–
Related Theory

(Colourless)
(3) X + ZnSO4(aq) → XSO4(aq) +Zn
(4) X + AgNO3(aq) → XNO3(g) + Ag
+ H2O(l) → 2KOH(aq)
(iii) 3Fe(s) + 4H2O(g) → Fe3O4(s) + 4H2(g)
(Pale green)
(2) X + CuSO4(aq) → XSO4(aq) + Cu
X displaces iron, iron is more reactive than
copper and silver that is why X is highly
reactive metal of all given metals.
Al2O3(s) + 6HCl(l) →
X will be able to displace sliver from its
silver nitrate solution as silver is less
reactive than copper even.
The increasing order of reactivities of these
metals X, Zn, Cu and Ag is as :
Ag < Cu < Zn < X
Observations:
(1) X + FeSO4(aq) → XSO4(aq) + Fe
Non-Metals
Non-Metals
are
electronegative
in
nature as they accept
electrons and form
negative ions.
S + 2e– → S2–
(2) Metals combine with Non-metals combine
oxygen to form basic with oxygen to form
or amphoteric oxides. acidic
or
neutral
oxides
which
are
2Cu + O2 → 2CuO
covalent compounds.
C(s) + O2(g) → CO2(g)
(3) Metals react with dil Non-metals do not
acids such as dil HCl displace
hydrogen
and dil H2SO4 to form from dilute acids as
salt and H2 gas.
they cannot supply
electrons to H+ ions.
Mg + 2HCl →
(s)
MgCl2(aq) + H2(g)
(4) Metals
react with Non-metals do
water and produce react with water.
a metal hydroxide or
oxide and H2.
not
2K(s) + 2H2O(l) →
2KOH(aq) + H2(g) + Heat
(5) Metals
react with
chlorine to form metal
chlorides,
which
are
electrovalent
compounds.
Non-metals
form
covalent
chlorides
which are generally
volatile
liquids
or
gases.
Ca2+ + 2Cl– → CaCl2 P4(s) + 6Cl2(g) →
4PCl3(g)
Science Class X
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Metals
(2) In this reaction, iron oxide loses oxygen
and hence reduces to iron whereas
aluminium gains oxygen and oxidizes
to form aluminium oxide.
Non-Metals
(6) A few active metals Non-metals combine
like Na, K and Ca can with hydrogen to form
force the hydrogen covalent hydrides.
atom
to
accept N + 3H
→
2(g)
2(g)
electrons to form
2NH3(g)
salts called hydrides.
As both oxidation and reduction
reactions
are
taking
place
simultaneously, it is a redox reaction.
(Any 3 of 6 points can be written to get full
marks)
(3) It is also an exothermic reaction as a
large amount of heat is evolved in this
reaction. The heat given out is so much
that it is used to join railway tracks or
cracks in machine parts
79. Ore A is a carbonate ore as it gives carbon
82. (A)
The electrons dot structure for sodium,
2Na(s) + H2(g) →
2NaH(s)
dioxide on heating. The steps involved in
extraction of A are:
(1) Calcination: The carbonate ore is heated
strongly in the limited supply of air to
produce metal oxide.
D
ACO3 → AO + CO2
(2) Reduction to metal: The oxide ore is
reduced with C (coke).
AO + C → A + CO
Ore B is a sulphide ore as it gives sulphur
dioxide on heating. The following steps are
involved in its extraction:
(1) Roasting: The sulphide ore is heated
strongly in the excess of air to produce
metal oxide.
D
2BS + 3O2 → 2BO + 2SO2
(2) Reduction: Oxide of metal B is reduced by
carbon to obtain the corresponding metal.
BO + C → B + CO2/CO
Carbon
Metal
80. (A) Here, A is iron (III) oxide granules and B
is aluminium powder. On mixing them
physically together, no reaction takes
place. However, when the mixture is
heated, a highly exothermic reaction,
known as thermite reaction, takes place.
The equation for the chemical reaction
taking place is:
Heat
Fe2O3(s) + 2Al(s) → 2Fe(l) + Al2O3(s) + Heat
In this reaction, aluminium metal displaces
iron from iron oxide and iron is in molten
form as it is a highly exothermic reaction.
The product C is therefore iron metal in
molten form and the other product D is
aluminium oxide which floats over molten
iron.
(B) The above reaction is:
(1) A displacement reaction as aluminium
being more reactive than iron displaces
iron from iron oxide,
oxygen and magnesium is shown below:
.
.. .. ... .
Na ...O.... Mg
.
Related Theory

G.N. Lewis introduced a simple method to represent
the valence electrons by dots or small crosses
around the symbol of the atom. These symbols are
known as dot symbols or Lewis symbols.
Element
Symbol
Atomic
Electronic
Valence
Number Configuration Electrons
Sodium
Na
11
2, 8, 1
1
Oxygen
O
8
2, 6
6
12
2, 8, 2
2
Magnesium Mg
(B) Formation of sodium oxide:
Atomic number of sodium (Na) = 11
Its electronic configuration = 2, 8, 1
Atomic number of oxygen (O) = 8
Its electronic configuration = 2, 6
Each sodium atoms can lose only one
electron and attains stable configuration
like that of Neon (2, 8)
Na →
Na+ + e–
Sodium
ion
But each oxygen atom requires two
electrons to attain stable configuration of
neon (2, 8).
O
+ 2e– 2–
O
So, two atoms of sodium will lose two
electrons (i.e., one each)
Na
O
(Na+)2
O
2–
Na
Formation of magnesium oxide:
Atomic number of magnesium (Mg) = 12
Electronic configuration = 2, 8, 2
It loses two electrons from its valence shell
and acquires electronic configuration of
neon (2, 8) and form Mg2+ ion.
Metals and Non-Metals
15
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2+
–
Mg ×
× → Mg + 2e
2–
Mg
Magnesium
ion
Atomic number of oxygen = 8
Its electronic configuration = 2, 6
O
+ 2e– 2–
O
Oxide ion
It gains two electrons to acquire the stable
configuration of neon (2, 8) and becomes
oxide ion (O2–)
16
+
O
(Mg2+) O
(C) The ions present in sodium oxide compound
(Na2O) are sodium ions (2Na+) and oxide
ions (O2–).
Ions present in Magnesium oxide
compound (MgO) are magnesium ions
Mg2+ and oxide ions (O2–).
Science Class X
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ANSWER SHEET
SELF PRACTICE
2. (a) only single bonds
Related Theory
Explanation: H—N—H
|
H
Ammonia

A molecule of ammonia (NH3) has only
single bonds. The atomic number of N is 7. Its
electronic configuration is [2, 5] so it needs 3
more electrons to complete its octet. Since the
valency of nitrogen is 3, one atom of nitrogen
combines with three atoms of hydrogen to
form the molecule of ammonia. An atom of
nitrogen can form three covalent bonds.
6. (b) Unsaturated, as sooty deposit represents
unburnt hydrocarbons.
Explanation: Because they do not undergo
complete combustion, saturated hydrocarbons
produce a clear flame, but unsaturated
hydrocarbons produce a yellow flame with a
lot of black smoke. Because there are unburnt
carbon particles in the flame so, it forms a
yellow colour.
7. (b) oxidising agent
Explanation: In this reaction, alkaline
KMnO4 acts as an oxidising agent. It adds
oxygen to ethanol that results in the formation
of ethanoic acid. An oxidising agent adds oxygen
or removes hydrogen from other substances.
13. (c)
Explanation: The chemical formula of benzene
is C6H6. Benzene is the simplest aromatic
compound, which consists of six carbon atoms
bonded in a hexagonal ring. Each carbon atom
is bonded to one hydrogen atom and two
carbon atoms. Benzene molecule contains
alternate single and double bonds.
17. (d)
Explanation: In structures (a), (b) and (c), all the
carbon atoms are attached by covalent bonds
in a continuous straight chain.
1
1
Propane and chloropropane are saturated
hydrocarbons which contain only single bonds.
H H H
H H H
| | |
| | |
H—C—C—C—H
H—C—C==C—H
| | |
|
H
H H H
Propyne
3
4
5
2
3
4
5
6
H3C—CH2—CH2—CH2—CH2—CH3
Explanation: Alkenes and alkynes are
unsaturated hydrocarbons as they have
double and triple covalent bonds between
carbon atoms respectively.
Propane
H
|
H—C—CC—H
|
H
2
H3C—CH2—CH2—CH2—CH2
6|
CH3
10. (d) (II) and (III)
A hydrocarbon in which two carbon atoms are
connected by a ‘double bond’ or a ‘triple bond’ is
called an unsaturated hydrocarbon.
Propene
H H H
| | |
H—C—C—C—Cl
| | |
H H H
Chloropropane
1
CH3
2|
3
4
5
H2C—H2C—H2C—CH2
6|
CH3
In structure (d), —CH3 group is attached to the
second carbon atom of the chain forming a
branch. Hence, compound in structure (d) is a
branched chain hydrocarbon.
1
CH3
2
3
4
5
CH—CH2—CH2—CH3
H3 C
Related Theory

A branched chain hydrocarbon contains some side
chains which are bonded with parent carbon chain.
Carbon and its Compounds
17
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19. (a) propanal
32. (c) low melting and boiling point
Explanation: Carbon compounds have low
melting and boiling point due to the weak force
of attraction between the molecules in the
compounds of the carbon which makes it easy
to break the bond in the molecules thus they
have low melting and boiling point.
Explanation: 3 2 1
CH3—CH2—CHO
There are three carbon atoms in the chain. So,
the name of the basic carbon chain is propane.
The functional group present in aldehyde
(—CHO). So, the suffix used is ‘al’. The name
of the carbon chain is modified by replacing
the final ‘e’ with ‘al’. So, the name of the given
compound is propanal.
33.
Substance used
for Oxidation of
Ethanol
22. (a) rain
Explanation: Soaps work on soft water. Rain
water is distilled water which is appropriate
for the cleansing purpose. In rain water, micelle
formation takes place effectively which is
appropriate for cleansing action of soap.
(a) Alkaline KMnO4
CH3—CH2OH
40. (b) Both (A) and (R) are true but (R) is not the
correct explanation of (A).
Explanation: The compounds C2H4, C3H6, C4H8
belong to the same homologous series as the
successive compounds differ by CH2 unit in
their molecular formula and also difference in
their atomic masses is 14 unit.
Gradation in physical properties such as
melting and boiling points and solubility in a
particular solvent is also seen with an increase
in molecular mass in any homologous series.
25. (c) (I) and (II)
42. (d) (A) is false but (R) is true.
Explanation: Any two adjacent homologous
differ by 1 carbon atom and 2 hydrogen
atoms in their molecular formula. e.g., CH3OH,
CH3CH2OH. These are first two adjacent
homologous-Methanol and ethanol. The
difference between them is —CH2.
Related Theory

Isomers are molecules that contain the same
number of atoms of the same elements but differ in
structural arrangement and properties. Isomers have
the same chemical formula. Structural isomers have
the same molecular formula but different parent
chain of carbon atoms in the molecule.
Related Theory

Difference in the molecular masses of any two
adjacent homologues is 14 µ. For example the
molecular mass of methanol (CH3OH) 32 µ and that
of its next higher homologue ethanol (CH3CH2OH) is
46 µ. So the difference in molecular masses is 46 –
32 = 14 µ.

All compounds of a homologous series show similar
chemical properties.

The members of a homologous series show a
gradual change in their physical properties with
increase in molecular mass.
28. (c) esters of long chain fatty acids.
Explanation: Soaps are sodium or potassium
salts of long chain fatty acids. When
triglycerides in fat/oil react with aqueous
NaOH or KOH, they are converted into soap
and glycerol. This is called alkaline hydrolysis
of esters.
18
CH3COOH
Explanation:
Carbon
forms
covalent
compounds with other atoms as it can neither
gain 4 electrons to complete its octet nor can
it lose 4 electrons to attain nearest inert gas
configuration due to energy considerations.
Test tubes P and Q contain sodium and
potassium salts which do not react with soap.
So, soap solution forms lather.
Explanation: The chemical formula of butane
is C4H10. Structural formulae in other options
(III) and (IV) show only 8 hydrogen atoms. (I)
and (II) are isomers of butane as they have
the same molecular formula but different
structures. Structure (I) is n-butane and
structure (II) is iso-butane.
Alkaline KMnO4 + Heat
Or acidified K2Cr2O7 + Heat
38. (c) (A) is true but (R) is false.
The hardness in water is caused by calcium
and magnesium salts and soap reacts with the
calcium and magnesium salts to form scum.
As test tubes R and S contain calcium and
magnesium salts, no lather will be formed on
adding soap solution to these test tubes.
CH3COOH
Explanation: Ethanol undergoes oxidation
in the presence of oxidizing agents such as
alkaline KMnO4 or acidified K2Cr2O7 to form
ethanoic acid.
24. (a) P and Q
Explanation: Soap does not form lather in
hard water as it forms an insoluble substance
(scum) in hard water.
Compound formed
on Oxidation of
Ethanol
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44. (d) (A) is false but (R) is true.
Explanation: Carbon has 4 electrons in
its outermost shell. So, it needs 4 more
electrons to complete the octet and obtain
noble gas configuration. If it is to gain or
lose electrons:
(1) It could gain four electrons and forms
C4– anion. But it would be difficult
for the nucleus with six protons to hold
on to ten electrons.
OH
|
(vi) CH3— C—CH2—CH3
|
CH3
(vii)
CH3 OH
|
|
CH3— CH—CH—CH3
They are classified as isomers as all of
them have the same molecular formula.
47. (B) (i) The name of the ester having formula
HCOOC2H5 is ethyl methanoate.
(2) It could lose four electrons and forms
C4+ cation. But it would require a huge
amount of energy to remove four
electrons leaving behind a carbon
cation with six protons in the nucleus
holding on to that two electrons.
(ii) The molecular formula of alcohol is
C2H5OH (ethanol) and of acid is H
COOH (methanoic acid)
This is why carbon shows the tendency
to share electrons with carbon atoms
or with other elements.
(C) The reverse of esterification is saponification.
Alkaline hydrolysis of an ester to give the
salt of the corresponding carboxylic acid
and the alcohol is called saponification.
CH3 COOC2 H5 + NaoH → CH3 COO Na +
C2 H5 OH
45. (B) The bonds formed by carbon with other
elements are strong as the size of carbon
atom is quite small which enables its
nucleus to hold on to the shared pairs of
electrons strongly.
48. (C) The ingredient which is added to the
aqueous solution to make the soap come
out of solution is salt. When we add
common salt to the solution, the solubility
of the soap present in it decreases, due
to which all the soap separates out from
the solution in the form of a solid. Though,
most of the soap separates out on its own
but some of it remains in solution which is
precipitated out by the use of common salt.
46. (C) Isomers of C5 H11 OH
H H H H H
| | | | |
(i) —C—C—C—C—C—OH
| | | | |
H H H H H
H H H H
| | | |
—H—C—C—C—C—OH
(ii)
| | | |
H H | H
|
H—C—H
|
H
H
|
H—C—H
|
H | H
| | |
(iii) H—C—C—C—OH
| | |
H | H
|
H—C—H
|
H
H H H OH H
| | | | |
(iv) H —C—C—C—C—C— H
| | | | |
H H H H H
H H OH H H
| | | | |
(v) H —C—C—C—C—C— H
| | | | |
H H H H H
H COO H + HO C2 H5 → H COOC2 H5 +
H2O
49. (A) (b) CnH2n+2
Explanation: CnH2n+2 is the general formula
for saturated hydrocarbons.
Saturated
hydrocarbons
are
the
compounds of carbon and hydrogen which
have a single covalent bond between two
carbon atoms. These are also known as
alkanes and have the general formula given
by CnH2n+2, where n is the number of carbon
atoms.
Related Theory

The first five members of the alkanes is given in the
following table:
Name
Molecular Formula
Methane
CH4
Ethane
C2H6
Propane
C3H8
Butane
C4H10
Pentane
C5H12
Carbon and its Compounds
19
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(B) (d) methane
Explanation: Methane is the main
component of natural gas and is a
naturally occurring hydrocarbon gas
mixture consisting primarily of methane,
but commonly including varying amounts
of other higher alkanes, and sometimes
a small percentage of carbon dioxide,
nitrogen, hydrogen sulphide, or helium.
H
x.
C
x.
x.
H
x.
H
H
(E) (b) 3
Explanation: Structural isomers are the
compounds with identical molecular
formula but different structures.
The molecular formula of pentane is C5H12,
as it is an alkane with five carbon atoms. It
has three isomers as shown:
H H H H H
| | | | |
H—C—C—C—C—C—H
| | | | |
H H H H H
H H H H H
| | | | |
H—C—C—C—C—C—H
| | |
|
H H H
H
to ethanol it has been observed that the
purple colour of potassium permanganate
first becomes colourless and then does not
change on adding further.
(C) (a) sodium ethoxide and hydrogen
Explanation:
CH3 CH2 OH + Na → CH3 CH2ONa +
(sodium ethoxide)
1
H2
2
(Hydrogen)
(D) (b) Methanol
Explanation: Ethonal is added with
methonal to prevent its misuse as a
beverage.
(E) (a) CH2 == CH2
Explanation: On heating ethanol at 443 k
with conc. H2SO4 in dehydrotion of ethanol
to produce ethene.
Conc.H2 SO4
CH3CH2OH 
 CH2=CH2 + H2O
443 K
52. (C) Formation of bond between carbon and
chlorine atoms.
|
H—C—H
|
H
H
|
H—C—H
H
H
|
|
|
|
H—C — C — C — H
|
|
H
H
Carbon has four electrons in the valence
shell and need four more electrons to
complete its octet by acquiring the stable
electronic configuration of the nearest noble
gas, neon. Chlorine’s atomic number is 17
and its electronic configuration is 2, 8, 7, M
shell needs only one electron to complete
its octet by acquiring the electronic
configuration of the nearest noble gas,
argon. In order to complete its octet, carbon
shares its four valence electrons with one
electron each of four chlorine atoms to
form a molecule of carbon tetrachloride.
The four shared pairs of electrons form C-Cl
single covalent bonds.
H—C—H
|
H
50. (B) The melting point of pure ethanoic acid is
very low. On cooling, it freezes to form ice
like flakes and it looks like a glacier. Due to
this property, it is called glacial ethanoic
acid or glacial acetic acid.
(C) When an ester is treated with alkali (NaOH)
the ester is converted back to alcohol and
sodium salt of carboxylic acid.
CH3COOC2H5(aq)
Ethyl Ethanoate
NaOH
20
in which the carbon atoms have only single
covalent bonds between them.
Formula of any one saturated hydrocarbon is
CH4, C2H6, C3H8, C4H10, C5H12
(Any 1 of 5 can be written to get full marks)
C2H5OH(aq) +
Ethanol
CH3COONa(aq)
Sodium ethanoate
This process is called saponification process
as there is ester for the preparation of soap.
51. (B) (c) (I), (II) and (IV)
56. A saturated hydrocarbon is a hydrocarbon
Explanation: When a 5% solution of
alkaline potassium permanganate is added
Related Theory

The general formula of saturated hydrocarbons or
alkanes is CnH2n+2, where n is the number of carbon
atoms in one molecule.
Science Class X
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The saturated hydrocarbons are not very reactive.


protons in its nucleus holding on to just two
electrons.
The saturated hydrocarbons generally give a clean
flame. This is because the percentage of carbon is
comparatively low which gets oxidized completely
on combustion.
(2) Carbon cannot form C4– anion by gaining
four electrons, as it would be difficult for
the nucleus with six protons to hold on to
ten electrons.
Examples of compounds formed by carbon:
(1) Methane (CH4)
(2) Ethene (C2H4)
(3) Propyne (C3H4)
(4) Ethanol (C2H5OH)
(Any 1 of 4 examples can be written to get full
marks)
A hydrocarbon in which the two carbon atoms are
connected by a double bond or a triple bond is called
an unsaturated hydrocarbon. Alkenes and Alkynes
are unsaturated hydrocarbons
59. Covalent compounds are poor conductors of
electricity because these compounds do not
have any charged particles as they are made
by sharing of electrons.
Related Theory

Ionic compounds conduct electricity in molten state
or in solutions. A solution of ionic compound contains
ions which move to the opposite electrodes when
electricity is passed through the solution. Ions move
freely and conduct electricity.
Related Theory

61. Benzene is a cyclic unsaturated hydrocarbon.
Depending upon the number of electron pairs
shared, there can be single, double or triple covalent
bond.
—
H
C —
—
H —
—H
C—
C
—
H
—
—
C
H
Explanation:
Benzene
is
unsaturated
hydrocarbon as it has more than one double
bond and all the carbon atoms are arranged in
the form of a ring. Its molecular formula shows
it has 6 carbon atoms and 6 hydrogen atoms
and it has 3 carbon-carbon double bonds, 3
carbon-carbon single bonds and 6 carbonhydrogen single bonds.
63. Molecular formula of first two members of the
homologous series having functional group –Cl
is CH3Cl and C2H5Cl.
Related Theory

Homologous series is a series of carbon compounds
in which the hydrogen in a carbon chain is replaced
by the same functional group.
The IUPAC name of the compound CH3Cl is
Chloromethane and of C2H5Cl is Chloroethane.

Covalent compounds are usually liquids or gases,
have usually low melting and boiling points, are
usually insoluble in water but soluble in organic
solvents and do not conduct electricity.
74. 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.
Carbon exhibits catenation much more than
silicon. In-fact no other element exhibits the
property of catenation to the extent seen in
carbon compounds.
Silicon forms compounds with hydrogen
which have chains of upto seven or eight
atoms, but these compounds are very reactive.
The carbon-carbon bond is very strong and
hence stable. This gives us a large number of
compounds with many carbon atoms linked to
each other.
Carbon has a valency of four and it is capable
of bonding with four other atoms of carbon or
atoms of some other mono-valent elements..
—
C
—
C
—C—
—
—
—
H
Carbon can share 1, 2 or 3 electron pairs with other
carbon atoms or with atoms of other elements to
achieve noble gas configuration.
The functional group Cl is a halogen which belongs
to Group 17 of the modern periodic table and other
members of the halogen functional group are Br, I.
The bonds formed by carbon atoms are very
strong and do not break easily so carbon
compounds are stable.
73. Carbon forms covalent compounds with other
77. (A) CH3COCH2CH2CH2CH3 : Ketone (>C=O)
(1) Carbon cannot form C4+ cation by losing
four electrons, as it would require a large
amount of energy to remove four electrons
leaving behind a carbon cation with six
or
(B) CH3CH2CH2COOH : Carboxylic acid (—COOH)
or —C—OH
==
atoms by sharing electron pairs because of the
following reasons:
O
Carbon and its Compounds
21
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Shared pair
(C) CH3CH2CH2CH2CHO : Aldehyde (—CHO)
or
R
R
(D) CH3CH2OH : Alcohol (—OH)
R—R
Related Theory

The shared pair constitutes a single
covalent bond between two ‘R’ atoms and
is represented by a single line (–). ‘R’ shares
only one electron so its valency is 1.
In case of element ‘Q’, it has one valence
electron. It is a metal and it can attain its
stable configuration by losing one electron
and form ionic or electrovalent bond.
‘S’ has already its stable configuration; it
cannot gain or lose or share the electron
and its valency is zero so it is an inert gas.
A functional group in an organic compound is an
atom or a group of atoms bonded together in a
unique fashion, which is usually the site of chemical
reactivity in an organic molecule.
78. (A) P and R
[CBSE Marking Scheme Term-2 SQP 2022]
Explanation: The number of electrons are
either lost or gained or shared by one atom
of an element to achieve the nearest inert
gas electronic configuration, which gives us
the valency of the element.
Element
Electronic
Valence
Valency
Configuration Electrons
P
2, 6
6
–2
Q
2, 8, 8, 1
1
+1
R
2, 8, 7
7
–1
S
2, 8, 8
8
0
Related Theory

The two pairs of electrons on each P atom and three
pairs of electrons on each R atom are not involved
in bond formation and are called unshared pairs or
lone pairs. The elements P and R are oxygen and
chlorine atoms.
(B) Carbon has a valency four or tetravalency
and catenation
[CBSE Marking Scheme Term-2 SQP 2022]
Explanation: Carbon is a non-metal with
atomic number (Z) equal to 6. Its electronic
configuration is 2, 4. Thus, carbon has four
valence electrons. It does not take part in
the ionic bond formation as it can neither
gain nor lose four electrons. Therefore,
carbon completes its octet by sharing
four electrons present in the valence shell
with the electrons of other atoms. This
means that carbon is tetravalent or has
a valency of four. Carbon has the unique
ability to form bonds with other atoms of
carbon, gives rise to large molecules and
the property is called catenation, Carbon
can react with P and forms CP2.
From the table, it is clear that P and R are
non-metals. We know covalent bonds are
formed between non-metals by sharing
the electrons. In case of ‘P’ we see the
formation of a double bond between
between two P atoms. An atom of ‘P’ has 6
electrons in L shell and it requires two more
electrons to complete the octet as that of
the nearest gas neon so each atom of ‘P’
shares two electrons with another atom of
‘P’ to give the structure P2.
Two shared pairs
P
P
P
C
P
P=P
The two electrons contributed by each
‘P’ atom give rise to two shared pairs of
electrons so a double bond is formed
between two P atoms and ‘P’ is divalent i.e.
P2. P shares two electrons so its valency is
2.
In case of ‘R’, it has 7 electrons in its M
shell and it requires one more electron to
complete its octet as that of the nearest
gas argon. So two ‘R’ atoms share their
electrons to form a molecule of ‘R’ i.e., R2.
The electron dot structure of ‘R’ molecule is:
22
P=C=P
Carbon can also react with R and forms
CR4.
83. (B) C4H8 and C5H10 are homologues as they:
(i) “- CH2-”
(ii) Differ in 14u molecular mass
(iii)Same functional group
(iv) Same general formula
(Any two reasons)
[CBSE Marking Scheme Term-2 SQP 2022]
Science Class X
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Detailed Answer: A homologous series is a
group of organic compounds having similar
structures and similar chemical properties
in which the successive compounds differ
by – CH2 group. The general fromula of
the homologous series of alkenes is CnH2n
where n is the number of carbon in one
molecule of alkene.
(1) C4H8 and C4H10 can be represented by
the general formula CnH2n.
(2) C4H8 and C5H10 are two adjacent
homologues which differ by 1 carbon
atom and 2 hydrogen atoms in their
molecular formulae i.e., —CH2.
(3) The difference in the molecular masses
of these two homologues is 14 u.
(4) These two homologues show similar
chemical properties.
(5) The members of homologous series
show a gradual change in their physical
properties with increase in molecular
mass. For example, as the number of
carbon atoms per molecule increase,
the melting points, boiling points and
densities of its members also increase
gradually.
87. (A) Ethanol can be converted to ethene by
dehydration of ethanol in the presence of
concentrated sulphuric acid at 170°C. The
reaction is named as dehydration reaction.
90. (A) Carboxylic acid is ethanoic acid:
Carboxylic acid with molecular formula
C2H4O2 is acetic acid or ethanoic acid
(CH3COOH) having the structure as:
H
O
|
H—C—C
|
O—H
H
(B) Alcohol is ethanol:
The given alcohol forms acetic acid on
oxidation with alkaline KMnO4, followed by
acidification. Therefore, it must be ethanol
with structure CH3—CH2—OH.
Alkaline KMnO4
CH3CH2OH →
CH3COOH
Acidification
(C) X is ethyl ethanoate:
The reaction of ethanoic acid with
ethanol in the presence of a few drops of
conc. H2SO4 is an esterification reaction
that forms an ester, ethyl ethanoate
(CH3COOC2H5).
H SO
2
4
CH3—COOH + C2H5OH →
Ethanoic acid
CH3—COOC2H5 + H2O
Ethyl ethanoate
92. (A) (i) The number of covalent bonds in one
molecule of ethanol is 8.
Conc. H2SO4
CH3CH2OH →
CH2==CH2 + H2O
160°-170°C
Ethanol
Ethene
(B) Propanol can be converted to propanoic
acid by oxidation in the presence of alkaline
potassium permanganate and heat i.e., by
the oxidation of propanol using an oxidising
agent such as alkaline KMnO4. It is named
as oxidation reaction.
Alk. KMnO4
CH3CH2CH2OH →
CH3CH2COOH
Heat
Propanol
Propanoic acid
88. (1) Test 1 (Litmus Test) Take two strips of blue
litmus paper. Place a drop each of the
alcohol and carboxylic acid on these strips
separately. The blue litmus paper turns red
in the case of carboxylic acid and remains
unaffected in the case of alcohol.
(2) Test 2 (Sodium hydrogen carbonate test /
sodium carbonate test)
A pinch of sodium hydrogen carbonate
or sodium carbonate is added, to both
separately. If brisk effervescence with the
evolution of a colorless gas is observed, it
indicates the presence of carboxylic acid.
If no change is observed then it confirms
the presence of the alcohol.
(3) Test 3 (Ester test or any other suitable
test)
(Any two)
[CBSE Marking Scheme 2015]
Ethanol
H H
| |
H—C—C—O—H
| |
H H
(ii) The acid that is used to convert ethanol
to ethene is concentrated sulphuric acid
(H2SO4).
(B) Concentrated sulphuric acid removes
water form ethanol, that is why it is called
dehydrating agent.
conc. H2 SO4
CH3 – CH2 OH 

443 K
H
H
C =C
H
H
+ H2O
(C) As you can observe from the table, the
volume of gas collected decrease with
the passage of time. The decrease in the
collection of volume of gas suggest that
the reaction slowed down because sodium
has been used up.
(D) The gas released is hydrogen gas.
95. Compound
C with molecular formula
C2H4O2 contains two oxygen atoms, so it can
be either ester or carboxylic acid.
Since it reacts with sodium metal to form
compound R and evolves a gas which burns
with a pop sound, therefore, it should be a
carboxylic acid which forms sodium ethanoate
and hydrogen gas with sodium metal.
Carbon and its Compounds
23
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2CH3COOH + 2Na → 2CH3COONa + H2
(C)
(R)
The gas which burns with a pop sound is
hydrogen gas. Reaction of ethanoic acid with
alcohol in the presence of an acid (conc. H2SO4)
forms sweet smelling ester. So, compound S,
that is formed due to the reaction of ethanoic
acid and methanol (A), is methyl ethanoate
with molecular formula C3H6O2 and structural
formula CH3COOCH3.
Conc. H2SO4
CH3COOH + CH3OH →
(C)
(A)
CH3COOCH3 + H2O
(S)
Adding NaOH to C, gives R and water.
CH3COOH + NaOH → CH3COONa + H2O
(R)
S on treatment with NaOH solution gives back
R and A.
CH3COOCH3 + NaOH → CH3COONa + CH3OH
(R)
(R)
(A)
Hence, compound
C = Ethanoic acid (CH3COOH),
R = Sodium ethanoate (CH3COONa),
A = Methanol (CH3OH) and
S = Methyl ethanoate (CH3COOCH3).
24
98. Combustion of compound C forms two moles
of CO2 and three moles of H2O. It must contain
two carbon atoms and six hydrogen atoms
with molecular formula C2H6, that is ethane.
Compound C is ethane (C2H6).
Ethane i.e., compound C is formed by the
addition of one mole of hydrogen to compound
B in the presence of Ni catalyst. So, compound
B must be unsaturated hydrocarbon with
two carbon atoms, that is ethene (CH2==CH2)
and obtained by heating compound A with
concentrated H2SO4 which shows it to be an
alcohol.
So, compound A could be C2H5OH (Ethanol).
Hot conc. H2SO4
C2H5OH → C2H4 + H2O
(A)
(B)
Ni
C2H4 + H2 → C2H6
(B)
(C)
2C2H6 + 7O2 → 4CO2 + 6H2O + Heat and light
(C)
Hence;
A = Ethanol (CH3CH2OH),
B = Ethene (CH2==CH2) and
C = Ethane (CH3—CH3)
Science Class X
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ANSWER SHEET
SELF PRACTICE
4. (b) II and IV
portion of leaf inside the bottle having NaOH
did not prepare any food. The portion of leaf
which was outside the bottle having NaOH
showed the positive test with iodine solution as
that portion has made starch in five hours by
taking carbon dioxide from the atmosphere so
this activity suggests that CO2 is necessary for
the process of photosynthesis.
Explanation: Chloroplasts absorb sunlight and
are found in the palisade mesophyll cell (II) and
guard cell (IV). The majority of chloroplasts are
concentrated in palisade cells to absorb as
much sunlight as possible for photosynthesis.
6. (c) Valves ensure that the blood does not flow backwards.
Explanation: One of the functions of blood is
to transport oxygen, carbon dioxide, enzymes,
hormones etc. from one part of body to
another so this statement is incorrect.
Human heart has four chambers—2 Atria and
2 ventricles so statement (b) is incorrect.
Valves prevent backward flow of blood as they
are located at each end of two ventricles and
act as one way inlet.
Both oxygen–rich and oxygen-deficient blood
does not get mixed in the human heart as it
has four chambers. The left side chamber-left
atrium and left ventricle carry only oxygenated
blood. Whereas right atrium and right ventricle
have only deoxygenated blood and they do
not get mixed in the process of circulation.
Hence, this statement is incorrect.
Related Theory

Our heart has 4 valves.
(1) Tricuspid valves: Present between right atrium
and right ventricle.
(2) Pulmonary valve: Present between
ventricle and the pulmonary artery.
right
35. (c) (A) is true but (R) is false.
Explanation: Transpiration is the process by
which plants lose extra water off their surface
with the help of the stomata of the leaves.
The plants usually lose water during very hot
season. As a result, the temperature of the
plant decreases and it cools down. However,
transpiration does not help in translocation of
food in the plants.
Related Theory

Transpiration is an important plant process. The
three important tasks that are accomplished due to
transpiration are:
(1) Decrease in temperature of the plant: When
the plants lose water, the temperature of the
plant decreases and therefore, the plant cools
down.
(2) Transpirational pull: When plants lose water, an
empty space is created in the plants. As a result,
the minerals and water from the root are pulled
up. This process is called transpirational pull.
(3) Plant turgidity: Transpiration helps plants in
maintaining the turgidity of the plants.
38. (c) (A) is true but (R) is false.
Explanation: The hearts of amphibians have
three chambers: two atria and one ventricle.
Because of how the withdrawals occur
between the atria, the mixing of oxygenated
and deoxygenated blood is kept to a minimum.
Since amphibians don't require a lot of energy,
they can withstand some amounts of blood
that is both oxygenated and deoxygenated
mixing.
(3) Mitral valve: Present between the left atrium
and the left ventricle.
(4) Aortic value: Present between left ventricle and
aorta.
23. (b) (II) and (III)
Explanation: Plants need nitrogen for the
process of protein synthesis. However, they
cannot utilise nitrogen as it is. It is converted
into inorganic form such as nitrate and nitrite
and in organic form as urea by the action of
various microorganisms. It is then taken up by
the plants for various processes.
30. (b) Carbon dioxide is necessary for preparing
carbohydrate
Explanation: When the student places the
plant in light for 5 hours, The portion of
leaf outside the bottle prepared food by
the process of photosynthesis. Whereas the
41. (d) (A) is false but (R) is true.
Explanation: Oxygenated blood flows in
pulmnary vein and arteries have narrow
lumen.
44. (B) Insulin
Related Theory

A person who is diabetic, his body either does
not produce enough insulin or is not able to use it
properly.
Life Processes
25
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45. (C) It is necessary to separate the oxygenated
blood from mixing with deoxygenated
blood as mammals and birds have high
energy needs because they constantly
require energy to maintain their body
temperature.
47. (A) In test tube A, Lime water turned milky
immediately and in test tube B: Lime water
turned milky after a long time Lime water turns milky when carbon
dioxide gas is passed through it as a white
precipitate of calcium carbonate is formed
when lime water (calcium hydroxide)
reacts with carbon dioxide gas.
Lime water turns milky in test tube A as
the exhaled air is rich in carbon dioxide.
Whereas, lime water takes a lot of time to
turn milky in test tube B as the amount of
carbon dioxide present in atmospheric air
is very less as compared to exhaled air and
hence, carbon dioxide is produced after a
long time.
(B) We can say that the amount of carbon
dioxide is more in exhaled air and very less
in the atmosphere.
(i) This activity demonstrates the process
of fermentation.
(ii) The
products
formed
during
fermentation
are
ethyl
alcohol
(ethanol), carbon dioxide and energy.
(C) ATP (Adenosine Triphosphate) is the energy
currency in the living organisms. It is
produced during cellular respiration. ATP is
used to fuel all the activities in the cell. In
these processes, ATP is broken down giving
rise to a fixed amount of energy which can
drive the endothermic reactions taking
place in the cell.
49. (B) (b) To know how much oxygen is flowing in
the blood.
Explanation: Oxygen measurement is
must since it indicates the flow of the blood
inside our body is appropriate or not so
as to ensure the smooth function of the
circulatory system.
(C) (d) I, II and III or either of these
Explanation: COVID-19 influences blood
clotting and lead to lack of oxygen severe
inflammation also takes place during the
infection of COVID-19.
50. (D) (c) transpiration pull
Explanation:
The transpiration pull
becomes the major driving force in the
movement of water in the xylem during
day when the stomata are open and helps
in the absorption and upward movement
of water and minerals dissolved in it
from roots to the leaves. It also helps in
26
temperature regulation. The effect of root
pressure in transport of water is more
important at night.
(E) (c) hypotonic
Explanation: As the external medium is
hypotonic a solution containing lower
concentration of solutes as compared to
another cell) the root cells in contact with
the soil actively take up ions. This creates
a difference in the concentration of these
ions between the root and the soil. Water
therefore, moves into the root from the soil
to eliminate this difference.
51. (B) (b) Urea
Explanation: Glucose, amino acids, some
salts and excess of water are reabsorbed
from the filtrate the blood capillaries in
tubular region. The primary toxic waste in
our blood filtered by the kidney is urea.
(E) (b) Bowman’s capsule
52. (B) (c) Plasma, haemoglobin and iron
Explanation: Haemoglobin consists of the
Haemo group where the Fe atom is bonded
as the central atom in the complex and is
present in the blood and the blood flows in
the plasma. Thus, it is the combination of
plasma, haemoglobin and iron.
(E) (a) more in arteries and less in vein
Explanation: Since the arteries carry the
pure or the oxygenated blood from the
heart to the different body parts while
the veins carry the deoxygenated or the
impure blood from the different body parts
to back to the heart thus the force that
blood exerts against the wall of a vessel is
more in arteries and less in vein.
53. (C) (b) mucus
Explanation: The inner lining of the
stomach is protected by the mucus so as to
protect it from the corrosive action of the
hydrochloric acid.
(D) (c) Temperature can influence the action of
an enzyme
Explanation: As is evident from the graph
with the increase in the temperature the
enzyme activity is also influenced by the
temperature. Thus this point is valid in the
context of the graph.
62. If the pancreas of a person stops functioning,
the digestion of starch, fats and proteins will
be affected as pancreatic amylase, lipase
and trypsinogen secreted by pancreas acting
on them respectively will not be secreted
anymore.
67. The blood has platelet cells which circulate
around the body and help to clot the blood
at the time of injury. In the absence of
platelets, the process of clotting will be
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affected. Platelets play the important role of
coagulation of blood. No blood coagulation
would take place in the absence of platelets.
This would be a dangerous situation in case of
an injury, as it would result in excessive blood
loss and can even prove lethal for the person.
73. Respiration is the process of oxidation of food
substances that release carbon dioxide; it takes
place throughout the day. Photosynthesis
is the process of fixing the atmospheric
carbon dioxide into organic compounds
(carbohydrates) with release of oxygen gas as
by-product. Photosynthesis takes place during
the day in the presence of Sunlight only.
The CO2 released during respiration is
absorbed by plants for photosynthesis during
day causing net release of oxygen only. If a
plant is releasing carbon dioxide and taking
in oxygen during the day, it means that either
rate of photosynthesis is too slow to counter
balance the release of CO2 by respiration or
photosynthesis is not taking place at all.
78. (B) Plants are fixed at a place and do not show
any locomotion. Plants are less active.
Plants have a large portion of dead cells
in many tissues. Their cells do not need
to be supplied with materials so quickly.
Therefore, plants have low energy needs
and respire at a slow rate.
81.
Glucose
(6-carbon molecule)
In cytoplasm
Pyruvate
molecule glucose takes place in the cytoplasm
of cells of all organism and we obtain a 3carbon molecule compound called pyruvate.
Further pathway for breakdown of this
pyruvate depends on the amount of oxygen
available which are given ahead:
(1) Anaerobic Respiration: It takes place in
the absence of oxygen. In this process,
the pyruvate is converted into ethanol
and carbon dioxide. For example, in yeast
during fermentation.
(2) Lack of Oxygen: When there is a lack
of oxygen in our body, the pyruvate is
converted into lactic acid, which is a
3-carbon molecule, and energy. This takes
place in our muscles during vigorous
exercise or activity.
n
xy ge
of o
e
c
n
e
)
Abs
east
(in y
→


Lack of oxygen
(1) Arteries
(2) Veins
(3) Capillaries
One important feature of each type is given
below:
Arteries:
(1) These carry oxygenated blood from the
heart to different organs.
(2) They have thick elastic walls since blood
emerging from the heart is under high
pressure.
(3) These do not have any valves
Veins:
(1) These carry dexygenated blood from
different organs to the heart.
(2) These are thin-walled blood vessels as
blood is no longer under pressure.
(3) These have valves to ensure that blood
flows only in one direction.
Capillaries:
(1) These are one-cell thick vessles and hairlike blood vessels.
(2) These help in exchange of food materials,
gases and waste.
(3) These do not have valves
(Any 1 of 3 points can be written to get full
marks)

→




→
(in our muscle cells)
85. Blood vessels are of three kinds:
89. The first step in breakdown of the 6-carbon
(3-carbon molecule)
Pres
enc
e of
oxy
(in m
gen
ito c
hon
dria
)
(3) In order to protect this surface, it is usually
placed within the body, so there have to be
passages that will take air to this area. In
addition, there is a mechanism for moving
the air in and out of this area where the
oxygen is absorbed.
Carbon dioxide +
Lactic acid + Energy
Water + Energy Ethanol + Carbon (3-carbon molecule)
dioxide + Energy
(2-carbon molecule)
[CBSE Marking Scheme 2019]
84. The common characteristic features of the
organs performing respiration in terrestrial
organisms are:
(1) These organs have a structure that
increases the surface area which is in
contact with the oxygen-rich atmosphere.
(2) This surface is very fine and delicate since
the exchange of oxygen and carbon dioxide
has to take place across this surface.
Related Theory

Aerobic respiration takes place in the presence of
oxygen in mitochondria and pyruvate is broken
down to give carbon dioxide, water and energy.
Life Processes
27
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
The release of energy in aerobic respiration is much
more than anaerobic respiration.
 The
energy released during respiration is used to
synthesize a molecule called ATP, which is the energy
currency for most cellular processes.
90. (A)
Cloudy days will cause the rate of
photosynthesis to decrease due to the low
availability of sunlight.
(B) No rainfall in the area will cause the rate of
photosynthesis to decrease due to the low
availability of water.
(C) Good manuring in the area will cause the
rate of photosynthesis to increase due to the
improved availability of minerals necessary
for plant growth and development.
(D) The blockage of stomata due to dust
will cause the rate of photosynthesis to
decrease. This is because the CO2 required
for photosynthesis enters the leaves
through the stomata. If the stomata get
blocked due to dust, less CO2, or no CO2
would enter the leaves or leaf and the rate
of photosynthesis will drop.
93. (A) Importance
of nutrition: All
organisms need nutrients (food):
(1) To build up their body molecules
(2) To repair their worn out tissues.
(3) To get energy for doing work.
(4) To maintain various life processes.
living
Related Theory

Nutrition is the process of intake of nutrients and
its utilization by an organism in various biological
activities.
(B) Rhythmic contraction and expansion
movement of muscles lining various organs
of alimentary canal pushes the partially
or completely digested food forward in
the track. This rhythmic contraction and
relaxation movement is called peristaltic
movement. This occurs throughout the gut.
(C) Small intestine in herbivores is longer
than in carnivores as herbivores are grass
eating animals. Longer, small intestine of
herbivores help in digestion of cellulose
which is present in the grass. Meat is
digested easily so carnivores have shorter
small intestine.
(D) Mucus is secreted by gastric glands along
with hydrochloric acid and pepsin in the
stomach. If mucus is not secreted by the
gastric glands, hydrochloric acid would
corrode the inner lining of stomach and
may cause gastric ulcers.
28
94. Human respiratory system:
Passage of air: In human beings, air is inhaled
through the nostrils. The air passing through
the nostrils is filtered by fine hairs that line the
passage. The passage is also lined with mucus
which helps in this process. From here, the air
passes through the throat and into the lungs.
Rings of cartilage are present in the throat.
These ensure that the air-passage does not
collapse.
Gaseous exchange: Within the lungs, the
passage divides into smaller and smaller tubes
which finally form balloon-like structures which
are called alveoli. The alveoli provide a surface
where the exchange of gases can take place.
The walls of the alveoli contain an extensive
network of blood-vessels.
Role of diaphragm: When we breathe in, we
lift our ribs and flatten our diaphragm and
the chest cavity becomes larger as a result.
Because of this, air is sucked into the lungs and
fills the expanded alveoli.
Function of rib muscles and alveoli: The blood
brings carbon dioxide from the rest of the body
for release into the alveoli and the oxygen in
the alveolar air is taken up by the blood in the
alveolar blood vessels to be transported to all
the cells in the body. Respiratory pigment.
(haemoglobin) takes up oxygen from the air
in the lungs and carries it to tissues which are
deficient in oxygen before releasing it. Carbon
dioxide is more soluble in water than oxygen
is, and hence is mostly transported in the
dissolved form in our blood.
96. (A)
Excretory system in human beings (also
known as urinary system) consists of:
(1) a pair of kidneys
(2) a pair of ureters
(3) urinary bladder
(4) urethra
(B)The basic filtration unit in the kidneys is the
nephron. Major processes involved in the
formation of urine are:
Ultra filtration of substances
Selective reabsorption of substances in the
initial filtrate such as glucose, amino acids,
salts and a large amount of water.
Filtration units in kidneys: Nephrons
↓
Each nephron has cup shaped structure
(Bowman’s capsule)
↓
which leads into bundle of blood capillarilies
called glomerulus
↓
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Ultra filtration of blood in glomerulus
↓
Selective reabsorption of useful substances
like glucose, amino acids, water and salts
from filtrate in tubular part of nephron.
↓
Secretion of water and salts from blood
capillaries into the tubular parts of nephron
↓
Liquid urine passes to collecting duct of
kidney
↓
Urine carried to ureter
↓
Urine passes to urinary bladder
Urine stored in the bladder for sometime and
ultimately passes out of body through urethra
and urethral opening.
(2) Saliva contains the enzyme salivary
amylase that breaks some starch into
sugar.
(B) HCl in stomach:
(1) The hydrochloric acid creates an acidic
medium which facilitates the action of
the enzyme pepsin.
(2) It also kills germs or pathogens.
(C) Bile juice:
(1) It makes the acidic food coming from
the stomach alkaline, so as to enable
the action of pancreatic enzymes.
(2) It contains bile pigments and bile salts
which carry out emulsification of fats.
(D) Villi:
(1) These are finger-like projections
present in the inner lining of the small
intestines which increase the surface
area for absorption.
Related Theory

Human urine contains about 95% water and 5%
nitrogenous substances and a few other substances.
The amount of water reabsorbed during selective
reabsorption depends on how much of dissolved
waste is there to be excreted. We can control the
urge to pass out urine because urinary bladder is
muscular, so it is under our nervous control.
(2) In the large intestine villi absorb water
from the unabsorbed food.
101. (A) Oxygenated : B/D/F
[B = left ventricle/D = aorta/F = left auricle/
pulmonary vein]
98. Functions of the human heart are listed
below:
(1) The human heart is a muscular organ that
pumps oxygenated blood throughout the
body.
(2) It controls heart rate.
(3) It maintains blood pressure
(4) It pumps hormones and other vital
substances to different parts of the body.
Double circulation is necessary as the
separation of oxygenated and deoxygenated
blood allows a more efficient supply of
oxygen to the body cells as mammals have
high energy needs.
100. Function of given substances in
(A) Saliva:
(1) Saliva moistens the food so it can be
swallowed easily.
(B)
Deoxygenated: A/C/E [A = right ventricle/C
= pulmonary artery/E = right auricle/vena
cava](any two)
Pulmonary
artery
(Deoxygenated
blood)
Lungs
6
Main vein
(Vena cava)
(Deoxygenated
blood)
Right
atrium
1
Pulmonary
vein
(Oxygenated
blood)
3
Main artery
(Aorta)
(Oxygenated
blood)
Left
atrium
5
V2
V1 2
Left
Right
ventricle ventricle
Heart
4
Body organs
(1 mark should be deducted if the
arrows are not correctly marked)
[CBSE Marking Scheme 2019]
Life Processes
29
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ANSWER SHEET
SELF PRACTICE
3. (c) insuffiicient growth of the body
Explanation: Pituitary dwarfism is the
condition that affects the normal functioning
of the pituitary gland, caused by insufficient
amount of growth hormone in the body.
6. (c) (II) and (III)
Explanation: The fore brain (not hind brain)
is the main thinking part of the brain. It has
regions which receive sensory impulses from
various receptors. Separate areas of the fore
brain are specialised for hearing, smell, sight
and so on.
All the involuntary actions, including blood
pressure, salivation and vomiting, are
controlled by the medulla in the hindbrain.
Cerebellum is responsible for the precision of
voluntary actions and maintaining the posture
and balance of the body.
13. (c) rapid cell divisions in tendrillar cells that are away from the support
Explanation: Pea plants climb up other plants
or fences by means of tendrils. These tendrils
are sensitive to touch. When they come in
contact with any support, the part of the
tendril in contact with the object does not
grow as rapidly as the part of the tendril away
from the object. This causes the tendril to circle
around the object and thus, cling to it.
14. (b) chemotropism
Explanation: Chemotropism is the growth
of pollen tubes towards ovules. Pollen grain
responds to stimulus by growing a pollen tube
in downward direction, towards the ovule for
fertilization.
22. (b)
Spinal cord Message to
Brain
(CNS)
↓
Receptors (Skin)
↓
Sensory neuron
↓
Spinal cord
↓
Motor neuron
↓
Effector (Muscle)
↓
Response (Hand withdrawn)
A stimulus received by the receptors present
on the skin, is transmitted to the sensory nerve
which carries it to the central nervous system
(the spinal cord and brain). A motor neuron
carries the message from the central nervous
system to the effector which is muscle on the
skin and therefore, the withdrawal of hand
takes place.
26. (a) Both (I) and (III)
Explanation: Movement happens at a point
different from the point of touch. The plants
also use electrical-chemical means to convey
information from cell to cell. There are no
specialised tissues in plants for the conduction
of information. Plant cells change shape by
changing the amount of water in them.
31. (c) (A) is true but (R) is false.
Explanation: Just like animals, plants also
convey information from cell to cell. The plants
use electrical-chemical means to convey
this information from cell to cell, but unlike in
animals, there is no specialised tissue in plants
for the conduction of information.
38. (A) Thyroid gland is situated at the front of the
neck, just below the larynx (Adam’s apple).
Sensory neuron
39. (A) Each and every food item sets off an
Motor
neuron
Relay neuron
E ector = muscle in arm
Receptors = heat/pain
receptors in skin
30
Explanation: The sequence of events that occur
when we touch a hot plate are as follows:
Heat (Stimulus)
aroma that travels along and reaches
the olfactory receptors. These olfactory
receptors are present in the sense organnose. The olfactory receptor cells are
present in very large numbers and are
grouped together within a small area in
the back of nasal cavity. These olfactory
receptors detect smell.
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the signals are passed on to a motor
neuron. In vertebrates, most sensory
neurons do not pass directly into the brain,
but synapse in the spinal cord does. The
spinal cord acts as a link between spinal
nerves and the brain.
Related Theory

Our sense of smell is responsible for about 80% of what we taste without our sense of smell, our sense of taste is limited to only five distinct sensations:
Sweet, Salty, Sour, Bitter and “Umami” or Savoury sensation.
(B) We have got different types of taste buds
present on the tongue. Taste buds are
sensory taste receptors i.e., gustatory
receptors that help to perceive the taste.
We can taste the two drinks and identify
their tastes with the help of gustatory
receptors.
Smell and taste are closely linked. The
gustatory receptors present on the tongue
identify taste and olfactory receptors
present in the nose identify smell. Both
sensations are communicated to the brain.
The brain interprets and processes the
information and recognises the flavour of
two drinks.
(C) (i) When we have cold, we are not able to
taste and smell our food normally. Our
nose gets blocked during cold. On seeing
and tasting the food, the olfactory
receptors present on the tongue are not
stimulated but gustatoreceptors get
stimulated on coming in contact with
food. As smell and taste are closely
related, we will not be able to fully
appreciate the taste of food.
(ii) The olfactory receptors are present in
the nose of the human body.
The olfactory receptors sense smell.
40. (B) TSH levels in pregnant women need to be
monitored because high TSH levels and
hypothyroidism can increase the chances
of miscarriages.
Related Theory

Pregnancy places a physiological demand on
the thyroid gland which increases 40% in size and increases production of thyroxine and other hormones. Effects in any of these function can have negative implications for the long-term health of both the mother and child. Low or high TSH levels
may even affect baby’s brain development.
41. (A) The main centre for such reflex actions is
spinal cord.
Explanation: Reflex action is an involuntary
and nearly instantaneous movement in
response to a stimulus. A reflex action
involves a very simple nervous pathway
called a reflex arc. A reflex arc starts off
with receptors being excited on sensing a
stimulus. They, then send signals along a
sensory neuron to the spinal cord, where
42. (B) (a) Pituitary gland
Explanation: The pituitary gland is
known as the “master gland” because
its hormones control other parts of the
endocrine system, namely the thyroid
gland, adrenal glands, ovaries, and testes.
(C) (d) Both (II) and (IV)
Explanation: Iodine is required by the
thyroid gland to synthesize the hormone
thyroxin,
which
regulates
protein,
carbohydrates and fat metabolism in the
body. Deficiency of iodine in our diet may
cause goitre.
43. (C)If we take insufficient amount of iodine in
our diet, we are likely to have abnormal
enlargement of the thyroid gland also
known as goitre and the symptoms can
be swollen neck. We should use iodised
salt in our diet as iodine is essential for
the synthesis of hormone thyroxin which
regulates carbohydrate, protein and fat
metabolism in our body so as to provide
the best balance for growth.
44. (D) (c) cerebellum
Explanation: Cerebrum is the main thinking
part of the brain. The medulla controls
many involuntary actions including
blood pressure, salivation and vomiting.
Cerebellum is responsible for precision
of voluntary actions and maintaining the
posture and balance of the body.
Caution

Students frequently confuse the cerebellum and the cerebrum. The cerebrum, which is made up of right and left hemispheres, is the largest part of the brain.
It accounts for almost 80% of the total weight of the human brain. The cerebellum is the final component of the brain. The cerebellum is beneath the cerebrum.
(E) (a) forebrain
Explanation: The fore brain is the main
thinking part of the brain. It has regions
which receive sensory impulses from
various receptors. There are separate
areas in the forebrain which are specialised
for hearing, smell, sight and so on. There
are separate areas of association where
this sensory information is interpreted by
putting it together with information from
other receptors as well as with information
that is already stored in the brain.
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45. (B)Cerebellum is the part of brain which is
responsible for walking in a straight line.
It is located behind at the back of the
brain. Standing up or sitting down is also
controlled by cerebellum.
54. Auxin promotes growth. It is synthesised at the
shoot tip of the plant body.
60. The people were suffering from the disease
goitre which is caused due to the deficiency
of iodine in our diet. Iodine is necessary for the
thyroid gland to make the hormone, thyroxine
which regulates carbohydrate, protein and fat
metabolism in our bodies.
(C)
Pituitary gland
Thyroid gland
61. Parts of human brain which perform the
specified functions are:
(A) Sensation of feeling full: Forebrain
(B) Vomiting: Medulla (Hindbrain)
(C) Picking up a pencil: Cerebellum (Hindbrain)
(D) Riding a bicycle: Cerebellum (Hindbrain)
Pancreas
Related Theory
Testes

(i) Pancreas
(ii) Testes
(iii) Pituitary
(iv) Thyroid gland
Caution

Students often make mistakes while labelling the pituitary gland and mark it in other body parts other than the brain. The pituitary gland is a small gland at the base of the brain that is about the size of a pea.
Nerve fibres and blood vessels connect the pituitary gland to the hypothalamus (a part of the brain that controls the pituitary gland).
46. (B) (a) afferent neuron
Explanation: Afferent neuron carries the
information from the peripheral nervous
system to the central nervous system.
(D) (a) Part: Peripheral nervous system;
Function: Consists of sensory organs.
Explanation: The peripheral nervous
system consists of the nerves and ganglia
outside the brain and spinal cord.
47. (C) (a) positive geotropism
Explanation: The plant roots grow in the
direction of gravity and therefore exhibit
positive geotropism whereas plant shoots
grow opposite to the direction of gravity
and exhibit negative geotropism.
(D) (b) Gibberellin
Explanation:
Gibberellin
activates
vegetative growth of the embryo by
promoting the utilization of reserve food
which is stored in the endosperm.
The main thinking part of the brain is the forebrain
and has separate areas for hearing, smell, taste, etc.
 The midbrain controls reflex movements of the head,
neck and trunk in response to visual and auditory stimuli and also controls the reflex movements of the eye muscles, change in pupil size and shape of the eye lens.
 The hindbrain has three parts: Cerebellum, Pons and Medulla.
The cerebellum is responsible for precision of voluntary actions such as walking in a straight line and also maintains the posture and balance of the body.
Pons lies just above the medulla and takes part in
regulating respiration.
Involuntary actions such as blood pressure, vomiting,
coughing and salivation are controlled by the medulla in the hindbrain.
66. (A)Adrenal glands are located one on top of
each kidney.
(B) Testes are present only in males which
secrete the male sex hormone, testosterone.
68. (A) Pineal gland (B) Pituitary gland
(C) Thyroid gland (D) Thymus gland
Pineal gland
Pituitary gland
Thyroid gland
Thymus
49. The two components of central nervous
system in humans are brain and spinal cord.
32
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75. The hormones are secreted by the ductless/ endocrine gland.
Gland
Location of Gland
(A)
Pituitary It is located at the bottom Growth hormone secreted by pituitary gland
gland
of the hypothalamus at regulates growth and development of the body
the base of the brain
(B)
T h y ro i d It is located in the neck Thyroxine hormone secreted by thyroid gland
gland
region.
regulates carbohydrate, protein and fat metabolism
in the body.
(C)
Pancreas
It is located just below Insulin hormone secreted by pancreas lowers the
the stomach in the body. blood sugar level.
Related Theory

Function of hormones secreted by these gland
Pituitary gland is called the master gland of the body. A person having deficiency of growth hormone would be short in height and this leads to dwarfism
on the other hand, a person having too much growth hormone becomes very tall and this leads to gigantism.
77. Whenever a cheetah sees a prey, he moves
towards the prey at a very high speed.
This happens due to seeing a prey is receiving
the stimulus from the external environment.
This information is detected by the specialised
tips (dendritic tip) of some nerve cells and sets
off a chemical reaction that creates electrical
impulse.
This impulse travels from dendrite to cell body
of a neuron and then along the axon it reaches
nerve endings. At the end of the axon the
electrical impulse sets off the release of some
chemicals. These chemicals cross the synapse
(gap between two neurons) and start a similar
electrical impulse in the next neuron and to
neuromuscular junction.
As the nerve impulse reaches the junction,
neurotransmitter, a chemical is released which
diffuses across synapse to stimulate dendron
of other neuron. The nerve impulses from
axon of one neuron to the dendrites of next
neuron pass through a synapse as a chemical
message. Hence, a cheetah rushes towards his
prey.
Axon of
motor junction
Neuromuscular
Junction
Sarcolemma
Presynaptic
terminal
Synaptic
vesicles
Synaptic
cleft
Muscle
fiber
Capillary
Mitochondrion
Myofibrills
Postsynaptic
membrane
Neuromuscular Junction
During this event, contraction and expansion
of muscles take place. This contraction and
expansion is caused by the proteins released
in the muscles. (Troponin and Tropomyocin
are proteins which change the configuration in
the muscles and cause them to contract and
expand).
79. (A)Plant cells change their shape in response
to an external stimulus by changing the
amount of water in them which results in
swelling or shrinking.
The information is communicated from
one cell to the other by electrical-chemical
means.
Plants have a hormonal system that
regulates and coordinates their activities.
Light, touch, gravitational force, and other
stimuli elicit responses in plants. They exhibit
two distinct types of movements: growthdependent and growth-independent. The
leaves of the sensitive or 'touch-me-not'
plant begin to fold up and droop when
touched. This information is communicated
from cell to cell using electrical-chemical
methods. In order for movement to occur,
some cells must change their shape
(by changing the amount of water in
them). This type of growth-independent
movement is known as nastic movement.
(B) Name of hormone to:
(i) increase the height of a dwarf plant:
Auxins/ Gibberellins
(ii) cause rapid cell division in fruits and
seeds: Cytokinin
80. Nervous and hormonal systems together
perform the function of control and
coordination in human beings.
The nervous system consists of nerves or
neurons which control and coordinate all the
functions in the body. Nervous system consists
of sensory receptors which collect information
from the external environment in the form
of stimuli and then send them to the brain
in the form of electric impulse. In the brain,
information is interpreted and instructions are
sent to the effector organ, i.e., muscles which
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reveal responses. Nerves need assistance from
the hormones because they cannot reach each
and every part of the body.
Coordination in humans is brought about by
the secretions of endocrine glands. Endocrine
glands are ductless glands which secrete
chemical substances called hormones directly
into the blood. An organ which responds to
such a hormone is known as target organ.
34
Hormonal control is mainly based on a feedback
mechanism and tells the body to either pace
up or slow down as per the situation.
Nervous control, on the other hand, is more
of a direct control. Both of them complement
each other. Thus, it can be said that the nervous
and hormonal systems together perform the
function of control and coordination in human
beings.
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ANSWER SHEET
SELF PRACTICE
1. (a) (I) and (II)
Explanation: In the process of asexual
reproduction, only one parent is involved
and there is no fusion of the male and the
female gametes. As a result, the offsprings so
produced are genetically similar, having same
chromosome number to their parents and are
called clones.
4. (b) halving of chromosomes during gamete
formation
Explanation: The gametes have half the
number of chromosomes as compared to
that of normal body cells. Reduction division
(meiosis) takes place during gamete formation,
which halves the number of chromosomes in
both male and female gametes.
So, when a male gamete combines with a
female gamete during sexual reproduction,
then the new cell, zygote will have a normal
amount of DNA, i.e., original chromosome
number (as in parent) is restored.
6. (b) B and D
Explanation: Male and female gametes are
formed from the same parent plant in selfpollination (shown by “A” and “C”), and the
progeny produced almost exactly resemble the
parent plant.
As they grow on the same plant, this indicates
that the genetic makeup of the male and female
flowers is the same.
Cross-pollination, also known as xenogamy
(represented by the letters “B” and “D”), allows
two genetically different plant characteristics
from the same species to fuse. Due to the
mixing of genetically diverse gametes, it causes
genetic recombination and variability in plants.
7. (c) E→C→B→D→A
Explanation: The sperm fertilizes the egg in
the fallopian tube. The fertilized egg, the zygote
gets implanted in the lining of the uterus and
starts dividing and become embryo. Foetus is
a stage of embryo which nearly resembles a
human being. The embryo gets nutritive from
mother’s blood with the help of a special tissue
called placenta. The development of the child
inside the mother’s body takes approximately
nine months. The child is born as a result of
rhythmic contraction of the muscles in the
uterus.
10. (b) Sexual maturation
Explanation: These are the changes which
start taking place in early teen age years. All
these changes take place slowly, over a period
of months and years. They don not happen
all at the same time in person, nor do these
changes happen at an exact age. Each of these
changes does not becomes complete quickly.
11. (b) Parent cell will lead to the formation of
two daughter cells of equal sizes.
Explanation: The nucleus of Amoeba divides
into two equal parts which is followed by
the division of the cytoplasm, resulting in the
formation of two identical daughter cells. The
division of nucleus is called karyokinesis and
division of cytoplasm is referred as cytokinesis.
14. (d) It is a multicellular organism that breaks
into pieces that grows into new individuals.
Explanation: Spirogyra is a simple multicellular
organism which simply breaks up into smaller
pieces upon maturation. These pieces or
fragments grow into new individuals. However,
it is not true for the complex multicellular
organisms as these organisms are not a
random collection of cells. Specialised cells
are organised into organs and tissues are
organised into organs which are then placed at
definite positions in the body. In such a carefully
organised situation, cell by cell division would
impractical.
15. (d) R—Q
Exaplanation: R - Fallopian tube/oviduct
fertilization takes place
Q - Uterus. Zygote gets down and implants on
to the uterine wall.
P - Vagina or birth canal. It is the entry point
for sperm.
S - Ovary where ova are produced. Ovary
also releases hormone oestrogen which
is responsible for the secondary sexual
characteristics.
How do Organisms Reproduce?
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Caution
Related Theory

Menarche is the first menstrual cycle between the
age of 10 and 12 years in most girls. Ovulation is
the release of mature egg from one of the ovaries
every month in a female when a sperm fertilizes an
egg after ovulation, the fertilized egg travels down
into uterus where it gets implanted. A successful
implanation results in pregnancy which last for 40
weeks.
19. (b) Both (I) and (IV)
Explanation: When a girl is born, the ovaries
already contain thousands of immature eggs.
The two oviducts unite into an elastic baglike structure known as the uterus. The uterus
opens into the vagina through the cervix.
20. (c) Name of part: R is Ureter; Function: Passage
for sperms
Explanation: R is urethra and forms a common
passage for both the sperms and urine.
22. (d) Both (I) and (IV)
Explanation: Part labelled 2 is the stigma on
which the pollen grains land. The male germ
cell travels through the pollen tube (part
labelled 3) after pollination.
26. (a) A chromosome contains hundreds of genes,
which are composed of DNA.
Explanation: Genes are the DNA segments
that contain the instructions for making a
specific protein that operates in one or more
types of cells in the body. Chromosomes are
cellular structures that house a person’s genes.
Chromosomes, which are found in the cell
nucleus, contain genes.
Caution

Students usually make mistakes as they do not
understand the relationship between DNA, a gene,
and a chromosome. The nucleus is found inside the
cell. Inside the nucleus is the chromosome, which
contain genes.

Students should know that the placenta provides
nutrition to the developing embryo in both plants
and animals.
34. (a) Both (A) and (R) are true and (R) is the
correct explanation of (A).
Explanation: Organisms look similar because
they have similar body designs. As the DNA
in the cell nucleus is the information source
for making proteins and if this information
source is changed, different proteins will
be made which will lead to altered body
designs. Therefore, the creation of a DNA
copy by chemical reactions is a basic event in
reproduction.
37. (c) (A) is true, but (R) is false.
Explanation: Vegetative propagation is a
method of asexual reproduction in some
higher plants in which a new plant develops
from the vegetative parts of a plant such
as root, stem or leaf. It makes possible the
propagation of plants such as banana, jasmine
etc. that have lost the capacity to produce
seeds.
42. (B) (i) The organism seen on the bread is
Rhizopus and the thread like structures
are the hyphae of the bread mould. The
mode of reproduction in Rhizopus is spore
formation.
(C) The
disease kala-azar is caused by
Leishmania and it reproduces by binary
fission.
The disease malaria is caused by
Plasmodium and it reproduces by multiple
fission.
Differences
between
the
modes
of reproduction in Leishmania and
Plasmodium:
Asexual
Reproduction in
Leishmania
Asexual
Reproduction in
Plasmodium
Binary
fission It results in the
results
in
the production of many
production of two individuals.
individuals.
33. (d) (A) is false but (R) is true.
Explanation: The embryo is implanted in
the lining of the uterus where it continues to
grow and develop organs to become foetus.
The embryo gets nutrition from the mother’s
blood with the help of a special tissue called
the placenta, which is a disc embedded in the
uterine wall.
36
Nucleus of parent Nucleus of parent cell
cell divides only divides repeatedly to
once.
form a large number
of cells.
43. (B) The male gamete is formed in the anther
of a flower and female gamete is formed
in ovary of a flower.
44. (B) The different varieties of mango on the
same plant can be grown by grafting
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method of vegetative propagation in
which parts of two plants are joined in
such a way that grow as one plant.
(C) Advantages of growing plants by
vegetative propagation:
(1) The new plants produced are exactly
like the parent plants.
(2) Plants which do not form viable seeds
can be grown by this method.
(3) This method is a rapid, cheaper and
convenient method for growing
plants.
(4) Survival rate of new plants is almost
100% but it is upto 10% when raised
by seeds.
45. (A) Common signs of sexual maturation in
boys and girls:
(i) Hair growth in armpits and the genital
area between the thighs, which also
becomes dark in colour.
(ii) Thinner hair appear on legs and arms
as well as on the face.
(iii) Pimples begin to develop as skin
becomes oily.
(iv) Become conscious and aware of their
own bodies and those of others in new
ways.
(Any two)
(C) The factors that
population are:
(1) Birth rate
(2) Death rate
(3) Immigration
(4) Emigration
determine
size
of
(Any two)
Related Theory

The population size is actually population density,
the number of individuals per unit area. The size of
human population is a cause for concern for many
people because increase in population size makes it
harder to improve everybody’s standard of living.
46. (C) If there are no bee pollinator, we may
lose all the plants which are pollinated
by bees as well as the animals who eat
such plants. It means a world without
pollinators would not have diversity of
fresh produce and the plant would die.
The pollen begins to germinate, grows
pollen tube through the style towards the
egg cell and the pollen tube reaches the
ovary. Female germ cells are in the ovary.
In the ovary, male and female germ cells
unite to form a zygote. The zygote divides
several times to form an embryo within
the ovule. The ovule develops a tough
coat which changes to seed and ovary
grows rapidly which changes to fruit.
47. (B) Unwanted
pregnancies
can
terminated by surgical method.
be
Methods of contraception:
(1) Mechanical barrier methods such as
use of condom
(2) Surgical method such as tubectomy
in females and vasectomy in males
(3) Chemical methods such as Oral and
vaginal pills
(4) Intra Uterine Contraceptive Devices
such as copper –T
(Write any two to get full marks)
48. (A) Umbilical cord is connected to a special
tissue called placenta which is disc like
structure embedded in the uterine wall.
The foetus gets nutrition from the mother’s
blood with the help of placenta. Placenta
provides a large surface area for glucose
and oxygen to pass from the mother to
the foetus. The foetus also produces waste,
which is removed by mother’s blood
through the placenta.
(B) The development of the child inside the
mother’s body takes approximately nine
months. The child is born as a result of
rhythmic contractions of the muscles in the
uterus.
Gestation Period: It is the period of
development during which a foetus
develops beginning with fertilization and
ending at birth. The average length of
human gestation is 40 weeks or a month
or 280 days from the first day of woman’s
last menstrual period.
(C) If the egg/ ovum is fertililzed by a sperm, it
travels down the fallopian tube and then
pregnancy occurs.
The fertilized egg attaches to a soft, thick
and spongy lining of the uterus. Then
the placenta is formed which transfers
nutrition and oxygen to the foetus from
mother. Similarly the placenta transfers
waste material in to the mother’s blood to
be removed.
49. (A) (b)Sugar solution provides nutrition to the
yeast cells.
Explanation: Water does not provide any
energy to the yeast cells. So, yeast cells
fail to multiply in water due to inadequate
energy in its cells. Sugar provides energy
to them to carry out reproduction by
multiplying rapidly.
It divides a sexually by formation of buds.
A small bud is formed on the parent cell.
The nucleus of the parent cell divides
and enters the daughter cell. The bud
grows and detaches itself from the parent
How do Organisms Reproduce?
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cell and grows into a new cell. A sexual
reproduction also occurs by fission in some
yeasts.
(B) (c)budding
Explanation: In yeast, asexual reproduction
occurs mostly through budding, however
some species of yeast also reproduce
through binary fission.
(C) (a) (I) and (III)
Explanation: The rapid spreading of bread
mould on slices of bread is due to presence
of a large number of spores in the air and
also due to the presence of moisture and
nutrients.
(D) (c)Yeast: In this process, a small outgrowth
in the form of a bud grows on the parent
cell; Bread Mould. It has sporangia
which are the reproductive structure on
the hyphae and which produce many
spores.
Explanation: The thread like structures
that develop on the bread are the hyphae
of the bread mould (Rhizopus). They are not
reproductive parts. On the other hand, the
tiny blob-on-a-stick structures are involved
in reproduction. The blobs are sporangia,
which contain cells, or spores, that can
eventually develop into new rhizopus
individuals. The spores are covered by thick
walls that protect them until they come
into contact with another moist surface
and can begin to grow. At maturity the
sporangium bursts open and the spores are
released. When conditions are favourable
for growth the spores give rise to new
individuals.
(E) (b) Yeast and Hydra
Explanation: Yeast and Hydra reproduce
by budding whereas Rhizopus reproduces
by spore formation and Spirogyra
reproduces by fragmentation.
50. (B) Neem or Peepal produce seeds and grow
sexually by seeds. Vegetative propagation
is possible only in those plants which
have lost the capacity to produce seeds
or produce non viable seeds.
(C) A new plant develops from the vegetative
parts of a plant in the given plants:
Plant
38
Part of Plant used
for Vegatative
Propagation
Bryophyllum
Leaf
Potato
Stem
Dahlia
Root
Onion
Stem
Sweet potato
Root
Mint
Stem
Disadvantages:
Plants grown by vegetative propagation
have less vigour than the plants grown by
seeds. Undesirable characters or disease
contracted by the parent plant is also
transmitted to new plants.
51. (A) The steps followed to observe a permanent
slide of Amoeba under microscope are
given below.
(1) Fix the slide on the stage carefully.
(2) Adjust the microscope to low power
and focus.
(3) Adjust the diaphragm and mirror of
the microscopes so that sufficient light
enters to illuminate the slide.
(4) Finally adjust the microscope to high
power and focus.
(B) All three organisms Amoeba, Leishmania
and Plasmodium multiply by fission.
Amoeba divides into two daughter cells in
any plane. Leishmania forms two daughter
cells in a specific plane i.e., longitudinally
while Plasmodium divides to form many
daughter cells by multiple fission.
(C) P - Cell wall
Q - Pyrenoid
R - Chloroplast
S - Nucleus
Functions - cell wall (P) Makes the
spirogyra slippery to touch.
Explanation: Spirogyra reproduces by
fragmentation in which the body of
Spirogyra when matures, break into many
smaller pieces or fragments due to strong
water current. These different pieces later
grow into new individuals.
52. (C) (c) Paramecium and Amoeba
Explanation: Plasmodium divides by
multiple
fission
whereas
Amoeba,
Paramecium and Leishmania divide by
binary fission.
(D) (d)Each new generation will end up having
twice the amount of DNA that the
previous generation had.
Explanation: DNA copying gives rise to
variations. However, in sexual reproduction,
the amount of DNA in an offspring will be
the same as in the previous generation
because during gamete formation, meiosis
or reduction cell division takes place. During
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meiosis, chromosome number is halved
in gametes so that after fertilization,
the original number of chromosomes is
restored in the offspring.
53. (A) The period of adolescence when the
reproductive organs become functional, is
termed as puberty. Puberty is accompanied
by development of secondary sexual
characters both in males and females. The
puberty begins between ages 12 and 16
for boys. Puberty causes physical changes
both in boys and girls.
(B) (1) Thick hair growth in new parts of this
body such as armpits and the genital
area between the thighs.
(2) Thinner hair growth on legs and arms
as well as on face.
(3) The skin becomes oily and pimples may
begin to develop.
(4) The boys and girls become conscious and
aware of their own bodies and those of
others in new ways.
(C) (i) The human male and female bodies
show sexual maturation at puberty
due to the release of hormones in their
bodies. Oestrogen and testosterone
are the hormones released in females
and males respectively which are
responsible for secondary sexual
characters.
(ii) Sexual maturation at puberty begins at
different ages in the humans depending
on genetic and environmental factors. In
some people, the maturation happens
early and quickly, while in others it can
happen very slowly.
(E) (c)Menarche
Explanation:
The
commencement
of menstruation at puberty is called
menarche and marks the beginning of
reproductive life of a woman.
57. Medha observed after a week, that new
leaves and stem have grown from the cut
celery part. This shows that the whole celery
plant can reproduce from the cut celery part.
This is vegetative propagation which means
new plants can grow from the vegetative
part i.e., stem. This method is a very fast and
economical and all the celery leaves and stems
are genitally similar to the parent plant.
59. The organ producing sperms in humans is
Testes and that producing ova is Ovary.
Related Theory
 The testes are the male’s primary reproductive
organs which perform dual functions of production
of male gamete or sperm and the secretion of male
sex hormone, testosterone which is responsible
for the development of male secondary sexual
characters such as body hair, muscles and bone
mass.
 Ovaries are the female’s primary reproductive
organs which perform dual functions of production
of female gamete or ovum and the secretion of
female sex hormones, estrogen and progesterone.
61.
54. (B) (i) The following can be used for regulation
of child birth in a family:
Barrier methods : Use of condoms,
diaphragm, cervical cap.
 Chemical methods : Use of pills.
 Surgical methods: Vasectomy in male
and tubectomy in females.
(C) Male gonads : Testes
Female gonads : Ovaries
Functions of gonads
(1) Male Gonads (Testes): They produce
sperms and secrete male hormone –
testosterone
(2) Female gonads (Ovaries): They produce
ovum and secrete female hormone –
estrogen and progesterone.

55. (D) (b)Cervix
Explanation: The upper portion of the
uterus is broader, while its lower portion
is narrower, called cervix. The cervix opens
into the vagina which is a tubular structure
and also called ‘birth canal’.
Types of
fission
Name of Organism
(A)
Multiple fission
Plasmodium, yeast
(B)
Binary fission
Amoeba, Paramecium
Related Theory
 Binary fission is an asexual method of reproduction
in which the parent organism splits to form two new
organisms.

Multiple fission is also an asexual method of
reproduction in which the parent organism splits to
form many new organism at the same time.
62. Leaf
Related Theory

Vegetative propagation is a type of asexual
reproduction in which vegetative part of a plant
(root, leaf and stem) grows up into a new plant.
66. (A) Night-blooming flowers are white as white
colour shines in the dark and attracts the
insects at night. The day blooming are
brightly coloured as bright colour of flowers
attracts the insects during the day for
pollination. Insects are pollinating agents.
How do Organisms Reproduce?
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(B) Pumpkin bears only unisexual flowers. It
contains either male reproductive organ
i.e. stamens or have female reproductive
organ i.e. carpels it means two different
types of flowers are formed. Male flowers
do not set into fruits. They produce only
pollen grains, female flowers have carpels
which receive the pollen grains on their
stigma and after fertilization develop into
fruits. Hence, male flowers do not form
fruits where as female flowers change into
fruits.
atmosphere), the thread like structures
(hyphae) are developed which are nonreproductive parts. Thin stems having
blob like structures called sporangia
which contain spores enclosed in a
protective covering are formed. Spores
are asexual reproductive units which
would reproduce into more number of
bread mould when they get appropriate
conditions.
.......
67. The ratio of chromosome numbers between
egg and its zygote is 1:2.
Both sperm and egg contain half number of
chromosome, i.e., 23 chromosomes. Sperm is
genetically different from the egg in the way
that it contains either X or Y chromosome,
whereas an egg always contains an X
chromosome.
.......
.......
.......
Hyphae
(Non-reproductive parts)
74. Labelled diagram showing the stage in
binary fission of Amoeba in which its nucleus
elongates and a constriction appears in its cell
membrane is drawn below:
Caution

.......
Sporangium
(Reproductive part)
....
..........
Spore
Students usually get confused with the number
of chromosomes and make mistakes. Male and
female gametes contain haploid number of
chromosomes while zygote contains diploid number
of chromosomes.
Cell membrane
Nucleus
69. Reproduction is essential for the survival of
a species on Earth. Living organisms produce
more organisms of their kind to maintain their
species. The process of reproduction ensures
the continuity of life on Earth. It also leads to
the evolution of the species in the long run.
Cytoplasm
Constriction
Related Theory

72. (A) The greenish black powdery mass on
a stale piece of bread is due to bread
mould Rhizopus which reproduces by
spore formation.
[CBSE Marking Scheme Term-2 SQP 2022]


Detailed Answer:
Spore formation is an asexual mode of
reproduction in bread mould (Rhizopus).
In spore formation, parent plant produces
hundreds of microscopic spores in the
sporangia which are reproductive parts.
When the sporangium bursts open, the
spores spread into the air.
(B) Hyphae or thread like structures are
the vegetative parts and tiny blob
like structures or sporangia are the
reproductive parts.
[CBSE Marking Scheme Term-2 SQP 2022]
Detailed Answer: When the spores
get
a
suitable
substratum
and
appropriate conditions (warm and humid
40
Binary fission is the type of asexual reproduction
in which two individuals are formed from a single
parent and the parental identity is lost.
The first step in binary fission is elongation and
subsequent division of nucleus into two nuclei i.e.,
nuclear division.
Division of nucleus is followed by cytoplasmic
division. The final stage is the formation of two
daughter cells whereby identity of parent Amoeba is
lost.
75. A diagram showing the different parts of an
embryo of a gram seed is drawn below:
Plumule
Cotyledon
Radicle
Related Theory



The fleshy round structure on the embryonic axis
that contains reserve food material is known as
cotyledon.
The radicle is present on the micropylar end of the
embryonic axis and is the future root.
Plumule is present on the opposite extreme and is
the future shoot.
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76. Variations which are favourable increase
Nucleus
the chances of survival of the species. If an
organism can withstand a higher temperature,
then the variation goes on accumulating in its
future generations. Hence, these organisms
can survive sudden rise in the temperature.
This ensures the survival of the species.
But other organisms (variants) without this
variation may not survive due to sudden rise
in temperature. So, variation is beneficial to the
species, but not necessarily for individual.
Cytoplasm
Cell
membrane
Pseudo podia
Amoeba
(First stage)
Elongation of Nucleus
(Second stage)
Related Theory

Fission: It is a type of asexual reproduction in which a
single parent cell divides into two or more daughter
cells.
Fission
Related Theory

Variation: The differences in the traits shown by the
individuals of a species and also by the offsprings of
the same parents are know as variations.

Variations also arise due to inaccuracy in DNA
copying.

80. (A) The given diagram is of human female
reproductive system.
Name of the parts are given below:
(1) Fallopian tube/Oviduct
(2) Ovary
(3) Uterus
(4) Cervix
(5) Vagina
(B) Contraception: These are the techniques
which have been developed to prevent and
manage pregnancy.
Advantages of adopting contraceptive
methods:
(1) Avoiding frequent and unwanted
pregnancy.
(2) Keeping population and hence birth
rate under control.
(3) Helps in keeping proper gap between
two pregnancies.
(4) Helps in preventing the transfer of
sexually transmitted diseases.
(Any 3 of 4 points can be written to get
full marks)
The first two stages of binary fission in Amoeba
are shown:
Multiple
Parent cell split to
form more than two
new organism.
e.g., Plasmodium
83. (A) The
new organisms or offsprings
produced by one parent through asexual
reproduction are called clones. The
clones possess exact copies of the DNA
or genes of their parents and hence show
remarkable similarity to the parent or one
another. Asexual reproduction does not
involve gamete formation by meiosis and
fertilization of male and female gametes;
the offsprings are genetically similar to the
single parent.
(B) Sexual reproduction includes gamete
formation through meiosis and fertilization.
The gametes are special type of cells called
reproductive cells which contain only half
the amount of DNA or half the number of
chromosomes as compared to the normal
body cells of an organism. So, when a male
gamete combines with a female gamete
during sexual reproduction, then the new
cell zygote will have the normal amount of
DNA.
For example: Human sperm has 23
chromosomes and the human egg also
has 23 chromosomes. So, when a sperm
and an egg fuse together, then the zygote
formed will have 46 chromosomes.
81. Binary
fission is the type of asexual
reproduction in which two individuals are
formed from a single parent and the parental
identity is lost.
The first step in binary fission is elongation and
subsequent division of nucleus into two nuclei
i.e., nuclear division.
In binary fission, the identity of parental cell
is lost. After fission, parental cells are called
as daughter cells. The process starts with
elongation of nucleus.
Binary
Parent cell split
to form two new
organism.
e.g., Amoeba
85. Differences between modes of reproduction in
unicellular and multicellular organisms:
(1)
Unicellular
Multicellular
Organisms
Organisms
Sexual reproduction Sexual reproduction is
not seen normally.
normally seen.
How do Organisms Reproduce?
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(2)
(3)
(4)
Simple cell division Simple cell division
leads to formation of may or may not
next progeny.
lead to formation of
progeny.
No gamete formation Gamete
formation
takes place.
takes place.
Some types asexual
reproduction
are
binary fission and
miltipie fission.
Some types of asexual
reproduction
are
budding, regeneration,
fragmentation etc.
Its female reproductive part is known as
carpel or pistil.
(B) Diagram of longitudinal section of flower
showing the process of germination of
pollen on stigma along with the following
labelings:
(i) Male germ cell
(ii) Female germ cell
(iii) Ovary
(iv) Pollen tube
Pollen grain
Stigma
(Any 3 of 4 points can be written to get full
marks)
Male germ cell
86. When an organism reproduces sexually, it
Pollen tube
produces gametes through a special type of
division called meiosis or reductional division,
in which the original number of chromosome
becomes half. In sexual reproduction, male
gametes and female gametes are combined
and form zygote, and the original number of
chromosomes is restored.
Male gametes(n) + Female gametes(n) → Zygote(2n)
(23)
(23)
(46)
In asexual reproduction, only mitotic divisions
are involved and the chromosome number
remains same.
2n
mitotic division
2n
2n
Example: Humans have 46 chromosome or 23
pair of chromosomes.
When the gametes formed through meiosis, the
chromosomes number becomes half. Thus, an
organism maintains a constant chromosome
number through several generations.
87.
(B)
Budding in Hydra
Budding in Yeast
Buds produced are
multicellular
Buds produced are
unicellular
Buds get detached
from parent body
as soon as the
baby Hydra is fully
mature.
Buds may remain
attached to the
parent body till
3 - 4 buds are
developed.
(Write any one difference)
88. (A) A flower that contains both the male and
female reproductive structures (stamen
and pistil) is called a bisexual flower or
hermaphrodite.
Example: Rose, Hibiscus
42
Ovary
Female germ cell
(C) Pollination is just the transfer of pollen
grains from the anther of stamen to the
stigma of pistil. It is carried out by agents
like wind, insect etc.
For pollination to happen there is no need
of fertilisation of gametes i.e. their union.
Fertilisation is fusion of male and female
gamete and this process is facilitated
by pollination. By pollination pollens are
brought to eggs so that they can fuse and
fertilisation can take place.
Therefore, pollination may occur without
fertilisation but fertilisation will not take
place without pollination.
90. (B) Mrs. Raman gave advice for tubectomy
to Mrs. Sehgal because she knows that a
small family is a happy family. If a couple
has only two children, it can provide good
food, clothes, and education to each child.
This will keep parents as well as children
happy. Mrs Raman also knows that having
fewer children also keeps the mother in
good health.
(C) No, tubectomy does not prevent the spread
of sexually transmitted diseases.
(D) Since, there will be no physical barrier
during sexual contact, there is every
chance of a person (with tubectomy)
getting infected with sexually transmitted
diseases during sexual contact with an
infected person.
The
need
of
adopting
contraceptive
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methods is to avoid getting infected with
sexually transmitted disease and for
birth control. These are also used to avoid
unwanted pregnancies.
91. (A) Married
methods:
couples
use
contraceptive
(8)The rhythmic contraction of uterus
muscles gradually pushes the baby out
of the mother’s body through vagina
which is called parturition. This is how a
child is born.
93. (A) (i)—Ureter
(1)To avoid unwanted pregnancy
(ii)—Seminal vesicle
(2)To have proper time gap between two
children.
(iii)—Urethra
(iv)—Vas deferens
(3)To limit the number of children in the
family.
(4)To prevent transmission of many of
sexually transmitted diseases.
Seminal vesicle (ii)
Ureter
(i)
Bladder
(Any two)
(B) In human female reproductive system,
there are two ovaries. One egg is produced
every month by one of the ovaries during
ovulation:
(1) The ovum/egg goes into the oviduct or
fallopian tube.
(2)The sperms released by the human
male during copulation or mating are
deposited at the top of the vagina close
to the cervix of uterus. The sperms
travel through the uterus to the top of
fallopian tube.
(3)The fertilization of the ovum or egg
takes place in the oviduct and a zygote
is formed.
(4)The zygote divides rapidly as it moves
down slowly in the oviduct and forms
a hollow ball of hundreds of cells
called embryo. The close attachment
of embryo to the uterine wall is called
implantation.
(5)After implantation, a number of
developmental changes takes place in
embryo as well as in the wall of uterus.
When body features are seen clearly,
the developing embryo is called foetus.
(6)A disc-like special tissue develops
between the uterine wall and the
foetus which is called placenta. The
exchange of nutrients, oxygen and
waste products between the embryo
and the mother takes place through
the placenta.
(7)The time period from the fertilization
up to the birth of baby is called
gestation. The average gestation period
in humans is about nine months or 40
weeks or 280 days from the first day of
the last menstruation.
Vas deferens
(iv)
Urethra
(iii)
Testis
(B) Hormone secreted by testis: Testosterone.
It brings about changes in appearance seen
in boys at the time of puberty (secondary
sexual characteristics).
(C) Functions of (ii) (Seminal vesicle): Along
the path of vas deferens, seminal vesicle
along with prostrate gland pour their
secretions which make the transport
of sperms easier and this fluid provides
nutrition to sperms.
Functions of (iii) (Urethra): Urethra carries
urine from the bladder and sperms from
vas deferens through the penis.
94. (A) Male and female sex cells are formed in
gonads. Male sex cell are formed in testes.
Male sex cell are also called sperms or
male germ cells.
Female sex cells are also called ovaries
or female germ cell. One egg is produced
every month by one of the ovaries.
(B) Fertilization is the fusion of male germ cell
and female germ cell to give rise to new
individual organism. Fertilization takes
place in fallopian tube or oviduct.
(C) Fertilized egg implanted itself in the uterus
and develops into an embryo.
(D) In males, a small portion of the sperm duct
or vas deferens is removed by surgical
How do Organisms Reproduce?
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operation. The two cuts ends are then
ligated with threads. This prevents the
sperms from coming out. This procedure is
called vasectomy.
(3) The pollen tube normally enters the ovule
through a small opening called micropyle.
(4) Inside the ovule, the pollen tube releases
two male gametes into the embryosac
which contains the egg.
In female, a small portion of the fallopian
tubes or oviducts are removed by surgical
methods. The cut ends are then ligated or
tied with threads. It prevents the egg or
ovum to enter the fallopian tube.
(5) Fertilization of male and female gametes
takes place inside the embryosac.
(6) After fertilization, the zygote divides
several times to form an embryo within
the ovule.
Related Theory

want a female child.
97. Differences
between the two types of
pollination that occurs in plants, namely, self
pollination and cross-pollination which is given
below:
S.
No.
(1)
(2)
(3)
(4)
Self-Pollination
Cross-Pollination
The transfer of
polle grains from
the anther of
a flower to the
stigma of the
same flower or
another flower of
the same plant.
The transfer of
pollen grains
from the anther
of a flower to
the stigma of
another flower of
a different plant of
the same species.
It occurs in
It occurs
flowers which are between flowers
genetically same. which may
be genetically
different.
Self-pollination
increases genetic
uniformity
and decreases
variations.
Cross-pollination
decreases genetic
uniformity
and increases
variations.
Does not require
pollinators for
transfer of pollen
grains.
Requires
pollinators for
transfer of pollen
grains.
(Any 2 of 4 points can be written to get full
marks)
Following events take place when a pollen
lands on a suitable stigma:
(1) Pollen grains which are deposited on the
stigma form tubes called pollen tubes.
(2) One pollen tube grows through the style
and reaches the ovary where the ovules
are located.
44
(7) The ovule develops a tough coat and is
gradually converted into a seed, which
contains the food store.
Both vasectomy and tubectomy are also known
as methods of sterilization. These do not affect
the normal life. Surgery is also used for medical
termination of pregnancy (MTP). However, this
method is generally used by people who do not
(8) The ovary grows rapidly and ripens to
form a fruit
99. (A) Mature ovum is released from the ovary
during ovulation which occurs in the
middle of the menstrual cycle. The ovum
travels to the fallopian tube of the female
reproductive system. If mating takes place
during this time. One of the millions sperms
released will travel to the cervix of the
uterus and fusion may take place.
(B) Distinguishing feature between a sperm
and or ovum.
S. No.
Sperm
Ovum
(1)
Sperm is the male Ovum in the female
gamete produced in gamete,
produced
the testis of a male in the ovary of the
female.
It is smaller in size.
(2)
It does not contain It contains the foodfood-store for the store for the growing
growing embryo.
embryo.
(3)
It is motile.
It is larger in size.
(4)
It is non-motile.
Head
Nucleus
Tail
Nucleus
Ovum
Sperm
(Write any 2 points)
100. (A) The functions of the various parts of the
human male reproductive system are given
below:
(i) Testis: The formation of germ-cells
or sperms takes place in the testes.
It secretes the male hormone, called
testosterone, which regulates the
formation of sperms and brings about
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changes in appearance in boys at
puberty.
(ii) Vas deferens: It is a long tube which
arises from each testes and carries
sperms into organs called seminal
vesicles, where the sperms get
nourished and stored.
(B) Regeneration is a process in which an
organism is broken/ cut into pieces, these
pieces may grow into separate individuals.
Diagram
Planaria:
showing
regeneration
in
(iii) Urethra: It is a common passage for
both urine and spermatic fluid. The vas
deferens unites with a duct coming
from the urinary bladder to form
urethra.
(iv) Prostate: It is a gland that secretes an
alkaline fluid which gives lubrication
and nutrition for the sperm.
How do Organisms Reproduce?
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ANSWER SHEET
SELF PRACTICE
shows up only if it continues to be inherited by
the offspring for several generations.
2. (c) Law of segregation and Law of dominance
Explanation: In F1 generation, only red colour
is expressed which is called the dominant
character and it follows the law of dominance.
Red and White color trait express themselves in
definite proportion in F2 generation and follow
law of segregation.
Law of dominance: This law states that when
two alternative forms of a trait are present in
an organism, only one factor expresses itself
in F1 progeny and is called dominant while the
other that remains masked is called recessive.
15. (b) 4 tall plants and 1 medium height plant.
Explanation: Medium height plants are not
obtained when we cross a tall pea plant with
a short pea plant. They will be either tall (TT or
Tt) or short (tt).
18. (a) one
Explanation: One pair called the sex
chromosomes is present in the zygote of
humans. It is not always being a perfect
pair. Women have a perfect pair of sex
chromosomes, both called X (XX).
Law of segregation states that alleles of a pair
segregate from each other during gamete
formation.
Related theory
 Law of independent
assortment: This law states
that the two factors of each character assort or
But men have a mismatched pair, in which one
is a normal-sized X while the other is a short
one called Y (XY).
20. (a) Both (A) and (R) are true and (R) is the correct
explanation of (A).
separate out independent of factors of other
characters at the time of gamete formation and get
[CBSE Marking Scheme SQP 2020]
randomly arranged in offsprings.
Explanation : Violet colour of flower is
dominant.
5. (d) All of the above
Explanation: When tall pea plants having
round seeds were crossed with dwarf plants
having wrinkled seeds, the F1 progeny were
all tall with round seeds. On self-pollination
of F1 plants, the F2 progeny consisted of both
parental characters - Tall plants having round
seeds as well as dwarf plants with wrinkled
seeds. There were also new combinations
like tall plants with wrinkled seeds and dwarf
plants with round seeds in the F2 progeny.
Genotype of white flowers = v v
Genotype of violet flowers = V V
When a genetist crossed pea plant having
violet flowers with a pea plant having white
flowers.
vv
VV
V
V
v
v
Vv
Vv
Vv
Vv
10. (a) All variations in a species have equal
chances of survival.
Explanation: Variation means certain changes
which occur in sexually reproducing organisms
because of errors in DNA copying.
Variations are beneficial for species because
they increase the adaptability of an organism
to its changing environmental conditions.
The variation produced in organisms during
successive generations gets accumulated in
the organism. The significance of variations
46
All violet flowers
All violet flowers
Genotype
of
F
progeny=Vv
1
Genotype of F progeny
Vv
1
This indicates that both the traits were
inherited in F1 progeny. According to law of
dominance, the dominant colour expresses so
all the flowers are violet.
Science Class X
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26. (A) (c) Both (I) and (IV).
Explanation: Free earlobes are the most
common form of lobes found. This type
of earlobe is often large and hangs below
the point of attachment to the head. This
happens due to the influence of a dominant
allele. If the parents’ genes get expressed
by the dominant allele, then the child will
be born with free earlobes. The attached
earlobes are not rare, but are also not
commonly found. Earlobes of such type are
small in size and do not have hangs. They
are attached directly to the side of the head.
This kind of lobe’s structural formation is
due to the absence of the dominant allele
in the chromosomes. The recessive allele is
expressed instead in the chromosomes to
form an attached earlobe.
(C) (d) gene
Explanation: The shape of earlobe in an
individual is determined by genes.
The primary form of the gene that
determines the shape of the earlobe is
known as an allele. An allele is a gene
that is found at a specific position on
a chromosome. All genes in our body
have two copies, one from each parent.
When alleles combine, some exert a
‘stronger’ influence down than the others.
The stronger allele is responsible for the
dominant traits.
(E) (a) Earlobes of either parent may be
attached or free.
Explanation: As free earlobes is dominant
and attached earlobe is recessive, parent
having free earlobes can be FF or Ff. But
parent having attached earlobe will be ff.
Since child has attached earlobe, the
genotype will be ff.
So, the following genotypes are possible
for a child having attached earlobes.
(1) One parent Ff and other parent ff.
(2) One parent Ff and other parent also Ff.
27. (A)Dominant character : Tall height
t
Tt
Tall
F1 generation
On
Sel ng
×
Tt
Tt
t
T
Gametes
t
T
T
t
T
TT
Tall
Tt
Tall
t
Tt
Tall
tt
Short
F2 generation
Phenotypic ratio: Tall plants : Dwarf Plants
3
:
1
Genotypic ratio:
TT
:
Tt : tt
1
:
2
: 1
28. (B) Traits like ‘T’ are called dominant traits,
while those that behave like ‘t’ are called
recessive traits./Alternatively accept the
definition of dominant and recessive traits
with examples of T and t respectively /
Alternatively accept the law of Dominance
with examples of T and t.
[CBSE Marking Scheme Term-2 SQP 2022]
Explanation: The F1 generation possess
one factor of inheritance from each parent
plant which was carried in gametes and is
dominant trait.
(C) Out of 800 plants 600 plants will be tall
and 200 plants will be small 1 TT : 2Tt : 1tt
[CBSE Marking Scheme Term-2 SQP 2022]
Explanation: Since all the plants in the
F1 generation have the factors Tt, so all
of them are tall. When F1 progeny was
allowed to be self pollinated, both the
parental traits were expressed in definite
proportion in F2 generation.
Self Pollination
of F1 generation
Gametes
Recessive character : Short height
(C) 1 part Homozygous Tall : 2 parts
Heterozygous Tall : 1 part Homozygous
Dwarf.
Dwarf
tt
T
Gametes
F2
generation
×
Tt
t
T
Tt
t
T
T
t
T
TT
Tall
Tt
Tall
t
Tt
Tall
tt
Short
(B) Phenotype ratio = 3 : 1
Genotypic ratio = 1 : 2 : 1
×
Tall
TT
Parents
Heredity
47
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Phenotypic ratio 3 : 1
600 Tall
3
200 Short
1
800
4
Genotypic ratio
(D) Gametes produced by YyRR parent: YR,
yR. So, the percentage of yR gamete is
(1/2) × 100 = 50%
44. Let purple trait be represented by: PP
1 : 2: 1
T T : Tt : tt
Parents
PP × pp
↓
F1 Pp × Pp
↓
30. (B)Sexual reproduction involves exchange
of genetic materials of male and female
gametes to form zygote.
F2
32. (A)(d) BBss and bbSS
Explanation: It is given that one parent is
homozygous black and spotted and the
other parent is homozygous brown, solid
mouse. As it is also given that black is
dominant over brown and solid is dominant
over spotted, therefore genotype of black
and spotted mouse will be BBss and that of
brown and solid mouse will be bbSS.
(C)The genotype of mice of F1 generation
is BbSs. When mice of F1 generation are
crossed to get mice of F2 generation, we
have to find the genotype and phenotype
after filling the gametes in the square
below:
Gametes
BS
Bs
bS
bs
BS
BBSS
BBSs
BbSS BbSs
Bs
BBSs
BBss
bbSs
bS
BbSS
BBSs
bbSS bbSs
bs
BbSs
Bbss
bbSs
Bbss
Bbss
Black, solid: 9 (BBSS, BBSs, BbSS, BbSs)
Black, spotted: 3 (BBss, Bbss)
Brown, solid: 3 (bbSS, bbSs)
Brown, spotted: 1 (bbss)
(Selfing)
Gametes
P
p
P
PP
Pp
p
Pp
pp
Visible characters of F1 progeny all flowers
are purple coloured and in F2 progeny 3
are purple coloured and 1 is white coloured
flower.
[CBSE Marking Scheme 2019]
45. A small population of a species faces a greater
threat of extinction than a larger population.
This is because a small population has lesser
number of individuals in a species. Fewer
individuals in a species impose extensive
inbreeding among them. This leads to
decrease in amount of accumulated variation
as compared to larger population. Variations
help in survival of population under changed
environmental conditions.
51. In Mendel’s experiment with pea plants, when
he cross bred a tall pea plant with a short pea
plant, he found that the first generation (F1)
was of only tall plants. In the F1 progeny, no
short plants were obtained. However, in the
F2 generation, both tall and short plants were
obtained in the ratio 3 :1 respectively. In the F2
generation, both traits - tallness and shortness
were inherited where the tall character was
the dominant trait and short character was
the recessive trait.
(E) (c) 25%
Explanation: The genotype of F1 mice will
be BbSs and that of brown, spotted mice
is bbss. The gametes formed will be BS,
Bs, bS and bs from BbSs and bs from bbss.
When these are crossed, we get BbSs, Bbss,
bbSs and bbss in equal ratios. Therefore,
percentage of progeny having black coat
with solid pattern will be 25%.
Sel ng of F1
33. (A)Plants with round and yellow coloured
seeds.
(B) YYRR and yyrr
(C) Gametes are an organism’s reproductive
cell/sex cell. These will be only one types
of gametes produces. Types of gametes
produced will be Yr type of gametes.
48
F2
Conclusion: Reappearance of dwarf character
in F2 generation proves that the dwarf trait
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was inherited but not expressed in the F1
generation.
57. Round green
Wrinkled yellow
New combinations are produced because of
the independent inheritance of seed shape
and seed colour trait.
[CBSE Marking Scheme Term-2 SQP 2022]
Caution

53.
Students should be careful while drawing crosses
and Punnett squares. They also get confused in
between the ratio of tall plants and dwarf plants, try
to clear your concept and practice similar crosses.
S.
No.
Dominant trait
Recessive trait
(1)
The trait which
appears in the
F1 Progeny, is
dominant.
The trait which
remains
hidden
or which does
not appear in the
F1 Progeny is the
recessive trait.
(2)
It appears more
in number in F2
generation.
e.g., Free earlobes.
It appears less
in number in F2
generation.
e.g.,
Attached
earlobes
Detailed Answer: A cross between two plants
having two pairs of contrasting characters
is called dihybrid cross. One pea plant with
round yellow seeds (RRYY) in which RR are
the dominant genes for round shape whereas
YY are the dominant genes for yellow colour.
The other pea plant with wrinkled green seeds
(rryy) in which rr are the recessive genes for
wrinkled shape whereas yy are the recessive
genes for green colour. When two contrasting
pairs of traits (RRYY) and (rryy) were crossed,
the F1 progeny having (RrYy) round and
yellow seeds were formed. In F1 generation
only dominant characters were expressed.
The other characters (recessive) were not
lost even when they are not expressed. When
F1 plants were self-pollinated, four types of
combinations were seen. These included two
parental types and two new combinations.
In Mendel’s cross between round and wrinkled
pea plant, the percentage of the round plants
in the F2 generation is 75%.
Explanation: Monohybrid cross
round and wrinkled pea plants
RR
(Round)
R
rr
(Wrinkled)
R
r
r
between
Parent
Generation
Parents
Gametes
Round yellow × Wrinkled green
seeds
seeds
RRYY
rryy
Gametes
Ry
Rr
Rr
R
r
Rr
R
RR
Rr
r
Rr
rr
Rr
F1 generation
F1 progeny
All round
F1
Self pollination
RrYy
of F1
progeny
Three round (75%) and one wrinkled (25%)
55. The major function of the DNA is to store
information and pass it to offspring. It also
directs the synthesis of proteins, which are
necessary for a cell to perform its functions.
The part of DNA that provides information for
protein synthesis is called gene. Proteins control
specific characteristic or trait of an organism.
For example, a plant species has gene for the
characteristic called ‘tallness’. Now, the gene
for tallness will give orders to the plant cells
to make a lot of plant growth hormones. Due
to the formation of excess of plant growth
hormones, the plant will grow tall.
If the plant has genes for dwarfness, then
plant growth hormones production will be low.
As a result, the plant will not grow tall and will
remain short. The above examples explain how
proteins control the characteristic.
ry
RrYy
(Round, Yellow)
F1
×
RY
Ry
RrYy
rY
ry
Gametes
RY
Ry
rY
ry
RY
Ry
RY
Ry
rY
ry
rY
ry
RRYY
RRYy
RrYY
RrYy
Round
yellow
Round
yellow
Round
yellow
Round
yellow
RRYy
RRyy
RrYy
Rryy
Round
yellow
Round
green
Round
yellow
Round
green
RrYY
RrYy
rrYY
rrYy
Round
yellow
Round
yellow
Wrinkled Wrinkled
yellow
yellow
RrYy
Rryy
rrYy
Round
yellow
Round
green
Wrinkled Wrinkled
yellow
green
Heredity
rryy
49
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Progeny
Round : Round : Wrinkled : Wrinkled
yellow green yellow green
9
:
3
:
3
:
1
(2)All tall plants were produced in the F1
generation.
(3) When F1 tall plants were self-pollinated,
Mendel got both tall and short plants in
the ratio of 3 Tall : 1 Short.
(4)This clearly indicated that tall character
is dominant over short character which
although present would not be expressed
in F1 generation.
Phenotypic ratio
90 Round, yellow :
9
30 Round, green :
3
30 Wrinkled, yellow :
3
10 Wrinkled, green :
1
160 seeds
16
We can conclude that round and yellow seeds
are dominant characters. Occurence of new
combinations show that genes for round
(seeds shape) and yellow seeds (seed colour)
are inherited independently of each other.
Related Theory

Based on hybridisation experiment on garden pea
plant, Mendel proposed the laws of inheritance.
(i) Law of Dominance
(iii) Law of Independent Assortment.
59. (A)The phenotype of F1 progeny: All the
plants will have round and yellow seeds.
The round shape and yellow colour of
seed is dominant over the wrinkled shape
and green colour of seed in pea plant as
observed by Mendel.
Suppose R and Y denote dominant trait and
r and y denote recessive trait. Genotype
of the parent plant with round and green
seeds – RRyy.
RRyy
Gametes
F1 generation
(Round green)
rrYY
(Wrinkled yellow)
Ry
rY
RrYy
Round and Yellow
In F2 generation, we would have two more
new combinations other than the parent
plants which are plants having round and
green seeds and wrinkled green seeds.
The cross shows round and yellow seeds
in the F1 generation. It occurs because
dominant traits (round and yellow) express
themselves, whereas recessive traits
(wrinkled and green) get suppressed.
62. (A) (1)Mendel crossed pure tall pea plants
with pure short pea plants
50
Short
tt
×
F1 generation
Tt
All tall plants
F1
Tt
F2 generation
(ii) Law of Segregation
Parents
Tall
Tt
F1
Tt
×
Tt
Tt
Tt
3 tall
(either explanation or figure)
tt
1 short
(B) When pea plants with two different
characteristics like plants with round and
green seeds and the plants with wrinkled
and yellow seeds; were bred with each
other, the F1 generation had plants
with round and yellow seeds (dominant
character).
On self-pollination of F1 generation plants,
F2 generation obtained was a mixture of
round yellow, round green, wrinkled yellow
and wrinkled green in the ratio 9:3:3:1,
thus, showing that the traits are inherited
independently.
63. (A)Mendel carried out an experiment to study
inheritance of two traits in garden pea to
see the interaction and basis of inheritance
between them. He also concluded that
traits segregate during gamete formation
and finally he gave three laws.
(1) Law of Segregation
(2) Law of Dominance
(3) Law of Independent Assortment
(B) Mendel observed that in F1 generation,
feature of only one parental type appear.
The features of other parents were not
expressed. He called the first one which
appeared as dominant features/character
and the other features which did not
appear called them as recessive.
The characters are not lost even when they
are not expressed.
Science Class X
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When F1 offsprings were allowed to be
self pollinated, both the parental traits
were expressed in definite proportion in F2
generation.
64. (A)A green stemmed rose plant denoted
by GG and a brown stemmed rose plant
denoted by gg are allowed to undergo a
cross with each other.
He had started with two combinations of
characteristics and two new combinations
of characteristics appeared in F2 generation.
From the F2 generation of a dihybrid cross
Mendel postulated that inheritance of
factors which control a particular trait
in an organism are independent of the
other. This is called law of Independent
Assortment.
GG
G
g
g
Gametes
Gg
Gg
Gg
Gg
F1 generation
(All Green stemmed Rose Plants)
(i) The stems of all the rose plant in their
F1 progeny was found to be green.
When a single pair of contrasting
characteristics homozygous green
stemmed and homozygous brown
stemmed plants are crossed, only
dominant character was expressed
and recessive did not. The phenomenon
of appearance of only one of the two
contrasting traits in F1 generation is
termed as dominance.
r – wrinkled
Y – yellow
y – green
RRYY
RY
rryy
RY
ry
RrYy
RY Ry
Gametes
F1 generation
RrYy
RrYy
rY ry
RY Ry
rY ry
Gametes
RY
Ry
rY
ry
RY
RRYY
RRYy
RrYY
RrYy
Ry
RRYy
RRyy
RrYy
Rryy
rY
RrYY
RrYy
rrYY
rrYy
ry
RrYy
Rryy
rrYy
rryy
Plants with
Plants
Plants with Plants with
round and with round wrinkled
wrinkled
yellow
and green and yellow and green
coloured
coloured
coloured
coloured
seeds
seeds
seed
seeds
:
Gg
Cell division
G
g
GG
Gg
×
Cross
Pollination
Gg
(From F2 generation)
G
g
Gg
gg
Gametes
RrYy
F2 generation:
9
(ii) F1 plants are self pollinated to produce
the F2 generation.
Parents
ry
Cell division
G
(C) Ratio obtained in the F2 generation in the
above mentioned experiment = 9 : 3 : 3 : 1
Explanation: R – Round
gg
×
Cross
Pollination
3
:
3
:
1
Green stemmed
Rose plant
F2 generation
Brown stemmed
Rose plant
In F2 progeny, the percentage of brown
stemmed plants is 25%.
When F1 offsprings were allowed to
self pollinate, both the parental traits
were expressed in definite proportion in
F2 generation.
(iii) In F2 generation, 1 plant
genotype GG, 2 plants
genotype 2 Gg and 1 plant
geno type gg were produced.
genotypic ratio is,
GG : Gg : gg
1: 2 :1
having
having
having
So, the
Ratio of GG and Gg in F2 progeny is
1 : 2.
Heredity
51
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ANSWER SHEET
SELF PRACTICE
3. (d) 3 times
9. (d) (I), (II) and (III)
Explanation: Magnification produced by a
spherical mirror gives the relative extent to
which the image of an object magnified with
respect to the object size. It is expressed as the
ratio of the height of the image to the height
of the object.
m =
height of theimage hi
height of theobject h0
Given: Height of flask h = 15 cm
Height of image hi = 45 cm
m = 45 = 3
Explanation: Convex mirrors are diverging
mirror. These mirrors have their focus behind
the mirrors. They produce erect and virtual
images. The images produced are smaller than
the objects.
11. (c) Concave mirror of focal length 24 cm
Explanation: Concave mirror forms real image
of distant object, the sun, at its principal focus,
which is in front of the mirror. Focal length is
the distance between pole of the mirror and
the focus, focal length = 24 cm.
Option (a) is incorrect as convex mirror always
forms a virtual and erect image of the object,
which will be behind the mirror.
15
Related Theory

Magnification (m) is also related to object distance
(u) and image distance (v).

Magnification (m) =
hi
v
=–
h
u
Option (b) and (d) are incorrect as convex lens
forms a real image behind the lens.
14. (d) (IV)
Explanation: In (I), the incident ray is along the
normal, so no refraction of light will take place.
6. (b)
In (II), the two pins on the incident ray are
placed too close to each other.
Explanation: When a light ray is passed through
a glass slab obliquely, then, the emergent ray
is parallel to the direction of the incident ray.
The extent of bending of the ray of light at the
opposite parallel faces AB (air-glass interface)
and CD (glass-air interface) of the rectangular
glass slab is equal and opposite. This is why
the ray emerges parallel to the incident ray.
E
F
i1 N
Air
A
B
O
Glass
N’ r1
(1) The angle of incidence should lie between
30° and 60°.
(2) There should be a minimum gap of 5 cm
between the two pins on the incident ray.
(3) The point of incidence of the incident ray
air-glass interface should be on the left
side such that the emergent ray can be
easily located within the board.
Glass slab
most: f = 10 cm;
M L
D Air
O’ G
M’
52
Therefore, (IV) is the best experimental set-up
for tracing the path of a ray of light through
the glass slab because in this setup, the
following conditions are being met:
29. (a) Convex Lens that will Converge Light the
i2
C
In (III), the angle of incidence appears to be
more than 60°.
r2
H
P
Convex Lens that will Converge Light the
least: f =40 cm;
Reason: A convex lens of shorter focal
length bends the light rays through large
angles
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Explanation: A convex lens of short focal length
bends the light rays through large angles, by
focusing them closer to the optical centre and
hence has greater converging power. Similarly,
a convex lens of longer focal length bends light
rays through smaller angles and hence has
lesser converging power.
Focal point
Principle
axis
Focal
length
(a) Lens having short focal length
Focal
length
F
Focal point
(b) Lens having long focal length
having short focal length
Image Distance (v) cm: +40 cm:
Focal length (f ) cm: +15 cm.
1
1
−
10 ( −30 )
=
1
1
4
+
=
10 (30 ) 30
Since, image formed is at the back of the mirror
at 7.5 cm from the mirror, it must be virtual and
erect.
h
−v
m= i =
ho
u
hi
−7.5
=
4
30 cm
30. (d) Object Distance (u) cm: –24 cm;
=
v = 30 cm = 7.5 cm
4
Principle
axis
Focal
length
Explanation: Object size ho = + 4 cm
object distance u = –30 cm
focal length of diverging mirror f = + 10 cm
image distance v = ?
image size hi = ?
Using mirror formula,
1 1 1
= +
f
v u
11 1 1
== −
vv f u
F
Focal point
38. (c) +1.0 cm
hi =
Explanation: It is given that the candle (object)
is placed at 26 cm and the lens at 50 cm.
Therefore, u = object distance = –(50 – 26)
= –24 cm.
Similarly, image distance
v = (90 –50)cm = + 4 0 c m
To calculate the focal length of the lens,
we use the lens formula,
7.5 × 4
= 1.0 cm
30
43. (d) (A) is false but (R) is true.
Explanation: The speed of light in a medium
is inversely proportional to the refractive index
of the medium. The medium in which speed
of light is more is said to be optically rarer
medium while the medium in which speed of
light is less is optically denser. Speed of light in
an optically rarer medium (air) is greater than
the speed of light in a denser medium (glass).
46. (a) Both (A) and (R) are true and (R) is the
1 1 1 1
1
1
1
8
= − =
−
=
+
=
f v u 40 ( −24 ) 40 24 120 .
Therefore, focal length, f = +15 cm.
36. (d) 110°
Explanation: Initially, angle of incidence =
angle of reflection = 90o/2 = 45o.
On rotating the plane mirror by 10o about O in
the anti-clockwise direction, the position of the
incident ray does not change but the position
of the normal does.
So, angle of incidence = 45o + 10o = 55o.
By the Second Law of Reflection,
angle of reflection = angle of incidence = 55o.
Thus, angle between incident and reflected
rays is 110o.
correct explanation of (A).
Explanation: A concave lens has a virtual
focus as rays appear to diverge from the
principal focus located on the same side of the
lens. As per the sign convention, its focal length
is negative. As power is the reciprocal of focal
length, power of a concave lens is negative.
48. (C)At C i.e., centre of curvature
49. (D) (d) all mirrors irrespective of their shape.
Explanation: The laws of reflection hold
good for light reflected from any smooth
surface i.e., all mirrors regardless of its
shape.
50. (D) (a) real, inverted, diminished image is
formed on a screen placed at the
opposite side of the lens.
Light: Reflection and Refraction
53
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Explanation: When an object is kept
beyond 2F1, the image formed is real and
inverted and it is diminished, that is, size of
image is smaller than the size of the object.
As it is a real image, it can be obtained on
a screen placed on the opposite side of the
lens.
Convex lens only forms real and inverted
image of the object. The eye piece of
second lens (L2) is also a convex lens as it
forms inverted image with respect to the
original object and it magnified the real
image formed by the objective lens.
(B)(i)The value will be more than 1 and
Sign would be Negative. As the image
formed is magnified. Size of image is
greater than the size of the object
(E) (d) image becomes larger in size.
Explanation: We observe that as the object
is brought closer to the focus F1 from the
left, the real and inverted image becomes
larger in size.
i.e. hi > ho
As m =
52. (C) (d) First Student: Image will be formed
behind the mirror; Second Student:
Image will be formed behind the
mirror.
hi
,m>1
ho
When the image is real and inverted,
the image lies below the principal
axis. Therefore, height of image (hi)
is negative. As height of object (ho) is
always positive,
h
therefore, m = i = negative.
ho
Explanation: A concave mirror forms an
erect and virtual image of an object only
when the object is placed between the pole
and focus of the mirror as shown below:
Related Theory

When image is of same size as the object, i.e. hi = ho
h
m= i =1
ho
When image is smaller than the object, i.e. hi<ho
h
m = i , m<1
ho
Further, as the image formed is virtual, it
cannot be obtained on a screen. Image
formed is enlarged and behind the mirror.
As, focal length of the mirror = 20 cm =
40/2 cm,
therefore, range of object distance = 0 cm
to less than 20 cm.
If object is placed at focus (20 cm), it will
form a real, inverted and highly magnified
image of the object.
(D) (b) Real and inverted image will be formed
40 cm in front of the mirror of same
size as object.
Explanation: As the object is kept at 40
cm or 2f, image will also be formed at 2f.
It will be a real and inverted image will be
formed in front of the mirror and image
will be of same size as object.
When image is virtual and erect, image lies above
principal axis. So, hi is positive as height of object (ho)
is always positive.
(ii) The value will be more than 1 and Sign
would be Positive. In lens (L2) the image
formed is magnified, inverted but virtual.
Magnified means m > 1
Since image formed is virtual, m = +ve
54. (A) As the object is moved away from the
focus, the image moves closer to the focus,
till it becomes a point sized image when
object is at infinity. Size of image decreases
when object is moved away from the focus.
(B)
As the image formed by a convex
mirror is always virtual and diminished,
magnification m produced by a convex
mirror is always smaller than + 1. This is
because magnification m =
height of image
height of object
and since height of image < height of
object, m < +1.
55. (C) (i) As the ray of light travels along the
53. (A) Both lenses will be convex objective lens
(L1) forms a real, inverted and magnified
image of the given object.
54
same path without any bending. We can
conclude that the ray of light is incident
normally. Angle of incidence, angle of
refraction and angle of emergence all are
equal to 0o in case of normal incidence.
Science Class X
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A'
M
l=0
A
F2
O
e =0
2F2
F1 B
B' 2F1
C1
C2
N
75. (A) An incident ray parallel to the principal
(ii) The refractive index of a medium is
the ratio of speed of light in vacuum to
the speed of light in the medium.
3 × 108
Refractive index of glass =
= 1.5
2 × 108
and,
Refractive index of water
axis, after reflection appear to diverge from
the principal focus is case of convex mirror.
=
D
A
B
3 × 108
= 1.33
2.25 × 108
X
Therefore, refractive index of glass is
more than refractive index of water.
Related Theory

The magnification produced by a plane mirror is
always +1.

The plus sign indicates that the image is erect and
magnitude 1 indicates that size of image is equal to
the size of the object for all positions of the object.
71. To project the image of a candle flame on the
∠i = ∠r
∠i = Angle of incidence
∠r = Angle of reflection
Explanation: A straight line XP is principal
axis. AB is incident ray which is parallel to
principal axis XP of a convex mirror MM’.
The ray of light gets reflected at point B
on the mirror and goes in the direction BD
and it appears to be coming from the focus
F of convex mirror. According to the laws of
reflection, ∠i = ∠r. Therefore, the incident
and reflected rays make equal angles with
the principal axis.
(B) An incident ray passing through the
principal focus of a concave mirror, after
reflection, will emerge parallel to the
principal axis.
A
In
cid
en
tR
ay
LLL
LLLLLLL
Principal axis
X
C Norm
P
ay
al R
D
i
r
LL
Reflected Ray
LL
L
f = +15 cm
Since, the object distance from optical centre
of a convex lens is less than the focal length,
hence the object lies between focus F1
and optical centre O. Image formed will be
enlarged, virtual and erect and on the same
side of the lens as the object.
C
LL
73. Given : u = –10 cm
F
LL
walls of the school laboratory, he should use a
concave mirror as only a concave mirror forms
a real image (which can be obtained on a
screen) for all positions of object except when
the object is placed between the pole and
focus of the mirror. A convex mirror and plane
mirror, on the other hand, always form virtual
images.
P
M'
70. Characteristics of the images formed by plane
mirrors are:
(1) The image formed by a plane mirror is
virtual and erect.
(2) The image is of the same size as the object.
(3) It is formed as far behind the mirror as the
object is in front of it.
(4) The image is laterally inverted (inverted
sideways).
M
r
i
P
Pole
B
Concave mirror
∠i = ∠r
∠i = Angle of incidence
∠r = Angle of reflection
(C) An incident ray coming obliquely to the
principal axis towards a point (Pole of the
mirror) on the convex mirror is reflected
obliquely.
Light: Reflection and Refraction
55
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Inci
A
den
t
X
i
Principal axis
d
ecte
Refl
D
Speed of light in medium x = 3 × 108 m sec–1
on substituting the value we get,
Ray
r
Ray
P
F
C
Focus Centre of
curvature
Convex mirror
∠i = ∠r
∠i = Angle of incidence
∠r = Angle of reflection
to medium 1 is given by the ratio of the speed
of light in medium 1 and the speed of light in
medium 2.
Given,
Refractive index of medium ‘x‘ with respect to
2
medium ‘y‘ =
3
f=?
P=
1
f
f=
1
1
=
P
− 2.5 D
= – 0.4 m = – 40 cm
v = – 10 cm
...(i)
Refractive index of medium ‘y‘ with respect to
4
3
Speed of light in medium z
4
=
Speed of light in medium y
3
...(ii)
To calculate:
Refractive index of medium ‘z‘ with respect to
medium ‘x‘ = ?
nzx =
Hence, the speed of light in medium ‘y‘ is
2 × 108 m sec–1
f = – 40 cm
Speed of light in medium y
2
nxy =
=
Speed of light is medium x
3
xyz =
2 × 3 × 108
= 2 × 108 m sec–1
3
84. (A) P = –2.5 D
82. The refractive index of medium 2 with respect
medium ‘z‘ =
vy =
Speed of light in medium x
Speed of light in medium z
From (i) and (ii) equations,
32
3 3
9
= × =
43
2 4
8
Hence, refractive index of medium ‘z‘ with
9
respect to medium ‘x‘ is .
8
nzx =
8
speed of light in medium ‘x‘ = 3 × 10 m/s
speed of light in medium ‘y‘ = ?
nx =
Speed of light in air
c
Or
Speed of light in medium x
vx
ny =
Speed of light in air
c
=
Speed of light in medium y
vy
u=?
1
1 1
= −
f
v u
1
1 1
= −
u
v f
=
1
1
−
− 10 − 40
1
1
+
=
− 10 40
−4+1
=
40
−3
=
40
u=
−40
3
= – 13.33 cm
The object should be placed – 13.33 cm
from the lens for the formation of an image
at a distance of 10 cm from the lens.
(B)
M
A
V
Refractive index of medium ‘x‘ with respect to
‘y‘ =
2
3
2F1
vy
2
c
=
×
3
vx
c
Or
56
v
2
= y
3
vx
F1
B B'
N
86. (A)When an object is placed in front of the
lens between optical centre and principal
focus of a convex lens, the image is formed
beyond 2F1 (on the same side of the object).
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A'
M
A
O
B' 2F1
C1
F2
2F2
F1 B
C2
N
AB is the object and A’B’ is the image.
The image formed is enlarged, virtual and
erect so the value of magnification will be
greater than 1 and its sign will be positive.
(B) When an object a placed anywhere infront
of a concave lens.
When we place an object between infinity
and optical centre (O) of the concave lens,
the image will be formed between focus
(F1) and optical centre (O) on the same side
of the lens.
M
A
A'
2F1
B
F1
B'
O
N
AB is the object and A’B’ is the image the
image formed is diminished, virtual and
erect. So the sign of magnification is positive
(+) and the magnification will be less than 1.
(C) When the object is placed at 2F1 of convex
lens, the image is formed at 2F2 on the
other side of the lens.
=
−5+3
−2
=
75
75
75
= – 37.5 cm
−2
v=
∴ Image distance, v = – 37.5 cm
A screen should be placed at 37.5 cm
infront of the mirror.
(B) Size of image: Using the expression for
linear magnification.
h2
v
m= h = −
u
1
h’ =
−vh
u
−(37.5) × 4.0
=
( −25)
∴
h2 = – 6 cm
The height of image is 6 cm i.e. it is
magnified. The minus sign shows that the
image is below the principal axis. Therefore,
it is real and inverted.
(C) Scale is given as :
5 cm = 1 cm
FP = 3 cm
CP = 6 cm
AB = 0.8 cm
A’B’ = 1.2 cm
AB’ = 7.5 cm
M
M
A
B
B'
B
2F1
O
F2
2F2
A
C
P
F
C2 B'
F1
A'
C1
A'
N
The image formed is real, inverted and of
the same size.
87. (A) Given: Height of object, h1 = + 4 cm
Object distance, u = –25.0 cm
Focal length of a concave mirror of,
f = –15.0 cm
Using the mirror formula,
1
1 1
= +
f
v u
1
1 1
= −
v
f u
1 = 1 − 1
− 15
− 25
v
N
89. (A)Diagrams showing the pole, Centre of
Curvature and Principal axis of the spherical
mirrors, concave mirror and convex mirror
are drawn below:
Hollow sphere
Principal axis
Radius of
curvature
C
Center of
curvature
P
Pole
Concave mirror
Light: Reflection and Refraction
57
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To find the focal length f of the lens, we use the
lens formula,
1
1 1
= −
v u
f
Hollow sphere
Radius of
curvature
P
Pole
Principal axis
1
1
−
=
25 − 25
1
1
2
=
+
=
25
25
25
C
Center of
curvature
Concave mirror
(i) Pole: The centre of the reflecting
surface of the spherical mirror is a
point called the pole.
(ii) Centre of curvature: It is the centre
of the hollow sphere of glass of which
the reflecting surface of the spherical
mirror is a part.
(iii) Principal axis: It is the straight
imaginary line passing through the
centre of curvature and pole of a
spherical mirror.
(B) (i) The type of mirror is a concave mirror
as image formed is magnified, virtual
and erect, and behind the mirror. This is
possible when object lies between pole
and principal focus of a concave mirror.
(ii) Ray diagram showing the formation of
image.
A'
F
D
LLLLLLLL
C
B
B'
LL
Y
E
LL
A
LL
L
L
L
X
Here AB represents the object ‘P’ and
A’B’ represents its image ‘Q’ given in
question.
Related Theory

A convex mirror always forms a virtual erect but
diminished image of the object and a plane mirror
forms a virtual, erect image of same size as the
object.
90. As magnification, m, of the image formed
by the lens is -1, the lens is a convex lens as
concave lens always produces a virtual and
erect image of the object.
It is given that image distance, v = 25 cm.
1st method: Find position of object and focal
length of lens:
To find position of object or object distance u,
we will use the formula for magnification.
v
m= .
u
25
–1=
u
∴
u = – 25 cm.
58
f=
∴
25
cm = 12.5 cm
2
So, it is a convex lens having focal length 12.5 cm.
2nd method: Find position of object and focal
length of lens:
It is given that magnification is –1. That is,
image formed is real and inverted and of same
size as object. This is true only in case of a
convex lens when object is placed at 2F. In that
case, image is also formed at 2F but behind the
lens.
∴
2f = 25
⇒
f = 12.5 cm
Also,
u = – 25 cm
(Any 1 of the 2 methods can be written to get
full marks)
If the object is now displaced 15 cm towards
the lens, the object will lie at 10 cm from the
optical centre. So, object now lies between
optical centre and focus of the lens. To find
position of image, we will use lens formula.
1
1 1
= −
f
v u
1
1 1
=
+
v
f
u
1
1
=
−
12
.
5
10
2
1
=
−
25 10
2
= −
100
∴
v = – 50 cm
Image will be virtual, erect and formed 50 cm
in front of the lens.
Ray diagram:
A
B'
2F1
F1 B
O
F2
2F2
91. (A)Yes, even when one half of the convex lens
is covered with a black paper, the complete
image of the object will be formed.
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When the upper half of the lens is covered:
In this situation, rays of light coming from
the object will be refracted by the lower
half of the lens as shown in fig (a). These
rays meet at the other side of the lens to
form the image of the given object. When the lower half of the lens is covered:
In this situation, rays of light coming from
the object will be refracted by the upper
half of the lens. These rays meet at the
other side of the lens to form the image of
the given object, as shown in the figure (b).
We will get a sharp image but the
brightness of the image will be less now.
Ray diagram:
B
F
B'
92. (A)A real, inverted and same size image as
that of object formed by the concave mirror
will form an image of magnification –1. It is
possible only when the object is placed at
C (R = 2f). Hence, for the object distances of
20 cm and 30 cm, concave mirrors ‘A’ and
‘B’ will form the real, inverted and same
size images as that of the object therefore,
the mirrors ‘A’ and ‘B’ will form an image of
magnification –1.
A'
F
No light ray
transmitted
(b)
(B) Convex lens can be used as a magnifying
glass.
C
F
A'
C
covered with
black paper
B
A
F
A
B'
(b)
Explanation:
Focal length
Object distance
Position of object
Position of image
Size of image
fA 10 cm
10 cm
At F
At infinity
Highly enlarged
fA 10 cm
20 cm
At C
At C
Same size
fA 10 cm
30 cm
Beyond C
Between F and C
Diminished
fB 15 cm
10 cm
Between P and F
Behind the mirror
Enlarged
fB 15 cm
20 cm
Between F and C
Beyond C
Enlarged
fB 15 cm
30 cm
At C
At C
Same size
fC 20 cm
10 cm
Between P and F
Behind the mirror
Enlarged
fC 20 cm
20 cm
At F
At infinity
Highly enlarged
fC 20 cm
30 cm
Between F and C
Beyond C
Enlarged
(B) Concave mirror ‘C’ of focal length 20 cm
will be preferred to be used for shaving
purpose / make up. This is because when
we bring our face within its focal length it
forms a virtual, erect and enlarged image
of our face.
(C) Ray diagram for image formation by mirror ‘B’
(i) For object distance 10cm.
M
A
C
F
B
i
r
E
P
A’
(ii) For object distance 20 cm.
M
E
B'
A
O
B F
P
C
N
A'
B’
N
Light: Reflection and Refraction
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ANSWER SHEET
SELF PRACTICE
2. (c)violet and blue light get scattered more
than the lights of all other colours by the
atmosphere
Explanation: When sunlight passes through
the atmosphere, the fine particles in air scatter
the blue colour (shorter wavelengths) more
strongly than red colour. The scattered blue
light enters our eyes and the sky appears blue.
6. (d) r < v
[CBSE Marking Scheme Term-1 SQP 2021]
Explanation: The amount of deviation of
light depends on its colour as the refractive
index of glass is different for different colours.
Since, violet colour deviates the most and red
deviates the least, the angle of refraction r < v.
7. (d)
gets deviated and bends towards the
thicker part (base) of the prism.
Related Theory

The molecules of air and other fine particles in the
atmosphere are smaller in size than the wavelength
of visible light. These are more effective in scattering
light of shorter wavelengths at the blue end than
light of longer wavelengths at the red end.
3. (b)the ciliary muscles will contract
Explanation: When a person is seeing an
object closer to his eyes, the ciliary muscles
contract. This increases the curvature of the
eye lens. The eye lens then becomes thicker.
Consequently the focal length of the eye
lens decreases. This enables us to see nearby
objects clearly.
Related Theory

The curvature of the eye lens can be modified to
some extent by the ciliary muscles. The change in
curvature of the eye lens can thus change the focal
length.
4. (c) ∠i, ∠r and ∠A
Explanation: ∠i is the angle of incidence, which
is the angle between the incident ray and the
normal at the point of incidence and is marked
correctly.
∠r is the angle of refraction, which is the angle
between the refracted ray and the normal at
the point of incidence and is marked correctly.
∠e is the angle of emergence, which is the
angle between the emergent ray and the
normal at the point of emergence but is
wrongly marked here.
∠A is the angle between the two lateral faces
of a prism and is correctly marked.
∠D is the angle of deviation, which is the angle
between incident ray and emergent ray, but is
wrongly marked here.
So, ∠i, ∠r and ∠A are correctly marked here.
60
Related Theory

In a prism, the ray of light from air into glass bends
towards the normal. The ray of light from glass to air
bends away from the normal. In both cases, when a
ray of light passes through a prism, it bends towards
the base (thicker part) of the prism.
9. (b)
Atmosphere
Sun
Explanation: Sunlight travels at the speed of
light is an electro magnetic wave and it does
not require any medium for its propagation so
light can travel in vacuum of space. When this
light energy reaches the earth’s surface, some
of it is transferred to the earth’s surface. In fig
(b) only the light of the sun reaches the earth’s
atmosphere so option (b) is correct.
11. (d) (I) and (III)
Explanation: Light rays get refracted
whenever they travel at angle into a medium
with a different refractive index. In options (I)
and (III) the light ray enters the prism from the
air and the light ray moves from the prism into
the air, the refraction take place.
12. (a) are relaxed and the lens becomes thinner
Explanation: The focal length of the eye lens
increases when eye muscles are relaxed and
the lens becomes thinner. The ability of the eye
to adjust the focal length and form image of
near and far objects is called accommodation.
When an object is at infinity, the ciliary muscles
are relaxed and the focal length increases and
a clear image is produced. When an object
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is comparatively nearer, we find the cliliary
muscles need to tighten and the focal length
decreases with a clear image formation on the
retina.
14. (b) Rainbow formation
Explanation: When light from the sun falls on
the droplets of water suspended in air the light
ray gets refracted as well as dispersed. The
dispersed ray on striking the other surface of
the droplet suffers total internal refraction the
ray further suffers refraction and emerge out
as the band of colours in the form of a rainbow.
15. (b) (II)
[CBSE Marking Scheme Term-1 SQP 2021]
Explanation: When white light is incident on
a prism, it splits into its component colours
VIBGYOR. The colour of sky, namely blue, will
be seen as third colour from the top of the
spectrum if the prism is inverted as shown in
figure (ii).
=−
=
1
1
+
50 25
−1 + 2
50
1
50
Þ
f = +50cm = +0.5m.
=
As power, P =
1
,
f ( inm )
therefore, power of lens = + 2.0 D
It is a convex lens of power +2.0 D.
Corrective lens required for seeing distant
objects clearly:
Far point of the defective eye = 75 cm.
Here, u = - ∞, v = - 75 cm = - 0.75 m.
1 1 1
= −
v u
f
=−
100  1 
− −
75  ∞ 
=−
4
3
Þ
f = –0.75m.
Therefore, P = - 1.33 D
It is a concave lens of power – 1.33 D.
21. (b) Both (I) and (III)
18. (b)
Explanation: In the bright light, the pupil
becomes smaller to reduce the light entering
the eye. This happens when iris contracts the
pupil to allow less light to enter the eye. The
pupil of an eye acts like a variable aperture
whose size can be varied with the help of the
iris.
19. (d)
For Reading Purpose: Convex lens of
power +2.0 D ; For Seeing Distant Objects:
Concave lens of power –1.33 D
Explanation: Corrective lens required for
reading purpose:
Near point of defective eye = 50 cm.
Therefore, u = –25 cm, v = –50 cm.
Using lens formula,
1 1 1
= −
v u
f
=−
1  1
− −
50  25 
Explanation: The colours marked from 1 to 7
are :
(1) Red (colour of blood)
(2) Orange
(3) Yellow (colour of gold metal)
(4) Green (colour of leaves in plants)
(5) Blue (colour of sky)
(6) Indigo
(7) Violet
The violet colour is deviated the most while red
is deviated the least.
26. (c) (A) is true, but (R) is false.
Explanation: When the objects are observed
through hot air, the objects appear to be
moving slightly as the light refracts due to
atmospheric refraction. This refraction occurs
because hotter air is optically rarer than the
cooler air above it and has a refractive index
slightly less than that of the cooler air.
29. (A) The phenomenon is “atmospheric refraction”.
The two observations are:
Twinkling of stars, advance sunrise, delayed
sunset and apparent position of stars.
(Write any two to get full marks)
The Human Eye and the Colourful World
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(B) The total time difference on the duration of
day on earth = 4 minutes = 2 minutes (due
to advanced sunrise) + 2 minutes (due to
delayed sunset)
The angle of emergence is the angle
between the emergent ray FS and the
normal MM’ drawn at the point of
emergence = ÐSFM.
A
30. (B)The band of coloured components of white
light is known as spectrum of light. The
cause for formation of spectrum of light is
that different colours of light bend through
different angles with respect to the incident
ray as light passes through a prism.
N
i
Q
31. (A) A rainbow would not be visible on the
surface of the moon to the astronauts
as moon does not have atmosphere as
well as water. We know that a rainbow is
formed when the atmosphere contains a
large numbers of tiny water suspended
in air. These tiny water droplets act as
small prisms. They refract and disperse the
incident sunlight, then reflect it internally
and finally refract it again when it comes
out of the water droplets.
(B) At the time of sunset and sunrise, The
sun is near the horizon, appears to be
oval shaped due to the phenomenon of
atmospheric refraction. Light rays coming
from the lower end of the sun near the
horizon pass throught thicker layers of air
than the light rays from the upper edge
of the sun. Consequetly light rays from
the lower edge are diviated more and the
vertical diameter of the sun appears to be
diminished in size. So, the air appears to be
flattered i.e. of the oval shape.
At noon, the sun appears circular in shape
as the light rays normally pass through the
earth’s surface and go straight without any
refraction.
(C) Among all the colours, red is scattered
the least by smoke or fog. This is because
the wavelength of red colour is the
largest. Thus, it can be easily seen from
a distance. Since the wavelength of red
light is maximum in the spectrum, it’s
penetration power in the air is maximum
and so we can see red colour from farther
distances. thus, danger signal uses red
colour.
32. (B) (a) Angle of Incidence (i): PEN; Angle
of Refraction (r):
Emergence(e): SFM.
62
FEN’;
Angle
P
The angle of refraction is the angle
between the refracted ray EF and the
normal NN’ drawn at the point of incidence
= ÐFEN’.
E
B
PE – Incident ray
EF – Refracted ray
FS – Emergent ray
∠A – Angle of the prism
r
N’
M
D
M’
F
e
R
C
S
∠ i – Angle of incidence
∠ r – Angle of refraction
∠ e – Angle of emergence
∠ D – Angle of deviation
(C) (d) emergent ray and incident ray
Explanation: Unlike a glass slab where
emergent ray is parallel to the incident
ray, in a glass prism the emergent ray is
not parallel to the incident ray. The angle
of deviation (denoted by D) is the angle
between the emergent ray and incident
ray.
(D) (b)The refracting surfaces are inclined at
an angle.
Explanation: As the two refracting surfaces
AB and AC are not parallel to each other,
the emergent ray is also not parallel to the
incident ray. Whereas, in a glass slab, the
two refracting surfaces are parallel to each
other which explains why the emergent
ray is parallel to the incident ray.
33. (A) The lens L1 produces a parallel beam of
light when strong source of light is kept at
its focus. The lens L2 converges a parallel
beam of light and we get a sharp image of
the circular hole on the screen.
(C) As the sulphur particles begin to form,
we will observe blue light from the three
sides of the glass tank. This is due to
scattering of short wavelengths by minute
colloidal sulphur particles. The colour of
the transmitted light from the fourth side
of the glass tank facing the circular hole
will be orange red at first and then bright
crimson red colour on the screen.
of
Explanation: The angle of incidence is the
angle between the incident ray PE and the
normal NN’ drawn at the point of incidence
= ÐPEN.
G
H
The observations can be explained by
scattering of light. When the sulphur
particles begin to form, a colloidal solution
is formed in the glass tank due to which
short wavelengths of light are scattered by
the minute colloidal sulphur particles.
34. (D) (a) refractive index in medium is gradually
changing
Explanation: The atmospheric refraction
occurs in a medium of gradually changing
refractive index. Usually in the atmosphere
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air above is cooler than the air below it. So,
light rays coming from the sun or stars on
entering the earth’s atmosphere, undergo
refraction continuously before they of
reach the earth.
(E) (b) it bends towards the normal
Explanation: When star light enters the
earth’s atmosphere, undergoes refraction
continuously as the refractive index of the
medium is changing. Air above is warmer
as compared to air below it. So, light
travels from a rarer medium to a denser
medium due to which it bends towards the
normal.
Star
Apparent
star position
Ray path
Refractive index
increasing
35. (A) Piyush is suffering from myopia. Myopia is
also known as near-sightedness Piyush
can see nearby objects clearly but cannot
see distant objects clearly. for example his
blackboard.
Causes of Myopia:
(i) Elongation of the eye ball
(ii) Excessive curvature of eye lens
36. (B) (a) Blue, Green, Red
Explanation: The colours in ascending
order of wavelength are: Violet, Indigo,
Blue, Green, Yellow, Orange and Red. The
wavelength of violet colour is the least and
that of red is the greatest.
(D) (b) the different colours in the white light
bend towards the normal at different
angles on entering prism.
Explanation: When white light enters a
prism, different colours undergo refraction
and as they enter a denser medium, they
bend towards the normal at different
angles.
(E) (c) vacuum
Explanation: The velocity of waves of
all colours is different when light passes
through an optical medium as different
colours have different wavelengths and
refractive index depends on the medium.
However, when white light passes through
vacuum, all colours will have same
velocity.
37. (A) (c) Very fine particles scatter mainly red
light.
Explanation: When sunlight passes
through the atmosphere, the fine particles
in air scatter the blue colour as it shorter
wavelength. The danger signal lights are
red in colour as it is least scatter by fog or
smoke and can be seen in the same colour
at a distance.
(C) (b) Both (II) and (III)
Explanation: The colour of water in deep
sea and blue colour of sky are due to
scattering of light by the particles present
in sea water and atmosphere respectively.
Twinkling of stars is due to atmospheric
refraction whereas formation of rainbow
involves refraction, dispersion and total
internal reflection of light.
(D) (a) V
ery Fine Particles: Blue; Larger Particles:
Red ; Large Enough Particles: White
Explanation: The colour of the scattered
light depends on the size of the scattering
particles. Very fine particles scatter mainly
blue light while particles of larger size
scatter light of longer wavelengths. If the
size of the scattering particles is large
enough, then, the scattered light may even
appear white.
38. The sky would have looked dark if the earth
had no atmosphere as there would not be no
scattering.
40. There is no sky in the outer space and there is
no atmosphere containing air to scatter light.
The astronaut finds sky to be dark instead of
blue.
41. At noon, the sun is directly over head, the
sunlight passes through much smaller portion
of earth‘s atmosphere. The scattering is much
less and the sun looks white.
45. The phenomenon shown in the above diagram
dispersion. Speed of violet light inside the prism
is slowest and that of red is highest. Hence,
deviation of violet light is maximum and that
of red is minimum
When a beam of white light is passed through
a glass prism, it splits up into seven colours.
Different colours of light are characterised by
their different wavelengths. All the colours
travel in air/vaccum with the same speed, but
their speeds in any other refracting medium
(like glass or water) are different. As refractive
c
indix of glass n =
and v is different for
v
different colours, so ‘n’ is different for different
colours. As vviolet > vred, therefore nviolet > nred.
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It means refractive index of glass for violet
colour is more that the refractive index of
glass for red colour. Hence, deviation suffered
by violet colour is greater than the deviation
suffered by the red colour on passing through
the prism that is why violet colour is at the
lower end of the visible spectrum and red
colour is at the upper end of the spectrum.
49. The observation on the basis of which Newton
concluded that sunlight is made up of seven
colours was the emergence of white light
from the other side of a second identical and
inverted prism placed alongside the first prism
and through which all colours of the spectrum
were allowed to pass.
54. (A) Visible spectrum is the band of coloured
Correction for hypermetropic eye
58. (A) The colour of scattered light depends on
the size of the scattering particles.
Fine particles of small size scatter
mainly blue light (blue light has smaller
wavelength as compared to red light
The particles of larger size scatter mainly
the red light.
When the size of scattering particles is
quite large, all wavelengths are scattered
equally and the scattered light appears
white.
components of a white light beam.
(B) Red light is scattered the least by air
molecules and has longer wavelength. It
travels the long distance.
(C) The given setup will behave like a glass
slab, resulting in recombination of the
seven colours to produce white light there
will not be any dispersion.
56. (A) Myopic eye:
N'
N
(B) The hot air is lighter (less dense) than the
cooler air above it and has a refractive
index slightly less than that of the cooler
air. Since the physical conditions of the
refractive medium (air) are not stationery,
the apparent position of the object as seen
through the hot air, fluctuates or wavers.
So the wavering is an effect of atmospheric
refraction of light by earth’s atmosphere.
(C) Refraction
Far point of a myopic eye
O’
A
P2
R
R
V
O
White
light
Myopic eye
P1
V
Screen
White
light
A
59. (A) Snell’s law of refraction of Light: The ratio
of sine of angle of incidence to the sine of
angle of refraction is a constant, for the
light of a given colour and for the given pair
of media. This law is also known as Snell’s
law of refraction. (This law is true for angle
0 < i < 90º)
Correction for myopia
(B) Hypermetropic eye:
sin i
= Constant
sin r
N
Near point of a hypermetropic eye
This constant value is called the refractive
index of second medium with respect to
the first.
Related Theory
N

N'
Hypermetropic eye
64
Snell’s law of refraction of light is a second law of
refraction which was experimentally discovered by a
Dutch mathematician W. Snell and is, therefore, also
known as Snell’s law of Refraction.
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(B)
S.
No.
E
F
(3)
i2 N
Air
A
O
Glass
B
N' r
2
Glass
slab
O'
Prism splits the
incident white light
into a band of colours.
(Any one)
(i) When
a
narrow
beam
of
monochromatic light passes through a
a glass slab, it deviates from the actual
path but the direction of incident ray
and the emergent ray are parallel to
one another.
L
D Air
G
M' r2
H
P
Explanation: Emergent ray is parallel to the
direction of the incident ray. The extent of
bending of the ray of light at the opposite
parallel faces AB (air-glass interface) and CD
(glass-air interface) of the rectanglular glass
slab is equal and opposite and that is the
reason for the emergent ray to be parallel
to the incident ray. The light ray is shifted
sideward slightly. This sideward shift in the
path of a light ray on emergence from the
glass slab is called lateral displacement.
(ii) When a narrow beam of white light
passes through a glass slab, the white
light does not split it into its constituent
colours. The direction of incident ray
and the emergent ray of light are
parallel to each other.
(B) Glass Prism:
(i) When a narrow beam of mono
chromatic light passes through a
glass prism, it deviates from the actual
path but the direction of incident say
and the emergent ray of light are not
parallel to each other.
Related Theory
 Lateral displacement depends on:
 The thickness of glass slab
 The refractive index of glass slab
 The angle of incidence.
60. Differences between a glass slab and a glass
(ii) When a narrow beam of white light
passes through a glass prism, the
white light splits into its constituent
seven colours. The incident ray and the
emergent ray of light are not parallel
to each other.
prism.
(2)
Slab does
not split the
incident white
light into
a band of
colours.
(A) Glass Slab :
M
C
(1)
Glass Prism
i2
S.
No.
Glass Slab
Glass Slab
Glass Prism
A glass slab A glass prism has
has six parallel two triangular bases
refracting
and three rectangular
lateral surfaces. These
surfaces.
surfaces are inclined
to each other at some
suitable angle, which is
called angle of prism.
The ray of light
incident on one
surface of the
slab undergoes
two refractions
and emerges
from the other
surface of slab
in a direction
parallel
to
the
incident
ray. However,
it is slightly
displaced
laterally.
The deviation of a ray
on passing through
a prism depends on
angle of prism and
angle of incidence and
material of the prism.
Related Theory


Light which consists of only one colour or single
wavelength is called monochromatic light.
The white light of sun gives us various colours of
rainbow.
61. The defect of vision that arises due to gradual
weakening of ciliary muscles and diminishing
flexibility of the eye lens is presbyopia.
Most people find it difficult to see nearby
objects comfortably and clearly without using
corrective lenses. This is corrected by using
convex lens of appropriate power.
However, sometimes people may suffer from
both myopia and hypermetropia. This can be
corrected by using a bifocal lens.
A bifocal lens consists of two parts – the upper
part is concave lens and lower part is convex
lens. The upper part is for viewing distant
objects and the lower part facilitates near
vision.
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62. (C)Values displayed by Sudhir: Concerned,
In other words, when the light is dim, iris
expands the pupil and when the light is
bright, Iris contracts the pupil.
helpful, compassionate
Values displayed by teacher: Concerned,
Knowledgeable



Related Theory
Two causes of myopia are: (i) Excessive curvature of
the cornea (ii) Elongation of the eyeball
Myopia is also called short-sightedness as a person
is unable to see distant objects clearly but able to
see nearby objects distinctly.
To find focal length of the lens,
we use the relation =
1
Using this formula,
f ( in m ) .
1
m . = – 2m.
0.5
f(in m) = –
69. A beam of white light falling on a glass prism
gets split up into seven colours as shown in the
diagram:
(A) The colours at position marked 3 and 5 are
yellow and blue respectively. On, the other
hand, student has identified them as blue
and yellow. Hence the statement made by
the student is incorrect.
ht
e Lig
Whit
64. A person is suffering from hypermetropia or
long sightedness if he uses a convex lens for
correction of his vision.
Glass Prism
Power P of the lens = + 0.5 D
1
As,
P =
,
f ( in m )
Red
Ora
Yel nge
l
Gre ow
Blu en
e
Ind
ig
Vio o
let
The spliting of light into its component
colours is called dispersion. The band of
the coloured components of a light beam
is called its spectrum. Different colours of
light bend through different angles with
respect to the incident ray, as they pass
through the prism. The red light bends the
least while the violet the most.
where, f = focal length of the corrective lens,
1
m = +2m.
0.5
Focal length of lens = + 2 m or + 200 cm.
f =
66. (A)When we see a distant object the muscles
(B) (i) Position 7 is the position of violet colour,
which corresponds to the colour of a
solution of potassium permanganate.
(B) When we see a nearby object the ciliary
muscles contract, this increases the
curvature of the eye lens and it becomes
thicker consequently the focal length of the
eye lens decreases.
(ii) Position 1 is the position of red colour,
which corresponds to the colour of
‘danger’ or stop signal lights.
relax the lens become thin and its focal
length increases.
A normal eye is unable to clearly see the
objects placed closer than 25 cm because
the ciliary muscles of eyes are unable to
contract beyond a certain limit.

Related Theory
The ability of the lens to adjust is called power of
accommodation. The change in the curvature of the
eye lens can thus change its focal length.
Here, N = 25 cm and N’ = 1 m or 100 cm
68. (A) The parts are iris and pupil.
Between the cornea and the lens we have
a muscular diaphragm called iris, in which
a small hole is present which is called pupil.
Iris is the coloured part that we see in the
eye. The size of the pupil varies with the
help of iris. In dim light, the size of the pupil
increases with the help of Iris, so that more
light enters the eye. While in bright light,
the size of the pupil decreases, so that less
light enters the eye.
66
70. (B) X in the given figure is Iris. It controls the
size of the pupil.
(C) (i) When we increase the distance of an
object from the eye, the image distance
from the eye lens in the normal eye
remains unchanged.
(ii) When we decrease the distance of an
object from the eye, image distance
from the eye lens the normal eye
remains same.
The normal eye has the ability to adjust
focal length of its lens using ciliary
muscles. So when we change the
object distance within specified limits,
the image distance remains same.
71. (A) Functions of parts of human eye:
(i) Cornea: It is a thin membrane on the
front surface of the eyeball through
which light enters the eye.
(ii) Iris: Iris is a dark muscular assembly
which lies behind the cornea and
controls the size of the pupil.
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(iii) Crystalline (Eye) lens: It provides
the finer adjustment of focal length
required to focus objects lying at
different distances on the retina of the
eye.
(iv) Ciliary muscles: These are the muscles
that change the shape or curvature of
the eye lens so that nearby as well as
distant objects can be focused.
(v) Retina: It is a delicate membrane
containing a large number of light
sensitive cells which generate electrical
signals when illuminated.
(B) As the person is unable to see distinctly the
objects closer than 1 m, which is greater
than 25 cm or the near point of a normal
person, he is suffering from hypermetropia.
73. (A) Issac Newton was the first to use a glass
prism to obtain the spectrum of white
light. He tried to split various colours
of the spectrum of white light by using
another similar prism, he could not get
any more colours. Thus he proved that
sunlight is made of seven colours.
sensitive receptors of the retina are called rods
and cones. When light falls on these receptors,
they send electrical signals to the brain
through the optic nerve. The space between
the retina and eye lens is filled with another
fluid, called vitreous humor.
Working: The light coming from an object
enters the eye through the cornea and pupil.
The eye lens converges these light rays to form
a real, inverted and diminished image on the
retina.
The light sensitive cells of the retina get
activated with the incidence of light and
generate electric signals. These electric signals
are sent to the brain by the optic nerves and
the brain interprets the electrical signals in
such a way that we see an image which is
erect and of the same size as the object.
Power of accommodation: We are able to see
nearby as well as distant objects due to the
eye’s ability to adjust its focal length, which is
known as power of accommodation. Relaxation
of muscles makes the lens thinner and its focal
length increases to make us see the distant
objects clearly. Contraction of ciliary muscles
increases the curvature of the eye lens and
makes the eye lens thicker. Consequently, the
focal length of the eye lens decreases. This
enables us to see nearby objects clearly.
White light
beam
Glass prism
V
Wh
it
spe e light R
ctru
m
Ciliary muscles
(B) Atmospheric Refraction: It is the refraction
of light by the earth’s atmospheric layers
having varying refractive indices.
Two natural phenomena:
(1) Twinkling of stars,
(2) Advanced sunrise and delayed sunset
74. Structure of the human eye:
Cornea: The front part of the eye is covered by
a transparent spherical membrane, called the
cornea. Light enters the eye through cornea.
The space behind the cornea is filled with a
liquid called aqueous humor.
Iris: Just behind the cornea is a dark coloured
muscular diaphragm, which has a small
circular opening in the middle. It controls the
size of the pupil.
Crystalline lens
Aqueous
humour
Retina
Optic nerve
Pupil
Iris
Cornea
Vitreous homour
75. (B) (i) Given, P = –1.2 D
f=
1
P
(f = Focal Length, P = Power of lens)
1
−1.2D
f=
f= −
10
12
f = –0.83 m
(ii) The lens is diverging because the
power is negative.
Pupil: The pupil is the small circular opening
of iris. The pupil regulates and controls the
amount of light entering the eye.
Retina: The inner back surface of the eye
ball is called retina. It is a semi-transparent
membrane which is light sensitive and is
equivalent to the screen of a camera. The light
2F1
F1
O
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2F2
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ANSWER SHEET
SELF PRACTICE
7. (d) increases heavily
2. (a) 5A
Explanation: A short circuit is a low resistance
connection between two conductors. With low
resistance in the connection a high current
is present which delivers a large amount
of energy in a short time. Hence during a
short circuit, the electric current in the circuit
increases heavily.
Explanation: Material is same so resistivity will
be same. As resistivity depends on the nature
of material.
Given:
Case I
Case II
length = l
length l’ = 2.5 l
Area = A
Area A’ = ?
Resistance = R
Resistance R’ = 0.5 R
ρl
R=
A
ρl '
R’ =
A'
A=
ρl
R
...(i)
A’ =
A’ =
ρl '
R'
Explanation: We know that:
Combined Resistance
‘R‘ = R1 + R2
= 10 + 40 Ohm
= 50 Ohm
...(ii)
ρ × 2.5 l
0.5 R
Putting values of l’ and R’ in (ii) :
A’ =
10. (c) 50 Ohm
Hence the value of X is 50 Ohm.
11. (a) volt-ampere
Explanation: Power = Voltage × Current.
17. (a) 2 W
Explanation: When a number of resistances
are connected in series the resultant resistance
is equal to the sum of the individual resistances
and is greater than any individual resistance.
Four resistors are given:
ρl 2.5
×
R 0.5
A’ = A ×
2.5
= 5A
0.5
R=
Hence, conductor of length 2.5 l and resistance
0.5 R of the same material has 5A as cross
sectional area.
\
68
1 + 1 + 1 + 1 + 1
1
R = R1 R2 R3 R 4 R5
⇒
1 = 1 + 1 + 1 + 1 + 1
R
15 15 15 15 15
⇒
5
1
R = 1 5 = 25
⇒
1
R = 25 Ω
R = R1 + R2 + R3 + R4
1 1 1 1
+ + +
2 2 2 2
1+1+1+1
=
2
=
1
5. (b) 25
Ω
Explanation: When resistors are connected in
parallel combination the total resistance will
always be lesser than the least resistance in
the circuit. Thus equivalent resistance obtained
1
by connecting five resistors of resistance 5 Ω
each, parallel to each other.
⇒
1
W
2
4
=
=2W
2
Hence, the maximum resistance which can be
made using four resistors each of resistance
1
W is 2 W.
2
Caution

Students usually waste their time in finding how
to connect resistors in order to find maximum and
minimum resistance. They should know that all
resistors should be connected in parallel and in
series to obtain minimum and maximum resistances
respectively.
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19. (a) Resistance: Resistance of C will be the least;
Nature: C is the best conductor of electricity
Explanation: The relation between resistance
l
R and resistivity is given by R = ρ , where l is
A
the length of the conductor and A is the area
of cross section. It is given that the four wires
have equal lengths and area of cross section.
Therefore, resistance will be least for C whose
resistivity is the least and therefore it is the best
conductor. Resistance will be the greatest for D
whose resistivity is the greatest among A, B, C
and D, and therefore, D is an insulator.
27. (d) nature of the material
Explanation: The resistivity of a material is
constant for a particular temperature at a
constant temperature. Resistivity of material
does not depend on length, thickness and
shape of the material. It only depends on the
temperature.
Resistivity: It is the amount of resistance offered
by a conductor of unit length and unit area of
cross-section. SI unit of resistivity is ohm m
(Ω m).
29. (d) (A) is false but (R) is true.
Explanation: Fuse used in electric circuits uses
Joule’s law of heating. Fuse protects circuits
and appliances by stopping the flow of any
unduly high electric current. The fuse is placed
in series with the device.
31. (b) Both (A) and (R) are true but (R) is not the
correct explanation of (A).
Explanation: In a purely resistive circuit,
the source energy is expended to overcome
resistance to the flow of electrons due to which
the energy is dissipated in the form of heat.
This is the heating effect of electric current
and amount of heat H produced in time t for a
steady current I is given by, H = VIt
Whereas, the statement that the potential
difference across the ends of a conductor is
directly proportional to the current flowing
through it is the Ohm’s law, which does not
explain the heat dissipated.
34. (a) Both (A) and (R) are true and (R) is the
correct explanation of (A).
Explanation: When two charged bodies at
different electric potentials are connected by a
metal wire, then electric charges will flow from
the body at a higher potential to the one at
lower potential. This will continue to happen
till they both acquire the same potential.
36. (A) Heater A has higher resistance
Heater A,
P = 100W
V = 220V
P = VI
∴
I=
100
P
=
= 0.45A
220
V
∴
R=
220
V
=
= 489Ω
0.45
I
Heater B,
∴
P = 150W
V = 220V
P = VI
P
I=
V
150
I=
= 0.68A
220
R=
V
I
R=
220
= 323Ω
0.68
37. (A) A battery helps to maintain a potential
difference across a conductor.
The symbol of battery is:
+
–
(B) The resistance of a conductor is
proportional to its length and inversely
proportional to its cross-sectional area.
l
R =r
A
If the fuse wire is thick, the cross sectional
area would increase and the resistance
across the wire would decrease. So even
a high current flows through the fuse,
it would not melt and it would act as a
normal electric contact in the circuit or
wiring. So the fuse wires are made thin.
(C) For diagram (I)
Resistance of A and B in series
R1 = RA + RB
= 2 + 3 = 5W
Resistance of C and D in series
R2 = RC + RD
= 4 + 5 = 9W
Resistance of R1 and R2 in parallel
1
1
1
=
+
RP
R1 R 2
1 1 14
= + =
5 9 45
RP =
45
= 3.21W
14
For diagram (II)
Resistance of A, B and C in series
R1 = RA + RB + RC
= 6 + 7 + 8 = 21W
Electricity
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Resistance of R1 in parallel with D,
1
1
1
=
+
RP
R1 R D
1
1 10
=
+ =
21
9 63
RP =
63
= 6.3W
10
38. (B) (c) Both (I) and (III)
Explanation: In a series combination of
resistors the current is the same in every
part of the circuit or the same current
through each resistor.
(D) (b) V = V1 + V2 + V3
Explanation: The total potential difference
across a combination of resistors in series
is equal to the sum of potential difference
across the individual resistors. That is,
V = V1 + V2 + V3
(B) (c) regulating the resistance through the
regulator.
Explanation: The fan regulator is a variable resistor in which current is varied by
varying the resistance in accordance with
Ohm’s law,
V = IR.
In a variable resistor, resistance is varied
by varying the length of the conductor as
(E) (c) The total current is halved
Explanation: Resistance is the obstruction
to the flow of electrons through a
conductor. When resistance is doubled,
the potential difference remains the
same. According to Ohm’s law, V = IR, and
therefore, I is halved.
39. (A) (d) Both (II) and (III)
Explanation: As can be seen from the graph,
the electric potential is very less when
r > 2.0 m. However, as the point charge Q
is brought closer to the given charge q, the
electric potential increases. Moreover, as
both charges are positive charges, more
and more energy is required to bring the
two charges close to each other due to the
strong force of repulsion between the two
charges.
(C) (b) 60 J
Explanation:
The
relation
between
potential difference between two points,
W
work done and charge is V =
.
Q
Therefore, Work done = VQ = 12 X 5 J = 60 J
(E) (a) 15 V
Explanation:
The
relation
between
potential difference and work done is given
by
W
V=
Q
Therefore,
120
V=
= 15V
8
70
R ∝l .
When the circular knob is rotated, it changes the resistance on the output terminals
and the current flowing through the fan is
thus controlled.
(D) (b) the length of the conductor is increased
Explanation: The resistance of a conductor
is directly proportional to its length and
inversely proportional to the area of cross
section. That is R ∝ l;
1
R ∝ l; R ∝ and by Ohm’s
A
law,
I=
V
R
⇒
I∝
1
R
This means that current will decrease
when resistance increases which is possible
when length of the conductor increases.
(E) (b) Both (I) and (III)
Explanation: The graph between current
(I) and potential difference (V) is a straight
line passing through the origin. The slope of
the above graph gives the resistance of the
conductor as V = IR
V
⇒
R=
by Ohm’s law.
I
41. (A) When the resistance of a resistor becomes
half of its initial value, the current becomes
double as per Ohm’s law.
Therefore, by Joule’s law of heating,
2  R
R
H = 2I   t = 4I2 t = 2I2Rt .
2
 2
( )
40. (A) (d) Position of Regulator Knob: 5; Effect on
Fan Speed: Fastest fan speed;
Explanation: Minimum resistance in
circuit
Explanation: When the regulator knob is
at 1, the speed of fan is slowest as the resistance in the fan circuit is the maximum.
When the knob is at 5, speed is fastest as
the resistance is least. Resistance is directly
proportional to the length of the conductor. By varying the length of the conductor,
resistance is varied and by Ohm’s law, current is also varied. At position 3, fan speed
is medium as resistance is almost half of its
maximum value.
This shows that the heating effect in the
resistor will become two times.
Science Class X
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(B) (i) The rating of ideal fuse can be
calculated by first finding the current
flowing through the electric iron. We
know that the electric power P is given
by P = VI, which means that
P 1000
I= =
= 4.54 A
V
220
(ii) To calculate the heat generated,
we will use the formula
H = Pt = 1000 × 30J
= 3.0 × 104J or 30KJ.
42. (A) The power rating of the electrical dryer is
Voltage, V
In this case, a 5A fuse must be used.
Further, I = 0 when V = 0, as the graph passes
through the origin.
Resistance of the resistor can be determined
by finding the slope of the VI graph as we know
from Ohm’s Law that V = IR, where R is the
resistance of the resistor.
V
\
R=
I
V
1800 W.
Therefore, the electrical energy consumed
in one day
E = Pt = 3600 Wh
= 3.6 kWh = 3.6 units
As June has 30 days, the electrical units
consumed in the entire month = 3.6 × 30
= 108 kWh or units
= 108 × 3.6 X 106 J
= 3.88 × 108 J.
The cost of electricity for the month of
June = ` 5.50 X 108 = ` 594.00
43. (B) (ii) Y is the best insulator as it has highest
value of resistivity. A bad conductor has
high resistivity.
46. The electric potential difference (V) between two
points in an electric circuit carrying some current
as the work done (W) to move a unit charge (Q)
from infinity to point A.
W
V=
Q
W = QV
Related Theory
Current, I
R = Slope of VI graph
R=
49. The filament is made of tungsten metal. The
metal offers resistances, work is done against
the current releasing heat energy. This is based
an Joule’s law of heating. Due to the resistance
offered by the filament. The filament heats
up and starts glowing, converting electrical
energy to light energy.
52. The graph obtained after plotting the values
of V and I is a straight line passing through
the origin which shows that the potential
difference (V) and the current flowing (I) vary
linearly with each other thereby verifying
Ohm’s law.
∆V
∆I
53. As pointer of both the ammeter and voltmeter
do not coincide with the zero marks on the
scales when circuit is open, it indicates zero
error in both the instruments. This zero error
should be subtracted from the readings taken
when circuit is closed.
To get correct readings using these instruments,
first we should find out the least counts of both
the instruments, i.e., the minimum value which
can be accurately measured.
Zero error = Initial reading (in open circuit) =
Pointer reading (in open circuit) × Least count
To get the actual reading, subtract zero error
or the initial reading of the open circuit to the
reading when you perform the experiment.
Actual reading = Final reading - Initial reading
(zero error)
55. (A) Given: Two resistors each of 8 W resistance
are connected in parallel.

SI unit of the above quantities are:
W = Joules (J)
V = Volts (V)
Q = Coulomb (C)
I
1
1 1
1
= + =1+
RP
8 8
8
=
2
= 4W
8
A resistor of 4 W resistance is connected in
series.
Rs = 4W
R = 4W + RP = 4W + 4W = 8W
Given, voltage in the circuit (V) = 8V.
So, current flowing through 4W,
I=
V
R
I=
8V
= 1A
8Ω
Electricity
71
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(B) Potential difference (V’) across 4W:
V’ = R I
V’ = 4 × 1
(B) We know that,
(i) Resistivity of wire is independent of its
length or/and area of cross-section.
Therefore, increasing the length of wire
6
by
times results in no change in
5
= 4V
57. An ideal ammeter is one which has zero
resistance. But that is not possible. Therefore,
the resistance of an ammeter should be as
close to zero as possible. If it is non-zero and
substantial, it will affect the current flowing
through the circuit.
This is because an ammeter is connected in
series in the circuit for the measurement of
electric current.
59. The expression P = I2R is used for calculating
electric power when only current I and
resistance R are known, whereas P = V2/R is
used for calculating power when voltage V and
resistance R are known.
61. The two conducting wires are of same material,
same lengths and diameters. Therefore, their
resistances will also be equal.
Let the resistance in each wire be R.
So in series,
Rs = R + R = 2R
1 1
2
1
In parallel,
= + =
R R
R
Rp
Or,
Rp =
R
2
We know that heat produced is given by,
H=
V2
×t
R
I
= 25
A
R =r
resistivity.
(ii) Where length of wire increases or
decreases, the volume of wire remains
the same but the area of wire changes
as well
For old wire, V = Al ...(i)
6
For new wire, l’ = l
5
V = A’ × l’
V = A’ ×
V
A
1
V = A' × 6l 5
\
A’ =
5
A
6
Pl
A
Now R’ = ρ
R’ = ρ
= ρ
6l 5
5A 6
36l
25 A
l 36
×
A 25
= ρ
V2
Hp = R × t
p
= R ×
Hp
 V2   V2 
× t ÷ 
× t
= 
 R s   R p 
Hs
=
Hp
Rp
Rs
=
1
R /2
=
4
2R
Hs : Hp = 1 : 4
63. (A)A distinguishing feature between the
resistance and resistivity of a conductor:
Resistance of a conductor depends on the
length and area of cross-section of the
conductor. On the other hand, resistivity
depends upon the properties of the metal
used in conductor. Unit of resistance is ohm
and unit of resistivity is ohm-meter.
6
l ...(iii)
5
Dividing both
V2
Hs = R × t
s
Hs
72
36
25
Putting value of R = 25
We get,
R’ =
36
× 25 = 36 Ohm.
25
64. (A) Mathematical expression for Joule’s law of
Heating:
H = I2Rt
Joule’s law of heating implies that heat
produced in a resistor is:
(i) directly proportional to the square of
Current.
(ii) directly proportional to the resistance
for a given current.
(iii) directly proportional to the time
for which current flows through the
resistor.
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Explanation :
(B) Given, Q = 96000 C
t = 2h
V = 40 V
Q
Current I =
t
I=
96000
2 × 3600
( 1 hour = 60 × 60)
40
I=
A
3
Related Theory
By applying Ohm’s law, we can calculate
resistance
V
R=
I
R=
P = 2.0 kWh
T = 3 hours
Cost of 1 kWh (1 unit) of electrical energy
from board = ` 3.5
E =P×T
= 2.0 × 3 = 6 kWh
Cost of buying electricity from main
electricity board = 6 × 3.50 = ` 21.00
40 × 3
=3W
40

The rate of which electric energy is consumed or
dissipated in an electric circuit is power. Commercial
unit of electric energy is kWh which is 1 unit of
electrical energy.
68. (A) Resistance of a conductor depends on the
following factors:
(1) length of the conductor
(2) area of cross section of the conductor
(3) nature of material of the conductor
(4) temperature of the conductor
2
Heat Produced H = I Rt
2
 40 
H= 
 × 3 × 2 × 60 × 60
 3 
40 40
=
×
× 3 × 2 × 60 × 60
3
3
= 3,840,000 J
or3840 kJ
The heat produced is 3840 kJ
Alternate method:
H = Power × Time
Q
= V×I×t=V×  ×t
t 
= V × Q = 40 × 96000 J
= 3840000J
= 3840 kJ
Related Theory

The heating effect of electric current is utilised in:
(I) Working of electrical heating devices such as
electric kettle, electric iron, toaster, oven etc.
(II) electric bulbs for producing light.
(III) electric fuse for protecting household wiring and
electrical appliances.
65. (A) Least count of the voltmeter =
1.5
V = 0.15
10
(B) Reading shown by the voltmeter = 1.5 + 2
× 0.15 = 1.8 V
(C) If R = 20 W, V = 1.8 V,
1.8
V
=
20
R
A = 0.09 A.
current I =
67. (A) E = P × T So, E = 3 × 2 = 6 kWh
Cost of buying electricity from the main
electricity board = 6 x 3.50 = ` 21.0
[CBSE Marking Scheme Term-2 SQP 2022]
(B) If we take two similar wires of same
length and same diameter, one of copper
metal and other of nichrome alloy, we will
find that the resistance of nichrome wire
is about 60 times more than that of the
copper wire. This shows that the resistance
depends on the nature of material of the
conductor.
69. The overall current needed = 9A. The voltage
is 12V
Hence by Ohm’s law
V = IR,
The resistance for the entire circuit
R=
12 4
= W.
9 3
R1 and R2 are in parallel.
Hence,
R=
(R1R 2 )
(R1 + R 2 )
=
4R 2
(4 + R 2 )
4
4R
=
3
4 + R2
R2 = 2Ω
Explanation: Given R1 = 4W, R2 = ?
I = 9A
V = 12 V
According to Ohm‘s law,
R = VI
R =
12
4
=
9
3
R1 and R2 are connected in parallel,
1
1 1
+
=
R1 R 2
RP
Electricity
73
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(ii) Let I2 be the current flowing through
3
1 1
= +
4
4 R2
1
2
3 1
= − =
4 4
R2
4
or
or
R2 = 2W
Out of two resistors 2W and 3W, the student
should choose 2W resistor.
70. (B) Let Resistance of Space 5 and 4 be R ohms
respectively
Resistance of Space 1 = 2 R ohms
Resistance of Space 2 = 30 ohms
Resistance of Space 3 = 20 ohms
Current = 22 A
V = 220 V
V
Total Resistance =
I
1
1
1
1
1
1
+
+
+
+
=
R1 R 2 R 3 R 4 R 6
R eq
1
1
1
1 1 1
+
+
+ + =
R eq
2R 30 20 R R
1
30 + 2R + 3R + 60 + 30
=
R
60 R
eq
1
150 + 5R
=
R eq
60 R
Req =
60R
60R
10R
R
60 R
220
=
150 + 5R 22
= 10(150 + 5R)
=1500 + 50R
= 1500
= 150W.
72. (A)
Resistors in the circuit are connected in
parallel combination.
R1
R2
+
+
V
–
–
–
A
+
(B) (i) Let I1 be the current flowing through
R1 = 10 Ohm
Given
Now
74
R2 = 15 Ohm
V = 5V
I1R1 = V
V
5
I1 = R =
= 0.5 A
10
1
Given
Now
V = 5V
I2R2 = V
V
5
I2 = R =
= 0.33 A
15
2
(C) This can also be done like this source
resistors are connected in parallel
1 1
1
= R +R
R
1
2
=
1
1 3+2 5
1
+
=
=
or Ω
10 15 30 30 6
R=6Ω
As per Ohm’s law V = IR
I=
V
5
=
= 0.83 A
R
6
73. Let Req be the net resistance of the combination of three bulbs in parallel, then,
4.5
V
Req = I = 3 = 1.5 Ω
All bulbs are identical, so they must have the
same resistance.
Let R be the resistance of each bulb and all
bulbs are connected in parallel, hence,
1
1 +1 +1
R eq = R R R
1
3
R eq = R
R = 3 × Req
= 3 × 1.5 = 4.5 Ω
4.5
V
I = R = 4.5 = 1 A
Current through each bulb:
(A) If B1 gets fused, the current in B2 and B3 will
remain unaffected, as voltage across bulbs
B2 and B3 remains the same. Hence, glow
of the other two bulbs will not be affected.
(B) When bulb B2 gets fused, the current
through B2 will be zero and the current in
B1 and B3 will remain 1A.
Now, net current,
A = A1 + A 2 + A 3
= 1 + 0 + 1 = 2A
Thus, current in ammeter, A1 = 1 ampere
Current in ammeter, A2 = 0
Current in ammeter, A3 = 1 ampere
Total current in ammeter, A = 2 ampere.
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(C) Power dissipated when all three bulbs
glow together,
P=V×I
P = 4.5 × 3 = 13.5 W
(4.5) 2
V2
= 1.5 = 13.5 W
R eq
P=
P = 13.5 W
74. (A) Resistance of first lamp
R1 =
=
V2
P
220 × 220
= 484 W
100
R2 =
=
V
P
220 × 220
= 4840 W
10
R= ρ
l
A
R’ = ρ
( l / 2)
2A
 l
 ρ A 
1
× 6 as R = 6 W =
4
 l
 ρ A 
R’ = 1.5 W
New resistance of the wire will be 1.5 W
(B) To get a total resistance of 3 W from
three resistors A, B and C, two resistors of
resistances 2 W each should be connected
in parallel. Their equivalent resistance is :
Current drawn from the line i.e.
V
I= R
P
220
= 0.5 A
440
Hence, the total current through the circuit
= 0.5 A
1
1
1
+
=
R1 R 2
R
1
1 1
2
= + =
=1W
R
2 2
2
R =1W
(B) Resistance of Resistor X(R1) = 2 W
This combination of equivalent resistance
1W should be connected in series with
the resistor of resistance 2W. So that the
equivalent resistance becomes 1W + 2W = 3W
Resistance of resistor Y(R2) = 3 W
R
5
=
V
5
= 1 A
Voltage across the 3 W resistor in series
V = IR
V = 1 × 3 = 3 V
Voltage across the 3 W resistor in the
series combination of resistors is 3 V.
4840
= 440 W
11
=
I=
=
10 + 1
11
=
=
4840
4840
1
= 4
1
1
1
=
+
Rp
484 4840
RP =
Total current in the circuit
1
length becomes
and cross-section area
2
becomes 2A.
1
1
1
= R +R
RP
1
2
Or
V = IR
When the wire of length l and crosssectional area A is doubled on itself, its
2
Since the two lamps are connected in
parallel, the equivalent resistance is given
by:
=2+3=5W
77. (A) Given, R = 6 W
Resistance of second lamp
Alternate method: Power dissipated,
(ii) Resistance in series RS = R1 + R2
V = 5V
(i)
2
2
3
2Ω
2Ω
3
2Ω
1Ω
2Ω
+
–
+
3Ω
–
Electricity
75
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78. (C) The relationship between brightness and
resistance of lamps connected in parallel is
inversely proportional. Therefore, the lamp
with the least resistance will glow the
brightest.
According to Ohm’s law
V = IR
V
So
R=
I
Lamp A, R =
60
= 20 W
3
Lamp B
R=
60
= 15 W
4
Lamp C R =
60
= 12 W
5
Lamp D R =
60
= 20 W
3
The lamp C has the least resistance and
has highest current flowing through it, so it
will glow the bright.
79. (B) (i)
Given, power = 400W, time = 10 hr.
Electricity consumed by refrigerator
= Power × Time
= 400 × 10
= 4000 Wh
= 4 kWh
(ii) Given, power = 80W, time = 12 hr.
Electricity consumed by two fans
= Power × Time
= 80 × 2 × 12
= 1920 Wh
= 1.92 kWh
(iii)Given, power = 18W, time = 6 hr.
Electricity consumed by six electric
tubes
= Power × Time
= 6 × 18 × 6
= 648 Wh
= 0.648 kWh
80. (A) (i) The two resistors that are connected in
series are R1 and R2
(ii) The one combination of resistors in
parallel are R4 and R5
Note : R1 and R2 are in parallel with R3.
(iii) The equivalent resistance RT of the
combination R1, R2 and R3.
R1 and R2 are connected is series
RS = R1 + R2
= 20 + 40
= 60 W
76
RS and R3 is parallel,
RT =
1
1
+
R S R3
=
1
1
+
60 30
1+2 3
=
=
60 60
(C)
RT =
60
= 20 W
3
R4 = 40 W R5 = 120 Ω
1
1
1
= R +R
RP
4
5
=
1
1
+
40 120
120
= 30 W
4
RP =
3+1
4
=
120 120
I = 0.25 A
By applying Ohm’s law,
V = IR
= 0.25 × 30
= 7.5 V
The
potential
difference
across
the
combination of R4 and R5 is 7.5 V.
81. (A) Meaning of the statement “The resistance
of a conductor is one ohm”: If the potential
difference across the two ends of a
conductor is I, V and the current through is
1A, then the resistance R of the conductor
is 1 W.
1Volt
1 Ohm =
1Ampere
(B) Electric power: The rate at which electric
energy is dissipated or consumed in an
electric circuit.
The rate of doing electric work or the
amount of wore done in one second is
called the electric power.
W(work done)
P=
...(ii)
t(time)
We know,
W = I2Rt...(i)
Substituting the (i) in (ii) equation
\
P=
I2Rt
t
Or
P = I2R
From ohm’s law V = IR
V
I=
R
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P=
V V
× ×R
R R
83.
4
2
V
=
R
Hence, the expression relating electric
power, potential difference and resistance is:
P=
V2
R
(C) Given,
I =5A
V = 220 V
Resistance required in the circuit
V
R=
I
I
+
n=
V=6V
Using the above formula, total resistance
of the circuit R = 20 + 4 = 24W.
(B) Current through the circuit is found by
applying Ohm’s law.
132
n
132
Or n = 3
44
each are connected in parallel.
1
1
1
+
=
RP
R1 R 2
1 1 2
+ =
6 6 6
6
=3W
2
\
RP =
RP and resistor having 2W resistance are
connected in series.
RS = RP + R
=3+2=5W
(ii)
Req = 5 W
V = 6V
Applying Ohm’s law,
V = IR
I =
V
R
6
= 1.2 Å
5
V
R
(C) The potential difference across the series
combination of resistances is equal to the
sum of the potential difference across
individual resistance, where V1 is the
potential difference across R1 and V2 across
R2.
82. (C) (i) Two resistors having 6 W resistance
=
I=
6
1
=
A=
A
24
4
= 0.25 A
Hence, 3 resistors each of resistance 132
W are connected in parallel to get the
required to carry 5A on a 220 V line.
( (
(A) The two resistances are connected in series
and equivalent resistance of resistors
connected in series is given by R = R1 + R2.
220
= 44 W
5
Resistance of each resistor = 132 W
If n resistors, each of resistance R
(132 W) are connected in parallel to get the
required resistance R, then
r
R=
n
44 =
– A +
In the given question, let R1 = resistance of
electric lamp = 20 Ohm
Let R2 = resistance of conductor = 4 Ohm.
=
–
6V
(i)
V1 = IR1
R1 = 0.25 × 20 = 5V
(ii) Similarly, V2 = IR2 = 0.25 x 4 = 1 V
(D) Power, P of the lamp = V1I = 5 × 0.25 =
1.25W, where V1 is the potential difference
across the lamp and I is the current flowing
through it.
Related Theory


Power can also be found out using the formula P =
VI = (IR) I = I2R, where I is the current flowing through
the lamp.
V2
V 
Or, using P = V   =
, where V is the potential
R
 R
difference across the lamp.
84. The three resistors are connected in series. The
effective resistance of 3 resistors R1, R2 and R3
connected in series is given by :
Reqv = R1 + R2 + R3
The effective resistance = 5 + 8 + 12 = 25 W
Electricity
77
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Total current flowing I in the circuit is given by
V
6
Ohm’s law, I =
=
A
R
25 .
As the current flowing through the resistors
connected in series is same, the current flowing
through the Ammeter = I = 0.24 A.
Reading of ammeter = 0.24 A.
The potential difference V is equal to the
sum of the potential difference V1, V2 and V3
across the individual resistors R1, R2 and R3
respectively. That is,
V = V1 + V2 + V3
As the voltmeter is connected across the
12 Ohm resistor or R3, we will calculate the
potential difference V3 across R3.
Using Ohm’s law, V3 = IR3 = 0.24 × 12V = 2.88V.
Reading of Voltmeter = 2.88 V
87.Given circuit diagram:
125 V
+
–
20 W
30 W
50 W
(A) Equivalent resistance : Resistances in series.
R = R1 + R2 + R3
= 20 W + 30 W + 50 W = 100 W
(B) Total current from the power supply : Total
current is determined by the voltage of the
power supply and the equivalent resistance
of the circuit.
125
V
I=
=
= 1.25 A
R
100
(C) Current through each resistor will be 1.25 A
Current is constant through resistors connected
in series.
(D) Voltage drop across each resistor,
V1 = IR1 = 1.25 A × 20 W = 25 V
V2 = IR2 = 1.25 A × 30 W = 37.5 V
V3 = IR3 = 1.25 A × 50 W = 62.5 V
In a series circuit they should equal the
voltage increase of the power supply.
Vtotal = V1 + V2 + V3
125 V = 25 V + 37.5 V + 62.5 V
125 V = 125 V
(E) The power dissipated in each resistor,
P1 = V1 × I1
= 25 V × 1.25 A = 31.250 W
P2 = V2 × I2
= 37.5V × 1.25A = 46.875 W
P3 = V3 × I3
= 62.5V × 1.25A = 78.125 W
In a series circuit, the element with the greatest
resistance consumes the most power.
78
88. (A) R
esistors connected in series: In a series
combination of resistors the current is the
same in every part of the circuit or the
same current flows through each resistor,
i.e., there is only one path for the flow of
current. When several resistors are joined
in series, the resultant resistance of the
combination RS equals the sum of their
individual resistances, R1, R2, R3.
RS = R1 + R2 + R3
arallel Combination of resistors: In a
P
parallel circuit each resistor is placed in
its own separate branch. A parallel circuit
provides multiple paths for the current to
flow.
The reciprocal of the equivalent resistance
of a group of resistors joined in parallel is
equal to the sum of the reciprocals of the
individual resistors.
1
1 1
1
=
+
+
Rp
R1 R 2 R 3
(B) To find total resistance in the circuit.
Let RA be the value of total resistance in
first combination i.e. 5W and 20W.
Then
1
1 1
=
+
RA
R1 R 2
1
1 1
4 +1
5
R A = 5 + 20 = 20 = 20
or RA = 4 W
Let RB be the value of resistance in this
combination.
Then,
1
1
1
1
=
+
+
RB
R3 R4 R5
1
1
1
1
=
+
+
RB
15 20 10
4+3+6
=
60
13
=
60
or RB =
60
W
13
Total resistance across the circuit will be:
RA + RB = 4 W +
60
W = 8.6 W
13
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ANSWER SHEET
SELF PRACTICE
5. (c) The
pattern of the magnetic field
associated with the solenoid is different
from the pattern of the magnetic field
around a bar magnet.
Explanation: Only Statement (c) is incorrect
because the pattern of magnetic fields due to
solenoid and bar magnet is almost same.
N
Explanation: The earth wire is usually
connected to a metal plate deep in the
earth near the house and is used as a safety
measure for those appliances that have a
metallic body. The metallic body is connected
to the earth wire, which provides a lowresistance conducting path for the current.
Thus, it ensures that any leakage of current
to the metallic body of the appliance keeps
its potential to that of the earth, and the user
may not get a severe electric shock.
S
Solenoid Bar Magnet

Explanation: The direction of magnetic field
around a current carrying conductor is given
by Right hand thumb rule. In this case, the
direction of current will be opposite to the
direction in which negative charge is moving.
Therefore, by applying right hand thumb rule,
we find that the magnetic field is in clockwise
direction.
19. (c) prevent electric shock
S
N
17. (a) clockwise direction
Related Theory
A solenoid is a coil of many circular turns of
insulated copper wire wrapped closely in the form
of a cylinder. It is used in inductors, electromagnets,
antennas.
10. (c) it will increase in strength only
Explanation: When a soft iron bar is
introduced within a current-carrying solenoid,
the magnetic field inside the solenoid increases
because the iron bar magnetises since it
quickly gains magnetic characteristics when
current is flowing around it and loses magnetic
qualities when the current is switched off.
13. (b) 2
Explanation: By applying Fleming’s left-hand
rule, we can determine that the force acting on
the wire is parallel to both the magnetic field
and the wire’s current. As a result, there are
only two options for the force’s direction: either
upward or downward.
14. (d) Both (I) and (IV)
Explanation: The magnetic field is strongest
where the magnetic field lines are the closest,
which is near the poles at A. Magnetic field
lines are weakest where magnetic field lines
are farthest, which is at distance away from
the magnet at D.
21. (a) Both (A) and (R) are true and (R) is the
correct explanation of (A).
Explanation: The force or a charged particle
moving in a uniform magnetic field always acts
in the direction perpendicular to the direction
of motion of the charge. As work done by a
magnetic field on the charge is zero, W = FS cosθ.
So, the energy of the charged particle does not
change.
23. (c) (A) is true but (R) is false
Explanation: The direction of force acting
on a current carrying conductor placed in a
magnetic field depends upon the direction
of current and magnetic field. Whereas, the
magnitude of force is highest when direction
of current is perpendicular (not parallel) to
direction of magnetic field.
25. (C) Vikram is right when he says that the bulb
acts like a normal fuse, when it fuses due to
break in its filament, a break in the filament
means a break in the path of the current
between the terminals, similarly eletric
fuse interrupts the flow of current in an
electric circuit. A bulb has a filament made
of tungsten having high melting point i.e.
3380º. The bulb is usually filled with the
chemically inactive nitrogen and argon
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gases to prolong the life of the filament.
When an electric current passes through
a thin metal filoment, the filament gets
heated and starts glowing and emits light.
26. (C) (c) Both (a) and (b)
Explanation: A fuse in a circuit acts as
a safety divice. As soon as the current
flowing is a circuit exceeds the rated value,
the fuse wire melts due to Joule’s law of
heating and the electric circuit is broken.
(C) In India frequency of AC supply
f = 50 Hz
1
1
Time period T = =
s
f
50
As current changes its sign after nth interval
t=
T
2
t=
1
1
1
×
=
s
2 50 100
(D) (a) in series
Explanation: The fuse wire is placed
in series with the electric circuit or the
appliance. If a current larger than the
specified value flows through the circuit,
the temperature of the fuse wire increases
beyond its melting pant. As a result, the
fuse wire melts and breaks the circuit.
(E) (c) Joule’s law of heating
Explanation: The electric fuse works on
the principle of Joule’s heating that heat
produced due to excess current flowing
through it melt the fuse wire which cut off
the circuit and the appliance is prevented
from damage.
27. (B) Overloading occurs when a large number
of electrical appliances of high power
rating are connected in a single socket,
sometimes, overloading may also occur
due to an accidental hike in supply voltage.
Due to overloading, current in a circuit
increases even behind the current rating
of the circuit. As a result, a large amount
of heat is produced due to Joule’s heating
effect which may damage or burn out the
applicances.
(C) Following characteristic features should
be considered while making a residential
building:
(i) All the electrical points must be joined
in parallel only.
(ii) Two separate circuits should be used,
one of 5 A for bulbs, tubes, fan etc.
and another of 15 A for geysers, air
conditioners, refrigerators etc.
(iii) Separate switch should be used for
each appliance.
(iv) Fuse of appropriate current rating
should be used.
(v) Proper earthing should be done.
28. (A) If too many appliances are connected to
the same socket, it may lead to overloading
due to which the live and neutral wires
may come in contact and cause electric
fires.
80
or
t = 0.01 sec.
In USA
t=
T
1
×
s
2 120
T=
2
1
=
s
120 60
Frequency of AC,
1
f=
T
1
= 60 Hz
1
60
Frequency of AC supply in US is 60 Hz and
in India, it is 50 Hz.
=
29. (B) (c) Fleming’s left hand rule
Explanation: The direction of force
experienced by the current carrying loop in
the above meter is given by Fleming’s left
hand rule.
(C) (c) 90°
Explanation: The magnitude of the force is
the highest when the direction of current
is at right angles to the direction of the
magnetic field.
30. (C)
31. (C)The crowding of the iron fillings at the ends
of the magnet indicate position of two
magnetic poles N and S of bar magnet.
It also indicates that the strength of
magnet is maximum at poles (where it is
crowded.
32. (A) The direction of force acting on a current
carrying conductor kept in a magnetic
field depends on the direction of current
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flow in the conductor and the direction of
magnetic field.
(B) The angle between a current carrying
conductor and magnetic field is 90­º for
the force acting on a conductor due to
magnetic field which will be maximum.
Explanation: The direction of magnetic
field lines are correctly shown in figure (b)
above as the magnetic field lines originate
from the North pole of a bar magnet
and terminate at the South pole of the
magnet. And no two magnetic field lines
intersect each other. We observe that the
magnetic field lines are intersecting each
other in figures (a) and (c). And in figure
(d), the direction of field lines is not shown
correctly as explained above.
33. (A)Kumar must bring a magnetic compass
needle near the wire and see near which
wire needle shows deflection. If the
needle gets deflected, the wire is current
carrying wire. If needle does not show any
deflection when brought near the wire, the
wire carries no current. Only the current
carrying wire shows deflection in the
needle of magnetic compass.
(C)In domestic circuits, mostly 5 A and
15 A rating of fuse are used. You must have
noticed that a rating of 15 A fuse is used
for electric iron, geyser, refrigerator etc.
Whereas a fuse of rating 5 A is used for
bulbs, tubelights, fans, CFLs etc. 5 A (current
carrying capacity) fuse means if there is a
flow of more than 5 A current through the
fuse, it will blow off saving the circuit from
over current.
Same is the case with 15 A fuse. By
choosing 5 A and 15 A fuse, we may be
fitting, wiring of appropriate rating for
appliances. If the fuse, with a defined rating,
is replaced by one with a larger rating, then
the fuse wire will not burn even when a
current greater than safe limit is flowing.
As a result the electrical appliances will be
damaged.
34. (A) (c) Both (I) and (IV)
Explanation: The iron filings arrange
themselves in a pattern as shown in the
figure given in the activity as the iron filings
experience a force due to the influence of
the magnet. It is the magnet that exerts
a force as it has a magnetic field around
it and the iron filings experience this force
and arrange themselves in a pattern.
(D)(b)Increases as the needle is moved
towards the poles
Explanation: The deflection in the compass
increases as the needle is moved towards
the pole as the strength of the magnetic
field is maximum near the poles.
(E) (b)
37. A current carrying conductor kept in a
magnetic field experiences maximum force
when direction of current is at right angles to
the direction of the magnetic field according to
Fleming’s left hand rule.
38. Magnetic field lines will be concentric circles
in the plane of the paper and in anti-clockwise
direction which can be found out by applying
Right hand thumb rule.
40.
S
N
–
+
battery
41. If the direction of the current in the straight
wire is changed, the direction in the magnetic
field line will be observed.
53. Solenoid is a closely bound cylindrical coil of
insulated metallic wire. A current carrying
freely suspended solenoid behaves like a
magnet and when suspended freely, it rests in
north-south direction.
A current carrying solenoid behaves like a bar
magnet with fixed polarities at its ends. The
end of the current carrying solenoid at which
the current flows anticlockwise behaves as a
north pole while that end at which the direction
of current is clockwise behaves as a south
pole. The direction of magnetic field is always
perpendicular to the direction of current flow
and the magnitude of the magnetic field
inside a solenoid is directly proportional to the
current flowing through the solenoid.
Thus, when the current through the solenoid
is reversed, the direction of magnetic field is
reversed.
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55. Schematic diagram of common domestic
circuits is shown below:
A fuse is a safety device in a circuit that
prevents damage to the appliances and the
circuit due to overloading. A fuse works on
Joule’s heating effect and is connected in series
with the electric devices.
Every fuse has a specific rating; it means that a
specific fuse will allow only a specific amount
of current through itself. When the current
exceeds the limit, then the fuse will heat up
and melt.
Overloading occurs when the live wire and the
neutral wire come into direct contact. In such
a situation, the current in the circuit abruptly
increases. This is called short-circuiting. The
use of an electric fuse prevents the electric
circuit and the appliance from a possible
damage by stopping the flow of unduly
high electric current. The Joule heating that
takes place in the fuse melts it to break the
electric circuit. Overloading can also occur due
to an accidental hike in the supply voltage.
Sometimes overloading is caused by connecting
too many appliances to a single socket.
The burnt out fuse should be replaced by
another fuse of identical rating because if
a fuse of a lower rating is used, then it will
blowup and the electric appliances will not
work. If a fuse of a higher rating is used, then in
case of short circuit and overloading, the fuse
will not blow up and excess amount of current
may harm the appliances or the circuit may
get damaged.
82
56. (A) The current is in the east-west direction.
Applying the right-hand thumb rule, we
get that the direction of magnetic field at a
point above the wire is from south to north
or anticlockwise direction. The direction of
magnetic field at a point directly below
the wire is north to south or clockwise
direction.
(B) (i) According to Fleming’s left-hand rule,
hold the forefinger, the centre finger
and the thumb of your left hand at
right angles to one another.
Adjust your hand in such a way that
the forefinger points in the direction
of the magnetic field and the centre
finger points in the direction of current,
then the direction in which the thumb
points, gives the direction of the force
acting on the conductor.
(ii) According to Maxwell’s right hand
thumb rule: Imagine that you are
holding the current-carrying wire in
your right hand so that your thumb
points in the direction of current, then
the direction in which your fingers
encircle the wire will give the direction
of the magnetic field lines around the
wire.
57. (A) An electromagnet is a temporary strong
magnet. Its magnetism is only for the
duration of current passing through it. The
polarity and strength of an electromagnet
can be changed.
Uses of electromagnet:
Electromagnets are used:
(1) in electrical appliances like electric bell,
electric fan etc.
(2) in electric motors and generators.
(3) in radios, television, microphone etc.
(4) in separating iron from non-magnetic
material.
(5) in magnetising steel bars.
(Any two)
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ANSWER SHEET
SELF PRACTICE
1. (d) 5000 kJ
Explanation: 90% of the energy captured (10%
law) from the previous trophic level is lost to
the environment, leaving only 10% available to
the next trophic level.
In this food chain, at the fourth trophic level,
only 5KJ energy is available to the snake.
⇒ Energy available to snake = 5 kJ
⇒ Energy available to frog = 50 kJ
⇒ Energy available to grasshopper = 500 kJ
⇒ Energy available to grass = 5000 kJ.
10% of
5000 kJ
1st trophic level
5000 kJ
Grass
10% of
→
Grasshopper 
500 kJ
→
2nd trophic level
500 kJ
10% of
Snake
50 kJ
3rd trophic level 4th trophic level
50 kJ
5 kJ
Frog
→
So, the energy available at the producer level
will be 5000 kJ.
3. (c) chemical energy
Explanation: The 10% of energy available for
transfer to the next level in an ecosystem is in
the form of chemical energy. The producers
capture the energy present in sunlight and
convert it into chemical food energy, which is
passed further onto other trophic levels
7. (d) bacteria and fungi
Explanation: The microorganisms comprising
bacteria and fungi break down the dead
remains and waste materials of living
organisms. These microorganisms are known
as decomposers. Example of decomposers
are: bacteria and fungi. They break down
the complex organic substances into simple
inorganic substances that go into the soil. This
means that they are recycling the waste of the
environment. Recycling increase the fertility of
the soil as well as help in cleansing the area.
9. (a) Decrease in energy at higher trophic levels
Explanation: There is a loss of energy as we go
from one trophic level to the next, which limits
the number of trophic levels in a typical food
chain. As a result, organisms continue to lose
energy as trophic levels go up.
The trophic levels are limited to 4, as beyond
that level, the organisms will struggle to survive.
10. (c) T1
Explanation: At each trophic level, most of the
energy available is utilised for respiration and
excretion and other life processes and only ten
percent of the available energy is passed on to
the next level.
Because only 10% of the available energy can
be passed on to the next trophic level, higher
trophic levels have substantially less energy
content and the number of trophic levels in
a food chain is limited. The lower the trophic
level, higher will be energy content.
Hence, the greatest amount of energy is
expected in trophic level T1.
Hence, T1 > T2 > T3 > T4.
11. (d) they are made of non-biodegradable
materials
Explanation: Disposable plastic plates should
not be used because they are made of nonbiodegradable materials that cannot be broken
down into its basic simpler compounds by
micro-organisms. So, they are an environmental
concern as they become pollutants and create
disposal problems.
14. (c) Both (II) and (III)
Explanation: The flow of energy in an
ecosystem is unidirectional and is from
producer to primary consumer, primary
consumer to secondary consumer and from
secondary consumer to tertiary consumer.
15. (b) Both (II) and (III)
Explanation: All food chains are of varying
length and complexity, as a food chain may
consist of three or four or even five organisms.
Food chains, however, consist of only three or
four steps as the loss of energy at each step is
so great that very little usable energy remains
after four trophic levels. There are generally
a greater number of individuals at the lower
trophic levels of an ecosystem, the greatest
number is of the producers. As each organism
is generally eaten by two or more other kinds
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of organisms which in turn are eaten by
several other organisms, the relationship can
be shown as a series of branching lines called
a food web instead of a straight line food chain.
18. (d) Shark
Explanation: As the chemical pollutants
like DDT are non-biodegradable, their
concentration increases as we go up a
food chain. Therefore, the organisms at
the highest trophic levels will have the
maximum concentration of harmful chemical
pollutants. In this food chain, as Shark is
at the topmost trophic level, it will have
the maximum concentration of chemical
pollutants
20. (c) Composting takes place only in the
presence of oxygen.
Explanation: Composing is an aerobic process
which takes place in the presence of oxygen.
this is also called aerobic biodegradation.
Related Theory

Compost making is a decomposition process whic
can be completed anywhere from two weeks to two
years depending on the materials used, the size of
the pile and how often it is truned.
21. (d) (A) is false but (R) is true.
Explanation: The maximum concentration of
chemicals and pesticides occurs at the highest
trophic level or at the top level of the food
chain. This is known as biological magnification
and happens because the harmful chemicals
enter our food chain and being nonbiodegradable gets accumulated progressively
at each trophic level.
27. (a) Both (A) and (R) are true and (R) is the
correct explanation of (A)
Explanation: The substances that cannot be
broken down by the action of micro-organisms
are called non-biodegradable substances. e.g.,
plastic, galss etc.
30. (C) Trophic levels.
31. (A) The methods of waste disposal are:
Recycling,
preparation
of
compost,
incineration, land fill etc.
(B) Incinerators are used to destroy waste
from hospitals, household waste, chemical
waste etc. Incineration greatly reduces
the volume of the waste. This is because a
large amount of ash is left behind which is
disposed off by landfill.
(C) The reasons for generation of heaps of
garbage are:
(1) Improvement in our lifestyle have
resulted in greater amounts of waste
generation.
(2) Changes in the attitude also have a
role to play with more and more things
we are becoming disposable.
(3) Changes in packaging have resulted
in much of our waste becoming nonbiodegradable.
These waste will impact the environment
and cause soil contamination. The waste
will also impact human health. A heap of
waste may generate greenhouse gases like
methane which can cause global warming.
32. (A) The various levels or steps in a food chain
are called trophic levels. They represent
the successive stages of nourishment like
primary producer, primary consumer,
secondary consumer, tertiary consumer
etc.
(B) An example of food chain:
Plants → Grasshopper → Frog → Snake →
Kite
whereas they are present in ponds or
lakes.
Car
niv
o
Lion
Herbivores
Deer
Plants
Grass
Producers
Pyramid of numbers showing 3 step food
chain. The base of this pyramid shows
producers and the top of this pyramid is
84
In this example, an insect (for example,
grasshopper eats plants). Insect is then
eaten by the frog which is in turn eaten by
the snake. The snake is then eaten by the
kite.
34. (C) (d) decomposers are absent in an aquarium
re
The various steps in a food chain are
formed by carnivores. As we go higher and
higher in each trophic levels, the number of
organisms goes on decreasing.
Explanation: An aquarium is an artificial
ecosystem and the organisms present
in it are maintained in an artificially
created and restricted environment. In
the aquarium the uneaten food as well
as the waste generated by the fishes
mixes with the water and is left untreated
as decomposers are absent. The waste
materials thus, accumulate in the water
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making it toxic. Hence, an aquarium has to
be cleaned after regular intervals.
On the other hand, a pond or a lake
ecosystem is a natural ecosystem and the
types of organisms present in it maintain
balance. The decomposers present in ponds
and lakes breakdown the dead remains
and waste products of organisms and thus
keeps the water clean. Therefore, ponds
and lakes do not require physical cleaning.
(D) (a) They would eat all the smaller fishes
thereby generating lot of waste.
Explanation: Predator fishes are fishes
that eat other fishes. If we add such species
together, eventually there will be only
predator fishes as they would eat all the
smaller fishes. To avoid this we should not
mix a predatory fish with prey fish.
built on a landfill site.
(D) (c) Both (I) and (II)
(E) (b) A: Phytoplankton; B: Zooplankton; C:
Small fish
Explanation: Various trophic levels are
connected through food chains. For example,
in an aquatic food chain, phytoplanktons
are the producers, zooplanktons are the
primary consumers, and small fish is the
secondary consumer and large fish is the
tertiary consumer.
35. (B) (b) All types of wastes are placed in a
Explanation: Municipal authorities all
over the country are encouraging people
to segregate household wastes and put
them in separate bins. In Delhi, wet wastes
from kitchen is put in green coloured
bins whereas dry wastes are put in blue
coloured bins.
(C) (d) Burning
Explanation: Since biodegradable or
organic wastes like vegetable peels, waste
food, leaves, dead flowers, and egg shells
can be recycled, they are converted into
manure by burying them in compost pits
and the process is called composting.
Anaerobic digestion can be used to treat
food and similar wet organic wastes. It
takes place in a closed container, excluding
oxygen. It is clean and relatively odour-free.
It produces a nutrient-rich solid material
called digestate and biogas containing
methane and CO2.
Explanation: Sewage is treated at sewage
treatment plants, which may be located
close to where the sewage is created
or an “on-site” system (in septic tanks,
biofilters or aerobic treatment systems).
Alternatively, sewage can be collected and
transported by a network of pipes and
pump stations to a municipal treatment
plant. Sewage treatment is the process of
removing contaminants from domestic and
municipal wastewater, containing mainly
household sewage plus some industrial
wastewater.
(E) (a) source reduction and reuse
single bin.
Large areas used for waste disposal are
called landfills. Landfill is another method
to manage huge amount, of biodegradable
waste. In a landfill, garbage is buried
in such a way that it does not damage
the environment. Garbage buried inside
landfills stay here for a long time as it
decomposes very slowly. After a landfill
is full, it can be converted into a park. For
example, Indraprastha Park in New Delhi is
Explanation: Industrial waste is the waste
produced by industrial activity which
includes any material that is rendered
useless during a manufacturing process
such as that of factories, mills, and mining
operations. As industrial wastes may
contain hazardous substances, it is best
for the local industries to treat the waste
at the source itself or reuse certain wastes
before releasing the waste.
37. (A) Grasses/Plants act as producers. They
convert light energy into chemical
energy by the process of photosynthesis.
All animals are directly or indirectly
dependent on plants for food and energy.
Hence, organisms of the first trophic level
(grasses/plants) are considered to be of
primary importance. The organisms of this
level would have the lowest concentration
of pesticides in their bodies due to
biomagnification.
38. (B) (d) ultraviolet radiation acting on oxygen
molecule.
Explanation: Ozone at the higher levels of
the atomosphere is a product of ultraviolet
radiations acting on oxygen (O2) molecules.
The higher energy UV radiations split
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apart some molecular oxygen (O2) into
free oxygen (O) atoms. These atoms then
combine with the molecular oxygen to
form ozone as shown:
the energy of sunlight that falls on their
leaves and convert it into food energy. The
energy that is captured by the autotrophs
does not go back to the solar input As it
moves progressively through the various
trophic levels, it is no longer available to
the previous level as the flow of energy in
an ecosystem is unidirectional as shown
below.
UV
O2 → O + O;
O + O2 → O3
(D) (b) Car
with
AC,
refrigerator,
extinguisher, aerosol sprays
fire
Top
carnivores
Explanation: The depletion of ozone layer
is linked to synthetic chemicals such as
chlorofluorocarbons (CFCs) which are used
as refrigerants, aerosol sprays and in fire
extinguishers.
Carnivores
Herbivores
(E) (d) Tides
Explanation: Ozone
layer
depletion
causes increased incidence of ultraviolet
radiations on the surface of the Earth and
this causes skin cancer, cataract and other
eye damage. Tides are caused due to the
gravitational pull of the moon on the earth.
39. (A) (c) biological magnification
Explanation: When pesticides and other
non-biodegradable
chemicals
enter
the food chain, these get accumulated
progressively at each trophic level and this
is known as biological magnification
(D) (a) DDT is non-biodegradable and hence
found in largest concentration in
tertiary consumers.
Explanation: DDT is a non-biodegradable
pesticide and when sprayed, it finds its
way to soils from where it enters the food
chain. The concentration of DDT increases
progressively as we go up the trophic level
due to biological magnification.
(E) (c) pollution
Explanation: Biological magnification is
a result of pollution caused by spraying
pesticides etc., which pollute the water and
soil.
40. (A) (b) unidirectional
Explanation: When green plants are eaten
by primary consumers, a great deal of
energy is lost as heat to the environment,
some amount goes into digestion and in
doing work and the rest goes towards
growth and reproduction. An average of
10% of the food eaten is turned into its own
body and made available for the next level
of consumers.
41. (A) 10% of the net primary productivity of
terres trial ecosystem is eaten and digested
by herbivores. According to 10% law, only
10% of the energy entering a particular
trophic level of organisms is available for
transfer to the next higher trophic level.
(B) The flow of energy in a food chain is
unidirectional. Green plants capture 1% of
sunlight and convert it into food energy to
herbivores. The remaining 90% of energy
is used in various life processes (digestion,
growth, reproduction etc.) by plants. As per
10 percent law, herbivores transfer only
10% of energy to carnivores as shown in
the given pyramid but it does not go back
from herbivores to producers.
Primary
Consumers
rgy
Producers
100J
ne
Herbivores
10J
ne
ei
86
Explanation: The green plants in a
terrestrial ecosystem capture about 1% of
as
(E) (d) All of the above
cre
(C) (c) A part of it is captured and passes to
the herbivores.
Sunlight
De
Explanation: The flow of energy is
unidirectional in an ecosystem. The energy
that is captured by the autotrophs does
not revert back to the solar input and the
energy which passes to the herbivores
does not come back to autotrophs.
Producers
1000J
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(C) (i) Flow of energy
Third trophic level/ secondary consumer/ small
carnivore.
55. As frogs are primary herbivores, they feed on
(ii) Flow of matter
small insects such as flies and mosquitoes. This
way, they help in keeping the population of
mosquitoes under control. However, with the
decrease in number of frogs in a village, the
population of female anopheles mosquitoes
increased as they no longer were eaten by
frogs, which caused an increase in the number
of malaria patients in the village.
58. Autotrophs are the organisms that can make
Caution

Students usually make mistakes while drawing
food chains and trophic levels and indicate wrong
organism as example at given trophic level. Be sure
whether that organism belongs to the trophic level
as per their eating habits or not.
42. Biodegradable and non-biodegradable wastes
should be discarded in two different dustbins
because of the following reasons:
(1) The biodegradable wastes kept in a
separate bin can be dumped directly in
a pit for composting or for dumping in
landfills.
(2) The useful part of non-biodegradable
wastes kept in a different bin can be
separated for recycling or reusing them.
(3) Non-useful part of wastes can be disposed
off in such a way that it does not harm
the people or pollute the environment in
anyway.
(4) Segregated waste is cheaper to dispose
off because it does not require as much
manual or mechanical sorting as mixed
waste.
(Any 2 of 4 points can be written to get full
marks)
53. Food chain is a sequence of who eats whom in
an ecosystem. No, lizards are a part of many
food chains. Without them food chain will be
disrupted.
Grass → grass hopper → frog → snake →
peacock.
their own food from carbon dioxide and
water under the action of sunlight and in the
presence of chlorophyll. Example: All green
plants.
Heterotrophs are the organisms which
cannot make their own food by the process of
photosynthesis and are dependent on others
for food. Example: All animals. Decomposers
are the organisms that decompose the
complex molecules present in the dead
remains of plants and animals. Example:
Bacteria, Fungi.
63. (1) Large jar filled with water, oxygen, food
and aquatic plants and animals.
(2) Oxygen/oxygen pump.
(3) Fish food.
(4) Aquatic plants/Producers provide O2
during photosynthesis.
(5) Aquatic animals/Consumers release CO2
for the process of photosynthesis.
(6) Decomposers are also important for
natural cleaning of the aquarium.
[CBSE Marking Scheme 2019]
66. Methods to reduce the problem of waste
disposal:
(1) Segregation of waste should be done
by separating biodegradable waste
substances
from
non-biodegradable
substances.
(2) By recycling solid wastes like paper, plastic
and metals etc, i.e., they are reprocessed
or melted and remoulded to make new
articles.
(3) By composting biodegradable domestic
wastes such as fruit and vegetable peels,
leaves of potted plants can be converted
into compost and used as a manure.
(4) By reducing and reusing of non
biodegradable substances.
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(5) By minimizing the use of disposable items
which are non-biodegradable.
(Any 3 of 5 points can be written to get full
marks)
The energy available to the snakes will be
available as 10% of 200 J. Thus, the energy
available to the snake is 20 J.
10% law states that during transfer of
energy from one trophic level to the next
trophic level, only about 10% of energy is
available to the higher trophic level.
To summarise:
68. (A) A terrestrial food chain comprising of four
trophic levels:
Grass → Grasshopper → Frog → Snake
(B) If we kill all the organisms in one trophic
level the transfer of food energy to next
level will stop.
Organisms of previous trophic level will
also increase.
For example: If all herbivores in an
ecosystem are killed:
There will be no food available for the
carnivores of that area.
Consequently they will also die or will shift
to other areas.
Populations of producers will also increase
in absence of herbivores causing imbalance
in the ecosystem.
(C) Consider the same food chain as we have
made (A):
Grass → Grass hopper → Frog → Snake
In this food chain, second trophic level is
grass hopper and the energy available at
this trophic level is 2000 J.
According to 10% law, 10% of energy will
be available to frog (Third trophic level)
which is 200 J.
%
%
Grass → Grass hopper10

→ Frog 10

→
First
Trophic
level
Second
trophic
level
2000 J
Third
trophic
level
200 J
Snakes
Fourth
trophic
level
20 J
69. (A) (i) Food chain using the above information
Spinach → Caterpillar → Black bird →
Eagle
(ii) The arrows in a food chain represent
transfer of energy from producers to the
consumers
(B) Impact on the food chain after spinach is
sprayed with pesticides.
(1) Only some caterpillars survive
(2) Less food for the black birds and they
may die
(3) Less food for eagle and eagles may die
(4) Pesticides in surviving caterpillars
transfer to consumer and biomagnification in consumers
(C) (i)
Organism
Producer
Herbivore
Secondary
Consumer

Frog

Grasshopper
Maize plant
Tertiary
consumer


Snake
(ii) 90% of the energy captured from the
previous trophic level is lost to the
environment, leaving only 10% of energy
available to the next trophic level.
In this food chain chain, snake is at the
fourth trophic level, only 8kj of energy is
available to the snake.
(1) Energy available to snake = 8 kJ
(2) Energy available to frog = 80 kJ
(3) Energy available to grasshopper = 800 kJ
88
Grass
10.% of
8000 kJ
Grasshopper
8000 kJ
(1st Trophic level)
10% of
8000 kJ
8000 kJ
(2nd Trophic level)
Frog
10% of
80 kJ
80 kJ
(3rd Trophic level)
Snake
8 kJ
(4th Trophic level)
(iii) Major consequence of dramatic decling in
population of frog:
(1) Decreasing frog population will trigger
(2) Grasshopper will damage the producer
(maize plan population) affection
number of food chains and food webs.
Science Class X
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(3) Snake population will also go down
having implications on eagle population.
(4) Feeding patterns may also change
drastically.
71. (A) The use of polythene and plastic should
be reduced in daily life because of the
following reasons:
(1) They cause land pollution and water
pollution as they are non-biodegradable
substances.
(2) Burning of plastic and polythene
releases toxic gases and causes air
pollution.
(3) Plastic and polythene bags cause
harm to the life of animals when they
consume it along with food.
(4) Plastic and polythene bags can cause
blocking in sewage line.
(5) The surfaces of tiny fragments of plastic
and polythene may carry diseasecausing organisms and act as a vector
for diseases in the environment.
(Any 2 of 5 points can be written to get full
marks)
(B) Students would have avoided use of plastic
and polythene in following ways:
(1) By carrying tiffin and water in steel
containers thereby avoiding the use of
plastic.
(2) By carrying their items in cloth bags,
jute bags or paper bags instead of
using polythene bags.
(3) By boycotting all such items such as
plastic plates and spoons, juices served
in plastic cans and stationery items etc.,
made of plastic or polythene which are
sold in school canteen.
(Any 2 of 3 points can be written to get full
marks)
73. (A) (i) In the given food web, the producer
is cactus as it is the first trophic level.
Producers always occupy the first
trophic level.
(ii) The primary consumers in the desert
food web are - insect, small lizard and
rat.
(Any one will get you full marks)
Related Theory

Primary consumers are herbivores and they comprise
2nd trophic level.
(iii) Secondary consumers in the above
figure are: scorpion, large lizard, snake,
falcon, fox.
(Any one will get you full marks)
Secondary consumers prey upon herbivores
(B) Food chain containing a snake - Cactus →
Rat → Snake
(C) Impact of pesticides on the food web is
shown in figure
(1) Kills insects
(2) less food for scorpion / large lizard.
(3) Less number of large lizards
(4) Less food for fox / falcon.
(5) Bioaccumulation in the top predator
falcon / fox
(6) Damage to offsprings
bioaccumulation.
due
to
(7) Less damage to cactus from insects.
(8) More food for small lizard / rats
(9) Change in populations of secondary
consumers.
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