1
Ans 1.
CARBON AND ITS COMPOUNDS
H H H
|
|
|
H—C—C—C—H
|
|
|
H H H
½
The name of the hydrocarbon is propane.
½
Ans 2.
0.02%.
1
Ans 3.
(a) 5 to 8% solution of acetic acid in water in called vinegar. It is used as a preservative in
pickles.
½+½
(b) Carbon compounds have low melting and boiling point because force of attraction
between carbon compounds are not very strong.
1
Ans 4.
When an ester reacts with water in presence of a base, a salt of carboxylic acid, and an
alcohol are produced. Such a reaction is called saponification.
1
For example : When ethyl ethanoate is heated with a solution of sodium hydroxide,
sodium ethanoate and ethanol are produced.
½
Boil
 CH3COONa + C2H5OH.
CH3COOC2H5 + NaOH 
Ans 5.
½
The name of the compound formed is Ethene and its structural formula is CH2 = CH2.
½
CH3CH2OH
Conc. H 2 SO4


 443K
CH2=CH2 + H2O
1½
This reaction is dehydration reaction. In this reaction conc. H2SO4 acts as dehydrating
agent.
1
Ans 6.
Since C gives addition reactions with Br2 and H2, so it is unsaturated hydrocarbon. It gives
an alcohol with phosphoric acid and water i.e., CH3CH2OH. So C is ethene. Thus, A is
ethanol which on oxidation with alkaline KMnO4 gives ethanoic acid. So we can conclude
that A is ethanol, B is ethanoic acid and C is ethene.
CH3CH2OH
(A)
CH3CH2OH
(A)
Ans 7.


CH3COOH
(B)
C2H4
(C)
 Phosphoric acid and water
CH3CH2OH
(A)
3
(a) Bonds which are formed by sharing of an electron pair between two atoms are known
as covalent bonds. Seven covalent bonds are present in ethane.
1+1
(b) Characteristic features of covalent compounds :
(i) They have low melting and boiling points.
½
(ii) They are poor conductor of electricity.
½
(c) Atomic no. of oxygen : 8
½
Electronic configuration : 2, 6
CHEMICAL
SUBSTANCES
NATURE
½
AND
ITS
BEHAVIOUR
P-5
O
O
O O
O O
O O
1

Ans 1.
Hydrogenation.
1
Ans 2.
Sodium ethanoate, carbon dioxide and water.
1
Ans 3.
(a) (i) Methanoic acid, (ii) Butanone.
½+½
(b) Carbon generally forms compounds by covalent compounds because carbon can
neither donate nor accept four electrons for completing its octet. So, it shares its four
electrons with other atoms forming covalent bonds.
1
Ans 4.
Ans 5.
Alk . KMnO 4
(a) CH3CH2OH 

 CH3COOH + H2O
1
(b) Na2CO3 + 2CH3COOH  2CH3COONa + CO2 + H2O
1
The oxidizing agents used for the conversion of ethanol to ethanoic acid are alkaline
potassium permanganate (KMnO4) and acidified potassium dichromate (K2Cr2O7).
1
Test
Ethanol
Ethanoic acid
1. Litmus Test
No change in colour of litmus
solution.
Blue litmus solution turns
red.
2. Reaction with
sodium carbonate
No brisk effervescence
Brisk effervescence due to
evolution of CO2.
1+1
Ans 6.
(a) Difference between alkanes and alkenes
Alkanes
Alkenes
General formula
CnH2n+2
(n= no. of carbon atoms)
CnH2n
(n= no. of carbon atoms)
Naming
All the members end with 'ane'
All the members end with 'ene'
Combustion
Burns with a clean flame
Burns with a sooty flame
H
|
H—C—H
|
H
H H
|
|
C= C
|
|
H H
Methane
Ethene
1+1
(b) Alkanes are saturated hydrocarbons which always burn with a clean blue flame
because complete combustion takes place in sufficient oxygen to give CO2 and H2O
with the liberation of large amount of heat and light.
1
Ans 7.
P-6
(a) Vegetable oil is an unsaturated compound and will decolourise bromine water, while
butter (an animal fat) is a saturated compound which does not decolourise bromine
water.
1
SCIENCE – X
TERM – 2
Ans 8.
(b) Ethanoic acid.
1
(c) Soap reacts with magnesium and calcium salts of hard water to form scum.
1
(a) Ethane (C2H6).
½+½
(b) Homologous series : A series of compounds in which the same functional group
substitutes for hydrogen in a carbon chain is called a homologous series.
C4H10
:
Butane
1
C2H5OH
:
Ethanol
1
(c) Being tetravalent carbon, atom is neither capable of losing all of its four valence
electrons nor it can easily accept four electrons to complete its octet. If carbon were to
gain or lose electrons :
(i) It could gain four electrons forming C4- (anion). But it would be difficult for the nucleus
with six protons to hold on to ten electrons, i.e., four extra electrons.
(ii) It could lose four electrons forming C4+ (cation). But it would require a large amount of
energy to remove four electrons leaving behind a carbon cation with six protons in its
nucleus holding on to just two electrons.
Carbon overcomes this problem by sharing its valence electrons with other atoms of
carbon or with atoms of other elements. That is why carbon form compounds mainly
by covalent bonding.
2

Ans 1.
H H
|
|
H — C — C — OH
| |
H H
1
Ans 2.
The products obtained on complete combustion of ethanol are carbon dioxide and water
along with the release of heat and light.
1
Ans 3.
(a) Functional group is an atom or a group of atoms in a carbon compound that gives the
molecule its characteristic physical and chemical properties. It is the site of reactivity
in an organic compound. For example :
(i) Alcohol is — OH
(ii) Carboxylic acid is — COOH.
1
(b) When soap reacts with hard water, the minerals present in water reacts with soap and
form a white curdy substance known as scum. It reduces the cleansing ability of soap.
1
Ans 4.
Ans 5.
(a) Colour of KMnO4 disappears because it takes part in the oxidation of ethanol.
1
(b) Ethanol is oxidized to produce ethanoic acid.
1
Hydrogenation : The process in which unsaturated compounds reacts with hydrogen in
the presence of nickel (as a catalyst) to form saturated compounds is called hydrogenation.
1
This reaction is commonly used in the hydrogenation of vegetable oils. Vegetable oils have
long unsaturated carbon chains, which are converted into vegetable ghee i.e., saturated
fatty acids.
1
CHEMICAL
SUBSTANCES
NATURE
AND
ITS
BEHAVIOUR
P-7
R
R
C= C
R
Ans 6.
R
R R
|
|
H—C—C—H
|
|
R R
1
In hard water, soaps reacts with calcium and magnesium salts which are present in hard
water and form insoluble substances called scum.
1
Detergents are ammonium or salphonate salts of long chain carboxylic acids. The charged
end of these compounds do not form insoluble precipitates with calcium and magnesium
ions in hard water.
2
Ans 7.
(i) In favour of negative response :
(a) It is an excellent industrial solvent.
(b) It is used in many ways such as-in medicines and ornamentation.
(c) It is used as a disinfectant.
In favour of positive response :
(a) It is the reason behind deaths of many people.
(b) Many adolescents get affected out of it and become addicted
(c) Is being misused even where it is of important use.
3
(ii) Initiatives :
(a) Drive to increase awareness among people by the help of / skit / role plays /
drama / article writing.
(b) Chart preparation and slogan writing.
2

A. Choose the Correct Options :
Ans 1.
(C) Alcohol is inflammable.
Ans 2.
(A) Olive oil and sodium hydroxide.
Ans 3.
(A) Form white precipitate called scum.
Ans 4.
(A) Water loving head and water repelling tail.
Ans 5.
(C) Soap is sodium salt of fatty acid.
Ans 6.
(B) Water is hard due to Ca and Mg ions.
Ans 7.
(C) A is hard water.
Ans 8.
(C) Excess of soap solution.
Ans 9.
(D) In all the test-tube, clear solution is observed.
Ans 10.
(A) Substance X will be sodium hydrogen carbonate.
B. Answers the following :
Ans 1.
(1 marks each)
(2 marks each)
(a) Handle ethanoic acid carefully.
(b) Add only small amount (0.01 g) of NaHCO3 or Na2CO3 to ethanoic acid to control the
intensity of CO2 evolved.
1+1
Ans 2.
Neutralization reaction.
It is used in the form of vinegar, which is a dilute solution of ethanoic acid in water.
1+1
P-8
SCIENCE – X
TERM – 2
Ans 3.
(a) Stir the soap solution carefully so that it does not spill out.
(b) Let the soap set and float on the spent lye before removing it from the beaker.
1+1
Ans 4.
Ans 5.
Ans 1.
Ans 2.
Ans 3.
Ans 4.
Ans 5.
Ans 6.
+
-
Ca(HCO3)2 (aq) + 2Na stearate (aq)  Ca(stearate)2 (s) + 2NaHCO3 (aq)
Sodium stearate
Scum
(soap)
CaSO4 (aq) + 2Na+ stearate- (aq)  Ca(stearate)2 (s) + Na2SO4 (aq)
Scum
Yes, Hydrogen ions provided by HCl makes water hard.
H+ + Na+ stearate–  Steric acid + Na+
1+1
Silicon, No they are very reactive.
½+½
C4H8, C3H6.
½+½
(a) 2C2H5OH + 2Na  2C2H5ONa + H2
½+½
‘X’ is ethanol and Y is hydrogen.
1
(b) CH3CH2OH  CH2 = CH2 + H2O
°
Carbon fibers are made by heating synthetic fibers such as polyacrylonitrile at 3000 C in
the absence of oxygen. It is used for making sports goods, space craft etc.
2
The sodium salt of long chain benzene sulphonic acid or the sodium salt of long chain
alkyl hydrogen sulphate which has cleansing property in water is known as synthetic
detergent. For example, Sodium-n-dodecyl sulphate [CH3 – (CH2)10 – CH2 – SO4 – Na+]. 2
(i) Methane
H
H
|
H
C
H
H—C—H
|
H
1½
H
(ii) Ethane
H H
|
|
H—C—C—H
|
|
H H
Ans 7.
1+1
H
H
H
C
C
H
H
H
1½
(i) The name of the carboxylic acid is ethanoic acid.
Structure
H
O
| ||
H — C — C — OH
|
H
1
(ii) The name of the alcohol is ethanol.
Structure
H H
|
|
H — C — C — OH
|
|
H H
1
CHEMICAL
SUBSTANCES
NATURE
AND
ITS
BEHAVIOUR
P-9
(iii) The compound ‘X’ formed is ethyl acetate (C2H5COOCH3).
Structure
H O
H H
| ||
| |
H—C—C—O—C—C—H
|
| |
H
H H
1
Ans 8.
(a) Butyne
(b) Ethanol
(c) Propanone
1
1
1
Ans 9.
(a) The inlets for air in stove get blocked which leads to yellow flame.
(b) This problem is harmful for our environment as incomplete combustion results in the
formation of oxides which are major pollutants of our environment.
(c) For preventing this situation gas or stove burners should be cleaned time to time.
1+1+1
Ans 10.
(a) The compounds that contain the same molecular formula but different structures are
called isomers. The isomers of a compound have different physical properties.
1
(b) Two possible isomers with the molecular formula C3H6O are acetone and propanal.
Acetone is the simplest ketone whose IUPAC is prop-2-one. 1
(c) Their structures are as follow :
O
H O
H
||
(1) CH3 — C — CH3  H
1½
C
C
C
H
H
O
||
(2) CH3 — CH2 — C — H  H
H
H
H
O
C
C
C
H
H
1½
H

Ans 1.
Acetic acid or vinegar.
1
Ans 2.
C4H10 and C5H12.
Ans 3.
(a) Reactions between unsaturated hydrocarbons with simple substances to form a single
saturated product are addition reactions.
1
½+½
(b) Those reactions in which one or more hydrogen of a saturated hydrocarbon is replaced
by an atom or a group of atoms are substitution reactions.
1
Ans 4.
Soaps
1. Soaps are sodium salts of long chain
carboxylic acids.
Detergents
Detergents are sodium salts of long chain
benzene sulphonic acids.
2. The ionic group in soaps is –COO–Na+. The ionic group in soaps is SO3– Na+ or
SO4– Na+ .
3. Soaps are not useful when water is
hard.
P-10
Detergents can be used for washing
purposes even when water is hard.
SCIENCE – X
TERM – 2
4. Soaps are biodegradable.
Some of the detergents are nonbiodegradable.
5. Soaps have relatively weak cleansing
action.
Detergents have a strong cleansing action.
(any four) ½+½+½+½
Ans 5.
(a) Catalyst is a substance that cause a reaction to occur or proceed at a different rate
without being affected.
CH2 = CH2 + H2  CH3 – CH3
1
(b) C2H6 and C4H10.
Ans 6.
½+½
(a) Ethanol burns in the presence of oxygen to form CO2 and H2O.
CH3CH2OH (l) + 3O2 (g)  2CO2 (g) + 3H2O (l) + heat.
1
(b) Ethanol, when heated with conc. H2SO4 at 443 K or Al2O3 at 623 K undergoes
dehydration, i.e. loses water molecule to form alkene.
CH3CH2OH + (Hot conc. H2SO4)  C2H4 + H2O
Ethanol
1
Ethene
(c) Ethanol reacts with sodium metal to form sodium ethoxide and hydrogen gas.
2C2H5OH
Ethanol
Ans 7.
+
2Na
Sodium

2C2H5ONa
Sodium ethoxide
+
H2
Hydrogen
1
(a) CH3COOC2H5 + NaOH  C2H5OH + CH3COONa
(b) CH3COOH + NaHCO3  CH3COONa + H2O + CO2
(c) CH4 + Cl2  CH3Cl + HCl
Ans 8.
1+1+1
(a) Ethanol reacts with ethanoic acid in the presence of acid catalyst to give ethyl ethanoate.
CH3COOH + CH3CH2OH  CH3COOCH2CH3 + H2O.
(b) When ethanol reacts with sodium, sodium ethoxide is produced along with the
liberation of hydrogen.
2CH3CH2OH + 2Na  2CH3CH2ONa + H2
Ans 9.
1½ + 1½
Two examples of covalent compounds are ethanol and ethanoic acid.
½+½
Difference between the properties of Covalent and Ionic compounds :
Ionic compounds
CHEMICAL
Covalent compounds
1. The bonding in ionic compounds
involves a metal and a non-metal.
These compounds contain bonding
between non- metals.
2. They involve complete transfer of
electrons while formation of compounds. The bonds are formed due
to electrostatic force of attraction.
They have sharing of electrons in
constituent atoms.
3. They are extremely polar .
They are either non-polar or have less
polarity.
4. These compounds have definite shape
and are generally crystalline solid.
Shape is not definite and generally
found in liquid or gaseous state. However, they contain the characteristic
shape of the molecule with covalent
bond.
5. They have high m.pt and b.pt due to
strong electrostatic force of attraction.
They have comparatively low m.pt and
b.pt.
SUBSTANCES
NATURE
AND
ITS
BEHAVIOUR
P-11
6. They are good conductor of electricity
in molten form.
They are bad conductor of electricity.
7. They are soluble in polar solvent.
They may be soluble in both the polar
and non-polar solvents.
8. They contain stronger bond due to
ionic bonding. e.g., NaCl, KCl, etc.
They have comparatively weak bonding
due to covalent nature. e.g., CH4, CO2
etc.
(any four) 4

Ans 1.
Ethane.
1
Ans 2.
H H
|
|
H — C — C — OH
|
|
H H
1
Ans 3.
Saturated Organic Compounds
Unsaturated Organic Compounds
1. These organic compounds contain
single carbon-carbon covalent bond.
These organic compounds contain at least
one double or triple covalent bond.
C
C
or
— C  C —
2. Due to the presence of all single covalent bonds, these compounds are less
reactive.
Due to the presence of double and triple
bonds, these compounds are more reactive.
3 Saturated compounds undergo substitution reactions. Example :
CH4 + Cl2  CH3Cl + HCl
Unsaturated compounds undergo addition
reactions. Example :
C2H5 + Cl2  C2H5Cl2
Chloromethane
4. The number of hydrogen atoms is more
when compared to its corresponding
unsaturated hydrocarbon.
Ans 4.
C == C
Ethene
1,2 dichloroethane
The number of hydrogen atoms is less when
compared to its corresponding unsaturated
hydrocarbon.
(any two) 1+1
(a) Carbon forms large number of compounds due to the following reasons :
(i) Catenation : Carbon forms bond with other atoms of carbon.
(ii) Tetravalency : Carbon share four electrons with other atoms.
½+½
(b) On the other hard some compounds are saturated because they contain single bond
only between two carbon atoms but some compounds are unsaturated because in
them, hydrocarbons valency of carbon is satisfied by double or triple bonds.
1
Ans 5.
In order to form ionic bond, carbon atom either has to lose four electrons to form C4+ ion or
gain four electrons to form C4 – ions.
The loss or gain of four electrons is not easy for carbon since energy needed is very high.
2
Ans 6.
(a) C3H4, C2H4 will undergo addition reactions because they are unsaturated compounds.
1
(b) C3H4 – Alkyne
1
C2H4 – Alkene
P-12
1
SCIENCE – X
TERM – 2
Ans 7.
Ans 8.
(a) A : CH2 = CH2
B : CH3 – CH3
C : CH3 – CH2 – Cl
Ethene
Ethane
Chloroethane
1½
(b) CH2 = CH2 + H2  CH3 – CH3
(A)
(B)
1
Type of reaction : Addition reaction.
½
(a) The force of attraction between the molecules of covalent compounds are not so strong
as that of ionic compounds. So, they have low melting and boiling points.
1
(b) Esters are organic compounds which have sweet smell.
Activity :
Aim : To demonstrate esterification process using ethanol and acetic acid.
1
Materials Required : Beakers, water, test-tube, ethanol, ethanoic acid, conc. H2SO4.
Procedure :
(i) Take 2 ml of ethanol in a test-tube.
1
(ii) Add 2 ml of ethanoic acid into it.
(iii) Add few drops of conc. H2SO4.
(iv) Warm it in a beaker containing water.
(v) Observe the smell of the products formed.
Observation : Pleasant fruity smelling compound (called ester) is formed.
1
Chemical Reaction :
CH3COOH (l) + C2H5OH (l)  CH3COO2H5 + H2O
Ethanoic acid
Ethanol
Ethyl ethanoate Water
1
Conclusion : Carboxylic acid reacts with alcohol in presence of conc. H2SO4 which act
as a dehydrating agent to from esters.
Ans 1.
Ans 2.
Ans 3.
Ans 4.
Ans 5.
The products obtained on complete combustion of ethanol are carbon dioxide and water
along with the release of heat and light.
1
Butanal.
½
CH3 – CH2 – CH2 – CHO.
½
We can distinguish between an alcohol and a carboxylic acid on the basis of their reaction
with sodium carbonate and sodium hydrogen carbonate. Carboxylic acids reacts with
sodium carbonate and sodium hydrogen carbonate to evolve CO2 gas that turns lime
water milky.
Alcohol, on other hand, do not react with sodium carbonate and sodium hydrogen
carbonate.
2
Group of atoms which decides the chemical properties of organic compounds is called the
functional group of an organic compound.
2
(a) In tincture of iodine and cough syrup.
½+½
Conc. H SO
2
4
(b) CH3CH2OH 
 CH2 = CH2 + H2O..
Ans 6.
1
Conc. H2SO4 acts as a dehydrating agent.
1
‘A’ is saturated hydrocarbon i.e., propane.
‘B’ is unsaturated hydrocarbon i.e., CH2 = CH – CH3, which will undergo addition
reaction.
½+½
e.g. CH2 = CH – CH3 + Br2  CH2Br – CHBr – CH3
1
(1, 2 – dibromo-propane)
CHEMICAL
SUBSTANCES
NATURE
AND
ITS
BEHAVIOUR
P-13
Vegetable oil + H2  Vegetable ghee
(unsaturated)
(saturated)
i.e., R — C == C — R + H2  R — C — C — R
|
|
|
|
R
R
R
R
Ans 7.
1
(a) Saturated hydrocarbons burn with clean flame and do not decolourise brown colour
of bromine. Unsaturated hydrocarbons burn with sooty flame and decolourises brown
colour of bromine.
1
(b) Ethanol burns in air to form carbon dioxide and water.
CH3CH2OH + 3O2  2CO2 + 3H2O
1
(c) Reaction between methane and chlorine is considered a substitution reaction because
one atom of hydrogen from methane is replaced by one atom of chlorine and forms
chloromethane.
1
Ans 8.
Heating 'X' Compound- Ethanol at 443K with excess conc. Sulphuric acid results in the
formation of a compound 'Y' called Ethene and if ethene is added with one mole. of H2 to
form 'Z' compound called Ethane is combustion results in 2 mole. of CO2 & 3 mole of H2O.
The chemical equations for the reactions :
Conc. H SO
443K
2
4
 CH2 == CH2
CH3CH2OH 
Nickel Catalyst
CH2 == CH2 + H2 
CH3 — CH3
H
2
2C2H6 + 7O2  4CO2 + 6H2O + Heat + Light
5

A. Choose the correct option :
Ans 1.
(A) Noble gas
Ans 2.
(C) Triple Covalent bond
Ans 3.
(D) Propyne
Ans 4.
(A) CnH2n + 2
Ans 5.
(B) C14H26
Ans 6.
Ans 7.
(B) C2H2
Ans 8.
(A) It is used as antifreeze in car radiators
Ans 9.
(B) Ethanol
Ans 10. (B) CH3COOH
B. Answer the following :
Ans 1.
Chemical name – Ethanol
Chemical formula – CH3CH2OH
Ans 2.
Chemical name – Ethanoic acid
Chemical formula – CH3COOH
Ans 3.
Functional group in compound A – (—OH) Alcohol
Functional group in compound B – (—COOH) Carboxylic acid
(2 marks each)
Alk. KMnO4

 CH3COOH + H2O
Ans 4.
CH3CH2OH
Ans 5.
Oxidation reaction.
P-14
(1 mark each)
1+1
1+1
1+1
2
2
SCIENCE – X
TERM – 2
2
PERIODIC CLASSIFICATION OF ELEMENTS
Ans 1.
There are 18 vertical columns in the modern periodic table and they are called as groups.
½+½
Ans 2.
There are 7 horizontal rows in the modern periodic table and they are called as periods.
½+½
Ans 3.
Radius of Y is bigger than that of X.
1
In 'X' the number of shells is two while in Y it is three.
1
(a) Electronic configuration of element 'Y' with atomic number 3 : 2, 1.
½
Ans 4.
Electronic configuration of element 'A' with atomic number 17 : 2, 8, 7.
½
Therefore the formula of the compound : YA.
(b) Electronic configuration of the element with atomic number 10 : 2, 8
Valency : Zero.
Ans 5.
(a)
19K (Potassium)
½
½
is the element that has one electron in the outermost shell.
Electronic configuration = 2, 8, 8, 1
(b)
4Be (Berylium) and 20Ca (Calcium) belong to
the same group, because they have same
number of valence electrons.
Be = 2, 2 and Ca = 2, 8, 8, 2
(c) Berylium (Be) and Fluorine (F) have same period i.e., second period which has 2 shells
(K and L) whereas Potassium (K) and Calcium (Ca) also belongs to same period
i.e., fourth period which has 4 shells (K, L, M and N).
1+1+1
Ans 6.
The early attempts to classify elements were made on the basis of atomic masses of the
elements. Modern basis is the atomic number. This is better because now it is established
that atomic number of an element is the fundamental property. Further are certain anomalies
in Mendeleev’s periodic table such as the position of isotopes and misplacements of
certain elements, e.g. Argon is placed (Atomic mass = 40, Atomic number = 18) before
potassium in Mendeleev’s periodic table (Atomic mass = 39, Atomic number = 19)
3
Ans 7.
(a) Hydrogen gas
1
(b) By carbon dioxide gas
1
(c) Because sodium reacts violently with water and air.
1
(d) Lead, copper or gold
1
(e) By painting iron articles
1

Ans 1.
(i) They are unreactive.
(ii) They have zero valency.
½+½
Ans 2.
Properties of elements are periodic function of their atomic numbers i.e., the number of
protons or electrons present in the neutral atom of an element.
1
Ans 3.
(a) 1 pm = 10–12 m.
Atomic radius = 37 × 10–12 m.
CHEMICAL
SUBSTANCES
NATURE
1
AND
ITS
BEHAVIOUR
P-15
(b) In a group, atomic size : increases down the group.
½
In a period, atomic size : decreases in moving from left to right.
Ans 4.
Nitrogen
:
K shell-2
L shell-5
Phosphorus
:
K shell-2
L shell-8
½
½
M shell-5
½
Phosphorus will be less electronegative because its electrons are further away from
the nucleus and thus nucleus can be easily lost.
1
Ans 5.
(a) The position of elements X in the modern periodic table is group number 17 and period
number 3. The position of element Y in the modern periodic table is group number 2
and period number 4. Electronic configuration of element X = 2, 8, 7.
It has 3 shells so period number is 3.
Halogens are kept in group 17.
Electronic configuration of element Y = 2, 8, 8, 2.
It has 4 shells so period number is 4.
The valence shell has 2 electrons so the group number is 2.
(b) Li is less reactive due to more nuclear attraction of electrons.
(c) K has the largest atomic radius due to presence of more shells.
Ans 6.
Ans 7.
1+1+1
(a) Number of electrons in the outermost orbit of X = 1 and Y = 2.
½
(b) Valency of X = 1 and Y = 2.
½
(c) Metal X is more metallic than Y.
½
(d) Atomic size of X is bigger than that of Y.
½
(e) Chloride XCl ; YCl2
½
(f) Sulphate X2SO4 ; YSO4
½
(a) To study the properties of elements and to keep the elements with similar properties
together.
1
(b) Chemical properties of elements and atomic number.
1
(c) Metals lie on extreme left, metalloids lie in the middle and non-metals lie on the right
side.
2
(d) They should be placed in the same slot, since they have the same number of electrons.
1

Ans 1.
Elements in a group have similar electronic configuration. So, all of them have 1 valence
electron.
1
Ans 2.
Second period.
Ans 3.
(a) Mg : 2, 8, 2
1
Al : 2, 8, 3
½+½
rd
Ans 4.
(b) 3 period; because electrons in these atoms are filled in K, L, M shells.
1
(a) Properties of elements are a periodic function of their atomic numbers.
1
(b) Periods 7 and groups 18.
Ans 5.
½+½
(a) Noble gases are so called because they are inert gases and they do not react easily.
(b) Helium (He), Argon (Ar).
(c) They are placed in a separate group because they have complete doublet or octet
configuration.
1+1+1
P-16
SCIENCE – X
TERM – 2
Ans 6.
(a) A and B, C and D have same period.
(b) A and C have one valence electron, so they have same group.
(c) C is more reactive because C will be placed below A in the periodic table and reactivity
increases down the group.
1+1+1
Ans 7.
(a) On moving from left to right, atomic size decreases.
½
Atomic number increases, number of protons and electrons increases, extra electron
increase in same shell.
½
Extra proton increases the effective nuclear charge.
½
Electrons are attracted closer to the nucleus.
½
(b) Metallic character of elements decreases across a period as the tendency to lose
electron decreases.
1
Ans 8.
Na (Sodium), Mg (Magnesium) and Al (Aluminium) are metals. Si (Silicon) is a metalloid.
P (Phosphorus), S (Sulphur) and Cl (Chlorine) are non-metals.
2
As we go from left to right across the third period, metallic character decreases and nonmetallic character increases.
2
The most metallic element in the third period is Na and the most non-metallic element is C.
1

Ans 1.
Oxygen is divalent in nature. The valency of Magnesium in Magnesium oxide is + 2. The
formula of Magnesium bromide is MgBr2, since Bromine has valency equal to one.
1
Ans 2.
These are two isotopes of the same element and have atomic number Z = 1. Therefore, both
have one proton.
1
Ans 3.
(a) Atomic radius decreases.
1
(b) Metallic character decreases.
1
(i) Atomic size
½
(ii) Valency or combining capacity
½
(iii) Metallic property
½
(iv) Non-metallic property.
½
Ans 4.
Ans 5.
(a) The valency of Q is 3 as its valence shell has three electrons in it.
(b) Elements P and Q are metals as they have 2 electrons in their valence shell and they
are positively charged ions whereas elements R and S are non-metals as they gain
electrons to complete their octet.
(c) P and Q will form basic oxides as they are metals.
Ans 6.
1+1+1
(a) Z
½
(b) Halogens
½
(c) Magnesium and nitrogen
Ans 7.
½+½
(d) Silicon
½
(e) X has bigger size than P because X has less effective nuclear charge.
½
(a) Neon has two shells, both of which are completely filled with electrons (2 electrons in
K shell and 8 electrons in L shell).
1
CHEMICAL
SUBSTANCES
NATURE
AND
ITS
BEHAVIOUR
P-17
(b) Magnesium has the electronic configuration : 2, 8, 2.
1
(c) Silicon has a total of three shells, with four electrons in its valence shell (2 electrons in
K shell, 8 electrons in L shell and 4 electrons in M shell).
1
(d) Boron has a total of two shells, with three electrons in its valence shell (2 electrons in
K shell and 3 electrons in L shell).
1
(e) Carbon has twice as many electrons in its second shell as in its first shell (2 electrons
in K shell and 4 electrons in L shell).
1

Ans 1.
The two elements X and Y will show same chemical properties because they have same
number of valence electrons in group. They form positively charged ions by losing one
electron.
½+½
Ans 2.
Positively charged ions are called cations and negatively charged ions are called anions.
½+½
Ans 3.
Mg, P, Cl, Ar.
1
Increase in atomic number means increase in nuclear charge which tends to pull the
electrons closer to the nucleus and reduces the size of the atom.
1
Ans 4.
It will be 7 in X as well as in Z.
1
The reason being that number of electrons in the outermost shell in the elements in same
group is same.
1
Ans 5.
Ans 6.
(i) From top to bottom in group - metallic character increases.
1
(ii) From left to right in a period - metallic character decreases.
1
(a) 9, because atomic number increases by one in going from one element to the next.
1
(b) 'X', because of less effective nuclear charge.
1
(c) 'Y' has smaller atomic size because new shells are added when we go down the group.
1
Ans 7.
(a) 12.
½
(b) 2.
½
(c) Be.
½
Because Be has two electrons in its valence shell. Given element also has two electrons
in its valence shell. Since valence electrons determine the chemical property, hence the
given element has same chemical properties as that of Be.
1½
Ans 8.
(a) They have same number of valence electrons.
1
(b) K, because metallic character decreases along the period and increases down the
group.
½+½
Ans 9.
(c) 17.
1
(a) Formula of these compounds are XSO4, X3(PO4)2.
2
(b) It belongs to group 2nd and 3rd period.
2
(c) It will form ionic compounds as it can lose two electrons to aquire stable electronic
configuration.
1

P-18
SCIENCE – X
TERM – 2
Ans 1.
Hydrogen should be placed in group I, since it has only one electron in its outermost shell.
1
Ans 2.
The metallic character of a metal increases with increase in the size of its atom.
1
Ans 3.
It is three in Na, Al and P.
1
Ans 4.
Elements with the same number of occupied shells are placed in same period.
1
B, C, N, O.
1
In a period, as the atomic number increases, size of the atom decreases. This is because,
increase in nuclear charge tends to pull electrons closer to nucleus and reduces the size of
the atom.
1
Ans 5.
Order of atomic number of elements
A < C < B.
1
Because atomic size generally decreases along a period. So, B has the highest atomic
number followed by C and A.
1
Ans 6.
Ans 7.
(a) Number of electrons in the outermost orbit of X = 1 and Y = 2.
½
(b) Valency of X = 1 and Y =2.
½
(c) Metal X is more metallic than Y.
½
(d) Atomic size of X is bigger than that of Y.
½
(e) Chloride XCl ; YCl2
½
(f) Sulphate X2SO4 ; YSO4
½
(a) X - 2, 5 : Group 15
½
Y - 2, 8 : Group 18
½
Z - 2, 8, 4 : Group 14
½
(b) Y is a noble gas which is unreactive and has complete outermost shell.
½
Z is metalloid and possesses semi-metal, properties of both metals and non-metals.
½
Ans 8.
(c) X and Y belong to the same period i.e., IInd period.
½
(i) 16.
1
(ii) A is smaller in size than B. Atomic size decreases on moving from left to right in a
period.
1
(iii) O. Both are of the same group in the periodic table.
1

Ans 1.
(a) Magnesium
: 2, 8, 2.
½
(b) Calcium
: 2, 8, 8, 2.
½
Ans 2.
Hydrogen and Helium.
Ans 3.
(a) Atomic number.
½+½
½
(b) Down the group, the effective nuclear charge experienced by the valence electron is
decreasing because the outer most electrons are further away from the nucleus.
Therefore, they can be lost easily. Hence metallic character increases down the group.
1½
CHEMICAL
SUBSTANCES
NATURE
AND
ITS
BEHAVIOUR
P-19
Ans 4.
(a) C and D
Ans 5.
(i) The valency of X is one. It has tendency to loose one electron. Therefore, X will form
cation.
1
(ii) Y; because atomic radius decreases across a period due to increase in the electrostatic
force between electrons and nucleus.
1
Ans 6.
(a) Since element ‘A’ belongs to group 15 and has configuration (2, 8, 5) or 5 valence
electrons. It is a non -metal.
1
(b) Tendency to gain electrons decreases down the group as effective nuclear charge
experienced by valence electrons decreases.
1
Ans 7.
(b) B
(c) C
(d) A
½+½+½+½
(c) Anion.
1
(a) Y
(b) X
½
½
(c) Z
½
(d) 2
½
(e) Y : non-metal
Z : metal
½
½

Ans 1.
Ans 2.
Metals are placed on the left side while non-metals are placed on the right side of the
periodic table.
½+½
Valency is same i.e., 1.
1
Ans 3.
(a) Third period and group 16.
1
(b) Size increases down the group as new shells are added.
1
(a) 3
(b) 1
½
½
(c) PQ
½
(d) S
½
Ans 4.
Ans 5.
(a) X has more metallic character than Y.
Oxide of Y
Oxide of X
: X2O
Chloride of X :
XCl
:
Chloride of Y :
YO
½+½
YCl2
½+½
Ans 6.
(a) Borderline elements which are intermediate in properties between metals and nonmetals are called metalloids.
1
(b) Boron, silicon, germanium and arsenic.
½+½+½+½
Ans 7.
(a) Because it exhibits properties of both metals and non-metals.
(b) Germanium and boron.
Ans 8.
1
½+½
(c) Middle, Right side of metalloids.
1
(a) Element ‘A’ is a metal because it is present in group I (1 valence electrons) and can lose
electron easily.
1
(b) Element C has larger size than 'B' because it has more number of shells than 'B'.
Also, C lies in the third period and has three shells, whereas B lies in second period
and has two shells.
1
(c) 'B' being an element of 3rd group has three valence electrons, therefore its valency is 3.
1
P-20
SCIENCE – X
TERM – 2
Ans 1.
(i) Organisation makes our life simple, easy and systematic.
(ii) As classification and organization help us in our daily life, in the same way,
classification of elements has made the study of elements easier and simple.
We can know about the properties of elements because of this classification.
Associated Value : The learner will be motivated to adopt well- disciplined and organized
life.
3
Ans 2.
(i) By sensitizing people about adulteration.
(ii) Small amount of pebbles/impurities are mixed in pulses.
(iii) When we wash the pulses, the impurities float on the water.
Associated Value : The learner will become more vigilant towards Food adulteration and
they started to check this menance.
3
Ans 3.
(a) In the periodic table, elements are placed according to their electronic configuration. If
an element has only one shell in its electron configuration, it is placed in the first
period. If the element has two shells then it is placed in the second period, and so on.
Vertical columns in the periodic table are called groups. There are eighteen groups and
in a group all the elements have same number of valence electrons.
1+1
(b) They will be placed at the same slot as their atomic number, valence electron and
valency are same and have same chemical properties.
1
Ans 4.
(a) Since each element contains the same number of valence electrons, n in the outermost
shells of their atoms. Therefore X, Y and Z belong to the same group.
(b) The atomic radii increase from X to Y and Y to Z. So, metallic character increases from
X towards Z.
(c) X, Y and Z have melting points above 25°C (room temperature). Therefore X, Y and Z
are solid at room temperature.
(d) The ionization potential decreases from X to Y and Y to Z, because they belong to the
same group in the descending order.
5

A. Choose the correct option :
(1 mark each)
Ans 1.
(D) Increasing atomic mass
Ans 2.
(A) John Newlands
Ans 3.
(B) 18 groups
Ans 4.
(D) Ni
Ans 5.
(C) He
Ans 6.
(A) Li < Na < K < Rb < Cs
Ans 7.
(B) K
Ans 8.
(A) Chalcogens
Ans 9.
(B) Be, Mg, Ca
Ans 10.
(A) 2nd group, 3rd period
B. Fill in the Blanks :
Ans 1.
(1 × 10)
Atomic masses
CHEMICAL
SUBSTANCES
NATURE
AND
ITS
BEHAVIOUR
P-21
Ans 2.
Atomic masses
Ans 3.
Groups
Ans 4.
Smaller
Ans 5.
Increases
Ans 6.
Lesser
Ans 7.
Argon
Ans 8.
Seven
Ans 9.
Shells
Ans 10.
Dobereiner.

P-22
SCIENCE – X
TERM – 2
3
HOW DO ORGANISMS REPRODUCE ?
Ans 1.
DNA copying is necessary during reproduction because it leads to the transmission of
characters from parents to offspring and variations.
1
Ans 2.
Because hundreds of biochemical reactions occur during preparation of DNA copies. Few
of them are liable to run and form a different product and hence they are not identical to the
original. Thus, this gives rise to variations.
1
Ans 3.
(a) Leishmania reproduces by binary fission and Plasmodium reproduces by multiple
fission.
(b)
Binary Fission
Multiple Fission
1. Nucleus divides into two daughter Nucleus divides into many daughter
nuclei.
nuclei.
2. Constriction is formed in the cell Division of cytoplasm occurs very late. Thus
after nuclear division.
many cells are produced simultaneously.
3. Only two daughter cells are pro- Many new daughter cells are produced.
duced.
4. It takes place in favourable conditions.
It takes place in adverse conditions.
1+1
Ans 4.
(i) Vegetative propagation is the only known method of multiplication of seedless plants,
which gives a genetically uniform population.
(ii) Seeds and fruits are of uniform quality, size, taste and aroma and have a good quality
of variety.
1+1
Ans 5.
Two functions are :
(i) Testis : It produces sperms and secretes male sex hormones called testosterone.
1
(ii) Ovary : It produces ovum and secretes female sex hormones called estrogen and
progesterone.
1
Ans 6.
Disease caused by virus—AIDS, Genital Warts.
Disease caused by bacteria—Gonorrhea, Syphilis.
1
Prevention of Diseases :
(a) By use of devices made of plastic, metal or a combination of the two which is inserted
into uterus.
(b) By using contraceptive devices.
(c) By educating people and maintaining hygiene.
(d) By avoiding multiple sexes.
2
Ans 7.
The pollination stimulates the ovary to grow. It is essential for fertilization and seed
development. Pollination also prevents ovary abscission. Pollen grains contain small
amount of auxin, which together with a limited amount of additional auxin form the
carpellary tissues that can support the initial growth of ovary. Subsequent fruit growth
requires normal seeds which synthesise auxin, gibberellins and cytokinins. Thus, seeds
play a key role in fruit development.
3

THE
WORLD
OF
LIVING
P-23
Ans 1.
Bryophyllum.
1
Ans 2.
Each pollen grain produced two male gametes.
1
Ans 3.
Sperms contain two types of chromosomes X—chromosome and Y—chromosome.
Egg contains one type of chromosomes only X—chromosome.
1+1
Ans 4.
Yes, people terminate pregnancy to avoid birth of an unwanted child. This is being
misused in sex selective abortion of female foetus called female foeticides.
2
Ans 5.
Hydra reproduces by budding using the regenerative cells. A bud develops as an outgrowth
in hydra, due to repeated cell division at one specific site. When fully mature, the bud
detaches itself from the parent body and develops into new independent individuals.
Tentacles
3
Bud
Ans 6.
(a) A special tissue that develops between the uterine wall and the embryo (foetus) is
called placenta. In placenta, the villi of the embryo comes in contact with blood
sinuses of the uterus. This provides a large surface area for the exchange of materials
between the embryo and the mother. Glucose, amino acids, oxygen and other useful
materials pass from the mother to the embryo. Waste substances and carbon dioxide
generated by the embryo are transferred to the mother ’s blood through the placenta.
(b) The four ways of preventing pregnancy are :
(i) Mechanical barrier method of contraception
(ii) Hormonal method of contraception
(iii) Chemical method of contraception
(iv) Reversible sterilization
Advantages of using such preventive measures are :
(a) It provides awareness to both male and female regarding fertility regulation.
(b) Reduction in the number of children to the limit of enjoying family life as well as
rearing the children with better resources as and when they are available. 1 + 2 + 1

Ans 1.
Ans 2.
During cell division in unicellular organisms, the nucleus of the parent cell divides only
once to form two daughter nuclei along with the cytoplasm that undergoes cleavage. In
this way, two daughter cells are formed from one single parent.
1
Oviduct or Fallopian tube.
1
Ans 3.
(a) Testis produce sperm.
(b) Testis produce male sex hormone called testosterone.
Ans 4.
1+1
Difference between pollen grain and ovule :
S. No.
(i)
(ii)
Pollen Grain
Produced by anthers.
Contains male germ cells.
Ovule
Produced by ovary.
Contains female germ cells.
1+1
P-24
SCIENCE – X
TERM – 2
Ans 5.
There are three parts of carpel :
(a) Ovary—It contains the ovule.
(b) Style—It exposes the stigma for pollination.
(c) Stigma—It is sticky and receives the pollen grains during pollination.
Ans 6.
Ans 7.
1+1+1
(i) Fusion of a male gamete with a female gamete is known as fertilization.
1
(ii) (a) It promotes diversity of characters in the offsprings.
½
(b) It results in the new combination of genes brought together in gamete.
½
(c) It increases genetic variation.
½
(d) It plays a prominent role in the origin of new species.
½
(a)
Ovary
Fallopian tube
Endometrium
Uterus
Cervix
Vagina
Vulva
(b) (i) If embryo moves down to reach the uterus. The embedding of embryo in the thick
inner lining of the uterus is called implantation. Then, a special tissue develops
between the uterine wall and the embryo (foetus) called placenta, where the
exchange of nutrients, oxygen and waste products takes place.
(ii) If the zygote is not formed, the uterine wall, endometrium and corpus luteum
breaks down by the discharge of blood, mucus and broken endometrial, lining
from the vagina causing menstruation.
3+1+1

A. Choose the correct option :
(1 mark each)
Ans 1.
(D) Shoot arises from plumule.
Ans 2.
(C) The diagram shows germination of seed.
Ans 3.
(D) For showing germination, seeds of gram, water, bowl and wet clothes are required.
Ans 4.
(A) Cotyledons store food.
Ans 5.
(B) Gram seed is dicotyledonous.
Ans 6.
(B) II, III and IV are correct statements.
Ans 7.
(C) III and IV are correct observations.
Ans 8.
(A) After binary fission, two daughter nuclei are formed.
Ans 9.
(D) Bud always forms a chain of bud and then detaches from the parent cell.
Ans 10.
(B) Two nuclei in centrally constructed amoeba, one in yeast cell, one in bud.
THE
WORLD
OF
LIVING
P-25
B. Answer the following :
Ans 1.
(2 marks each)
(a) Take a petri dish. Place damp cotton on it.
(b) Put the dicot seeds on the petri dish. Allow it to germinate.
(c) Place the germinated seed on the watch glass.
(d) Open the two cotyledons with a forcep and needle.
(e) Observe the structure of seed.
(f) Also observe the embryonal axis with a magnifying glass and label its parts.
Ans 2.
2
Albuminous seed :
(a) Seeds having copious amount of endosperm. e.g., Wheat.
1
Exalbuminous seed :
(b) Seeds in which endosperm is used up. e.g., Grass.
Ans 3.
1
(a) Conditions for the seed germination should be optimum i.e. warmth, moisture and air.
(b) Care should be taken to separate two cotyledons so that the embryonal axis is intact.
1+1
Ans 4.
Significance of Asexual reproduction :
(1) Requirement of one parent to produce offspring so no need to migrate to other place to
reproduce.
(2) Numerous offspring can be produced in a short span of time.
(3) Minimize the use of energy & time.
Ans 5.
1+1
Yes. Binary fission and budding do not involve any variations, also these involve single
parent.
1+1

Ans 1.
The function of petals is to attract insects for pollination and to protect the reproductive
organs, which are at the centre of the flower.
1
Ans 2.
IUCD : Intra-Uterine Contraceptive Devices.
½
HIV : Human Immuno Deficiency Virus.
½
Ans 3.
(a) After fertilization, ovules become seeds and ovary forms the fruit.
½+½
(b) Pollination is the transfer of pollen grains from anther to the stigma of a flower.
Fertilization is the fusion of male and female gametes.
Ans 4.
½
½
Syphilis, Gonorrhoea, AIDS, Warts. (any two)
½+½
These infectious diseases spread from one person to another by sexual contact with an
infected person.
Ans 5.
P-26
Symptoms of STDs :
½
Burning sensation at urination, urethral discharge, sores in genitals.
½
VAS DEFERENS :

its a part in male reproductive system.

it futhur connects with the urethra .

the sperms are carried through the vas deferens.

the seminal vesicles and prostrate gland add their secreation to the sperms in the vas
deferens.
SCIENCE – X
TERM – 2
FALLOPIAN TUBE :
Ans 6.

its a part in female reproductive system.

it furthur connects with a bag like structure called uterus.

the ovum are carried through the fallopian tube.

the fertilization takes place in fallopian tube.
1+1
No, it cannot be identical to the original cell.
It has variation.
It becomes the basis for organic evolution.
Ans 7.
1+1+1
(a) Hair growth on face, chest, armspit and genital area / body becomes muscular / voice
becomes deep and coarse / penis occasionally begins to become enlarged and erect.
(any two)
1
(b) Formation of sperms needs lower temperature than the normal body temperature.
Hence, testis lie outside the body cavity in the scrotum.
1
(c) Urethra is a common duct for the passage of both urine and semen.
Ans 8.
1
It is an asexual method of reproduction. In this process, vegetative part of a plant body
such as stem, leaves, bulbs, tubers are used for growing new plants by cutting, grafting
and layering which are identical to parents.
2
It is used to grow :
(a) Plants that have lost the capacity to produce seeds.
(b) Plants which are genetically similar enough to the parent plants.
Ans 9.
½+½
(a) Sexual Reproduction :
(1) Two parents take part.
(2) Variation occurs in offspring.
(3) Fertilization takes place.
(4) Gametes are involved.
(5) Mixing of hereditary material.
Asexual Reproduction :
(1) Single parent.
(2) Offspring are genetically identical to each other and to their parent.
(3) No Internal fertilization or External fertilization.
(4) No gametes.
(5) No mixing of hereditary material.
(b) Sexual reproduction promotes diversity of characters in the offsprings. It result in new
combinations of genes brought together in the gamete and this reshuffling increases
genetic variation. It plays a prominent role in the origin of new species. The sexual
mode of reproduction incorporates the process of combining DNA from two different
individuals during reproduction.
3+2

Ans 1.
The time period from the development of foetus inside the uterus till birth is called
gestation period.
1
Ans 2.
No, because copper T, will not prevent contact of body fluids. Thus, it will not protect her
from sexually transmitted diseases.
1
THE
WORLD
OF
LIVING
P-27
Ans 3.
Ans 4.
(a) Oviduct or Fallopian tube.
1
(b) Ovaries.
1
Radicle : Part of germinating seed that gives rise to root.
Plumule : Part of germinating seed that gives rise to shoot.
Ans 5.
Ans 6.
1+1
(a) Uterus / womb.
1
(b) Vagina.
1
(a) Self-pollination is the transfer of pollen grains from anthers to the stigma of the same
flower or another flower of the same plant.
1
Occurs in bisexual flowers.
Cross pollination is the transfer of pollen grains from anthers to the stigma of another
flower borne on another plant of the same species.
It occurs in unisexual as well as bisexual flowers.
1
(b) The pollen tube which carries male gamete, travels through the style to reach the
ovary. The ovary contains ovules. Each ovule has an egg cell. The fusion of male
gamete and female gamete, called fertilization, gives rise to the zygote.
The zygote is capable of growing into a new plant.
Ans 7.
Ans 8.
1
(a) Both produces gametes and sex hormones.
1
(b) Contains body plant (embryo) which maintains the species of a particular plant.
1
(c) To carry sperm from testis, which enter penis for their release.
1
(a) Hydra/ Planaria.
1
(b) Placenta is a specialized tissue that helps the embryo in getting nourishment. It is a
disc that is embedded in the uterine wall.
2

Ans 1.
(i) It helps in the production of female gamete or ovum.
(ii) It secretes female sex hormones, estrogen and progesterone.
½+½
Ans 2.
Fertilization takes place in oviducts (fallopian tubes) and implantation takes place in
uterus.
½+½
Ans 3.
Unisexual flowers : Possess either male or female reproductive organs.
Bisexual flowers : Possess both male and female reproductive organs.
1+1
Ans 4.
Menstruation is the shedding off the uterine lining along with the degenerated ovum
through the vagina as blood and mucus. When egg is not fertilized, then the inner lining of
the uterus is sheded off.
2
Ans 5.
Amoeba : Occurs in any plane.
1
Leishmania : Occurs in definite orientation.
1
(a) Rose / Sugarcane, Grapes.
1
(b) (i) Many organisms can be produced at the same time.
1
Ans 6.
(ii) Organisms are able to tide over unfavourable conditions.
Ans 7.
P-28
1
(a) Spore formation takes place in Rhizopus. Rhizopus consists of fine thread like
projection called hyphae. It has a knob like structure which is involved in reproduction
called sporangia, containing spores, that develop into new Rhizopus.
2
SCIENCE – X
TERM – 2
Sporangium
Sporangiospores
Sporangiophore
Coenocylic hyphoe
Rhizoids
(b) More number of spores are produced which can easily help the Rhizopus to spread.
1
Ans 8.
(a) In binary fission, two daughter cells are formed and in multiple fission, many daughter
cells are formed simultaneously.
1
Plasmodium which is a malarial parasite reproduces by multiple fission.
(b) (i) Plants bear flowers and fruits earlier.
(ii) Desirable traits can be propagated due to genetic similarity.
Ans 9.
(a) (i) One egg is discharged every month into the fallopian tube.
1
½
½
1
(ii) The wall of uterus thickens and becomes spongy to receive embryo, if fertilization
occurs. If there is no fertilization, the thickened lining of uterus breaks down,
along with blood and mucus causing menstruation.
1
(b) Provides large surface area for exchange of materials between the embryo and the
mother. For example glucose, respiratory gases and removal of wastes occurs through
it.
1
(c) A process by which an organism produces young individuals of its own species.
Continuity of the species generation after generation is maintained.
1
1

Ans 1.
The lining of the uterus becomes spongy and thick to nourish the embryo but when the egg
it not fertilized, the lining is not needed and it breaks down and comes out through vagina
in blood and mucus.
1
Ans 2.
Prostate glands provides secretion which contains chemical for motality of sperm. Seminal
vesicles contain fructose, protein and other chemicals that are necessary for nourishing
and stimulating sperms.
1
THE
WORLD
OF
LIVING
P-29
Ans 3.
Ans 4.
Bacterial Infection : Gonorrhoea or warts.
½
Viral Infection : AIDS or syphilis.
½
It can be prevented by barrier methods, e.g., using condoms.
1
Regeneration : The process by which new organisms develop from the body parts of
certain organisms. Such organisms can be cut into a number of pieces and each piece
grows into separate individual.
1
Reproduction : In this process, growth of the individual organism takes place by sexual or
asexual method.
½
Ans 5.
Two organisms which grow by regeneration are Hydra and Planaria.
½
Asexual reproduction : Generations are created from a single individual.
½
Sexual reproduction : New generations are created from two individuals, out of which one
is male and other is female.
½
Examples of asexual reproduction :
(a) Amoeba, (b) Hydra.
Ans 6.
Ans 7.
½+½
(i) It helps in transfer of pollen grains from anther to stigma.
1
(ii) Helps in evolution and survival.
1
(iii) Prevents fertilization - Barrier method.
1
(a)
Plumule
Radicle
Cotyledon
Micropyle
1½
Testa
(b) (i) Radicle : Future root, (ii) Plumule : Future, (iii) Cotyledon (food store).
Ans 8.
1½
Placenta is a disc-shaped structure between mother and embryo.
1
Functions :
Ans 9.
(i) Provides large surface area for glucose and O2 to pass from mother to embryo.
1
(ii) Removal of wastes from embryo.
1
This will prevent himself from contracting various STD’s as well as spreading of diseases.
Associated Value : The learner will be motivated to maintain a good reproductive health
as well.
3

Ans 1.
(i) Grafting in rose & Layering in Jasmine.
½+½
(ii) (a) It is cheaper, easier and more rapid method of propagation.
(b) Better quality of plants can be maintained.
P-30
½
½
SCIENCE – X
TERM – 2
Ans 2.
(c) Produce seeds with prolonged period of dormancy.
½
(d) Traits or characters of the parent plant are preserved.
½
Significance of vegetative propagation are :
(i) It is rapid, easier and cheaper method of multiplication.
(ii) It helps to multiply the seedless plants.
(iii) The plants with poor seed viability or prolonged dormancy can be multiplied by this
process.
(iv) It can grow plants in unfavourable conditions.
(iv) It helps to preserve the characters of variety.
(vi) Grafting is the best economic method.
Ans 3.
3
Approach of society is baseless. Sex of child is determined by type of chromosome present
in sperm (X or Y) that fuses with ovum at the time of fertilization.
Associated Value : The learner will be able to improve their mindsets that help them to
discontinue various social malaise viz gender inequality, female foeticide etc.
3
Ans 4.
1. Absolutely not.
Human approach/behavior
One should be aware about the mode of transmission of disease, danger of sharing needles
etc.
Associated Value : The learners will understand that only a generous and improved
mindset of society can help the people to fight HIV/AIDS related problems.
3
Ans 5.
(i) Three methods of contraception are :
(a) Mechanical Barrier Methods.
1
(b) Chemical Methods.
1
(c) Surgical Methods.
1
(ii) (a) Discussion of benefits of family planning by people.
½
(b) Organising seminar.
½
(c) Making poster.
½
(d) Observing the problems due to enhanced population.
½

A. Choose the correct option :
Ans 1.
(C) DNA
Ans 2.
(A) Leishmania
Ans 3.
(B) Vegetative propagation
Ans 4.
(C) Anther and filament
Ans 5.
(D) Both stamens and carpels
Ans 6.
(B) Ovary
Ans 7.
(C) Produce male gametes and hormone
Ans 8.
(D) Fallopian tubes
Ans 9.
(C) Mechanical method
Ans 10.
(B) Vas deferens
THE
WORLD
OF
LIVING
(1 mark each)
P-31
B. Fill in the Blanks :
Ans 1.
evolution
Ans 2.
tubectomy
Ans 3.
Gynoecium
Ans 4.
seminal vesicles
Ans 5.
two
Ans 6.
Zygote
Ans 7.
bacteria
Ans 8.
Vasectomy
Ans 9.
root
Ans 10.
evolution
(1 × 10)

P-32
SCIENCE – X
TERM – 2
4
HEREDITY AND EVOLUTION
Ans 1.
Ans 2.
Ans 3.
Ans 4.
Ans 5.
Ans 6.
Ans 7.
Ans 8.
Speciation is the evolution of reproductive isolation among once—interbreeding
populations i.e. the development of one or more species from an existing species.
1
Two factors that could lead to speciation are :
(i) Genetic drift , (ii) Reproductive isolation.
½+½
Male individual has 46 chromosomes but because the gametes are always haploid i.e., they
have half the number of chromosomes; sperms will be haploid (23 chromosomes). Female
individual also contains only 23 chromosomes in egg. It is the fusion of the sperm and egg
which leads to an offspring’s with 46 chromosomes.
2
(i) If we dig into the earth and start finding fossils, it can be assumed that the fossils
closer to the surface are more recent to those found in the deeper layers.
(ii) By detecting the ratios of different isotopes of the same element in the fossil material.
1+1
(i) A self-pollinating plant species.
1
(ii) A species reproducing asexually.
1
(i) Variation refers to the difference in the characters or traits among the individuals of a
species.
1
(ii) Variation enabls the organisms to adapt themselves in changing environment. It forms
the basis of heredity. They form raw materials for evolution and development of new
species.
2
Lamarck theory is called the theory of inheritance of acquired characters. It says that the
organisms are exposed to new needs due to a change in the environment. They acquire
same characters during their lifetime to adapt themselves to the new needs. These characters
are passed on to the next generation. For example, the long neck of giraffe is due to needs
to catch the leaves of long tress.
A. Weismann rejected the theory of Lamarck, stating the adaptations, if based in the
somatic cells, cannot be inherited.
3
(i) Traits A characteristic feature is called trait.
(ii) Inherited Traits are the traits which are transferred from parents to young ones.
Acquired Traits are the traits which an organism acquires in life time.
(iii) Formation of new species from the existing ones is called speciation.
Factors which could lead to the rise of new species are : geographical isolation, genetic
drift, natural selection.
1+2+2

Ans 1.
All the plants would be tall plants (Tt).
Ans 2.
Comparing the DNA of different species gives a direct estimate of how much the DNA has
changed during the formation of these species. Thus, it helps us to a great extent in tracing
the evolutionary pathways.
1
Ans 3.
F1 generation refers to the offspring resulting immediately from a cross of the first set of
parents.
1
THE
WORLD
OF
LIVING
1
P-33
on the other hand, F2 generation refers to the offspring resulting from a cross among the
members of F1 generation.
1
Ans 4.
The genes controlling a particular trait separate from each other during gamete formation.
Hence gamete is always pure as far as contrasting characters are considered and will
possess only one gene set.
2
Ans 5.
No, the information provided is not enough to tell whether blood group A or O is
dominant.
1
Every character is controlled by a pair of alleles. And here it is not mentioned whether the
man and woman were homozygous or heterozygous for their traits.
1
Ans 6.
These traits are acquired traits. Acquired traits are those variations which an individual
develops during its life time due to the effect of environmental factors, use and disuse of
organs and conscious efforts. The traits affect the structure and functioning of cells, tissues
and organs without influencing the genetic material. Acquired traits are also called
somatic variations as their influence is restricted to somatic cells. The germ cells are not
affected. Acquired traits are, therefore, non-inheritable. They disappear with the death of
the individual e.g. muscular body of an athlete, learning of music.
3
Ans 7.
(i) Decomposition of organisms.
1
(ii) (a) The age of fossils can be estimated by fossil dating.
1
(b) The study of fossils is important because :
(i) It helps to analyse racial history of plants and animals.
½
(ii) It helps to measure the geological time.
½

Ans 1.
The pair of chromosomes involved in determining the sex of an individual, e.g., X and Y
chromosomes in humans.
1
Ans 2.
(i) Relative method, (ii) Carbon dating.
Ans 3.
In asexually reproducing organisms, there is no mixing of genes of two different individuals
since only one partner is involved.
1
½+½
There are very minor variations generated in such organisms due to small inaccuracies in
DNA copying.
1
Ans 4.
Characteristics are details of appearance of behaviour. In other words, a particular form or
a particular function. Presence of four limbs in humans in an example of characteristics.
2
Ans 5.
(i) The phenotype of F1 progeny is Tt because F1 or hybrid plants were not intermediate
between the two alternate forms of a character. They resembled only one parent. In a
cross between tall and dwarf of pea plants, the F1 plants were all tall.
(ii) The phenotypic ratio of F2 progeny is 3 : 1.
(iii) F1 progeny resembled only one parent which is dominant in nature but in F2 progeny
phenotypically three plants are tall and one plant is dwarf. But according to genotype
ratio one plant is pure tall, two are hybrid tall and one plant is pure dwarf.
1+1+1
Ans 6.
P-34
The science of heredity and variation is called genetics. Mendel conducted breeding
experiments in a garden pea plant (Pisum sativum) with two different contrasting characters.
He found that only one character appeared in first generation but both the characters
reappeared in the subsequent generation. On the basis of results of his experiments, he put
forward the various principles of inheritance. He also suggested that each character of pea
plant is controlled by ‘factor’.
3
SCIENCE – X
TERM – 2
Ans 7.
(i) This is to show that each trait is influenced by both paternal and maternal DNA.
1
(ii) No.
1
This is because all the F1 progeny plants show the genetic makeup Pp, which results
in purple flowers.
1½
(iii) Dominant trait is a genetic trait that is expressed in a person who has only one copy of
that gene.
1½

A. Choose the correct option :
(1 mark each)
Ans 1.
(B) They have same origin as they both are edible roots.
Ans 2.
(C) Variation.
Ans 3.
(C) Wings of insect and wings of bat.
Ans 4.
(D) All of these because all the pairs show same origin.
Ans 5.
(C) Similarities in appearance and function but different in structure.
Ans 6.
(C) Same origin and different function shows homology.
Ans 7.
(C) Jointed appendages of A and segmented body without appendages.
Ans 8.
(A) Cone of gymnosperm and flower of angiosperm.
Ans 9.
(D) Both have common ancestors.
Ans 10. (D) Wings of housefly.
B. Answer the questions :
(2 marks each)
Ans 1.
Theory : Two different ways these similarities in anatomical structures can be classified is
either analogous structures or homologous structures. While both of these categories have
to do with how similar body parts of different organisms are used and structured, only one
is actually an indication of a common ancestor somewhere in the past.
2
Ans 2.
Homologous organs found in animals–wings of birds and flippers of whale are homologous
structures. They both are modifies forelimbs but perform different function. Wings are
meant for flying whereas flippers help in swimming.
2
Ans 3.
Analogous organs found in plants–Thorns found in Pyracantha and spines found in
Berberis vulgaris provide protection to plants but both have different origin. Thorns are
the modified stem and spines are modified leaves.
2
Ans 4.
Organs which remain in the body which do not perform any function but were part of an
organ or working part which was lost during evolution.
e.g.- tail bone below our pelvic bone was the part of the bones supporting the tail when our
ancestors were primates.
2
Ans 5.
Homologous Organs :
(1) Have the same structural design and developmental origin.
(2) But they have different functions and appearance.
Example : The forelimb of a frog, a man, a lizard and a frog seem to be built from the same
basic design of bones, but they perform different functions.
Analogous Organs :
(1) Have different basic structural design and developmental origin.
(2) But they have similar appearance and perform similar functions.
Example : The wings of birds and bats look similar. But in birds wings are covered by
feathers all along the arm but the wings of bats are skin folds stretched between elongated
fingers.
2
THE
WORLD
OF
LIVING
P-35
Ans 1.
In human beings, the sex of the individual is largely genetically determined.
1
Ans 2.
The eyes of Planaria are just eyespots which detect light.
1
Ans 3.
The experiences of an individual may only affect the somatic cells, but will not change the
DNA of the germ cells.
1
Ans 4.
Change in non-reproductive tissues cannot be passed on to the DNA of the germ cells.
Therefore, the experiences of an individual during its lifetime cannot be passed on to its
progeny.
1
Traits acquired by an organism during its lifetime are known as acquired traits.
1
These traits are not inherited because they do not cause any change in the DNA of the
organism.
1
Ans 5.
The field of evolution and classification are interlinked in the following manner :
The more characteristics two species have in common, the more closely they are
related. Classification of species is a reflection of their evolutionary relationship. 1

The more number of characteristics shared by two organisms more, is the probability
of their having common ancestors.
1
Thus, classification of an organism is the reflection of its evolutionary path.

Ans 6.
Ans 7.
Ans 8.
(i) Genetic Drift : If due to some reasons migrant population mixes up with local
population and reproduces the progenies, it will result in the genes of migrant
population entering a new population.
1½
(ii) Natural Selection : If the DNA changes are severe enough, such as change in the
number of chromosomes, then eventually the green cells of the two groups cannot fuse
with each other. Or a new variation emerges in which mating can occur only be within
the similar group.
1½
(a) Male individual has 46 chromosomes but because the gametes are always haploid i.e.,
they have half the no. of chromosomes; sperms will be haploid (23 chromosomes).
Female individual also contains only 23 chromosomes in egg. It is the fusion of this
sperm and egg which leads to an offspring with 46 chromosomes.
1½
(b) Geographical isolation will not be a major factor in the speciation of a self-pollinating
plant species. It involves the same flower / plant from where male and female gametes
are formed.
1½
During breeding, farmers selected for a very short distance between leaves among the wild
cabbage plants and grew the cabbage we eat.
1
Selecting the features of arrested flower led to the development of broccoli.
1
On the other hand, cauliflower was bread by selecting sterile flowers.
1

Ans 1.
The only progressive trend in evolution seems to be that more and more complex body
designs have emerged over a period of time.
1
Ans 2.
9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green.
Ans 3.
In crossing, if two or more traits are involved, their genes assort independently, irrespective
of the combinations present in the parents.
1
Ans 4.
P-36
1
So, new combinations of genes appear in the offsprings leading to new traits.
1
Genes are located on chromosomes in linear sequence and at fixed positions.
1
Chemically genes are made up of nucleic acids which constitute DNA.
1
SCIENCE – X
TERM – 2
Ans 5.
As the population of the given species reproduce asexually there would be only very
minor differences generated due to small inaccuracies in DNA copying, so trait B is likely
to be arisen earlier as it is present in 60% of the population.
1
Trait A which exists in 10% of the population may be originated due to variations.
Ans 6.
1
(i) Geographical isolation : Due to some physical barriers such as mountain, sea etc. the
already separated subgroups gets further separated which affects gene flow.
(ii) Genetic drift : Random change in gene frequencies by chance alone in small population,
although it does not show any survival advantages.
(iii) Natural selection : In this new suitable traits are selected to live in the prevailing
environment.
1+1+1
Ans 7.
Monohybrid Cross.
White Flowers
WW
Purple Flowers
×
ww
Ww
White Flowers
1
(i) Genotypes of F2 progeny are
WW, Ww, ww
1
(ii) Ratio between White : Purple flowers are :
3 :1
Ans 8.
1
Mendel showed that the traits of the garden pea may be dominant or recessive in the
following manner :
(i) He took tall a plant and a short plant. The produced progeny from them and calculated
percentage of tall or short progeny.
1
(ii) In the F1 progeny, all plants were tall i.e., only one of the parental trait was seen.
½
(iii) In F2 progeny of the F1, all the plants are not tall. Instead one quarter of them are short.
½
(iv) Thus both traits i.e., tallness and shortness are inherited in the F1 plants but only one
is expressed dominantly and other trait remains recessive.
1

Ans 1.
Farmers carried out artificial selection of plants with slightly larger leaves in the wild
cabbage and came up with a leafy vegetable called kale.
1
Ans 2.
Long tentacles.
Ans 3.
Natural selection, genetic drift, variations and geographical isolation can lead to speciation
in sexually reproducing organisms.
½×4=2
Ans 4.
No.
1
1
Geographical isolation prevents gene flow between populations of a species whereas
asexual reproduction generally involves only one individual. In an asexually reproducing
organism, variations can occur only when the copying of DNA is not accurate. Therefore,
geographical isolation cannot prevent the formation of new species in an asexually
reproducing organism.
1
Ans 5.
Analogous organs are one such evidence that is used to determine how close two species
are related.
The presence of feathers in dinosaurs and birds indicates that they are evolutionarily
related.
THE
WORLD
OF
LIVING
P-37
Dinosaurs had feathers not for flying but instead these feathers provided insulation to
these warm-blooded animals. However, the feathers in birds are used for flight.
This proves that reptiles and birds are closely related and that the evolution of wings
started in reptiles.
2
Ans 6.
(i) (a) vertebrate, (b) invertebrate.
½+½
(ii) If we dig into a rock / earth then it is reasonable to suppose that the fossil found in the
upper layers must be of more recent origin than the fossil found in the deeper layers.
1
The second way by detecting the ratios of different isotopes of the same element in the
fossil material.
1
Ans 7.
RRYY

RY
×

RrYy – F1


Round green

Round yellow
9
Ans 8.
Ans 9.
rryy – Parents

ry – Gametes
:
3
:
1

Wrinkled
yellow
3

Wrinkled – F2
green
:
1
½×4=2
(i) (a) The pink colour of flower is the dominant trait whereas the recessive trait in
white colour.
(b) 3 : 1 (Pink : White)
(ii) Gene.
2
1
Species ‘X’ is more closely related with species ‘Y’.
1
The more closely related species will have more characteristics in common. Hence species
X and Y are closely related.
1
Eukaryotic organisms are further classified on the basis of whether they are unicellular or
multi-cellular.
1
This basis marks a very fundamental difference in body design, because of specialisation
of cell types and tissues.
1
Homologous organs help to identify an evolutionary relationship between apparently
different species.
1

Ans 1.
Ans 2.
Favourable variation helps a species to adapt to change in their environment and they
promote survival of the species.
1
This is because the decrease in weight will not cause a change in the DNA of germ cells,
due to which this change cannot be inherited.
1
Ans 3.
Natural selection selects those variations in a population which give a survival advantage
and helps the better population to fit their environment better.
1
On the other hand, genetic drift can alter gene frequencies in small populations and
provide diversity without any survival benefits.
1
Ans 4.
(i) Micro-evolution : Evolution resulting from small specific genetic changes that can
lead to a new sub-species.
1
(ii) Fossils : Fossils are preserved traces of living organisms, that got buried deep inside
the earth millions of years ago.
1
P-38
SCIENCE – X
TERM – 2
Ans 5.
During sexual reproduction, a female gamete (egg) fuses with a male gamete (sperm)
which are haploid to form zygote. Zygote is diploid and contains 23 chromosomes from
mother and 23 from father.
1
In this way, an equal genetic contribution of male and female parents is ensured in the
progeny.
1
Ans 6.
(i) Genetics is the branch of biology the deals with the study of heredity and variations.
1
(ii) Gregor Johann Mendel, garden pea.
½+½
(iii) Garden pea plants were easily available / they grow in one season / fertilization was
easy.
(any two) ½+½
Ans 7.
(i) Gene.
½
It is the carrier of genetic information from one generation to another.
(ii) The traits that are obtained from parents are inherited traits.
Ans 8.
½
½
Example : Fused and free ear lobes.
½
The traits that develop during lifetime of an individual are acquired traits.
½
Example : Muscular body of a wrestler.
½
(a) The evolution of feathers in certain dinosaurs had nothing to do with flight.
1
In them, the feathers carried out the function of providing insulation in cold weather.
1
(b) Inherited traits are characters that are transferred from one generation to another. 1
Acquired traits on the other hand are developed during the life time of an individual
and cannot usually be transferred to future generations.
1
(c) Sex chromosome is either a pair of chromosomes, usually designated X or Y, in the
germ cells of most animals, that combine to determine the sex and sex-linked
characteristics of an individual.
1

Ans 1.
(i) No.
(ii) Acquired trait
(iii) Acquired characters are in non-reproductive tissue. They can be acquired by exercise.
Change in non-reproductive tissue cannot be passed on to the DNA of the germ cells.
½+½+2
Ans 2.
(i) If any natural calamity occurs and kills these small number of surviving tigers, they
can become extinct resulting in the loss of genes forever.
(ii) Small number will lead to little recombination and lesser variations that are very
important for giving better survival chances to the species.
(iii) Less number of species means lesser extent of diversity and lesser number of traits
which reduce the chances of adaptability with respect to the change in the environment.
1+1+1
Ans 3.
THE
The two uses of fossils are :
(i) Racial history of plants.
½
(ii) Past climatic conditions of earth.
½
WORLD
OF
LIVING
P-39
Study of fossils provide evidence in favour of organic evolution because :
(i) Fossils helps to identify an evolutionary relationship between apparently different
species.
1
(ii) The fossils present in the bottom rocks are simple while the most recent fossil found in
the upper strata are highly complex. This geographical succession completely agrees
with the concept of evolution.
1
Ans 4.
DNA - Deoxy Ribonucleic Acid.
It contains the cell nucleus.
It contains information source for making proteins.
Cells uses chemical methods to build copies of their DNA which separate with cell
division.
Since DNA copying is not exact so variation occurs which is advantagious to species.
1+1+1+1+1
Ans 5.
(i) Two characters. They are shape of seed and colour of seed.
1
Round shape and yellow colour of seed are the dominant characteristics.
1
Wrinkled shape and green colour of seed are the recessive characteristics.
1
(ii) Cellular DNA is the information source for making proteins in the cell.
1
(iii) Snails can change their sex during their lifetime.
1

Ans 1.
In most cases when organisms die, their bodies decompose and become lost.
1
But every once in a while, the body or at least some parts of the body may be in an
enviornment that does not decompose completely.
1
It is through such preserved traces of living organisms that we get fossils.
For example : If a dead insect gets caught in hot mud, it will not decompose quickly, the
mud will eventually harden and retain the impression of the body parts of the insect. 1
Ans 2.
Fossils : The remains and / or impressions of organisms that lived in the past.
½
(i) Theory of inheritance of acquired characters :
(a) The fossils that we find close to the surface are more recent than the one we find in
deeper layers.
½
(b) Dating fossils by detecting the ratio of different isotopes of the same element.
½
(c) Homologous organs have same basic structural design.
(ii) (a) Analogous organs have different basic structural design.
½
½
(b) Homologous organs perform different functions.
(c) Analogous organs perform similar functions.
Ans 3.
½
J. B. S. Haldane suggested that life must have developed from the simple inorganic
molecules which were present on earth soon after it was formed. He speculated that the
conditions on earth at that time could have given rise to more complex organic molecules
that were necessary for life. The first primitive organisms would arise from further chemical
synthesis.
2
Later, Stanley L. Miller and Harold C. Urey conducted experiment to find out about the
origin of organic molecules. They assembled an atmosphere similar to that thought to exist
P-40
SCIENCE – X
TERM – 2
on early earth (molecules like ammonia, methane and hydrogen sulphide, but no oxygen
over water). This was maintained at a temperature just below 100°C and sparks were
passed through the mixture of gases of stimulate lightning.
2
At the end of a week, 15% of the carbon had been converted to simple compounds of carbon
including amino acids which make up protein molecules.
1
This is how life originated from inanimate matter.

A. Choose the correct option :
Ans 1.
(C) evolutionary relationship
Ans 2.
(A) Homo habilis
Ans 3.
(B) characteristic
Ans 4.
(A) fossil found near the earth surface is more recent
Ans 5.
(D) skin folds stretched between elongated fingers
Ans 6.
(B) take one chromosome from each pair
Ans 7.
(B) RrYy
Ans 8.
(D) both amino acids and simple sugars
Ans 9.
(B) mother
Ans 10.
(B) they are present in separate chromosomes.
B. Fill in the Blanks :
(1 mark each)
(1 × 10)
Ans 1.
dominant
Ans 2.
homologous
Ans 3.
Mendel
Ans 4.
dihybrid
Ans 5.
Central Asia
Ans 6.
X-chromosome
Ans 7.
Fossils
Ans 8.
hatching of male
Ans 9.
convergent
Ans 10.
hetrogametic.

THE
WORLD
OF
LIVING
P-41
5
LIGHT : REFLECTION AND REFRACTION
Ans 1.
0 (Zero).
1
Ans 2.
Refractive index of atmosphere is different at different altitudes because the air density
changes with altitudes.
1
Ans 3.
E
Q
N
i
A
B
F
r
N1
G
D
C
e
H
Refraction through a glass slab
The emergent ray CD is parallel to the incident ray AB, but it has been laterally displaced
by a perpendicular distance CN with respect to the incident ray.
2
Ans 4.
vg = ( 3 x 108) / 1.5
1
8
= 2 × 10 m/sec
Ans 5.
1
(a) Refractive index of alcohol > refractive index of water.
So, alcohol is optically denser than water.
When a ray of light enters from water to alcohol, it bends towards the normal :
1½
Incident
ray
Rarer
medium
i
r
Denser
medium
Refracted
ray
(b) Refractive index of diamond with respect to glass is given by
Refractive index of diamond/Refractive index of glass = 2.42/1.50 = 1.61 Ans 6.
(i) Reason is lateral displacement.
1½
1
(ii) Lateral displacement depends upon thickness of the slab, the incident and refraction
angles.
1
P-42
SCIENCE – X
TERM – 2
(a) If white light source is used with parallel side, the ray will pass through the focus.
(b) Light rays will not pass through the non-parallel side.
Ans 7.
1
(i) Snell's law of refraction : The ratio of the 'sine' of the angle of incidence to the 'sine' of
the angle of refraction is constant, i.e., = constant, for the light of a given colour and for
the given pair of media.
2
(ii) u = – 25 cm , f = – 15.0 cm , h0 = 4 cm
Using mirror formula,
1
1
+
v
u
1
1
+
v
25
1
1
–
v
25
1
v
=
1
f
=
1
 15
=
1
 15
1
1
+
15
25
75
v =–
cm
2
h
v
m = i =–
u
ho
= –
 75 
hi
= –  2  25 
4


h i = – 6 cm
Screen should be placed at a distance of 75/2 cm on the left side from the pole in order
to obtain a sharp image.
Size of the image : 6 cm.
Nature of the image : Real and inverted.
3

Ans 1.
Light will bend away from its normal direction.
Ans 2.
Angle of incidence and thickness of slab.
Ans 3.
Ratio of the height of image to the height of object is magnification.
1
½+½
M (mirror) = h1/h = – v/u
M (lens) = h1/h = v/u
M = Magnification
h = Height of object
h 1 = Height of image
v = Image distance
u = Object distance.
Ans 4.
1+½ +½
Image of same size in plane mirror.
Image is enlarged in concave mirror.
Image is diminished in convex mirror.
Image in all the three is virtual.
NATURAL
PHENOMENA
½ +½ +½ +½
P-43
Ans 5.
In this case of convex lens, the size of the object is bigger than the size of the image. Thus
we can conclude that the image formed is real and inverted and also the image formed is
on the other side. According to the sign convention, for lenses the image distance on the
other side will be positive . Therefore v = 12 cm
For real and inverted image, the magnification will be negative. Therefore m = – 2/3
m=
v
u
2
12
=
3
u
u=
12  3
= – 18 cm
2
1
1
1
–
=
f
v
u
1
1
1
–
=
 18
f
12
23
1
=
f
36
36
= 7.2 cm
5
Therefore the focal length is 7.2 cm.
f=
Ans 6.
3
We look our face in each mirror, turn by turn.
(i) If the image formed is of same size as our face but laterally inverted for all positions,
then it is a plane mirror.
(ii) If the image formed is erect and enlarged initially but gets inverted as the face is moved
away, then it is a concave mirror.
(iii) If the image formed is erect and smaller in size for all positions, then it is a convex
mirror.
1+1+1
Ans 7.
(i) Ability of a lens to converge or diverge light rays is termed as power of a lens.
Power = 1/f (m)
½+½
SI unit of power is Dioptre.
(ii) Lens of focal length 15 cm is of larger power because power is inversely proportional
to the of focal length.
1
(iii) Height of the object h = + 2.0 cm;
Focal length f = + 10 cm;
object-distance u = –15 cm;
Image-distance v = ?;
Height of the image h’ = ?
Since
or
1
1
1
–
=
f
v
u
1
1
1
=
+
f
v
u
1
1
1
1
1
=
+
=–
+
v
10
15
10
(  15)
P-44
SCIENCE – X
TERM – 2
or,
2  3
1
1
=
=
v
30
30
v = + 30 cm.
The positive sign of v shows that the image is formed at a distance of 30 cm on the other
side of the optical centre. The image is real and inverted.
Magnification m = h'/h = v/u
or, h’ = h (v/u)
Height of the image, h’ = (2.0) (+30/–15) = – 4.0 cm
Magnification m = v/u
or, m = (+30 cm) / (–15 cm) = – 2
The negative signs of m and h’ show that the image is inverted and real. It is formed
below the principal axis. Thus, a real, inverted image, 4 cm tall, is formed at a distance
of 30 cm on the other side of the lens. The image is two times enlarged.
3

Ans 1.
If the rays of light do not actually meet after reflection or refraction, but appear to meet
when produced backwards, than that point constitutes virtual image.
1
Ans 2.
Normal incidence implies, i = 0°. Hence, from the second law of reflection, r = 0°.
Ans 3.
The power of a lens is defined as the reciprocal of its focal length (f) expressed in metres. SI
unit of power is dioptre.
One dioptre is defined as the power of a lens whose focal length is 1 metre.
Ans 4.
1
1+½ +½
(i) According to the Snell's law–
The ratio of the sine of angle of incidence to the sine of angle of refraction is constant, i.e.,
sine of angle of incidence (i)/sine of angle of refraction (r) = constant.
This constant is called refractive index (n).
n = sin i/sin r
1
(ii) Refractive index is a number which measures the bending of light when passing from
one medium to another. It is the ratio of velocity of light in vacuum to the velocity of
light in the medium.
Mathematically,
Refractive index, µ = Velocity of light in vacuum/Velocity of light in medium
Ans 5.
1
(i) A convex mirror always forms an erect, virtual and diminished image of an object
placed anywhere in front of it. Thus, convex mirrors enable the driver to view much
larger traffic behind him than would be possible with a plane mirror.
(ii) A concave mirror is used as a shaving or make-up mirror because it forms an erect and
enlarged image of the face when it is held closer to the face.
1½ + 1½
Ans 6.
(a) Such a ray falls normally on the mirror. So its angle of incidence is 0°.
(b) The concave mirror with smaller aperture forms the sharper image because it is free
from spherical aberration.
(c) No change. The focal length of a convex mirror does not depend on the nature of the
medium.
1+1+1
Ans 7.
NATURAL
(i) When light falls on a surface and gets back into the same medium, it is called
reflection. A highly polished surface or mirror reflects the light.
2
PHENOMENA
P-45
(ii) (a) Concave mirrors are used as shaving mirrors and in headlight of automobiles.
(b) Convex mirror are used as rear-view mirrors in vehicles.
(c) Plane mirrors are used to see our virtual image.
1+1+1

A. Multiple choice questions :
(1 mark each)
Ans 1.
(A) Larger the distance, better and sharper focus can be achieved.
Ans 2.
(D) Inverted, sharp and real image is formed by convex lens. Distance between mirror and
screen is the focal length.
Ans 3.
(D) The emergent ray bends at an angle to the direction of incident ray.
Ans 4.
(C) Correct path shown.
Ans 5.
Ans 6.
(C) 30 cm from the optical center on the other side of the lens.
(C) Blurred image is seen when separation between the screen and mirror is less than the
focal length.
Ans 7.
(C) Convex lens, lens holder, screen with stand, a measuring scale.
B. Answer the following :
(2 marks each)
Ans 1.
(a) Use a very sharp tipped pencil to draw thin lines to represent incident and refracted
rays.
Ans 2.
Ans 3.
(b) The convex lens drawn should be thin and of small aperture. (This is required for
obtaining the distinct image.)
1+1
(a) The glass slab should be perfectly rectangular with all its faces smooth.
(b) While fixing the pins P1 and P2 or the pins P3 and P4, care should be taken to maintain
a distance of about 5 cm between the two pins. This would help in tracing the direction
of incident ray and that of emergent ray with greater accuracy.
1+1
(a) While viewing the collinearity of pins and images, the eye should be kept at a distance
from the pins so that all of them can be seen simultaneously. The collinearity of pins
fixed on one side of the glass prism and the images of pins on the other side could also
be confirmed by moving the head slightly to either side while viewing them. All the
pins and images of pins would appear to move together if they are collinear.
(b) The pins P1, P2, P3 and P4 fixed on the paper may not be exactly perpendicular (or
vertical) to the plane of paper. It is therefore desirable to look at the feet of the pins or
their images while establishing their collinearity. That is why the position of each pin
is marked with pointed tip of the pins on the paper.
1+1

Ans 1.
1½
Nature of the image : Virtual and erect.
P-46
½
SCIENCE – X
TERM – 2
Ans 2.
Given : object distance (u) = – 5 cm , focal length (f) = 10 cm
1/v + 1/u = 1/f
1/v = 1/f – 1/u
1/v = 1/10 – 1/(– 5)
1/v = 1/10 + 1/5
1/v = (1 + 2) / 10
1/v = 3/10
v = 10/3
v = 3.33 cm
Ans 3.
Ans 4.
Therefore the image formed is virtual and errect and 3.33 cm behind the mirror.
2
(i) Convex lens.
1
(ii) Image formed is virtual.
1
(a) n1 > n2, lens behaves as diverging lens
1½
(b) n1 = n2, no refraction occurs.
1½
Ans 5.
(i)
1½
NATURAL
PHENOMENA
P-47
Ans 6.
Ans 1.
Ans 2.
(ii) (a) f = –15
v = –10
u=?
lens formula 1/f = 1/v – 1/u
1/u = 1/v – 1/f
1/u = 1/ –10 – 1/–15
1/u = –1/10 + 1/15
1/u = (–3+2)/30 (taking LCM)
1/u = –1/30
u = – 30
Thus, object should be placed at a distance of 30 cm..
magnification = v/u
m = –10/–30
m = 1/3
m = 0.33
(b) Virtual, erect and diminished.
1½
(i) Reciprocal of focal length or degree of convergence or divergence is a power of lens.
S.I. unit is dioptre.
½+½
(ii) v = 50 cm
Since a same sized, real and inverted image is formed, therefore the object must be
placed at the centre of curvature i.e., at 2F1
and since the image is formed at 2F2, hence the distance b/w 2F1 and O is 50 cm.
Thus the needle is placed 50 cm in front of the lens.
R = 50 cm
f = 50/2 = 25 cm
f = 0.25 m
P = 1/f
P = 1/0.25
P = 100/25
P = 4D
2

Speed of light decreases in the presence of a medium e.g., when a light ray enters in water
from air, its speed decreases.
1
Angle of incidence is the angle (i) made by the incident ray with the normal. Angle of
refraction is the angle (r) made by the refracted ray in the second medium with the normal.
1
Ans 3.
1½
Nature of the image : Virtual and erect.
P-48
½
SCIENCE – X
TERM – 2
Ans 4.
u = –12 cm, m = – 4 m = –v /u
– 4 = –v / –12
v = – 4 × 12
v = – 48
2
Ans 5.
The light ray bends towards the normal when a ray of light travels from an optically rarer
medium to an optically denser medium. Since water is optically denser than air, a ray of
light travelling from air into water will bend towards the normal.
2
Ans 6.
(a) (i) Power =+5D
So the lens is convex
P = 1/f
f = 1/5 m
f = 20 cm
So in each case
1/v – 1/u = 1/f
1/v = 1/u + 1/f
v = uf/(u + f)
v = 20 u/(20 + u)
Now u varies from –18, –20, –22, –30 cm
So,
v = –180 cm
If u = –18 cm
so, magnification is (v/u) = +10
Similarly
v = +220 cm, if u = –22 cm
so, magnification (v/u)=-10
v = +60 cm ,if u = –30 cm
so, magnification is (v/u) = –2
If u = –20 cm
v = infinity
so the magnification = – infinity
Now magnification is positive in 1st case
where u = –18 cm
all the other images can be obtained on screen
1
(ii) At 22 cm and 30 cm image is received on a screen.
1
(b) Film projectors and telescopes.
Ans 7.
1
(a) (i) +2D indicates that the lens is converging lens (convex lens)
p = 1/f.....where f in meter.....so focal length f = 50 cm
(ii) – 4D indicates that the lens is diverging lens (concave lens)
similarly the focal length is –25 cm
NATURAL
PHENOMENA
P-49
Ans 8.
Ans 1.
Ans 2.
Ans 3.
Ans 4.
Ans 5.
Ans 6.
(b) To find the image for convex lens
u = –100 cm
f = 50 cm
v = (f x u)/(f – u) = 100 cm
magnification is 1
For concave lens
u = –100 cm
f = –25 cm
v = – (f × u)/(f + u)
v = –20 cm
3
(i) Laws of Refraction of Light :
(a) The incident ray, the refracted ray and the normal to the interface of two transparent
media at the point of incidence, all lie in the same plane.
(b) 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 a given pair of media.
2
(ii) R = + 3 m, u = – 5 m, v = ?, h' = ?
v = 1·15 m.
Magnification m = + 0·23
The Image formed is virtual, erect and smaller in size by a factor of 0·3.
3

Refractive index being a ratio of two similar quantities and has no unit.
1
The size of the image formed keeps on increasing as the object is brought closer towards
the convex lens.
1
v = + 12 cm, f = +10 cm.
1/v – 1/u = 1/f
1/12 – 1/u = 1/10
u = – 60 cm.
2
(i) The incident ray, the refracted ray and the normal all lie in the same plane.
(ii) The ratio of sine of angle of incidence in the first medium to the sine of angle of
refraction in the second medium is a constant and is known as refractive index of the
second medium with respect to the first medium.
1+1
(i) Image distance = Object distance
½
(ii) Size of the image = Size of the object
½
(iii) Image is laterally inverted
½
(iv) Image is always virtual and erect.
½
(a) v = 3, u or v = 30
½
– 1/30 – 1/10 = 1/f
f = – 7·5 cm
½
(b)
1
P-50
SCIENCE – X
TERM – 2
Ans 7.
Ans 8.
Ans1.
Ans 2.
Ans 3.
Ans 4.
Ans 5.
Ans 6.
f = 20 cm, u = ?, m = 2
m = v/u  2 = v/u  v = 2u
1/f = 1/v – 1/u
1/20 = 1/2u – 1/u  u = –10 cm
v = – 20 cm
3
Refractive index of a medium with respect to vacuum is called absolute refractive index.
1
Refractive index of Bw.r.t. A = nb/na
1
Higher the optical density, lesser is the velocity of light in the medium.
1

Yes, all distances are measured from the optical centre of the lens. Conventionally, the
object is placed on the left side of the lens.
1
Speed of light is different in different media, As the medium changes, the light has to
choose a path of minimum time. Hence, the direction of the light changes. This phenomenon
is known as refraction of light.
1
v = 8·57 cm.
Nature of the image : Virtual.
2
For the same angle of incidence in media P, Q and R, the angles of refraction is minimum for
the medium R. Hence velocity of light would be minimum in R.
1+1
alfl = afl /aal
= 1.65/1.36 = 1.21
Refractive index of dense flint glass with respect to alcohol is 1·21.
2
(a) When n1 > n2
1
Light goes from rarer to denser medium  it diverges.
(b) When n1 = n2
There is no change in the medium  no bending or refraction occurs.
(c) When n1 < n2
Light goes from denser to rarer medium  It converges.
1
1
Ans 7.
Image Real and equal in size to object
Image Virtual and enlarged
1½+1½

Ans 1.
Never try to catch or hold anything in deep water, as it appears nearer then its actual
position.
Associated Value : The learners will be encouraged to obey rules strictly to avoid life risk
during recreational activities like boating etc.
3
NATURAL
PHENOMENA
P-51
Ans 2.
i = 0° and r = 0°
8
½+½
8
nx = c/vx = 3 × 10 /1.5 × 10 = 2
Ans 3.
2
Yes/Self introspection is essential.
Associated Value : The learners will be able to tackle the adolescent challenges developed
due to enormous physical and emotional changes in their life during this stage.
3
Ans 4.
(a) u = 30 cm, v = 30 cm
This is possible if the object is placed at 2 f
 2f = 30 cm, f = 15 cm
1
(b) u = 15 cm, v = 70 cm is incorrect. This is because, if the object is at focus then image is
formed at infinity.
1
(c) In (iii) case, because object is at the centre of curvature.
Ans 5.
1
(i) This means that the ratio of speed of light in air and speed of light in diamond is 2.42.
1
(ii)
1+1
(iii) Virtual image produced by concave mirror is magnified, than produced by plane
mirror of the same size while the virtual image produced by convex mirror is
diminished.
1
(iv) Real.
1

Ans 1.
(i) (a) Shaving mirror
–
Concave mirror
½
(b) Rear view mirror
–
Convex mirror
½
(c) Reflector in search-lights
–
Concave mirror
½
(ii) (a) Real image can be obtained on screen but virtual image cannot be obtained.
½
(b) Reflected / Refracted rays actually meet where real image is formed while for
virtual they only appear to meet.
½
(c) Real image is always inverted while virtual image is always erect.
Ans 2.
½
u = – 30 cm, f = + 20 cm, h1 = 4.0 cm
v = 60 cm
or,
Ans 3.
h2 = – 8 cm
2+1
(i) u = – 12 cm
f = + 15 cm
Using mirror formula, v = 6·7 cm
½+½
Magnification
½+½
m = 2·5 cm
As the needle is moved farther from the mirror, image moves to the focus and the size
of image goes on decreasing.
½+½
P-52
SCIENCE – X
TERM – 2
(ii) Concave mirrors are used in solar furnaces as they concentrate solar energy in the
focal plane and help in attaining high temperatures.
1
(iii) When one half of a convex lens is covered with a black paper, the lens will produce a
complete image of the object but the intensity of the image is reduced because rays from
the top portion of the lens only, are refracted and form the image.
1

A. Choose the correct option :
Ans 1.
(B) Two
Ans 2.
(C) 5.0 ft.
Ans 3.
(B) 0°
Ans 4.
(C) It bends away from the normal
Ans 5.
(B) It bends towards the normal
Ans 6.
(C) Frequency
Ans 7.
(B) Point-sized
Ans 8.
(C) Concave mirrors
Ans 9.
(A) 66.7 cm, converging
Ans 10.
(B) intensity of image will be less
B. Answer the following :
Ans 1.
O
Ans 2.
Same sized
Ans 3.
deviation
Ans 4.
increase
Ans 5.
Centre
Ans 6.
Convex
Ans 7.
in the centre of
Ans 8.
Principal axis
Ans 9.
30 cm
Ans 10.
200 cm.
(1 mark each)
(1 × 10)

NATURAL
PHENOMENA
P-53
6
HUMAN EYE AND COLOURFUL WORLD
Ans 1.
Ciliary muscles.
1
Ans 2.
25 cm to infinity.
1
Ans 3.
A spectrum is the band of distinct colours we obtain when the white light is split by a
prism. We can recombine the component of white light by passing them through a prism
placed upside down near the given prism. When we pass white light through two
identical prisms held side by side with their refracting edges in opposite directions; the
first prism disperses white light but the second prism recombines them. Thus light
emerging from second prism is white.
2
Ans 4.
2
Ans 5.
Myopic eye can be corrected by using concave lens. A concave lens of suitable power will
diverge the rays a little before they enter the eye. This results in focusing the rays on the
retina. The defect is hence removed.
1
1
P-54
SCIENCE – X
TERM – 2
Ans 6.
(i) Because least distance of distinct vision is 25 cm, and book is kept at a distance of 10
cm.
1
(ii) Given that,
f = 5 cm
Let distance of distrint vision = Minimum distance of object to clear vision = Object
distance = u = – 25 cm,
Use lens formula,
1
1
1
–
=
f
v
u
1
1
1
–
=
v
5
 25
1
1
1
+
=
v
25
5
51
1
1
1
4
= –
=
=
v
5 25
25
25
v=
Ans 7.
25
cm
4
25
1
v
=+ 4 =
Magnification m = +
u
4
 25
Here, image distance, v = – 75 cm 2
object distance, u = –25 cm So, the focal length will be 1/f = 1/v – 1/u or 1/f = 1/25 – 1/75
or 1/f = 0.027
Thus, f = 37.03 cm = 0.37 m
The power of the lens would be, P = 1/f
So, P = 1/0.37 = 2.70 Dioptre
3

Ans 1.
(i) Least deviation – Red.
½
(ii) Maximum deviation – Violet.
½
Ans 2.
The pupil helps to control the amount of light entering the eye.
1
Ans 3.
The angle between the extended incident ray and the emergent ray is called the angle of
deviation.
1
This is because the different colours travel through a glass prism at different speeds.
Ans 4.
1
(i) It is the ability of the eye lens to adjust its focal length, according to the distance of an
object from it.
1
(ii) In aged people, the flexibility of the eye muscles decreases and the power of
accommodation reduces.
1
NATURAL
PHENOMENA
P-55
Ans 5.
Rainbow is typically observed soon after a rain shower. It is caused due to dispersion of
sunlight by raindrops suspended in air.
1
Water drops act like prisms. Rainbow is obtained in the direction opposite to that of the
sun.
1
Water drops, refract the sunlight, disperse, reflect internally and then refract it again. It is
due to dispersion of light that the eye sees different colours in a series in the sky. This is
rainbow.
Rainbow formation
Ans 6.
1
(i) Incorrect
3- turmeric (yellow)
½+½
1- chilli powder (red)
(ii) Copper sulphate—blue colour—position—5
Green—position—4
½+½
(iii) Incorrect, violet bends the most.
Ans 7.
½+½
Don’t Rub eyes.
Wash them properly and softly (or Any other suggestions)
Associated Value : The learners will follow the tips to maintain the good eye care.
3

Ans 1.
The person is suffering from hypermetropia.
1
Ans 2.
It enables us to see and perceive the object in our vicinity.
1
Ans 3.
The process of splitting up of white light into its constituent colour as it passes through a
refracting medium is known as dispersion of light.
1
It bends towards the base of the prism after passing through it.
Ans 4.
1
The nearest point to the eye at which an object is visible distinctly is called the near point
(25 cm).
1
The maximum distance upto which the normal eye can see things is called the far point
(infinity).
1
Ans 5.
(i) Reciprocal of focal length expressed in metres is called power. SI unit is dioptre.
1
(ii) P = 1/f
f = 1/p =1/400
P-56
SCIENCE – X
TERM – 2
1/v – 1/u = 1/f
1/40 – (–1/u) = 1/400
1/u = 1/400 – 10/400
1/u = – 9/400
u = – 400/9
u = – 44.4
2
Ans 6.
The sun rays have to travel through a larger atmospheric distance. As b < r, most of the
blue light and shorter wavelengths are removed by scattering. Only red colour, which is
least scattered is received by our eye and appears to come from sun. Hence the appearance
of sun at sunset or sunrise or full noon near the horizon may look reddish.
3
Ans 7.
(a) Student is suffering from myopia.
½
½
Two possible causes are :
(i) Due to excessive curvature of lens.
½
(ii) Elongation of eye ball.
½
(b)
1
(c) We can initiate other people to participate in the camp.
As a human being we should also register our eyes for donation after death.
1+1

Ans 1.
Blue.
1
Ans 2.
In the hypermetropic eye, the light focuses behind the retina instead of focussing directly
on the retina. 1
Ans 3.
This is because of scattering of light.
1
Light from the sun near the horizon passes through larger distance in the earth's atmosphere
before reaching our eyes. So, most of the blue light and shorter wavelengths are scattered
away by the particles. Therefore, the light that reaches our eyes is of shorter wavelengths.
This gives rise to the reddish appearance. But during the day shorter wavelength are
scattered more by molecules of air. Scattered blue light enters our eyes.
1
NATURAL
PHENOMENA
P-57
Ans 4.
(i) Dispersed white light consists of seven colours of different wavelengths.
These components of light travel with different speeds in glass and get refracted by
different angles. Due to this, splitting of white light takes place.
1
Ans 5.
(ii) Violet, Indigo, Blue, Green, Yellow, Orange, Red.
1
(i) Night blindness.
1
(ii) (a) Proper diet provide Vitamin A to reduce this defect.
1
(b) Taking care of eyes for proper functioning of rod-shaped cells.
Ans 6.
1
(i) The eye lens of human eye is made up of a crystalline lens of variable focal length. The
focal length of human eye can be changed by the action of ciliary muscles to form the
images of the objects at different positions on the retina. This can be achieved by
relaxing the ciliary muscles for distant object and contracting the ciliary muscles for
nearby object.
2
(ii) In the morning, the sunlight covers a larger distance and thick layers of atmospheres,
so except the red colour light, most of the other colours are scattered into the atmosphere.
As only red colour light reaches our eye, the morning sun appears to be red.
2
(iii) Due to the absence of atmosphere, no scattering occurs and sky appears dark.
1

Ans 1.
The splitting of light into its component colours is called dispersion of light.
1
Ans 2.
Size of the particles of the medium through which it is passing.
1
Ans 3.
(i) Hypermetropia / far-sightedness / long-sightedness.
½
By using a convex lens.
½
(ii) (a) Increase in focal length of eye lens.
½
(b) Shortening of the eyeball.
Ans 4.
½
Myopia
½
Concave lens
½
f =1/P = 1/(– 2.5) =– 0.4 m.
Ans 5.
1
(a) The distance upto which normal eye can see clearly without putting any strain on
muscles of eye.
1
(b) Controls the size of the pupil.
Ans 6.
1
The various phenomena observed in nature due to scattering of light are as following :
(i) Sun appears red near the horizons (during sunrise and sunset) and white when seen
overhead.
(ii) Bluish colour of the sky.
(iii) Visible path of light as it enters a dark room.
(iv) Danger signals or stop signals are usually red.
(v) Blueness of distant mountains.
(vi) Sunlight filtering through clouds.
Ans 7.
½×6=3
(i) In the region above the fire, the hot air is lighter than the cool air above it. Further the
refractive index of hotter region is less than the cooler region and it is not static.
1
These physical conditions are not steady. Subsequently the image when seen through
the hot region is not stationary. The apparent position of the object fluctuates. This
gives the wavering effect to the image.
1
P-58
SCIENCE – X
TERM – 2
(ii) The phenomenon of obtaining a spectrum of colours by passing white light through
the prism is known as dispersion.
½
Dispersion occurs because the refraction of different colours travel with different
speeds in a refracting medium. Speed of violet colour is least and that of the colour is
the most. Hence the refractive index of the medium is largest for violet colour and least
for the red colour.
1
(iii) As a result, the refraction or bending of violet colour is the maximum and that of red
colour is minimum. This difference in the extent of bending of different colour of light
causes dispersion of white light into its constituent colours as they emerge out of
prism. Red colour bends the least and violet bends the most.
1½

Ans 1.
Due to change in atmospheric conditions, density changes so position keeps on changing.
1
Ans 2.
To regulate the amount of light entering the eye.
1
Ans 3.
(i) Presbyopia.
1
(ii) He shall have to use both kinds of lenses. Convex lens for long-sightedness and
concave lens for short-sightedness.
1
Ans 4.
Eye lens is made up of fibrous material. Its curvature can be changed by ciliary muscles
which changes its focal length. When muscles are relaxed, the lens becomes thin and focal
length increases and the eye is able to see distant objects. Similarly when ciliary muscles
contract, focal length decreases and eye is able to see nearby objects.
1
The minimum distance at which objects can be seen distinctly is called distance of distinct
vision. This is 25 cm.
1
Ans 5.
(a) Myopia
1
(b)
1
Ans 6.
Retina contains light sensitive cells known as rod and cones. These cells get activated
upon illumination and generate electrical signals or pulses.
1
The electrical signals are sent to the brain through optic nerves.
1
In the brain, the signals are processed, interpreted and the objects in front of the eye are
perceived.
1
Ans 7.
(i) Presbyopia.
½
(ii) v = – 50 cm u = – 25 cm
NATURAL
PHENOMENA
½
P-59
Using lens equation, 1/f = 1/v – 1/u 1/f = –1/50 – 1/(– 25)
1/f = –1/50 + 1/25 1/f = 1/50 P = 1/f (in m) = 100/50 = 2 D
2
(iii)
1+1

Ans 1.
The ability or the property of the eye lens to adjust its focal length in order to be able to focus
both near and distant objects is known as the power of accommodation.
1
Ans 2.
It is like a camera having a lens system forming an inverted, real image on the light
sensitive screen, retina inside the eye.
1
Ans 3.
The molecules of air and other fine particles in the atmosphere have size smaller than the
wavelength of the visible light.
1
These are more effective in scattering light of shorter wavelength at the blue end than the
light of longer wavelength at the red end. Thus, the blue colour is due to the scattering of
sunlight through fine particles in air.
1
Ans 4.
In bright light, the size of the pupil is small to control the amount of light entering the eye.
1
When we enter a dim room, it takes some time so that the pupil expands and allows more
light to enter and helps to see things clearly.
1
Ans 5.
(i) It is the minimum distance to which, normal eye can see two similar object distinctly.
1
(ii) Eye lens is comparatively thicker, while reading a book.
Ans 6.
1
Here, image distance, v = –75 cm object distance, u = –25 cm So, the focal length will be 1/f = 1/v – 1/u or 1/f = 1/25 – 1/75
or 1/f = 0.027
thus, f = 37.03 cm = 0.37 m
P-60
SCIENCE – X
TERM – 2
The power of the lens would be, P = 1/f
Ans 7.
so, P = 1/0.37 = 2.70 Dioptre
3
(i) Spectacle having bifocal lens.
1
(ii) u = , v = – 150 cm.
Using lens formula,
1/f = 1/v – 1/u
1/f = 1/v – 0
f = v  f = – 1.50 m.
Power of lens
P = 1/f(m) = 1/(–1.5) = – 0·67 D
2
Nature of lens : Concave.
(iii)
2

Ans 1.
The focal length of eye lens cannot decrease below a certain minimum limit.
1
Ans 2.
Defect : Myopia, Nature of lens : Concave/diverging.
Ans 3.
At the time of sunrise or sunset, the sun is near the horizon. The rays from the sun have to
travel a much larger part of atmosphere to reach the earth.
2
Ans 4.
(i) Parallel rays of different colours deviate differently while passing through a glass
prism because angle of refraction of different colours are different. Hence difference in
refractive index takes place.
1
(ii) (a) Total reflection, (b) refraction.
Ans 5.
½+½
½+½
Continuously changing atmosphere refract light from the stars by different amount from
one moment to the other. When atmosphere refracts more starlight towards us, the stars
appears to be bright and when the atmosphere refracts less starlight, then the stars appear
to be dim.
1
However the planets are nearer to us than the stars, they appears to be comparatively
bigger to us. So, they cannot be considered as a point source, hence no twinkling is seen.
1
Ans 6.
NATURAL
When the sun is overhead at noon, then the light coming from the sun has to travel a
relatively shorter distance through the atmosphere to reach us. During this shorter journey
of sunlight, only a little of the blue colour of the white light is scattered. Since the light is
coming from the overhead, sun has almost all its component colours in the right proportion,
therefore the sun in the sky overhead appears white to us.
3
PHENOMENA
P-61
Ans 7.
The boy uses spectacles of focal length of –50 cm. Negative sign of focal length means he is
using concave lens. Thus he has myopic vision defect. The power of a lens is given by
power of a lens in diopter = 1/focal length of the lens in meter = 1/(– 0.5) = –2 Diopter. 3
Ans 8.
(i) Myopia, short-sightendness.
1
(ii) Lens defect (increased thinness), Eye ball defect (shortening)
2
(iii) Frendship, concern for each other, value and balanced diet.
2

Ans 1.
(i)
1
(ii) (a) Elongation of the eye ball
(b) Decrease in focal length of eye lens.
(iii) u = , v = – 1·5 m
P = 1/f = 1/(–1.5)
= – 0·66 D
½
½
Ans 2.
(i) Atmospheric refraction
(ii) Scattering of light
(iii) Atmospheric refraction.
1
1
1
Ans 3.
Atmospheric refraction causes the sun to be visible even when it is below the horizon. At
the time of sunset or sunrise, the light coming from the sun is highly oblique.
1
½
½
The rays of light from the sun below the horizon pass through a greater distance in the
atmosphere and gets bent to an observer such that they seem to be coming from above the
horizon.
1
Thus, even when the sun’s position is actually just below the horizon before the sunrise or
after the sunset, the sun is visible to us.
1
Ans 4.
Yes, (Independent view of student).
Associated Value : The learners will be motivated/inspire to appreciate the idea of organ
donation in their life.
3
Ans 5.
(i) Presbyopia.
1
(ii) (a) Weakening of ciliary muscles.
1
(b) Reducing ability of the lens to change the curvature.
1
(iii) (a) To make people aware of eye diseases.
1
(b) To tell people to take proper and balanced diet.
1

Ans 1.
The phenomenon of change in the direction of propagation of light caused by the large
number of particles present in the atmosphere is called scattering of light.
Example : The path of beam of light becomes visible through a colloidal solution due to
scattering of light.
2
P-62
SCIENCE – X
TERM – 2
Ans 2.
It is due to scattering of light. Light near the horizons passes thicker layers of air and larger
distance of atmosphere. Hence most of the blue light is scattered away and longer
wavelengths reach our eyes, giving rise to reddish colour of sun.
2
Ans 3.
Myopia (Reason).
Get the proper checkup for the eyes.
Should wear spectacles as recommended by doctor.
Associated Value : The learners will be able to handle eye sight disorder if any in their life.
3
Ans 4.
Yes.
Yes (Independent view of a student.)
Associated Value : The learners will understand the fact that the urban people will have to
keep their city clean & pollution free if they wish to experience some beautiful natural
events viz twinkling of stars, rainbow etc.
3
Ans 5.
(a) The person is suffering from hypermetropia.
1
2
Two possible causes :
(i) Greater focal length of the lens.
½
(ii) Eye ball becoming smaller.
½
(b) Eye donation advertisements are important as :
(i) They make the people aware about donation of organs after their death.
(ii) Sympathetic nature towards others.
½+½

NATURAL
PHENOMENA
P-63
A. Choose the correct option :
Ans 1.
(A) Rarer to denser medium
Ans 2.
(C) Atmospheric refraction
Ans 3.
(A) Elongation of eye-ball
Ans 4.
(B) –0.66 D
Ans 5.
(B) Angle of deviation
Ans 6.
(C) Inverted
Ans 7.
(D) Fovea-centralis
Ans 8.
(D) All of these
Ans 9.
(C) least scattered by fog or smoke
Ans 10.
(B) Black
(1 mark each)
B. Fill in the Blanks :
Ans 1.
short
Ans 2.
far point
Ans 3.
accommodation
Ans 4.
ciliary muscles
Ans 5.
Red
Ans 6.
scattering
Ans 7.
10
Ans 8.
greater
Ans 9.
atmospheric refraction
Ans 10.
gathering.
(1 × 10)

P-64
SCIENCE – X
TERM – 2
7
MANAGEMENT OF NATURAL RESOURCES
Ans 1.
Ans 2.
Ans 3.
Ans 4.
Ans 5.
Ans 6.
Ans 7.
Ans 8.
(a) By segregating biodegradable and non-biodegradable waste.
½
(b) By reusing and recycling the non-biodegradable waste.
½
Without the regular grazing by sheep, the grass first grows very tall and then falls over
preventing fresh growth.
1
Water harvesting at the community level is capturing, collection and storage of rain water
and surface run off for filling either small water bodies or recharging ground water. This is
carried out through water shed management, check dams, earthern dams, roof top
harvesting and filter wells in flood drains.
Benefits :
(i) It ensures water availability in non-rainy season.
(ii) Water becomes available for drinking as well as irrigation.
1+1
(i) By switching off unnecessary lights and fans to save electricity.
(ii) By repairing leak taps to save water.
(iii) Plastic, paper, glass and metal items and recycling these materials to make required
things.
(iv) The used envelopes can be reversed and can be used again instead of throwing away.
½+½+½+½
While the environment is preserved, the benefits of controlled exploitation should also go
to the local people. Forest resources ought to be used in a manner that is both
environmentally and developmentally sound, a process in which decentralised economic
growth and ecological conservation go hand in hand. Human intervention has been very
much part of the forest landscape.
3
(i) As their formation is extremely slow, hence these resources will be exhausted.
1
(ii) Two causes are :
(a) Loss of vegetation cover and dumping of urban waste.
1
(b) Diversion for high water demanding crops and pollution from industrial effluents.
1
(i) Coal and Petroleum.
½
Formed from the degradation of biomass millions of years ago.
½
(ii) (a) Preservation of biodiversity.
½
(b) Forest products.
½
Reasons for depletion :
(a) Indiscriminate falling of trees.
½
(b) Destroying biodiversity to produce one particular kind of trees.
½
(i) Reusing is even better than recycling because the process of recycling uses some
energy. In the ‘reuse’ strategy, you simply use things again and again, without using
any energy for generating something new.
2
(ii) (a) Clearing of forest land for agriculture.
(b) Building of roads through forests.
(c) Falling of large number of trees for furniture.
1×3=3

NATURAL
RESOURCES
P-65
Ans 1.
It is a group of bacteria found in human intestine whose presence in water indicates
contamination by disease causing micro-organisms.
1
Ans 2.
Mining causes pollution because of the large amount of the slag which is discarded for
every tone of the metal extracted.
1
Ans 3.
They are in no way dependent on forests, but have a considerable stay in their management.
They recognize the need to preserve biodiversity as a whole.
2
Ans 4.
(i) Fossil fuels are conventional sources of energy that will not last longer.
(ii) To reduce air pollution.
Ans 5.
1+1
(i) These are non-renewable and causes pollution.
1
(ii) They contribute to global warming.
1
Ans 6.
The natural resources are limited. Due to increasing human population, the demand and
exploitation of natural resources have increased. The management of natural resources
requires a long-term perspective so that the natural resources last for the generations to
come. The management should also ensure equal distribution of resources so that all can
be benefitted from the development of these resources.
3
Ans 7.
(i) Petroleum and coal are composed of carbon, hydrogen, nitrogen and sulphur.
½+½+½+½
(ii) (a) Local people gets benefit from forest produce.
(b) Wild life and nature can be conserved.
Ans 8.
½+½
(i) (a) The natural habitats of wild animals and birds should be preserved by establishing
national parks and sanctuaries.
1
(b) To conduct a periodic surveys of national parks and bio-reserves.
1
(ii) (a) Hunting of tigers should be banned.
1
(b) A compaign like paper dipping should be arranged for the awareness of the
people.
1
(c) The survival of captive-bred animals in the wild life should be proper.
1

Ans 1.
Forest.
1
Ans 2.
Plastic bottles can be reused and also can be recycled to make required things.
1
Ans 3.
Our natural resources are limited. With the rapid increase in human population, due to
improvement in health care, the demand for all resources is also increasing. Sustainable
management is necessary to provide the economic well -being to the present and the future
generations and to maintain a healthy environment and life support system.
Reuse can be practiced by use, i.e., as we can reuse plastic bottles and envelopes.
Ans 4.
2
(i) When combustion takes place, oxides of sulphur, nitrogen and carbon monoxide are
formed, which are poisonous at high concentrations.
1
(ii) CO2 is a greenhouse gas which leads to global warming.
Ans 5.
1
Coal and petroleum were formed from the degradation of biomass millions of year ago.
1
Other elements present are hydrogen, nitrogen, sulphur and oxygen.
P-66
SCIENCE – X
½+½
TERM – 2
Ans 6.
If resources are used in accordance with short term aims, present generation will be able to
utilize the resources properly for growth and development.
But if we use resources with long term aims, future generations will also be able to utilze
resources for fulfilling its need. Thus, it would be better to use our natural resources with
a long-term perspective so that it could be used by the present generation as well as
conservation for future use.
1+1+1
Ans 7.
We must conserve our forest because :
(i) They are source of raw materials for our industries.
(ii) They provide us with wood to make furniture and buildings.
(iii) They are source of medicine.
(iv) They provide firewood and food. (any three)
1½
Causes of deforestation are :
(i) Overgrazing.
(ii) Building roads and dams.
(iii) Natural calamities like forest fire. (any three)
Ans 8.
1½
(i) By 1983, the value of the previously worthless Sal forests in the Arabari forest range of
Madnapur district was estimated to be about 12.5 crores.
1
(ii) Rain water harvesting.
1
(iii) (a) Reduced availability of forest products.
(b) Deterioration in the quality of soil.
(c) Reduction in the sources of water.
1×3=3

Ans 1.
(i) Roof top rain water harvesting.
½
(ii) Watershed management.
½
Ans 2.
Coliform.
1
Ans 3.
The three R’s to save environment are :
Reduce, Reuse and Recycle.
1
Reuse : In Reuse strategy, we can simply use things again and again, without using any
energy for generating something new.
½
Examples : Instead of throwing away used envelopes, we can reverse them and use them
again.
½
Ans 4.
Ans 5.
Ans 6.
Coal and Petroleum.
½+½
These fuels were made by decomposition of biomass over millions of years ago.
1
(i) Water spreads to recharge wells.
½
(ii) Provides moisture to vegetation over a wide area.
½
(iii) Does not provide breeding ground to mosquitoes.
½
(iv) Protected from contamination.
½
(i) Conservation of forest into National Parks and Sanctuaries.
(ii) Take help of local people in conserving forest.
(iii) Do not allow the destruction of forests for making roads, dams and hotels etc.
1+1+1
NATURAL
RESOURCES
P-67
Ans 7.
The advantages are :
(i) Watershed management only increases the production and income of the watershed
community.
(ii) It also migrates droughts and floods.
(iii) It increases the life of the downstream dam and reservoirs.
Ans 8.
Ans 9.
1+1+1
(i) Khadins and Tanks.
1
(ii) Bandharas and Tals.
1
(iii) Ahars and Pynes.
1
(i) The benefits of recycling natural resources are :
(a) Prevents wastage of potentially useful materials.
1
(b) Reduces the consumption of fresh raw materials.
1
(c) Reduces energy usage.
1
(d) Decrease environmental pollution.
1
(ii) Government of India has recently instituted an Amrita Devi Bishnoi National Award
for wildlife conservation in the memory of ‘Amrita Devi Bishnoi’.
1

Ans 1.
Sardar Sarovar Dam.
1
Ans 2.
Plants carry out photosynthesis, which converts the energy of the sun into a form which
can be used by the rest of the living world.
1
Ans 3.
(i) Displace large number of peasants and tribals without adequate compensation or
rehabilitation.
1
(ii) They swallow up huge amounts of public money.
Ans 4.
1
(a) Chipko movement was started in early 1970s in village Reni in Garhwal by the women
of Uttarakhand to stop felling of forest trees of their area.
1
(b) We should conserve forests because they maintain biological diversity, provide food
and safeguard the future of the tribals besides providing ecological balance of
ecosystems.
1
Ans 5.
It is necessary to conserve forest and wildlife because :
(i) Forest provides us oxygen and causes rainfall.
(ii) Forest prevents soil erosion.
(iii) Forest provides us medicines and raw materials for industries.
Ans 6.
1+1+1
Fossil fuels like coal and petroleum have to be used very carefully due to the following
reasons :
(i) They are present in extremely limited quantity.
(ii) Their combustion produces harmful gases such as oxides of nitrogen and sulphur
and a green house gas i.e., carbon dioxide.
(iii) The huge reservoirs of carbon present in fossil fuels will be converted into carbon
dioxide leading to increased global warming.
1+1+1
Ans 7.
P-68
Sustainable development is the use of components of biological diversity in a way that
does not interfere with the natural functioning of ecological processes and benefits the
present generation. This maintain its potential to meet the needs and aspiration of the
future generations. For sustainable development, conservation and management of the
natural resources are very necessary.
3
SCIENCE – X
TERM – 2
Ans 8.
(a) Fuels formed from the decomposition of ancient animal and plant remains millions of
years ago and which provides energy on combustion are called fossil fuels.
1
For example : coal and petroleum.
1
(b)
Khadin system of water harvesting
3

Ans 1.
The conservation of forest and wildlife is necessary to restore the ecological balance.
1
Ans 2.
Since plastic is non-biodegradable, it cannot be degraded by the micro-organisms and
hence cannot be recycled using the method of composting.
1
Ans 3.
Water harvesting is a practice of collecting rain water in safe storage places from where it
can be used throughout the year, both for drinking and irrigation purposes.
1
The two water harvesting structures are :
Ans 4.
(i) Khadin system in Rajasthan.
½
(ii) Kulhs in Himachal Pradesh.
½
(i) Air pollution is caused by burning fossil fuels.
½
(ii) CO2 produced by burning fossil fuels produces greenhouse effect.
½
(iii) They are non-renewable sources of energy.
½
(iv) The oxides of carbon, nitrogen and sulphur that are released on burning fossil fuels
are acidic oxides which lead to acid rain.
½
Ans 5.
(i) The people living in or around the forests who are directly dependent on forest
product.
(ii) The Forest Department of the government which owns the land and controls the
resources from forests.
(iii) The industrialists who use the forest products, but are not dependent on the forest of
a particular area.
(iv) The wildlife and nature enthusiasts who want to conserve nature in its present form.
4× ½=2
Ans 6.
In order to check soil erosion one may take the following steps :
(1) By preventing intensive cropping.
(2) Implementing terrace farming in hilly region.
(3) By doing afforestation.
(4) By proper drainage in the given region.
NATURAL
RESOURCES
3
P-69
Ans 7.
The difference between national park and sancturies are :
National Park
Sancturies
1. Its size is generally small ranging from
100 to 500 square kilometer.
2. Its boundary is fixed.
3. Its main aim is to provide habitat for
particular wild animals.
Ans 8.
Ans 9.
Its size ranges from 500 to 1000 square
kilometer.
Its boundary is not fixed.
Its aim is to provide habitat for general
species.
3
The major sources of water pollution are given as follows :
(1) Pollution through discharge of industrial waste in rivers.
(2) Sewage discharge from urban cities.
(3) Spilling of oil by ships in sea.
(4) Use of excessive fertilizers.
(5) Use of excessive herbicides.
(6) Use of excessive pesticides.
3
(i) (a) It does not evaporate.
(b) It spreads out to recharge wells and provides moisture for vegetation over a wide
area.
(c) It does not provide breeding grounds for mosquitoes such as by human and
animal waste.
1×4=4
(ii) The local people and their future generations are greatly dependent on the forest
products for various needs such as food, shelter, medicines etc. Hence they are
concerned about the welfare of the forests in their area.
1

Ans 1.
Surangams in Kerala and Kattas in Karnataka.
½+½
Ans 2.
Carbon dioxide is the greatest contributor of global warming.
1
Ans 3.
(i) Ground water level increases.
1
(ii) Ground water keeps the layers of soil above it moist and prevents loss of water by
evaporation.
So that they can last for generations to come since their source is limited. To ensure equal
distribution of natural resources to all people.
1
Ans 4.
(i) Since it contains different species of plants, animals, bacteria, fungi, etc.
(ii) Loss of biodiversity leads to loss of ecological stability.
Ans 5.
Nitrogen and sulphur.
½+½
Carbon monoxide which is highly poisonous.
Ans 6.
1
1
1
(1) Growing plants and trees in the open area in the school.
(2) Arrangement for water harvesting
(3) Reporting any kind of water leakage in the school.
Associated Value : The learners will be motivated to act like a eco-club members and
participate actively in its all environment saving activities.
3
Ans 7. The main reasons behind the pollution in river Ganga are given as follows :
(1) The Ganga runs its course of over 2500 km from Gangotri in the Himalayas to Ganga
Sagar in the Bay of Bengal. It is being turned into a drain by more than a hundred
P-70
SCIENCE – X
TERM – 2
towns and cities in Uttar Pradesh, Bihar and West Bengal that pour their garbage and
excreta into it.
Ans 8.
(2) Largely untreated sewage is dumped into the Ganges every day.
(3) The pollution caused by other human activities like bathing, washing of clothes and
immersion of ashes or unburnt corpses.
(4) Industries contribute chemical effluents to the Ganga’s pollution load.
(5) Large amount of tourist activities taking place along the banks of river ganga are also
responsible for its pollution.
(any 3 points) 3
Stakeholders have the various purposes :
(i) Local people need large quantities of firewood, small timber and also forest products. (ii) The forest department of the Government for the proper maintenance of the forest
resources. (iii) The industrialists depends for raw materials. Ans 9.
Ans 10.
(iv) The wildlife and nature lovers need conservation of natural resources in its perfect
form.
3
Rise in the population has increased the demand for resources like food, water, etc. In
order to meet these demands of the increasing population, resources are over- exploited.
Since these resources are limited, therefore, sustainable management is necessary, if these
resources need to be continually used by us as well as by the future generation. 3
(i) The local people obtain large quantities of firewood, timber and thatch from the
forests.
1
(ii) Bamboo is used to make slats for huts and baskets for collecting and storing food
materials.
1
(iii) Implements for agriculture, fishing and hunting are largely made of wood.
1
(iv) Forests are sites fishing and hunting.
1
(v) Also they gather fruits, nuts and medicines from the forests.
Ans 1.
1

Ans 3.
Ans 4.
Giving people control over their local water resources ensures that mis-management and
over-exploitation of the water resources can be reduced or removed.
1
The government of India has recently instituted the ‘Amrita Devi Bishnoi Wildlife Protection
Award’ for individual or communities from rural areas that have shown extraordinary
courage and dedication in protecting wildlife.
1
(i) Carbon, (ii) Hydrogen, (iii) Nitrogen, (iv) Sulphur.
½+½+½+½
(i) Use tubelights or CFLs rather than bulbs.
Ans 5.
(ii) Switch off electrical appliances when not in use.
(iii) In winters we use extra clothing rather than heaters.
(i) Flooding of neighbouring areas.
Ans 6.
(ii) Displacement of locals.
(iii) Increased soil salinity.
(1) Yes.
Ans 2.
(any 2 points) 1+1
(any 2 points) 1+1
1
(2) Global warming, degradation of environment health hazards.
1
Associated Value : The learners will appreciate the fact that rational use of natural
resourses is their responsibilities towards saving has nature for future generations as
well.
1
NATURAL
RESOURCES
P-71
Ans 7.
Ans 8.
Ans 9.
Water harvesting can be done in a variety of ways :
— Capturing runoff water from rooftops. — Collecting runoff water from local catchments. — Detaining seasonal floodwaters from local streams. — Conserving water through watershed management.
3
Unequal distribution of water can lead to several problems like
(1) Supply of poor quality of water which may lead to health hazards.
(2) Rise of conflicts between the people.
(3) Increase in environmental issues like drought and floods.
(4) Effects the survival of plants and animals species.
3
(i) Coliform : A group of bacteria found in human intestines, whose presence in water
indicates contamination by disease-causing micro-organisms.
(ii) Recycling : The act of processing used or abandoned materials for creating new
products.
(iii) Sustainable development : A pattern of resources used for obtaining economic and
social growth of the present generation while preserving the resources for the needs of
future generations.
(iv) Chipko Andolan : A grass root level movement in which the villagers used to hug the
forest trees and prevent their mass felling by the contractors.
(v) Watershed management : A scientific method of developing land and water resources
to increase the biomass production without causing ecological imbalance. 1 × 5 = 5

A. Choose the correct option :
Ans 1.
(A) Floods and droughts
Ans 2.
(B) Remediation
Ans 3.
(B) West Bengal
Ans 4.
(B) Carbon dioxide
Ans 5.
(A) Biodiversity hotspots
Ans 6.
(A) Reduce, Recycle, Reuse
Ans 7.
(B) Himanchal Pradesh
Ans 8.
(D) Indira Gandhi Canal
Ans 9.
(A) Sugarcane
Ans 10. (A) Non-renewable
B. Fill in the Blanks :
Ans 1.
Fuel
Ans 2.
'Rare'
Ans 3.
Carbon monoxide
Ans 4.
Katta
Ans 5.
Oxides
Ans 6.
Dams
Ans 7.
Eris
Ans 8.
Irrigation
Ans 9.
Canals
Ans 10. 25%
(1 mark each)
(1 × 10)

P-72
SCIENCE – X
TERM – 2
8
OUR ENVIRONMENT
Ans 1.
Biodegradable wastes are degraded by saprophytic organisms or decomposers but nonbiodegradable waste are not degraded by decomposers, so they should be kept in separate
dustbins.
1
Ans 2.
Industrial effluents, domestic sewage, emission from vehicles causes pollution.
Ans 3.
An improper disposal of wastes means addition of pollutants into the environment—air,
water and soil. They will harm living beings, human assets and human beings. For
example, passage of sewage into water body will cause entrophication, stink development
of sludge, killing of animals and source of water borne pathogens.
2
Ans 4.
The great Himalayan National Park contains within its reserved area, alpine meadows
which were grazed by sheeps in summer. But now without regular grazing by sheep, the
grass first grows very tall and then falls over, preventing fresh growth.
2
Ans 5.
(i) Gardening and planting trees.
½
(ii) Use of gunny bags / paper bags in place of polythene.
½
(iii) Use of compost and vermicompost in place of fertilizers.
½
(iv) Seperation of biodegradable and non-biodegradable substances.
½
Ans 6.
1
(i) Green plants have chlorophyll in their leaves due to which they can synthesize their
own food in sunlight.
1
(ii) Consumers are the organisms which depend on others for their survival, or food.
Consumers are herbivores, carnivores or omnivores. Man is omnivore which depend
on others for its food either plants or flesh.
2
Ans 7.
(a) An aquarium is an artificial and incomplete ecosystem in contrast to a pond / lake
which is natural, self-sustaining and complete ecosystems.
1
(b) The harmful byproducts are gases such as SO2 and NO2. They cause extensive air
pollution and are responsible for acid rain.
1
(c) Crop fields are man-made and some biotic and abiotic components are manipulated
by humans.
1
Ans 8.
(i) The full form of :
(a) UNEP : United Nations Environment Programmes.
1
(b) CFCs : Chlorofluoro Carbons.
1
(ii) Organisms can grouped as producers, consumers and decomposers according to the
manner in which they obtain their sustenance from the environment.
1
(iii) Two problems that would arise in absence of decomposers in an ecosystem are :
(a) Decomposition of garbage as well as dead plants and animals will not take place.
1
(b) Natural replenishment of soil will not take place.
1

Ans 1.
Chlorofluoro carbons.
It depletes ozone layer.
NATURAL
RESOURCES
½+½
P-73
Ans 2.
To minimize the use of chemicals / pesticides in agriculture.
1
Ans 3.
(i) Energy from sun flows into autotrophs then to heterotrophs and then to decomposer.
1
(ii) The energy does not revert from autotrophs to solar input.
Ans 4.
1
(a) Number of deer increases which will result in less amount of grass leading to soil
erosion.
(b) Food available for lions would be less. Amount of grassland will increase.
Ans 5.
1+1
(i) Biomagnification is maximum in fourth trophic level i.e., fish.
1
(ii) (a) Caused by the use of several pesticides and chemicals for crop protection and
storage.
(b) It harms the trophic level and causes toxicity and death.
(c) It can be prevented by minimising the use of chemicals in food productivity.
1+1
Ans 6.
Impact of removing all organisms of a trophic level is different for different trophic levels.
1
For examples, if we remove producers from a food chain, no organism will get food, if we
remove herbivores from a food chain, then carnivores will die and producers will also die
due to competition for space and nutrients. If all the organisms of a trophic level are
removed, it will create an imbalance in the ecosystem.
2
Ans 7.
Waste generated in our houses daily :
(i) Kitchen waste, (ii) Plastic bags, (iii) Vegetable/fruit peels/rinds.
1½
Measures for disposal :
(i) Segregation of biodegradable and non-biodegradable wastes.
(ii) Safe disposal of plastic bags and paper wastes for recycling.
(iii) Prepare a compost pit for kitchen wastes.
1½

Ans 1.
Biological magnification.
1
Ans 2.
Oxygen in the presence of Ultraviolet rays splits oxygen molecule into two oxygen atoms.
1
Ans 3.
Organisms which feed upon small carnivores and constitute the fourth trophic level.
Examples—hawk, lion.
Ans 4.
1+1
Plastic bags are non-biodegradable wastes, they are not degraded, they begin to stink,
emitting foul gases spoil the beauty of places and contaminate soil, water and air with
toxins.
Cloth bags are used instead of plastic bags because they are stronger, more durable and
washable.
1+ 1
Ans 5.
P-74
Substance which can be degraded and disposed off naturally by saprophytic organisms
or decomposers are called biodegradable, e.g., organic remains, garbage, sewage, livestock
waste. Substances which cannot be degraded by saprophytes are known as nonbiodegradable. They are mostly man-made articles like pesticides, plastic, polythene etc.
Biodegradable articles are formed naturally in biosphere. Decomposer organisms feed on
them by secreting digestive juices and absorbing the solubilised substances. Biogenetic
nutrients are released in the process called mineralization.
3
SCIENCE – X
TERM – 2
Ans 6.
(i) Ecosystem is a self-contained segment of nature which consists of a distinct biotic
community and specific abiotic environment, both interacting and exchanging
materials between them.
The two components are biotic and abiotic factors.
2
(ii) An aquarium is an artificial system which is also incomplete due to absence of
producers, food chains and decomposers. There is no recycling and self- cleaning.
However, a pond or a lake is a self- sustained, natural and complete ecosystem where
there is perfect recycling of nutrients.
1
Ans 7.
(i) (a) It is exhausting our fossil fuels.
1
(b) It increases deforestation.
1
(ii) Planting trees, reducing the use of non-renewable sources of energy, saving water,
proper garbage disposal, less use of chemicals, cleaning society.
3

Ans 1.
Ozone is a molecule formed by three atoms of oxygen, which shields the surface of the
earth from ultraviolet radiations.
1
Ans 2.
Abiotic and biotic components.
Ans 3.
(i) Noise and air pollution are increasing.
½
(ii) Waste leads to water pollution.
½
(iii) SO2, NO2 etc. emitted by the industries are toxic.
½
½+½
(iv) Radioactive radiations emitted by nuclear power stations are toxic to living organisms.
½
Ans 4.
In a food web, each organism is generally eaten by two or more other kinds of organisms
which in turn are eaten by several other organisms.
So, a food web is a complex network of many interconnected food chains and feeding
relationships.
2
Ans 5.
Cause of Concern : Ozone layer present in the stratosphere has thinned out by about 8%
over the equator and more so over the Antarctica where a big ozone hole appears every
year. This has increased the level of UV-B radiations reaching the earth by 15-20 %. These
radiations are causing increased number of skin cancers, cataracts and reduced immunity
in human beings. These causes increased incidence of blindness of animals, death of
young ones, reduced photosynthesis, higher number of mutations and damage to articles.
2
Steps to Limit Damage :
(i) Ban on the production and use of halons.
(ii) Ban on production and use of chlorofluoro carbons.
½+½
Ans 6.
(i) They cause biomagnification and increase pollution.
(ii) They make environment unclean.
(iii) They kill useful micro-organisms.
1
1
1
Ans 7.
Organisms that feed directly or indirectly on producers and cannot synthesized their own
food from inorganic sources are called consumers.
2
Herbivores, Carnivores, Omnivores and Parasites are various categories of consumers.
1
Ans 8.
NATURAL
(i) Omnivores, (ii) Decomposers, (iii) Producers, (iv) Food web, (v) Biological magnification.
1×5=5
RESOURCES
P-75
Ans 1.
Polythene bags and plastic containers.
Ans 2.
If all the snakes of food chain are killed, the peacocks belonging to the next level will also
die. Also the population of rats in the preceding level will highly increase.
1
Ans 3.
Effects of ozone depletion on health can be :
Ans4 .
Ans 5.
½+½
(i) Skin cancer
½
(ii) Damage to eyes
½
(iii) Effect on immunity
½
(iv) Can change the structure of DNA.
½
Z — Tertiary consumers
1
X — Primary consumers
1
Decomposers break down the complex organic matter and help in recycling it. It help in
natural replenishment of the soil.
1
Bacteria and fungi are two decomposers.
Ans 6.
½+½
When green plants are eaten by herbivores, a great deal of energy is lost in the form of heat
to the environment.
1
Some amount goes into digestion and in doing work and the rest goes towards growth and
reproduction.
1
The remaining is only an average of 10% of the food eaten that is turned into its own body
and made available for the next level of consumers.
1
Ans 7.
Plastic is a non-biodegradable substance.
1
Hence a plastic bag will persist in the environment and lead to several environmental
problems.
1
It would be better to use paper bags or cloth bags instead of polythene bags.
Ans 7.
1
The flow of energy is unidirectional. The energy that is captured by the autotrophs does
not revert back to the solar input and the energy which is passed to herbivores does not
come back to autotrophs. So, as it moves progressively through various trophic levels it is
no longer available to the previous levels.

The green plants in a terrestrial ecosystem capture about 1% of the sunlight that falls
on them and convert it into food energy.

When green plants are eaten by primary consumers, a high amount of energy is lost as
heat to the environment, some goes into digestion and rest into growth and
reproduction. Only 10% of the energy is made available for the next level of consumers.

The loss of energy at each level of the food chain is so great that only very little usable
energy remains at the last level.

Thus, there are generally a greater number of individuals at the lower trophic level of
an ecosystem and less number of tertiary or quaternary consumers.
5

Ans 1.
They use solar energy to convert raw materials into organic compounds by the process
called photosynthesis.
1
Ans 2.
Ecosystem is the functional unit of the environment comprising of the living and nonliving components.
1
P-76
SCIENCE – X
TERM – 2
Ans 3.
(a) Decomposers are those micro-organisms that obtain energy from the chemical
breakdown of dead organisms or animals or plant wastes.
1
(b) Decomposers break down complex organic substances into simpler inorganic
substances that go into the soil and are used up by the plants.
1
Ans 4.
There is only 10% flow of energy from one trophic level to the next higher level. The loss of
energy at each step is so great that very little usuable energy remains after fourth or fifth
trophic levels. Hence, only 4 to 5 trophic levels are present in each food chain.
2
Ans 5.
Ozone is a triatomic molecule of oxygen (O3). It forms a protective blanket over the earth’s
atmosphere and its depletion causes harmful effects on human. It absorbs the UV radiation
coming from the sun.
2
Ans 6.
(i) Biological Magnification : The increase in the concentration of a chemical per unit
weight of the organisms with the successive rise in trophic level.
1
(ii) Water  Plankton  Fish  Fish eating bird
(0.2ppm.) (5.0ppm) (240ppm)
Concentration in Fish would be approximately 240 ppm.
Ans 7.
1
If all the waste generated would be biodegradable, this will also create problem as the
process of decompositon is quite slow, so waste will start accumulating.
1
Biodegradable waste would affect the environment in the following two ways :
(i) Some biodegradable substances like cowdung blocks the sewage pipes and produce
foul smell, thus pollutes surrounding environment.
1
(ii) Some biodegradable substance like fruit and vegetable wastes serve the breeding site
of flies and mosquitoes which spread various diseases in surrounding environment.
1
Ans 8.
(i) Living organisms require energy to perform various activities such as growth,
reproduction, metabolism etc.
1
(ii) Ultimate source of energy – Sun
1
(iii) Consumers, both primary and secondary, use chemical form of energy.
1
(iv) In an ecosystem, energy flows in the form of food.
1
(v) Unidirectional flow, i.e., from producers to consumers.
1

Ans 1.
Chlorofluoro carbons.
1
Ans 2.
Hawks.
1
Ans 3.
Four examples of abiotic factors are :
(i) Temperature, (ii) Rain, (iii) Soil, (iv) Wind.
Ans 4.
Ans 5.
(a) Primary consumer.
1
(b) 10 kJ.
1
Non-biodegradable substances persist in the environment for a long time and cause
environment pollution.
They may also harm the various members of the ecosystem.
Ans 6.
½+½+½+½
2
According to 10% Law.
Plants have 20,000 J energy available.
Plants transferred their 10% energy to deer i.e.
20,000 × (10/100) = 2000 J
NATURAL
RESOURCES
1
P-77
Deer received 2000 J energy from the plants and now it transfer 10% energy of itself to lion.
2,000 × (10/100) = 200 J
1
Hence, lion will have 200 J energy from deer.
Ans 7.
1
Water  Planktons  Small fish  Large fish
Some poisonous chemicals enter the water bodies and the water animals are affected by it
as these chemicals enter the bodies of these animals through food chain. The phytoplanktons
absorb these harmful chemicals from water and they enter the food chain at producer level
and then transferred to next trophic level. Phytoplanktons were eaten up by zooplanktons.
Zooplanktons were eaten up by small fishes. Hence, the chemicals were transferred from
phytopankton to zooplankton, from zooplankton to small fishes and from small fishes to
large fishes. The maximum concentration of harmful chemicals is occupied by the top level
consumers in a food chain.
3
Ans 8.
The amount of Ozone in the atmosphere began to drop sharply in the 1980s.
1
This decrease is due to the synthetic chemicals like chlorofluoro carbons which are used
as refrigerants and in fire extinguishers.
1
In 1987, United Nations Environment Programme (UNEP) succeeded in forgings an
agreement to freeze CFC production at 1986 levels.
1

Ans 1.
Ans 2.
These micro-organisms break down the complex organic substances into simple inorganic
substances that go into the soil and are used once more by plants.
1
Grass
 Grasshopper 
Frogs
(4000 J)
(400 J)
(40 J)

Snakes
(4 J)
So, for snakes and frogs, 4 J and 40 J energy will be available by 10% law respectively.
½+½
Ans 3.
At the higher atmospheric levels, the higher energy UV radiations split apart some molecular
oxygen (O2) into free oxygen (O) atoms. These atoms then combine with molecular oxygen
(O2) to form ozone (O3).
2
Ans 4.
The UV rays are highly damaging the organisms. For example, it is known to cause skin
cancer in human beings.
1
This is done by the ozone layer in the higher levels of the Earth’s atmosphere.
Ans 5.
1
Biotic factors are the living components of the environment which interact with each other
as well as with the abiotic factors of the ecosystem.
Example : Plants and animals.
2+½+½
Ans 6.
When any organism dies, it is eventually eaten by detrivores (like vultures, worms and
crabs) and broken down by decomposers, mostly bacteria and fungi. These micro-organisms
break down the dead remains and waste products of plants and animals. They are the
decomposers as they break-down the complex organic substances into simple inorganic
substances that go into the soil and release nutrients for recycling in to the food chain. 3
Ans 7.
(i) These organisms are fungi and they belong to the group of decomposers.
(ii) Role of decomposers in nature :
Ans 8.
P-78

Clean the environment by acting on dead and decaying material.

Replenish soil by converting complex organic matter into simpler form.
3
The trophic level of an organism is the position it holds in a food chain. A food chain
usually consists of 4-5 levels.
SCIENCE – X
TERM – 2
(1) Primary producers ; the organisms that make their own food from sunlight, forms the
base of every food chain. These organisms are called as autotrophs.
(2) Primary consumers are animals that eat primary producers; they are also called
herbivores (plant-eaters).
(3) Secondary consumers eat primary consumers. They are carnivores (meat-eaters) and
omnivores (animals that eat both animals and plants).
(4) Tertiary consumers eat secondary consumers.
(5) Quaternary consumers eat tertiary consumers.
Food chains "end" with top predators, animals that have little or no natural enemies.
Example : Food chain in a Grassland biome.
3
Grass
Primary Producers

Grasshopper
Primary Consumers

Rates
Secondary Consumers

Snakes
Tertiary Consumers

Hawk
Quaternary Consumers
A. Choose the correct option :
Ans 1.
(B) Garden
Ans 2.
(C) Very little usable energy is left after the fourth step
Ans 3.
(A) 1 - c, 2 - b, 3 - a, 4 - d
Ans 4.
(B) First trophic level
Ans 5.
(D) Decomposers
Ans 6.
(C) Stratosphere
Ans 7.
(D) United Nations Environment Programme
Ans 8.
(A) Green, Blue
Ans 9.
(A) Persist in environment for long time
Ans 10.
(B) CFCs
B. Fill in the blanks :
Ans 1.
decomposers
Ans 2.
biotic
Ans 3.
artificial
Ans 4.
10%
Ans 5.
food chain
NATURAL
RESOURCES
(1 mark each)
(1 × 10)
P-79
Ans 6.
food web
Ans 7.
1%
Ans 8.
trophic level
Ans 9.
14%
Ans 10.
biomass

P-80
SCIENCE – X
TERM – 2