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Page 1
1
Class: X
Science (086)
Marking Scheme 2018-19
Time allowed: 3 Hours Maximum Marks: 80
Q No SECTION A Marks
1. Glucose/Amino acids 1
2. Coliform bacteria 1
SECTION B
3.
Mg - 12 2, 8, 2
Cl - 17 2, 8, 7
MgCl
2
Solution of Magnesium Chloride conduct electricity because an ionic compound
dissociates into ions when dissolved in water and hence conduct electricity.
2
4. (Fertilization results in formation of zygote).
Zygote divides several times, to form an embryo. The ovule develops a thick coat
and is into seed. The ovary grows rapidly and ripens to form the fruit.
2
5.
We have n =
? v =
=
m/s = 2 x 108 m/s
% decrease =
x 100% = 66.7%
OR
From Snell’s law
Since c and are constant therefore
Therefore velocity of light is maximum in medium C.
2
6.
A white precipitate is formed.
Na
2
SO
4
(aq)
+
BaCl
2
(ag) ? 2NaCl (aq) + BaSO
4
( ?)
Doubled Displacement reaction
It is a reaction in which there is an exchange of ions between the reactants.
2
SECTION C
7.
Plaster of Paris CaSO
4 .
½H
2
O
Calcium Sulphate Hemihydrate
Preparation
373K
CaSO
4.
2H
2
O CaSO
4 .
½H
2
O + 1½ H
2
O
(Gypsum)
3
v
c
r
i
n ? ?
sin
sin
i sin
v r ? sin
Page 2
1
Class: X
Science (086)
Marking Scheme 2018-19
Time allowed: 3 Hours Maximum Marks: 80
Q No SECTION A Marks
1. Glucose/Amino acids 1
2. Coliform bacteria 1
SECTION B
3.
Mg - 12 2, 8, 2
Cl - 17 2, 8, 7
MgCl
2
Solution of Magnesium Chloride conduct electricity because an ionic compound
dissociates into ions when dissolved in water and hence conduct electricity.
2
4. (Fertilization results in formation of zygote).
Zygote divides several times, to form an embryo. The ovule develops a thick coat
and is into seed. The ovary grows rapidly and ripens to form the fruit.
2
5.
We have n =
? v =
=
m/s = 2 x 108 m/s
% decrease =
x 100% = 66.7%
OR
From Snell’s law
Since c and are constant therefore
Therefore velocity of light is maximum in medium C.
2
6.
A white precipitate is formed.
Na
2
SO
4
(aq)
+
BaCl
2
(ag) ? 2NaCl (aq) + BaSO
4
( ?)
Doubled Displacement reaction
It is a reaction in which there is an exchange of ions between the reactants.
2
SECTION C
7.
Plaster of Paris CaSO
4 .
½H
2
O
Calcium Sulphate Hemihydrate
Preparation
373K
CaSO
4.
2H
2
O CaSO
4 .
½H
2
O + 1½ H
2
O
(Gypsum)
3
v
c
r
i
n ? ?
sin
sin
i sin
v r ? sin
2
Precaution
Gypsum should not be heated above 373 K otherwise it will form CaSO
4.
OR
Sweet tooth leads to tooth decay. Which is caused by the action of Bacteria on food particles
remaining in the mouth and acid is formed. The pH of the mouth falls below 5.5 and the
tooth enamel dissolves resulting in cavities Toothpastes are generally basic, they neutralise
the excess acid produced in the mouth and prevent tooth decay.
8.
X – 2, 8, 6
a) Since ‘X’ has three energy shells and period number of an element is equal to the number
of energy shells, X belongs to 3
rd
period.
b) X has 6 valence electrons it belongs to group 16.
c) Valency will be 2. To acquire noble gas configuration it will gain 2 electrons.
3
9.
i) less intake of Iodine (in the diet)
ii) will lead to gigantism
iii) timely secretion of testosterone
3
10.
Just before Starch test – Pale yellow
Just after Starch test – Blue black
Chlorophyll
6CO
2
+ 6H
2
O C
6
H
12
O
6
+ 6O
2
Sunlight
O
2
is obtained from water (H
2
O), as splitting of water results in formation of Hydrogen
(used for making glucose) and oxygen (by-product).
3
11. m = -
-1 =
?u = - 30 cm
Using mirror formula: u = -30 cm, v = -30 cm
=
+
=
+
f = -15 cm
Now u’ = - 10 cm, f = - 15 cm
Using mirror formula
=
_
For ray diagram refer NCERT Page 166 Figure 10.7.
OR
Power of a lens is the degree of convergence of divergence of light rays achieved by a lens.
Lens L
1
: f
1
=
=
= +10 cm ; Convex lens
Lens L
2
: f
2
=
=
= +20 cm ; Convex lens
Lens L
3
: f
3
=
=
= -10 cm ; Concave lens
Lens L
2
will form a virtual and magnified image of an object placed at 15 cm from the
convex lens because concave lens can never form virtual and magnified image of an object
and convex lens form such image only when the object is placed between the optical centre
and principle focus of the convex lens.
For diagram refer NCERT Page 181 figure 10.16 (f).
3
Page 3
1
Class: X
Science (086)
Marking Scheme 2018-19
Time allowed: 3 Hours Maximum Marks: 80
Q No SECTION A Marks
1. Glucose/Amino acids 1
2. Coliform bacteria 1
SECTION B
3.
Mg - 12 2, 8, 2
Cl - 17 2, 8, 7
MgCl
2
Solution of Magnesium Chloride conduct electricity because an ionic compound
dissociates into ions when dissolved in water and hence conduct electricity.
2
4. (Fertilization results in formation of zygote).
Zygote divides several times, to form an embryo. The ovule develops a thick coat
and is into seed. The ovary grows rapidly and ripens to form the fruit.
2
5.
We have n =
? v =
=
m/s = 2 x 108 m/s
% decrease =
x 100% = 66.7%
OR
From Snell’s law
Since c and are constant therefore
Therefore velocity of light is maximum in medium C.
2
6.
A white precipitate is formed.
Na
2
SO
4
(aq)
+
BaCl
2
(ag) ? 2NaCl (aq) + BaSO
4
( ?)
Doubled Displacement reaction
It is a reaction in which there is an exchange of ions between the reactants.
2
SECTION C
7.
Plaster of Paris CaSO
4 .
½H
2
O
Calcium Sulphate Hemihydrate
Preparation
373K
CaSO
4.
2H
2
O CaSO
4 .
½H
2
O + 1½ H
2
O
(Gypsum)
3
v
c
r
i
n ? ?
sin
sin
i sin
v r ? sin
2
Precaution
Gypsum should not be heated above 373 K otherwise it will form CaSO
4.
OR
Sweet tooth leads to tooth decay. Which is caused by the action of Bacteria on food particles
remaining in the mouth and acid is formed. The pH of the mouth falls below 5.5 and the
tooth enamel dissolves resulting in cavities Toothpastes are generally basic, they neutralise
the excess acid produced in the mouth and prevent tooth decay.
8.
X – 2, 8, 6
a) Since ‘X’ has three energy shells and period number of an element is equal to the number
of energy shells, X belongs to 3
rd
period.
b) X has 6 valence electrons it belongs to group 16.
c) Valency will be 2. To acquire noble gas configuration it will gain 2 electrons.
3
9.
i) less intake of Iodine (in the diet)
ii) will lead to gigantism
iii) timely secretion of testosterone
3
10.
Just before Starch test – Pale yellow
Just after Starch test – Blue black
Chlorophyll
6CO
2
+ 6H
2
O C
6
H
12
O
6
+ 6O
2
Sunlight
O
2
is obtained from water (H
2
O), as splitting of water results in formation of Hydrogen
(used for making glucose) and oxygen (by-product).
3
11. m = -
-1 =
?u = - 30 cm
Using mirror formula: u = -30 cm, v = -30 cm
=
+
=
+
f = -15 cm
Now u’ = - 10 cm, f = - 15 cm
Using mirror formula
=
_
For ray diagram refer NCERT Page 166 Figure 10.7.
OR
Power of a lens is the degree of convergence of divergence of light rays achieved by a lens.
Lens L
1
: f
1
=
=
= +10 cm ; Convex lens
Lens L
2
: f
2
=
=
= +20 cm ; Convex lens
Lens L
3
: f
3
=
=
= -10 cm ; Concave lens
Lens L
2
will form a virtual and magnified image of an object placed at 15 cm from the
convex lens because concave lens can never form virtual and magnified image of an object
and convex lens form such image only when the object is placed between the optical centre
and principle focus of the convex lens.
For diagram refer NCERT Page 181 figure 10.16 (f).
3
3
12.
R
1
=
=
= 484 ?
R
2
=
=
= 242 ?
In series:
R
S
= R
1
+ R
2
= 484 + 242 = 726 ?
?I
S
=
=
=
A = 0.30 A
In parallel:
=
+
=
+
R
p
=
?
I
P
=
=
=
A = 1.36 A
3
13.
R
a
= ?
R
b
= ? (
) = 9
= 9 R
a
R
c
= ?
=
=
R
a
Hence R
b
>R
a
>R
c
?
a
= ?
b
= ?
c
because all the three conductors are of same material.
3
14. Biogas is a mixture of methane, carbon dioxide, hydrogen, hydrogen sulphide.
Following steps are involved in obtaining biogas:
i) Mixing (Slurry of cattle-dung and water)
ii) Digesting (decomposition of cattle-dung by anaerobic bacteria)
iii) Formation of biogas
iv) Residue left after the formation of biogas.
3
15. Damaging as it is a deadly poison.
Beneficial as it shields the surface of the earth from UV radiations of the Sun.
By not using synthetic chemicals like CFCs, that deplete O
3
layer.
OR
? Flow of energy is unidirectional.
? Terrestrial plants take about 1% of the Sun’s energy and change it to
chemical energy.
? A great deal of energy is -lost as heat/ used for digestion/doing
work/growth and reproduction.
? Only 10% of organic matter present at each trophic level (and available to
next trophic level).
? Food chains are mainly of 3-4 trophic levels (because of 10% law) .
? The number of producers are maximum (the number reduces in subsequent
trophic levels).
? Food webs are more common (as compared to isolated food chains).
? Biological magnification can be observed. (Any three)
3
SECTION-D
16. a) Diagrams (activity 3.5 fig. 3.1)
Procedure
Observation - Heat is transferred from one end of metal wire to the free end of wire
which melts the wax and pin falls. Shows metals conduct heat.
4
Page 4
1
Class: X
Science (086)
Marking Scheme 2018-19
Time allowed: 3 Hours Maximum Marks: 80
Q No SECTION A Marks
1. Glucose/Amino acids 1
2. Coliform bacteria 1
SECTION B
3.
Mg - 12 2, 8, 2
Cl - 17 2, 8, 7
MgCl
2
Solution of Magnesium Chloride conduct electricity because an ionic compound
dissociates into ions when dissolved in water and hence conduct electricity.
2
4. (Fertilization results in formation of zygote).
Zygote divides several times, to form an embryo. The ovule develops a thick coat
and is into seed. The ovary grows rapidly and ripens to form the fruit.
2
5.
We have n =
? v =
=
m/s = 2 x 108 m/s
% decrease =
x 100% = 66.7%
OR
From Snell’s law
Since c and are constant therefore
Therefore velocity of light is maximum in medium C.
2
6.
A white precipitate is formed.
Na
2
SO
4
(aq)
+
BaCl
2
(ag) ? 2NaCl (aq) + BaSO
4
( ?)
Doubled Displacement reaction
It is a reaction in which there is an exchange of ions between the reactants.
2
SECTION C
7.
Plaster of Paris CaSO
4 .
½H
2
O
Calcium Sulphate Hemihydrate
Preparation
373K
CaSO
4.
2H
2
O CaSO
4 .
½H
2
O + 1½ H
2
O
(Gypsum)
3
v
c
r
i
n ? ?
sin
sin
i sin
v r ? sin
2
Precaution
Gypsum should not be heated above 373 K otherwise it will form CaSO
4.
OR
Sweet tooth leads to tooth decay. Which is caused by the action of Bacteria on food particles
remaining in the mouth and acid is formed. The pH of the mouth falls below 5.5 and the
tooth enamel dissolves resulting in cavities Toothpastes are generally basic, they neutralise
the excess acid produced in the mouth and prevent tooth decay.
8.
X – 2, 8, 6
a) Since ‘X’ has three energy shells and period number of an element is equal to the number
of energy shells, X belongs to 3
rd
period.
b) X has 6 valence electrons it belongs to group 16.
c) Valency will be 2. To acquire noble gas configuration it will gain 2 electrons.
3
9.
i) less intake of Iodine (in the diet)
ii) will lead to gigantism
iii) timely secretion of testosterone
3
10.
Just before Starch test – Pale yellow
Just after Starch test – Blue black
Chlorophyll
6CO
2
+ 6H
2
O C
6
H
12
O
6
+ 6O
2
Sunlight
O
2
is obtained from water (H
2
O), as splitting of water results in formation of Hydrogen
(used for making glucose) and oxygen (by-product).
3
11. m = -
-1 =
?u = - 30 cm
Using mirror formula: u = -30 cm, v = -30 cm
=
+
=
+
f = -15 cm
Now u’ = - 10 cm, f = - 15 cm
Using mirror formula
=
_
For ray diagram refer NCERT Page 166 Figure 10.7.
OR
Power of a lens is the degree of convergence of divergence of light rays achieved by a lens.
Lens L
1
: f
1
=
=
= +10 cm ; Convex lens
Lens L
2
: f
2
=
=
= +20 cm ; Convex lens
Lens L
3
: f
3
=
=
= -10 cm ; Concave lens
Lens L
2
will form a virtual and magnified image of an object placed at 15 cm from the
convex lens because concave lens can never form virtual and magnified image of an object
and convex lens form such image only when the object is placed between the optical centre
and principle focus of the convex lens.
For diagram refer NCERT Page 181 figure 10.16 (f).
3
3
12.
R
1
=
=
= 484 ?
R
2
=
=
= 242 ?
In series:
R
S
= R
1
+ R
2
= 484 + 242 = 726 ?
?I
S
=
=
=
A = 0.30 A
In parallel:
=
+
=
+
R
p
=
?
I
P
=
=
=
A = 1.36 A
3
13.
R
a
= ?
R
b
= ? (
) = 9
= 9 R
a
R
c
= ?
=
=
R
a
Hence R
b
>R
a
>R
c
?
a
= ?
b
= ?
c
because all the three conductors are of same material.
3
14. Biogas is a mixture of methane, carbon dioxide, hydrogen, hydrogen sulphide.
Following steps are involved in obtaining biogas:
i) Mixing (Slurry of cattle-dung and water)
ii) Digesting (decomposition of cattle-dung by anaerobic bacteria)
iii) Formation of biogas
iv) Residue left after the formation of biogas.
3
15. Damaging as it is a deadly poison.
Beneficial as it shields the surface of the earth from UV radiations of the Sun.
By not using synthetic chemicals like CFCs, that deplete O
3
layer.
OR
? Flow of energy is unidirectional.
? Terrestrial plants take about 1% of the Sun’s energy and change it to
chemical energy.
? A great deal of energy is -lost as heat/ used for digestion/doing
work/growth and reproduction.
? Only 10% of organic matter present at each trophic level (and available to
next trophic level).
? Food chains are mainly of 3-4 trophic levels (because of 10% law) .
? The number of producers are maximum (the number reduces in subsequent
trophic levels).
? Food webs are more common (as compared to isolated food chains).
? Biological magnification can be observed. (Any three)
3
SECTION-D
16. a) Diagrams (activity 3.5 fig. 3.1)
Procedure
Observation - Heat is transferred from one end of metal wire to the free end of wire
which melts the wax and pin falls. Shows metals conduct heat.
4
4
b) Ore HgS cinnabar
Roasting 2HgS (s) + 3O
2
(g) ? 2HgO(S) + 2SO
2
(g)
?
Reduction 2HgO(s) ? 2Hg(l) + O
2
(g)
17.
2CH
3
COOH + 2Na ? 2CH
3
COONa + H
2
O
(A) (B)
CH
3
COOH + C
2
H
5
OH CH
3
COOC
2
H
5
+ H
2
O
(C) ( D)
CH
3
COOC
2
H
5
+ NaOH Acid C
2
H
5
OH + CH
3
COONa
A - CH
3
COOH
B - CH
3
COONa
C - C
2
H
5
OH
D - CH
3
COOC
2
H
5
OR
a) Carbon has electronic configuration 2, 4. It could gain four electrons forming C
-4
anion or
lose 4 electrons to form C
+4
cation . Both are not possible due to energy considerations.
Carbon overcomes this problem by sharing electrons and forming covalent compounds.
Two reasons for forming large number of compounds:
1) Catenation
2) Tetra valency
b) Formation of NH
3
molecule
N – 2, 5
H – 1
Three hydrogen atoms each share their 1 electron with
nitrogen to form three covalent bonds
and make an ammonia molecule (NH
3
). ammonia molecule.
5
18.
a) page 137, figure 8.11
Correct diagram with correct labelling , correctly matched with the following parts:-
i) Fallopian Tube/Oviduct
ii) Uterus
iii) Vagina
b) People prefer use of condoms as it prevents STDs/gives privacy to the user. Condoms
help create a mechanical barrier preventing meeting of sperms and ovum.
5
19.
i) Natural selection
ii) Genetic drift
5
Page 5
1
Class: X
Science (086)
Marking Scheme 2018-19
Time allowed: 3 Hours Maximum Marks: 80
Q No SECTION A Marks
1. Glucose/Amino acids 1
2. Coliform bacteria 1
SECTION B
3.
Mg - 12 2, 8, 2
Cl - 17 2, 8, 7
MgCl
2
Solution of Magnesium Chloride conduct electricity because an ionic compound
dissociates into ions when dissolved in water and hence conduct electricity.
2
4. (Fertilization results in formation of zygote).
Zygote divides several times, to form an embryo. The ovule develops a thick coat
and is into seed. The ovary grows rapidly and ripens to form the fruit.
2
5.
We have n =
? v =
=
m/s = 2 x 108 m/s
% decrease =
x 100% = 66.7%
OR
From Snell’s law
Since c and are constant therefore
Therefore velocity of light is maximum in medium C.
2
6.
A white precipitate is formed.
Na
2
SO
4
(aq)
+
BaCl
2
(ag) ? 2NaCl (aq) + BaSO
4
( ?)
Doubled Displacement reaction
It is a reaction in which there is an exchange of ions between the reactants.
2
SECTION C
7.
Plaster of Paris CaSO
4 .
½H
2
O
Calcium Sulphate Hemihydrate
Preparation
373K
CaSO
4.
2H
2
O CaSO
4 .
½H
2
O + 1½ H
2
O
(Gypsum)
3
v
c
r
i
n ? ?
sin
sin
i sin
v r ? sin
2
Precaution
Gypsum should not be heated above 373 K otherwise it will form CaSO
4.
OR
Sweet tooth leads to tooth decay. Which is caused by the action of Bacteria on food particles
remaining in the mouth and acid is formed. The pH of the mouth falls below 5.5 and the
tooth enamel dissolves resulting in cavities Toothpastes are generally basic, they neutralise
the excess acid produced in the mouth and prevent tooth decay.
8.
X – 2, 8, 6
a) Since ‘X’ has three energy shells and period number of an element is equal to the number
of energy shells, X belongs to 3
rd
period.
b) X has 6 valence electrons it belongs to group 16.
c) Valency will be 2. To acquire noble gas configuration it will gain 2 electrons.
3
9.
i) less intake of Iodine (in the diet)
ii) will lead to gigantism
iii) timely secretion of testosterone
3
10.
Just before Starch test – Pale yellow
Just after Starch test – Blue black
Chlorophyll
6CO
2
+ 6H
2
O C
6
H
12
O
6
+ 6O
2
Sunlight
O
2
is obtained from water (H
2
O), as splitting of water results in formation of Hydrogen
(used for making glucose) and oxygen (by-product).
3
11. m = -
-1 =
?u = - 30 cm
Using mirror formula: u = -30 cm, v = -30 cm
=
+
=
+
f = -15 cm
Now u’ = - 10 cm, f = - 15 cm
Using mirror formula
=
_
For ray diagram refer NCERT Page 166 Figure 10.7.
OR
Power of a lens is the degree of convergence of divergence of light rays achieved by a lens.
Lens L
1
: f
1
=
=
= +10 cm ; Convex lens
Lens L
2
: f
2
=
=
= +20 cm ; Convex lens
Lens L
3
: f
3
=
=
= -10 cm ; Concave lens
Lens L
2
will form a virtual and magnified image of an object placed at 15 cm from the
convex lens because concave lens can never form virtual and magnified image of an object
and convex lens form such image only when the object is placed between the optical centre
and principle focus of the convex lens.
For diagram refer NCERT Page 181 figure 10.16 (f).
3
3
12.
R
1
=
=
= 484 ?
R
2
=
=
= 242 ?
In series:
R
S
= R
1
+ R
2
= 484 + 242 = 726 ?
?I
S
=
=
=
A = 0.30 A
In parallel:
=
+
=
+
R
p
=
?
I
P
=
=
=
A = 1.36 A
3
13.
R
a
= ?
R
b
= ? (
) = 9
= 9 R
a
R
c
= ?
=
=
R
a
Hence R
b
>R
a
>R
c
?
a
= ?
b
= ?
c
because all the three conductors are of same material.
3
14. Biogas is a mixture of methane, carbon dioxide, hydrogen, hydrogen sulphide.
Following steps are involved in obtaining biogas:
i) Mixing (Slurry of cattle-dung and water)
ii) Digesting (decomposition of cattle-dung by anaerobic bacteria)
iii) Formation of biogas
iv) Residue left after the formation of biogas.
3
15. Damaging as it is a deadly poison.
Beneficial as it shields the surface of the earth from UV radiations of the Sun.
By not using synthetic chemicals like CFCs, that deplete O
3
layer.
OR
? Flow of energy is unidirectional.
? Terrestrial plants take about 1% of the Sun’s energy and change it to
chemical energy.
? A great deal of energy is -lost as heat/ used for digestion/doing
work/growth and reproduction.
? Only 10% of organic matter present at each trophic level (and available to
next trophic level).
? Food chains are mainly of 3-4 trophic levels (because of 10% law) .
? The number of producers are maximum (the number reduces in subsequent
trophic levels).
? Food webs are more common (as compared to isolated food chains).
? Biological magnification can be observed. (Any three)
3
SECTION-D
16. a) Diagrams (activity 3.5 fig. 3.1)
Procedure
Observation - Heat is transferred from one end of metal wire to the free end of wire
which melts the wax and pin falls. Shows metals conduct heat.
4
4
b) Ore HgS cinnabar
Roasting 2HgS (s) + 3O
2
(g) ? 2HgO(S) + 2SO
2
(g)
?
Reduction 2HgO(s) ? 2Hg(l) + O
2
(g)
17.
2CH
3
COOH + 2Na ? 2CH
3
COONa + H
2
O
(A) (B)
CH
3
COOH + C
2
H
5
OH CH
3
COOC
2
H
5
+ H
2
O
(C) ( D)
CH
3
COOC
2
H
5
+ NaOH Acid C
2
H
5
OH + CH
3
COONa
A - CH
3
COOH
B - CH
3
COONa
C - C
2
H
5
OH
D - CH
3
COOC
2
H
5
OR
a) Carbon has electronic configuration 2, 4. It could gain four electrons forming C
-4
anion or
lose 4 electrons to form C
+4
cation . Both are not possible due to energy considerations.
Carbon overcomes this problem by sharing electrons and forming covalent compounds.
Two reasons for forming large number of compounds:
1) Catenation
2) Tetra valency
b) Formation of NH
3
molecule
N – 2, 5
H – 1
Three hydrogen atoms each share their 1 electron with
nitrogen to form three covalent bonds
and make an ammonia molecule (NH
3
). ammonia molecule.
5
18.
a) page 137, figure 8.11
Correct diagram with correct labelling , correctly matched with the following parts:-
i) Fallopian Tube/Oviduct
ii) Uterus
iii) Vagina
b) People prefer use of condoms as it prevents STDs/gives privacy to the user. Condoms
help create a mechanical barrier preventing meeting of sperms and ovum.
5
19.
i) Natural selection
ii) Genetic drift
5
5
iii) Law of Dominance
iv) Acquired characters are not inherited
v) Gene flow
OR
a) Body or its parts that are not decomposed/preserved traces of organisms.
(to begin with new line)
? Upon digging the earth, the fossils that are found closer to the surface are
more recent than the fossils in deeper layers.
? By detecting the ratios of different isotopes of the same element in the fossil
material.
b) i) by selecting very short distances between leaves.
ii) by selecting sterile flowers.
20.
(a) Power of accommodation: It is the ability of the eye lens to adjust its focal length.
Ciliary muscles of eye are responsible for change in its focal length.
(b) Myopia
Causes : i) excessive curvature of the eye lens
ii) elongation of eyeball
This defect can be corrected by using a concave lens of suitable power.
For ray diagram refer NCERT Page 189 figure 11.2 (a), (b) and (c)
5
21.
(i) Refer NCERT Page 233, Activity 13.8
(ii) a) The galvanometer needle deflects momentary in one direction because when the key
is closed ,magnetic field lines around coil-2 increases momentary that causes induced
current in coil-2.
b) The galvanometer needle deflects momentary but in opposite direction because when the
key is opened ,magnetic field lines around coil-2 decreases momentary that causes induced
current in coil-2.
OR
Electric generator for principle, diagram, working Refer NCERT page 236, point 13.6.
5
SECTION E
22.
The colour of dilute solution of Sodium hydroxide turns pink or adding Phenolphthalein as
NaOH is a base .When excess of HCl is added the final mixture becomes colourless due to
neutralisation of base with an acid.
OR
Metals in increasing order of reactivity - Copper, iron ,zinc and magnesium
1. Color of the solution changes from blue to green
2. Reddish brown deposits on iron filings.
2
23.
a) NaHCO
3
CH
3
COOH + NaHCO
3
? CH
3
COONa + H
2
O + CO
2
(1)
b) CO
2
on passing through lime water turns milky.
2
24.
Yeast
Budding – The parent (yeast) cell produces a small protuberance that grows to form a bud.
The nucleus of the parent (yeast) cell divides, such that the daughter nucleus moves into the
daughter cell and the process continues to form a chain of buds.
OR
In absence of KOH- CO
2
released by germinating seeds is not absorbed, partial vacuum is
not created in the conical flask, air pressure in the flask is not reduced, water level does not
rise in the delivery tube.
2
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FAQs on Class 10 Science: CBSE Sample Question Paper Solutions (2018-19) - CBSE Sample Papers For Class 10
| 1. What are the benefits of solving CBSE sample question papers for Class 10 Science? |  |
Ans. Solving CBSE sample question papers for Class 10 Science helps students to understand the exam pattern and marking scheme. It also helps them in identifying their weak areas and working on them. Additionally, it improves time management skills and boosts confidence before the actual exam.
| 2. How can I access CBSE sample question papers for Class 10 Science? | |
Ans. CBSE sample question papers for Class 10 Science can be accessed through the official website of CBSE (cbse.nic.in) or various educational websites that provide study material and resources for students.
| 3. Are CBSE sample question papers for Class 10 Science enough to prepare for the board exams? | |
Ans. CBSE sample question papers for Class 10 Science are a great resource for exam preparation, but they should not be relied on solely. Students should also refer to their textbooks, notes, and other study materials to get a thorough understanding of the subject.
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Ans. Yes, solving CBSE sample question papers can improve grades in Class 10 Science as it helps in revision and practice. It also helps in identifying the mistakes and rectifying them before the actual exam, which can lead to better scores.
| 5. How can I use CBSE sample question papers effectively for Class 10 Science exam preparation? | |
Ans. To use CBSE sample question papers effectively for Class 10 Science exam preparation, students should solve them in a simulated exam environment, i.e., adhering to the time limit and attempting questions with full concentration. They should analyze their performance after every attempt and work on their weak areas. Additionally, they should seek guidance from their teachers or mentors for effective utilization of these resources.
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