CBSE Class 10 Science Question Paper 2020 Set 31-2-1 Solutions

Series : JBB/2 SET-1 Paper Code No: 31/2/1
Strictly Confidential: (For Internal and Restricted use only)
Secondary School Examination-2020
Marking Scheme – SCIENCE
(SUBJECT CODE :086) (PAPER CODE –31/2/1 )
General Instructions: -

1. You are aware that evaluation is the most important process in the actual and
correct assessment of the candidates. A small mistake in evaluation may lead
to serious problems which may affect the future of the candidates, education
system and teaching profession. To avoid mistakes, it is requested that before
starting evaluation, you must read and understand the spot evaluation
guidelines carefully.Evaluation is a 10-12 days mission for all of us.
Hence, it is necessary that you put in your best effortsin this process.
2. Evaluation is to be done as per instructions provided in the Marking Scheme.
It should not be done according to one’s own interpretation or any other
consideration. Marking Scheme should be strictly adhered to and religiously
followed. However, while evaluating, answers which are based on latest
information or knowledge and/or are innovative, they may be assessed
for their correctness otherwise and marks be awarded to them. In class-
X, while evaluating two competency based questions, please try to
understand given answer and even if reply is not from marking scheme
but correct competency is enumerated by the candidate, marks should
be awarded.
3. The Head-Examiner must go through the first five answer books evaluated by
each evaluator on the first day, to ensure that evaluation has been carried out
as per the instructions given in the Marking Scheme. The remaining answer
books meant for evaluation shall be given only after ensuring that there is no
significant variation in the marking of individual evaluators.
4. Evaluators will mark( √ ) wherever answer is correct. For wrong answer ‘X”be
marked. Evaluators will not put right kind of mark while evaluating which
gives an impression that answer is correct and no marks are awarded. This
is most common mistake which evaluators are committing.
5. If a question has parts, please award marks on the right-hand side for each
part. Marks awarded for different parts of the question should then be totaled
up and written in the left-hand margin and encircled. This may be followed
strictly.
6. If a question does not have any parts, marks must be awarded in the left-hand
margin and encircled. This may also be followed strictly.
7. If a student has attempted an extra question, answer of the question
deserving more marks should be retained and the other answer scored out.
8. No marks to be deducted for the cumulative effect of an error. It should be
penalized only once.
9. A full scale of marks 0-80 has to be used. Please do not hesitate to award full
marks if the answer deserves it.
10. Every examiner has to necessarily do evaluation work for full working hours
i.e. 8 hours every day and evaluate 20 answer books per day in main subjects
and 25 answer books per day in other subjects (Details are given in Spot
Guidelines).
11. Ensure that you do not make the following common types of errors committed
by the Examiner in the past:-
 Leaving answer or part thereof unassessed in an answer book.
31/2/1 Page 1 of 9

 Giving more marks for an answer than assigned to it.
 Wrong totaling of marks awarded on a reply.
 Wrong transfer of marks from the inside pages of the answer book to the
title page.
 Wrong question wise totaling on the title page.
 Wrong totaling of marks of the two columns on the title page.
 Wrong grand total.
 Marks in words and figures not tallying.
 Wrong transfer of marks from the answer book to online award list.
 Answers marked as correct, but marks not awarded. (Ensure that the right
tick mark is correctly and clearly indicated. It should merely be a line.
Same is with the X for incorrect answer.)
 Half or a part of answer marked correct and the rest as wrong, but no
marks awarded.

12. While evaluating the answer books if the answer is found to be totally
incorrect, it should be marked as cross (X) and awarded zero (0)Marks.

13. Any unassessed portion, non-carrying over of marks to the title page, or
totaling error detected by the candidate shall damage the prestige of all the
personnel engaged in the evaluation work as also of the Board. Hence, in
order to uphold the prestige of all concerned, it is again reiterated that the
instructions be followed meticulously and judiciously.

14. The Examiners should acquaint themselves with the guidelines given in the
Guidelines for spot Evaluation before starting the actual evaluation.

15. Every Examiner shall also ensure that all the answers are evaluated, marks
carried over to the title page, correctly totaled and written in figures and
words.

16. The Board permits candidates to obtain photocopy of the Answer Book on
request in an RTI application and also separately as a part of the re-
evaluation process on payment of the processing charges.

31/2/1 Page 2 of 9

Series : JBB/2 SET-1 Paper Code No: 31/2/1
MARKING SCHEME- CLASS X SCIENCE (2019-20)
QUESTION PAPER CODE : 31/2/1
S. Value Points/Expected Answer MARKS TOTAL
N MARKS
O
SECTION A
1. Due to weak intermolecular forces. 1
2 Two / Lithium and Beryllium 1
3. (a) Are deeper hot regions of earth’s crust where molten rocks are 1
formed.
(b) New Zealand / United States of America / China/Indonesia, ½+½
Philippines / Turkey/ New Mexico. (Any two)
(c) Electromagnetic Induction. 1
(d) In case of A.C. transmission of power/electricity takes place 1 4
without much loss of energy.
4. (a) In the neck region 1
(b) Thyroxine regulates carbohydrate, proteins and fat metabolism in the 1
body./ It promotes growth of body tissue.
(c) Excess of secretion of throxine in the body /overactivity of the thyroid 1
gland 1 4
(d) Can be controlled by including iodised salt in our diet.
(or any other relevant answer)
5. (b) / B,C and D
OR
5. (d) /Opaque eye lens 1
6. (c) / 8Ω 1
7. (d) / R 2 > R1 > R 3 1
8. (c)/ Sugarcane and rice
OR
8. (c) / Carbon monoxide 1
9. (b) / Maharashtra 1
10. (d) / x= Physical state of KClO3 and KCl
y = Reaction condition
z= Physical state of O2 1
11. (b) / Clove oil 1
12. (b) / Group 13 period 2
OR
12. (b) / X2 Y 1
13. (a) / Both (A) and (R) are true and (R) is the correct explanation of the
assertion. 1
14. (d) / (A) is false, but (R) is true. 1
SECTION B
15. (i) A = CaO / Quick lime/ Calcium oxide ½
B = Ca(OH)2 / Slaked lime / Calcium hydroxide ½

(ii) CaO + H2 O → Ca(OH)2 + heat or energy 1

31/2/1 Page 3 of 9

(iii) Combination reaction ½+½ 3
Exothermic reaction
16. (i) 2 formula units of CaSO4 /Calcium sulphate share 1 molecule of
water of crystallization. 1

(ii) due to its alkaline nature . 1
(iii) CuSO4 .5H2O  CuSO4 + 5H2O
(Blue) (white) 1
/ Due to loss of water of crystallization.

OR
(i)

1

(ii) Wet litmus paper 1
(iii) HCl solution , it is due to the formation of H + ion on in the water
/ H3 O+ (Hydronium ions) ½+½ 3
17. (a) A ½
(b) D ½
(c) B ½
(d) C ½
(e) A2 C ½
(f) B & D
½ 3
18. (a) Grass → Grass hopper → Frog → Snake 1
(Or any other relevant example)
(b) Transfer of food energy to the next higher level will not take place , then
the organisms of the upper trophic levels will be affected , increase in the 1
population of the organisms belonging to the previous trophic level /
imbalance in the food chain.

10% Law 10% law
c 2000J 200 J 20J
1
II Trophic III Trophic IV Trophic
Level Level Level

[If calculation of the amount of energy is not shown , deduct ½ mark .]
OR ½
(a) (i) O2 ½
(ii) O3 ½
(iii) Breathing /Respiration ½
(iv) Absorbs harmful ultra violet (UV) radiations.
uv
½
(b) O2 O+O ½ 3
O + O2 → O3

19. Secretions Functions
(a) mucus (d)Protects the inner lining of stomach
from the acid / softening of food
31/2/1 Page 4 of 9

(b) HCl(Hydrochloric acid) (e)Provides the acidic medium for
action of enzyme / Kill the germs. ½x6
(c) Pepsin (f) Digest proteins
(Note : a,b and c may in any order but there function must match / be given 3
along with the secretion.
20. (i) Homologous organs: Mammals have forelimbs as do birds, reptiles and 1½
amphibians .The basic structure of the limbs is similar though it has been
modified to perform different functions in various vertebrates. Therefore
these are homologus organs. 1½
(ii) Fossils: Study of fossils of Archeopteryx / Dinosaurs show the presence
of feathers used for insulation in cold weather but later became useful for
flight. So birds have evolved from reptiles. 3
21  Chromosomes are thread like structures present in nucleus containing 1
genetic material / DNA
 Number of chromosomes are reduced to half during gametes / germ cell 1
formation .
After fertilization of germ cells the number of chromosomes is 1
maintained in progeny. 3
22 (i)

1

(ii)

1

(iii)

1 3

23 (i) Galvanometer (G) shows deflection (for very short time) 1
(ii) Galvanometer (G) shows deflection for a very short time in
opposite direction to the previous observation. 1
Common Reason: Due to variation in current flowing through coil 1,
magnetic field associated with coil 2 changes. Due to which an induced 1
current will generate consequently galvanometer shows momentry 3
defelection.
24 (a) (i) Size of eyeball decreases
(ii) Focal length of eye lens is too long / Power of eye lens decreases. ½+½
(b) Diagrams :

1

31/2/1 Page 5 of 9

Hypermetropic Eye

1

Corrected Eye

OR
(a) Small size particles scatter shorter wavelength (violet) or large sized 1
particles scatter larger wavelength (Red).
(b) Due to variation in physical condition of hot air. 1
(c) Diagram

1

(Splitting of white light is essential )
3

SECTION C
25 (a)
 Metals high up in reactivity series cannot be obtained from
their compounds by heating with carbon as carbon can not
reduce the oxides of these elements while those in the middle
of the reactivity series are extracted first by converting their
sulphides or carbonates into oxides and then reducing by 1
Carbon .
 It is because these metals have high affinity for oxygen than ½
Carbon .
 Electrolytic reduction ½
 Sodium is obtained from its molten chloride by passing
electricity.
 at Cathode : Na+ + e-  Na ½ +½
at Anode : 2Cl Cl2+ 2e-
(b)

2

OR
(i) 2Cu + O2  2CuO 1
(ii) Al2O3 + 6HCl  2AlCl3 + 3H2O 1
(iii) 2K + 2H2O  2KOH + H2 1
Heat 1
(iv) 2HgS + 3O2 2HgO + 2SO2 1 5
(v) AI2O3 + 2NaOH  2NaAlO2 + H2O

31/2/1 Page 6 of 9

26 (a)
 Homologous series is a group of compounds which have the same
functional group , same general formula and where to 1
successive member differ by – CH2 in the molecular formula
Example : CH3-OH , CH3 -CH2-OH
Functional group : -OH , General Formula : CnH2n OH 1

(b) Esterification :
The reaction of carboxylic acid with an alcohol in the presence of
H2SO4 yields an ester.
Conc .H 2 SO 4
CH3COOH+C2H5OH CH3COOC2H5 + H20
1½
Heat Ester
(If word equation given award full marks)

Addition Reaction :
A reaction in which two or more atoms are added across a double or
triple bond in presence of catalyst is called addition reactions.
Pt ./Pd or Ni 1½
CH2=CH2+H2 CH3-CH3 5

27 (a)
 Nephron 1
 Structure : Cluster of blood capillaries / glomerulus is
associated with cup shaped structure called Bowman’s 1½
capsule, which leads to coiled tubular part of Nephron.
Function : Collects the filterate and reabsorbs useful substances like
glucose,amino acids, salts and water from filterate and forms urine. 1½

(b) Amount of excess water in the body ½
Amount of wastes dissolved ½ 5
28 (a)
 Chemical Method ½
 Barrier Method ½
 Surgical Method ½

(b) Increase in female foeticide / Declining child sex ratio 1
(Any One)
Benefit : Maintaining male-female sex ratio for a healthy society ½

(c) Bacterial  Gonorrhoea ½
Syphilis ½

Viral  Warts ½
AIDS ½
OR

(a) (i) Ovary Production of female germ cell/egg 1
Production of hormone – estrogen
(Any one)
(i) Oviduct Site of fertilization 1

(b) (i) Thickening of the uterus lining 1
(ii) Wall of uterus breaks/Menstruation occurs. 1

(c) Providing the nutrition / O2/to the developing embryo /foetus or 1
removal of waste from the fetus. 5

31/2/1 Page 7 of 9

29 l
(a) R=𝜌 A
½
lx 30
Rx = ρ =
Ax π × 2 × 2

ly 10 ½
Ry = ρ =ρ
Ay π×1×1

R x ρ. lx Ay
= ×
Ry Ax ρ. ly

30 π×1×1
= × ½
π×2×2 10
Rx 3
=
Ry 4
½
Rx : Ry = 3:4

(b) Here I = 500 mA ; Resistance of lamp = R1

Resistance of the conductor (R2) = 10 Ω

Pot. Difference (V) = 6 volt

Net resistance of the circuit ½
V
R=R1+R2=
I

6
R1+10 Ω=0.5
1½
R1+10 Ω = 12 Ω
5
R1= (12-10) Ω
1
=2 Ω
30. (a)
(i) Real and magnified

1½

Object distance must be between 10 to 20 cm
(ii) Virtual and magnified
1½

Object distance must be less than 10 cm
(b) f = 10 cm ; u = - 10 cm

1 1 1 ½
+ =
v u f

1 1 1
v
= f −u

31/2/1 Page 8 of 9

1 1 1
v
= 10 − −10
1
1 1 1
= +
v 10 10

1 1 ½
= ∴ v = 5cm
v 5

OR
(a) (i) Ability of a lens to converge or diverge light rays/reciprocal 1
of focal length of lens.
(ii) It is a point on principal axis at which light ray parallel to 1
principal axis converges after reflection.
1 1 1
(b) (i) for spherical lens : v − u = f ½
1 1 1 ½
(ii) for spherical mirror : v + u = f
(c)

2

Distance of object (BO) = 10 cm
Focal length (OF1 ) = 15 cm 5

(If the distance in the diagram are not marked , deduct ½ marks)

31/2/1 Page 9 of 9