CBSE Class 10 Science Question Paper 2020 Set 31-5-1 Solutions

Strictly Confidential: (For Internal and Restricted use only)
Secondary School Examination-2020
Marking Scheme – SCIENCE
(SUBJECT CODE: 086) (PAPER CODE : 31/5/1 )
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.
• Giving more marks for an answer than assigned to it.
• Wrong totaling of marks awarded on a reply.

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• 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.

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Series: JBB/5 SET-1 Paper Code No: 31/5/1

MARKING SCHEME-CLASS X SCIENCE (2019-20)
QUESTION PAPER CODE :31/5/1
S.NO Value Points/Expected Answer MARKS TOTAL
MARKS
SECTION A
1. No charged particles/ions 1 1
2 All are metalloids/Shows the properties of metals and non-metals
OR
Properties of elements are a periodic function of their atomic number 1 1
3. (a) Cells which convert solar energy to electrical energy/electricity 1
(b) Voltage – 0.5 to 1V ½
Electricity –0.7W ½
(c) India receives great amount of solar energy throughout the year. 1
(d) Advantages :- No moving parts/require little maintenance /work quite
satisfactorily without any focusing device/can be set up in remote and ½+½
inaccessible areas. 4
(Any Two)
4. (a) Thyroid stimulating hormone. 1
(b) It stimulates / regulates thyroid gland to produce thryroid hormone or
thyroxine. 1
(c) Because high and low TSH level may increase the chances of 1
miscarriage. 1 4
(d) Proper medication is required.
5. (C) / remains unchanged 1 1
6. (B) / 10−3 A and 10−6 A respectively 1 1
7. (A) / 5A 1 1
8. (D) /I , II and III
OR
(D) / Reduce 1 1
9. (B)/ Chipko Movement 1 1
10. (B) / Decomposition & Redox 1 1
11. (C)/ Green 1 1
12. (B) / XY2
OR
(B ) / (C)
Group 16 and period 3 /Group 17 and period 3
(Note- Both are correct, marks to be awarded for any one) 1 1
13. (iv) / (A) is false, but (R) is true 1 1
14. (ii) / Both (A) and (R) are true, but (R) is not the correct explanation of 1 1
the assertion(A)
SECTION B
15. (a) ‘M’ is magnesium /Mg ½
‘N’ is Magnesium oxide / MgO ½
(b) 2Mg + O2 → 2MgO 1
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(c) ‘M’ undergoes oxidation because oxygen is added to it/ Loss of 2 ½ +½ 3
electrons
16. (a) Anode- Oxygen
Cathode- Hydrogen 1
(b) Because one molecule of water contains two atoms of hydrogen and
one atom of oxygen/ 2H2O2H2+O2 1
(c) Electrolysis of water will not take place 1

OR
(a) Chemical Name – Sodium Carbonate decahydrate
Common Name – Washing Soda
Chemical Formula - Na2 CO3 . 10H2 O ½ ×3

(b) NaCl + H2 O + CO2 + NH3 → NH4 Cl + NaHCO3

Heat
2NaHCO3 Na2 CO3 + H2 O + CO2

Na2 CO3 + 10H2 O → Na2 CO3 . 10H2 O
1
(c) It helps in removing permanent hardness./ It forms insoluable Ca or
½ 3
Mg salts in the form of scum
17. (a) Li ,K ½
(b) Mg ½
(c) C ½
(d) K ½
(e) S ½
(f) Al ½ 3
18.  Trophic level - Each step or level of a food chain forms a trophic 1
level 1
 Grass → Insect → Frog →Snake/Hawk / Correct Diagram
(any other)
 Because it moves progressively through the various trophic levels 1
and is no longer available to the previous level from producers to
consumers.

OR

(i) Aquatic ½
(ii) Abiotic ½
(iii) Air/Water/Soil/Temperature /Non-living ½
(iv) Living organism/plants and animals ½ 3
(v) Definition – All the interacting organisms in an area together
with the non living constituents of the environment form an 1
ecosystem /interaction between biotic and abiotic components.
19. (a) Exchange of gases. 1
(b) Because amount of oxygen dissolved in water is fairly low as
compared to the air 1

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(c) (i) Pyruvate ½ 3
(ii) Carbon dioxide ½
20. (a) Because Tallness is the dominant trait
(b) The recessive character is expressed in the F2 generation when two 1
copies of the recessive trait are present together/(tt).
(c) In the F2 progeny , the dominant character is also expressed along 1
with the recessive character in ratio of 3:1 respectively. 3
1
21 (a)
 Secretions from seminal vesicle. 1
 22+X and 22+Y ½+½
(b) (i) Female-XX ½
(ii) Male – XY ½ 3
22 (a)

1

(b)

1

(c)

1

(Note : Deduct ½ marks overall if no arrows are shown)
3
23 (a) (i) Momentary deflection in the needle of the galvanometer to the left /
right. ½
(ii) Momentary deflection in the needle of the galvanometer but in
the opposite direction. ½
(iii) No deflection ½
(b) Electromagnetic induction. ½
(c) Motion of a magnet with respect to coil induces an electric current in the
coil which lasts so long as the motion is taking place / change in 1 3
magnetic field around a coil produces an induced current in it.
24 (a) Myopia/Short sightedness ½
(b) Concave/Diverging lens. ½
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(c)
 Excessive curvature of eye lens
 elongation of eye ball ½+ ½
1
(d) P(D) = f(m)
1 10 2 1
P(D) = = = = −0.4D
−2.5(m) −25 −5

(Deduct ½ mark if unit is not mentioned)
OR
(a) The Red colour is least scattered by fog or smoke, hence visible from a 1
long distance.
(b) Because in the absence of atomosphere there is no scattering of light. 1
(c) Because of atmospheric refraction, the sun appears above the horizon
even after actual sunset. 1 3
SECTION C
25 For ore X Calcination/ Heating in limited supply of air/absence of air. ½

heat 1
ZnCO3 (s) ZnO(s)+CO2 (g)
½
For Ore Y Roasting/Heating in excess of air.

heat
2ZnS s + 3O2 g 2ZnO(s)+2SO2 (g) 1

The metal oxide is reduced by using suitable reducing agent such as carbon. 1
ZnO s + C s → Zn(s) + CO(g) 1

(Note – Any other example can be taken)
OR
(a)Figure

1

 Impure copper is made the anode and thin strip of pure copper is made
½
the cathode.
 A solution of acidified copper sulphate is taken as electrolyte ½
( Note : Labelled diagram is to be awarded full marks)

On passing the current the pure metal from the anode dissolves into the 1
electrolyte and equivalent amount of pure metal is deposited on the cathode.

(b)
 By filling the gaps with molten iron formed in the reaction of
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Fe2O3 with aluminum powder. ½
 Thermit process/reaction ½
 Fe2 O3(s)+2Al(s) 2Fe(l)+Al2O3(s)+Heat 1 5
26 (a) When two or more organic compounds have same molecular formula
but different structural formula , then the compounds are called isomers 1
and this phenomenon is called isomerism

1+1

Butane Iso-Butane
(b) Because ‘X’ is an unsaturated carbon compound 1
(c) Oxidising agent. 1 5
27 (a) Because ventricles have to pump blood to various distant organs of the
body 1
(b) Because their energy requirement is low 1
(c) In aquatic vertebrates the blood goes only once through the heart during
one cycle while in terrerstrial vertebrates it goes through the heart two
times during each cycle. 1
(d) Because transpirational pull is greater during day time. 1
(e) To prevent the backflow of the blood /blood flows only in one direction 1 5
28 (a)
 AUreter ½
 B Seminal Vesicle ½
 CUrethra ½
 D Vas deferens ½

(b) Testosterone : 1
Role
 Regulates the formation of sperms
 Changes in appearance of boys at the time of puberty. ½ +½
(c) Function of ‘B’
 Providing nutrition and transportation to sperms. ½
Function of ‘C’
½
 Serves as a common passage to both sperms and urine.
OR
(a)
½+½
 Regeneration- the lost body part can be regenerated.
 Budding – a complete small individual develops on the parent body ½+½
during favourable conditions.
 Spore Formation – Spores are covered with thick wall that helps to ½+½
overcome unfavourable conditions.
(b) Buds produced in the notches along the leaf margins develop into new 1
plants.
(c) Advantages :
 Propagation of flowerless plants.
 Genetically similar to the parent plant. ½+½
 Plants raised by vegetative propagation bear flowers and fruits 5
earlier than those produced from seeds.
(Any two)

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29 P ½
(a) I1 = V1

100 W 10
I1 = 220 V = 22 A ½

P2 10 1
I2 = = = A ½
V 220 22

I = I1 + I2

10 1 11
= + A= A = 0.5 A 1
22 22 22

(b) (i)

½

½

(ii) Net R=R1+R2=2+3=5 Ω

I=𝑅
V 5
= 5 = 1A 1
𝑛𝑒𝑡

∴ Voltage across 3Ω resistor ∶
½ 5
∴ V = 1 × 3 = 3V
30 (a) .

2

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(Note –Deduct ½ mark if arrows are not shown)

Speed of light in air 3×10 8 3 ½, ½, ½
(b) nga = = = = 1.5
Speed of light in glass 2×10 8 2

1
(c) f (m) =P (D) ½
1 1 −10
f= = = = −0.4m ½+½
P −2.5D 25D

(Note –Deduct ½ marks if unit is not mentioned)
OR
1
(a) f(m) = P(D)
½

1 −10
f= = = −0.4m = −40cm
−2.5D 25D 1

f=-40 cm v =-10 cm u=?

1 1 1
− =
v u f ½

1 1 1
− =
−10 cm u −40 cm
1 1 1
− = +
u −40 10 ½
−1 + 4 3
= =
40 40
40
∴ u = − 3 =−13.3 cm 1

(b) Since the power is –ve , the lens used is concave / diverging ½

1

OA = v = -10cm ; OB = u =-13.3 cm ; OF = f = -40cm
5

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