CBSE Class 10 Question Paper 2026 Solution Science

FOR CBSE CLASS 10 EXAM PREPARATION

CBSE Class 10 2026
Question Paper
Solution · Science
EXAM YEAR TYPE SUBJECT

CBSE Class 10 2026 Question Paper Solution Science

Notes · Sample Papers · Previous Year Papers · Mock Tests

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Marking Scheme
Strictly Confidential
(For Internal and Restricted use only)
Secondary School Examination, 2026 (Xth)
SUBJECT NAME: Science (Q.P. CODE /Set No. 086/31-4-1)

General Instructions : -

1 You are aware that evaluation is the most important process in the actual and correct
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assessment of the candidates. A small mistake in evaluation may lead to serious problems

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which may affect the future of the candidates, education system and teaching profession. To

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avoid mistakes, it is requested that before starting evaluation, you must read and understand
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the spot evaluation guidelines carefully.
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2
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“Evaluation policy is a confidential policy as it is related to the confidentiality of the
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examinations conducted, evaluation done and several other aspects. Its leakage to
public in any manner could lead to derailment of the examination system and affect the
life and future of millions of candidates. Sharing this policy/document to anyone,
publishing in any magazine and printing in Newspaper/Website, etc. may invite action
under various rules of the Board and IPC.”
3 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
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be assessed for their correctness otherwise and due marks be awarded to them. In

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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
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enumerated by the candidate, due marks should be awarded.
l a
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4 The Marking scheme carries only suggested value points for the answers.
These are in the nature of Guidelines only and do not constitute the complete answer. The
students can have their own expression and if the expression is correct, the due marks should
be awarded accordingly.
5 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. If there is any variation, the same should be zero after deliberation
and discussion. 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.
Evaluators will mark ( √ ) wherever answer is correct. For wrong answer CROSS ‘X’ be
6
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m marked. Evaluators will not put right (✓) while evaluating which gives an impression that
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answer is correct and no marks are awarded. This is most common mistake which

e m7 evaluators are committing.
las
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If a question has parts, please award marks on the right-hand side for each part. Marks
a
g
awarded for different parts of the question should then be totalled up and written in the left-
a 8
hand margin and encircled. This may be followed strictly.
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.
9 If a student has attempted an extra question, answer of the question deserving more marks
should be retained and the other answer scored out with a note “Extra Question”.

om
PAGE 1 {31-4-1}
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Page 1 of 36

10 No marks to be deducted for the cumulative effect of an error. It should be penalized only
once.
11 A full scale of marks 80 (example 0 to 80/70/60/50/40/30 marks as given in Question Paper)
has to be used. Please do not hesitate to award full marks if the answer deserves it.

12 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).This is in view of the reduced
syllabus and number of questions in question paper.
13 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 totalling of marks awarded on an answer.
● Wrong transfer of marks from the inside pages of the answer book to the title page.
● Wrong question wise totalling on the title page.
● Wrong totalling of marks of the two columns on the title page.
● Wrong grand total.
● Marks in words and figures not tallying/not same.
● 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.
14 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.
15 Any unassessed portion, non-carrying over of marks to the title page, or totalling 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.
16 The Examiners should acquaint themselves with the guidelines given in the “Guidelines for
Spot Evaluation” before starting the actual evaluation.
17 Every Examiner shall also ensure that all the answers are evaluated, marks carried over to
the title page, correctly totalled and written in figures and words.
18 The candidates are entitled to obtain photocopy of the Answer Book on request on payment
of the prescribed processing fee. All Examiners/Additional Head Examiners/Head
Examiners are once again reminded that they must ensure that evaluation is carried out
strictly as per value points for each answer as given in the Marking Scheme.

PAGE 2 {31-4-1}

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MARKING SCHEME
SCIENCE (Subject Code-086)
(PAPER CODE: 31/4/1) (10-04-86K)
Q.No. EXPECTED OUTCOMES/VALUE POINTS Marks Total
Marks
SECTION – A
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BIOLOGY
1.
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(C) / By breaking down the nutrients of bread outside the body and then 1
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absorbing them.
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2.
a s
(C) / Formation of fruit.
(B)l/ (iv)
g
a1
1 1

a(B)g / (i) and (iv)
3. 1
4. 1 1
5. (A) / (i), (ii) and (iii) 1 1
6. (B) / (i) and (iv) 1 1
7. (D) / Contraction of Left Ventricle. 1 1
8. (C) / Assertion (A) is true but Reason (R) is false. 1 1
9. (A) / Both Assertion(A) and Reason(R) are true, and Reason (R) is correct 1 1
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explanation of Assertion(A).
•
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10. Necessary- It helps in transport of water and minerals from roots to 1

•
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leaves/ helps in temperature regulation.
a
Evil- It results in loss of water.
ag from one trophic level is transferred to the
1 2
11. • Only 10% of the energy 1
next level. Most of the energy is lost as heat.
• So very little usable energy will remain after four trophic levels./ 1
10% 10%
Grass → Grasshoper → Frog
(Producer ) (
Primary
) (
Secondary
)
1000k Cal. consumer 100k Cal. consumer 10k Cal.
(Any other suitable food chain) 2

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12 (A)
•
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As his pancreatic duct is blocked, enzymes for digestion will not be
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1

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transported to small intestine / The blockage will cause difficulty
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las • Insulin secretion is already affected.
a g 1

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(B)
• Reflex action ½
• An automatic and quick response to a stimulus which does not involve ½
thinking.

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• Stimulus → Receptor → sensory neuron → relay neuron (spinal cord) 1 2
→ Motor neuron → effector (muscle) response.
13. (a) Because adrenal gland secretes adrenaline hormone that enables the 1
animal body to face a scary situation.
(b) To prevent female foeticide. 1
(c) Plants raised by vegetative propagation can bear flowers and fruits
earlier than those produced from seeds /
Such methods also make possible the propagation of plants such as
banana, orange, rose and jasmine that have lost the capacity to
produce seeds / 1
All plants produced are genetically similar enough to the parent
plant to have all its characteristics. 3

14. (a) Parents : RR  rr
(Red flowers) (White flowers) ½

Gametes :
½
F1: (All red)

Rr  Rr ½

Gametes:
F2
½

(b)
Phenotypic ratio/ Showing external look of plants : 33 : 11
3 : 1 ½
(Red) (White)
Genotypic Ratio : 1: 2: 1
RR : Rr : rr ½
3
15. (a) Due to hydrotropism, the roots grow towards area of water 1
availability so that they can absorb water and minerals.

(b) Auxins promote cell elongation / Auxin regulates the tropic 1
responses like bending (growth) of stem towards light.

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(c)
Movement by sensitive plants Tropic Movements by plants
Non-directional Directional 1
Fast Slow 1
Growth independent movement Growth related
(Any two points, any other difference)
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OR
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(c)
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l as Movement of Roots Movement of Shoots
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Gravity: Grows towards gravity / Gravity: Grows away from gravity /
a Positive geotropism
Light: Grows away from light /
Negative geotropism
Light: Grows towards light/ Positive
Negative phototropism phototropism 1 4

16. (A)

(a)

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(b)
• Absorption of light energy by chlorophyll.
• Conversion of light energy to chemical energy and splitting of water
molecule into hydrogen and oxygen.
• Reduction of carbon dioxide to carbohydrates. 2

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(c)
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a OR

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(B)
(a)

Diagram 1
Labelling of any 4 parts ½x4

(b) Filtration of blood to remove wastes. 1
(c) It is a process by which useful substances like glucose, amino acids,
vitamins, salts and most of water are reabsorbed from filtrate, back 1
into the blood. 5

SECTION – B
Chemistry
17. (B)/ 2 : 1 1 1
18. (C)/ Exothermic Reaction 1 1
19. (B)/ Hydrochloric acid 1 1
20. (A)/ The hydrophobic tail of the soap molecule is in the interior of the cluster, 1 1
whereas hydrophilic end is on the surface of the cluster.
21. (D)/ Magnesium 1 1
22. (D)/ Steel 1 1
23. (A)/ By adding acid to water with constant stirring. 1 1
24. (A)/ Both A and R is true and R is correct explanation of A. 1 1
25. (a) A → Sodium hydrogen carbonate/ baking soda/ NaHCO3 ½
B → Sodium carbonate/ washing soda/ Na2CO3 ½

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1
(b) • CaSO4  2 H2O ½
• 373K / 100C ½ 2

26. (a) Decomposition of silver chloride (AgCl) to form silver (Ag) 1

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(b) Decomposition of lead nitrate releases NO2 / Nitrogen dioxide

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1

. c om e m .
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e prevent the oxidation of fats in food items. l as
(c)
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ag
1 3
27. g
a(i) (a) Carbonate ore: Calcination
(A)

½

(b) Sulphide ore : Roasting

½

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(ii)

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• Anode: Impure copper rod. ½
• Cathode: Pure copper strip. ½
OR
(B)
(i)

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1

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Nitric acid is a strong oxidising agent and oxidises H2 gas produced to

as water.
g la 1 3

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28. (a) Universal indicator 1
(b) A/ (pH = 3) 1

(c)
• Strong acids give more H+ ions in water. / Strong acids pH 1-2 1
• Weak acids give less H+ ions in water. / Weak acids pH 6-7 1

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OR

(c) (i) When pH<5.6 1
(ii) Weak acid: acetic acid, formic acid ½+½ 4
29. (A)
(i) Carbon cannot form C4+ cation because removal of four electrons is
energetically not possible. Carbon cannot form C4− anion because nucleus 2
with six protons cannot hold ten electrons.
(ii)
• A series of compounds in which the same functional group substitutes 1
for hydrogen in carbon chain is called a homologous series.
• Because of higher molecular mass of C4H10 than C3H8 or C2H6.
1
(iii) /
Ethanoic acid / acetic acid CH3COOH is formed /
1
acidified K2Cr2O7 +Heat
CH3-CH2OH → CH3COOH

OR
(B)
(i) I. Propanal 1
II. Propyne 1
(ii) I. Ester group 1
II.
1

(iii) Soaps reacts with calcium and magnesium salts present in hard water 1 5
to form scum (insoluble substance).
SECTION – C
Physics

30. (A) / Iris and pupil 1 1
31. (C) / Presbyopia 1 1
32. (A) / Both Assertion (A) and Reason (R) both are true, and Reason (R) is the 1 1
correct explanation of Assertion
33. (A)
P
• I= ½
V
2000
I=
200

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I = 10 A ½
• Current passing through electric heater is 10A which is much more
than rated value (4A) of fuse. Hence fuse will melt and break the 1
circuit. So, it cannot be used.
OR

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(B)
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An electromagnet is formed by wrapping a current carrying insulated
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copper wire in the form of coil around a magnetic material like soft
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iron core. / 1
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l•as By placing a magnetic material like soft iron as a core material inside
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g
the current carrying solenoid.
a Strength of electromagnet can be increased by increasing the
(i) Number of turns of coil
½
½ 2
(ii) Current flowing through the coil.
34. (a)
1 Volt is the potential difference between two points in a current carrying
conductor when one Joule work is done to move a charge of 1Coulomb from
1J
one point to the other. / 1V = 1C 1
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W
(b) V=
Q

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½
W=QxV
W = 1 6  10−19  100 l as
W = 1 6  10−17 J ag ½
2
35. (a) (i) Speed of light in medium A > Speed of light in medium B. ½
(ii) Media B and C both have same optical density. ½
(b)

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s e g la
g la a
a 2
3
36. Rainbow is caused by dispersion of sunlight by tiny water droplets, present in
the atmosphere. The water droplets act like small prisms. They refract and
disperse the incident sunlight, then reflect it internally, and finally refract it 2
again when it comes out of the raindrop.

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1

3

37. (a)
• The rod AB gets displaced from its original position ½
• Because it experiences a force when placed in external magnetic field ½
(b)
• Fleming’s Left Hand Rule 1
• Stretch the thumb, fore finger and middle finger of your left hand such
that they are mutually perpendicular. If forefinger points in the
direction of magnetic field, second finger in the direction of current,
then the thumb will point in the direction of motion or force acting on 1
the conductor. 3

38. (a) Real / magnified. 1
(b) Converging / Concave mirror, Plane mirror 1
(c)
• Convex lens 1
• It magnifies the image formed by the curved mirror 1
OR
(c) The plane mirror redirects/ reflects the light rays coming from the curved
mirror towards eyepiece. 2 4

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39. (A) Let I be the total current flowing through the circuit and
I1 and I2 be the currents flowing through 4 (R) and R2 resistors.
(i) (I) Potential difference across R2 is same as that of across 4 resistor
as they are connected in parallel.
∴ V (across R2) = I1 R
m
½

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= 1.5  4
. co
. c = 6V ½
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em Current flowing through I = I – I l as
las(II) 2 1

ag
ag V
= 2.0 – 1.5 = 0.5 A ½

R= 2
I2
6
𝑅2 =
0.5

R2 = 12  ½
(III) Potential difference across 2 resistor

m
V=IR
V = 2 x 2 = 4V
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e
Potential difference across R1 = 12 – (6 + 4) = 2V
s
½

g l a
𝑅 =a
𝑉
1 𝐼
2
𝑅1 = 2  R1 = 1 
½

(ii) Resistance of conductor depends on
• Length of the conductor / R  l ½
1
• Area of cross section / R α
m
c. o
A ½
• Rl
m
m .co Rα
1

s e m
la
A

s e 𝑙
g
la
Rα
g
A
a
a R=ρ
𝑙
A
1
Where  = Resistivity (a proportionality constant)

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OR

(B)
𝑉 ½
(i) 𝐼=
𝑅
220
𝐼 = 55
½
I=4A
½
Power of electric iron P = VI
P = 220  4
½
P = 880 W

(ii) E=Pxt ½

Energy (3 bulbs) = 3  100  5
= 1500
= 1.5 kWh
Energy (electric heater) = 1.0  0.5
= 0.5 kWh
½
Total energy consumed (1 day) = 1.5 + 0.5 = 2 kWh
Total energy consumed (30 days) = 30  2
= 60 kWh
= 60 units
Total cost = Units  Rate
= 60  3.60
= ₹ 216 1

(iii) The resistivity of an alloy is generally higher than that of its
constituent metals. / Alloys do not oxidise (burn) readily at high 1
temperatures. 5

PAGE 12 {31-4-1}

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.
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s em l a
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Marking Scheme
Strictly Confidential
(For Internal and Restricted use only)
Secondary School Examination, 2026 (Xth)
SUBJECT NAME : Science (Q.P. CODE /Set No. 086/31-4-2)

General Instructions: -

m
1
om
You are aware that evaluation is the most important process in the actual and correct assessment of
. co
c m
the candidates. A small mistake in evaluation may lead to serious problems which may affect the future
. s e
of the candidates, education system and teaching profession. To avoid mistakes, it is requested that
m
e l a
before starting evaluation, you must read and understand the spot evaluation guidelines carefully.
s g
“Evaluationgl
a a of the
2
a conducted, evaluation done and several other aspects. Its leakage to public in
examinations
policy is a confidential policy as it is related to the confidentiality

any manner could lead to derailment of the examination system and affect the life and future of
millions of candidates. Sharing this policy/document to anyone, publishing in any magazine
and printing in Newspaper/Website, etc. may invite action under various rules of the Board and
IPC.”
3 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
m
.co
information or knowledge and/or are innovative, they may be assessed for their correctness
otherwise and due marks be awarded to them. In Class-X, while evaluating two competency-

e m
based questions, please try to understand given answer and even if reply is not from marking

l as
scheme but correct competency is enumerated by the candidate, due marks should be awarded.

4 ag value points for the answers.
The Marking scheme carries only suggested
These are in the nature of Guidelines only and do not constitute the complete answer. The students
can have their own expression and if the expression is correct, the due marks should be awarded
accordingly.
5 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. If there is any variation, the same should be zero after deliberation and discussion. 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.
m
6
m
Evaluators will mark ( √ ) wherever answer is correct. For wrong answer CROSS ‘X’ be marked.
.co
.co e m
Evaluators will not put right (✓) while evaluating which gives an impression that answer is correct and
s
em a
no marks are awarded. This is most common mistake which evaluators are committing.
s g l
g la different parts of the question should then be totalled up and written
7 If a question has parts, please award marks on the right-hand side a in thepart.left-hand
for each Marks awarded for
a encircled. This may be followed strictly.
margin and

8 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.
9 If a student has attempted an extra question, answer of the question deserving more marks should be
retained and the other answer scored out with a note “Extra Question”.

om
PAGE 1 {31-4-2}
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Page 13 of 36

10 No marks to be deducted for the cumulative effect of an error. It should be penalized only once.

11 A full scale of marks 80(example 0 to 80/70/60/50/40/30 marks as given in Question Paper) has to
be used. Please do not hesitate to award full marks if the answer deserves it.

12 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). This is in view of the reduced syllabus and number of
questions in question paper.
13 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 totalling of marks awarded on an answer.
● Wrong transfer of marks from the inside pages of the answer book to the title page.
● Wrong question wise totalling on the title page.
● Wrong totalling of marks of the two columns on the title page.
● Wrong grand total.
● Marks in words and figures not tallying/not same.
● 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.

14 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.
15 Any unassessed portion, non-carrying over of marks to the title page, or totalling 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.
16 The Examiners should acquaint themselves with the guidelines given in the “Guidelines for Spot
Evaluation” before starting the actual evaluation.
17 Every Examiner shall also ensure that all the answers are evaluated, marks carried over to the title
page, correctly totalled and written in figures and words.
18 The candidates are entitled to obtain photocopy of the Answer Book on request on payment of the
prescribed processing fee. All Examiners/Additional Head Examiners/Head Examiners are once
again reminded that they must ensure that evaluation is carried out strictly as per value points for
each answer as given in the Marking Scheme.

PAGE 2 {31-4-2}

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.
.co s e m

s em l a
a ag
MARKING SCHEME
SCIENCE (Subject Code-086)
(PAPER CODE : 31/4/2) (10-04-86K)

Q.No. EXPECTED OUTCOMES/VALUE POINTS Marks Total
Marks
m
om SECTION – A . co
. c em
e m
BIOLOGY
as
l 1
1.
as
(C) / Formation of fruit 1
ag
(B) / (iv)gl
2.
a 1 1
3. (A) / (i), (ii) and (iii) 1 1
4. (B) / (i) and (iv) 1 1
5. (D) / Contraction of Left Ventricle. 1 1
6. (B) / Grass and tree 1 1
7. (A) / Entry of water into guard cells. 1 1
m
8.

m .co
(A) / Both, Assertion (A) and Reason (R)are true, and Reason (R) is correct 1 1

e
explanation of Assertion (A).
9.
l
(D) / Assertion (A) is false and Reason (R)asis true. 1 1
10. (A) a g
• As his pancreatic duct is blocked, enzymes for digestion will not be 1
transported to small intestine / The blockage will cause difficulty in
digestion of proteins, carbohydrates and fats.
• Insulin secretion is already affected. 1

OR
(B)
m½
m • Reflex action
.co
m .co • An automatic and quick response to a stimulus which does not involve
s em ½

e la
thinking.

las • Stimulus → Receptor → sensory neuron → relay neuron (spinal cord) →
ag 1

ag Motor neuron → effector (muscle) response.
2
11. • Necessary- It helps in transport of water and minerals from roots to 1
leaves / It helps in temperature regulation.
• Evil - It results in loss of water. 1
2

om
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Page 15 of 36

12. (a) uv
O 2 ⎯⎯→ O + O
1
O + O2 → O3 /
(Ozone)
UV radiations split molecular oxygen into free oxygen and these atoms
combine with oxygen to form ozone.
(b) It can cause diseases like skin cancer in human beings. 1
(or any other harmful effect) 2

13. (a) Parents : RR  rr
(Red flowers) (White flowers) ½ ½

Gametes :
½ ½
F1 (All red)

Rr  Rr ½ ½

Gametes :
F2
½ ½

(b)
Phenotypic ratio/ Showing external look of plants : 33 : 11 ½ ½
3 : 1
(Red) (White) 3
Genotypic Ratio : 1: 2: 1
½ ½
RR : Rr : rr 3

14. (a) Some waste is stored in leaves that fall off/wastes may be stored in
cellular vacuoles/other wastes are stored as resins and gums in old xylem/ 1
may excrete some waste into soil via roots.
(b) Mammals and birds have high energy needs to maintain their body
temperature. Therefore, mixing of blood is prevented by separation of
right and left side of heart and thus they have double circulation. 2

3

PAGE 4 {31-4-2}

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.co s e m

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15. (a) Due to hydrotropism, the roots grow towards area of water availability so 1
that they can absorb water and minerals.

(b) Auxins promote cell elongation / Auxin regulates the tropic responses 1
like bending (growth) of stem towards light.

m
co
(c)

o m
Movement by sensitive plantsTropic Movements by plants
m .
. c
Non-directional Directional 1
s e
Fast em a
s l
1 ag
Slow

g a independent movement Growth related
lGrowth
a (Any two points and any other difference)

OR
(c)

Movement of Roots Movement of Shoots
Gravity: Grows towards gravity / Gravity: Grows away from gravity / 1
m
.co
Positive geotropism Negative geotropism
Light: Grows away from light / Light: Grows towards light/ Positive
em
Negative phototropism phototropism 1

l as 4

16. (A)
ag
(a)

1+1

m
m .co
m .co s e m
s e g la
g la a
a (b)
• Absorption of light energy by chlorophyll.
• Conversion of light energy to chemical energy and splitting of water
molecule into hydrogen and oxygen. 2
• Reduction of carbon dioxide to carbohydrates

om
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(c)
1

OR

(B)
(a)

1
Diagram
½x4
Labelling of any 4 parts

1
(b) Filtration of blood to remove wastes.
(c) It is a process by which useful substances like glucose, amino acids,
vitamins, salts and most of water are reabsorbed from filtrate, back into 1
5
the blood.
SECTION-B
CHEMISTRY
17. (D)/ Brown fumes of NO2 1 1
18. (A) / The hydrophobic tail of the soap molecule is in the interior of the cluster, 1 1
whereas hydrophilic end is on the surface of the cluster.
19. (C) / Exothermic Reaction 1 1
20. (B) / Hydrochloric acid 1 1
21. (C) / Iron 1 1
22. (A) / By adding acid to water with constant stirring. 1 1

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23. (D) / Steel 1 1
24. (A) / Both A and R is true and R is correct explanation of A. 1 1
25. (a) Ca (OH) 2 + Cl2 → CaOCl2 + H2O / 1

m
om
(Bleaching powder)
. co
. c m
e2
NaCl + Hm
e2O + CO2 + NH3 → NH4Cl + NaHCO3 l as
s
(b)
g
1

g la (Baking soda) a
26. (a) a
• Substance oxidised = Na ½
• Substance reduced = O2 ½
(b) Double displacement/Precipitation reaction 1
(c) Silver chloride decomposes into silver and chlorine by sunlight. 1 3
27. (A)
m
.co
(i) (a) Carbonate ore: Calcination
½
e m
Sulphide ore : Roasting la
s
ag
(b)

½

(ii)

1

(iii)
m
c. o
m • Anode: Impure copper rod.
m .co ½

m
• Cathode: Pure copper strip.
s e ½

s e OR
g la
g la (B)
a
a (i)
1
1

(ii) Nitric acid is a strong oxidising agent and oxidises H2 gas produced to
1
water.
3

om
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28. (a) Universal indicator 1
(b) A/ (pH = 3) 1

(c)
• Strong acids give more H+ ions in water. 1
• Weak acids give less H+ ions in water. 1
OR

(c) (i) When pH < 5.6 1
(ii) Weak acid: acetic acid, formic acid ½+½ 4
29. (A)
(i) Carbon cannot form C4+ cation because removal of four electrons is
energetically not possible. Carbon cannot form C4− anion because nucleus with 2
six protons cannot hold ten electrons.
(ii)
• A series of compounds in which the same functional group substitutes for 1
hydrogen in carbon chain is called a homologous series.
• Because of higher molecular mass of C4H10 than C3H8 or C2H6.
1
(iii) /
Ethanoic acid / acetic acid CH3COOH is formed /
acidifiedK2Cr2O7 +Heat 1
CH3-CH2OH → CH3COOH
OR
(B)
1
(i) I. Propanal
1
II. Propyne
1
(ii) I. Ester group
II.
1

(iii) Soaps reacts with calcium and magnesium salts present in hard water to
1
form scum (insoluble substance).
5
SECTION-C
PHYSICS
30. (C) / Presbyopia 1 1
31. (A) / Iris and pupil 1 1

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32. (A) / Both Assertion (A) and Reason (R) both are true, and Reason (R) is the 1 1
correct explanation of Assertion
33. (A)
P
• I= ½
V
m
co
2000
I=
om .
200
I = 10 A
. c ½
em
• e m l as
l as
Current passing through electric heater is 10A which is much more than
rated value (4A) of fuse. Hence fuse will melt and break the circuit. So, it 1 g
a
ag
cannot be used.
OR
(B)
• An electromagnet is formed by wrapping a current carrying insulated
copper wire in the form of coil around a magnetic material like soft iron
1
core. /
By placing a magnetic material like soft iron as a core material inside the

m
current carrying solenoid.

.co
• Strength of electromagnet can be increased by increasing the

m
(i) Number of turns of coil

s e ½

a
(ii) Current flowing through the coil.
l
½ 2
34. (i) ag
• Voltmeter ½
• Used to measure potential difference ½
(ii)
• Variable resistance / Rheostat ½
• Used to change the resistance in the circuit. / Regulates current without
changing the voltage source. ½ 2
m
m .co
35. (a)

c. o •• The rod AB gets displaced from its original position
s e m ½

em (b) a
Because it experiences a force when placed in external magnetic field
l
½

las a g
ag •
•
Fleming’s Left Hand Rule
Stretch the thumb, fore finger and middle finger of your left hand such that
1

they are mutually perpendicular. If forefinger points in the direction of
magnetic field, second finger in the direction of current, then the thumb
will point in the direction of motion or force acting on the conductor. 1 3

om
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36. (a) (i) Speed of light in medium A > Speed of light in medium B. ½
(ii) Media B and C both have same optical density. ½

(b)

2

(Deduct ½ mark for not showing direction of ray of light) 3
37. (i)
• Myopia 1
• Excessive curvature of eye lens / Elongation of eye ball 1
(ii)

1
3
38. (a) Real / magnified. 1
(b) Converging / Concave mirror, Plane mirror 1

(c)
• Convex lens 1
• It magnifies the image formed by the curved mirror 1
OR
(c) The plane mirror redirects/ reflects the light rays coming from the curved
2 4
mirror towards eyepiece.

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39. (A) (i)
• Since V-I graph is a straight line passing through the origin / V ∝ I, it
follows Ohm’s Law. 1
• R = Slope of V-I graph
0.8−0.4 0.4
R= = = 4Ω
0.2−0.1 0.1

m
(Note: Resistance can be calculated by taking any other two points from the 1

om
graph. For the point 2.2V and 0.6A answer will vary)
. co
. c em
m
e and 7 resistors are in series, R = 3 + 7 = 10Ω l as
(ii) (I)
as3
l R is in parallel combination with 10
s
ag
ag s

= +
1 1 1
𝑅𝑝 10 10
Rp = 5 1
Other two 5 resistors are in series with Rp.
Net R = 5 + 5 + 5 = 15 1
V ½
(II) Total current, I =
R
5
I=
m
.co
15
1 ½
em
I= A
3

l as
OR

(B) (i)
ag
• Power consumed by a device that carries 1 A of current when operated at
a potential difference of 1 V / If one joule energy is consumed in one 1
second then power of instrument is said to be 1 watt / 1 W = 1 volt × 1
ampere
• P = VI 1
(ii)

o m ½
c
E=Pxt
m .
c. o Energy (3 bulbs) = 3  100  5
s e m
s em = 1500
g la
l a a
ag
= 1.5 kWh
Energy (electric heater) = 1.0  0.5
= 0.5 kWh
½
Total energy consumed (1 day) = 1.5 + 0.5 = 2 kWh
Total energy consumed (30 days) = 30  2

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= 60 kWh
= 60 units
Total cost = Units  Rate
= 60  3.60
= ₹ 216 1

(ii) 1 kW h = 1000 watt × 3600 second
= 3.6 × 106 watt second 1 5
= 3.6 × 106 joule (J)
-o0o-

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.
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s em g l a
a Strictly Confidential a
Marking Scheme

(For Internal and Restricted use only)
Secondary School Examination, 2026 (Xth)
SUBJECT NAME : Science (Q.P. CODE /Set No. 086/31-4-3)

General Instructions : -

m
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
om . co
c m
which may affect the future of the candidates, education system and teaching profession. To

m .
avoid mistakes, it is requested that before starting evaluation, you must read and understand
s e
s e
the spot evaluation guidelines carefully.
g l a
2
l a
“Evaluation policy is a confidential policy as it is related to the confidentiality of the
g a
a
examinations conducted, evaluation done and several other aspects. Its leakage to
public in any manner could lead to derailment of the examination system and affect the
life and future of millions of candidates. Sharing this policy/document to anyone,
publishing in any magazine and printing in Newspaper/Website, etc. may invite action
under various rules of the Board and IPC.”
3 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
m
.co
be assessed for their correctness otherwise and due marks be awarded to them. In
Class-X, while evaluating two competency-based questions, please try to understand
e m
s
given answer and even if reply is not from marking scheme but correct competency is

l a
enumerated by the candidate, due marks should be awarded.
4
ag
The Marking scheme carries only suggested value points for the answers.
These are in the nature of Guidelines only and do not constitute the complete answer. The
students can have their own expression and if the expression is correct, the due marks should
be awarded accordingly.
5 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. If there is any variation, the same should be zero after deliberation
and discussion. 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.
m
.co
6 Evaluators will mark ( √ ) wherever answer is correct. For wrong answer CROSS ‘X’ be
m
.co m
marked. Evaluators will not put right (✓) while evaluating which gives an impression that

m
answer is correct and no marks are awarded. This is most common mistake which
s e
s e evaluators are committing.
g la
g la 7
a
If a question has parts, please award marks on the right-hand side for each part. Marks

a 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.
8 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.
9 If a student has attempted an extra question, answer of the question deserving more marks
should be retained and the other answer scored out with a note “Extra Question”.

om
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10 No marks to be deducted for the cumulative effect of an error. It should be penalized only
once.
11 A full scale of marks 80 (example 0 to 80/70/60/50/40/30 marks as given in Question Paper)
has to be used. Please do not hesitate to award full marks if the answer deserves it.

12 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).This is in view of the reduced
syllabus and number of questions in question paper.
13 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 an answer.
● 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/not same.
● 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.
14 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.
15 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.
16 The Examiners should acquaint themselves with the guidelines given in the “Guidelines for
Spot Evaluation” before starting the actual evaluation.
17 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.
18 The candidates are entitled to obtain photocopy of the Answer Book on request on payment
of the prescribed processing fee. All Examiners/Additional Head Examiners/Head
Examiners are once again reminded that they must ensure that evaluation is carried out
strictly as per value points for each answer as given in the Marking Scheme.

PAGE 1 {31- 4- 3}

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MARKING SCHEME
SCIENCE (Subject Code-086)
(PAPER CODE : 31/4/3) (10-04-86K)

Q.No. EXPECTED OUTCOMES/VALUE POINTS Marks Total
Marks
m
co
SECTION – A

. c om Biology
e 11m .
1. (B) / (iv)
e m
1
s
la 1
2.
l as
(C) / By breaking down the nutrients of bread outside the body and then absorbing them.
ag
1

g
3. (C) / Formation of fruit 1
4. a
(D) / Mammals 1 1
5. (D) / Organic substance from leaves to the other parts of the plant. 1 1
6. (D) / Contraction of Left Ventricle. 1 1
7. (B) / (i) and (iv) 1 1
8. (A) / Both Assertion (A) and Reason(R) are true and Reason(R) is correct explanation 1 1
of Assertion (A).
9. (C) / Assertion(A) is true, but Reason(R) is false. 1 1
10.
m
• Decomposers are organisms that break down dead and decaying organic matter
into simpler inorganic substances.
• Roles:
m .co 1

➢ Environment clean up s e
l a ½

ag
➢ Nutrient recycling ½ 2
(or any other relevant point)
11. • Food:
▪ P – Glucose ½
▪ Q – Starch ½
• Raw materials: CO2, H2O ½
• Conditions: Presence of sunlight, chlorophyll ½
2
12. (A)
m
c
•
om
As his pancreatic duct is blocked, enzymes for digestion will not be transported
m.co 1

. e
to small intestine / The blockage will cause difficulty in digestion of proteins,

e m carbohydrates and fats.
las
las • Insulin secretion is already affected.
ag 1

ag OR
(B)
• Reflex action
½
• An automatic and quick response to a stimulus which does not involve thinking.
½
• Stimulus → Receptor → sensory neuron → relay neuron (spinal cord) → Motor
neuron → effector (muscle) response. 1 2

om
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13. (a)
Parents: TTPP X ttpp
tall purple short white

Gametes: TP tp
↓
F1 TtPp

Selfing of F1
TtPp  TtPp 1

F2
gametes TP Tp tP tp
TP TTPP TTPp TtPP TtPp
Tp TTPp TTpp TtPp Ttpp
tP TtPP TtPp ttPP ttPp
tp TtPp Ttpp ttPp ttpp

tall and purple : 9
tall and white : 3
1
short and purple : 3
short and white : 1
1 3
(b) In F2 progeny probability of Tall, White = 3/16
14. (a) Because adrenal gland secretes adrenaline hormone that enables the 1
animal body to face a scary situation.
(b) To prevent female foeticide. 1
(c) Plants raised by vegetative propagation can bear flowers and fruits earlier than
those produced from seeds /
Such methods also make possible the propagation of plants such as banana,
orange, rose and jasmine that have lost the capacity to produce seeds /
All plants produced are genetically similar enough to the parent plant to have all 1
its characteristics. 3

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15. (a) Due to hydrotropism, the roots grow towards area of water availability so that 1
they can absorb water and minerals.
(b) Auxins promote cell elongation / Auxin regulates the tropic responses like 1
bending (growth) of stem towards light.
(c)
Movement by sensitive plants Tropic Movements by plants
m
co
Non-directional Directional
om .
1
Fast
. c Slow
e m
e m
Growth independent movement Growth related 1
l as
l as (Any two points and any other difference)
ag
ag
OR
(c)

Movement of Roots Movement of Shoots
Gravity: Grows towards gravity / Gravity: Grows away from gravity / 1
Positive geotropism Negative geotropism
Light: Grows away from light / Light: Grows towards light/ Positive
m
.co
Negative phototropism phototropism
1 4
16. (A)
e m
l as
ag
(a)

1+1

m
m
c. o•• Absorption
(b)
m .co
of light energy by chlorophyll.
s e
s em Conversion of light energy to chemical energy and splitting of
l a
water
g
molecule

la a
into hydrogen and oxygen.

ag
• Reduction of carbon dioxide to carbohydrates 2
(c)
1

OR

om
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(B)
(a)

Diagram 1
Labelling of any 4 parts ½x4

(b) Filtration of blood to remove wastes. 1
(c) It is a process by which useful substances like glucose, amino acids,
vitamins, salts and most of water are reabsorbed from filtrate, back into the 1
blood. 5
SECTION – B
Chemistry
17. (B)/ Hydrochloric acid 1 1
18. (A)/ The hydrophobic tail of the soap molecule is in the interior of the cluster, whereas 1 1
hydrophilic end is on the surface of the cluster.
19. (D)/ Magnesium 1 1
20. (C)/ Exothermic Reaction 1 1
21. (B)/ 2 : 1 1 1
22. (C)/ Lead and Tin 1 1
23. (B)/ Sodium Hydrogen Carbonate 1 1
24. (A)/ Both A and R is true and R is correct explanation of A. 1 1
25. (a)

1
Heat
(b) 2NaHCO3 → Na2 CO3 + H2 O + CO2
1 2
(Deduct ½ mark for no/ incorrect balancing)

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26. (A)
(i) (a) Carbonate ore: Calcination
½

m
(b) Sulphide ore: Roasting

om . co
c m
½

m . s e
(ii)
s e g l a
g l a a
1
a
(iii) ½
• Anode: Impure copper rod. ½
• Cathode: Pure copper strip.
OR
m
.co
(B)
(i)
e m
l a s 1

ag 1
1 3
(ii) Nitric acid is a strong oxidising agent and oxidises H2 gas produced to water.
27. (a)
• Green colour of the crystal changes brown. ½
• Colourless gases with odour of burning sulphur. ½
(any other observation)
(b) Copper changes to copper (II) oxide which is black in colour./

m
om .co
1

m . c s em
s e (c) l a
ag /
(Award full marks if chemical equation with colour changes is written)

g la Calcium oxide (quick lime) and colourless gas (CO2) is evolved)
a 1
3

om
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28. (a) Universal indicator 1
(b) A/ (pH = 3) 1
(c)
• Strong acids give more H+ ions in water. 1
• Weak acids give less H+ ions in water. 1
OR

(c) (i) When pH<5.6 1 4
(ii) Weak acid: acetic acid, formic acid ½+½
29. (A)
(i) Carbon cannot form C4+ cation because removal of four electrons is energetically
not possible. Carbon cannot form C4− anion because nucleus with six protons cannot 2
hold ten electrons.
(ii)
• A series of compounds in which the same functional group substitutes for 1
hydrogen in carbon chain is called a homologous series.
• Because of higher molecular mass of C4H10 than C3H8 or C2H6.
1
(iii) Ethanoic acid / acetic acid / CH3COOH is formed /
1

OR
(B)
(i) I. Propanal 1
II. Propyne 1
(ii) I. Ester group 1
II.
1

(iii) Soaps reacts with calcium and magnesium salts present in hard water to form 1
scum (insoluble substance).
5
SECTION – C
Physics
30. (A) / Iris and pupil 1 1
31. (C) / Presbyopia 1 1
32. (A) / Both Assertion (A) and Reason (R) both are true, and Reason (R) is the correct 1 1
explanation of Assertion

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33. (A)
𝑃
• 𝐼= ½
𝑉
2000
𝐼=
200
I = 10 A
m
½
• Current passing through electric heater is 10A which is much more than rated
om . co
. c
value (4A) of fuse. Hence fuse will melt and break the circuit. So, it cannot be
1
e m
used.
e m l as
l a s OR
ag
(B)
g
• Anaelectromagnet is formed by wrapping a current carrying insulated copper
wire in the form of coil around a magnetic material like soft iron core. /
By placing a magnetic material like soft iron as a core material inside the current 1
carrying solenoid.
• Strength of electromagnet can be increased by increasing the
½
(i) Number of turns of coil
½ 2
(ii) Current flowing through the coil.
34. (i)
m
The potential difference, V, across the ends of a given metallic wire in an electric

m .co
circuit is directly proportional to the current, I flowing through it, provided its 1
temperature remains the same
s e
(ii)
l a
ag
1

2

35. Rainbow is caused by dispersion of sunlight by tiny water droplets, present in the
atmosphere. The water droplets act like small prisms. They refract and disperse the
m
.co
incident sunlight, then reflect it internally, and finally refract it again when it comes out 2
m
of the raindrop.

m .co s e m
s e g la
g la a
a 1

3

om
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36. (a)
• The rod AB gets displaced from its original position ½
• Because it experiences a force when placed in external magnetic field ½
(b)
• Fleming’s Left Hand Rule 1
• Stretch the thumb, fore finger and middle finger of your left hand such that they
are mutually perpendicular. If forefinger points in the direction of magnetic
1
field, second finger in the direction of current, then the thumb will point in the
direction of motion or force acting on the conductor. 3

37. (a) f = + 2m u = − 8m
1 1 1 ½
v
+ =
u f
1 1 1
= −
v f u
1 1 1
= + ½
v 2 8
8 1
v = = 1.6 m
5
The image is formed 1.6m behind the mirror
(b)

1

3
38. (a) Real / magnified. 1
(b) Converging / Concave mirror, Plane mirror 1
(c)
• Convex lens 1
• It magnifies the image formed by the curved mirror 1
OR
(c) The plane mirror redirects/ reflects the light rays coming from the curved
mirror towards eyepiece. 2
4

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.
.co s e m

s em l a
a ag
39. (A) (i)

m
om . co
. c e m
e m l as
a s
As in parallell combination, potential difference applied across each resistor is same. ag
ag through each resistor R , R and R is
1
Current flowing 1 2 3
V V V ½
I1 = , I2 = , I3 =
R1 R2 R3

If resistance of parallel combination is Rp, then current drawn from battery is
V
I=
Rp

Current, I = I1 + I2 + I3 ½
V V V V
m
.co
= + +
Rp R1 R2 R3
1
=
1
+
1
e m+
1

s the potential difference applied across
Rp R1 R2 R3
1

l a
ag
(ii) As all the resistors are connected in parallel,
each resistor is 12V
V = IR ½
V 12 ½
I1 = = = 6A
R1 2
12
I2 = = 3A ½
4
12
I3 = = 2A ½ 5
6

m
.co
OR
m
c.(i)o Resistivity is the resistance offered by a wire of unit length and unit area mof cross
(B)

s e
s em g l a
la
1
a
section in an electric circuit.

ag (ii)
½
• Resistivity of conducting wire will remain same.
• Resistivity is the characteristic property of the material. ½
½
• Resistance will remain same

om
PAGE 1 {31- 4- 3}
. c
. c e m
m as
se
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𝑙
• R = 𝜌𝐴

If l′ = 2l A′ = 2A
𝑙′
R′ = ρ ′
A
2𝑙
R′ = ρ ½
2A

R′ = R
(iii)
2
H = I Rt 1

H
I=√
Rt

100
I=√
4×1 1 5
I = 5A

-oOo-

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