Karnataka SSLC Question Paper 2025 Answer Key Science

Government of Karnataka
Karnataka Secondary Education Examination Board

Question Paper
ANSWER KEY

CCE RF/RR/PF/PR A
PÜ®ÝìoPÜ ÍÝÇÝ ±ÜÄàûæ ÊÜáñÜᤠÊÜåèƯ|ì¿á ÊÜáívÜÈ, ÊÜáÇæÉàÍÜÌÃÜí, ¸æíWÜÙÜãÃÜá & 560 003
KARNATAKA SCHOOL EXAMINATION AND ASSESSMENT BOARD,
MALLESHWARAM, BENGALURU – 560 003

ÊÜÞa…ì/H²ÅÇ… 2025 ÃÜ ±ÜÄàûæ - 1
MARCH/APRIL 2025 EXAMINATION - 1

ÊÜÞ¨ÜÄ EñܤÃÜWÜÙÜá
MODEL ANSWERS

—⁄MOÊfi}⁄ —⁄MSÊ¿ : 83-E CODE NO. : 83-E
…Œ⁄æ⁄fl : …e¤k´⁄
Subject : SCIENCE
(»èñÜ ËþÝ®Ü, ÃÜÓÝ¿á®Ü ËþÝ®Ü ÊÜáñÜᤠiàÊÜ ËþÝ®Ü / Physics, Chemistry & Biology )
(ÍÝÇÝ A»Ü¦ì / ÍÝÇÝ ±Üâ®ÜÃÝÊÜ£ìñÜ A»Ü¦ì / TÝÓÜX A»Ü¦ì /
TÝÓÜX ±Üâ®ÜÃÝÊÜ£ìñÜ A»Ü¦ì )
( Regular Fresh / Regular Repeater / Private Fresh / Private Repeater )
(»èñÜ ËþÝ®Ü / Physics )
( BMW«ŒÈ »⁄·¤®⁄¥¿»⁄fl / English Medium )
¶´¤MO⁄ : 02. 04. 2025 ] [ V⁄¬Œ⁄r @MO⁄V⁄◊⁄fl : 80

Date : 02. 04. 2025 ] [ Max. Marks : 80

PART – A
( Physics )

Qn.
Value Points Total
Nos.

I. Multiple choice questions : 3×1=3

1. Convex mirrors are used in
(A) torches (B) rear view mirrors of vehicles

(C) search-lights (D) shaving mirrors

Ans. :

(B) rear view mirrors in vehicles 1

CCE RF/RR/PF/PR(A)/101/1837 (MA)-PHY [ Turn over

83-E (PHY) 2
Qn.
Value Points Total
Nos.

2. The sun is visible to us about two minutes before the actual
sunrise and about two minutes even after the actual sunset
because of atmospheric
(A) refraction of light (B) reflection of light
(C) scattering of light (D) dispersion of light
Ans. :
(A) refraction of light 1
3. A current carrying rod is placed between the poles of a
horse-shoe magnet. For the maximum displacement of the
rod the angle between the direction of electric current and
the direction of magnetic field must be
(A) 0° (B) 45°
(C) 90° (D) 180°
Ans. :
(C) 90° 1
II. Answer the following questions : 3×1=3
4. Draw a symbol diagram of a resistor used in an electric
circuit.
Ans. :

1
5. “Two magnetic field lines do not intersect each other.”
Justify.
Ans. :
If they did, it would mean that at the point of intersection
the compass needle would point towards two directions
which is not possible. 1
6. Observe the circuit diagram given below. Mention the
direction of magnetic field that forms around the point P
and around the point Q.

CCE RF/RR/PF/PR(A)/101/1837 (MA)-PHY

3 83-E (PHY)
Qn.
Value Points Total
Nos.

Ans. :
1
Anticlockwise at the point P.
2
1
Clockwise at the point Q.
2

[ Give marks for mentioning the direction in the figure ] 1
III. Answer the following questions : 2×2=4

7. Explain Newton’s experiment that shows the recombination
of spectrum of white light.
Ans. :

Newton used a glass prism to obtain the spectrum of
white light.
When he placed a second identical prism in an inverted
position with respect to the first prism and allowed the
spectrum to pass through it, a beam of white light
emerges.
This shows the recombination of spectrum of white
light. 2

8. Write any two differences between near-sighted ( Myopic )
and far-sighted ( Hypermetropic ) eyes.
Ans. :
Myopia Hypermetropia
Can see nearby objects Can see distant objects
clearly / cannot see clearly / cannot see
distant object distinctly nearby objects distinctly
The image of a distant The image of a nearby
object is formed in front object is formed behind
of the retina the retina
The far point is nearer The near point is farther
than infinity away from the normal
The focal length of the The focal length of the
eyeball is too short / the eyeball is too long / the
eyeball elongates eyeball becomes too
small

CCE RF/RR/PF/PR(A)/101/1837 (MA)-PHY [ Turn over

83-E (PHY) 4
Qn.
Value Points Total
Nos.

This defect can be This defect can be
corrected by using a corrected by using a
concave lens of suitable convex lens of suitable
power power
( Any two ) 1+1 2

IV. Answer the following questions : 3×3=9

9. Draw the ray diagram of image formed when an object is
kept at 2 F1 of the convex lens. With the help of the ray
diagram write the position and nature of the image.
( F 1 : Principal focus of the lens )

Ans. :

Ray diagram of image formation when the object is kept at
2 F2 of the convex lens.

For ray diagram — 2
1
The image is formed at 2 F2
2
1
Real and inverted image is formed.
2 3

10. a) State Ohm’s law.
b) Write the factors on which the resistance of a
conductor depend.
Ans. :

a) The potential difference across the ends of a metallic
wire (resistor) in an electric circuit is directly
proportional to the current flowing through it, provided
its temperature remains the same. 1

CCE RF/RR/PF/PR(A)/101/1837 (MA)-PHY

5 83-E (PHY)
Qn.
Value Points Total
Nos.

b) The resistance of a conductor depends on the following
factors :
1
its length
2
1
its area of cross-section
2
1
the nature of its material
2
1
temperature.
2 3

11. a) What factor helps for determining the relative strength
of a magnetic field ?
b) What is solenoid ? How is an electromagnet prepared
by using it ?
OR
In domestic electric circuits,
a) What is the function of earth wire ?
b) What precautions should be taken to avoid
overloading ?
Ans. :

a) The degree of closeness of the field lines. 1
b) A coil of many turns of insulated copper wire wrapped
closely in the shape of a cylinder is called a solenoid. 1
An electromagnet is prepared by placing a piece of
magnetic material like soft iron inside a solenoid of
strong magnetic field / inside a current carrying
solenoid. 1 3

OR

a) If any leakage of current occurs in the electrical
appliances, keeps their potential difference the same as
to that of the earth / provides a low resistance
conducting path for the current. 1

b) Fuse should be used in the electric circuit
Should avoid the direct contact of live wire and
neutral wire.

CCE RF/RR/PF/PR(A)/101/1837 (MA)-PHY [ Turn over

83-E (PHY) 6
Qn.
Value Points Total
Nos.

Avoid the damage of insulation of wires
Should repair any faults in the electrical
appliances
Should not connect too many appliances to a
single socket. ( Any four ) 1
×4=2 3
2

V. Answer the following questions : 2×4=8

12. a) How ammeter and voltmeter are connected in an
electric circuit ? Mention their function.
b) “In domestic electric circuit, the electrical appliances
are generally connected in parallel.” Give reasons.
Ans. :
1
a) Ammeter is connected in series.
2
1
Voltmeter is connected in parallel.
2
1
Ammeter measures electric current.
2

Voltmeter measures potential difference between
1
two points.
2

b) A parallel circuit divides the current through the
electrical gadgets.
When one component fails, the circuit will not
break. The other components work.
The total resistance is less in this circuit.
More helpful in cases of different current is
required for different electrical appliances.
( Any two ) 1+1 4

13. The radius of curvature of a spherical mirror is 36 cm. Find
its focal length. A candle of 5 cm length is placed at a
distance of 27 cm in front of a convex mirror of the same
focal length. Mention the position and nature of the image
and also find the size of the image formed.
OR
The focal length of a spherical mirror is 10 cm. Find its
radius of curvature. An object of 4 cm size is placed at

CCE RF/RR/PF/PR(A)/101/1837 (MA)-PHY

7 83-E (PHY)
Qn.
Value Points Total
Nos.

20 cm in front of a concave mirror of the same focal length.
At what distance from the mirror should a screen be placed
in order to obtain a sharp image of the object ? Mention the
nature of the image formed and also find the size of the
image.
Ans. :

Given,
Radius of curvature R = 36 cm
Object distance u = – 27 cm
Height of the object h = 5 cm
Image distance v = ?
Height of the image h l = ?
R 1
Focal length f =
2 2

36 1
f = = 18 cm
2 2

1 1 1 1
Since + = ( or )
v u f 2

1 1 1
= −
v f u
1  1  1 1 1
= −   = +
18  − 27  18 27 2

1 3+2 5
= =
v 54 54
54 1
v = = 10·8 cm
5 2

The distance of the image is 10·8 cm
hl −v h ( −v ) 1
= ( or ) h l =
h u u 2

5 ( − 10 ⋅ 8 ) 1
hl = = 2 cm
− 27 2

∴ Height of the image is 2 cm
1
∴ The image is virtual and erect.
2 4

OR

CCE RF/RR/PF/PR(A)/101/1837 (MA)-PHY [ Turn over

83-E (PHY) 8
Qn.
Value Points Total
Nos.

Focal length f = – 10 cm
Object distance u = – 20 cm
Object height h = 4 cm
Image distance v = ?
Image height h l = ?
Radius of curvature = ?
1
Radius of curvature R = 2 f
2
1
= 2 ( 10 ) = 20 cm
2
1 1 1
+ =
v u f
1 1 1 1
( Or ) = −
v f u 2

1  1  1 1 1
= −  = − +
− 10  − 20  10 20 2

1 − 2 +1 1
= = −
v 20 20
1
v = – 20 cm
2

The screen should be placed at 20 cm in front of the mirror.
hl −v +h ( −v ) 1
= ( or ) h l =
h u u 2

4 ( − ( − 20 ) )
=
− 20
4 × 20
=
− 20
1
h l = – 4 cm
2

∴ Image size is – 4 cm
1
∴ The image is real and inverted.
2 4

CCE RF/RR/PF/PR(A)/101/1837 (MA)-PHY

CCE RF/RR/PF/PR A
PÜ®ÝìoPÜ ÍÝÇÝ ±ÜÄàûæ ÊÜáñÜᤠÊÜåèƯ|ì¿á ÊÜáívÜÈ, ÊÜáÇæÉàÍÜÌÃÜí, ¸æíWÜÙÜãÃÜá & 560 003
KARNATAKA SCHOOL EXAMINATION AND ASSESSMENT BOARD,
MALLESHWARAM, BENGALURU – 560 003

ÊÜÞa…ì/H²ÅÇ… 2025 ÃÜ ±ÜÄàûæ - 1
MARCH/APRIL 2025 EXAMINATION - 1

ÊÜÞ¨ÜÄ EñܤÃÜWÜÙÜá
MODEL ANSWERS

—⁄MOÊfi}⁄ —⁄MSÊ¿ : 83-E CODE NO. : 83-E
…Œ⁄æ⁄fl : …e¤k´⁄
Subject : SCIENCE
(»èñÜ ËþÝ®Ü, ÃÜÓÝ¿á®Ü ËþÝ®Ü ÊÜáñÜᤠiàÊÜ ËþÝ®Ü / Physics, Chemistry & Biology )
(ÍÝÇÝ A»Ü¦ì / ÍÝÇÝ ±Üâ®ÜÃÝÊÜ£ìñÜ A»Ü¦ì / TÝÓÜX A»Ü¦ì /
TÝÓÜX ±Üâ®ÜÃÝÊÜ£ìñÜ A»Ü¦ì )
( Regular Fresh / Regular Repeater / Private Fresh / Private Repeater )
( ÃÜÓÝ¿á®Ü ËþÝ®Ü / Chemistry )
( BMW«ŒÈ »⁄·¤®⁄¥¿»⁄fl / English Medium )
¶´¤MO⁄ : 02. 04. 2025 ] [ V⁄¬Œ⁄r @MO⁄V⁄◊⁄fl : 80

Date : 02. 04. 2025 ] [ Max. Marks : 80

PART – B
( Chemistry )
Qn.
Value Points Total
Nos.

VI. Multiple choice questions : 2×1=2

14. Correct statement related to the soaps among the following
is, soaps
(A) easily give excess of foam in hard water
(B) form insoluble precipitate in hard water
(C) easily clean oils like dirt in hard water
(D) are sodium salts of sulphonic acid

CCE RF/RR/PF/PR(A)/101/1837 (MA)-CHE [ Turn over

83-E (Chem.) 2
Qn.
Value Points Total
Nos.

Ans. :

(B) form insoluble precipitate in hard water 1
Heat
15. CuO + H 
2
→ Cu + H 2 O

In this chemical reaction,
(A) hydrogen is reduced to form water
(B) exchange of ions took place between the reactants
(C) copper oxide is oxidised to form copper
(D) copper oxide is reduced to form copper
Ans. :
(D) copper oxide is reduced to form copper 1
VII. Answer the following questions : 2×1=2
16. Mention any two measures for preventing corrosion of iron.
Ans. :
Painting
Oiling
Greasing
Galvanising
Making alloys
Chromium plating
( any two ) 2× 1
2 1
17. How is concentrated acid diluted ?
Ans. :
By adding the acid slowly to the water with constant
stirring. 1

VIII. Answer the following questions : 3×2=6

18. What is rancidity ? Mention any two methods of preventing
rancidity.

Ans. :

When fat and oil containing food materials oxidise, then
their smell and taste change. This is called rancidity. 1

CCE RF/RR/PF/PR(A)/101/1837 (MA)-CHE

3 83-E (Chem.)
Qn.
Value Points Total
Nos.

Methods of preventing rancidity :

Adding substances which prevent oxidation
( antioxidants )

Keeping fried food materials in airtight containers

Flushing bags of chips with nitrogen gas.
( any two ) 2× 1
2 2
Give reason :
19.
a) Zinc oxide is called as an amphoteric oxide
b) Sodium metal is stored in kerosene.
OR
Give reason :
a) Gold is used to make jewellery
b) Ionic compounds in the solid state do not conduct
electricity.

Ans. :

a) Zinc oxide reacts with both acids and bases to produce
salt and water. 1

b) Sodium metal reacts violently with water and
atmospheric oxygen but does not react with
kerosene. 1 2

OR

a) Lustrous

Has ductile property

Has malleable property
Least reactive. ( any two ) 2× 1
2

b) Movement of ions in the solid is not possible due
to their rigid structure.

Free ions will not form. ( any one ) 1 2

CCE RF/RR/PF/PR(A)/101/1837 (MA)-CHE [ Turn over

83-E (Chem.) 4
Qn.
Value Points Total
Nos.

Complete the following oxidation reaction and name the
20.
functional group that is found in the product :

Ans. :
1
Alkaline KMnO 4
2

1
Acidified K 2 Cr 2 O 7
2

1
→ CH 3 COOH
2

1
– COOH / Carboxylic acid.
2 2

IX. Answer the following questions : 2×3=6
Draw the diagram of the arrangement of apparatus used to
21.
show the electrolytic refining of copper. Label the following
parts :
i) Cathode
ii) Impurities.

Ans. :

Cathode

Impurities

Diagram — 2
1 1
Parts — +
2 2 3

CCE RF/RR/PF/PR(A)/101/1837 (MA)-CHE

5 83-E (Chem.)
Qn.
Value Points Total
Nos.

22. Write the chemical equations for the following reactions.
i) Quicklime is reacted with water
ii) Zinc pieces are added to copper sulphate solution
iii) Sodium chloride solution is added to silver nitrate
solution.
OR
Balance the following chemical equations :
i) H2 + O2 → H2O

ii) Na 2CO3 + HCl → NaCl + H2O + CO2

iii) N2 + H2 → NH3

Ans. :
i) Calcium oxide + Water → Calcium hydroxide
CaO + H 2 O → Ca (OH) 2

( any one ) 1
ii) Copper sulphate + Zinc → Zinc sulphate + Copper
CuSO 4 + Zn → ZnSO 4 + Cu

( any one ) 1
iii) Sodium chloride + Silver nitrate → Silver chloride
+ Sodium nitrate
NaCl + AgNO 3 → AgCl + NaNO 3 .

( any one ) 1 3
OR
1 1
i) 2H 2 + O 2 → 2H 2 O +
2 2
1 1
ii) Na 2 CO 3 + 2HCl → 2NaCl + H 2 O + CO 2 +
2 2
1 1
iii) N2 + 3H2 → 2NH3 +
2 2 3

X. Answer the following question : 1×4=4
a) NaOH, Ca (OH) 2 , H 2 and Cl 2 materials are given to
23.
you. By using which of these materials you can prepare
bleaching powder ? Write the chemical name and one
use of the bleaching powder.

CCE RF/RR/PF/PR(A)/101/1837 (MA)-CHE [ Turn over

83-E (Chem.) 6
Qn.
Value Points Total
Nos.

b) i) How do you identify a solution as basic solution
by using blue litmus paper ?
ii) Under what condition does a farmer treat the soil
of his field with slaked lime ?

Ans. :
1
a) Ca (OH) 2
2
1
Cl 2
2
1
CaOCl 2 / Calcium oxychloride
2

— for bleaching cotton and linen in the textile
industry
— for bleaching wood pulp in paper factories
— for bleaching washed clothes in laundry
— as an oxidising agent in many chemical
industries
— as disinfectant to make drinking water free
from germs.
1
( any one )
2

b) i) The blue litmus paper does not change its colour
in basic solution. 1
ii) When acidic property of soil increases
When pH value of soil decreases
When the concentration of H + / H3O+ ions

in soil increases.
( any one ) 1 4

XI. Answer the following question : 1×5=5
a) Write any two differences between saturated and
24.
unsaturated carbon compounds.
b) What are structural isomers ? Write the structural
isomers of butane.

CCE RF/RR/PF/PR(A)/101/1837 (MA)-CHE

7 83-E (Chem.)
Qn.
Value Points Total
Nos.

c) Write the electron dot structure of methane molecule.

Ans. :

a)
Saturated carbon Unsaturated carbon
compounds compounds
Have single bond Have double or triple
between carbon-carbon bonds between carbon-
atoms. carbon atoms.
Less reactive More reactive
Undergo substitution Undergo addition and
reaction substitution reactions.
Give blue flame on Give yellow / red flame
combustion on combustion

( any two ) 2×1

b) Carbon compounds with same molecular formula but
different structures are called structural isomers. 1

1 1
+
2 2

c)

1 5

CCE RF/RR/PF/PR(A)/101/1837 (MA)-CHE [ Turn over

CCE RF/RR/PF/PR A
PÜ®ÝìoPÜ ÍÝÇÝ ±ÜÄàûæ ÊÜáñÜᤠÊÜåèƯ|ì¿á ÊÜáívÜÈ, ÊÜáÇæÉàÍÜÌÃÜí, ¸æíWÜÙÜãÃÜá & 560 003
KARNATAKA SCHOOL EXAMINATION AND ASSESSMENT BOARD,
MALLESHWARAM, BENGALURU – 560 003

ÊÜÞa…ì/H²ÅÇ… 2025 ÃÜ ±ÜÄàûæ - 1
MARCH/APRIL 2025 EXAMINATION - 1

ÊÜÞ¨ÜÄ EñܤÃÜWÜÙÜá
MODEL ANSWERS
—⁄MOÊfi}⁄ —⁄MSÊ¿ : 83-E CODE NO. : 83-E
…Œ⁄æ⁄fl : …e¤k´⁄
Subject : SCIENCE
(»èñÜ ËþÝ®Ü, ÃÜÓÝ¿á®Ü ËþÝ®Ü ÊÜáñÜᤠiàÊÜ ËþÝ®Ü / Physics, Chemistry & Biology )
(ÍÝÇÝ A»Ü¦ì / ÍÝÇÝ ±Üâ®ÜÃÝÊÜ£ìñÜ A»Ü¦ì / TÝÓÜX A»Ü¦ì /
TÝÓÜX ±Üâ®ÜÃÝÊÜ£ìñÜ A»Ü¦ì )
( Regular Fresh / Regular Repeater / Private Fresh / Private Repeater )
( iàÊÜ ËþÝ®Ü / Biology )
( BMW«ŒÈ »⁄·¤®⁄¥¿»⁄fl / English Medium )
¶´¤MO⁄ : 02. 04. 2025 ] [ V⁄¬Œ⁄r @MO⁄V⁄◊⁄fl : 80

Date : 02. 04. 2025 ] [ Max. Marks : 80

PART – C
( Biology )
Qn.
Value Points Total
Nos.

XII. Multiple choice questions : 3×1=3

25. The information source for making proteins in the cells is
(A) Gene (B) Chromosome

(C) DNA (D) Ribosome

Ans. :

(C) DNA 1

CCE RF/RR/PF/PR(A)/101/1837 (MA)-BIO [ Turn over

83-E (Bio) 2
Qn.
Value Points Total
Nos.

26. Blood sugar level increases : Undersecretion of insulin
hormone :: Swelling of the neck : ..................... .
(A) Undersecretion of thyroxine hormone
(B) More secretion of insulin hormone
(C) More secretion of thyroxine hormone
(D) Excess intake of iodine.
Ans. :
(A) Undersecretion of thyroxine hormone 1
27. Type of asexual reproduction that occurs in spirogyra is
(A) Budding (B) Regeneration
(C) Binary fission (D) Fragmentation
Ans. :
(D) Fragmentation 1
XIII. Answer the following questions : 3×1=3
28. What is the function of ozone layer ?
Ans. :
It shields the surface of the earth from ultraviolet (UV)
radiation coming from the sun. 1
29. What is neuron ?
Ans. :
The structural and functional unit of nervous system /
nerve tissue 1
30. The events that occur during photosynthesis are given
below. Write these events in correct order.
i) Splitting of water molecules into hydrogen and oxygen
ii) Absorption of light energy by chlorophyll
iii) Conversion of carbon dioxide to carbohydrates
iv) Conversion of light energy to chemical energy.
Ans. :
ii) Absorption of light energy by chlorophyll
iv) Conversion of light energy to chemical energy
i) Splitting of water molecules into hydrogen and oxygen
iii) Conversion of carbon dioxide to carbohydrates. 1

CCE RF/RR/PF/PR(A)/101/1837 (MA)-BIO

3 83-E (Bio)
Qn.
Value Points Total
Nos.

XIV. Answer the following questions : 3×2=6
31. How do auxins promote the growth of tendrils of climbing
plants around a support ?
OR
How does our body respond when adrenaline is secreted
into the blood ?
Ans. :
When tendrils come in contact with any support,
auxins diffuse to the part of tendril which is away from
the support. 1
Due to this, the part of tendril away from the support
will grow faster than the part of tendril near to the
support and causes the tendril to circle around the
support. 1 2
OR

The heart beats faster, resulting in supply of more
1
oxygen to our muscles.
2

The blood to the digestive system and skin will be
reduced due to the contraction of muscles around
1
small arteries in these organs.
2
1
This diverts the blood to our skeletal muscles.
2

The breathing rate also increases because of the
1
contractions of the diaphragm and the rib muscles.
2 2
32. Construct a food chain using the organisms ; snake, frog,
grass and grasshopper. Which organism has more
accumulation of harmful chemicals in this food chain ?
Ans. :

Grass → Grasshopper → Frog → Snake 1
Snake 1 2

33. Draw the diagram showing the germination of pollen on
stigma and label ‘ovary’.

CCE RF/RR/PF/PR(A)/101/1837 (MA)-BIO [ Turn over

83-E (Bio) 4
Qn.
Value Points Total
Nos.

Ans. :

Ovary

Diagram — 11
2
1
Part —
2 2

XV. Answer the following questions : 4 × 3 = 12

34. What is the role of the following enzymes in human
alimentary canal ?
i) Trypsin
ii) Amylase
iii) Lipase
Ans. :

i) Digests proteins present in food. 1
ii) Breaks down starch which is a complex molecule to
give simple sugar. 1
iii) Breaks down emulsified fats. 1 3

CCE RF/RR/PF/PR(A)/101/1837 (MA)-BIO

5 83-E (Bio)
Qn.
Value Points Total
Nos.

35. Justify the following statements :
a) Sexual type of reproduction leads to more variations.
b) In woman’s uterus the role of placenta is significant for
the development of foetus.
OR
a) How does menstruation occur in women ?
b) In male reproductive system the testes are located
outside the abdominal cavity in scrotum. Why ?
Ans. :
a) Sexual reproduction has the involvement of DNA
molecules from two different organisms.
( The combination of male and female gametes
with different genes takes place ) 1
The variations in each generation of population of
organisms increase due to new combination of
genes. 1

b) Placenta
Provides a large surface area for glucose and
1
oxygen to pass from the mother to the embryo.
2

Removes waste substances generated by the
developing embryo by transferring into the
1
mother’s blood.
2 3

OR
a) If the egg is not fertilized, it lives for about one
day. 1
Thick and spongy lining of the uterus slowly
breaks and comes out through the vagina as blood
and mucous.
( Menstruation takes place ). 1

b) Sperm formation requires a lower temperature than the
normal body temperature. 1 3

CCE RF/RR/PF/PR(A)/101/1837 (MA)-BIO [ Turn over

83-E (Bio) 6
Qn.
Value Points Total
Nos.

36. Draw the diagram showing the structure of human brain.
Label the following parts :

i) Hypothalamus

ii) Medulla

Ans. :

Hypothalamus

Medulla

Diagram — 2
1 1
Part — +
2 2 3

37. Round green colour ( RRyy ) seeds producing pea plant is
crossed with wrinkled yellow colour seeds ( rrYY ) producing
pea plant. Show the result of F2 generation with the help of

a checker board and write the ratio of varieties of plants.

OR

a) Explain the sex determination of a child in human
beings.

b) How can the traits of organisms be decided as either
dominant or recessive ?

CCE RF/RR/PF/PR(A)/101/1837 (MA)-BIO

7 83-E (Bio)
Qn.
Value Points Total
Nos.

Ans. :

RRyy × rrYY
( round, green ) ( wrinkled, yellow )

Gametes : Ry rY

RrYy F 1 generation

( round, yellow )
RrYy × RrYy
( F1 ) ( F1 )

RY Ry rY ry
RY RRYY RRYy RrYY RrYy
Ry RRYy RRyy RrYy Rryy
rY RrYY RrYy rrYY rrYy
ry RrYy Rryy rrYy rryy
To checker board — 2
The ratio of varieties of plants is = 9 : 3 : 3 : 1 1 3
OR
a) One odd pair of sex chromosomes found in father.
1
They are X and Y chromosomes.
2
1
But, mother has only XX chromosomes.
2

The child received X haploid chromosome from
father also receives X chromosome from mother,
therefore XX pair of chromosome represents girl
1
child.
2

The child received Y haploid chromosome from
father, also receives X chromosome from mother,
therefore XY pair of chromosome represents boy
1
child.
2

OR

CCE RF/RR/PF/PR(A)/101/1837 (MA)-BIO [ Turn over

83-E (Bio) 8
Qn.
Value Points Total
Nos.

Parents : Father Mother
XY XX

Gametes : X Y X X

Zygote : XX XX XY XY
Gender : Girl Girl Boy Boy 2
b) The traits that express / appear in the organisms are
dominant traits. The traits that are invisible / hidden /
less appearing traits are considered as recessive
traits. 1 3
XVI. Answer the following question : 1×4=4
38. a) Explain briefly the role of haemoglobin pigments
present in our blood.
b) What are the different strategies of excretion found in
plants ?
Ans. :
a) Haemoglobin pigments give red colour to the blood
and have high affinity towards oxygen. 1
These supply dissolved oxygen to all the cells of
the body through the blood circulation. 1
( In turn help in the production of energy )
b) Get rid of excess water by transpiration
Waste products of plants are stored in cellular
vacuoles
Leaves fall off that store waste products in them.
Waste products are stored as resins and gums in
old xylem.
Plants excrete some waste substances into the soil
around them.
( Any four ) 4× 1
2 4

CCE RF/RR/PF/PR(A)/101/1837 (MA)-BIO

CCE RF/RR/
A PF/PR
PÜ®ÝìoPÜ ÍÝÇÝ ±ÜÄàûæ ÊÜáñÜᤠÊÜåèƯ|ì¿á ÊÜáívÜÈ,
ÊÜáÇæÉàÍÜÌÃÜí, ¸æíWÜÙÜãÃÜá – 560 003
KARNATAKA SCHOOL EXAMINATION AND ASSESSMENT BOARD,
MALLESHWARAM, BENGALURU – 560 003
ÊÜÞa…ì/H²ÅÇ… 2025 ÃÜ ±ÜÄàûæ - 1
MARCH/APRIL 2025 EXAMINATION - 1
ÊÜÞ¨ÜÄ EñܤÃWÜ ÜÙÜá
MODEL ANSWERS
ÓÜíPæàñÜ ÓÜíTæ : 83-K CODE NO. : 83-K
ËÐÜ¿á ËþÝ®Ü
Subject : SCIENCE
»èñÜ ËþÝ®Ü, ÃÜÓÝ¿á®Ü ËþÝ®Ü ÊÜáñÜᤠiàÊÜ ËþÝ®Ü / Physics, Chemistry & Biology
ÍÝÇÝ A»Ü¦ì / ÍÝÇÝ ±Üâ®ÜÃÝÊÜ£ìñÜ A»Ü¦ì / TÝÓÜX A»Ü¦ì / TÝÓÜX ±Üâ®ÜÃÝÊÜ£ìñÜ A»Ü¦ì
Regular Fresh / Regular Repeater / Private Fresh / Private Repeater
PܮܰvÜ ÊÜÞ«ÜÂÊÜá / Kannada Medium
»èñÜ ËþÝ®Ü / Physics
©®ÝíPÜ 02. 04. 2025 ] [ WÜÄÐÜu AíPÜWÜÙÜá : 80
Date : 02. 04. 2025 ] [ Max. Marks : 80
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
»ÝWÜ & A
( »èñÜ ËþÝ®Ü )
I. ŸÖÜá&BÁáR ±ÜÅÍæ°WÜÙÜá 3×1=3
1. ²à®Ü ¨Ü±Üì|WÜÙÜ®Üá° CÊÜâWÜÙÜÈÉ E±ÜÁãàXÓÜáñݤÃæ
(A) pÝa…ìWÜÙÜá (B) ÊÝÖÜ®ÜWÜÙÜ ×®æã°ào ¨Ü±Üì|WÜÙÜá
(C) ñܱÝÓÜOÝ ©à±ÜWÜÙÜá (D) ûèÃÜ ¨Ü±Üì|WÜÙÜá
EñܤÃÜ (B) ÊÝÖÜ®ÜWÜÙÜ ×®æã°ào ¨Ü±Üì|WÜÙÜá 1

CCE RF/RR/PF/PR(A)/101/1836 (MA) PHY [ Turn over

83-K 2
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
2. ÓÜã¿áì®Üá ÊÝÓܤÊÜ ÓÜãÁãàì¨Ü¿áQRíñÜ GÃÜvÜá ¯ËáÐÜWÜÙÜ Êæã¨ÜÆá
ÖÝWÜã ÊÝÓܤÊÜ ÓÜã¿ÞìÓܤ¨Ü GÃÜvÜá ¯ËáÐÜWÜÙÜ ®ÜíñÜÃÜÊÜä ®ÜÊÜáWæ
WæãàaÜÄÓÜÆá PÝÃÜ| ÊÝ¿ááÊÜáívÜƨÜÈÉ®Ü
(A) ¸æÙÜQ®Ü ÊÜQÅà»ÜÊÜ®Ü (B) ¸æÙÜQ®Ü ±ÜÅ£¶ÜÆ®Ü
(C) ¸æÙÜQ®Ü aܨÜáÃÜáËPæ (D) ¸æÙÜQ®Ü ÊÜ|ìË»Üg®æ
EñܤÃÜ (A) ¸æÙÜQ®Ü ÊÜQÅà»ÜÊÜ®Ü 1

3. Jí¨Üá PÜá¨ÜáÃæ ÇÝÙÝPÜꣿá PÝíñÜ¨Ü «ÜáÅÊÜWÜÙÜ ®ÜvÜáÊæ
˨ÜáÂñ…¯ÜÅÊÝÖÜËÃÜáÊÜ Jí¨Üá ÓÜÇÝPæ¿á¬Üá® CÄÓÜÇÝX¨æ. ÓÜÇÝPæ¿á
Óݧ¬Ü¯ÜÆÉoÊÜâ WÜÄÐÜuÊÝWÜÆá ˨ÜáÂñ…¯ÜÅÊÝÖÜ¨Ü ©PÜáR ÊÜáñÜᤠPÝíñÜûæàñÜŨÜ
©PÜáRWÜÙÜ ®ÜvÜáÊæ CÃܸæàPÝ¨Ü Pæãà®ÜÊÜâ
(A) 0º (B) 45º
(C) 90º (D) 180º
EñܤÃÜ (C) 90º 1
II. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ EñܤÄÔ 3×1=3

4. Jí¨Üá ˨ÜáÂñ… ÊÜáívÜƨÜÈÉ E¯ÜÁãàXÓÜáÊÜ Ãæãà«Ü¨Ü bÖ殿á bñÜÅÊܬÜá®
ŸÃæÀáÄ.
EñܤÃÜ :

1
5. GÃÜvÜá PÝí£à¿á ŸÆÃæàTæWÜÙÜá Jí¨Ü¬æã®í¨Üá dæà©ÓÜáÊÜâ©ÆÉ. D
ÊÝPÜÂÊÜ®Üá° ÓÜÊÜá¦ìÔ.
EñܤÃÜ :
Jí¨Üá ÊæàÙæ dæà©Ô¨ÜÃæ, dæà©ÓÜáÊÜ ¹í¨ÜáË®ÜÈÉ ©PÜãÕb¿á ÓÜãi¿áá
GÃÜvÜã ©PÜáRWÜÙÜñܤ ¯¨æàìÎÓÜᣤ¨æ Gí¨Ü¥Üì B¨ÜÃæ A¨Üá ÓÝ«ÜÂËÆÉ. 1
6. PæÙÜWæ PæãqrÃÜáÊÜ ÊÜáívÜÆ bñÜÅÊÜ®Üá° WÜÊÜá¯Ô. ÊÝÖÜPÜ ñÜí£¿á P ¹í¨ÜáË®Ü
ÓÜáñܤÆã ÖÝWÜã Q ¹í¨Üá訆 ÓÜáñܤÆã EípÝWÜáÊÜ PÝíñÜûæàñÜÅ¨Ü ©PÜR®Üá°
£ÚÔ.

CCE RF/RR/PF/PR(A)/101/1836 (MA) PHY

3 83-K
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ

EñܤÃÜ :
 P ¹í¨ÜáË®ÜÈÉ A±ÜŨÜüOÝPÝÃÜÊÝXÃÜáñܤ¨æ. ½
 Q ¹í¨ÜáË®ÜÈÉ ±ÜŨÜüOÝPÝÃÜÊÝXÃÜáñܤ¨æ. ½

bñÜÅ ŸÃæ¨Üá, ©PÜR¬Üá® WÜáÃÜá£Ô¨ÜÃÜã AíPÜ ­àvÜáÊÜâ¨Üá 1
III. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ EñܤÄÔ 2×2=4

7. ¹Ú Ÿ|¡¨Ü Ãæãà×ñÜ¨Ü ¯Üâ¬ÜÃ… ÓÜíÁãàg¬æ¿á¬Üá® ñæãàÄÓÜáÊÜ
¬ÜãÂo¬…ÃÜÊÜÃÜ ¯ÜÅÁãàWÜÊܬÜá® ËÊÜÄÔ.
EñܤÃÜ :

 ¬ÜãÂo¬…ÃÜÊÜÃÜá Jí¨Üá WÝi¬Ü ¯ÜorPÜÊܬÜá® E¯ÜÁãàXÔ ¹Ú ¸æÙÜQ¬Ü
Ãæãà×ñÜ樆 Üá° ±Üvæ¨ÜÃÜá.
 ®ÜíñÜÃÜ AÊÜÃÜá ÓÜÊÜáÃÜã²¿Þ¨Ü ÊÜáñæã¤í¨Üá ±ÜorPÜÊÜ®Üá° Êæã¨ÜÆ®æà
¯ÜorPÜPæR ËÃÜá¨Üœ ©QR¬ÜÈÉ ñÜÇæPæÙÜWÝX CÄÔ Ãæãà×ñÜÊܬÜá® A¨ÜÃÜ
ÊÜáãÆPÜ ÖÝÀáÔ¨ÝWÜ ¹Ú¿á ¸æÙÜQ®Ü QÃÜ|ÊÜâ ÖæãÃÜÖæãËá¾ñÜá.
 C¨Üá ¹Ú ¸æÙÜQ¬Ü Ãæãà×ñÜ¨Ü ¯Üâ¬ÜÃ… ÓÜíÁãàg¬æ¿á¬Üá®
ñæãàÄÓÜáñܤ¨æ.
2

CCE RF/RR/PF/PR(A)/101/1836 (MA) PHY [ Turn over

83-K 4
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
8. ÓÜËáà±Ü¨ÜêÑr ÊÜáÁãಿá ÖÝWÜã ¨ÜãÃܨÜêÑr Öæç¯ÜÃ… ÊæápæãÅà°¿á
PÜ|á¡WÜÙÜ ¬ÜvÜáÊÜ| ¿ÞÊÜâ¨Ý¨ÜÃÜã GÃÜvÜá ÊÜÂñÝÂÓÜWÜÙܬÜá® ŸÃæÀáÄ.
EñܤÃÜ :
ÓÜËáà±Ü¨ÜêÑr ÊÜáÁãà²¿á ¨ÜãÃܨÜêÑr Öæç±ÜÃ… ÊæápæãÅಿá
 ÓÜËáà±Ü¨Ü ÊÜÓÜá¤WÜÙ®
Ü Üá° ÓܳÐÜrÊÝX  ¨ÜãÃÜ¨Ü ÊÜÓÜá¤WÜÙ®Ü Üá° ÓܳÐÜrÊÝX
¬æãàvÜŸÖÜá¨Üá / ¨ÜãÃÜ¨Ü ÊÜÓÜá¤WÜÙÜ®Üá° ¬æãàvÜŸÖÜá¨Üá / ÓÜËáà±Ü¨Ü ÊÜÓÜá¤WÜÙÜ®Üá°
ÓܳÐÜrÊÝX ®æãàvÜÇÝWܨÜá ÓܳÐÜrÊÝX ®æãàvÜÇÝWܨÜá
 ¨ÜãÃÜ¨Ü ÊÜÓÜá¤Ë¬Ü ¹íŸÊÜâ  ÓÜËáà¯Ü¨Ü ÊÜÓÜá¤Ë¬Ü ¹íŸÊÜâ
Ãæq®Ý¨Ü ÊÜááí»ÝWܨÜÈÉ ÃÜã±Üâ Ãæq®Ý¨Ü ×í»ÝWܨÜÈÉ ÃÜã±Üâ
WæãÙÜáÛñܤ¨æ WæãÙÜáÛñܤ¨æ
 ¨ÜãÃÜ¨Ü ¹í¨Üá A®ÜíñÜQRíñÜ  ÓÜËáà±Ü ¹í¨Üá ñÜáí¸Ý
ÖÜ£¤ÃܨÜÈÉÃÜáñܤ¨æ ¨ÜãÃܨÜÈÉÃÜáñܤ¨æ
 PÜ~¡®Ü ÊÜáÓÜãÃÜ¨Ü ÓÜíWÜÊÜá¨ÜãÃÜÊÜâ  PÜ~¡®Ü ÊÜáÓÜãÃÜ¨Ü ÓÜíWÜÊÜá¨ÜãÃÜÊÜâ
ñÜáí¸Ý bPÜR¨ÝXÃÜáÊÜâ¨Üá / ñÜáí¸Ý E¨ÜªÊÝXÃÜáÊÜâ¨Üá /
PÜ|á¡WÜávæx¿áá E¨ÜªÊÝWÜáÊÜâ¨Üá PÜ|á¡WÜávæx¿áá A£ bPÜR¨ÝWÜáÊÜâ¨Üá
 ÓÜãPܤ ÓÝÊÜá¥ÜÂì¨Ü ¯ÊÜá°  ÓÜãPܤ ÓÝÊÜá¥ÜÂì¨Ü ²à®Ü
ÊÜáÓÜãÃÜÊÜ®Üá° E±ÜÁãàXÔ D ÊÜáÓÜãÃÜÊÜ®Üá° E±ÜÁãàXÔ D
¨æãàÐÜÊÜ®Üá° ÓÜįÜwÓÜŸÖÜá¨Üá. ¨æãàÐÜÊÜ®Üá° ÓÜįÜwÓÜŸÖÜá¨Üá.
¿ÞÊÜâ¨Ý¨ÜÃÜã GÃÜvÜá 1 + 1 2
IV. D PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ EñܤÄÔ 3×3=9
9. ²à®Ü ÊÜáÓÜãÃÜ¨Ü ÊÜááí¨æ 2 F 1 ®ÜÈÉ ÊÜÓÜá¤Êæäí¨Ü®Üá° CÄÔ¨ÝWÜ EípÝWÜáÊÜ
¯ÜÅ£¹íŸ¨Ü ÃæàTÝbñÜÅÊܬÜá® ŸÃæÀáÄ. ÃæàTÝbñÜÅ¨Ü ÓÜÖÝ¿á©í¨Ü
¯ÜÅ£¹íŸ¨Ü Óݧ¬Ü ÊÜáñÜᤠÓÜÌ»ÝÊÜÊܬÜá® ŸÃæÀáÄ.
[ F 1 ÊÜáÓÜãÃÜ¨Ü ±ÜÅ«Ý®Ü ÓÜíWÜÊÜá ]
EñܤÃÜ :

CCE RF/RR/PF/PR(A)/101/1836 (MA) PHY

5 83-K
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
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¯ÜÅ£¹íŸ¨Ü ÃæàTÝbñÜÅ.

ÃæàTÝbñÜÅPæR & 2
 ¯ÜÅ£¹íŸÊÜâ 2 F2 ®ÜÈÉ EípÝWÜáñܤ¨æ. ½
 ÓÜñÜ ÊÜáñÜᤠñÜÇæPæÙÜWÝ¨Ü ¯ÜÅ£¹íŸÊÜâ EípÝWÜáñܤ¨æ. ½ 3
10. a) KÊÜå…¬Ü ­¿áÊÜáÊܬÜá® ­ÃÜã°Ô.
b) ÊÝÖÜPÜ¨Ü Ãæãà«ÜÊÜâ AÊÜÆí¹ÔÃÜáÊÜ AíÍÜWÜÙܬÜá® ŸÃæÀáÄ.
EñܤÃÜ :
a) Ô§ÃÜÊÝ¨Ü ñݯÜÊÜެܨÜÈÉ Ë¨ÜáÂñ…ÊÜáívÜƨÜÈÉ®Ü ÇæãàÖܨÜ
ñÜí£¿á (Ãæãà«ÜPܨÜ) ®ÜvÜáË®Ü Ë»ÜÊÝíñÜÃÜÊÜâ A¨ÜÃÜ ÊÜáãÆPÜ
ÖÜÄ¿ááÊÜ Ë¨ÜáÂñ…¯ÜÅÊÝÖÜPæR ¬æàÃÜ A¬Üá¯ÝñܨÜÈÉÃÜáñܤ¨æ. 1

b) ÊÝÖÜPÜ¨Ü Ãæãà«ÜÊÜâ PæÙÜX®Ü AíÍÜWÜÙÜ®Üá° AÊÜÆí¹Ô¨æ.
 A¨ÜÃÜ E¨Üª ½
 A¨ÜÃÜ AvÜxPæã¿á¤ ½
 A¨ÜÃÜ ±ÝÅPÜê£PÜ WÜá| ½
 ñݱÜÊÜÞ®Ü ½ 3

CCE RF/RR/PF/PR(A)/101/1836 (MA) PHY [ Turn over

83-K 6
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
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11. a) PÝíñÜûæàñÜÅ¨Ü ÓݱæàûÜ ÍÜQ¤¿á®Üá° ¯«ÜìÄÓÜÆá ÓÜÖÝ¿áPÜÊÝWÜáÊÜ AíÍÜ
¿ÞÊÜâ¨Üá

b) ÓæãÇæ¬Ý¿å…x Gí¨ÜÃæà¬Üá C¨Ü¬Üá® E¯ÜÁãàXÔPæãívÜá ˨ÜáÂñ…PÝíñÜÊܬÜá®
ÖæàWæ ñÜ¿ÞÄÓÜáñݤÃæ
A¥ÜÊÝ

WÜêÖÜ Ë¨ÜáÂñ… ÊÜáívÜÆWÜÙÜÈÉ,

a) »ÜãÓÜí±ÜPÜì ñÜí£¿á PÝ¿áìÊæà®Üá

b) KÊÜÃ…ÇæãàwíW…A®Üá° ñܲ³ÓÜÆá ¿ÞÊÜ ÊÜáá®æ°aÜcÄPæ¿á PÜÅÊÜáWÜÙÜ®Üá°
ñæWæ¨ÜáPæãÙÜÛ¸æàPÜá
EñܤÃÜ :

a) PÝí£à¿á ŸÆÃæàTæWÜÙÜ ¨ÜorOæ ±ÜÅÊÜÞ|. 1

b) AÊÝÖÜPÜ Öæã©Pæ CÃÜáÊÜ ñÝÊÜáÅ¨Ü ñÜí£¿á A®æàPÜ ÓÜáÃÜáÚWÜÙÜ®Üá°
Jñæã¤ñݤX ÓÜᣤÃÜáÊÜ ÔÈívÜÃ… BPÝÃÜÊܬÜá® ÓæãÇæ¬Ý¿å…x
G®Üá°ÊÜÃÜá. 1

ÖæaÜác PÝíñÜûæàñÜÅ Öæãí©ÃÜáÊÜ / ˨ÜáÂñ… ¯ÜÅÊÜ×ÓÜᣤÃÜáÊÜ
ÓæãÇæ¬Ý¿å…x¬Ü JÙÜ»ÝWܨÜÈÉ Jí¨Üá PÝí£à¿á ÊÜÓÜá¤ÊݨÜ
Êæá¨Üá Pܹº|¨Ü ñÜáívÜ®Üá° CÄÔ Ë¨ÜáÂñ…PÝíñÜÊÜ®Üá°
ñÜ¿ÞÄÓÜáñݤÃæ. 1
3

A¥ÜÊÝ

CCE RF/RR/PF/PR(A)/101/1836 (MA) PHY

7 83-K
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a) ˨ÜáÂñ… E¯ÜPÜÃÜ|WÜÙÜÈÉ ÓæãàÄPæ EípݨÝWÜ AÊÜâWÜÙÜ
Ë»ÜÊÝíñÜÃÜ樆 Üá° »ÜãËá¿á Ë»ÜÊÝíñÜÃÜPæR ÓÜÊÜá®ÝXÓÜáñܤ¨æ. /
PÜwÊæá Ãæãà«ÜÊܬÜá® Öæãí©ÃÜáÊÜ Ë¨ÜáÂñ…¯ÜÅÊÝÖÜ¨Ü ¯Ü¥ÜÊܬÜá®
EíoáÊÜÞvÜáñܤ¨æ. 1

b)  ˨ÜáÂñ…ÊÜáívÜƨÜÈÉ ¶ÜäÂÓ… A¬Üá® ŸÙÜÓܸæàPÜá.
 ÓÜiàÊÜ ñÜí£ ÖÝWÜã ñÜoÓܧ ñÜí£WÜÙÜá ®æàÃÜ
ÓÜí±ÜPÜìÊÝWÜáÊÜâ¨Ü®Üá° ñܲ³Ó¸
Ü æàPÜá.
 ñÜí£WÜÙÜ Êæáà騆 AÊÝÖÜPÜ Öæã©Pæ
ÖݯWæãÙÜWÝWÜáÊÜâ¨Ü®Üá° ñܲ³ÓÜáÊÜâ¨Üá.
 ˨ÜáÂñ… E¯ÜPÜÃÜ|WÜÙÜÈÉ ¿ÞÊÜâ¨Ý¨ÜÃÜã ¨æãàÐÜ˨ܪÃæ
A¨Ü®Üá°ÓÜıÜwÓܸæàPÜá.
 ÖÜÆÊÜâ E¯ÜPÜÃÜ|WÜÙܬÜá® Jí¨æà ÓÝPæp…Wæ
hæãàwÓܸÝÃܨÜá. ½×4=2
¿ÞÊÜâ¨Ý¨ÜÃÜã ®ÝÆáR AíÍÜWÜÙÜá 3

V. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ EñܤÄÔ 2×4=8

12. a) ˨ÜáÂñ…ÊÜáívÜÆPæR BËá¾àoÃ… ÊÜáñÜᤠÊæäàÇ…rËáàoÃ… A¬Üá® ÖæàWæ
ÓÜí±ÜQìÓÜáñݤÃæ AÊÜâWÜÙÜ PÝ¿áìÊÜ®Üá° £ÚÔ.

b) WÜêÖÜ Ë¨ÜáÂñ…ÊÜáívÜÆWÜÙÜÈÉ, ˨ÜáÂñ… E¯ÜPÜÃÜ|WÜÙܬÜá® ÓÝÊÜÞ¬ÜÂÊÝX
ÓÜÊÜÞíñÜÃÜ PÜÅÊÜá¨ÜÈÉ ÓÜí±ÜQìÔÃÜáñݤÃæ. PÝÃÜ|WÜÙÜ®Üá° Pæãw.
EñܤÃÜ :

CCE RF/RR/PF/PR(A)/101/1836 (MA) PHY [ Turn over

83-K 8
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
a)  BËá¾àoÃ… A¬Üá® ÓÜÃÜ~ PÜÅÊÜá¨ÜÈÉ hæãàwÔÃÜáñݤÃæ. ½
 ÊæäàÇ…rËáàoÃ… A¬Üá® ÓÜÊÜÞíñÜÃÜ PÜÅÊÜá¨ÜÈÉ
hæãàwÔÃÜáñݤÃæ. ½
 BËá¾àoÃ… ˨ÜáÂñ…ÊÜáívÜƨÜÈÉ¬Ü Ë¨ÜáÂñ…¯ÜÅÊÝÖÜÊܬÜá®
AÙæ¿ááñܤ¨æ. ½
 ÊæäàÇ…rËáàoÃ… GÃÜvÜá ¹í¨ÜáWÜÙÜ ¬ÜvÜáË¬Ü Ë»ÜÊÝíñÜÃÜÊܬÜá®
AÙæ¿ááñܤ¨æ. ½

b)  ÓÜÊÜÞíñÜÃÜ ÊÜáívÜÆÊÜâ ˨ÜáÂñ… E¯ÜPÜÃÜ|WÜÙÜ ÊÜáãÆPÜ
˨ÜáÂñ… ¯ÜÅÊÝÖÜÊܬÜá® Ë»ÜiÓÜáñܤ¨æ.
 ¿ÞÊÜâ¨Ý¨ÜÃÜã Jí¨Üá ZoPÜ Ë¶ÜÆÊݨÜÃæ ÊÜáívÜÆÊÜâ PÜw¨Üá
ÖæãàWÜáÊÜâ©ÆÉ. EÚ¨Ü ZoPÜWÜÙÜá PÝ¿áì¯ÊÜì×ÓÜáñܤÊæ.
 D ÊÜáívÜƨÜÈÉ®Ü Joár Ãæãà«ÜÊÜâ PÜwÊæá CÃÜáñܤ¨æ.
 ˼¬Ü® ˨ÜáÂñ… E¯ÜPÜÃÜ|WÜÚWæ ˼¬Ü® ˨ÜáÂñ…¯ÜÅÊÝÖÜ
AWÜñÜÂËÃÜáÊÜ ÓÜí¨Ü»ÜìWÜÙÜÈÉ ÖæaÜác ÓÜÖÝ¿áPÜÊÝWÜáñܤ¨æ.
¿ÞÊÜâ¨Ý¨ÜÃÜã GÃÜvÜá AíÍÜWÜÙÜá 1+1 4

13. Jí¨Üá WæãàÚà¿á ¨Ü±Üì|¨Ü ÊÜPÜÅñÝ £ÅgÂÊÜâ 36 cm BX¨æ. C¨ÜÃÜ
ÓÜíWÜÊÜá¨ÜãÃÜÊÜ®Üá° PÜívÜá×wÀáÄ. C¨æà ÓÜíWÜÊÜá¨ÜãÃÜÊÜ®Üá° Öæãí©ÃÜáÊÜ
²à®Ü ¨Ü±Üì|¨Ü ÊÜááí¨æ 5 cm WÝñÜÅ¨Ü Jí¨Üá Êæáà|¨Ü Ÿ£¤¿á¬Üá® 27 cm
¨ÜãÃܨÜÈÉ CÄÔ¨æ. EípÝ¨Ü ¯ÜÅ£¹íŸ¨Ü Óݧ¬Ü ÊÜáñÜᤠÓÜÌ»ÝÊÜÊܬÜá® £ÚÔ
ÖÝWÜã ¯ÜÅ£¹íŸ¨Ü WÝñÜÅÊܬÜá® PÜívÜá×wÀáÄ.
A¥ÜÊÝ

CCE RF/RR/PF/PR(A)/101/1836 (MA) PHY

9 83-K
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
Jí¨Üá WæãàÚà¿á ¨Ü±Üì|¨Ü ÓÜíWÜÊÜá¨ÜãÃÜÊÜâ 10 cm BX¨æ. C¨ÜÃÜ
ÊÜPÜÅñÝ £ÅgÂÊÜ®Üá° PÜívÜá×wÀáÄ. C¨æà ÓÜíWÜÊÜá¨ÜãÃÜÊÜ®Üá° Öæãí©ÃÜáÊÜ
¯ÊÜá° ¨Ü±Üì|¨Ü ÊÜááí¨æ 4 cm WÝñÜÅ¨Ü Jí¨Üá ÊÜÓÜá¤ÊÜ®Üá° 20 cm
¨ÜãÃܨÜÈÉ CÄÔ¨æ. ÊÜÓÜá¤Ë¬Ü ÓܱÐÜr ¯ÜÅ£¹íŸÊܬÜá® ¯Üv¿
æ áÆá ¯ÜÃܨæ¿á¬Üá®
¨Ü¯Üì|©í¨Ü GÐÜár ¨ÜãÃܨÜÈÉ CÄÓܸæàPÜá ? EípÝ¨Ü ¯ÜÅ£¹íŸ¨Ü
ÓÜÌ»ÝÊÜÊܬÜá® £ÚÔ. ÖÝWÜã ¯ÜÅ£¹íŸ¨Ü WÝñÜÅÊܬÜá® PÜívÜá×wÀáÄ.
EñܤÃÜ :

ÊÜPÜÅñÝ £Åg R = 36 cm
ÊÜÓÜᤨÜãÃÜ u = – 27 cm
ÊÜÓÜá¤Ë®Ü WÝñÜÅ h = 5 cm
¯ÜÅ£¹íŸ¨Ü ¨ÜãÃÜ v = ?
¯ÜÅ£¹íŸ¨Ü GñܤÃÜ h l = ?

ÓÜíWÜÊÜá ¨ÜãÃÜ f = R ½
2
36
f = = 18 cm ½
2
1 1 1
  A¥ÜÊÝ ½
v u f
1 1 1
 
v f u
1  1  1 1
=      ½
18   27  18 27
1 32 5
 
v 54 54
54
v = = 10·8 cm ½
5

CCE RF/RR/PF/PR(A)/101/1836 (MA) PHY [ Turn over

83-K 10
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
¯ÜÅ£¹íŸÊÜ⠨ܯÜì|¨Ü ×í¨æ 10·8 cm ¨ÜãÃܨÜÈɨæ.
hl v h ( v )
 ( or ) h l = ½
h u u
5 (  10  8 )
hl = = 2 cm ½
 27

 ¯ÜÅ£¹íŸ¨Ü GñܤÃÜ 2 cm C¨æ.
 ¯ÜÅ£¹íŸÊÜâ Ëá¥Ü ÖÝWÜã ¬æàÃÜÊÝX¨æ. ½ 4

A¥ÜÊÝ

ÓÜíWÜÊÜá ¨ÜãÃÜ f = – 10 cm

ÊÜÓÜá¤Ë®Ü ¨ÜãÃÜ u = – 20 cm

ÊÜÓÜá¤Ë®Ü WÝñÜÅ h = 4 cm

¯ÜÅ£¹íŸ¨Ü ¨ÜãÃÜ v = ?

¯ÜÅ£¹íŸ¨Ü GñܤÃÜ h l = ?

ÊÜPÜÅñÝ £Åg = ?

ÊÜPÜÅñÝ £Åg R = 2 f ½

= 2 ( 10 ) = 20 cm ½
1 1 1
  (A¥ÜÊÝ)
v u f

1 1 1
  ½
v f u

CCE RF/RR/PF/PR(A)/101/1836 (MA) PHY

11 83-K
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
1  1  1 1
=      ½
 10   20  10 20

1  21 1
  
v 20 20

v = – 20 cm ½

±ÜÃܨæ¿á®Üá° ¨Ü±Üì|©í¨Ü 20 cm ¨ÜãÃܨÜÈÉ CÄÓܸæàPÜá.

hl v h ( v )
 (A¥ÜÊÝ) h l = ½
h u u

4 (  (  20 ) )
=
 20

4  20
=
 20

h l = – 4 cm ½

 ¯ÜÅ£¹íŸ¨Ü GñܤÃÜÊÜâ – 4 cm

 ¯ÜÅ£¹íŸÊÜâ ÓÜñÜ ÊÜáñÜᤠñÜÇæPæÙÜWÝX¨æ. ½ 4

CCE RF/RR/PF/PR(A)/101/1836 (MA) PHY [ Turn over

CCE RF/RR/
A PF/PR
PÜ®ÝìoPÜ ÍÝÇÝ ±ÜÄàûæ ÊÜáñÜᤠÊÜåèƯ|ì¿á ÊÜáívÜÈ,
ÊÜáÇæÉàÍÜÌÃÜí, ¸æíWÜÙÜãÃÜá – 560 003
KARNATAKA SCHOOL EXAMINATION AND ASSESSMENT BOARD,
MALLESHWARAM, BENGALURU – 560 003
ÊÜÞa…ì/H²ÅÇ… 2025 ÃÜ ±ÜÄàûæ - 1
MARCH/APRIL 2025 EXAMINATION - 1
ÊÜÞ¨ÜÄ EñܤÃWÜ ÜÙÜá
MODEL ANSWERS
ÓÜíPæàñÜ ÓÜíTæ : 83-K CODE NO. : 83-K
ËÐÜ¿á ËþÝ®Ü
Subject : SCIENCE
»èñÜ ËþÝ®Ü, ÃÜÓÝ¿á®Ü ËþÝ®Ü ÊÜáñÜᤠiàÊÜ ËþÝ®Ü / Physics, Chemistry & Biology
ÍÝÇÝ A»Ü¦ì / ÍÝÇÝ ±Üâ®ÜÃÝÊÜ£ìñÜ A»Ü¦ì / TÝÓÜX A»Ü¦ì / TÝÓÜX ±Üâ®ÜÃÝÊÜ£ìñÜ A»Ü¦ì
Regular Fresh / Regular Repeater / Private Fresh / Private Repeater
PܮܰvÜ ÊÜÞ«ÜÂÊÜá / Kannada Medium
ÃÜÓÝ¿á®Ü ËþÝ®Ü / Chemistry
©®ÝíPÜ 02. 04. 2025 ] [ WÜÄÐÜu AíPÜWÜÙÜá : 80
Date : 02. 04. 2025 ] [ Max. Marks : 80
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
»ÝWÜ & B
( ÃÜÓÝ¿á®Ü ËþÝ®Ü )
VI. ŸÖÜá&BÁáR ±ÜÅÍæ°WÜÙÜá 2×1=2

14. PæÙÜX®ÜÊÜâWÜÙÜÈÉ ÓÝŸã®ÜáWÜÚWæ ÓÜíŸí˜Ô¨Üíñæ ÓÜÄ¿Þ¨Ü ÖæàÚPæÁáí¨ÜÃæ,
ÓÝŸã®ÜáWÜÙÜá
(A) WÜvÜÓÜá ¯àÄ®ÜÈÉ ÓÜÃÝWÜÊÝX ÖæaÜác ®æãÃæ¿á®Üá° PæãvÜáñܤÊæ
(B) WÜvÜÓÜá ¯àÄ®ÜÈÉ PÜÃÜWÜ¨Ü JñܤÃÜÊÜ®Üá° EíoáÊÜÞvÜáñܤÊæ
(C) WÜvÜÓÜá ¯àÄ®ÜÈÉ GOæ¡¿áíñÜÖÜ PæãÙæ¿á®Üá° ÓÜáÆ»ÜÊÝX ÓÜÌaÜfWæãÚÓÜáñܤÊæ
(D) ÓÜÇæã#à¯P… BÊÜáÉWÜÙÜ Óæãàw¿áí ÆÊÜ|WÜÙÝXÊæ
EñܤÃÜ (B) WÜvÜÓÜá ¯àÄ®ÜÈÉ PÜÃÜWÜ¨Ü JñܤÃÜÊÜ®Üá° EíoáÊÜÞvÜáñܤÊæ 1

CCE RF/RR/PF/PR(A)/101/1836 (MA) CHE [ Turn over

83-K 2
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
15. D ÃÝÓÝ¿á¯PÜ QÅÁá¿áÈÉ,
(A) ÖæçvæãÅàg®… A±ÜPÜÐÜìOæWæãívÜá ¯àÃÜá EípÝX¨æ
(B) ±ÜÅ£ÊÜñÜìPÜWÜÙÜ ®ÜvÜáÊæ A¿Þ®ÜáWÜÙÜ Ë¯ÊÜá¿á EípÝX¨æ
(C) ñÝÊÜáÅ¨Ü BPæÕ„v… EñÜRÐÜìOæWæãívÜá ñÝÊÜáÅ EípÝX¨æ
(D) ñÝÊÜáÅ¨Ü BPæÕ„v… A±ÜPÜÐÜìOæWæãívÜá ñÝÊÜáÅ EípÝX¨æ
EñܤÃÜ (D) ñÝÊÜáÅ¨Ü BPæÕ„v… A±ÜPÜÐÜìOæWæãívÜá ñÝÊÜáÅ EípÝX¨æ 1
VII. D PæÙÜX®Ü ±ÜÅÍæ°Wæ EñܤÄÔ 2×1=2
16. Pܹº|¨Ü ÓÜíûÝÃÜ|ÊÜ®Üá° ñÜvæWÜoárÊÜ ¿ÞÊÜâ¨Ý¨ÜÃÜã GÃÜvÜá PÜÅÊÜáWÜÙÜ®Üá°
£ÚÔ.
EñܤÃÜ :
 Ÿ|¡ ÖÜaÜácÊÜâ¨Üá
 GOæ¡ ÓÜÊÜÃÜáÊÜâ¨Üá
 XÅàÓ… ÖÜaÜácÊÜâ¨Üá
 WÝÂÆ̯àPÜÃÜ|
 ËáÍÜÅÇæãàÖÜÊܮݰXÓÜáÊÜâ¨Üá
 PæãÅàËá¿áí Çæà±Ü®Ü ¿ÞÊÜâ¨Ý¨ÜÃÜã GÃÜvÜá 2½ 1
17. ÓÝÄàPÜêñÜ BÊÜáÉ樆 Üá° ÖæàWæ ÓÝÃÜÄPܤWæãÚÓÜÇÝWÜáÊÜâ¨Üá
EñܤÃÜ :
¯ÃÜíñÜÃÜ PÜÆáPÜáËPæÁãí©Wæ BÊÜáÉÊÜ®Üá° ¯«Ý®ÜÊÝX ¯àÄWæ ÓæàÄÓÜáÊÜâ¨ÜÄí¨Ü. 1
VIII. D PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ EñܤÄÔ 3×2=6

18. PÜÊÜáoáËPæ Gí¨ÜÃæà®Üá PÜÊÜáoáËPæ¿á®Üá° ñÜvæWÜoárÊÜ ¿ÞÊÜâ¨Ý¨ÜÃÜã
GÃÜvÜá Ë«Ý®ÜWÜÙÜ®Üá° £ÚÔ.
EñܤÃÜ :
PæãŸáº ÊÜáñÜᤠGOæ¡¿á®Üá° Öæãí©ÃÜáÊÜ BÖÝÃÜ ±Ü¨Ý¥ÜìWÜÙÜá EñÜRÐÜìOæWæãívÝWÜ
AÊÜâWÜÙÜ ÊÝÓÜ®æ ÊÜáñÜᤠÃÜáb Ÿ¨ÜÇÝWÜáñܤ¨æ. C¨Ü®Üá° PÜÊÜáoáËPæ G®Üá°ñæ¤àÊæ. 1

CCE RF/RR/PF/PR(A)/101/1836 (MA) CHE

3 83-K
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
PÜÊÜáoáËPæ¿á®Üá° ñÜvæWÜoárÊÜ Ë«Ý®ÜWÜÙÜá
 EñÜRÐÜìOæ¿á®Üá° ñÜvæWÜoárÊÜ ÊÜÓÜá¤WÜÙÜ®Üá° (±ÜÅ£ EñÜRÐÜìPÜWÜÙÜá)
ÓæàÄÓÜáÊÜâ¨Üá.
 PÜÄ¨Ü BÖÝÃÜ ±Ü¨Ý¥ÜìWÜÙÜ®Üá° WÝÚ ±ÜÅÊæàÎÓÜ¨Ü ÓÜíWÝÅÖÜPÜWÜÙÜÈÉ
ÍæàSÄÓÜáÊÜâ¨Üá.
 b±…Õ®Ü ±æäor|PæR ®æçpæãÅàg®… A¯ÆÊÜ®Üá° ÖÝÀáÓÜáÊÜâ¨Üá.
¿ÞÊÜâ¨Ý¨ÜÃÜã GÃÜvÜá 2½
2
19. PÝÃÜ| Pæãw
a) ÓÜñÜá訆 BPæÕ„v… A®Üá° E»Ü¿áÊÜ£ì BPæÕ„v… G®Üá°ñݤÃæ.
b) Óæãàw¿áí ÇæãàÖÜÊÜ®Üá° ÔàÊæáGOæ¡¿áÈÉ ÓÜíWÜÅ×ÔvÜáÊÜÃÜá.
A¥ÜÊÝ
PÝÃÜ| Pæãw
a) b®Ü°ÊÜ®Üá° B»ÜÃÜ|WÜÙÜ ñÜ¿ÞÄPæ¿áÈÉ ŸÙÜÓÜáÊÜÃÜá.
b) Z®ÜÔ§£¿áÈÉ A¿Þ¯P… ÓÜí¿ááPܤWÜÙÜá ñÜÊÜá¾ ÊÜáãÆPÜ Ë¨ÜáÂñ…
ÖÜÄ¿áÆá ¹vÜáÊÜâ©ÆÉ.
EñܤÃÜ :
a) ÓÜñÜá訆 BPæÕ„v…, BÊÜáÉ ÊÜáñÜᤠ±ÜÅñÝÂÊÜáÉWÜÙæÃÜvÜÃÜ hæãñæWÜã
±ÜÅ£ÊÜ£ìÔ¨ÝWÜ ÆÊÜ| ÊÜáñÜᤠ¯àÃÜ®Üá° Eñܳ£¤ ÊÜÞvÜáñܤ¨æ. 1

b) Óæãàw¿áí ÇæãàÖÜÊÜâ ¯àÃÜá ÊÜáñÜᤠÊÝñÝÊÜÃÜ|¨Ü
BQÕg®…®æãí©Wæ AñÜÂíñÜ ÃÜ»ÜÓÜÊÝX ÊÜ£ìÓÜáñܤ¨.æ B¨ÜÃæ
ÔàÊæáGOæ¡¿á hæãñæ ÊÜ£ìÓÜáÊÜâ©ÆÉ. 1
2
A¥ÜÊÝ
a)  ÖæãÙܲ¨æ

CCE RF/RR/PF/PR(A)/101/1836 (MA) CHE [ Turn over

83-K 4
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
 ñܮܠWÜá| Öæãí©¨æ
 PÜáo WÜá| Öæãí©¨æ
 AñÜÂíñÜ PÜwÊæá QÅ¿ÞÎàÆ
¿ÞÊÜâ¨Ý¨ÜÃÜã GÃÜvÜá 2½
b)  Z®ÜÊÜÓÜá¤WÜÙÜ PÜt| ÃÜaÜ®æÀáí¨ÝX A¿Þ®ÜáWÜÙÜ aÜÆ®æ
ÓÝ«ÜÂËÆÉ.
 ÓÜÌñÜíñÜÅ A¿Þ®ÜáWÜÙÜá EípÝWÜáÊÜâ©ÆÉ.
¿ÞÊÜâ¨Ý¨ÜÃÜã Jí¨Üá 1
2
20. PæÙÜX®Ü EñÜRÐÜì| QÅÁá¿á®Üá° ±Üä|ìWæãÚÔ ÊÜáñÜᤠEñܳ®Ü°¨ÜÈÉ
PÜívÜáŸÃÜáÊÜ QÅ¿ÞWÜáí±Ü®Üá° ÖæÓÜÄÔ.

EñܤÃÜ :

 ûÝÄà¿á KMnO 4 ½

 BËáÉà¿á K 2 Cr 2 O 7 ½

  CH 3 COOH ½

 – COOH / PݸÝìQÕÈP… BÊÜáÉ ½ 2
IX. D PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ EñܤÄÔ 2×3=6
21. ñÝÊÜáÅ¨Ü Ë¨Üá©̻Üg¯à¿á ÍÜá©œàPÜÃÜ|ÊÜ®Üá° ñæãàÄÓÜÆá E±ÜÁãàXÓÜáÊÜ
E±ÜPÜÃÜ|WÜÙÜ hæãàvÜOæ¿á bñÜÅÊÜ®Üá° ŸÃæÀáÄ ÊÜáñÜᤠPæÙÜX®Ü »ÝWÜWÜÙÜ®Üá°
WÜáÃÜá£Ô.
i) PÝÂ¥æãàv…
ii) PÜÍܾÆWÜÙÜá
EñܤÃÜ :

CCE RF/RR/PF/PR(A)/101/1836 (MA) CHE

5 83-K
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ

22. PæÙÜX®Ü QÅÁáWÜÚWæ ÃÝÓÝ¿á¯PÜ ÓÜËáàPÜÃÜ|WÜÙÜ®Üá° ŸÃæÀáÄ
i) ÓÜáor ÓÜá|¡ÊÜâ ¯àÄ®æãí©Wæ ÊÜ£ìÔ¨æ
ii) ñÝÊÜáÅ¨Ü ÓÜÇæ#àp… ¨ÝÅÊÜ|¨ÜÈÉ ÓÜñÜá訆 aÜãÃÜáWÜÙÜ®Üá° ÓæàÄÔ¨æ.
iii) Óæãàw¿áí PæãÉàÃæçv… ¨ÝÅÊÜ|ÊÜ®Üá° ¸æÚÛ¿á ®æçpæÅàp… ¨ÝÅÊÜ|PæR
ÓæàÄÔ¨æ.
A¥ÜÊÝ
PæÙÜX®Ü ÃÝÓÝ¿á¯PÜ ÓÜËáàPÜÃÜ|WÜÙÜ®Üá° ÓÜĨÜãXÔ
i) H2  O2  H2O

ii) Na2CO3  HCl  NaCl  H2O  CO2

iii) N2  H2  NH3
EñܤÃÜ :
i)  PÝÂÈÕ¿áí BPæÕ„v… + ¯àÃÜá → PÝÂÈÕ¿áí ÖæçvÝÅPæÕ„v…
 CaO + H 2 O → Ca (OH) 2

¿ÞÊÜâ¨Ý¨ÜÃÜã Jí¨Üá 1

CCE RF/RR/PF/PR(A)/101/1836 (MA) CHE [ Turn over

83-K 6
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
ii)  ñÝÊÜáÅ¨Ü ÓÜÇæ#àp… + ÓÜñÜá → ñÝÊÜáÅ + ÓÜñÜá訆 ÓÜÇæ#àp…
 CuSO 4 + Zn → ZnSO 4 + Cu

¿ÞÊÜâ¨Ý¨ÜÃÜã Jí¨Üá 1

iii)  Óæãàw¿áí PæãÉàÃæçv… + ¸æÚÛ¿á ®æçpæÅàp… → ¸æÚÛ¿á
PæãÉàÃæçv… + Óæãàw¿áí ®æçpæÅàp…
 NaCl + AgNO 3 → AgCl + NaNO 3 1 3
A¥ÜÊÝ
i) 2H 2 + O 2 → 2H 2 O ½+½
ii) Na 2 CO 3 + 2HCl → 2NaCl + H 2 O + CO 2 ½+½
iii) N2  3H2  2NH3 ½+½ 3
X. D PæÙÜX®Ü ±ÜÅÍæ°Wæ EñܤÄÔ 1×4=4

23. a) NaOH, Ca(OH)2, H2 ÊÜáñÜᤠCl2 ÊÜÓÜá¤WÜÙÜ®Üá° ¯ÊÜáWæ
¯àvÜÇÝX¨æ. CÊÜâWÜÙÜÈÉ ¿ÞÊÜ ÊÜÓÜá¤WÜÙÜ®Üá° ŸÙÜÔPæãívÜá ¯àÊÜâ
aÜÆáÊæ±Üâw¿á®Üá° ñÜ¿ÞÄÓÜáËÄ aÜÆáÊæ±Üâw¿á ÃÝÓÝ¿á¯PÜ
ÖæÓÜÃÜá ÊÜáñÜᤠA¨ÜÃÜ Jí¨Üá E±ÜÁãàWÜÊÜ®Üá° ŸÃæÀáÄ.
b) i) ¯àÈ Èo¾Ó… PÝWܨÜÊÜ®Üá° E±ÜÁãàXÔPæãívÜá Jí¨Üá
¨ÝÅÊÜ|ÊÜ®Üá° ±ÜÅñÝÂËáÉà¿á ¨ÝÅÊÜ|Êæí¨Üá ÖæàWæ
WÜáÃÜá£ÓÜáËÄ
ii) ¿ÞÊÜ ÓÜí¨Ü»Üì¨ÜÈÉ JŸº ÃæçñÜ ñܮܰ PÜêÑ »ÜãËá¿á ÊÜá~¡Wæ
AÃÜÚ¨Ü ÓÜá|¡ÊÜ®Üá° ¸æÃæÓÜáñݤ®æ
EñܤÃÜ :
a)  Ca (OH) 2 ½
 Cl 2 ½
 CaOCl 2 / PÝÂÈÕ¿áí BQÕ PæãÉàÃæçv… ½

CCE RF/RR/PF/PR(A)/101/1836 (MA) CHE

7 83-K
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
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 — Ÿpær PÝTÝì®æWÜÙÜÈÉ ÖÜ£¤ ÊÜáñÜᤠ®ÝÃÜ®Üá° aÜÆáÊæ ÊÜÞvÜÆá
— PÝWÜ¨Ü¨Ü PÝTÝì®æ¿áÈÉ ÊÜáÃÜ¨Ü £ÃÜáÙÜ®Üá° aÜÆáÊæ
ÊÜÞvÜÆá
— ÇÝíwÅ¿áÈÉ ñæãÙæ¨Ü Ÿpær¿á®Üá° aÜÆáÊæ ÊÜÞvÜÆá
— A®æàPÜ ÃÝÓÝ¿á¯PÜ PÝTÝì®æWÜÙÜÈÉ EñÜRÐÜì|PÝÄ¿ÞX
— PÜáw¿ááÊÜ ¯àÃÜ®Üá° QÅËáÊÜááPܤWæãÚÓÜÆá ÓæãàíPÜá®ÝÍÜPÜÊÝX
¿ÞÊÜâ¨Ý¨ÜÃÜã Jí¨Üá ½

b) i) ¯àÈ Èo¾Ó… PÝWܨÜÊÜâ ±ÜÅñÝÂËáÉà¿á ¨ÝÅÊÜ|¨ÜÈÉ Ÿ|¡
Ÿ¨ÜÈÓÜáÊÜâ©ÆÉ. 1

ii)  ÊÜá~¡®Ü BËáÉà¿á WÜá| ÖæaÝc¨ÝWÜ
 ÊÜá~¡®Ü pH ÊÜåèÆ PÜwÊæá¿Þ¨ÝWÜ
 ÊÜá~¡®ÜÈÉ H  / H3O A¿Þ®ÜáWÜÙÜ ÓÝÃÜñæ ÖæaÝc¨ÝWÜ.
¿ÞÊÜâ¨Ý¨ÜÃÜã Jí¨Üá 1
4
XI. D PæÙÜX®Ü ±ÜÅÍæ°Wæ EñܤÄÔ 1×5=5

24. a) ±Ü¿Þì±Ü¤ ÊÜáñÜᤠA±Ü¿Þì±Ü¤ PÝŸì®… ÓÜí¿ááPܤWÜÙÜ ®ÜvÜáÊÜ|
¿ÞÊÜâ¨Ý¨ÜÃÜã GÃÜvÜá ÊÜÂñÝÂÓÜWÜÙ®Ü Üá° ŸÃæÀáÄ.
b) ÃÜaÜ®Ý ÓÜÊÜÞíXWÜÙÜá Gí¨ÜÃæà®Üá ŸãÂpæà®…®Ü ÃÜaÜ®Ý ÓÜÊÜÞíXWÜÙÜ®Üá°
ŸÃæÀáÄ.
c) Ëáà¥æà®… A|á訆 CÇæPÝó®… aÜáQR ÃÜaÜ®Ý Ë®ÝÂÓÜÊÜ®Üá° ŸÃæÀáÄ.
EñܤÃÜ :

CCE RF/RR/PF/PR(A)/101/1836 (MA) CHE [ Turn over

83-K 8
±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
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a)
±Ü¿Þì±Ü¤ PÝŸì®… ÓÜí¿ááPܤWÜÙÜá A±Ü¿Þì±Ü¤ PÝŸì®… ÓÜí¿ááPܤWÜÙÜá
 PÝŸì®…&PÝŸì®… ±ÜÃÜÊÜÞ|áWÜÙÜ  PÝŸì®…&PÝŸì®… ±ÜÃÜÊÜÞ|áWÜÙÜ
®ÜvÜáÊæ HPÜŸí«ÜÊÜ®Üá° Öæãí©Êæ ®ÜvÜáÊæ ©ÌŸí«Ü A¥ÜÊÝ
£ÅŸí«ÜWÜÙÜ®Üá° Öæãí©Êæ
 PÜwÊæá QÅ¿ÞÎàÆÊÝXÊæ  ÖæaÜác QÅ¿ÞÎàÆÊÝXÊæ

 B¨æàÍÜ®Ü QÅÁáWÜÚWæ JÙܱÜvÜáñܤÊæ  ÓÜíPÜÆ®Ü ÊÜáñÜᤠB¨æàÍÜ®Ü
QÅÁáWÜÚWæ JÙܱÜvÜáñܤÊæ
 ¨Ü×Ô¨ÝWÜ ¯àÈ hÝÌÇæÀáí¨Ü  ¨Ü×Ô¨ÝWÜ Pæí±Üâ / ÖÜÙÜ©
EÄ¿ááñܤÊæ hÝÌÇæÀáí¨Ü EÄ¿ááñܤÊæ
¿ÞÊÜâ¨Ý¨ÜÃÜã GÃÜvÜá 21

b) Jí¨æà A|áÓÜãñÜÅ B¨ÜÃæ ˼®Ü° ÃÜaÜ®æ Öæãí©ÃÜáÊÜ PÝŸì®…
ÓÜí¿ááPܤWÜÙÜ®Üá° ÃÜaÜ®Ý ÓÜÊÜÞíXWÜÙÜá G®Üá°ÊÜÃÜá. 1

½+½
c)

1 5

CCE RF/RR/PF/PR(A)/101/1836 (MA) CHE

CCE RF/RR/
A PF/PR
PÜ®ÝìoPÜ ÍÝÇÝ ±ÜÄàûæ ÊÜáñÜᤠÊÜåèƯ|ì¿á ÊÜáívÜÈ,
ÊÜáÇæÉàÍÜÌÃÜí, ¸æíWÜÙÜãÃÜá – 560 003
KARNATAKA SCHOOL EXAMINATION AND ASSESSMENT BOARD,
MALLESHWARAM, BENGALURU – 560 003
ÊÜÞa…ì/H²ÅÇ… 2025 ÃÜ ±ÜÄàûæ - 1
MARCH/APRIL 2025 EXAMINATION - 1
ÊÜÞ¨ÜÄ EñܤÃWÜ ÜÙÜá
MODEL ANSWERS
ÓÜíPæàñÜ ÓÜíTæ : 83-K CODE NO. : 83-K
ËÐÜ¿á ËþÝ®Ü
Subject : SCIENCE
»èñÜ ËþÝ®Ü, ÃÜÓÝ¿á®Ü ËþÝ®Ü ÊÜáñÜᤠiàÊÜ ËþÝ®Ü / Physics, Chemistry & Biology
ÍÝÇÝ A»Ü¦ì / ÍÝÇÝ ±Üâ®ÜÃÝÊÜ£ìñÜ A»Ü¦ì / TÝÓÜX A»Ü¦ì / TÝÓÜX ±Üâ®ÜÃÝÊÜ£ìñÜ A»Ü¦ì
Regular Fresh / Regular Repeater / Private Fresh / Private Repeater
PܮܰvÜ ÊÜÞ«ÜÂÊÜá / Kannada Medium
iàÊÜ ËþÝ®Ü / Biology
©®ÝíPÜ 02. 04. 2025 ] [ WÜÄÐÜu AíPÜWÜÙÜá : 80
Date : 02. 04. 2025 ] [ Max. Marks : 80
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»ÝWÜ & C
( iàÊÜ ËþÝ®Ü )
XII. ŸÖÜá&BÁáR ±ÜÅÍæ°WÜÙÜá 3×1=3
25. iàÊÜPæãàÍܨÜÈÉ ±æäÅàqà®…WÜÙÜ ñÜ¿ÞÄPæWæ ÊÜÞ×£¿á ÊÜáãÆÊÜâ
(A) ÊÜíÍÜÊÝ× (B) ÊÜ|ìñÜíñÜá
(C) w.G®….G. (D) Ãæç¸æãàÓæãàÊÜå…
EñܤÃÜ (C) w.G®….G.
1

CCE RF/RR/PF/PR(A)/101/1836 (MA) BIO [ Turn over

83-K 2

±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
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26. ÃÜPܤ¨ÜÈÉ®Ü ÓÜPÜRÃæ ÊÜáor ÖæaÝcWÜáÊÜâ¨Üá C®ÜáÕÈ®… ÖÝÊæãàì®…®Ü PÜwÊæá
ÓÜÅËPæ PÜᣤWæ¿áá F©PæãÙÜáÛÊÜâ¨Üá ……………… .
(A) ¥æçÃÝQÕ®… ÖÝÊæãàì®…®Ü PÜwÊæá ÓÜÅËPæ
(B) C®ÜáÕÈ®… ÖÝÊæãàì®…®Ü ÖæaÜác ÓÜÅËPæ
(C) ¥æçÃÝQÕ®… ÖÝÊæãàì®…®Ü ÖæaÜác ÓÜÅËPæ
(D) A˜PÜ AÁãàw®… ÓæàÊÜ®æ
EñܤÃÜ (A) ¥æçÃÝQÕ®… ÖÝÊæãàì®…®Ü PÜwÊæá ÓÜÅËPæ
1
27. Ó泄ÃæãàWæçÃݨÜÈÉ PÜívÜáŸÃÜáÊÜ AÇæçíXPÜ ÓÜíñÝ®æãàñܳ£¤¿á Ë«Ü
(A) ÊæãWÜáYËPæ (B) ±Üâ®ÜÃÜáñݳ¨Ü®æ
(C) ©Ì ˨ÜÙÜ®Ü (D) ñÜáívÜÄPæ
EñܤÃÜ (D) ñÜáívÜÄPæ
1
XIII. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ EñܤÄÔ 3×1=3
28. KÃækãà®… ±Ü¨ÜÃÜ¨Ü PÝ¿áìÊæà®Üá
EñܤÃÜ :
C¨Üá ÓÜã¿áì¯í¨Ü ŸÃÜáÊÜ ®æàÃÜÙÝ£àñÜ ËQÃÜ|©í¨Ü »ÜãËá¿á
ÊæáàÇ澄¿á®Üá° ÃÜüÓÜáñܤ¨æ. 1
29. ®ÜÃÜPæãàÍÜ ®ÜãÂÃÝ®… Gí¨ÜÃæà®Üá
EñܤÃÜ :
®ÜÃÜÊÜäÂÖÜ¨Ü ÃÜaÜ®ÝñܾPÜ ÖÝWÜã PÝ¿ÞìñܾPÜ ZoPÜ. 1
30. ¨Üá£ÓÜíÍæÉàÐÜOæ¿áÈÉ gÃÜáWÜáÊÜ Zo®æWÜÙÜ®Üá° PæÙÜWæ ¯àvÜÇÝX¨æ. D
Zo®æWÜÙ®Ü Üá° ÓÜÄ¿Þ¨Ü PÜÅÊÜá¨ÜÈÉ ÊÜÂÊÜÓæ§WæãÚÔ ŸÃæÀáÄ
i) ¯àÄ®Ü A|áWÜÙÜá ÖæçvæãÅàg®… ÊÜáñÜᤠBQÕg®… BX
Ë»ÜiÓÜƳvÜáÊÜâ¨Üá
CCE RF/RR/PF/PR(A)/101/1836 (MA) BIO

3 83-K

±ÜÅÍæ°
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ii) PæãÉàÃæãàµÇ…¯í¨Ü ¸æÙÜQ®Ü ÍÜQ¤ ×àÃÜáËPæ
iii) PÝŸì®… væç BPæÕ„v… PݸæãàìÖæçvæÅàp… BX ±ÜÄÊÜ£ìÓÜƳvÜáÊÜâ¨Üá
iv) ¸æÙÜQ®Ü ÍÜQ¤¿áá ÃÝÓÝ¿á¯PÜ ÍÜQ¤¿ÞX ±ÜÄÊÜñÜì®æ¿ÞWÜáÊÜâ¨Üá
EñܤÃÜ :
ii) PæãÉàÃæãàµÇ…¯í¨Ü ¸æÙÜQ®Ü ÍÜQ¤ ×àÃÜáËPæ
iv) ¸æÙÜQ®Ü ÍÜQ¤¿áá ÃÝÓÝ¿á¯PÜ ÍÜQ¤¿ÞX ±ÜÄÊÜñÜì®æ¿ÞWÜáÊÜâ¨Üá
i) ¯àÄ®Ü A|áWÜÙÜá ÖæçvæãÅàg®… ÊÜáñÜᤠBQÕg®… BX
Ë»ÜiÓÜƳvÜáÊÜâ¨Üá
iii) PÝŸì®… væç BPæÕ„v… PݸæãàìÖæçvæÅàp… BX ±ÜÄÊÜ£ìÓÜƳvÜáÊÜâ¨Üá 1

XIV. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ EñܤÄÔ 3×2=6

31. Jí¨Üá B«ÝÃÜ¨Ü ÓÜáñܤ ŸÚÛ ÓÜÓÜÂWÜÙÜ PÜáwWÜÙÜ ¸æÙÜÊÜ~Wæ¿á®Üá° BQÕ®…WÜÙÜá
ÖæàWæ Eñæ¤àiÓÜáñܤÊæ
A¥ÜÊÝ
AwÅ®ÜÈ®… ÃÜPܤ¨ÜÈÉ ÓÜÅËPæ¿Þ¨ÝWÜ ®ÜÊÜá¾ ¨æàÖÜÊÜâ ÖæàWæ ±ÜÅ£QÅÀáÓÜáñܤ¨æ
EñܤÃÜ :

 ŸÚÛ¿á PÜáwWÜÙÜá B«ÝÃÜ¨Ü ÓÜí±ÜPÜìPæR Ÿí¨ÝWÜ BQÕ®…WÜÙÜá
B«ÝÃÜ©í¨Ü ¨ÜãÃܨÜÈÉÃÜáÊÜ PÜáw¿á »ÝWÜPæR ÖæaÜác
ËÓÜÃÜOæWæãÙÜáÛñܤÊæ. 1

 C¨ÜÄí¨ÝX, B«ÝÃÜ¨Ü PÜvXæ ÃÜáÊÜ PÜáw¿á »ÝWÜQRíñÜ
¨ÜãÃÜËÃÜáÊÜ PÜáw¿á »ÝWÜÊÜâ ÊæàWÜÊÝX ¸æÙæ¨Üá B«ÝÃÜ¨Ü ÓÜáñܤ
ÊÜêñݤPÝÃܨÜÈÉ AíqPæãÙÜáÛñܤÊæ. 1 2

A¥ÜÊÝ

CCE RF/RR/PF/PR(A)/101/1836 (MA) BIO [ Turn over

83-K 4

±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
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 ÖÜê¨Ü¿á¨Ü ŸwñÜ¨Ü ÊæàWÜÊÜâ ÖæaÝcWÜáñܤ¨æ ÊÜáñÜᤠÓÝ°¿ááWÜÚWæ
ÖæaÜác BQÕg®… ±ÜäÃæçPæ¿ÞWÜáñܤ¨æ. ½

 iàOÝìíWÜÊÜäÂÖÜ ÊÜáñÜᤠaÜÊÜáì¨ÜÈÉ®Ü ÓÜ|¡ A±Ü«ÜÊÜá¯WÜÙÜ ÓÜáñܤ
CÃÜáÊÜ ÓÝ°¿ááWÜÙÜ ÓÜíPÜáaÜ®æÀáí¨Ü D AíWÜWÜÚWæ ÖÜÄ¿ááÊÜ
ÃÜPܤÊÜâ PÜwÊæá¿ÞWÜáñܤ¨æ. ½

 C¨Üá ®ÜÊÜá¾ AÔ¤±ÜígÃÜ¨Ü ÓÝ°¿ááWÜÚWæ ÃÜPܤ¨Ü ÖÜÄ¿áËPæ¿á®Üá°
£ÃÜáXÓÜáñܤ¨æ. ½

 ±ÜPæRÆáŸá ÓÝ°¿ááWÜÙÜ ÊÜáñÜᤠÊܱæ¿á ÓÜíPÜáaÜ®ÜWÜÚí¨Ü
EÔÃÝo¨Ü WÜ£¿áã PÜãvÜ ÖæaÝcWÜáñܤ¨æ. ½ 2

32. ÖÝÊÜâ, Pܱæ³, ÖÜáÆáÉ ÊÜáñÜᤠËávÜñæ D iàËWÜÙÜ®Üá° ŸÙÜÔPæãívÜá Jí¨Üá
BÖÝÃÜ ÓÜÃܱÜÚ¿á®Üá° ÃÜbÔ. D BÖÝÃÜ ÓÜÃܱÜÚ¿áÈÉ ÖݯPÝÃÜPÜ
ÃÝÓÝ¿á¯PÜWÜÙÜ ÓÜíWÜÅÖÜÊÜâ ¿ÞÊÜ iàË¿áÈÉ ÖæaÝcXÃÜáñܤ¨æ

EñܤÃÜ :

 ÖÜáÆáÉ → ËávÜñæ → PÜ±æ³ → ÖÝÊÜâ 1

 ÖÝÊÜâ 1 2

33. ÍÜÇÝPÝWÜÅ¨Ü ÊæáàÇæ ±ÜÃÝWÜ¨Ü ÊæãÙæ¿ááËPæ¿á®Üá° ñæãàÄÓÜáÊÜ bñÜÅÊÜ®Üá°
ŸÃæÀáÄ ÊÜáñÜᤠAívÝÍÜ¿á ÊÜ®Üá° WÜáÃÜá£Ô.

EñܤÃÜ :

CCE RF/RR/PF/PR(A)/101/1836 (MA) BIO

5 83-K

±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ

bñÜÅPæR — 1½
»ÝWÜPæR — ½ 2
XV. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ EñܤÄÔ 4 × 3 = 12

34. ÊÜÞ®ÜÊÜ®Ü ià|ì®ÝÙܨÜÈÉ PæÙÜX®Ü Q|ÌWÜÙÜ ±ÝñÜÅÊæà®Üá
i) qŲծ…
ii) AÊæáçÇæàÓ…
iii) Çæç±æàÓ…
EñܤÃÜ :
i) BÖÝÃܨÜÈÉÃÜáÊÜ ±æäÅàqà®Ü®Üá° ià~ìÓÜáñܤ¨æ. 1

ii) ²ÐÜu GíŸ ÓÜíQà|ì A|áÊÜ®Üá° Ë»ÜiÔ ÓÜPÜRÃæ¿á®Üá° ¹vÜáWÜvæ
ÊÜÞvÜáñܤ¨æ. 1

iii) GÊÜáÈÕàPÜÃÜ|WæãívÜ PæãŸáºWÜÙÜ®Üá° Ë»ÜiÓÜáñܤ¨æ. 1
3
35. PæÙÜX®Ü ÖæàÚPæWÜÙÜ®Üá° ÓÜÊÜá¦ìÔ
a) ÇæçíXPÜ Ä࣠ÓÜíñÝ®æãàñܳ£¤¿áá ÖæaÜác ¼®Ü°ñæWÜÙÜ®Üá°
EíoáÊÜÞvÜáñܤ¨æ.
CCE RF/RR/PF/PR(A)/101/1836 (MA) BIO [ Turn over

83-K 6

±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
b) ÔŒà¿á WÜ»ÝìÍÜ¿á¨ÜÈÉ »ÜãÅ|¨Ü ¸æÙÜÊÜ~WæWæ gÃÝ¿áá 訆 ±ÝñÜÅ
ÊÜáÖÜñŲ̈ݪX¨æ.
A¥ÜÊÝ
a) ÔŒà¿áÃÜÈÉ MáñÜáaÜPÜÅÊÜâ ÖæàWæ EípÝWÜáñܤ¨æ
b) ÊÜÞ®ÜÊÜÃÜ ±ÜâÃÜáÐÜ ÓÜíñÝ®æãàñܳ£¤ ÊÜäÂÖܨÜÈÉ ÊÜêÐÜ|WÜÙÜá Q¸æãºpær¿á
ÖæãÃÜWæ ÊÜêÐÜ| bàƨÜÈÉÃÜáñܤÊæ. HPæ
EñܤÃÜ :
a)  ÇæçíXPÜ ÓÜíñÝ®æãàñܳ£¤¿áÈÉ GÃÜvÜá ˼®Ü° iàËWÜÙÜ
w.G®….G. A|áWÜÙÜ JÙÜWæãÙÜáÛËPæ C¨æ.
(¼®Ü°ÊÝ¨Ü ÊÜíÍÜÊÝ×WÜÙÜ®Üá° Öæãí©ÃÜáÊÜ WÜívÜá ÊÜáñÜá¤
Öæ|á¡ ÈíWÝ|áWÜÙÜ ÓÜíÁãàg®æ EípÝWÜáñܤ¨æ) 1

 ÊÜíÍÜÊÝ×WÜÙÜ ÖæãÓÜ ÓÜíÁãàg®æÀáí¨Ü iàËÓÜí¨Ü~¿á
±ÜÅ£ ²àÚWæ¿áÈÉ¿áã ¼®Ü°ñæWÜÙÜá ÖæaÝcWÜáñܤÊæ. 1

b) gÃÝ¿áá
 ñÝÀáÀáí¨Ü »ÜãÅ|PæR WÜáÉPæãàÓ… ÊÜáñÜᤠBQÕg®… ÖݨÜá
ÖæãàWÜÆá ËÍÝÆÊÝ¨Ü ÊæáàÇ澄¿á®Üá° J¨ÜXÓÜáñܤ¨æ. ½
 ¸æÙæ¿áᣤÃÜáÊÜ »ÜãÅ|©í¨Ü Eñܳ£¤¿Þ¨Ü ñÝÂgÂ
±Ü¨Ý¥ÜìWÜÙÜ®Üá° ñÝÀá¿á ÃÜPܤPæR ÊÜWÝìÀáÔ
ÖæãÃÜÖÝPÜáñܤ¨æ. ½ 3
A¥ÜÊÝ
a)  AívÜÊÜâ ¶ÜÈñÜWæãÙÜÛ©¨ÜªÃæ, A¨Üá Jí¨Üá ©®Ü¨ÜÊÜÃæWæ
Ÿ¨ÜáQÃÜáñܤ¨æ. 1

CCE RF/RR/PF/PR(A)/101/1836 (MA) BIO

7 83-K

±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
 ¨Ü±Ü³ÊÝ¨Ü Óܳíi®ÜíñÜÖÜ WÜ»ÜìPæãàÍÜ¨Ü JÙÜÓܤĿáá
¯«Ý®ÜÊÝX ¹ÃÜáPÜá ¹oár ÃÜPܤ ÊÜáñÜᤠÇæãàÙæ¿á ÃÜã±Ü¨ÜÈÉ
Áãà¯Àáí¨Ü ÖæãÃÜŸÃÜáñܤ¨æ. (MáñÜáaÜPÜÅ EípÝWÜáñܤ¨æ) 1

b) Ëà¿Þì|áWÜÙÜ Eñݳ¨Ü®æWæ ¨æàÖÜ¨Ü EÐÜ¡ñæXíñÜ PÜwÊæá
EÐÜ¡ñæ¿á AÊÜÍÜÂPÜñæÀá¨æ. 1 3
36. ÊÜÞ®ÜÊÜ®Ü Ëá¨ÜáÚ®Ü ÃÜaÜ®æ¿á®Üá° ñæãàÄÓÜáÊÜ bñÜÅÊÜ®Üá° ŸÃæÀáÄ. PæÙÜX®Ü
»ÝWÜWÜÙÜ®Üá° WÜáÃÜá£Ô
i) Öæç±æäà¥ÜÇÝÊÜáÓ…
ii) ÊæávÜáÇÝÉ
EñܤÃÜ :

bñÜÅPæR — 2
»ÝWÜPæR — ½ + ½ 3

37. ¨ÜáívÜ®æ¿á, ÖÜÔÃÜá Ÿ|¡¨Ü ¹àgWÜÙÜ®Üá° ( RRyy ) Eñݳ©ÓÜáÊÜ ŸpÝ~
ÓÜÓÜÂÊÜ®Üá° ÓÜáPÝR¨Ü, ÖÜÙÜ© Ÿ|¡¨Ü ¹àgWÜÙÜ®Üá° ( rrYY ) Eñݳ©ÓÜáÊÜ
ŸpÝ~ ÓÜÓܨæãí©Wæ ÓÜíPÜÃÜ|WæãÚÔ¨æ. F 2 ²àÚWæ¿á ¶ÜÈñÝíÍÜÊÜ®Üá°
aÜPÜRÃ… ¸æãàv…ì®Ü ÓÜÖÝ¿á©í¨Ü ñæãàÄÔ ÊÜáñÜᤠÓÜÓÜÂWÜÙÜ Ë«ÜWÜÙÜ
A®Üá±ÝñÜÊÜ®Üá° ŸÃæÀáÄ.

A¥ÜÊÝ

CCE RF/RR/PF/PR(A)/101/1836 (MA) BIO [ Turn over

83-K 8

±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
a) ÊÜÞ®ÜÊÜÃÜÈÉ ÊÜáWÜáÊæäí¨ÜÃÜ ÈíWÜ ¯«ÜìÃÜOæ¿ÞWÜáËPæ¿á®Üá°
ËÊÜÄÔ.
b) iàËWÜÙÜÈÉ WÜá|WÜÙÜ®Üá° ±ÜÅŸÆ A¥ÜÊÝ ¨ÜáŸìÆ Gí¨Üá ÖæàWæ
£àÊÜÞì¯ÓÜŸÖÜá¨Üá
EñܤÃÜ :
RRyy X rrYY
(¨ÜáívÜ®æ¿á, ÖÜÔÃÜá) (ÓÜáPÝR¨Ü, ÖÜÙÜ©)

ÈíWÝ|áWÜÙÜá Ry rY

RrYy F 1 ²àÚWæ

¨ÜáívÜ®æ¿á, ÖÜÙÜ©
RrYy X RrYy
( F1 ) ( F1 )

ÈíWÝ|áWÜÙáÜ RY Ry rY ry

RY RRYY RRYy RrYY RrYy

Ry RRYy RRyy RrYy Rryy

rY RrYY RrYy rrYY rrYy

ry RrYy Rryy rrYy rryy
aÜPÜRÃ… ¸æãàv…ìWæ 2

ÓÜÓÜÂWÜÙÜ Ë«ÜWÜÙÜ A®Üá±ÝñÜ 9:3:3:1 1 3
A¥ÜÊÝ
a)  ñÜí¨æ¿áÈÉ Jí¨Üá hæãàw ¸æÓÜ ÈíWÜ ÊÜ|ìñÜíñÜáWÜÙÜá
PÜívÜáŸÃÜáñܤÊæ. AÊÜâ ‘X’ ÊÜáñÜᤠ‘Y’ ÊÜ|ìñÜíñÜáWÜÙÝXÊæ. ½

CCE RF/RR/PF/PR(A)/101/1836 (MA) BIO

9 83-K

±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
 B¨ÜÃæ, ñÝÀá¿áÈÉ PæàÊÜÆ ‘XX’ ÊÜ|ìñÜíñÜáWÜÚÃÜáñܤÊæ. ½
 ñÜí¨æÀáí¨Ü HPÜ WÜá~ñÜ ÊÜ|ìñÜíñÜá ‘X’ A®Üá° ±Üvæ¨Ü
ÊÜáWÜáÊÜâ ñÝÀáÀáí¨ÜÆã ‘X’ ÊÜ|ìñÜíñÜáÊÜ®æ°à
±Üvæ¿ááÊÜâ¨ÜÄí¨Ü, ‘XX’ ÊÜ|ìñÜíñÜá hæãàw¿áá Öæ|á¡
ÊÜáWÜáÊÜ®Üá° ±ÜÅ£¯˜ÓÜáñܤ¨æ. ½
 ñÜí¨æÀáí¨Ü HPÜWÜá~ñÜ ÊÜ|ìñÜíñÜá ‘Y’ A®Üá° ±Üv¨
æ Ü
ÊÜáWÜáÊÜâ ñÝÀáÀáí¨Ü ‘X’ ÊÜ|ìñÜíñÜáÊÜ®æ°à
±Üvæ¿ááÊÜâ¨ÜÄí¨Ü, ‘XY’ ÊÜ|ìñÜíñÜá hæãàw¿áá WÜívÜá
ÊÜáWÜáÊÜ®Üá° ±ÜÅ£¯˜ÓÜáñܤ¨æ. ½
A¥ÜÊÝ

2

b) iàËWÜÙÜÈÉ ÊÜÂPܤÊÝWÜáÊÜ / WæãàaÜÃÜÊÝWÜáÊÜ WÜá|WÜÙÜ®Üá° ±ÜÅŸÆ
WÜá|WÜÙÜá, ÊÜÂPܤÊÝWÜ¨Ü / ÓÜá±Ü¤ÊÝ¨Ü / PÜwÊæá ÊÜÂPܤÊÝWÜáÊÜ
WÜá|WÜÙÜ®Üá° ¨ÜáŸìÆ WÜá|WÜÙÜá Gí¨Üá £àÊÜÞì¯ÓÜŸÖÜá¨Üá. 1
3

CCE RF/RR/PF/PR(A)/101/1836 (MA) BIO [ Turn over

83-K 10

±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
XVI. PæÙÜX®Ü ±ÜÅÍæ°Wæ EñܤÄÔ 1×4=4

38. a) ®ÜÊÜá¾ ÃÜPܤ¨ÜÈÉÃÜáÊÜ ×àÊæãàWæãÉ๮… ÊÜ|ìPÜWÜÙÜ ±ÝñÜÅÊÜ®Üá°
ÓÜíü±Ü¤ÊÝX ËÊÜÄÔ.

b) ÓÜÓÜÂWÜÙÜÈÉ PÜívÜáŸÃÜáÊÜ ËË«Ü ËÓÜgì®Ý ñÜíñÜÅWÜÙÜá ¿ÞÊÜâÊÜâ

EñܤÃÜ :

a)  ×àÊæãàWæãÉ๮… ÊÜ|ìPÜÊÜâ ÃÜPܤPæR Pæí±Üâ Ÿ|¡Ê®Ü Üá°
¯àvÜáñܤ¨æ ÊÜáñÜᤠBQÕg®… PÜvæWæ ÖæaÜác
BPÜÑìñÜÊÝXÃÜáñܤ¨æ. 1

 C¨Üá ÃÜPܤ¨ÜÈÉ ËÈà®ÜWæãívÜ BQÕg®… A®Üá°
ÃÜPܤ±ÜÄaÜÆ®æ¿á ÊÜáãÆPÜ ¨æàÖÜ¨Ü GÇÝÉ iàÊÜPæãàÍÜWÜÚWæ
±ÜäÃæçÓÜáñܤ¨æ. 1

D ÊÜáãÆPÜ ÍÜQ¤ Eñݳ¨Ü®æWæ ÓÜÖÝ¿áPÜÊÝX¨æ

b)  ÖæaÝc¨Ü ¯àÃÜ®Üá° ¸ÝÐܳËÓÜgì®æ¿á ÊÜáãÆPÜ ÖæãÃÜÖÝPÜáñܤ¨æ.

 ÓÜÓÜ ñÝÂg ±Ü¨Ý¥ÜìWÜÙÜá iàÊÜPæãàÍÜ¨Ü ÃÜÓܨݯWÜÙÜÈÉ
ÓÜíWÜÅÖÜÊÝWÜáñܤ¨æ.

 GÇæWÜÙÜÈÉ ñÝÂg ±Ü¨Ý¥ÜìWÜÙÜá ÓÜíWÜÅÖÜÊÝX AÊÜâ E¨ÜáÄ
ÖæãàWÜáñܤÊæ.

CCE RF/RR/PF/PR(A)/101/1836 (MA) BIO

11 83-K

±ÜÅÍæ°
±ÜÅÍÝ°®ÜáÓÝÃÜ ÊÜåèÆÂÊÜÞ±Ü®Ü AíPÜWÜÙÜá
ÓÜíTæÂ
 ÖÜÙæ¿á PæÕ„ÆíWÜÙÜÈÉ ñÝÂg ±Ü¨Ý¥ÜìWÜÙÜá ÃÝÙÜ ÊÜáñÜá¤
AíoáWÜÙÜ ÃÜã±Ü¨ÜÈÉ ÓÜíWÜÅÖÜÊÝWÜáñܤÊæ.

 ÓÜÓÜÂWÜÙÜá ñÜÊÜá¾ ÓÜáñÜ¤È®Ü ÊÜá~¡®ÜÈÉ PæÆÊÜâ ñÝÂgÂWÜÙÜ®Üá°
ËÓÜiìÓÜáñܤÊæ.

¿ÞÊÜâ¨Ý¨ÜÃÜã ®ÝÆáR 4½ 4

CCE RF/RR/PF/PR(A)/101/1836 (MA) BIO [ Turn over

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