SCIENCE (086)

MARKING SCHEME (2022-23):

Set - 1
Questions
SECTION - A

1 c. ii > i > iv > iii 1

2 b. 2, 2, 4 1
3 a. Reddish brown colour ferric oxide is formed. 1
4 c, Absorb moisture from the gas. 1
5 b. MgO + H2O → Mg (OH)2 1
6 c. Yellowish green 1
7 d. C4H8 1

8. (a) Large amount of water flows into guard cells 1

9. (d) Light is necessary for photosynthesis 1
10. (b) Tt and Tt 1
11. (c )The volume of thoracic cavity increases 1
12. (a) Dendrite– cell body— axon — nerve ending 1

13 . C

14. D

15. B

16. (d) :

17. (b) Both A and R are true but R is not the correct explanation of A 1
18. (c ) A is true but R is false 1
19. (a) Both A and R are true and R is the correct explanation of A 1

Q 20. C

SECTION - B
21. 2 NaCl(aq)+ 2H2O → 2 NaOH(aq)+ Cl2(g)+ H2
X → Cl2 1
Ca(OH)2+Cl2 → CaOCl2 + H2O
Y → Ca0Cl2 1
OR
X = CaCO3 ½
 Gas = CO2 ½

CaCO3 +H2SO4 → CaSO4 + H2O + CO2 1

22. Hormones reach each and every cell in the body. Electrical impulse is
limited only to those areas connected by nerves.
Once electrical impulse is generated the cell takes time to reset. Hence it is
short lived.
In chemical communication hormones help in sustained stimulation over a longer
period. (1X2 = 2marks)
23. Test tube B (1 mark) (2 marks)
Pepsin is a protein digesting enzyme
Pepsin requires an acidic medium to act on protein rich food.
So pepsin along with HCl will act on egg white for its digestion. (1 mark)
24. Alveoli of lungs have rich capillary network for efficient gas exchange.
Nephron in kidneys have rich capillary network for filtration of blood and reabsorption
of water and nutrients during urine formation. (1X2= 2 marks)

Q 25. (a) Violet (b) Red ½+1/2

(b) Due to difference in speed of light of different wave lengths 1.

OR

(i) Myopia 1

(ii) f = 1/-4.5 =-0.22m 1

(iii) Concave lens. 1

26. (a) Green plants capture 1% of solar energy that falls on the leaves and convert it
into food energy. 1% of 10000= 100 J (1 mark)
(b) According to 10% law only 10% of energy at each trophic level is transferred to
next trophic level. Grasshopper will get 10J and shrew being secondary consumer
will get 10% of 10 J = 1J. (1 mark) (1X2= 2 marks)

SECTION - C
27.
a. Fe2O3(s) + 2 Al(s) → 2 Fe(l) + Al2O3(s) 1
Displacement reaction or redox reaction ½
b. 2 AgBr(s) → 2 Ag(s) + Br2(g) 1
Photolytic decomposition reaction ½
28.
a. Plaster of Paris (1)
[373K]
 b. CaSO4.2H2O → CaSO4 . ½H2O + 1½ H2O (1)
c. Used as plaster for supporting fractured bones at the right position (1)

29. (a) 3 - pulmonary vein (½)

6 - vena cava (½)
(b) Part 2 is the capillary network in lungs. Here carbon dioxide from blood is
given into the lung alveoli and oxygen from alveoli diffuses into blood.
Part 5 is the capillary network in tissues. Here oxygen is given from blood
into tissues and carbon dioxide from tissues diffuses into the blood. (1 mark)
(c ) deoxygenated blood from tissues —-gills for oxygenation—---- oxygenated blood
to tissues. (1 mark)

OR
lack of oxygen
(a) Glucose —---------pyruvate—---------------- lactic acid + energy (1 mark)
in cytoplasm ( in muscle cells)

absence of oxygen
Glucose—---------pyruvate —----------------- ethanol + carbon dioxide + energy
in cytoplasm ( in yeast) (1 mark)
(b) At night there is no photosynthesis.The carbon dioxide released during
respiration is not utilised.So CO2 elimination is the major exchange activity.
During day CO2 generated during respiration is used up for photosynthesis.
Hence there is no CO2 release. Instead oxygen release is the major activity.
(1 mark)

30. (a) 0 -12 cm 1/2

(b) image larger (½)

Diagram (1)

© At C ( 24cm) (1)

31. (a) 20cm (½)

u=v at 2F (½)

(b) Sno.6 (½) …Because when object is between P & F image is virtual. (½)

( c ) Diagram (1)

32.(a) Definition (1)

(b) diagram (1)

 (c ) Enhances the power of electro magnet

OR

1. Magnetic field at P is into the plane of paper and at Q it is out of the plane ½+½

The strength of the magnetic field at Q will be larger as strength of the field α 1/r 1M

2. The direction of magnetic field at the centre of a current carrying circular loop is
pependicular to the plane of the loop 1M

33. uv
O2 —------.> O + O
O + O2 ------> O3 (2 marks)
Chloroflurocarbons in refrigerators have chlorine which will react with ozone and lead
to its depletion. (1 mark)
SECTION - D
34.
● Soaps are sodium or potassium salts of long chain carboxylic acids whereas
detergents are ammonium or sulphonate salts of long chain carboxylic acids. 1
● Soaps have two ends – one hydrophobic and the other hydrophilic. ½
● Micelle formation ½
● Figure ½
● Explanation 1
● Soaps form an insoluble substance called scum by the reaction with calcium and
magnesium salts in hard water. 1
● Problem can be solved with the use of detergents ½
OR
C2H5OH + 3 O2 → 2 CO2 + 2 H2O + Heat
1
2 C2H5OH + 2 Na → 2 C2H5ONa + H2 1
C2H5OH →(alk. KMnO4 + heat) CH3-COOH 1
CH3-COOH + C2H5OH → CH3-CO-O-C2H5OH + H2O 1
C2H5OH →(hot conc. H2SO4) CH2 = CH2 + H2O 1

35. (a) Errors during DNA copying

Fusion of gametes of two different individuals during sexual reproduction.(1mark)
(b) (i) A - pollen grain (½ mark)
(ii) pollen reaches B stigma through air, water or insects (½)
(iii) C is pollen tube that carries the male gametes to ovule for fertilisation with egg
(½)
(c ) chemotropic movement (½)
(d) drawing (1 mark)
(i) Radicle (½ mark)
(ii) cotyledon(½ mark)
 OR

(a) Drawing - (1 mark)

Planaria reproduce by regeneration.

When cut into 3 pieces, each piece grows into a complete organism.
There are specialised cells that increase in number to form mass of cells
From this mass of cells different types of cells are formed in an organised manner.
(1 mark)
[b] C - uterus [½]
D - cervix [½]
[c] implanted embryo continues dividing , placenta develops [1 mark]

[d] avoid pregnancy, birth control, prevent STD [ any two, 1 mark]

36. (i) Resistance In Series is always greater than resistance in

parallel. The slope of V-I graph gives resistance.

Hence, line B denotes resistance in parallel combination

and line A denotes resistance in series combination. 1

(ii) R Ω (1)
 (iii) Case 1

P= v2/R…..(1)

Case 2

P1 = (110)2/ R…(2)

From (1) & (2)

P1= 25 W (1 + 1+1/2)

(iv) remains same (1/2)

SECTION - E

37.
a. An aqueous solution of copper salt 1
b. First reaction at anode 1
Second at cathode 1

c. Water insoluble impurities

OR 1
Gold, Silver, any other relevant answer

38. [a] wrinkled seed, Law of dominance [1 mark]

[b] 750, 250 [1 mark]
[c] 50% [1 mark]
parent Rr X rr
gametes R r r
Rr rr
[round] [wrinkled] [1 mark]

OR
RrVV,
Parents RrVV X rr vv
Gametes RV, rV rv
F1 RrVv rrVv
[round, violet] [ wrinkled, violet] [1 mark]

39. (i) Law of refraction. (1)

 (ii) Difference in Optical density (1)

(iii) n = C/V

n=1.32 (2)

OR

No bending occurs ie i=0

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