FIITJEE

INDIAN OLYMPIAD QUALIFIER IN JUNIOR SCIENCE (IOQJS)

PART – 1 (NSEJS)
Held on: March 06, 2022
ANSWER KEY

1. C 2. C 3. D 4. A
5. C 6. C 7. D 8. C
9. C 10. D 11. C 12. B
13. D 14. C 15. B 16. C
17. A 18. A 19. C 20. C
21. B 22. D 23. B 24. B
25. ABCD 26. ABC 27. ABD 28. ACD
29. ABC 30. AB 31. AD 32. BD
33. B

FIITJEE Ltd., Punjabi Bagh Centre, 31, 32, 33 Central Market, West Avenue Road, Punjabi Bagh, New Delhi – 110026, Ph: 011 - 45634000
 HINTS & SOLUTIONS
1. C
3
 R 
1. Z  A 
R  u
3
  3
 1   u
Z  A   A 1  
u  R
 1  
 R
u
as  1 using Binomial
R
 3u 
Z  A 1  
 R
3A
Z uA
R
y  mx  c

2. C
2. P = gh
P1 = P2
P(9.8) 200 = P(3.7) h
h = 530 m

3. D
3. at extreme position, potential energy is maximum.

4. A
4. Wheatstone

2 4

Req = 3

3
I  1A
3
2 4

5. C
5. Wheel A Wheel B
 
16 teeth 24 teeth
6 Revolution 4 Revolution
per second per second
1
So in second wheel B completes 2 revolution in anticlockwise direction.
2
B
A

FIITJEE Ltd., Punjabi Bagh Centre, 31, 32, 33 Central Market, West Avenue Road, Punjabi Bagh, New Delhi – 110026, Ph: 011 - 45634000
 displacement of mass = 2(2r)
22 14
 22
  16 cm
7 11
increment in potential energy  mg (16)

6. C
6.

E = 5 × 108 × (4r2) …(i)

E = mc2 …(ii)
Star
Earth
r From eq. (i) and (ii)
m = 5.62 × 1013 kg

r = 300 light year

r = [300 × 9.46 × 1015] m

7. D
J
7. Energy of 50 person  50  100
S
 50  5000  4hr
E 
 100  1000
E = 40 kwh
E = 40 units

8. C
8. Synthesis of immunoglobulins is not a function of mature RBCs.

9. C
9. Amphibians and Reptiles are group of limbless animals.
 Order Apoda of class Amphibia carry limbless Amphibians (e.g.Ichthyophis)
 Snakes are limbless reptiles belongs to Order Ophidia.

10. D
10. Housefly, fruitfy, mosquitoes have one pair of wings.

11. C
11. Single stranded RNA viruses tend to mutate more rapidly than double stranded viruses. RNA
polymirase that copies the virus genes generally lacks proofreading skills which make RNA
viruses prone to high mutation rates – upto million times greater than DNA.

12. B
12. Aerenchyma is present in leaves & petioles of hydrophytes

13. D
13. Bryophytes are not well differentiated into roots, stem & leaves so this statement is false.

14. C
14. Carbon fixation in succulents plants takes place through crassulacean acid metabolism.

FIITJEE Ltd., Punjabi Bagh Centre, 31, 32, 33 Central Market, West Avenue Road, Punjabi Bagh, New Delhi – 110026, Ph: 011 - 45634000
15. B
15. If a flower emit fruity or musky fragrance (very strong) after sunset then it will be pollinated
by Bats. Bees or Butterflies pollinate plants where scent is high during the day while moths &
bats pollinate plants whose fragrance is greatest at night.

16. C
16. Coal a fossil fuel contains 0.3 to 5% of S
S  O2

 SO2
1mole SO2 = 6.022  1023  3 atoms
= 1.8066  1024 atoms

17. A
17. Stomach fluid has HCl, KCl, and NaCl.
Each of 0.01 M and volume of fluid is 2L.
Al(OH)3 is the antacid used to neutralize stomach acid HCl.
Moles of HCl present in 2 L of fluid = M  V
= 0.01  2
= 0.02 moles.
3HCl  Al  OH3   AlCl3  3H2O

3 mole needs 1 mole

1 mole needs 1 mole
3
1 0.02
0.02 mole   0.02 mole   78 g Al  OH3
3 3
= 0.52 g

18. A
18. Mixture of CaCO3 + CaO = 0.5 g
xg+yg
 x  y  0.5 g …(1)
CaCO3  s   CaO  s   CO2  g 


x x
mol   mol
100 100
x
=  56 gCaO
100
= 0.56 x g CaO
Given that
0.56 x + y = 0.434 …(2)
Solving equation (1) and (2)
0.066
x  0.15 g
0.44
y = 0.5 – 0.15 = 0.35 g
0.35
 % of y (CaO) =  100  70%
0.5

19. C
19. Order of metallic character is
O < C < Be < Li < Na

20. C
20. CH3COOH given is 50 ml of 0.1 M.
CH3 COOH H2O CH3 COO  Na
Weak acid

FIITJEE Ltd., Punjabi Bagh Centre, 31, 32, 33 Central Market, West Avenue Road, Punjabi Bagh, New Delhi – 110026, Ph: 011 - 45634000
 According to ostwalds dilution law, degree of dissociation increases with increase in dilution.
So, on dilution the number of ions will increase and the conduction.

21. B
21. Compound X + NaOH  Pungent gas which turns red litmus blue.
X is NH4Cl
X  aq  AgNO3  aq  
 AgCl  NH4NO3
ppt

 X is NH4Cl
NH4Cl (aq) is slightly acidic due to hydrolysis.

22. D
22. Equal volume of all gases contains equal number of molecules or moles under similar
conditions of temperature and pressure.
Given masses are equal
 Moles will be equal if their molecular masses are also equal.
xg xg
 N2  , CO 
28 g mol1 28 g mol1

23. B
23. 10 milimoles of each Na2SO4, MgSO4 and Al2(SO4)3 in three different container P, Q and R
respectively. And each of them are made as one litre solution.
Na2SO 4   2Na   SO 24
P 
1 moles gives 3 mole ions

MgSO 4  Mg2  SO 24


 Q
1 mole gives 2 mole ions

Al2  SO 4 3 
 2Al3   3SO24
R
1 mole gives 5 mole ions.
The order of osmotic pressure is R > P > Q or Q < P < R.

24. B
24. Y1
Y2
P2 X2
1
10m
6m
P1  (8, 0)
X1
(-2, 0) P1
P2

displacement if P1  10 m
displacement if P2  10 m
8m

90°

6m
10m
1
90°

FIITJEE Ltd., Punjabi Bagh Centre, 31, 32, 33 Central Market, West Avenue Road, Punjabi Bagh, New Delhi – 110026, Ph: 011 - 45634000
 8
tan 1 
6
4
tan 1 
3
1 = 53.2°

25. ABCD
c 3  108
25. frequency v  
 650  10 9
 4.6  1014 HZ
Energy of each photon  E = hv
= 6.625 × 1034 × 4.6 × 1014
= 3.4 × 1019J
If n is number of photon in each pulse then
Total energy of pulse  nE
nE
Power  P 
t
n  3.1 1019
0.6 
30  10 3
n = 5.9 × 1016
Total energy of one pulse = nE
 5.9 × 1016 × 3.1 × 1019
 1.1 × 1017 eV

26. ABC
26.
8/3 A 4A 4/3 A
 
16V
12
6

I1 = 8/3 A I2 = 4/3 A
2
8
1. Power in R1     6  42.6 W
3
I1
2. 2
I2
3. Total current  4 A
4. 6  & 12  both are in parallel combination

27. ABD
t  sin i  r  
27. 1. lateral shift 
(x) cosr
 sin  A  B   sin A cosB  cos A sinB
t  sinicos r  cos i sinr 
x
cos r
 cos i sinr 
x  t  sini  
 cos r 

2. For maximum value of x, i = 90°

FIITJEE Ltd., Punjabi Bagh Centre, 31, 32, 33 Central Market, West Avenue Road, Punjabi Bagh, New Delhi – 110026, Ph: 011 - 45634000
 t  sin  90  r  
x
cosr
cos r
xt
cos r
x=t

3. The angle of refraction and angle of incidence are related with refractive index in
accordance with Snell’s law, Hence lateral shift also depends on the refractive index

4. Refractive index increases with decrease in wavelength, so lateral displacement also

depends on the wavelength.

28. ACD
28. Option B is correct.

29. ABC
29. If front part of brain is affected then it will impair speaking ability, smell, and walking on
narrow path.

30. AB
30. The molecules which are primarily responsible for structure support and mortality are actin
and tubulin alpha.

31. AD
31. Types of bonding in dry ice which is solid CO2 are covalent and vander waals forces.

32. BD
32. Temperature of water increases on dissolving the compound. That means it is exothermic
process. So they are KOH and HCl.

33. B
33. X is soap solution of 1% w/v
O

 X is O Na

Y is detergent solution of 1% w/v

O

S O Na

O
X splits into two equal parts X1 and X2
Y splits into two equal parts Y1 and Y2
X1 + NaCl (g)  No effect
Y1 + NaCl (g)  No effect.
X 2  CaCl2  g   Pr ecipitation or formation of scum .
 Soap 
Hareness
salt
cau sin g

Y2 is detergent so does not form ppt with CaCl2

FIITJEE Ltd., Punjabi Bagh Centre, 31, 32, 33 Central Market, West Avenue Road, Punjabi Bagh, New Delhi – 110026, Ph: 011 - 45634000