Class: NSEJS NSEJS FINAL EXAM KEY Date: 17-11-19 Time ... · Date: 17-11-19 . Time: 2hrs : QP CODE.54 . Max. Marks: 240 : CODE.54 NSEJS INITIAL KEY 17-11-19 Q.No. 1 2 3 4 5 6 7 8

Class: NSEJS NSEJS FINAL EXAM KEY Date: 17-11-19

Time: 2hrs QP CODE.54 Max. Marks: 240

CODE.54 NSEJS INITIAL KEY 17-11-19

Q.No. 1 2 3 4 5 6 7 8 9 10

Ans. C B B D B D A C A D

Q.No. 11 12 13 14 15 16 17 18 19 20

Ans. B C B A D B B C D C

Q.No. 21 22 23 24 25 26 27 28 29 30

Ans. C A B B A C Block A B C

Q.No. 31 32 33 34 35 36 37 38 39 40

Ans. C B A A A C B C D A

Q.No. 41 42 43 44 45 46 47 48 49 50

Ans. C A B A B B C A C B

Q.No. 51 52 53 54 55 56 57 58 59 60

Ans. A D A C B C D A C D

Q.No. 61 62 63 64 65 66 67 68 69 70

Ans. D C C B A D D D B A

Q.No. 71 72 73 74 75 76 77 78 79 80

Ans. A C A A A A A D C B

NSEJS_FINAL_EXAM_SOLUTIONS_CODE.54_Dt.17-11-19

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BIOLOGY 01. Appearance of notochord in larval stage with chitinous exoskeleton in adult is a

feature of protochordates

03. Nissil bodies main function is to synthesize and release proteins that are important for neuronal growth and regeneration of axons

04. Saturated lipids in plasma membrane 05. (i) a fungal (iii) conjugation deficient bacteriam 06. Central dogma of DNA 07. Learning process through which strength of a behaviour is modified by

reinforcement 08. Initial length / final length

2

46

6 10 1.5 104 10

−

−×

= ××

09. Salt mixed for fermentation. Yeast granules were not activated prior to mixing with the flour 10. Prokaryotic cells are unicellular while eukaryotes are multicellular Histones are present in euarkyotes and absent in prokaryotes (i) and (iii) false 11. In planaria each piece develops into new organism. In asterias central disc along

with limb is used for regeneration. 12. The ability to produce on abundance of offspring needs all the factors 15. Nociceptors are for painful stimuli. 16. Wavelength 17. (54 and 27) Meiosis-II = Mitosis 18. Seed is not covered by ovary 19. Gibberllic acid 20. Mangrooves


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CHEMISTRY 21. Y is phosphorus because it forms two oxides, 2 3P O and 2 5P O , which when

dissolved in water forms weak acids, 3 3 3 4&H PO H PO respectively. Phosphorus

is observed in various allotropic forms like white, red, yellow, scarlet etc. and

phosphates are used in agriculture.

22. Fullerene is an allotropic form of carbon and represented as 60C , its structure

consists of hexagonal and pentagonal rings. In between any two carbon atoms

only one sigma bond is present.

23. 2 2 2 22CaCO HCl CaCl H O CO+ → + +

44 gm 2CO = 100 gm 3CaCO

Therefore 0.88 gm 2100 0.88 244

CO gm= × =

Percentage purity = 2 100 504× = %

24. + → + ↑2 2 22Zn NaOH Na ZnO H

+ → + ↑2 212

Al NaOH NaAlO H

26. 6 6 2 6 6 621378 291

106.578

3C H Cl C H Cl+ →

3 71 291× →

106.5 ?→

106.5 291 145.53 71

g× =×

27. P) 110H+ − = Q) 12 10H+ − = × R) 310OH− − = S) 310OH− − =

1pH = 0.699pH = 11pH < 1pH =

28. 22 2CuCl Cu Cl+ −→ +

Correct option should be Q P R S> > >

So, No change in concentration

29. Polyvinyl chloride and polyethene are thermoplastic


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30. 1) Vinegar-acidic

2) NaCl Neutral−

3) NaOH − Strong base

4) Baking soda-Weak base

32. No. of meq of HCl = No. of meq of ammonia + No. of meq of NaOH

150 1 602

x× = + ×

50 30 20x = − =

Wt. of ammonia = 20 17 0.34

1000× =

33. Hg Ga Li Ca< < <

Element M.P Hg 038.83 C− Ga 029.76 C Li 0180.5 C Ca 0842 C

34. 22 4 4 ,Na WO WO Na− +→

( ) 2 32 4 42

,Pb PO Pb PO+ −→

∴ ( )4Pb WO

35. 3 2 24 5 4 6NH O NO H O+ → +

36. 2 &CO NO are linear

2 2&NO N O are angular

37. 22 2CaCl Ca Cl+ −→ +

No. of moles of 244.4 0.4396101

CaCl = =

No. of moles of ions in 1 litre = 3 0.4396 1.3188× =

No. of ions in 1 litre = 23 231.3188 6.023 10 7.94 10× × = ×

No. of ions in 1 ml = 207.94 10×

38. 3 2, &Ne N Mg− + have ten electrons each

39. Equal volumes of all gases have same number of moles or molecules at STP

40. 3 2, ,AlCl MgCl LiCl are covalent in Nature


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PHYSICS

41. Conceptual

42. Total area ( ) 228 14 cm× , change = 420 µC

For ( ) 22228 14 2 77

× − × × change be

Then 28 14 42014 6 9

×=

×

g 90 C⇒ = µ

So, change for the proton asked is 45 C2g= µ .

43. 13iRε

=

2

21 1

gH i Rt tRε

= =

Given 1 2H H=

2 4niRε

=

( )2 2 2

22 2 4 4

4 4N NH i R t R tR Rε ε = = × =

2 2 2

29 36 64

N N NR Rε ε

⇒ = = = ∴ =

44. Let ‘m’ be the mass of ice block

So, when ice block is in the water

Amount of water displaced be ( )1 0 at 4W

mVP C

=

When ice melts completely, amount of water formed be 2V

( )2 0 at 0W

mVP C

⇒ =

As 0 0 at 0 at 4W WP C P C<

2 1 water level will riseV V> ⇒ .


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45. Given 2A DV V− = …(1)

6B DV V− = …(2)

8C DV V V− = …(3)

1AR K= Ω

( ) ( )2 1 4B AV V− ⇒ − =

3

4 22 10BR K−= = Ω×

( ) ( )3 2 2C BV V− ⇒ − =

3

2 12 10CR K−= = Ω×

, 33

2 2 1010

i A−= = ×

46. Conceptual

47. From graph from 0x = to 4x m= work done

1 4 4 852

w = × × =

From 4x m= to 5x m= workdone = ( ) ( )4 1 4J=

∴ from 0x = to 5mx = total work = 12J

From w.e.t ( ) 2112 6 02

V= × − 2m/sV⇒ =

48. gW V= −∆

( ) ( )1 cos 1 cosmgL V v mgL⇒− − θ = −∆ ⇒∆ = − θ

49. To acquire terminal velocity net force on ball should be zero

50. At max displacement 0KE =

PE ME=

2 31 4 102

A −× = ×

( )21 31 10 4 10 0.8 /2

K K N m− −× = × ⇒ =


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51. ( ) ( )211 10 502

S a a= =

( ) ( )21 21 20 2002

S S a a+ = =

2 2 1150 3S a S S= ∴ = .

52. for hydrogen of deuterium gm

ratio is ( )( )

/12 /

D

H

q mq m

=

( )/

mv vrBq B q m

= =

( )

( ) ( )/ 2 1 1/ 1 2

H H D

D DH

B q mr Vl B q m V

= × = × =

53. Conceptual

54. 2xf =

1 1 1v u f− =

( )

1 1 1v y f f− =− −

2

1 1 1 yv f y f fy f= + =

− −

2

yf fV

y−

=

Distance of image from focus = f v−

2 2 2

4fy f f xf

y y y−

= − = = .


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55. First position 3 3V V xx= − ⇒ =

−,

1 1 1 8cm3 6

xx x− = ⇒ =

Second position 2 2V V yy= − ⇒ =

−

1 1 1 9cm2 6

yy y− = ⇒ =−

1cmy x⇒ − =

So number of rotations = 10

56. from the diagram

From triangle BDQ 90 90 180r r gθ + + − = ⇒ =

From triangle PRQ 2 90 90 180 2i r i+ + − = ⇒ = θ

From PQDC ( ) 090 2 9 360i r− + + ∆ + θ =

3 1802θ

θ − = , 0 05 180 72 362

Aθ= ⇒θ = ⇒∠ = .

57. Sound wave produced by the string will be longitudinal in nature

58. Conceptual

59. 2 2.4xV

= …(1)

( )2

4.4d xV−

= …(2)

( ) ( )3.4 3.4 340 1.156km 1.16kmd dv= ⇒ = = ≈

60. ( )20 2 20 2 20 20 60100 200C x X C Z−

= ⇒ = + = + =


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MATHEMATICS

61. 2 5 3 0x x− + = 2 3 5α + = α

2 3 52x + = 2 3 5β + = β

( ) ( )6 8 86 83 3a a 6+ = α −β + α −β

( ) ( )6 2 6 23 3= α + α −β +β ( )7 75= α −β

6 8

7

3 5a aa+

=

62. Length of side of ABCD = a

Radius of the outer circle = a

the perimeter of the circle 2the perimeter of ABCD 4 2

aaπ π

= =

63.

2FG =

~ABE EFG∆

25

FGAB

=

~HEG GCB∆

2 2

3 3HE x xBC

= = ⇒ =

Area of triangle = 1 255 52 2× × =

64. ( )1________ 1x yz+ =

( )2 _______ 2y xz+ =

( )2 ________ 3z xy+ =

Equation (1)-Equation(2)


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( ) ( ) 0x y z y x− + − =

( ) ( )1 0x y z− − =

1,z x y= =

:1Case For x y= equations are convert into

( ) ( )2 1 2 _________ 4y yz y z+ = ⇒ + =

( )2 22 2 _________ 5z y z y+ = ⇒ = −

Put z in equation (4)

( )23 2y y− = 23 2y y− =

33 2y y− =

3 3 2 0y y− + =

( ) ( )21 2 0y y− + =

1 2So y or y= = −

: 2 1Case For z = equations are convert into

( )2 2x y x y+ = ⇒ = −

1 2xy + =

( )2 1so y y− =

22 1y y− =

2 2 1 0y y− + =

( )21 0y − =

1 1So y and x= =

So 2 solutions ( ) ( )1,1,1 2, 2, 2and − − −

65. − > ⇒ < − >2 16 0 4 & 4x x x

25 8 16x x+ = −

− − =2 5 24 0x x

( )( )8 3 0x x− + =

8 8x x= ⇒ = ±

Product = 64−


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66. 2008=2008+8

4 32000 2 5= ×

Number of factors of 2000 5 4 20= × =

In which 1,2,4,5,8 are also include but N>8

So, total 20 5 15− = values are possible

67. 2 25775 3 5 7 11 5 231 15 385 175 33= × × × = × = × = ×

455 is not a factor of 5475

68. 1 1 1a b cb c a

+ = + = +

( )1 1 .............. 1b ca bc b bc

−− = − =

( )1 1 .......... 2c ab ca c ac

−− = − =

( )1 1 ........ 3a bc ab a ab

−− = − =

Multiply these 3 relations

( )( )( ) ( )( )( )2 2 2

b c c a a ba b b c c a

a b c− − −

− − − =

2 2 2 1 1a b c abc= ⇒ = ±

Examples:

11, , 2 12

a b c abc= = − = − ⇒ =

11, , 2 12

a b c abc= − = = ⇒ = −

69. Lets the ages of children’s 9,8,7,6,4,3,2,1(as 9 is oldest age)

L.C.M of ages = 504

We need to calculate 4 digit number which is multiple of 504 and have each

digit twice

504 11 5544.× =

70. Let number is N

9 6 ,N x x y I += + ∈

21 12N y= +

9 6 21 12x y+ = +

3 7 12x y− =


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x 3 10 17 …….

y 1 4 7 …….

So number are 9 3 6, 9 10 6, 9 17 6.......× + × + × +

33,96,159....... nT

1111nT <

( )1 1111a n d+ − <

( )33 1 63 1111n+ − <

18.11.......n <

18n =

71. Given equation identify, so each coefficient zero

2 2 25 6 0, 3 2 0 & 0α − α + = α − α + = α − =

2∴α =

72. 1 1 1

x a x b c+ =

+ +

( )2

2 1x a bcx a b x ab

+ +=

+ + +

( ) ( )22x a b c x a b x ab+ + = + + +

( )2 2 0x a b c x ab bc ca+ + − + − − =

Sum of the roots is zero

2a b c+ =

Product = ( )ab c a b= +

( )2

2a b

ab+

= −

( )2 212

a b= − +

73. ( )1 1 3x n= + −

Sum = ( )2 1 3 9252n n + − =

[ ]3 1 25 74n n − = ×

25, 73n x= =


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74. The probability of getting a sum either 7 or 11 is 29

( ) 2 719 9

P E = − =

75.

tan H hx−

α =

tan H hx+

β =

tantan

H hH h

α −=

β +

( ) ( )tan tan

tan tanH h H hH h H h− − +α + β

=α + β − + +

( )tan tantan tan

h Hβ − α

=α + β

76.

; ; ;OA a OB b OC c OD d= = = =

4; 8; 32; 16ab bc cd da= = = =

( ) 12b a c+ =

( ) 48d a c+ =

( )( ) 60a c b d+ + =


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77. 3sec tan2

θ + θ =

2sec tan3

θ − θ =

1 2 3 13sec2 3 2 12 θ = + =

5sin

13θ =

78. 2− not possible

79. ( )( )51 2 3 2x p a x x Ax B= − + + +

Put 1x =

1 O A B= + +

Put 2x =

512 2A B= +

By solving 51 512 1& 2 2A B= − = −

80. The length of the side of triangle = 2 2 3+

Area = ( )23 2 2 3 4 3 64

+ = +

Documents

Class: NSEJS NSEJS FINAL EXAM KEY Date: 17-11-19 Time ... · Date: 17-11-19 . Time: 2hrs : QP CODE.54 . Max. Marks: 240 : CODE.54 NSEJS INITIAL KEY 17-11-19 Q.No. 1 2 3 4 5 6 7 8