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NTSE STAGE-I-2017-18 PAGE-1

NATIONAL TALENT SEARCH EXAMINATION-2017-18, JHARKHAND SCHOLASTIC APTITUDE TEST (SAT)_HINTS & SOLUTIONS

1. Sol. (3) 2. Sol. (4)

3. Sol. 1 1 2R r r

1 22

1 2

r rRr r

1 1 21 2

2 1 2

R (r r ) (r r )R r r

2

1 2

1 2

(r r )r r

4. Sol. (3) e–

IB

B

= out of the paper

using fleming left hand Rule, Force experienced by electron will be, upwards. 5. Sol. (1) 6. Sol. X-ray is an electromagnetic wave.

7. Sol. (1)

Dispersive power will be gretest for Red light for a medium. 8. Sol. (1) 9. Sol. (3)

10. Sol. 2

ratedrated

VPR

rated220 220R

100

2

rated

(110)PR

= 25 W

11. Sol. (3) u

v = 80 cm

f = ?

1 1 1v u f

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SAT (SOLUTION) NTSE STAGE-I-2016-17

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100 1 180

Pf

100 1 2580

P .

And lens is concave lens

12. Sol. Range sin2

sine is maximum when angle is 90°.

2 9045

13. Sol. (1) 14. Sol. (3)

Pb(NO3)2 + 2HCl PbCl2 + 2 HNO3

White ppt 15. Sol. (4)

Isomers of C4H8 = 4

i.e (i) 1- Butene, (ii) 2- Butene, (iii) 2- methyl -1- propene (iv)Cyclo butane

16. Sol. (3)

C-C bond length= 1.54 AO

17. Sol. (4) Only Terminal Alkynes give red precipitate with ammonical CuCl (i.e. 1- Butyne)

18. Sol. (2)

Among given options strongest reducing agent is LiAlH4.

Propionic acid Reduction 1- Propanol

19. Sol. (3) Oxidation of secondary alcohols give Ketones in one step.

20. Sol. (2) R-O-R (ether) is a Lewis base.

21. Sol. (2)

2NaNO2 + Dil. H2SO4 Na2SO4 + + NO2 + NO + H2O

NO2 & NO gases will not turn lime water milky.

22. Sol. (2)




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23. Sol. (1) F.P. depression is the process in which adding is solute to a solvent decreases the f.p. of the

solvent when we add solid particles in a liquid than it hinders the inter particle attraction so that f.p.

decreases.

24. Sol. (4) Alpha particle is He nuclei.

25. Sol. (1) Blue Zone.

26. Sol. (1) Base

27. Ans. 4 28. Ans. 2 29. Ans. 2 30. Ans. 3 31. Ans. 2 32. Ans. 2 33. Ans. 2 34. Ans. 4 35. Ans. 4 36. Ans. 2




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37. Ans. 3 38. Ans. 4 39. Ans. 4 40. Ans. 3 41. (3) Sol. Let n = 9M + 7

3n – 1 = 3 (9m + 7) = 27m + 21

= 27m + 18 + 3

= 9 ( 3m + 2) + 3 27 m + 21 then Remainder is 3

= (27m + 18 ) + 3

42. (4) Sol. 3 23 1x x x

So 33 1

1a – d a a d a

1a d .a. a d

2 2 1a a d

2 21 1a d

21 1d

2d

So a + d = 1 2




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43. (2)

Sol. 3 1 12 1 1 2 1k k k

2 1 1k k

2k

3 1 2 1k k

3 3 2 1k k

K = 3 –1 K =2

44. (2) Sol. k , k : 1 k , k :1

bka

bk 1a

qkp

qk 1p

cka

rkp

2 cka

2b ck

a(k 1) a

2 rk

p

2q rk

p(k 1) p

2

2 2kb c

aa (k 1)

2

2 2k q r

pp (k 1)

2

2 2k c a

a(k 1) b

2

2 2k r p

p(k 1) q

2

2 2k ca

(k 1) b a

2 2

k pr(k 1) q

= 2

cab

2 2ca prb q

2222 2

12

22 22 2

pr4ca b 1 4b 1 qD b 4ac bD q 4pr pr prq 1 q 1 4

q q

2

2bq

45. (1) Sol.




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tan P2

+ tan Q b2 a

P Q Ctan tan2 2 a

P + Q = 2

P Q

2 4

P Qtan tan2 2 4

P Qtan tan2 2 1P Q1 tan tan

2 2

ba 1c1a

b c1a a

c b 1a a c b a c = a + b

46. (1)

Sol.

1 11

22 2

2 1 3 1325 212 1

2

n a n ds nns na n d

1 1

2 2

122 3 13

5 21122

na dnnna d

So 1 232

n

So n = 47

1 1

2 2

23 3 47 13 123 5 47 21 2

a da d




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PAGE-7

47. (2)

Sol.

6030

60 30vt c 2v

DB

A

200 3

ABD

200 3302

tan ºvt v

1 200 323 vt v

2 200 3 600vt v

ABC

200 360tan ºvt

Vt = 200 By eq (1) & (2)

2 600vt v

200 2 600v

2v = 400

V = 200 m/minut

Or 200 6060

meterVminut

12000 12

60V k / hr

48. (2)

Sol.

sin x y a bsin x y a b

Apply C & D

sin x y sin x ysin x y sin x y

a b a ba b a b

So

2 22 2

sin xcos y acos xsin y b

tan x coty = a/b

of tan x atan y b




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PAGE-8

49. (1) Sol. Total out comes = 62 = 36 Favoulable out comes = (4,5) (5,4) (5,5) (6,3) (3,6) (6,4) (4,6) (5,6) (6,5) (6,6)

So 10 536 18

P

50. (3)

Sol.

1 23 2 5 2m m2 5 6 5

13

3m1

2m 3

m1m2 = –1 BAC 90

equation of line passing though A(5, 2) is :

2x + y – 12 = 0

2(5) + 2 – 12

10 + 2 – 12 = 0 This is the only equation with satisfies (5, 2) 51. (4)

Sol.

2 2(x 4) (y 0) 25

2 2x y 8x 16 25 0 2 2x y 8x 9 0

52. (2) Sol. a, b, c, d, e, f, g First (a + b + c + d) 4 x 8 = 32 Last (d + e+ f + g) 4 x 16 = 64 32 + 64 = 96 96 – (7 x 10) = 26




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53. (4) Sol. r2 = 10 m r1 = 5 m

Area increase 2 22 1r r–

4 4

3.14 100 – 254

= 58.875 m2 54. (4)

Sol.

2 212 3 5l .

144 12 25l .

12 5l . cm

2r 2 r

22 7 22 7 712.5 27 2 7 2 2

11 x 12.5 + 77

= 214.5 cm2

55. (1)

Sol. Volume of hollow sphere 3 34 R – r3

4 22 64 – 83 7

4 22 456 22 83 7 3

Now making cone so 21 4r h 22 83 3

1 22 44 4 h 22 83 7 3

4 22 8 7h 14 cm22 1 16




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56. Sol. (4)

352.5

5352.5

2rl rh

22 2252 3 53 2 52 5 37 7

. .

Solving = 9735 m2

57. Sol. (2)

30P (OB = OP = radius)

So 120BOP º

and 90OPT º (tangent)

Now Q + 120OPT º (Exterior angle property)

30º




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58. (3) Sol. Here OP2 – PR2 = OR2

So OR = 3 cm

OPR OTP (By AA criteria)

OP OR 5 3OT OP 5 VT 5

10VT3

So TP2 = TV x TM

2 10 40TP3 3

20TP cm3

59. Sol. (2)

A

B

Q

r

r

rPO Q

Let APO & 90OAP º

So Sin 12 2rr

30

So 60APB º

& PA = PB (tengent)

So its a Equailateral triangle




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60. Sol. (3)

x (90º –x)

(90º –x)x

DA B

C

So BDC CDA

BD DCCD DA

So 2CD BD AD




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PAGE-13

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