NSEJS-Solution-2012

HINTS & SOLUTIONS (YEAR-2012-13)_NSEJS (STAGE-I) ANSWER KEY Que s.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Ans

D

C

C

A

B

A

A

C

D

D

D

B

C

D

A

Que s.

21

22 23 24

25

26

27

28

29

30

31

32

33

34

35

36

37

Ans

Bonus

A

A

B

C

D

A

C

A

B

D

B

C

D

C

Que s.

41

42 43 44

45

46

47

48

49

50

51

52

53

54

55

56

57

Ans

B

A

D

C

B

D

C

A

A

D

A

D

B

C

Que s.

61

62 63 64

65

66

67

68

69

70

71

72

73

74

75

Ans

A

A

B

B

A

A

B

D

B

D

A

A

C

D

B

B

A

C

1.

Con. H2SO 4 C12H22O11     12C + H2O Charcoal (dehydration)

2.

 1 =  ,  2 =   1=  ,2=  m1 = m , m2 = m

R1 R2

3.

 1 1 A 2 A1  =   A 2 2 1 A2

2

2

R1  1  1    = R2   2  4

C 76 D

C

B 77 A

18 A 38

19 C

D

39

A ,B C 58 B 78 B

20

59 C

40 A 60 D

79

80

B ,C A

Given : x2 + xy + xz = 135 .......(1) y2 + yz + yx = 351 .......(2) z2 + zx + zy = 243 .......(3) On adding (1) (2) and (3) x2 + y2 + z2 + 2 (xy + yz + zx) = 729 ...... (4) x + y + z = 27 Eq (4) – Eq (1) y2 + z2 + 2yz + xy + zx = 594 (y + z)2 + x(y + z) = 594 (y + z) (x + y + z) = 594

594 27 y + z = 22 Similarly x + z = 14 and x + y = 18 Solving above equations x = 5, y = 13, z = 9 So x2 + y2 + z2 = 275 Ans. (c)

y+z=

A   . 2 . 2   A1  1 

density of both wire will be same then volume will be same then,

A

17

594 y + z = xyz

1  A 2  2  1 R1  . . = R2  2  A1  1  2

 1  R1   = R2  2 

16

4.

I = 1.2 I R1.2 R P (I)2 R  (I)2 R  100  P I2R

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

[(1.2)2 (1.2)  1]  100 1

13.

Area of ABC =

1 × 6 × 8 = 24 sq. unit 2

Area of AQR =

2 of area of ABR 6

= 72.8 % 5.

Let the number is in base x. 3x2 + 6x + 3 + x3 + 5x + 6 = x3 + 4x2 + 5x + 2 x3 + 3x2 + 11x + 9 = x3 + 4x2 + 5x + 2 x2 – 6x – 7 = 0 (x – 7) (x + 1) = 0 x = 7 or x = – 1 So, number is in base 7 Subtraction in base 7 : 654 – 456 = 165

Ans. (b)

–

6.

232 90 R

228 88 X

9.

 2 mole  K

10.

 1 mole  K M = m

 

– 2



228 90 Y

m = 2m M = m  = 2m  = 2M

1 4 of of area of ABC 3 10

=

1 2 16  × 24 = sq. unit. 3 5 5

area of CRP =

M.F. Lines will be double  2n

5 6 of of area of ABC 8 10

=

5 6 × × 24 = 9 sq. unit. 8 10

=

4 3 × × 24 6 8 = 6 sq. unit. area of PQR = area of ABC – (area of AQR + area of PRC + area of BQP) = 24 – (6 + 9 + 3.2) = 24 – 18. 2 = 5.8 sq. unit. Ans. (c)

30  36  30 96 = = 48 cm 2 2 48  ( 48  30)  ( 48  36)  ( 48  30 )

=

48  18  12  18 = 18 × 4 × 6 = 432 sq. cm. Circumradius =

14.

abc 4

B2D3 2B3+ + 3D2– Let solubility = s 2s 3s Ksp = [B3+]2[D2–]3 = [2s]2[3s]3 = 4s2 × 27s3 = 108s5

30  36  30 = 4  432 = 18.75 cm

(d) Direction of magnetic filed will be given by right hand thumb rule. According to this rule magetic field will be out of the plane of paper.

4 or area or APB 6

=

a = 30 cm, b = 36 cm, c = 30 cm S=

5 of area of BRC 8

=

area of BQP =

11.

Ans. 12.

=

16.

n= =

8 32

1 4

 1 mole = 6.023 × 1023 molecule

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 1/4 mole = 6.023 × 1023 ×

21.

1 4

1 mol 100

= 1.5 × 1023 molecules 17.

m=

c2

200  10 6  1.6  10 19 =

3  10 

8 2



1 (2 – a2 – b2) = cos (x + y) 2

19.

= 56 m 

24.

Heat required to convert 60 g ice at 0ºC into 60 g water at 0ºC = 60 × 80 = 4800 Heat given : 20 g water at 40º = 20 × 1 × 40 = 800 since heat requied is more then neat given so ice will not melt completely. So final temperatue = 0ºC

25.

Given

kg

sinx + siny = a ... (1) cosx – cosy = b ... (2) Adding the square of (1) & (2) sin2x + sin2 y + 2 sinx siny + cos2x + cos2y – 2 cosx cosy = a2 + b2 (sin2x + cos2 x) + (sin2y + cos2 y) + 2(sinx siny – cosx cos y) = a2 + b2 1 + 1 – 2 cos (x + y) = a2 + b2 2 – a2 – b2 = 2 cos (x + y)

50  0.1 50  0.1 22.4 mol CaCO3  ×  CO2 1000 1000 2

= 0.056 

m = 0.355 × 10–27 kg m = 3.55 × 10–28 kg 18.

50  0.1 mol 1000

 L.R. = HCl  2 mol HCl  22.4  CO2

Energy = 200 MeV = 200 × 106 × 1.6 × 10–19 Joule According to Einstein energy mass relation E = mc2 (m is reduced mass, c is speed of light) E

CaCO3 + 2HCl CaCl2 + H2O + CO2

1  xy = k (constant) x Now x increased by 25%

y 

Ans. (a) So,

Projectile motion so, parabolic path (C)

5 x y’ = k 4

5 x y’ = xy 4 4 y 5 So, % change to y is 80%.

y’ =

27.

20.

 AB + CD = BC + AD 10 + 30 = 2 AD  BC = AD AD = 20 cm In ADE AD2 = AE2 + DE2 (20)2 = 4r2 + (10)2 4r2 = 400 – 100 4r2 = 300 300 r2 = 4 r2 = 75  Area of circle = r2 = 75

2 2 2 2 2 + + + + 15 35 63 99 9999 =

1 1  1 1 1 1 1 1 1 1  3  5  5  7  7  9  9  11  ...... 99  101  

28.

Ans. (d)

Ans. (a)

=

1 1 – 3 101

=

101  3 98 = 3  101 303

H2SO4

Ans. (c)

2H+ + SO42–

2H2O 2H+ + 2OH– discharge potential of OH– is less So OH– discharge first At positive electrode 2OH–  H2O +

1 O + 2e– 2 2

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31.

Into boy’s hand since horizontal velocity will be same for both boy & ball.

32.

x

75 =

x

75 = a 1

75 x

y

45 =

z

 1 1 1     (x + y + z) > 9 x y z m>9 So, m = 10 and m = 12 is possible. Ans. (b)

15 = a

=a 35.

1 log 75 = log a x

= y

+

log3a

10 = 6

... (1)

45

=a

= 88.88  89

y = loga45 = 2 log3a + log5a

.... (2)

41.

1

=a

42.

Sixty sixth independence day was on 15 Aug. 2012. 15 Aug. 2012 is wednesday. Next wednesday on 15 Aug. will come in 2018 So after 6 years is the answer. Ans. (a)

43.

Sound waves moves from air to water and we know that speed of sound in water is greater than in air. So water is rarer medium for sound than air then sound wave is moving from denser to rare medium then angle of refraction is greater than angle of incidence.

log15 log a

3 5 z = log15 a = loga + loga Add (1) and (2)

x+y=

3 log5a

+

34.

... (3)

3 log3a

= 3[ log5a + log3a ] x + y = 3z 33.

20

t= g = = 2 sec. 10

1 log 15 = log a z

z=

time before v = 0 u

15 = a

15 z

m2 32

100 × 32 36

m2 =

1 log45 = loga y

z

m2 32

100 m = 2 36 32

45 = a 1 y

m2 m1

2 300 × = 600 0. 6

log 75 x= = log75 a log a 2 log5a

v1 t2 t1 × v 2 =

Ans. (b)

In all containers fluid is same so density is also same and height of liquid is same and pressure = dgh then, P1= P2= P3 = P4

44.

72012 (7 2 )1006 = 25 25 = (50 – 1)1006  Remainder = 1

Ans. (d)

x+y +z=1

1 1 1   =m x y z Applying A.M. & H.M.

45.

xyz 3 > 1 1 1 3   x y z

51.

X 2, 8, 18, 3 1s22s22p63s23p63d104s24p1 31

n

l

m

S

4

1

–1

+ – 1/2

K

 –

C N

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52.

w = k = = 53.

Power consumed by 10 is P1 = I2(10) Power consumed by parallel combination is

v =2+2×9  v = 20 m/s 1 (0.5)[(20)2 – 22] 2

P2 = I2 ×

Power is same, P1 = P2 100R 10 = 100  R 100 + R = 10 R

1 × 396 = 99J Ans. 4

Let the three consecutive natural no. are x, x + 1, x + 2 x  x  1 x  2 3 = x +1 Now : x (x +1)(x+2) = 124850054994 = 4997 × 4998 × 4999 x+ 1 = 4998  Ans. (d)

Their average =

54.

100R 100  R

100 = 11.11 1 9

R=

61.

x2 – px + q = 0 +=p  = q Given :  = m

  To find :

2N

m=

m 1  m2

55. By Ptolemy theorem AC × BD = AB × CD + BC × AD AC × 221 = 204 × 195 + 85 × 104 AC =

  2 1 2 

=

39780  8840 221

48620 221 = 220 Ans. (c)



=

 2  2

=

(  )2  2

=

p 2  2q

=



q

Ans. (a)

m r F1 62.

60° m

R

60.

60° r

m

F2 F1 =

Gm2 r2

= F2

Fnet = F12  F22  2F1F2 cos 60 10 I = 10  100R 100  R

=

2F2  2F2 

1 = 2

3F =

Gm2 3

r2

()

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63.

68.

This question is bonus in CODE 578 but It is correct in code 514 (Q.71) and Code 556 (Q. 78). a+b+c=1 a2 + b2 + c2 = 21 abc = 8 (1 – a) (1 – b) (1 – c) = (1 – a – b + ab) (1 – c) = 1 – a – b + ab – c + ac + bc – abc = 1 – (a + b + c) + (ab + bc + ac) – abc = 1 – 1 + ab + bc + ac – 8 = ab + bc + ac – 8 a+b+c=1 = a2 + b2 + c2 + 2(ab + bc + ca) = 1 = 21 + 2(ab + bc + ca) = 1 = 2(ab + bc + ca ) = – 20 ab + bc + ca = – 10 (1 – a) (1 – b) (1 – c) = – 10 – 8  = –18

Average O.S =

( 1)  (– 1) =0 2

or CaOCl2 2 – 2 + 2x = 0 2x = 0 x=0 74.

u = –x v = – (40 + x) m= 

v ( 40  x ) =  u x

Distance = 75 km Let speed of water current = x km/hr ATP 75 75 + = 16 3x 2x  x 75 75 + = 16 3x x

m=–3=–

x = 20 u = –20, v = – 60

1 [75 + 25] = 16 x

1 1 1   v u f

100 25 x= = 16 4 So, Speed of boat in still water = 2x

=2× 5 km/hr. 70.

1 1 1    20 60 f

25 = 12. 4

3  1 1  60 f f = – 15 cm

Ans. (a)

Eq. of line parallel to 4x + 3y = 5 is 4x + 3y = k.

75.

y x + =1 k/3 k/4 Given : x intercept = – 3

Cot2 (1 – 3sec + 2sec2) = 1  > 90º This is true when = 300º Cot2 300º(1–3sec300 + 2sec2300) 2

 1   1  3  ( 2)  2( 2)2 =    3



k =–3 4 k = – 12 Eq. is 4x + 3y = –12 or 4x + 3y + 12 = 0

=

1 [1– 6 + 8] 3

=

3 =1 3

Ans. (d) 71.

(NH4)2Cr2O7  Cr2O3 + N2 + 4H2O green

76.

O 73.

Ca

+1 Cl –1 Cl

40  x x



Ans. (c)

100 1.25 = – 80 cm

f=

1 1 1    u  80 u = 80 cm

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