answer key & solutions for national standard examination

ANSWER KEY & SOLUTIONS. FOR. NATIONAL STANDARD EXAMINATION ( JUNIOR SCIENCE) – 2013. PAPER CODE - 516. ANSWERS. 1. B. 2. D. 3. A. 4. D . 5. D. 6. B. 7...

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ANSWER KEY & SOLUTIONS FOR NATIONAL STANDARD EXAMINATION (JUNIOR SCIENCE) – 2013 PAPER CODE - 516 ANSWERS 1.

B

2.

D

3.

A

4.

D

5.

D

6.

B

7.

A

8.

C

9.

C

10.

Option not matching

11.

A

12.

A

13.

None of the answer is correct (The answer should be 56 J)

14.

C

18.

There can be 4 such points

19.

D

20.

23.

C

27.

15.

C

16.

A

17.

D

C

21.

B

22.

D

24.

C

25.

C

26.

C

B

28.

A

29.

C

30.

A

31.

C

32.

B

33.

B

34.

A

35.

D

36.

A

37.

D

38.

D

39.

A

40.

C

41.

B

42.

B

43.

C

44.

B

45.

A

46.

D

47.

C

48.

D

49.

D

50.

B

51.

C

52.

A

53.

C

54.

B

55.

C

56.

D

57.

C

58.

A

59.

C

60.

A

61.

C

62.

D

63.

B

64.

B

65.

B

66.

B

67.

B

68.

A

69.

A

70.

D

71.

B

72.

B

73.

D

74.

B

75.

D

76.

C

77.

C

78.

A

79.

B

80.

D

SOLUTIONS 1. Sol.

2. 2.

B (1) B1 will be inward, B2 will be outward, |B1| = |B2| (2) Both B1 and B2 will be inward, |B1| < |B2| (3) Both B1 and B2 will be inward, |B1| = |B2| (4) B1 will be inward, will be outward, |B1| < |B2| D l2  h2  r 2 l12  h2  4r 2

l22  h2  16r 2 rl  r

h

 r2

2



rl1  2r h2  4r 2 rl2  4r h2  16r 2 4 r h2  16r 2



 kr h2  r 2



16h  256r  k 2h2  k 2r 2 2

2

h

2r h2  4r 2  3r



h2  r 2

2

2

l2



4h  16r  9h  9r 7r 2  5h2 7 r 2  h2 5 7  7  16  r 2  256r 2  k 2  r 2  r 2  5  5  2 k  116 k is nearest to 11 2

l

2

r

3. Sol.

A 2 mole of O2 = 64 g .2 mole of O2 = 6.4 g

4. Sol.

D Wuchereria is the member of phylum nematoda (Aschelminthes)

5.

D

Sol.

I=

6.

B

6.

1

Q t Q = n (1.6 x 10–19) N = Number of charged particle.

1 1 1 1 1 1 1      .........   2 3 4 5 6 2012 2013 1 1 1 1 1 1 1     1       .........    2 3 4 5 6 2012 2013  

2r

1  1 1 1 1 2      .......  2012  2 4 6 8 1 1 1 1 1   1 1 1 1     1     ......      1  2  3  4  .........  1006  2 3 4 2012 2013     1 1 1 1     .........  1007 1008 1009 2013 7. Sol.

A

NH3

N H

H

H

SO32

S O

O

O

8. Sol.

C Because plastic is a non biodegradable agent.

9. Sol.

C    By conservation of linear momentum m1u1  m2 u2   m1  m2  v The sound energy produced  the loss in energy 1 1 1 m1u12  m2u22   m1  m2  v 2 = loss in energy 2 2 2

11. Sol.

A Alpha radiation

12. Sol.

A DNA and proteins are present in chromosomes.

13. Sol.

None of the answer is correct (The answer should be 56 J) K = W Total = 14 × 4 = 56 J Kf – Ki = 56 ( Ki = 0) C

14. 14.

b2  26  a2 x2 b2 x 2  2  a2  2b x 2 b x3   a2 x  26x x b a  1  .b  x a2  26 x a  1   a  x  .b  x a2  2b x2 











a  1  ab  bx  x a2  2b







x b  a  2b  ab  1  a

x

2

ab  a  1 a2  b

15. Sol.

C Eocell  EC  E A 0.75 = 0.34 - EA EA = -0.410 V

16. Sol.

A Deodar and Pinus are gymnosperm plants.

18. 18.

19. Sol.

OP = PA OA = OP OA = AP  A can be (9, 3),  3 5 3 5   3 5 3 5  5 5 , ,   ,   &  ,  6   2 6  2 6  2

A (x, x/3)

P (5, 0)

D 1 -    A    C 2

20. Sol.

C The females egg carries only one x chromosome (22 + x).

21. Sol.

B Like charges always repel each other while a charge body may attract an uncharged body due to polarisation. D

22. 22.

13a + 35b = 1000  a = 50, b = 10 m =a + b = 50 + 10 = 60 13x + 35y = 1000 y = 23, x = 15 n = x + y = 38 m + n = 60 + 38 = 98

24. Sol.

C It is the controlling centre of cell.

25. Sol.

C Distance = velocity x time 350 x 6 = 2100 m

26.

C

26.

r   A  P 1    100 

n

r   96,800 = P  1    100 

2

r   97,240  P  1  200  

(1) 4

(2) 4

r   1   2   97,240   200  2 1 96,800  r  1   100   

27. Sol.

B I and III only Poly alkene are highly flammable. This do not make its disposal difficult.

28. Sol.

A It is because of the reflex arc.

29. Sol.

C The change of magnetic flux linked with the coils.

30.

A 1 1 ac sinB  c 2 sin A 2 2 c sinB   a sin A AB sinB   BC sin 180  2B 

30.

AB sinB  BC sin 2B AB 1 1    sec B BC 2cosB 2 or sec B  1 

31. Sol.

Sec B 1

C

 4PCl3  g  P  g  6 Cl2  g   4 1 0 0 1- 4x x 6x 6 x   6x  KC  4 1  4x 

32. Sol.

B Placenta is responsible for the physiological activities like nutrition, respiration and excretion in foetus’s during pregnancy.

33.

B

Sol.

Ek = hf   h = plank’s constant

34.

A

r12 25 2  r22  r12 24 2 24r12  25r22  25r12

34.

49r12  25r22 r2 7 49 r2 2   2 r1 5 25 r1 r r 7 Ratio = 2  2  r1 r1 5

r1 r2

35. Sol.

D Boiling removes temporary hardness of water.

36. Sol.

A In angiosperm plants transportation of food takes place with the help of companion cells and sieve tubes.

37. Sol.

D Q = mct 96 = m (0.8) x 6

38.

D

38.

Putting x= a, we have x3 – 3ax2 + 3ax – a = 0 a3 – 3a3 + 3a2 – a = 0 a (–2a2 + 3a –1) = 0 a (2a2 – 3a + 1) = 0 a (a – 1) (2a – 1) = 0 1 a = 0, 1, 2  The 3 real roots are 0, 1,

39. Sol.

A C4H8 Butene H

H H

H

H-C=C-C-C-H H

CH3 CH3 - C = CH2

H

1 2

H

H

H

H

H-C-C=C-C-H H

H CH2 - CH 2 CH2

CH2

CH2 CH - CH3 CH2 40. Sol.

C Iodine is a core element for the formation of thyroxine hormone. It is abundantly available in the coastal area.

41. Sol.

B The gravity always acts downward

42.

B

42.

x2 + y2 + z2 = 1000 (x + y + z)2 = (50)2 x2 + y2 + z2 + 2 (xy + yz + zx) = 2500 2 (xy + yz + zx) = 2500 – 100 1500 xy + yz + zx  2 xy + yz + zx = 750  cost of y pencils + z pens + x note books = Rs. 750

43. Sol.

C Ag is placed below hydrogen in reactivity series.

44. Sol.

B It is adaptation according to their habitat.

45. Sol.

A The pressure at the same horizontal level remains same ha × (1.6) = hb (B) 26.6  1.6 B = 50 D

46. 46.

a2  b2 6 ab a b  6 b a a Let  x b 1 x 6 x

x 2  1  6x x 2  6x  1  0 f  5   25  30  1  4

f  6   36  36  1  1

f  x   0 if f (5) < f(x) < f(6)

5
C

Cl

Cl

Cl

CH3 - CH - Cl & CH - CH Position isomerism 48. Sol.

D Biological diversity is the variety and variability among living organisms and the ecological complexes in which they occur.

49. Sol.

D The acceleration is defined as a rate of change of velocity.

50.

B

50.

At 4’o clock, angle will be 120o We know that, 11   m  30h 2 11 o 120  m  120  h  4  2 11 240o  m 2 480 7 m  43 min. 11 11 = 43m 38s , time will be 4n 43m 38s

51.

C

Sol.

PV 

W RT Molar Mass

50.01 103 .5755  0.255   .0821 288 101325 M.M M.M  108



52. Sol.

A As per phylogenetic tree humans evolved from chimpanzees.

53. Sol.

C  Brightness  heat produced per unit time. 1 In parallel heat produced per unit time  R In series, heat produced per unit time  R.

54.

B

54.

4 digit numbers are 1000 to 9999. Least 4 digit multiple of 11 is 1001 and highest 4 digit multiple of 11 is 9999. 11 x 91 = 1001 11 x 909 = 9999 Among we will get 3 in the unit’s place only with products of 11 with 93, 103, 113, ……….,903. So there are 82 such numbers. 80  T  89

55. Sol.

C Solution 1: Solution 2:

Solution 3:

M1 = .100 mol L V1 = 1.00 mL 100 mL M V .1  1 M2  1 1   .001 V2 100 M2 V2 = M3V3 .001  10 = M3  100 M3 = .0001 pH = -log [H+] pH = -log[10–4] = 4

------ ([H+] = M3)

56. Sol.

D Antibiotics are provided specially for infectious diseases.

57.

C

Sol.

g 

g

ab

1 a

58.

2

 h  1    Re  Here, h = Re.

; h = height form the surface of earth.

 

1  1  b    a  b  1 1 = a  b     a b  x = a 

b>a Let b – a = y  a – b = –y b  a    y  y  x = (a – b) + ab ab

y  y since a < 1 ab So, x = + ve

59. Sol.

C Enthalpy of formation of an element in its standard state is zero.

60. Sol.

A Slope must first increase and then decrease. Also initial and final slope must be zero

61.

C

61.

m has 7 possibilities Possibilities are (7, 7, 1), (6, 6, 3), (5, 7, 3) (5, 6, 4), (4, 7, 4), (5, 5, 5) (7, 6, 2) n has 5 possibilities possibilities are : (3, 6, 7), (4, 5, 7), (6, 6, 4) (7, 7, 2), (5, 5, 6) m–n =7–5=2

62. Sol.

D Mole mass of C27H28Br2O5S = 624 100 No. of moles = = .160 mole 624 nsolute Molality = mass of solvant kg  Molality =

.160 = 0.201 mol kg–1 .7979

63. Sol.

B Refractive index  > 1 for light and  < 1 for sound.

64.

B

64.

xy 2  a3 ,yz 2  b3 ,zx 2  c 3 x 3 y 3 z3  a3b3c 3  xyz = abc xyz  yz 2  b3  abc

xy 2  z3  b3  abc z3 

b3  abc b 4  c  a3 a2

65.

B

Sol.

no. of moles =

66.

B

67.

B

3.4  .009 mole 342 No. of H-atom = .009  6.023 1023  22 =1.3  1023

68. A Sol. for question nos. 66 to 68: due to the water pollution which leads problem of Eutrophication 69.

A

Sol.

From the path of ray Deviation = 90°

70.

D

70.

5 41

2

45°

 x2

25  41  x 2 1025 = x2 x > 32 32 < x < 33 71. Sol.

B Cl

NH3 Pt

Cl

PtCl2(NH3)2  X – 2 + (0) =0 X = +2 72.

B

73.

D

NH3

O.No = +2 C.No = 4

74. B Sol. for question nos. 72 – 74: Pea is a winter season crop 75. D 1 Sol. K.E. = mv 2 2 1 = m (0)2  2(gsin )s  2 76. C 76.

There can be 6 such circles.

77. Sol.

C O is more electronegative element.

78. Sol. 79.

A Hydrostatic force is perpendicular to surface. B

79.

3x + 3y – 1 = 4x2 + y – 5 = 4x + 2y 3x + 3y – 1 = 4x + 2y x=y–1 y=x+1 (1) Again, 4x2 + y – 5 = 4x + 2y 4x2 – 4x – y – 5 = 0 4x2 – 4x – (x + 1) – 5 = 0 4x2 – 4x – x – 1 – 5 = 0

45°

4x2- – 5x – 6 = 0 4x2 – 8x + 3x – 6 = 0 (x – 2) (4x + 3) = 0 3 x = 2, x = 4 3 If x = Length of side of becomes negative which is not possible 4 So, x = 2 y=3  Side = 4x + 2y = 8 + 6 = 14 3 2  Area = 14   3  49 4 = 84 . 87  85 80.

D

Sol.

pOH = -log[OH–] = -log[2.5  10–2] = -log 2.5 + 2 log 10 = -.3118 + 2 pOH = 1.68 pH = .14 - 1.68 = 12.32