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 4r h2 16r 2 4 r h2 16r 2
kr h2 r 2
16h 256r k 2h2 k 2r 2 2
2
h
2r h2 4r 2 3r
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 = mct 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
ab
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
