NSEJS-Solution-2015

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HINTS & SOLUTIONS

NATIONAL STANDARD EXAMINATION IN JUNIOR SCIENCE

NSEJS_STAGE-I (2015-16)PAPER CODE : JS-532

1. Let a, a + d ,...... Given n = 10 a + a + d + a + 2d = 32 3a + 3d = 321a + d = 107 ...... (i)T8 + T9 + T10 = 405 3a + 24d = 405a + 8d = 135 ....(ii)Solving (i) and (ii) : d = 4 and a = 103

S10 = 2

10 [2(103) + (9)4] = 5 [242] = 1210

3. Floating conditionB = mgdwater

Vimmersed g = dwoodgVimmersed = 0.76 Voutside volume of wood = 0.24 VTo just immerse in water volume displaced bysteel ball should be 0.24 VSo,Msteel g = .24 Vgmsteel = .24 (a3)= .24 × 43 15.36 gram.

Ans (c)

5. a 5 b a5 + 4b must be divisible by 13If b = 0 a = 6b = 1 a = 3b = 2 No value of ab = 3 No value of ab = 4 a = 7b = 5 a = 4b = 6 a = 1b = 7 No value of ab = 8 a = 8b = 9 a = 5so, only 7 such numbers are possible.

7. T = 2 g

, f same

m v = 2m v1 v1 = 2v

vmax = aw A v

A1 = 2A

Ans (b)

8. [H+] = 10–8 M (from acid)

[H+] = 10–7 M (from water, as the solution is very

dilute)

Total [H+] = 10–8 + 10–7 M

= 11 × 10–8 M

pH = –log10 [H+]

= –[log1011 + log10 10–8 ]

= –[1.02 – 8]

= – [–6.98] = + 6.98

Thus, pH of the solution willbe greater than 6

but less than 7

9.A B

CD M

N

x

y

O

28 14

96

50

50


AB || CD

x = 22 4850 = 14

y = 22 1450 = 48

ABCD is a trapezium

Area of ABCD

= 21

(96 + 28) (x + y)

= 21

(127) 62

= 3844 cm2

11. Let a, a = d, a + 2d, ....... be the AP

a + a + d + a + 2d + a + 3d = 56 4a + 6a =56

2a + 3d = 28 ....... (i)

a + (n – 4)d + a + (n – 3)d + a + (n – 2)d + a

+ (n –1)d = 112

4a + (4n – 10)d = 112 2a + (2n – 5)d = 56 ..(ii)

a = 11 from (i) d = 2

from (ii) 2(11) + (2n – 5) 2 = 56

11 + 2n – 5 = 28 2n = 28 – 11 + 5

2n = 22 n = 11

13. CaCO3 + HCl CaCl2 + H2O + CO2

(A) gas

CO2 + Ca(OH)2 CaCO3 + H2O

(Solution B) (A)

HCl + KMnO4 MnO2 + KCl + Cl2

(C)

Cl2 + Ca(OH)2 CaOCl2 + H2O

(C) (B) (D)

Therefore A,B,C,D are CaCO3, Ca(OH)2, Cl2,

CaOCl2 respectively

16. P & V are same

So, n1T1 = n2T2

M5

× 300 = 225.0

× 290

M = 41.37

17. N = abc = 100a+ 10b + c

N’= cba = 100 c + 10b + a

N – N’ = 99a – 99 c

= 99 (a–c)

so, all such numbers are divisible by 99.

So GCD of all such numbers is 99.

19. = 11 10–6 /°C

L = L0 t

0LL

= t = 11 10–6 70

0LL

% = 11 10–6 70 100 = 77 10–3 =

0.077 %

Ans (a)

21. 792 = 23 × 32 ×11

5901 AB04 is divisible by 8,9,11

if it iss divisible by 9. then sum of digit

5 + 9 + 0 + 1 + A + B + 0 + 4 = multiple of 9

19 + A + B = 27, 36

A + B = 8,17

But according to option only 8 in given so A + B

= 8

23. 5A3A

6A

A

3A1.5A I=4amp

0.5A

Ans (c)


24. I = rR

E

IR + Ir = E

V = E – Ir

I = )120(3 =

71

amp

V = 3 – 71 1 =

720

= 2.857 V..

or

I = )120(3 =

71

amp

V = IR = 20 71

= 7

20 = 2.857 volt

Ans. (a)

25.

By kirchhof’s law

By KVL

–6 –2I – I + 9 = 0

I = 1 amp

Vp – 6 – 2I – VQ = 0

Vp – VQ = 6 + 2 1 = 8 volt

Ans. (a)

26. Equivalent voltage = 11

1619

=

215

volt

I = .rRV.Eq

eq = 2110

215

/=

221215

//

= 75

amp.

P = I2 R = 4925

× 10 = 49250

= 5.12 W

Ans (b)

27. Acidic oxides = CO2, SO2, P2O5, SO3

Basic oxides = MgO, CaO

Neutral oxides = N2O, NO, CO

Amphoteric = H2O, Al2O3, ZnO, PbO

28. Let the smallest Natural number =x

Let Quotient =

Remainder = 21

ATQ Dividend = Divisor × Quotient + Remain-

der

x ×15 = 63 ×Q + 21

x = 152163 Q

x = 5721 Q

if Q = 3

x = 57321

= 5763

x = 570

= 14

31. (I) Lift is moving with constant speed in

upward direction so, T– mg = 0

or T = mg constant.

(correct)

(II) Kinetic energy = 21

mv2 = constant

(correct)


(III) u = – r

GMM(not constant) (wrong)

(IV) Velocity is constant so acceleration of lift

is zero

(V) P.E. is changing and K.E. is constant so

mechanical energy will not be constant

(wrong)

Ans (d)

32. The reactivity order of the given metals is

Mg > Zn > Fe > Pb

33. arABC = ½ AB × CF .....(1)

arABC = ½ BC × AD .....(2)

arABC = ½ AC × BE .....(3)

(1) × (3) (arABC)2 = ¼ AB×CF×AC×BE

¼(AB×AC) (CF×BE)

=¼(409.6) (202.5) = 20736

ar ABC = 144

(1)×(2)×(3) we got

(arABC)3 = 81

×AB×CF×BC×AD×AC×BE

(144)3 = 81

×(AB×AC) × (CF×BE) × (AD×BC)

=2985984 × 81

×(409.6)(202.5)(AD×BC)

288 = AD × BC

35. Separation is minimum so speed of each car

will be same that will be minimum, so kinetic

energy will be minimum.

Ans. (d)

36. 7

4KMnO

H Mn+2

Transfer of 5electron

6

722 OCrK

H Cr+3

Transfer of total 6 electron

Ratio is 5 : 6

37. m = 338 – 288

m = 13 2 –12 2

m = 2

v = fuuf

= 6126

126)12)(6(

= 12 cm

39. f = – 12 , u = –6

v = f–uuf

=

126–12–6–

= 6126

= 12cm

Ans. (C)

40. methyl cyclobutane is

Molecular fromula is C5H10

41. 4 digit number = abcd= 1000 a + 100b + 10c + d= (986a) + 14a + (87b) + 13b + 10c + dno is divisible by 2914a + 13b + 10c + d = 29 n is the remaining

numbera + b + c + d = 29(13a + 12b + 9c) + (a + b + c + d) = 29n13a + 12b + 9c = 29m ...(i)9a + 9b + 9c + 9d = 29 9 ...(ii)equation (i) - equation (ii)4a + 3b – 9d = 29 (m – 9) the possible cases are4a + 3b – 9d = – 294a + 29 = 9d – 3b4a + 29 = 3 (3d – b)


a = 4, 7 only possiblea = 4 3d – b = 153d = b + 15b = 9, d = 8a = 74a + 3b – 9d = 04a = 9d – 3b4a = 3 {3d – b}when a = 9 no soluion for this12 = 3d – bd = 6, b = 6no. = 9686d = 7, b = 9no. 9947b = 6, 3 no solution4a + 3b – 9d = 293{b – 3d} = 29 – 4aa = 2, 5, 8 only possibleonly five solutions:759849887859a = 6, 3sol. not possible968657 = 3(3d – b)99473d – b = 193d = 19 + bb = 8, d = 9no. = 7859b = 5, d = 8no. 7598

43. 2Tcos = mg T = cos2

mg

cos T

Ans. (c)

47. According to definition of conservative force.

Ans. (a)

45. No. of triangle = 14C2 = 21314

= 91

48. Na2CO3 .H2O + HCl ml10

23 OHNaHCONaCl

‘x’ g 100 mL 0.25M 0.05 M NaOH

20 mLMeq. of NaOH = Meq. of final solution0.05 × 20 = 10 × N

Nfinal = 102005.0

= 0.1

N1V1 – N1V1 = NfVf

0.25 × 100 – 621000x

= 0.1 × 100

25 – 31x500

= 10

15 = 31x500

x = 5003115

= 10093

g = 930 mg

49. 8888888 * 8888888divisible by 11 64 – (48 + *) = 0 or divisible by 1164 – 48 – * = 0 or divisible by 1164 – * = 0 or divisible by 11 * = 5

51. In figure , this is the position of TIR, So angle ofincidence must be greater than critical angle.

Ans. (c)

53. (41)16 – (14)16

(412)8 – (142)8

an – bn is divisible by a – b

(412)8 – (142)8

is divisible by (41)2 – (14)2 = 1485

52. Aspirin C9H8O4 250 mg tablets paracetamol

C8H9NO2 500 mg tablets

Given (1) + 0.5% error in each tablet

‘x’ molecules extra in aspirin ‘y’

100 g 99.5 g

250 × 10–3 100

g105.99250 3–

= 248.75 × 10–3 extra = 1.25 g × 10–3

180 g NA


1.25g × 10–3 180

g1025.1N 3–A

Extra molecules = x

100 g 99.5 g

500 × 10–3g 100

5.9910500 3–

= 497.5 × 10–3 g

extra = 2.5 × 10–3 g

151 g NA

2.5 × 10–3 g 151

N105.2 A3–

extra molecules = y

yx

= 180

1025.1N 3–A

× NA105.2

1513–

= 5.218025.1151

= 2180151 = 360

151

360 x = 151 y

xy

y = 2.4 x

55. Acceleration due to gravity is uniform over the

body, then, centre of gravity and centre of mass

will coincide.

Ans. (c)

56. 4Na + O2 – 2Na2O

92 g 124 g

student took mass of Na as 11

44g 76 g

To prepare 1.24 g 76g Na2O 44g Na req.

1.24 g 764424.1

g Na

Na2O = 0.7179 g Na

So he took 0.7179 g Na

But in real 92 g Na give 124 g Na2O

0.7179g Na will give 921247179.0

g Na2O

= 0.9676 g Na2O

error in mass of Na2O = 1.24 – 0.9676

= 0.2724 g

% error = 24.1

1002724.0 = 21.97 %

~– 22%

57.

1431000000 3

1431999999 3

14313

= 1 Ans.

59. In equilibrium, under the effect of several forces

the sum of torques about any point must alway

be equal to zero.

Ans. (c)

61.43

+ 283

+ 703

+ 1303

+............+ 97003

= ?

41

3

+ 743 + 107

3 + 1310

3 +........+ 10097

3

= 411–4

+ 74

4–7

+ 1077–10

+ 1310

10–13

+.....+ 1009797–100

= 1 – 41

+41

– 71

+71

– 101

+ 101

– 131

+.......+ 971

– 1001

On simplification.

= 1 – 1001

= 10099

=0.99


62. g

If v constant

r

v)a(2

63. According to definition of newton 3rd law action

and reaction always act on different bodies and

there will be no time gap between action and

reaction

Ans (d)

64. Sodium peroxide Na2O2

Peroxide ion –22O

Structure : O – O

65. 102, 136, 170.....................986

a =102

d =34

an = 986

a + (n–1). d = 986

102 + (n – 1) × d = 986

(n–1) × 34 = 986 – 102

(n–1) = 34884

n –1 = 26

n = 27

sum = 2n

[a + an]

=2

27 [102 + 986]

= 14688

67. When force is always perpendicular to initialdirection of motion then path of projectile will

be parabolic.

Ans. (a)

68. N = 1s2 2s2 2p3

O = 1s2 2s2 2p4

C = 1s2 2s2 2p2

F = 1s2 2s2 2p5

The values of IEs show a sudden jump in 5th IE.

Configuration of C shows after losing 4electron

it will aquire configuration of He and thus renoval

of 5 electron will require much larger energy

69.

D C

BA

O

20

20

x

x

30

30

2020

By Appolonius theorem,

in DCB,

(30)2 + (20)2 = 2(202 + x2)

900 + 400 = 2 [400 + x2]

2

1300 = 400 + x2

650 = 400 + x2

x2 = 250

x = 5 10

BD = 10 10

71. Since |a| = rv 2

a v2 , it is the equation of parabola

Ans. (b)

73. 72015 = 7. (72014) = 7(72)1007

= 7.(50 – 1)1007

Remainder when divided by 25

= 7.(–1)1007

= – 7

= – 7 + 25

= 18


75. Medium y is denser, so speed will decrease,

fequency depends on source sowill decrease

by relation V = n

Ans. (c)

76. 6C(s)+ 3H2(g) C6H6 (l)

Now combustion of C(s)

6C(s)+ 6O2 6CO2 ; H1 = 6x ....(i)

3H2 + 23

O2 3H2O (l) ; H2 = 34 ....(ii)

C6H6 + 215

O2 3H2O ; 6CO2 ; H =Z

Reverse the reaction.

6CO2+ 3H2O C6H6 + 215

O2

H5= – Z......(iii)

After adding equation (i) , (ii) and (iii)

6C(s) + 3H2(g) C6H6 (l) : H

H = H , + H2 + H3

= 6x + 3y – z

77. A (p + q + r) ; B (q, r + p) ; C(r, p + q)

Area of (ABC) = |21

[p(r + p – p – q) + q(p + q

– q – r) + r(q + r – r – p)]|

= |21

[pr + pq – pq – qr + qr – pr]|

= 0

79. x = 6t2

dtdx

= V = 12t

at t = 0

initial velocity u = 0

2

2

dTxd

= a = 12, t = 0.5 sec.

V = u + at

V = 0 + (12) (0.5) = 6 m/s

Ans. (d)

80. In evaporation , water takes heat, overcomes

the intermolecular force of attraction and con-

verts into vapour phase.

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NSEJS-Solution-2015