NSEJS-Solution-2015

NATIONAL STANDARD EXAMINATION IN JUNIOR SCIENCE

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

HINTS & SOLUTIONS 1.

Let a, a + d ,...... Given n = 10  a + a + d + a + 2d = 32  3a + 3d = 321 a + d = 107 ...... (i) T8 + T9 + T10 = 405  3a + 24d = 405 a + 8d = 135 ....(ii) Solving (i) and (ii) : d = 4 and a = 103 S10 =

3.

Ans 5.

7.

a5b a5 + 4b must be divisible by 13 If b = 0 a=6 b=1 a=3 b=2 No value of a b=3 No value of a b=4 a=7 b=5 a=4 b=6 a=1 b=7 No value of a b=8 a=8 b=9 a=5 so, only 7 such numbers are possible.

, f same

m v = 2m v 1 v1 =

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

Floating condition B = mg dwater Vimmersed g = dwoodg Vimmersed = 0.76 V outside volume of wood = 0.24 V To just immerse in water volume displaced by steel ball should be 0.24 V So, Msteel g = .24 Vg msteel = .24 (a3) = .24 × 43 15.36 gram. (c)

 T = 2 g

v 2

v max = aw  A  v A1 =

A 2

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

96 N

A

9.

x

B 50

O y D

50 C

M 28

14

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AB || CD 16. x=

50  48

y=

50 2  14 2 = 48

2

2

= 14

P & V are same So, n1T1 = n2T2 5 0.25 × 300 = × 290 M 2

 ABCD is a trapezium  M = 41.37  Area of ABCD

=

17.

N = abc = 100a+ 10b + c N’= cba = 100 c + 10b + a

1 (96 + 28)  (x + y) 2

N – N’ = 99a – 99 c = 99 (a–c)

=

so, all such numbers are divisible by 99.

1 (127) 62 2

So GCD of all such numbers is 99.

= 3844 cm2 11.

19.

 = 11  10–6 /°C L = L0 t

Let a, a = d, a + 2d, ....... be the AP a + a + d + a + 2d + a + 3d = 56  4a + 6a =56  2a + 3d = 28 .......

L –6 L0 = t = 11  10  70

(i)

a + (n – 4)d + a + (n – 3)d + a + (n – 2)d + a + (n –1)d = 112 L –6 –3 L0 % = 11  10  70  100 = 77  10 =

4a + (4n – 10)d = 112  2a + (2n – 5)d = 56 ..(ii)  a = 11

from (i) d = 2 0.077 %

from (ii) 2(11) + (2n – 5) 2 = 56

13.

 11 + 2n – 5 = 28

 2n = 28 – 11 + 5

 2n = 22

 n = 11

Ans

(a)

21.

792 = 23 × 32 ×11

CaCO3 + HCl  CaCl2 + H2O + CO2

5901 AB04 is divisible by 8,9,11

(A)

if it iss divisible by 9. then sum of digit

gas

5 + 9 + 0 + 1 + A + B + 0 + 4 = multiple of 9 CO2 + Ca(OH)2  CaCO3 + H2O (Solution B)

19 + A + B = 27, 36

(A)

A + B = 8,17 But according to option only 8 in given so A + B

HCl + KMnO4  MnO2 + KCl + Cl2

=8

(C) Cl2 + Ca(OH)2  CaOCl2 + H2O (C)

(B)

CaOCl2 respectively

3A

(D)

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

1.5A

6A

23.

3A

5A A

Ans

(c)

CAREER POINT, 128, Shakti Nagar, Kota-324009 (Raj.), Ph. 0744-2503892

0.5A

I=4amp

24.

I=

15 / 2 Eq. V 15 / 2 5 I = R  r . = 10  1 = = amp. eq 21 / 2 7 2

E R r

IR + Ir = E V = E – Ir

I=

P = I2 R =

3 1 = amp (20  1) 7

25 250 × 10 = = 5.12 W 49 49

Ans

(b)

27.

Acidic oxides = CO2, SO2, P2O5, SO3 Basic oxides = MgO, CaO

1 20 V=3– 1= = 2.857 V.. 7 7

Neutral oxides = N2O, NO, CO Amphoteric = H2O, Al2O3, ZnO, PbO

or

I=

3 1 = amp (20  1) 7

28.

Let the smallest Natural number =x Let Quotient =  Remainder = 21

V = IR = 20  Ans.

1 20 = = 2.857 volt 7 7

ATQ Dividend = Divisor × Quotient + Remainder x ×15 = 63 ×Q + 21

(a)

x=

63  Q  21 15

x=

21  Q  7 5

25.

if Q = 3 By kirchhof’s law By KVL

x=

21  3  7 63  7 = 5 5

x=

70 = 14 5

–6 –2I – I + 9 = 0 I = 1 amp Vp – 6 – 2I – VQ = 0 Vp – VQ = 6 + 2  1 = 8 volt Ans.

(a)

26.

9  1  6  1 15 Equivalent voltage = = volt 1 1 2

31.

(I) Lift is moving with constant speed in upward direction so, T– mg = 0 or T = mg constant. (correct) (II) Kinetic energy =

1 mv 2 = constant 2 (correct)

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(III) u = –

GMM (not constant) (wrong) r

36.

7



H KMnO4   Mn

+2

Transfer of 5electron

(IV) Velocity is constant so acceleration of lift is zero

6



H K 2Cr2O7   Cr

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

+3

mechanical energy will not be constant

Transfer of total 6 electron

(wrong)

 Ratio is 5 : 6

Ans

(d)

32.

The reactivity order of the given metals is

37.

m = 338 –

288

m = 13 2 –12 2

Mg > Zn > Fe > Pb

m= 2 33.

arABC = ½ AB × CF .....(1) arABC = ½ BC × AD .....(2)

v=

uf ( 6)( 12) 6  12  = = 12 cm  6  12 6 uf

arABC = ½ AC × BE .....(3) (1) × (3) (arABC)2 = ¼ AB×CF×AC×BE

39.

f = – 12 , u = –6

¼(AB×AC) (CF×BE) v=

=¼(409.6) (202.5) = 20736

–6–12 6  12 uf = = = 12cm u–f – 6  12 6

ar ABC = 144 Ans. (C)

(1)×(2)×(3) we got

(arABC)3 =

(144)3 =

1 ×AB×CF×BC×AD×AC×BE 8

40.

methyl cyclobutane is

1 ×(AB×AC) × (CF×BE) × (AD×BC) 8

Molecular fromula is C5H10 =2985984 ×

1 ×(409.6)(202.5)(AD×BC) 8

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)

41.

4 digit number = abcd = 1000 a + 100b + 10c + d = (986a) + 14a + (87b) + 13b + 10c + d no is divisible by 29 14a + 13b + 10c + d = 29 n is the remaining number a + b + c + d = 29 (13a + 12b + 9c) + (a + b + c + d) = 29n 13a + 12b + 9c = 29m ...(i) 9a + 9b + 9c + 9d = 29  9 ...(ii) equation (i) - equation (ii) 4a + 3b – 9d = 29 (m – 9)  the possible cases are 4a + 3b – 9d = – 29 4a + 29 = 9d – 3b 4a + 29 = 3 (3d – b)

CAREER POINT, 128, Shakti Nagar, Kota-324009 (Raj.), Ph. 0744-2503892

a = 4, 7 only possible a = 4  3d – b = 15 3d = b + 15 b = 9, d = 8 a=7 4a + 3b – 9d = 0 4a = 9d – 3b 4a = 3 {3d – b} when a = 9 no soluion for this 12 = 3d – b d = 6, b = 6 no. = 9686 d = 7, b = 9 no. 9947 b = 6, 3 no solution 4a + 3b – 9d = 29 3{b – 3d} = 29 – 4a a = 2, 5, 8 only possible only five solutions: 7598 4988 7859 a = 6, 3 sol. not possible9686 57 = 3(3d – b) 9947 3d – b = 19 3d = 19 + b b = 8, d = 9 no. = 7859 b = 5, d = 8 no. 7598

43.

2Tcos = mg

T =

mg 2 cos 

48.

NaCl  NaHCO 3  H2O Na2CO3 .H2O + HCl   10ml

‘x’ g

100 mL 0.25M

 0.05 M NaOH 20 mL Meq. of NaOH = Meq. of final solution 0.05 × 20 = 10 × N 0.05  20 = 0.1 10 N1V1 – N1V1 = NfVf

Nfinal =

0.25 × 100 –

x  1000 = 0.1 × 100 62

25 –

500 x = 10 31

15 =

500 x 31

x=

15  31 93 = g = 930 mg 500 100

49.

8888888 * 8888888 divisible by 11  64 – (48 + *) = 0 or divisible by 11 64 – 48 – * = 0 or divisible by 11 64 – * = 0 or divisible by 11 *=5

51.

In figure , this is the position of TIR, So angle of incidence 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

  cos   T 

Given (1) + 0.5% error in each tablet

Ans.

(c)

‘x’ molecules extra in aspirin ‘y’

47.

According to definition of conservative force.

100 g  99.5 g

Ans.

(a)

250 × 10–3 

45.

No. of triangle = 14C2 =

14  13 = 91 2

250  99.5  10 –3 g 100

= 248.75 × 10–3 extra = 1.25 g × 10–3 180 g  NA

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1.25g × 10–3 

N A  1.25  10 –3 g 180

= 0.9676 g Na2O  error in mass of Na2O = 1.24 – 0.9676

Extra molecules = x

= 0.2724 g

100 g  99.5 g % error =

500  10 –3  99.5 500 × 10–3g  100

0.2724  100 = 21.97 % 1.24

–~ 22% –3

= 497.5 × 10 g  extra = 2.5 × 10–3 g

57.

10000003 143

151 g  NA

2.5 × 10–3 g 

2.5  10 –3  NA extra molecules = y 151



x N A  1.25  10 –3 151 = × y 2.5  10 – 3  NA 180

151 1.25 151 151 = = = 180  2.5 180  2 360

143



59.

13 = 1 Ans. 143

In equilibrium, under the effect of several forces the sum of torques about any point must alway be equal to zero.

360 x = 151 y y y = 2.4 x x

55.

999999  13

Ans.

(c)

61.

3 3 3 3 3 + + + +............+ = ? 9700 4 28 70 130

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

3 3 3 3 3 + + + +........+ 97  100 1  4 4  7 7 10 10  13

124 g

student took mass of Na as 11  44g

76 g

=

4 – 1 7 – 4 10 – 7 13 – 10 100 – 97 + 47 + + +.....+ 97  100 7  10 10 13 1 4

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

 1.24 g 

1.24  44 g Na 76

=1–

1 1 1 1 1 1 1 1 1 + – + – + – +.......+ – 97 100 4 4 7 7 10 10 13

On simplification.

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

=1–

1 99 = =0.99 100 100

0.7179  124 g Na2O 92

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62.

68.

 g

N = 1s2 2s2 2p3 O = 1s2 2s2 2p4

2

If v constant

 v  (a)   r  

C = 1s2 2s2 2p2 F = 1s2 2s2 2p5 The values of IEs show a sudden jump in 5th IE.

63.

According to definition of newton 3rd law action

Configuration of C shows after losing 4electron

and reaction always act on different bodies and

it will aquire configuration of He and thus renoval

there will be no time gap between action and

of 5 electron will require much larger energy

reaction Ans

(d)

64.

Sodium peroxide Na2O2

20

O 22 –

O

20

69.

20

x

30 By Appolonius theorem,

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

in DCB,

a =102

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

d =34

 900 + 400 = 2 [400 + x2]

A



a + (n–1). d = 986 102 + (n – 1) × d = 986

1300 = 400 + x2 2

 x2 = 250

884 34

(n–1) =

 x = 5 10

n –1 = 26

 BD = 10 10

n = 27

v2 Since |a| = r

71. sum =

n [a + an] 2

=

27 [102 + 986] 2

= 14688

B

 650 = 400 + x2

(n–1) × 34 = 986 – 102

a  v 2 , 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

Ans.

20

Structure : O – O

an = 986

67.

C

x

Peroxide ion

65.

30

D

When force is always perpendicular to initial direction of motion then path of projectile will

=–7

be parabolic.

= – 7 + 25

(a)

= 18

CAREER POINT, 128, Shakti Nagar, Kota-324009 (Raj.), Ph. 0744-2503892

at t = 0 75.

Medium y is denser, so speed will decrease, initial velocity u = 0 fequency depends on source sowill decrease by relation V = n

d2 x

Ans.

(c)

dT 2 = a = 12, t = 0.5 sec.

76.

6C(s)+ 3H2(g) C6H6 (l)

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

Now combustion of C(s) 6C(s)+ 6O2 6CO2 ; H1 = 6x ....(i) 3H2 +

3 O 3H2O (l) ; H2 = 34 ....(ii) 2 2

Ans.

(d)

80.

In evaporation , water takes heat, overcomes the intermolecular force of attraction and converts into vapour phase.

15 C6H6 + O2 3H2O ; 6CO2 ; H =Z 2

Reverse the reaction.

6CO2+ 3H2O C6H6 +

15 O 2 2

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) = |

1 [p(r + p – p – q) + q(p + q 2

– q – r) + r(q + r – r – p)]|

= |

1 [pr + pq – pq – qr + qr – pr]| 2

=0

79.

x = 6t2 dx dt = V = 12t

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