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