ENTHUSIAST, LEADER & ACHIEVER COURSE

TARGET : PRE-MEDICAL 2013

MAJOR TEST # 10
 NEET-UG DATE : 01 - 05 - 2013
FULL SYLLABUS
ANSWER KEY
Q. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
A. 2 4 1 2 4 4 3 1 3 2 3 1 1 3 4 3 2 3 1 4
Q. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
A. 2 3 4 1 2 4 3 1 1 3 3 4 4 3 2 1 2 2 4 2
Q. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
A. 2 2 1 1 1 3 2 3 2 3 3 3 1 3 3 3 3 2 1 3
Q. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
A. 3 1 2 4 1 1 3 2 1 1 1 1 2 2 4 4 2 4 4 1
Q. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
A. 4 2 2 1 3 4 3 1 3 3 3 1 4 4 1 1 3 2 1 1
Q. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
A. 4 3 3 2 1 1 3 3 2 2 2 3 1 4 4 4 2 4 2 1
Q. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
A. 3 1 2 3 4 2 4 3 2 4 2 4 2 2 3 1 1 3 2 2
Q. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
A. 1 2 2 2 4 4 2 2 3 3 4 2 2 1 1 2 4 4 3 4
Q. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
A. 4 3 4 1 3 1 3 2 3 1 4 2 2 4 1 2 2 2 1 4

HINT – SHEET
1. Lyman series produces U.V. radiation
n 1 T T  n 
Balmer series produces Visible radiation 4. n T  n  2 T  T  2  n 
Pachan series produces Infrared radiation
So correct answer is 4  3  6 
=2  = 0.02
600 
1 1
2. Active fraction at instant t2, t 2 / T 1/ 2

5. Here u = + 10 cm
2 3
v = + 15 cm
1 2 5cm
Active fraction at instant t1, t1 / T 1 / 2
 15cm
2 3

t 2 / T 1/ 2 t 2  t1
2
 t / T 1/ 2
 2  2 T1/ 2 = 21 1 1 1
21 By lens maker formula  
v u f
 t2 – t1 = T1/2 = 50 days
1 1 1
3. By using v 2 = u 2 + 2as where v = 0,    f = – 30 cm
15 10 f
u = 2 m/s, a = – g = – 5 m/s2 we have s = 0.4 m

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 MAJOR TEST
TARGET : PRE-MEDICAL 2013 (NEET-UG) 01–05–2013
6. Inet = Idisc – Iremoved I 3I
15. f' = f & Intensity  Area so I' = I – 
4 4
2
1 1  R 40
= (9M)R2 – M  = MR2 3 3
2 2 3 9 17. Ig =  30 , Ig' =  20
50 2950 50  R
7. Additional kinetic energy = TE2 – TE1
50 R 3
   50 + R = 4500
GMm  GMm  1 1 1  50 2950 2
=      GmM  
2R 2  2R 1  2 R 1 R 2   R = 4450

18. For given conditions mg = m2a = ka

1 2h
8. h  gt 2  g  2
2 t mg 2 10
a = =
k 200
g  h t  = 0.1 m = 10 cm
then  100   2  100 = e1 + 2e2
g  h t   (4 r 2 )T 4
19. Solar constant =
(4 R 2 )
9. Energy loss in C = energy stored in L
r 2( t  273 ) 4
1/ 2 =
1 1 1  C(V  V ) 2 2
R2
CV12  CV22  LI2  I =  
1 2

2 2 2  L 20. x = asint
a 
10. For transistor action the base region must be  = asint  t =
2 6
very thin and lightly doped & the emitter-base
FG 2  IJ t  T
junction is forward biased and base-collector 
HTK =
6
 t =
12
junction is reverse biased
q
21. total = curved + plane surfaces =
0
Y = AB = (NAND)
11. A B Y
q 1 F q  I
1
0
1
0
0
1
 + 2A =
0
 A =
2
GH  JK
0
0 1 1 22. Induced emf in primary coil
1 0 1
d d
EP =  (0  4 t )= 4volt
dt dt
12. option (c) and (d) are incorrect because Induced emf in secondary coil
option (c) is true only for spherically symm. ES N S E 1500
bodies option (d) radius of gyration is irrelevant   S   ES = 120volt
EP N P 4 50
with C.G.
LM FG V IJ OP LMV  FG V  1IJ OP
VA–VB = V  8  4
13.
m
Power = Fv = v  t  v = v2(Av)
23.
N H KQ – N H 4 KQ
V V V
= Av3 = (100)(2)3 = 800 W =– + =–  VB > VA Ans (4)
2 4 4
Q kA  T1  T2 
14.  b
t  24. v = at +
t c
 [c] = [t] = T ;
 A
k    T1  T2 
Q' 4 1 kA  T1  T2  [v ]
 = [v] = [at]  [a] = [t ] = LT–2 ;
t 4 16 
Q
 Q' = [b] = (LT–1)T = L
16

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 MAJOR TEST
PRE-MEDICAL : ENTHUSIAST, LEADER & ACHIEVER COURSE 01–05–2013
25. According to question and by using COME
31. Q4 = 4CV
GMm 1
– + m (fv)2 = 0 + 0
RR 2 FG 6 CIJ 6 CV
Q2 = H 11 K V =
11
GM 2GM
 fv = but v =
R R
Q2 6 CV 1 3
 Q = × =
4 11 4CV 22
2GM GM 1
Therefore f = f= 5
R R 2 33. PV = RT where  =
moles
32
34. U = µC V T & 0 = W +  U
 U = –6R ( W = 6R)
26. As voltage drop across 8= 2 8
FG R IJ 3
F VI 2 Therefore –6R = 1   1 T = RT
2 H K
= 4V GH P  R JK
 T = –4  Tfinal = (T – 4)K
Therefore voltage drop across 3 = 3V
[  4V is divided in ratio of resistances between u
35. Let time of flight be T then T = g
1 and 3]
( 3)2 Let h be the distance covered during last ‘t’
Hence power dissipated in 3 = = 3 watt second of its ascent
3
Velocity at point B = vB = u – g(T – t)
12400
27. Energy of photon =  3eV u
4100 
= u–g  g  t  = gt

 h = vBt – 1 gt
2
2
28. 2 1 2 1 2
 h = gt – gt = gt
2 2

dv
1 LM q q
1 3 1 2 q q
2 3 q q OP 36. Here
dt
= constant = a (say)

N
Ui = 4   ( 0. 4) ( 0. 3)  (0 5. )
0 Q Use v2 = u2 + 2as where
1 LM q q
1 3 q1 q2 q q
 2 3
OP s = 2 × 2r =80 m, u = 0, v = 80 m/s
Uf = 4  
0 N( 0. 4) 
( 0. 3) (01. ) Q    
37.  A  B .  A  B  = 0
1 FG 1 1 IJ
Therefore U=Uf–Ui= 4   q2q3
0 01
.

0. 5 H K  
 A2 – A . B +
  2
B.A – B = 0
q 2 q3 q3    
=   (10-2) = 4   (8q2)  A = B ( A . B = B .A )
0 0

 K = 8q2 v  vs
0 I 
= 0
FG e IJ 0 FG ev IJ 38. Source is stationary   = constant & f'=
v
29. B=
2R 2R H TK =
2R H 2 R K
FGs
f= 1 v f=
v IJ  1

 R2 =
 0 ev
 R 
v H K  1  5  f = 1.2f
4 B B

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 MAJOR TEST
TARGET : PRE-MEDICAL 2013 (NEET-UG) 01–05–2013

T2 W 56. For the cell reaction

39. Use  = 1 – = Fe + 2Fe3+  3Fe2+
T1 Q
40. For electromagnetic wave Anode reaction is
Fe  Fe2+ + 2e–
E cathode reaction is
2Fe3+ +2e–  2Fe2+
vwave =E × B ECell = ECathode – EAnode (E° is reduction potential)
= 0.771 – (–0.441)
B
46. We know that E Cell = 1.212V


oxidising nature  S.R.P. 61.

1 O O O O
Reducing nature 
S.R.P. F3 C – C – OH > Cl C3 – C – OH > H – C – OH > CH – C 3– OH
 In the given values, F 2 has highest S.R.P. –I effect –I effect No I-effect +I-effect
(more) (less)
therefore it is strongest oxidising agent.
 In the given values Iodine has least S.R.P. CHO COO CH 2–OH
therefore I– is strongest reductant
50% KOH
47. O–2 ions form CCP, therefore 4 O–2 ions are 62.  +
present per unit cell. Cl Cl Cl
 No. of tetrahedral voids = 8
63. Vitamin B complex
1 64. S N Reaction does not involve rearrangement
2

Tetrahedral voids occupied by A+2 =  8 = 2 5 4 3 2 1

4 65. H 3C—CH=CH—C 
CH
Also, no. of octahedral voids present = 4 Pent-3-en-1-yne
Octahedral voids occupied by B+ = 4 66.
 Formula of oxide could be = A2B4O4
or AB2O2
48. Orbital angular momentum of a p-electron is
h
given by = (  1)
2
h
= 1(1  1)
2 conc. NaOH
h 67. 2HCHO   HCOONa + CH3OH
= 2 (No formation of C–C bond)
2
1 h 68. (Nucleophilic substitution of aryl halide)
=
2   –M effect
X X
rB MA V /t 49 20
49.   B B   NO 2
rA MB VA / t A MB 10  <
MB = 12.25 u NO 2
wt 1000
(Nucleophilic substitution of alkyl halide) 
50. Molarity (M) = × stability of carbonium ion
mol.wt. vol (ml)
 (CH3)3C–X > (CH3)2CH–X
25.3 1000
= × 71. In the ClF3 , Cl atom is sp3d hybridised, having
106 250
= .955 mol/L of Na2CO 3 trigonal bipyramidal geometry, in which axial
and Na2CO 3  2Na+ + CO3 –2 bonds are longer than equitorial bonds.
therefor [Na+] = 2 × 0.955 = 1.910 M 77. Given ions
[CO3 –2 ] = 0.955 M (i) C2
2–
(ii) He2+ (iii) O2– (iv) NO
51. No of atoms = No. of molecules × atomicity –
Total e 14 3 17 15
= 0.1 × NA × 3 –
Bonding e 10 2 10 10
= 1.806 × 1023
Anti 4 1 7 5
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 MAJOR TEST
PRE-MEDICAL : ENTHUSIAST, LEADER & ACHIEVER COURSE 01–05–2013
bondinge–

5.1eV I1
Na  
 Na + hence 'e' gain
e gain enthalpy –5 – 1 
82.
10 4 2 1 10 7 10 5
B.O. enthalpy of Na+ is 5.1 eV
2 2 2 2
83. Due to +ve charge on central atom, absence of
=3 = 0.5 = 1.5 = 2.5
synergic bond makes. C– O bond strongest.
78. Down the group in Gr –16 hydrides F
F
M–H bond length increases (due to increases
84. In SF4 no. of ep = : S
in size)
F
Hence acidic nature increases
F
Hence Ka while pKa bp +  p = ep
79. because Na2Cr2O7 is hygroscopic hence give 4 +1=5
less priority. Hybridisation = sp3d
86. In isoelectronic series
80. On strong heating only Li gives normal oxide On increasing Zeff radius decreases
while other alkali metlas gives peroxide or super 87. In BeSO4 H.E. > L.E. ( Soluble in water)
oxide

CORRECT ION IN M AJOR T EST

Test D ate PH ASE
Q. 54
30-03-2013 M LP,M LQ,M LR,M LS
A. 2
Q. 58 86
06-04-2013 M LP,M LQ,M LR,M LS
A. 3,4 2
Q. 54 70
11-04-2013 Leader + Enthuse + Achie ver
A. B 3
Q. 52 84
20-04-2013 Leader + Enthuse + Achie ver
A. B 1,2
Q. 100 102 103 110 144
A. 3(H) B 1 B 3
25-04-2013 Leader + Enthuse + Achie ver
Q. 152
A. B
Q. 8 16 30 36 51
A. 3 4 4 3 1
28-04-2013 Leader + Enthuse + Achie ver
Q. 144 161
A. 3 3

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