1502-hs
TRANSCRIPT
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Q. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
A. 1 1 3 2 3 3 3 4 1 2 1 1 3 4 3 2 2 3 1 2 4 2 1 2 3
Q 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
A. 2 2 4 2 1 1 3 1 2 3 2 3 1 3 3 3 1 4 4 1 3 4 3 3 1
Q. 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75
A. 1 3 2 1 2 2 1 3 3 1 3 4 2 4 3 4 2 2 2 2 3 1 2 2 2
Q. 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
A. 1 3 2 4 3 3 2 1 4 4 4 2 3 3 4 3 3 4 3 4 3 4 4 2 2
Q. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125
A. 2 3 4 3 1 4 3 4 4 2 2 4 4 4 4 4 2 2 1 4 4 2 4 3 4
Q. 126 127 128 129 130 131 132 133 134 135 136 137138 139140 141 142 143 144 145 146147 148 149 150
A. 4 4 3 2 3 3 1 3 1 2 1 1 4 1 1 4 1 3 4 1 4 1 1 1 2
Q. 151 152 153 154 155 156 157 158 159 160 161 162163 164165 166 167 168 169 170 171172 173 174 175
A. 3 3 1 4 2 2 4 1 3 2 4 1 2 1 4 2 1 1 1 4 2 3 1 2 4
Q. 176 177 178 179 180 181 182 183 184 185 186 187188 189190 191 192 193 194 195 196 197 198 199 200
A. 3 3 4 2 1 3 3 3 3 4 2 3 2 2 3 3 2 4 1 1 4 2 4 2 3
HINT SHEET
DATE : 15 - 02 - 2013AIIMS (FULL SYLLABUS)
TARGET : PRE-MEDICAL 2013
ENTHUSIAST COURSE
MAJOR TEST # 08
1. Form mechanical energy conservation
1
2mV2
1
2m
V
2
2FHG
IKJ =
1
2Vx2
K =3
4
2
2
mV
x
2.r
l
where =1 minute
so,1 1
60 180 60
rad and l=3m
3mx r 10km
1
180 60
l
3. Heat gain by Ice = Heat loss by water
m; Lfus. + mi S(Tm0) = mS(50 Tm)
50 80 + 50 1 (Tm0) = 100 1 (50 Tm)
Tm= 6.67C4. On same level of third liquid.
Pressure on left hand side = Pressure on right hand
side
0+ 20g = p
0+ 10 (1.5 ) g + h(2) g
On solving h = 2.5 cm
5. L1= L 2
200L L 3L
100
Percentage increase in rotational energy
=2 1
1
E E100
E
=2
1
E1 100
E
=
2 2
2
2 2
1
L 9L1 100 1 100 800%
L L
6. Fringe width Therefore, and hence willdecrease 1.5 times when immersed in the liquid.
The distance between central maxima and 10th
maxima is 3cm in vacuum. When immersed in
liquid it will reduce to 2cm. Position of centralmaxima will not change while 10th maxima will
be obtained at y =4cm.
7. P3> P1 P3
P1
V1 V2
ADCITE
V
PWIT= Ve
WAD= Ve
W < O
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8. Range will become twice if velocity of efflux offer
becomes twice now as 2gh . Therefore h
should become 4 times or 40 m.
Thus an extra pressure equivalent to 30 m of water
should be applied
1 atm = 10.33 m of water,
30 m of water 3 atm9.
2
CM
Loss in PE = gain in rotational KE
mg
2
=
1
2I2=
1
2
2m
3
2
2
v
v = 3g
10. NCERT-I Pg. # 118
W =
(1,1)
(0,0) F,ds
Here ds dxi dyj dzk
W =(1,1)
2
(0,0)(x dy ydx)
=(1,1)
2
(0,0)(y dy xdx) (ax x = y)
W =
(1,1)3 2
(0,0)
y x 5J
3 2 6
11. PV = MM
RT
1 V =1
2
MR 298 ....... (i)
1.5 V =1 2
2 3
M M
R 298 .....(ii)
Equationi
ii 1
2
M 1
M 3
12.We = effective weight
fv
we
v
2we
wef 'v
fv= w
e
6rv1= w
e.....(i)
Inequilibrium 2 we w
e= 6rv
2.....(ii)
from (1) & (2) v1= v
2= 1 m/s
13. Percentage change in time period
T100%
T
= 1 100
2
[g = 0]
According to question 100
= 4%
T 100%T
=
1
2 4% = 2%
14. NCERT-I Pg. # 128
F S1/3
i.e, acceleration a s1/3
or vdv
ds= Ks1/3or v2 s2/3
or v s1/3
Now P = F.v
or P s1/3
.s1/3
or P s0
i.e., power is independent of s.
15. RFA A 1
MS v s r s
, s = same than RF1
r
16. Z1 2 2 + 3 1 = Z
2 2 1 + 5 1 = Z
C
Z1 Z
2= 4
A1 4 2 = A2 1 4 = ACA
1 A
2= 4
18. NCERT-I Pg. # 128
Power P = F.v
or P = (ma) v
a =P
mv
or vdv
ds=
P
mv
or v2dv = Pm
ds
orP
m
2
1
vs
2
0 v
ds v .dv
or 3 3
2 1
P 1(s) (v v )
m 3
or s =3 3
2 1
m(v v )
3P
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33. Since, image formed by mirror will be at
distance OM on right side, so image will be
formed for convex mirror at OMMP. So,
v = OM MP and object is at OP.
u = OP
36. For nodes sin 2x = 0
x = 0,1
2, ......
minimum length =1
2m
37. The ray is incident on
the interface of a rarer
medium (air) from a
denser medium (water)
and the angle of
airwater
Refractedray
Reflected
rayIncident
ray
incidence is less than the critical angle. The ray
will be partly refracted and partly reflected.
Also,the Snell's law n2sin
1= n
2sin
2tells that
the angle of refraction will be more than the angle
of incidence, on entering the air ray will bend
away from the normal [Fig] The angle between
the reflected and the refracted ray is less than
1802.
40. f =1 T
2 m
1 2 1
2 1 2
f T
f T
42. C2H
5NH
2+ CS
2+HgCl
2 C
2H
5N=C=S
43. For n = 3
Number of e= 2n2= 2(3)2= 18
+1/2 1/2
3s 3p 3d
46.
O O+Ph OH
CO2
O
Ph CH3* I +NaOH2
O
ONa PhCHI +3*
47. At the end point N1V
1= N
2V
2
V2= 7.5 mL
Therefore concentration of salt1
10= 0.1 M
at end point.
B+ + H2O BOH + H+
CC C C
14
h 12
b
k 10k
k 10
= 102
2 2
h
C 0.1k
1 1
102+ 1 = 0
1 1 40
20
= 0.27
[H+] = C= 0.1 0.27 = 0.027 M
49. Tollen's reagent used to detect terminal alkynes/
aldehydes.
50. PhC N PhCNH2
OOH/H O2
53.
CHO
CHO
CH OH2
COOOH
[Intramolecular cannizaro]
54. 1 L solution contain 0.01 mole of Co(NH3)
5SO
4]Br
& 0.01 mol [Co(NH3)
5Br]SO
4
1 L of mixture X + AgNO3(excess) AgBr
1 L mixture of + BaCl2(excess)BaSO
40.01 mole
55. NCERT, Part-I, Class-11th, Page No. # 167
57. RCOC H2 5
O
CH MgBr3RCCH3
O
CH MgBr3
H O2RCCH3
OH
CH3
58. If wt. of O2= x g
wt. of N2= 4x g
Moles of O2=
x
32
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Moles of N2=
4x
28
Number of molecules = Number of moles NA
59. H = E + (PV)
H = 30 + 2 (5 3) + 5 (4 2)
H = 44 L-atm
61.
CH CH2 3
Br2
h
CHCH3
Br
alc. KOH
CH=CH2
Br2
CH CH2
Br
Br
62. 14H++ 6Fe+2+ Cr2O
726Fe+3+ 2Cr+3+7H
2O
63. NCERT-11th,Part-I, Page No.# 162
Heat capacity, resistance & Enthalpy are mass
dependent so extensive properties.
66. In CrO2Cl
2
x + 2(2) + 2(1) = 0
x = +6
67. For neutralisation of strong acid with strong base
H = 57.3 Limiting reagent kJ
69. CH CHCOOH3
CH OH2
LiAlH4
CH CHCH OH3 2
CH OH2
70.
2
2
n aP
V
(V nb) = nRT
term 2
2
n areprents intermolecular force.
71. SO2(g) + NO
2(g) SO
3(g) + NO(g)
1 mol 1 mol 1 mol 1 mol
1 x 1 x 1 + x 1 + x
Q = 1 (Which is less than KC), so reaction is pro-
ceed in forward direction
KC=
2
2
(1 x)
(1 x)
; KC= 16
4 =1 x
1 x
x = 0.6 M
74. For ideal gas Z = 1
78. When n = 3
may be 0, 1, 2
For value of m to +including zero85. NCERT Page No. 177 Para = 2
92. NCERT Pg. # 317 Para-2 line 4-6
94. NCERT-XI Page No. 262 (H)
95. NCERT XI Page No. 88, IstPara
96. NCERT-XI, Page No. 115, Fig.98. NCERT-XII Page No. 228, 229, 231
100. NCERT-XII, Page No. 77
102. NCERT-XI Page No. 265 (H)
142. NCERT-11th, Part-I, Page No. # 146
144. Assertion is wrong because besides amounts, pres-
sure also depends on volume, however reason is
correct because both frequency and impact are
directly propartional to root mean square speed
which is T .
146. Volume of H2SO
4req. = x/2 mL
155. Assertion :-U = q + W
U = nCvdT = 0
W = Pext.
.dV = 0 {Pext.
= 0}
q = 0
Reason :- Acc. to K.T.G.
157. At equilibrium G = 0 (always) not G
If G < 0 Process will be spontanpowNCERT-11th, Part-I, Page No. # 178 & 179
161. NCERT Page No. 131 Line = 2
162. NCERT XI Page No. 75, IIIrdPara
168. NCERT XIIthPg. # (E) - 60, (H) - 67
172. NCERT Pg. # 325
173. NCERT-XII Page No. 168
174. NCERT-XI, Page No. 102, Last Para