2404 hs aiims
DESCRIPTION
hSTRANSCRIPT
HINT – SHEET
1. ˆ ˆˆ ˆ(a + 2b).(5a – 4b) = 0
5 – 4 ˆa.b + 10 ˆ ˆb.a – 8 = 0
6 a . b = 3(1) (1) cos = 3/6 = 1/2
= cos–1
1
2
2. In the one dimensional elastic collision with one
body at rest, the body moving initially comes to rest
& the one which was at rest earlier starts moving with
the velocity that first body had before collision.
so, if m & V0 be the mass & velocity of body,
the change in momentum = mV0 Fdt = mV
0
Fdt = mV0 F =
2mV0t
= 2N
4. A galvanometer can be converted into a voltmeter
of given range by connecting a suitable resistance
R in series of galvanometer. which is given by :
3
10025
10 10s
VR G
I
= 10,000 – 25 = 9975
5. Energy [E] = [ML2 T–2]
Velocity [V] = [LT–1]
Time [T] = [T]
[E] = [M] [LT–1]2
or [E] = [M] [V]2 or [M] = [E] [V]–2 [T]0.
ANSWER KEY
DISTANCE LEARNING PROGRAMME(Academic Session : 2015 - 2016)
LEADER TEST SERIES / JOINT PACKAGE COURSETARGET : PRE-MEDICAL 2016
Test Type : ALL INDIA OPEN TEST (MAJOR) Test Pattern : AIIMS
TEST DATE : 24 - 04 - 2016
HS - 1/60999DM310315030
Que 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 3 2 1 4 3 3 2 3 4 3 1 2 2 1 1 1 2 2 4 3
Que 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans. 1 2 2 2 4 4 3 3 2 4 1 4 4 2 4 3 4 3 3 2
Que 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
Ans. 2 2 3 2 1 4 4 2 3 1 3 3 1 1 3 3 4 3 1 1
Que 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
Ans. 4 1 2 2 3 3 2 4 2 2 4 4 2 4 3 1 3 1 2 2
Que 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
Ans. 2 1 1 2 2 2 4 3 4 3 2 2 1 2 3 2 2 4 1 4
Que 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans. 1 4 3 4 3 2 1 3 3 2 4 2 2 3 2 1 2 4 4 1
Que 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans. 1 4 4 3 2 1 1 4 4 1 3 1 3 1 4 4 4 1 1 1
Que 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans. 3 1 4 1 3 2 1 1 4 3 2 2 1 1 1 3 3 4 4 2
Que 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans. 4 3 3 4 1 2 2 1 1 1 1 3 3 3 1 3 1 3 1 3
Que 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200
Ans. 3 3 3 4 1 4 1 1 2 2 2 2 4 4 4 3 2 3 3 2
0999DM310315030HS - 2/6
ALL INDIA OPEN TEST/Pre-Medical /AIIMS/24-04-2016
6. TE = 1
2mv2
2
2
K1
R
= 1
2mv2
2 71
5 10
mv2
10.
30°
mgsin
fs
mg sin – fs = ma __(1)
fs R =
2mR
2 __(2)
a = R __ (3)
mg sin – fs = 2f
s
fs =
mgsin
3
=
mg
6
12.
(45–x)
(22.5–x)
I1 is the image formed by concave mirror.
For reflection by concave mirror
u = – x, v = – (45 – x), f = – 10 cm,
1 1 1
–10 –(45 – x) –x
1 x 45 – x
10 x(45 – x) x2 – 45 x + 450 = 0
x = 15 cm, 30 cm
but x = 30 cm is not acceptable because x < 22.5 cm.
13. Time to reach max height
t =
2 5
3.5sec2
1 2 3.5 5 6
h
h1
h2
h = h2 – h
1
h = 2 22 1 2 1 2 1
1 1a(t t ) a(t t )(t t )
2 2
= 1
7.5(2.5 – 1.5)(2.5 1.5)2
h = 15 m
14. P = V
V
But V = V tV
V
= t
P = B t t = P
B
15. E = dV
dx P
EE = –10x i
16. For refraction at first surface,
2 1
1v =
2 1
R ...........(1)
For refraction at second surface,
3 2
2 1v v =
23
R ..........(2)
Adding Eqs. (1) and (2), we get-
3
2v =
13
R or v
2 =
3
3 1
R
Therefore, focal length of the given lens system
is
3
3 1
R
HS - 3/60999DM310315030
ALL INDIA OPEN TEST/Pre-Medical /AIIMS/24-04-2016
17.15kg
45°
N µN+ = mg
2 2
N µN– = T = 50N
2 2
NN
45° 45° T
Mg
= 1+ µ 15×10
=1 – µ 50
1+ µ
= 31 – µ
µ = 1
2
18. P1V
1 + P
2V
2 = PV
3 3 31 2
1 2
4T 4 4T 4 4T 4r r r
r 3 r 3 r 3
;
2 2 21 2r r r
19. Eeq = 2V
req = r/4
I = 1.6 16
7.5 75
V
2V
1.6V
r/4
7.5I =
2
r7.5
4
16 2
r75 7.54
r = 7.5
20. Clearly after the removal of mica sheet
the new fringe width () is
´ = (2D)
dInitial fringe shift after the introduction of mica sheet is
initial shift = D
d (µ – 1) t
Equating the two 2D
d =
D
d (µ – 1) t
= (µ – 1) t
2 = 5892 Å
21. Gravitational field
V –4 4g = – = – = J/kg m
x 10 10
Work done in moving a mass of 2 kg from the
surface to a point 5 m above the surface,
W = mgh = (2kg) 4 J
10 kgm
(5m) = 4J
22. volume of sphere = 40
8= 5cc
loss in weight = Thrust force
20 g = 1 × v × g
v = 20 cc
so, internal cavity = 20cc–5cc = 15cc
23.x 2
60 40
x = 3
24. Kmax
= hc
KA =
A
hc
__(1)
KB =
B
hc
__(2)
A = 2
B
KA =
B
hc
2
__(3)
from eqn __(2)
KA = B
1(K )
2
KA =
BK
2 2
KA <
BK
2
25. Mechanical energy = kinetic energy + potential
energy E = K + U(x) where K = 1
2mv2
If K = 0 then E = U(x)
If F = 0 then dU x dU x
F = – = 0 = 0dx dx
0999DM310315030HS - 4/6
ALL INDIA OPEN TEST/Pre-Medical /AIIMS/24-04-2016
26. Rate of flow = v
square hole = 2 × 2gy
circular hole = r2 × 2g(4y)
2 2gy = 2×R2 2gy
r = 2
29. By applying work energy theorem
K.E = WS + W
ext
agent
0 = –1
2Kx2 + Fx
2Fx =
K
Work done =22F
K
30.T
T
=
1
2
=T
T
=
1
2× 2 × 10–6 × 10 = 1 × 10–5
% change = T
100T
= 1 × 10–3
31. C2 and C3 are in parallel
So V2 = V3 and C2 = C3
+ –
V
C1 2C = C2 1
Q = CV = const.
So voltage will be same
V1 = V2 = V3
32. R = Ro
t
Th1
2
5×10–6 = 64 × 10–5
t
31
2
7 t / 31 1
2 2
7 = t
3
t = 21 day
33. R = 2 m ; = π
4rad/sec2
0 = 0 & =
2
2m
2m
2 m
As we know
= 0t +
1
2 t2
t = 2
= 2sec.
Average velocity = Total displacement
Time
= 2
= 1m/s2
34. Thernal capacity H = ms
1 1 1
2 2 2
H m s
H m s =
3
1 1 1
2 2 2
r s
r s
3
1
2
H 2 1 1
H 1 2 3
1
12
35. XL = XC
So Z = R = 2
irms = rmsV 10025 2A
Z 2·2
= 0°
Pavg = I2rms R
= 2(25 2) 2
= 2500 W
36. = 0.96
iE = 7.2 mA
= c
E
i
i
iC = 0.96 × 7.2 mA
iE = i
B + i
C
37. Total energy = 2 2
1 2
P P
2m 2m
= 2
1 2
P 1 1
2 m m
= 2(1 80)
2
1 1
1 2
= 4.8 kJ
HS - 5/60999DM310315030
ALL INDIA OPEN TEST/Pre-Medical /AIIMS/24-04-2016
38.1 2 1 2
0kt 2
60 – 40 60 40
k 107 2
…(1)
40 –
7
=
40k 10
2
…(2)
20 40 2
40 20
20+= 160 – 45= 140 =28°C
40. 10 – 245 × 103×40×10–6–VB
= 0
10 – 9.8 = VB
VB
= 0.2 V
46. N2(g) + 3H
2(g) 2NH
3(g)
H = –90 = 6 390 N N 3 435
50. At low pressure,
2m
aP
V
(V
m) = RT
PVm +
m
aRT
V
m
m
PV a1
RT V RT
Z = 1 –m
a
V RT
54. Heat of formation reaction
S + O2 SO
2
58.CH2 CH2 CH2
CH –CH2 2
NO2
NO2 CHOH
–M–I –I
–I
(I) (II) (III) (IV)
–ve charge in I, II, III are resonance stabilised IV
is least stable
and stability of carbanion –M –I
–M > –I (–I of –NO2 > –I of CHO)
stability order I > II > III > IV
62. Most acidic hydrogen is most likely to takes part
in tautomerism, if after removing H –ve charge
delocalised to more electronegative atom.
O O O OH
N N N N
Ha
H
Tautomers
65. Ag2CO3 2Ag + CO2 + 1/2 O2
21.6g
moles of Ag =21.6
0.2mol108
moles of Ag2CO
3 = 0.1 mol
wt. of Ag2CO3 = 0.1 × 276 g
66.
OOH
IIF(2) (1)
(1) (2)
(1) High priority
(2) Low priority
Same priority groups are on opposite side
E configuration
69. KC =2 2
322
2 2
[SO ] (48 /80)
[SO ] [O ] 12.8 9.6
64 32
70.
Ph
HCl
Peroxide
Ph
Cl
Mechanism :(HCl does not show peroxide effect)
Ph Ph
PhPh
H Methyl shift
ClCl
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ALL INDIA OPEN TEST/Pre-Medical /AIIMS/24-04-2016
74. RMgX + Cl–NH2 R–NH
2
87. Module, Page No. 213, 214
90. NCERT XI Pg. # 249,250
91. NCERT Pg. # 131
94. NCERT XI Pg. # 197,198
103. NCERT Pg. # 249 (E)
104. NCERT XII Pg # 79, diagram 5.7 (E)
NCERT XII Pg # 87, diagram 5.7 (H)
107. NCERT Pg. # 257 (E)
108. NCERT XII Pg # 204 (E), 222(H)
111. NCERT Pg. # 233 (E)
112. NCERT XII Pg # 75, 77, 78 (E), 83, 84, 86(H)
115. NCERT Pg. # 263 (E)
119. NCERT Pg. # 243 (E)
129. Deviation = 180 - 2 (i) = 60
60º60º
60º
C
A
B60º 60º
60º
157.CH –CH–COOH3
CH –CH –CH3 2 3
CH3Soda lime
CH –CH –CH –COOH3 2 2 CH –CH –CH3 2 3
Soda lime
The reaction complete through carbanion
intermediate.
159. There are positively charged species that do not
acts as electrophile.
Ex. N
H4 does not act as electorphile.
162. Module, Page No. 186, 187
169. NCERT XI Pg. # 249
171. NCERT XI Pg. # 199
176. NCERT XII Pg # 76 (E), 84 (H)
178. NCERT XII Pg # 199 (E), 216 (H)
180. NCERT XII Pg # 197 (E), 215 (H)