2404 HS Aiims

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


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


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

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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

0999DM310315030HS - 6/6


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)

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2404 HS Aiims