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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