HINT – SHEET

DATE : 19 - 02 - 2014 AIIMS (FULL SYLLABUS)

TARGET : PRE-MEDICAL 2014

ENTHUSIAST COURSE

DATE : 10 - 01 - 2010MAJOR TEST # 06

ANSWER KEY

HS - 1/5Your Target is to secure Good Rank in Pre-Medical 201401CM213083

1. V= RT

P

= 5

2 8.31 300

32 1.01 10

V = 1.53 × 10-3 m3

V = 1.53 lit.

2.

60°

mg Mg

30°

mg sin 60 Mg sin 302 2

M

3m

3.

1 2

1 2

m m

m mT 2

k = 2

3m 3m

4k k

4. To find the magnetic field outside a thick

conductor, the current may be assumed to flow

along the axis. As points 1 , 2 , 3 are equidistant

from the axis, B1 = B

2 = B

3

5. RF

1

r(S,T,T

0 = same )

6. u2 = 5gR v2 = u2 – 2gR= 5gR – 2gR = 3gR

u

v

B

Tangential acceleration at B isa

t = g (downwards)

Centripetal acceleration at B is aC =

R

v2

= 3g

Total acceleration will be

a = 2t

2C aa = g 10

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Ans. 1 4 3 1 2 1 1 1 2 1 3 1 3 3 2 1 2 3 3 2

Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

Ans. 1 1 2 3 1 3 3 1 3 4 3 3 2 3 2 2 1 1 3 2

Que. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

Ans. 3 4 3 4 1 1 4 4 3 4 2 4 3 4 2 2 3 1 4 3

Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

Ans. 3 4 3 3 1 4 1 3 4 1 2 3 2 2 1 3 3 2 3 2

Que. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

Ans. 3 3 3 4 3 3 2 3 4 2 3 2 4 4 3 1 4 1 3 3

Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120

Ans. 1 3 1 4 4 2 2 3 4 1 3 3 2 2 4 3 1 1 2 4

Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140

Ans. 1 1 2 1 4 2 1 2 2 4 3 1 1 2 1 1 3 2 1 3

Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160

Ans. 1 3 2 2 3 4 1 3 1 3 3 4 3 1 3 3 3 4 1 2

Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180

Ans. 1 3 2 2 4 3 1 3 3 2 2 1 1 3 2 4 3 3 1 1

Que. 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200

Ans. 2 1 2 1 1 4 1 1 3 2 2 1 3 4 4 3 3 1 4 4

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MAJOR TEST : AIIMS

7. 4 trips means 32 m

4 m

v

t = v

d v =

t

d =

8.0

32= 40 m/s

v =

TT = v2 =

4

2.0 × (40)2

= 4

10162 = 80 N

9. C2 = A2 + B2 + 2AB cos C A B

and C = A + B

so, C2 = A2 + B2 + 2AB

than cos = 1 = 0°

11. P

Q

R

S

As we are are moving away from P toward

sheet S spacing between electric lines of

force is increasing. ER < E

Q In direction

of electric field potential decreases.

VR < V

Q

13. For adiabatic process : Slope : dT

dP =

1–

T

P

1

dP

dT

=

13/5

3/5

T

P =

3/2

3/5

T

P = 2.5

T

P

2

dP

dT

=

5/2

5/7

T

P = 3.5

T

P

3

dP

dT

=

3/1

3/4

T

P = 4

T

P

16. Given s

p

e20

e

s

p

I18

I

% = out

in

P100 90%

P

17.1

2

2x12

2 x

1 2mu 4

K 1 2mu 1

1

2

x

x

u 2

u 1

1

2

2g1

22 g

(u ) / 2gH 4

H (u ) / 2g 1

g1

g2

u 2

u 1

1 1

1 1

2 2 2 2

x y

x g1

x y2 x g

2u u

u uR gu uR u u

g

= 2 2 4

1 1 1

19. For maximum current, all switches must be

openSo current through battery

=24

2A1 6 4.5 0.5

VAB

= (1A) (1) = 1V

20. Lateral shift = t sin(i r) 3 sin 30

cos r cos30

3 .tan 30 1m

21. <v>space

= vdx

dx

But a = dv

vdx

so vdx = 2v

dva

and dx = v

dva

so <v>space =

2vdv

av

dva

=

22

12

1

v du 14= m/s

9vdv

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MAJOR TEST : AIIMS

22. K.E. + P.E. = 0

K.E. = P.E. = eGM m

R

= mgR

= 103 × 10 × 64 × 105

= 6.4 × 1010 J

23. is very low XL is very low

VL = I XL 0 and Vc V0

24. For zero deviation (µ1 – 1)A1 – (µ2 – 1)A2 = 0

(1.54 – 1)4 = (1.72 – 1)

= 3°

25.(n–1)mm

M

T1

T2T2 T1

Mg – T1 = Ma …(i)

T1 = nma …(ii)

From eq. (i) & (ii) a = Mg

(M nm)

T2 = ma = Mmg

(M nm)

26. From Pascal's law, pressure is changing at every

point by the same amount. Hence, buoyancy

remains the same. So, the part of the block

inside water remains the same.

27. Given circuit can be redraw as

2V1/2

R

for max power , R = 1

2

P = I2 R = 2

2 1

1 2

= 2 watt

28.

A

4

Bx

x +(4 )2 2

D

x = 2 2x 16 x n

2 2x 16 = (x + n)

x2 + 162 = x2 + n22 = 2nxput n = 1, 2, 3 & 4

29. F1 = mg sin 30 – f

= mg sin 30 – umg cos 30

30ºmg sin 30º

F1

f

F1 =

mg

4

F2 = mg sin 30 + f

= mg sin 30 + µmg cos 30

30º

mg sin 30º

F2

f

= 3mg

4

1

2

F 1

F 3

31. 198 = 1

V 900

R 900

............(i),

180 = 1

V900

2R 900

.............(ii)

divided Eq (i) by (ii)

and on solving it R1 = 100

Now put value of R1 in any one equation.

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MAJOR TEST : AIIMS

32.

2n 1 D

x2a

For red light : x = 4 1 D

2a

× 6500 Å

For other light x = 6 1 D

2a

× Å

x is same for each.

5 × 6500 = 7 × or 5

7× 6500 = 4642.8Å

33.37°mgsin

37°F

F – mgsin37° = ma

F = mgsin37° + ma = 3 × 10 ×3

5+ 3 × 2 = 24

v = u + at = 0 + 2 × 5 = 10

P = Fv = 240 watt

34. Given 1

2

2

1

or = 1 = 2

1 1

1 1 2

K

K

21

1

q 1K 1 K k

2

2 2

2 1 2

K

K

K2 = 1

2

1 K (q 1)K

35. iB = mg

e Bmg

R

2 2vBmg

R

v = 2 2

mgR

B

37. 2 mgd = 1

2mv2 +

1

2 (2m) v2 + (mg) d

2 mgd = 23 mgd

mv2 2

38.

v

30º30

º

= t

/3 = 2

t24

t = 4 sec

39. v = E 1500

B 0.40 = 3.75 × 103 m/sec.

40. After one hour

1/ T

0

0

A 1 1

23A 3

After next 3 hours means four hours from initial4 44 / T 1/ T

0

A 1 1 1 1

A 2 2 93

74. x yr 3r

yx

y x

dr3

r d

82. NCERT (XI) Pg # 164, 167, 168

83. NCERT Pg. # 269

86. NCERT (XII) Pg # 174

87. NCERT Pg. # 276

89. NCERT : XIth, Page no. 35, IIndpara and

Page no. 38, Ist para

90. NCERT XII, Page no. # 107, Para = 1

98. NCERT XII, Page no. # 132, Para = 1

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MAJOR TEST : AIIMS

106. NCERT (XI) Page No. # 249 (Last Para) (Eng.)

NCERT (XI) Page No. # 249 ()

(Hindi)

110. NCERT (XI) Page No. # 198(Last Para) (Eng.)

NCERT (XI) Page No. # 198 (&

199 ) (Hindi)

114. By yeast 2ATP from glucose + 2 ATP from

fructose i.e. 4 ATP from one sucrose

2 ATP + 2 ATP

4 ATP

118. NCERT Pg # 294

121. S = 1

M tan

SL

= 1 1

M tan60 M 3

SV

= 1

M tan30=

3

M

so, SV

= 3SL

123. A – True R – True but Not correctexplanationHeating of core is due to hysteresis loss.

134. A = 60° = 2

452

3060

2

Ai m

302

Ar2

22/1

2/1

30sin

45sin

rsin

isin

Both Assertion and Reason are true butReason is not the correct explanation ofAssertion.

161. NCERT XIth Page No. 20

171. NCERT (XI) Pg # 131

175. NCERT (XI) Page No. # 176 (2nd Last Para)

(Eng.)

NCERT (XI) Page No. # 176 (2nd )

(Hindi)

177. NCERT (XI) Page No. # 216 (Calvin cycle)

(Eng.)

NCERT (XI) Page No. # 217 ( )

(Hindi)