answer key 5

1. µ = PV

RT

µ =

3136 830 1.5 10

8.31 408

µ = 0.05 moles

2. = 2n = 2(14) = 28

= t

28 = (25)

= 28

25

R

v

a = 2R = 228 28

25 25

× 80 × 10–2 10 m/s2

4. At is dimensionless so [At] =1 [A] =[T–1]

Also [x] = A

B

so [B] = [L–1T–1] Therefore

3 32

1 1

A TLT

B L T

HINT – SHEET

DATE : 24 - 02 - 2014 AIPMT (FULL Syllabus)

TARGET : PRE-MEDICAL 2014

ENTHUSIAST COURSE

DATE : 10 - 01 - 2010MAJOR TEST # 07

ANSWER KEY

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5. e = Ns B2 sss sin t

= 1 × 0 es

µ iS

2R sin t

e = 7 44 10 1 10

2 0.1

sin t

i = 10e 2 10

R 2

= 10–10 sin t

6. Q = er × 4r2 T4 t

1

2

Q

Q =

2

1

2

r

r

4

1

2

T

T

=

2 416 400

4 800

1

2

Q 1

Q 1

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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MAJOR TEST

7. (m/2)v = 30 m/sec1

u = 20

Momentum in x-direction

Pi = Pf

mu = x x1 2

m mv v

2 2

mu = x2

mv

2

x2v 2u 40

momentum in y-directionPi = Py

0 = y y1 2

m mv v

2 2

y y2 1v v 30 m /sec

9. Apply principle of superposition

+ +

Electric field due to a uniformly charged sphere

= 0

R

12

; E

resultant =

0

RE

12

10. e = 2 1

t

=

NBA NBA

t

=

2NBA

0.5

= 4NBA

11. than P < Q

or than P = Q

12. I = output

input

75

100 =

E

12

E = 9 J and E = 21

mv2

I = 21

(1)v2

v = 18

13. Here change in lengths is

= (AC + BC) – 2

1/ 22

1/ 22 2

2

x2 x – 2 2 1

2 2

2

1x x2 1 – 2

2

x

A B

W

C

Strain 2

2

x

2 2

14. Draw the straight line V = E – Ir where E = 3

volt, r = 0·75 on 6I5

3

2

1

00 1 2

V3 4

4given figure. The

intersection of this

line with V – I plot for

the conductor gives

the required voltage

1.5V and current 2A

15. e = (v B)

= ˆ ˆ ˆˆ ˆ(4i 6 j 8k) (2i 4 j) 4k B

= 16 V17. Time period for half part:

1 2T 2 2 2sec.

g g

So 2° part will

be covered in a time t = T

2= 1 sec.

For the left 1° part : = 0 sin (t) 1° = 2°

2sin t

T

1 2sin t

2 2

6

= × t t = 1/6 sec.

Total time T

2 + 2t 1+2×

1

6=1+

1

3=

4

3 sec.

18. There will be excess pressure 4T

pR

inside

the soap bubble. As RB > R

A > R

C so

PC > P

A > P

B. Therefore the air will flow from

A and C towards B

24–02–2014PRE-MEDICAL : ENTHUSIAST COURSE

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MAJOR TEST

19.

20. Shortest possible distance between the object

and real image is 4f.

For that the minimum distance of image from

lens is 2f.

u = –[3f – (µ – 1)t]

v = 2f

1 1 1

v u f

f

1t

21. Process AB Isobaric & VProcess BC Isothermal & V PProcess CD Isochoric & V = constP,TProcess DA Isothermal & VP

22. vcm =

1 1 2 2

1 2

m v m v

m m=

8 12 4 0

8 4

23. Amplitude = 0.15 m

K.E. = 2 21

K(A x )2

= 2 21

400(0.15 0.1 )2

= 2.5 J

25. Initially

1 1 1F 20 cm

36 45 F

Finally,

1 1 1 1 1

48 F 36 45405

40 240

5 1.377

27. Li = I0 – mR2(v/R)

= I0 – mvR

LF = (I + mR2)

Li = LF

28. U2 = 2(A2 – x1

2) …(i)

V2 = 2(A2 –x22) …(ii)

a = 2 x1 …(iii)

b = 2x2

…(iv)

Equation (ii)–(i)

U2 – V2 = 2(x2

2 – x1

2) …(v)

Adding equation (3) + (4)

a + b = 2 (x1 + x2) …(vi)

Divide equation (v)/(vi)

x2 – x1 =

2 2U V

a b

29. Assume capacitor as dipole and use F = q E ,

E = 3

2kP

r, p = Qd

30. Fringe width, D

d

When the apparatus is immersed in a liquid

and hence is reduced µ times (µ = refractive

index).

10´ = (5.5)

or 10´ D

d

(5.5)

D

d

or10

5.5

or µ = 1.8

31. Maximum tension in rope = 360N

cos60 = 720 N

amax

= maxT mg 720 600

m 60

= 2 m/s2

38. Standard equation

y = A cos2 2 vt

xsin

By comparing this equation with given

equation

2 x x

20 = 40 cm

Distance between nodes = 2

= 20 cm

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MAJOR TEST

40. Released energy Q = mc2 in the reaction from

eq. conservation of linear momentum

1 2 1 2O P P P P

from K = 2P

2m also K

1 = K

2

so KE of one point

221 M mc

v2 2 2

v = c 2 m

M

42. If initial velocity is zero then particle moves

along the direction of total acceleration.

43. Acceleration of block, a

= F mg

m

1a F g

m

From graph ; slope = 1 1

m 2kgm 2

and

y-intercept; – g = –2 = 0.2

45. From conservation of linear momentum

1 15m mp p

so 1 15m m

91. NCERT-XIth Pg. # 18, 19

95. Water potential (w) = solute potential (s)

[At atmospheric pressure ]

External pressure = p = 7

At external pressure w = s + p

– 15 + 7 = – 8 pascals

(w) = (w)

[]

= p = 7

w = s + p

– 15 + 7 = – 8 NCERT XI Pg. # E-179 (water potential)

H-180 97. NCERT - XII : Pg. # 60

105. NCERT XI Pg. # E-198 (topic = boron )

H-198 =

Boron is required for pollen germination i.e.

pollen tube formation and pollen tube transfers

the male gametes to female gamete in flowering

plants.

106. NCERT pg. # 275

107. NCERT - XII : Pg. # 59

110. NCERT - XIIth : Pg. # 152 (E), 164 (H).

111. NCERT-XIth Pg. # 33, Para 3.1.3

113. NCERT Pg. # 115, 110

115. NCERT XI Pg. # E-202 (4th para)

H-202 116. NCERT Pg. # 275

117. NCERT - XII : Pg. # 60

118. NCERT - XII : Pg. # 27 (E), 28 (H)

120. NCERT XII Pg. # 281

121. NCERT-XIth Pg. # 32


125. NCERT XI Pg. # E-213

H-213

126. NCERT Pg. # 285

127. NCERT - XII : Pg. # 57-62


134. NCERT - XIth : Pg. # 149 Para. 4


H-236,237

140. NCERT XII Pg. # 276–281


143. NCERT XI Pg. # 132


145. NCERT XI Pg. # E-154 (1st para)

H-154 147. NCERT - XII : Pg. # 169




(Topic = photoperiodism)

H-251 (= )157. NCERT - XII : Pg. # 165, 168

159. NCERT - XIIth : Pg. # 150 (E), 162 (H).



164. NCERT - XIIth : Pg. # 103 Para. 4

166. NCERT 11th Pg. # 326 (E)

167. NCERT - XII : Pg. # 25 (E), 27 (H)

169. NCERT - XIIth : Pg. # 151 (E), 162 (H).

170. NCERT XII Pg. # 266-267

173. NCERT XII Pg.# 138,139

174. NCERT - XIIth : Pg. # 104 Para. 1

176. NCERT - XII : Pg. # 60

177. NCERT - XII : Pg. # 25 (E), 26 (H)

answer key 5

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