JEE Physics Motion In A Plane MCQs Set 22

Answer: (a) 50 km/h

Explanation:
1. Let the train velocity be \( (x,y) \). For my velocity 10 east, the relative velocity \( (x-10,\,y) \) is north-east, so \( x-10=y \).
2. For my velocity 30 east, the relative velocity \( (x-30,\,y) \) is north, so \( x=30 \). Hence \( y=20 \), and the train velocity is \( (30,20) \) km/h.
3. If I move north with speed s, the relative velocity is \( (30,\,20-s) \). For south-east the components must be equal and opposite: \( 20-s=-30 \), so \( s=50 \) km/h.

Teacher's Note:
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Answer: (b) \( \vec A\cdot\vec B\le 0 \)

Explanation:
1. Square both sides: \( A^2+B^2+2\vec A\cdot\vec B=A^2 \).
2. So \( \vec A\cdot\vec B=-\tfrac12B^2 \).
3. Since \( B^2\ge0 \), the dot product is never positive: \( \vec A\cdot\vec B\le0 \). It is zero only when \( \vec B=0 \), so B need not be zero, perpendicular or anti-parallel.

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Answer: (c) \( \vec a\times\vec b=0 \) and \( |\vec a|>|\vec b| \) must both be true

Explanation:
1. Same direction means \( \vec a+\vec b=k(\vec a-\vec b) \) with \( k>0 \), so \( (1-k)\vec a=-(1+k)\vec b \). Hence \( \vec a\) and \( \vec b \) are parallel and \( \vec a\times\vec b=0 \).
2. Write \( \vec a=\lambda\vec b \). Then \( \vec a+\vec b=(\lambda+1)\vec b \) and \( \vec a-\vec b=(\lambda-1)\vec b \); they point the same way only if \( (\lambda+1)(\lambda-1)>0 \), i.e. \( |\lambda|>1 \).
3. So \( |\vec a|>|\vec b| \) as well. Both conditions hold together.

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Answer: (d) \( \hat a\times\hat c=x(\hat b\times\hat c) \)

Explanation:
1. Snell's law with relative index x gives \( \sin i=x\sin r \).
2. The magnitudes are \( |\hat a\times\hat c|=\sin i \) and \( |\hat b\times\hat c|=\sin r \).
3. Incident and refracted rays bend on the same side of the normal, so both cross products have the same direction. Hence \( \hat a\times\hat c=x(\hat b\times\hat c) \).

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Answer: (a) \( A=\frac{2C}{\sqrt3+1} \)

Explanation:
1. Components perpendicular to C must cancel: \( A\sin30^\circ=B\sin45^\circ \), so \( A=\sqrt2\,B \).
2. Components along C add up to C: \( A\cos30^\circ+B\cos45^\circ=C \), i.e. \( \sqrt2B\cdot\frac{\sqrt3}{2}+\frac{B}{\sqrt2}=C \).
3. \( B=\frac{\sqrt2\,C}{\sqrt3+1} \) and therefore \( A=\sqrt2B=\frac{2C}{\sqrt3+1}=(\sqrt3-1)C \).

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Answer: (b) \( \frac{3a^2+b^2-c^2}{2} \)

Explanation:
1. \( \overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD} \) gives \( c^2=a^2+b^2+2\,\overrightarrow{AB}\cdot\overrightarrow{AD} \), so \( \overrightarrow{AB}\cdot\overrightarrow{AD}=\frac{c^2-a^2-b^2}{2} \).
2. \( \overrightarrow{DB}=\overrightarrow{AB}-\overrightarrow{AD} \), so \( \overrightarrow{DB}\cdot\overrightarrow{AB}=a^2-\overrightarrow{AB}\cdot\overrightarrow{AD} \).
3. \( =a^2-\frac{c^2-a^2-b^2}{2}=\frac{3a^2+b^2-c^2}{2} \).

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Answer: (c) 60°

Explanation:
1. Take P = 1 and Q = 2 with angle \( \theta \) between them. The resultant makes angle \( \alpha \) with P where \( \tan\alpha=\frac{Q\sin\theta}{P+Q\cos\theta}=\frac{2\sin\theta}{1+2\cos\theta} \).
2. Set this equal to \( \frac{\sqrt3}{2} \). Try \( \theta=60^\circ \): \( \frac{2(\sqrt3/2)}{1+1}=\frac{\sqrt3}{2} \).
3. The expression increases steadily for small angles and is uniquely satisfied there, so \( \theta=60^\circ \).

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Answer: (d) I, II and IV only

Explanation:
1. For 2, 3, 4 N: three vectors can sum to zero because 4 is less than 2 + 3, and the largest possible net force is 9 N. So the net force ranges from 0 to 9 N and the acceleration from 0 to 4.5 m/s². (I) and (IV) are possible, (III) is not (it needs 10 N).
2. For 1, 3, 5 N: the minimum net force is 5 - (1+3) = 1 N and the maximum is 9 N. An acceleration of 3 m/s² needs 6 N, which lies in the range, so (II) is possible.

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Answer: (a) \( \hat i \)

Explanation:
1. Let \( \vec b=(b_1,b_2,b_3) \). Then \( b_1+b_2+b_3=1 \).
2. \( \vec a\times\vec b=(b_3-b_2,\ b_1-b_3,\ b_2-b_1)=(0,1,-1) \), so \( b_3=b_2 \), \( b_1=1+b_2 \).
3. Substituting in the dot product: \( 1+3b_2=1 \), so \( b_2=0 \), and \( \vec b=\hat i \).

Teacher's Note:
1. W
2. r
3. i
4. t
5. e
6.
7. t
8. h
9. e
10.
11. u
12. n
13. k
14. n
15. o
16. w
17. n
18.
19. v
20. e
21. c
22. t
23. o
24. r
25.
26. i
27. n
28.
29. c
30. o
31. m
32. p
33. o
34. n
35. e
36. n
37. t
38. s
39.
40. a
41. n
42. d
43.
44. u
45. s
46. e
47.
48. b
49. o
50. t
51. h
52.
53. t
54. h
55. e
56.
57. d
58. o
59. t
60.
61. a
62. n
63. d
64.
65. c
66. r
67. o
68. s
69. s
70.
71. p
72. r
73. o
74. d
75. u
76. c
77. t
78.
79. c
80. o
81. n
82. d
83. i
84. t
85. i
86. o
87. n
88. s
89. .
90.
91. T
92. h
93. e
94.
95. d
96. o
97. t
98.
99. p
100. r
101. o
102. d
103. u
104. c
105. t
106.
107. g
108. i
109. v
110. e
111. s
112.
113. o
114. n
115. e
116.
117. e
118. q
119. u
120. a
121. t
122. i
123. o
124. n
125.
126. a
127. n
128. d
129.
130. t
131. h
132. e
133.
134. c
135. r
136. o
137. s
138. s
139.
140. p
141. r
142. o
143. d
144. u
145. c
146. t
147.
148. g
149. i
150. v
151. e
152. s
153.
154. t
155. h
156. e
157.
158. r
159. e
160. s
161. t
162. .

Answer: (b) \( \theta=53^\circ \), time 6 min

Explanation:
1. To go from A to C the resultant velocity must be at 45 degrees to the bank (AB = BC), so its components along and across the river are equal: \( 5\cos\theta+1=5\sin\theta \).
2. For \( \theta=53^\circ \): \( 5(0.6)+1=4=5(0.8) \). So \( \theta=53^\circ \).
3. Time \( =\frac{AB}{5\sin\theta}=\frac{0.4\ \text{km}}{4\ \text{km/h}}=0.1 \) h \( =6 \) min.

Teacher's Note:
1. T
2. h
3. e
4.
5. s
6. w
7. i
8. m
9. m
10. e
11. r
12. '
13. s
14.
15. r
16. e
17. s
18. u
19. l
20. t
21. a
22. n
23. t
24.
25. v
26. e
27. l
28. o
29. c
30. i
31. t
32. y
33.
34. m
35. u
36. s
37. t
38.
39. p
40. o
41. i
42. n
43. t
44.
45. a
46. l
47. o
48. n
49. g
50.
51. A
52. C
53. .
54.
55. E
56. q
57. u
58. a
59. l
60.
61. c
62. o
63. m
64. p
65. o
66. n
67. e
68. n
69. t
70. s
71.
72. a
73. l
74. o
75. n
76. g
77.
78. a
79. n
80. d
81.
82. a
83. c
84. r
85. o
86. s
87. s
88.
89. t
90. h
91. e
92.
93. r
94. i
95. v
96. e
97. r
98.
99. i
100. s
101.
102. t
103. h
104. e
105.
106. k
107. e
108. y
109.
110. c
111. o
112. n
113. d
114. i
115. t
116. i
117. o
118. n
119. ;
120.
121. t
122. h
123. e
124.
125. t
126. i
127. m
128. e
129.
130. d
131. e
132. p
133. e
134. n
135. d
136. s
137.
138. o
139. n
140. l
141. y
142.
143. o
144. n
145.
146. t
147. h
148. e
149.
150. c
151. r
152. o
153. s
154. s
155. -
156. s
157. t
158. r
159. e
160. a
161. m
162.
163. s
164. p
165. e
166. e
167. d
168. .

Answer: (c) \( \vec a\times\vec b=\vec b\times\vec c=\vec c\times\vec a\ne0 \)

Explanation:
1. Since \( \vec c=-\vec a-\vec b \): \( \vec a\times\vec b \) and \( \vec b\times\vec c=\vec b\times(-\vec a-\vec b)=\vec a\times\vec b \).
2. Similarly \( \vec c\times\vec a=(-\vec a-\vec b)\times\vec a=\vec a\times\vec b \). So all three are equal.
3. The vectors are at 120 degrees to each other, so the common value has magnitude \( \sin120^\circ\ne0 \).

Teacher's Note:
1. R
2. e
3. p
4. l
5. a
6. c
7. e
8.
9. o
10. n
11. e
12.
13. v
14. e
15. c
16. t
17. o
18. r
19.
20. b
21. y
22.
23. m
24. i
25. n
26. u
27. s
28.
29. t
30. h
31. e
32.
33. s
34. u
35. m
36.
37. o
38. f
39.
40. t
41. h
42. e
43.
44. o
45. t
46. h
47. e
48. r
49.
50. t
51. w
52. o
53.
54. a
55. n
56. d
57.
58. s
59. i
60. m
61. p
62. l
63. i
64. f
65. y
66. ;
67.
68. t
69. h
70. e
71.
72. c
73. r
74. o
75. s
76. s
77.
78. p
79. r
80. o
81. d
82. u
83. c
84. t
85.
86. o
87. f
88.
89. a
90.
91. v
92. e
93. c
94. t
95. o
96. r
97.
98. w
99. i
100. t
101. h
102.
103. i
104. t
105. s
106. e
107. l
108. f
109.
110. v
111. a
112. n
113. i
114. s
115. h
116. e
117. s
118. .
119.
120. T
121. h
122. r
123. e
124. e
125.
126. u
127. n
128. i
129. t
130.
131. v
132. e
133. c
134. t
135. o
136. r
137. s
138.
139. w
140. i
141. t
142. h
143.
144. z
145. e
146. r
147. o
148.
149. s
150. u
151. m
152.
153. f
154. o
155. r
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157.
158. a
159. n
160.
161. e
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163. u
164. i
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166. a
167. t
168. e
169. r
170. a
171. l
172.
173. a
174. r
175. r
176. a
177. n
178. g
179. e
180. m
181. e
182. n
183. t
184. ,
185.
186. s
187. o
188.
189. t
190. h
191. e
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193.
194. a
195. r
196. e
197.
198. n
199. o
200. t
201.
202. p
203. a
204. r
205. a
206. l
207. l
208. e
209. l
210. .

Answer: (d) -3/2

Explanation:
1. Square the sum: \( |\vec a+\vec b+\vec c|^2=0 \).
2. \( 1+1+1+2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=0 \).
3. So the required sum is \( -\tfrac32 \).

Teacher's Note:
1. S
2. q
3. u
4. a
5. r
6. i
7. n
8. g
9.
10. a
11.
12. v
13. e
14. c
15. t
16. o
17. r
18.
19. e
20. q
21. u
22. a
23. t
24. i
25. o
26. n
27.
28. i
29. s
30.
31. a
32.
33. q
34. u
35. i
36. c
37. k
38.
39. w
40. a
41. y
42.
43. t
44. o
45.
46. o
47. b
48. t
49. a
50. i
51. n
52.
53. d
54. o
55. t
56.
57. p
58. r
59. o
60. d
61. u
62. c
63. t
64.
65. r
66. e
67. l
68. a
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70. i
71. o
72. n
73. s
74. .
75.
76. T
77. h
78. e
79.
80. s
81. a
82. m
83. e
84.
85. t
86. r
87. i
88. c
89. k
90.
91. w
92. o
93. r
94. k
95. s
96.
97. f
98. o
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100.
101. a
102. n
103. y
104.
105. n
106. u
107. m
108. b
109. e
110. r
111.
112. o
113. f
114.
115. v
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117. c
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122.
123. w
124. i
125. t
126. h
127.
128. z
129. e
130. r
131. o
132.
133. s
134. u
135. m
136. .

Answer: (a) \( 5(\hat i+\hat j) \)

Explanation:
1. A vector has the same magnitude in every coordinate system. Here \( |\vec A|=\sqrt{9+16}=5 \).
2. Check the options: \( |5\hat i|=5 \), \( |5\hat j|=5 \), \( \left|\tfrac52(\hat i+\sqrt3\hat j)\right|=\tfrac52\cdot2=5 \), but \( |5(\hat i+\hat j)|=5\sqrt2 \).
3. So \( 5(\hat i+\hat j) \) is impossible.

Teacher's Note:
1. C
2. h
3. a
4. n
5. g
6. i
7. n
8. g
9.
10. t
11. h
12. e
13.
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15. o
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18. d
19. i
20. n
21. a
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23. e
24.
25. a
26. x
27. e
28. s
29.
30. c
31. h
32. a
33. n
34. g
35. e
36. s
37.
38. c
39. o
40. m
41. p
42. o
43. n
44. e
45. n
46. t
47. s
48.
49. b
50. u
51. t
52.
53. n
54. e
55. v
56. e
57. r
58.
59. t
60. h
61. e
62.
63. m
64. a
65. g
66. n
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68. t
69. u
70. d
71. e
72. .
73.
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75. o
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79. r
80. e
81.
82. m
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87. t
88. u
89. d
90. e
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92.
93. f
94. i
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96. s
97. t
98. .

Answer: (b) 14 and 2

Explanation:
1. \( \vec d=-(\vec a+\vec b+\vec c) \), so \( |\vec d| \) is the magnitude of the sum of three vectors.
2. Case 1: the maximum is \( 3+5+6=14 \) (all parallel).
3. Case 2: the largest vector 7 exceeds the sum of the other two (1 + 4 = 5), so the minimum is \( 7-5=2 \).

Teacher's Note:
1. T
2. h
3. e
4.
5. r
6. e
7. s
8. u
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17.
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25.
26. v
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30. o
31. r
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33.
34. i
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36.
37. g
38. r
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40. a
41. t
42. e
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45.
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47. h
48. e
49. n
50.
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52. h
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55.
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57. r
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59.
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63.
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65. a
66. r
67. a
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69. l
70. e
71. l
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73.
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75. t
76.
77. c
78. a
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80.
81. b
82. e
83.
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85. e
86. r
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93.
94. i
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98. h
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100.
101. l
102. a
103. r
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108.
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110. a
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112. n
113. i
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115. u
116. d
117. e
118.
119. d
120. o
121. e
122. s
123.
124. n
125. o
126. t
127.
128. e
129. x
130. c
131. e
132. e
133. d
134.
135. t
136. h
137. e
138.
139. s
140. u
141. m
142.
143. o
144. f
145.
146. t
147. h
148. e
149.
150. r
151. e
152. s
153. t
154. .

Answer: (c) 1, 2, 1

Explanation:
1. Three vectors can add to zero if the largest magnitude does not exceed the sum of the other two.
2. 2, 4, 8: 8 > 6. 4, 8, 16: 16 > 12. 0.5, 1, 2: 2 > 1.5. None of these works.
3. 1, 2, 1: 2 = 1 + 1, which is possible when the two unit vectors are parallel and the third is opposite.

Teacher's Note:
1. T
2. h
3. e
4.
5. t
6. r
7. i
8. a
9. n
10. g
11. l
12. e
13.
14. i
15. n
16. e
17. q
18. u
19. a
20. l
21. i
22. t
23. y
24. ,
25.
26. w
27. i
28. t
29. h
30.
31. e
32. q
33. u
34. a
35. l
36. i
37. t
38. y
39.
40. a
41. l
42. l
43. o
44. w
45. e
46. d
47. ,
48.
49. i
50. s
51.
52. t
53. h
54. e
55.
56. t
57. e
58. s
59. t
60. .
61.
62. E
63. q
64. u
65. a
66. l
67. i
68. t
69. y
70.
71. c
72. o
73. r
74. r
75. e
76. s
77. p
78. o
79. n
80. d
81. s
82.
83. t
84. o
85.
86. a
87. l
88. l
89.
90. t
91. h
92. r
93. e
94. e
95.
96. v
97. e
98. c
99. t
100. o
101. r
102. s
103.
104. l
105. y
106. i
107. n
108. g
109.
110. o
111. n
112.
113. o
114. n
115. e
116.
117. l
118. i
119. n
120. e
121. .

Answer: (d) 1/4 s

Explanation:
1. Perpendicular means \( \vec a\cdot\vec b=0 \): \( 8t\cdot t^2-t^2\cdot2=0 \).
2. \( 8t^3-2t^2=0 \Rightarrow 2t^2(4t-1)=0 \).
3. For t > 0 this gives \( t=\tfrac14 \) s. (They are never parallel for t > 0, since \( \vec a\times\vec b \) has magnitude \( 16t+t^4>0 \).)

Teacher's Note:
1. P
2. e
3. r
4. p
5. e
6. n
7. d
8. i
9. c
10. u
11. l
12. a
13. r
14.
15. n
16. e
17. e
18. d
19. s
20.
21. a
22.
23. z
24. e
25. r
26. o
27.
28. d
29. o
30. t
31.
32. p
33. r
34. o
35. d
36. u
37. c
38. t
39.
40. a
41. n
42. d
43.
44. p
45. a
46. r
47. a
48. l
49. l
50. e
51. l
52.
53. n
54. e
55. e
56. d
57. s
58.
59. a
60.
61. z
62. e
63. r
64. o
65.
66. c
67. r
68. o
69. s
70. s
71.
72. p
73. r
74. o
75. d
76. u
77. c
78. t
79. .
80.
81. D
82. i
83. s
84. c
85. a
86. r
87. d
88.
89. t
90. h
91. e
92.
93. t
94. r
95. i
96. v
97. i
98. a
99. l
100.
101. r
102. o
103. o
104. t
105.
106. t
107.
108. =
109.
110. 0
111. .

Answer: (a) \( 3\cos\theta\,\hat i+3\sin\theta\,\hat k \)

Explanation:
1. A rotation anticlockwise as seen from +y is a positive rotation about the y-axis (right-hand rule), and it takes \( \hat z \) towards \( \hat x \), so it takes \( \hat x \) towards \( -\hat z \).
2. Clockwise as seen from +y is the opposite sense, so \( \hat x \) turns towards \( +\hat z \).
3. The new vector is \( 3\cos\theta\,\hat i+3\sin\theta\,\hat k \), and it stays in the x-z plane.

Teacher's Note:
1. U
2. s
3. e
4.
5. t
6. h
7. e
8.
9. r
10. i
11. g
12. h
13. t
14. -
15. h
16. a
17. n
18. d
19.
20. r
21. u
22. l
23. e
24.
25. w
26. i
27. t
28. h
29.
30. t
31. h
32. e
33.
34. a
35. x
36. i
37. s
38.
39. o
40. f
41.
42. r
43. o
44. t
45. a
46. t
47. i
48. o
49. n
50. .
51.
52. S
53. e
54. e
55. n
56.
57. f
58. r
59. o
60. m
61.
62. t
63. h
64. e
65.
66. p
67. o
68. s
69. i
70. t
71. i
72. v
73. e
74.
75. e
76. n
77. d
78.
79. o
80. f
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82. t
83. h
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85.
86. a
87. x
88. i
89. s
90. ,
91.
92. a
93.
94. p
95. o
96. s
97. i
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99. i
100. v
101. e
102.
103. r
104. o
105. t
106. a
107. t
108. i
109. o
110. n
111.
112. i
113. s
114.
115. a
116. n
117. t
118. i
119. c
120. l
121. o
122. c
123. k
124. w
125. i
126. s
127. e
128. .

Answer: (b) 129.2 s

Explanation:
1. At time t the distances from the crossing are \( 1500-10t \) and \( 1800-15t \), so \( d^2=(1500-10t)^2+(1800-15t)^2 \).
2. For the minimum, \( \frac{d(d^2)}{dt}=0 \): \( -10(1500-10t)-15(1800-15t)=0 \), i.e. \( 325t=42000 \).
3. \( t=\frac{42000}{325}\approx129.2 \) s (both cars have still not reached the crossing at this time).

Teacher's Note:
1. M
2. i
3. n
4. i
5. m
6. i
7. s
8. e
9.
10. t
11. h
12. e
13.
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15. q
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24. t
25. h
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51.
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53. q
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57. e
58.
59. r
60. o
61. o
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63. .
64.
65. C
66. h
67. e
68. c
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70.
71. t
72. h
73. a
74. t
75.
76. b
77. o
78. t
79. h
80.
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95.
96. a
97. p
98. p
99. r
100. o
101. a
102. c
103. h
104. i
105. n
106. g
107.
108. t
109. h
110. e
111.
112. c
113. r
114. o
115. s
116. s
117. i
118. n
119. g
120.
121. a
122. t
123.
124. t
125. h
126. e
127.
128. a
129. n
130. s
131. w
132. e
133. r
134. .

Answer: (c) \( \frac{1}{\sqrt3}(-\hat i+\hat j-\hat k) \)

Explanation:
1. \( \vec A\times\vec B=\begin{vmatrix}\hat i&\hat j&\hat k\\2&3&1\\1&0&-1\end{vmatrix}=-3\hat i+3\hat j-3\hat k \).
2. Its magnitude is \( \sqrt{27}=3\sqrt3 \).
3. The unit vector is \( \frac{1}{\sqrt3}(-\hat i+\hat j-\hat k) \) (the negative of this is also valid).

Teacher's Note:
1. A
2.
3. v
4. e
5. c
6. t
7. o
8. r
9.
10. p
11. e
12. r
13. p
14. e
15. n
16. d
17. i
18. c
19. u
20. l
21. a
22. r
23.
24. t
25. o
26.
27. t
28. w
29. o
30.
31. g
32. i
33. v
34. e
35. n
36.
37. v
38. e
39. c
40. t
41. o
42. r
43. s
44.
45. i
46. s
47.
48. a
49. l
50. o
51. n
52. g
53.
54. t
55. h
56. e
57. i
58. r
59.
60. c
61. r
62. o
63. s
64. s
65.
66. p
67. r
68. o
69. d
70. u
71. c
72. t
73. .
74.
75. D
76. i
77. v
78. i
79. d
80. e
81.
82. b
83. y
84.
85. i
86. t
87. s
88.
89. m
90. a
91. g
92. n
93. i
94. t
95. u
96. d
97. e
98.
99. t
100. o
101.
102. g
103. e
104. t
105.
106. a
107.
108. u
109. n
110. i
111. t
112.
113. v
114. e
115. c
116. t
117. o
118. r
119. .

Answer: (d) \( \frac{1}{\sqrt2} \) m/s² towards north-west

Explanation:
1. \( \Delta\vec v=\vec v_f-\vec v_i=5\hat j-5\hat i \), which has magnitude \( 5\sqrt2 \) and points north-west.
2. \( a_{avg}=\frac{5\sqrt2}{10}=\frac{1}{\sqrt2} \) m/s² towards north-west.

Teacher's Note:
1. A
2. v
3. e
4. r
5. a
6. g
7. e
8.
9. a
10. c
11. c
12. e
13. l
14. e
15. r
16. a
17. t
18. i
19. o
20. n
21.
22. i
23. s
24.
25. t
26. h
27. e
28.
29. c
30. h
31. a
32. n
33. g
34. e
35.
36. i
37. n
38.
39. v
40. e
41. l
42. o
43. c
44. i
45. t
46. y
47.
48. d
49. i
50. v
51. i
52. d
53. e
54. d
55.
56. b
57. y
58.
59. t
60. i
61. m
62. e
63. ,
64.
65. a
66. n
67. d
68.
69. t
70. h
71. e
72.
73. c
74. h
75. a
76. n
77. g
78. e
79.
80. i
81. s
82.
83. a
84.
85. v
86. e
87. c
88. t
89. o
90. r
91.
92. d
93. i
94. f
95. f
96. e
97. r
98. e
99. n
100. c
101. e
102. .
103.
104. A
105.
106. c
107. h
108. a
109. n
110. g
111. e
112.
113. i
114. n
115.
116. d
117. i
118. r
119. e
120. c
121. t
122. i
123. o
124. n
125.
126. a
127. l
128. o
129. n
130. e
131.
132. g
133. i
134. v
135. e
136. s
137.
138. a
139.
140. n
141. o
142. n
143. -
144. z
145. e
146. r
147. o
148.
149. a
150. c
151. c
152. e
153. l
154. e
155. r
156. a
157. t
158. i
159. o
160. n
161. .