JEE Physics System Of Particles and Rotational Motion MCQs Set 37

Answer: (a) \( \frac{v}{3R} \)

Explanation:
1. The system is free, so it rotates about its centre of mass. The centre of mass lies at distance \( R/2 \) from the hoop centre towards the bullet.
2. Angular momentum about the centre of mass: \( L=mv\left(R-\frac R2\right)=\frac{mvR}{2} \).
3. \( I_{cm}=\left[mR^2+m\frac{R^2}{4}\right]+m\frac{R^2}{4}=\frac32mR^2 \). So \( \omega=\frac{mvR/2}{(3/2)mR^2}=\frac{v}{3R} \).

Teacher's Note:
1. F
2. o
3. r
4.
5. a
6.
7. f
8. r
9. e
10. e
11.
12. b
13. o
14. d
15. y
16.
17. u
18. s
19. e
20.
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63.
64. R
65. e
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75.
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88. p
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96.
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101.
102. s
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107.
108. f
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112. b
113. o
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116.
117. t
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120.
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122. o
123. o
124. p
125.
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127. n
128. d
129.
130. t
131. h
132. e
133.
134. e
135. m
136. b
137. e
138. d
139. d
140. e
141. d
142.
143. b
144. u
145. l
146. l
147. e
148. t
149. .

Answer: (b) \( \frac{MR^2}{2} \)

Explanation:
1. For a full ring, the moments about two perpendicular diameters are equal and each equals \( \frac12(MR^2) \). The semicircle has half the mass and half the moment: \( I_x=I_y=\frac{MR^2}{2} \).
2. Both axes are in the plane, so for an axis at angle \( \theta \): \( I=I_x\cos^2\theta+I_y\sin^2\theta=\frac{MR^2}{2} \). It is independent of \( \theta \).

Teacher's Note:
1. P
2. l
3. a
4. n
5. a
6. r
7.
8. b
9. o
10. d
11. i
12. e
13. s
14.
15. w
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19.
20. e
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24. l
25.
26. m
27. o
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34. a
35. b
36. o
37. u
38. t
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40. t
41. w
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43.
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45. e
46. r
47. p
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50. d
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54. l
55. a
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58. i
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60. -
61. p
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73. a
74. v
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78. h
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80.
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85.
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92.
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94. b
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98.
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101. e
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107. -
108. p
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113.
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115. x
116. i
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118.
119. t
120. h
121. r
122. o
123. u
124. g
125. h
126.
127. t
128. h
129. a
130. t
131.
132. p
133. o
134. i
135. n
136. t
137. .

Answer: (c) \( 4MR^2 \)

Explanation:
1. The removed disc has area one ninth of the whole, so its mass is M.
2. \( I_{whole}=\frac12(9M)R^2=\frac92MR^2 \). \( I_{hole}=\frac12M\left(\frac R3\right)^2+M\left(\frac{2R}{3}\right)^2=\frac{MR^2}{18}+\frac{4MR^2}{9}=\frac{MR^2}{2} \).
3. Remaining: \( \frac92MR^2-\frac12MR^2=4MR^2 \).

Teacher's Note:
1. F
2. o
3. r
4.
5. a
6.
7. c
8. u
9. t
10. -
11. o
12. u
13. t
14. ,
15.
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17. u
18. b
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25. t
26. h
27. e
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29. m
30. o
31. m
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34. t
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61.
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63. i
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65. c
66. e
67. ,
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70. a
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72. e
73. n
74.
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76. b
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78. u
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80.
81. t
82. h
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84.
85. s
86. a
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118. t
119. h
120. e
121. o
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124. m
125. .

Answer: (d) All four are correct

Explanation:
1. (I) Two rods along x at distance \( l/2 \): \( 2m\frac{l^2}{4} \); two rods perpendicular to the axis: \( 2\cdot\frac{ml^2}{12} \). Sum \( =\frac23ml^2 \).
2. (II) Each rod: \( \frac{ml^2}{12}+m\frac{l^2}{4}=\frac{ml^2}{3} \); four rods \( =\frac43ml^2 \). (III) Corner axis: \( \frac43ml^2+4m\cdot\frac{l^2}{2}=\frac{10}{3}ml^2 \).
3. (IV) About one side: the opposite rod gives \( ml^2 \), the two perpendicular rods give \( 2\cdot\frac{ml^2}{3} \), the rod on the axis gives 0. Total \( \frac53ml^2 \). All statements are correct.

Teacher's Note:
1. B
2. u
3. i
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5. d
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7. a
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9. c
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133.
134. \
135. (
136.
137. I
138. _
139. z
140. =
141. I
142. _
143. x
144. +
145. I
146. _
147. y
148.
149. \
150. )
151. .

Answer: (a) \( \frac{mv_0^3}{4\sqrt2\,g} \)

Explanation:
1. At the highest point the velocity is horizontal: \( v_x=v_0\cos45^\circ=\frac{v_0}{\sqrt2} \), and the height is \( H=\frac{v_0^2\sin^2 45^\circ}{2g}=\frac{v_0^2}{4g} \).
2. The perpendicular distance of the velocity line from the point of projection is H, so \( L=mv_xH=m\frac{v_0}{\sqrt2}\cdot\frac{v_0^2}{4g}=\frac{mv_0^3}{4\sqrt2\,g} \).

Teacher's Note:
1. A
2. n
3. g
4. u
5. l
6. a
7. r
8.
9. m
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11. m
12. e
13. n
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17.
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24. a
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30. e
31.
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34.
35. m
36. a
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41. i
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47. p
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64. r
65. p
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68. d
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70. c
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73. a
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76. d
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112. t
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116. r
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120. r
121. e
122. n
123. c
124. e
125.
126. p
127. o
128. i
129. n
130. t
131. .

Answer: (b) \( \frac{v_0^2}{7g} \)

Explanation:
1. Friction \( \mu mg \) acts forward: \( a=\mu g=\frac{2g}{7} \). Its torque gives \( \alpha=\frac{\mu mgR}{\frac25mR^2}=\frac{5\mu g}{2R} \).
2. Pure rolling on the moving belt needs the contact point to move with the belt: \( v+\omega R=v_0 \), i.e. \( at+\frac52at=v_0 \), so \( t=\frac{2v_0}{7a}=\frac{v_0}{g} \).
3. Distance \( =\frac12at^2=\frac12\cdot\frac{2g}{7}\cdot\frac{v_0^2}{g^2}=\frac{v_0^2}{7g} \).

Teacher's Note:
1. T
2. h
3. e
4.
5. c
6. o
7. n
8. d
9. i
10. t
11. i
12. o
13. n
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15. f
16. o
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20. u
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24. r
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101. o
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111.
112. t
113. h
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118. t
119.
120. i
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122.
123. a
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125.
126. r
127. e
128. s
129. t
130. .

Answer: (c) \( \frac{m\omega^2}{2l}\left(l^2-x^2\right) \)

Explanation:
1. The tension at x supplies the centripetal force for the part of the rod beyond x.
2. \( T=\int_x^l\frac ml\,\omega^2r\,dr=\frac{m\omega^2}{2l}\left(l^2-x^2\right) \).

Teacher's Note:
1. T
2. a
3. k
4. e
5.
6. a
7.
8. s
9. m
10. a
11. l
12. l
13.
14. e
15. l
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19. n
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27. e
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32.
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54. n
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56.
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70.
71. o
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75.
76. e
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78. e
79. r
80. y
81. t
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83. i
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86.
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107.
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111. e
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115. t
116. .

Answer: (d) II, III and IV only

Explanation:
1. About O the angular momentum has a horizontal component that rotates with the ball, so the vector changes direction: (I) is false. Its magnitude and vertical component stay the same: (III) and (IV) are true.
2. About C, \( \vec r \) and \( \vec v \) are both horizontal and the speed is constant, so \( \vec L=mr^2\omega \) along the vertical is constant: (II) is true.

Teacher's Note:
1. A
2.
3. v
4. e
5. c
6. t
7. o
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9.
10. i
11. s
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73.
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160.
161. p
162. a
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165.
166. o
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168.
169. L
170. .

Answer: (a) I and IV only

Explanation:
1. The velocity of P is \( u \) (translation) plus \( u \) tangential (rotation). If CP makes angle \( \beta \) with the downward vertical, \( v=2u\sin\frac\beta2 \), which ranges from 0 to 2u: (I) true.
2. CP horizontal: \( \beta=90^\circ \), \( v=2u\sin45^\circ=\sqrt2u \): (II) false, (IV) true.
3. CP at 60 degrees below horizontal: \( \beta=30^\circ \), \( v=2u\sin15^\circ\ne u \): (III) false. (v = u occurs at \( \beta=60^\circ \).)

Teacher's Note:
1. E
2. v
3. e
4. r
5. y
6.
7. p
8. o
9. i
10. n
11. t
12.
13. o
14. n
15.
16. a
17.
18. r
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21. l
22. i
23. n
24. g
25.
26. b
27. o
28. d
29. y
30.
31. m
32. o
33. v
34. e
35. s
36.
37. a
38. b
39. o
40. u
41. t
42.
43. t
44. h
45. e
46.
47. c
48. o
49. n
50. t
51. a
52. c
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55. p
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58. n
59. t
60. ,
61.
62. s
63. o
64.
65. i
66. t
67. s
68.
69. s
70. p
71. e
72. e
73. d
74.
75. i
76. s
77.
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79. r
80. o
81. p
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83. r
84. t
85. i
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90.
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92. o
93.
94. i
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98. d
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102. a
103. n
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120. a
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124. p
125. o
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127. n
128. t
129. .

Answer: (b) first increases and then decreases

Explanation:
1. No external torque acts, so \( L=I\omega \) is constant.
2. As the insect moves towards the centre, I decreases, so \( \omega \) increases. After crossing the centre it moves outwards, I increases, so \( \omega \) decreases again.

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

Answer: (c) \( \sqrt{(a+r\alpha)^2+(r\omega^2)^2} \)

Explanation:
1. Acceleration of P = acceleration of O + acceleration of P relative to O.
2. Relative to O: tangential \( r\alpha \) (horizontal, forward for a point above the centre of a clockwise-rotating disc) and centripetal \( r\omega^2 \) (vertical, towards O).
3. Horizontal: \( a+r\alpha \); vertical: \( r\omega^2 \). Magnitude \( =\sqrt{(a+r\alpha)^2+(r\omega^2)^2} \).

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

Answer: (d) 10/3 m/s

Explanation:
1. The jerk is impulsive and the only external impulse is at the hinge, so angular momentum about the pulley centre is conserved.
2. \( m(5)r=mvr+\frac12mr^2\cdot\frac vr \), so \( 5=\frac32v \) and \( v=\frac{10}{3} \) m/s.

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

Answer: (a) \( \sqrt{gR} \)

Explanation:
1. At the top the particle stays on the surface only if \( mg-N=\frac{mv^2}{R} \) with \( N\ge0 \).
2. It leaves immediately when \( N=0 \), i.e. \( v=\sqrt{gR} \). Any speed at least this large makes it leave at once.

Teacher's Note:
1. A
2.
3. b
4. o
5. d
6. y
7.
8. l
9. e
10. a
11. v
12. e
13. s
14.
15. a
16.
17. c
18. o
19. n
20. v
21. e
22. x
23.
24. s
25. u
26. r
27. f
28. a
29. c
30. e
31.
32. a
33. t
34.
35. t
36. h
37. e
38.
39. p
40. o
41. i
42. n
43. t
44.
45. w
46. h
47. e
48. r
49. e
50.
51. t
52. h
53. e
54.
55. n
56. o
57. r
58. m
59. a
60. l
61.
62. r
63. e
64. a
65. c
66. t
67. i
68. o
69. n
70.
71. b
72. e
73. c
74. o
75. m
76. e
77. s
78.
79. z
80. e
81. r
82. o
83. .

Answer: (b) \( \frac{r\omega_0}{2} \)

Explanation:
1. Friction acts at the contact point, so angular momentum about the contact point is conserved.
2. \( mr^2\omega_0=mvr+mr^2\omega \) with \( v=\omega r \): \( mr^2\omega_0=2mr^2\omega \), so \( \omega=\frac{\omega_0}{2} \).
3. Velocity of centre \( =\omega r=\frac{r\omega_0}{2} \).

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

Answer: (c) \( \frac72\mu mg \)

Explanation:
1. \( F-f=ma \) and \( fR=\frac25mR^2\cdot\frac aR \), so \( f=\frac25ma \).
2. Then \( F=\frac75ma=\frac72f \), i.e. \( f=\frac27F \).
3. No slipping needs \( f\le\mu mg \), so \( F\le\frac72\mu mg \).

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

Answer: (d) changes in direction but not in magnitude

Explanation:
1. The horizontal component of angular momentum rotates with the bob, so the direction keeps changing.
2. The speed and the geometry stay fixed, so the magnitude is constant.

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

Answer: (a) 2/3

Explanation:
1. Angular momentum is conserved: \( \frac12mr^2\omega_0=\left(\frac12+1\right)mr^2\omega \), so \( \omega=\frac{\omega_0}{3} \).
2. Final KE \( =\frac12\cdot\frac32mr^2\cdot\frac{\omega_0^2}{9}=\frac13\left(\frac12\cdot\frac12mr^2\omega_0^2\right) \). One third remains, so two thirds is lost as heat.

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

Answer: (b) more than 3 but less than 6

Explanation:
1. \( \alpha=\frac{FR}{I}=\frac{2(20t-5t^2)}{10}=4t-t^2 \). Starting from rest, \( \omega=2t^2-\frac{t^3}{3} \).
2. \( \omega=0 \) again at \( t=6 \) s, when the direction reverses.
3. \( \theta=\int_0^6\omega\,dt=\left[\frac{2t^3}{3}-\frac{t^4}{12}\right]_0^6=144-108=36 \) rad \( =5.7 \) rotations.

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

Answer: (c) \( \frac g2 \)

Explanation:
1. \( mg-T=ma \) and \( TR=mR^2\cdot\frac aR \), so \( T=ma \).
2. \( mg=2ma \), so \( a=\frac g2 \).

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

Answer: (d) \( \frac{m_2L}{m_1+m_2} \)

Explanation:
1. Work equals \( \frac12I\omega_0^2 \), so I must be minimum: \( I=m_1x^2+m_2(L-x)^2 \).
2. \( \frac{dI}{dx}=2m_1x-2m_2(L-x)=0 \) gives \( x=\frac{m_2L}{m_1+m_2} \), the centre of mass position.

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

Answer: (a) 6.66 kg m² s⁻²

Explanation:
1. \( v=54\ \text{km/h}=15 \) m/s, so \( \omega=\frac{15}{0.45}=33.3 \) rad/s.
2. \( \alpha=\frac{33.3}{15}=2.22 \) rad/s², and \( \tau=I\alpha=3\times2.22=6.66 \) kg m² s⁻².

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

Answer: (b) 7

Explanation:
1. For ring radius R the triangle side is \( \sqrt3R \). Each rod: \( \frac{m(\sqrt3R)^2}{12}+m\left(\frac R2\right)^2=\frac{mR^2}{2} \), so \( I_{cm}=\frac32mR^2 \) and \( k^2=\frac{R^2}{2} \).
2. For rolling: \( f=\frac{Mg\sin\theta\,(k^2/R^2)}{1+k^2/R^2}=\frac{Mg\sin\theta}{3} \), and \( N=Mg\cos\theta \).
3. \( \mu_{min}=\frac{f}{N}=\frac13\tan\theta \), so \( x=1,\ y=3 \) and \( x+2y=7 \).

Teacher's Note:
1. T
2. h
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4.
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69.
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81.
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90. m
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115. R
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126. .

Answer: (c) 7

Explanation:
1. Rolling: \( v_0^2=\frac{2gh}{1+k^2/R^2} \). Sliding: \( v^2=2gh \).
2. \( \frac{v^2}{v_0^2}=1+\frac{k^2}{R^2}=\frac{25}{16} \), so \( \frac{k^2}{R^2}=\frac{9}{16} \) and \( k=\frac{3R}{4} \).
3. \( x=3,\ y=4 \), so \( x+y=7 \).

Teacher's Note:
1. T
2. h
3. e
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81. .

Answer: (d) 2

Explanation:
1. Height of P above the bottom is R, so the ball has descended \( 8R-R=7R \) to reach P. For a solid ball \( \frac12mv^2\left(1+\frac25\right)=mg(7R) \), so \( v^2=10gR \).
2. At P the only horizontal force is the normal reaction, which gives centripetal force: \( N=\frac{mv^2}{R}=10mg=10\ \text{N} \).
3. \( 5x=10 \), so \( x=2 \).

Teacher's Note:
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134. .

Answer: (a) 3

Explanation:
1. Toppling about the far base corner: \( F\cdot h=Mg\cdot\frac a2 \), where the height \( h=\frac{\sqrt3}{2}a \). So \( F=\frac{Mg}{\sqrt3} \).
2. No slipping: friction \( \mu Mg\ge F=\frac{Mg}{\sqrt3} \), so \( \mu\ge\frac1{\sqrt3} \).
3. Hence \( x=3 \).

Teacher's Note:
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2. i
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110. .

Answer: (b) 3

Explanation:
1. Torque about P: the vertical limb passes through P (zero); the horizontal limb gives \( mg\frac l2 \).
2. \( I_P=\frac{ml^2}{3}+\left[\frac{ml^2}{12}+m\left(l^2+\frac{l^2}{4}\right)\right]=\frac53ml^2 \).
3. \( \alpha=\frac{mgl/2}{5ml^2/3}=\frac{3g}{10l} \), so \( x=3 \).

Teacher's Note:
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137. .

Answer: (c) 1

Explanation:
1. For rolling, the friction needed is \( f=\frac13mg\sin\theta \) and \( N=mg\cos\theta \), so the required \( \mu=\frac{\tan\theta}{3}=\frac{\sqrt3}{3}=\frac1{\sqrt3} \).
2. Slipping starts when \( \frac{2-3x}{\sqrt3}=\frac{1}{\sqrt3} \), so \( x=\frac13 \) m. Hence \( P=1 \).

Teacher's Note:
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88. .

Answer: (d) 7

Explanation:
1. No rotation means the line of F passes through the centre of mass. Take masses proportional to length: bar mass \( \lambda l \), stem mass \( 2\lambda l \).
2. Measured down from AB, the centre of mass is at \( \frac{(2\lambda l)(l)}{3\lambda l}=\frac{2l}{3} \).
3. Distance from C \( =2l-\frac{2l}{3}=\frac43l \), so \( x=4,\ y=3 \) and \( x+y=7 \).

Teacher's Note:
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95. .

Answer: (a) 5

Explanation:
1. Central disc: \( \frac12MR^2+MR^2=\frac32MR^2 \).
2. The six outer centres are at distance 2R from the middle. Their squared distances from P are \( R^2(1,3,7,9,7,3) \), total \( 30R^2 \). Outer discs: \( 6\cdot\frac12MR^2+30MR^2=33MR^2 \).
3. Total \( =34.5MR^2=\frac{69}{2}MR^2 \), so \( 12x+9=69 \) and \( x=5 \).

Teacher's Note:
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