Download JEE MCQs for JEE Physics: Motion In A Straight Line
Access targeted multiple-choice questions for Motion In A Straight Line designed to align with the latest JEE academic syllabus for JEE Physics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.
Chapter-wise Objective Questions: Motion In A Straight Line
Access the complete set of multiple-choice questions for Motion In A Straight Line below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.
Multiple Choice Questions
(a) 4.0 m/s
(b) 5.0 m/s
(c) 5.5 m/s
(d) 4.8 m/s
Show Answer & Explanation
Answer: (a) 4.0 m/s
Explanation:
1. Speed over the second half = average of 4.5 and 7.5 (equal time intervals) = 6 m/s.
2. Let the total distance be 2d. Time = d/3 + d/6 = d/2.
3. Average speed = 2d / (d/2) = 4.0 m/s.
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(a) \( t=\frac{u}{g},\; h=\frac{u^2}{2g}+\frac{gT^2}{8} \)
(b) \( t=\frac{u}{g}+\frac{T}{2},\; h=\frac{u^2}{2g}-\frac{gT^2}{8} \)
(c) \( t=\frac{u}{g}+\frac{T}{2},\; h=\frac{u^2}{2g}+\frac{gT^2}{8} \)
(d) They never meet
Show Answer & Explanation
Answer: (b) \( t=\frac{u}{g}+\frac{T}{2},\; h=\frac{u^2}{2g}-\frac{gT^2}{8} \)
Explanation:
1. Equal heights: \( ut-\frac{1}{2}gt^2 = u(t-T)-\frac{1}{2}g(t-T)^2 \).
2. Simplifying gives \( uT=\frac{1}{2}g(2tT-T^2) \), so \( t=\frac{u}{g}+\frac{T}{2} \).
3. Substituting back, \( h=ut-\frac{1}{2}gt^2=\frac{u^2}{2g}-\frac{gT^2}{8} \).
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(a) 2 m
(b) 3 m
(c) 1 m
(d) 5 m
Show Answer & Explanation
Answer: (c) 1 m
Explanation:
1. Between A and B: \( v^2-(v/2)^2=2g(3) \), so \( \frac{3v^2}{4}=6g \) and \( v^2=8g \).
2. Above B the stone rises \( h=\frac{(v/2)^2}{2g}=\frac{v^2}{8g} \).
3. Therefore the maximum height above B is 1 m.
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(A) He must swim making an angle of 127° with the direction of flow of the river.
(B) He must swim perpendicular to the flow of the river.
(C) The length of the shortest possible path is 200 m.
(D) The length of the shortest possible path is 120 m.
(a) B and C only
(b) A and D only
(c) B and D only
(d) A and C only
Show Answer & Explanation
Answer: (d) A and C only
Explanation:
1. Swimmer speed (3) is less than river speed (5), so the shortest path is found when the swimming direction is perpendicular to the resultant velocity.
2. Then \( \cos\theta=-\frac{3}{5} \), so \( \theta=127^\circ \) with the flow.
3. Resultant speed \( =\sqrt{5^2-3^2}=4 \) km/h; cross-stream speed \( =3\sin127^\circ=2.4 \) km/h.
4. Path length \( =120\times\frac{4}{2.4}=200 \) m. So A and C are correct.
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(a) \( 2gH=nu^2(n-2) \)
(b) \( gH=(n-2)^2u^2 \)
(c) \( 2gH=n^2u^2 \)
(d) \( gH=(n-2)\,2u^2 \)
Show Answer & Explanation
Answer: (a) \( 2gH=nu^2(n-2) \)
Explanation:
1. Time to highest point = u/g, so total time \( t=\frac{nu}{g} \).
2. Displacement is \( -H=ut-\frac{1}{2}gt^2=\frac{nu^2}{g}-\frac{n^2u^2}{2g} \).
3. Hence \( 2gH=n^2u^2-2nu^2=nu^2(n-2) \).
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(A) \( |\vec a| \)
(B) \( \frac{\vec v\cdot\vec a}{v} \)
(C) the component of \( \vec a \) parallel to \( \vec v \)
(D) the component of \( \vec a \) perpendicular to \( \vec v \)
(a) A and B only
(b) B and C only
(c) A and D only
(d) D only
Show Answer & Explanation
Answer: (b) B and C only
Explanation:
1. Speed squared: \( v^2=\vec v\cdot\vec v \), so \( 2v\frac{dv}{dt}=2\,\vec v\cdot\vec a \).
2. Hence \( \frac{dv}{dt}=\frac{\vec v\cdot\vec a}{v} \), which is option B.
3. This is the projection of \( \vec a \) along \( \vec v \), which is option C. Options A and D are not equal to dv/dt in general.
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173. .
(a) \( a_0t_0\,\ell n(2) \)
(b) \( a_0t_0 \)
(c) \( \frac{a_0t_0}{\ell n(2)} \)
(d) \( \frac{a_0t_0}{2} \)
Show Answer & Explanation
Answer: (c) \( \frac{a_0t_0}{\ell n(2)} \)
Explanation:
1. The acceleration decays as \( a=a_0\,2^{-t/t_0}=a_0e^{-(\ln 2)t/t_0} \).
2. Terminal velocity \( =\int_0^\infty a\,dt=a_0\cdot\frac{t_0}{\ln 2} \).
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130. .
(A) The average speed of a particle in a given time is never less than the magnitude of the average velocity.
(B) It is possible to have a situation in which \( \left|\frac{d\vec v}{dt}\right|\ne 0 \) but \( \frac{d}{dt}|\vec v|=0 \).
(C) The average velocity of a particle is zero in a time interval. It is possible that the instantaneous velocity is never zero in the interval.
(D) The average velocity of a particle moving on a straight line is zero in a time interval. It is possible that the instantaneous velocity is never zero in the interval (infinite accelerations are not allowed).
(a) A and B only
(b) B, C and D only
(c) C and D only
(d) A, B and C only
Show Answer & Explanation
Answer: (d) A, B and C only
Explanation:
1. A: distance is at least the magnitude of displacement, so average speed is at least |average velocity|. True.
2. B: uniform circular motion has non-zero acceleration but constant speed. True.
3. C: one full circle gives zero average velocity while the velocity is never zero. True.
4. D: on a straight line, zero displacement needs the particle to turn back, and at the turning point the velocity is zero. False.
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62. c
63. a
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65. a
66. r
67.
68. o
69. n
70. e
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72.
73. (
74. s
75. p
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77. e
78. d
79. ,
80.
81. d
82. i
83. s
84. t
85. a
86. n
87. c
88. e
89. )
90. .
91.
92. O
93. n
94. e
95. -
96. d
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100. n
101. s
102. i
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104. n
105. a
106. l
107.
108. m
109. o
110. t
111. i
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113. n
114.
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116. o
117. r
118. c
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121.
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123.
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125. e
126. v
127. e
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129. s
130. a
131. l
132.
133. t
134. o
135.
136. r
137. e
138. t
139. u
140. r
141. n
142.
143. t
144. o
145.
146. t
147. h
148. e
149.
150. s
151. t
152. a
153. r
154. t
155. ,
156.
157. b
158. u
159. t
160.
161. t
162. w
163. o
164. -
165. d
166. i
167. m
168. e
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170. s
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175. l
176.
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183.
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192. .
(A) The particle comes to rest again at x = 4.
(B) The particle oscillates about x = 2.
(C) The maximum speed of the particle is 4 units.
(a) A and B only
(b) B and C only
(c) A and C only
(d) A, B and C
Show Answer & Explanation
Answer: (a) A and B only
Explanation:
1. \( v\frac{dv}{dx}=4-2x \) gives \( \frac{v^2}{2}=4x-x^2 \) (from rest at x = 0).
2. v = 0 again at x = 4, so A is true. The acceleration is zero at x = 2, which is the centre of the motion between 0 and 4, so B is true.
3. At x = 2: \( v^2=2(8-4)=8 \), so \( v_{max}=2\sqrt2 \), not 4. C is false.
Teacher's Note:
1. T
2. h
3. e
4.
5. f
6. o
7. r
8. c
9. e
10.
11. i
12. s
13.
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48.
49. =
50.
51. 2
52. ,
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60.
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81. m
82. o
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84. i
85. c
86.
87. a
88. b
89. o
90. u
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92.
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94.
95. =
96.
97. 2
98.
99. w
100. i
101. t
102. h
103.
104. a
105. m
106. p
107. l
108. i
109. t
110. u
111. d
112. e
113.
114. 2
115. .
116.
117. M
118. a
119. x
120. i
121. m
122. u
123. m
124.
125. s
126. p
127. e
128. e
129. d
130.
131. i
132. s
133.
134. a
135. m
136. p
137. l
138. i
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140. u
141. d
142. e
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144. t
145. i
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149.
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152. g
153. u
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157.
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159. r
160. e
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163. e
164. n
165. c
166. y
167. ,
168.
169. \
170. (
171.
172. 2
173. \
174. s
175. q
176. r
177. t
178. 2
179.
180. \
181. )
182. .
(a) 1/5
(b) 3/5
(c) 2/5
(d) 4/5
Show Answer & Explanation
Answer: (b) 3/5
Explanation:
1. Let the maximum speed be v. Acceleration stage: average speed v/2, time \( \frac{2\ell}{v} \).
2. Uniform stage: time \( \frac{2\ell}{v} \). Retardation stage: average speed v/2, time \( \frac{6\ell}{v} \).
3. Total time \( \frac{10\ell}{v} \), total distance \( 6\ell \), so average speed \( =0.6v \).
Teacher's Note:
1. I
2. n
3.
4. u
5. n
6. i
7. f
8. o
9. r
10. m
11. l
12. y
13.
14. c
15. h
16. a
17. n
18. g
19. i
20. n
21. g
22.
23. m
24. o
25. t
26. i
27. o
28. n
29.
30. b
31. e
32. t
33. w
34. e
35. e
36. n
37.
38. r
39. e
40. s
41. t
42.
43. a
44. n
45. d
46.
47. v
48. ,
49.
50. t
51. h
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53.
54. a
55. v
56. e
57. r
58. a
59. g
60. e
61.
62. s
63. p
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65. e
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68. i
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72. /
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74. .
75.
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77. d
78. d
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81. h
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102.
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114. d
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117. i
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120.
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127. h
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131. o
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144. .
(a) 500 m/s
(b) 110 m/s
(c) 55 m/s
(d) 650 m/s
Show Answer & Explanation
Answer: (c) 55 m/s
Explanation:
1. The change in velocity equals the area under the a-t graph.
2. Area of the triangle \( =\frac{1}{2}\times 11\times 10=55 \) m/s.
Teacher's Note:
1. T
2. h
3. e
4.
5. a
6. r
7. e
8. a
9.
10. u
11. n
12. d
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21. c
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31. -
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141. .
(a) 0
(b) 2 m
(c) 4.5 m
(d) 0.5 m
Show Answer & Explanation
Answer: (d) 0.5 m
Explanation:
1. The velocity becomes zero at \( t=4.5 \) s, after which the particle returns.
2. Velocity at t = 4 s is 1 m/s, so it moves \( \frac{1^2}{2\times2}=0.25 \) m further to the east before stopping.
3. From 4.5 s to 5 s it moves back \( \frac{1}{2}(2)(0.5)^2=0.25 \) m.
4. Distance covered \( =0.25+0.25=0.5 \) m (the displacement in this second is zero).
Teacher's Note:
1. D
2. i
3. s
4. t
5. a
6. n
7. c
8. e
9.
10. a
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180. .
(a) \( v\sqrt{\frac{W-F}{W+F}} \)
(b) \( v\sqrt{\frac{W}{W+F}} \)
(c) \( v\sqrt{\frac{W}{W-F}} \)
(d) \( v\sqrt{\frac{W+F}{W}} \)
Show Answer & Explanation
Answer: (a) \( v\sqrt{\frac{W-F}{W+F}} \)
Explanation:
1. Going up, the retardation is \( a_1=\frac{(W+F)g}{W} \), so \( h=\frac{v^2}{2a_1} \).
2. Coming down, the acceleration is \( a_2=\frac{(W-F)g}{W} \), so \( v'^2=2a_2h=v^2\frac{W-F}{W+F} \).
3. Thus \( v'=v\sqrt{\frac{W-F}{W+F}} \).
Teacher's Note:
1. A
2. i
3. r
4.
5. r
6. e
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174. e
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176. .
(a) 20 m
(b) 80 m
(c) 170 m
(d) 40 m
Show Answer & Explanation
Answer: (b) 80 m
Explanation:
1. First ball: \( h_1=\frac{1}{2}(10)(5)^2=125 \) m.
2. Second ball falls for 3 s: \( h_2=\frac{1}{2}(10)(3)^2=45 \) m.
3. Difference \( =125-45=80 \) m.
Teacher's Note:
1. B
2. o
3. t
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6. b
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9. l
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131. i
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133. e
134. .
(A) \( \frac{1}{x^3} \)
(B) \( \frac{-t}{x^3} \)
(C) \( \frac{1}{x}-\frac{t^2}{x^3} \)
(D) \( \frac{1}{x}-\frac{1}{x^2} \)
(a) A and B only
(b) B and D only
(c) A and C only
(d) C and D only
Show Answer & Explanation
Answer: (c) A and C only
Explanation:
1. Differentiate: \( 2x\dot x=2t \), so \( v=\frac{t}{x} \).
2. \( a=\frac{dv}{dt}=\frac{1}{x}-\frac{t}{x^2}\dot x=\frac{1}{x}-\frac{t^2}{x^3} \), which is option C.
3. Also \( \frac{1}{x}-\frac{t^2}{x^3}=\frac{x^2-t^2}{x^3}=\frac{1}{x^3} \), which is option A.
Teacher's Note:
1. T
2. w
3. o
4.
5. d
6. i
7. f
8. f
9. e
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73. ²
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80. =
81.
82. 1
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85. o
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90. p
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106. .
(a) \( -2\beta^2x^{-2n+1} \)
(b) \( -2n\beta^2x^{-2n-1} \)
(c) \( -2n\beta^2e^{-4n+1} \)
(d) \( -2n\beta^2x^{-4n-1} \)
Show Answer & Explanation
Answer: (d) \( -2n\beta^2x^{-4n-1} \)
Explanation:
1. \( a=v\frac{dv}{dx} \).
2. \( \frac{dv}{dx}=-2n\beta x^{-2n-1} \).
3. \( a=\beta x^{-2n}\times(-2n\beta x^{-2n-1})=-2n\beta^2x^{-4n-1} \).
Teacher's Note:
1. W
2. h
3. e
4. n
5.
6. v
7. e
8. l
9. o
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61. d
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69. d
70. d
71.
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73. h
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86. c
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113. p
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122.
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124. .
(a) \( h_1=\frac{h_2}{3}=\frac{h_3}{5} \)
(b) \( h_2=3h_1\text{ and }h_3=3h_2 \)
(c) h1 = h2 = h3
(d) \( h_1=2h_2=3h_3 \)
Show Answer & Explanation
Answer: (a) \( h_1=\frac{h_2}{3}=\frac{h_3}{5} \)
Explanation:
1. Starting from rest, distances in successive equal intervals are in the ratio 1 : 3 : 5.
2. So \( h_2=3h_1 \) and \( h_3=5h_1 \), giving \( h_1=\frac{h_2}{3}=\frac{h_3}{5} \).
Teacher's Note:
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(a) 24 m/s²
(b) 12 m/s²
(c) zero
(d) 6 m/s²
Show Answer & Explanation
Answer: (b) 12 m/s²
Explanation:
1. \( v=\frac{dx}{dt}=12-3t^2 \), which is zero at t = 2 s.
2. \( a=\frac{dv}{dt}=-6t \), so at t = 2 s, \( a=-12 \) m/s².
3. Retardation = 12 m/s².
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(a) \( (\text{distance})^{2} \)
(b) \( (\text{distance})^{-2} \)
(c) \( (\text{velocity})^{3/2} \)
(d) \( (\text{velocity})^{2/3} \)
Show Answer & Explanation
Answer: (c) \( (\text{velocity})^{3/2} \)
Explanation:
1. \( v=-(t+5)^{-2}=-x^2 \), so \( |v|=x^2 \).
2. \( a=2(t+5)^{-3}=2x^3 \).
3. Since \( x=|v|^{1/2} \), \( a=2|v|^{3/2} \), so a is proportional to (velocity)^{3/2}.
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126. .
(a) 55 m/s
(b) 40 m/s
(c) 60 m/s
(d) 75 m/s
Show Answer & Explanation
Answer: (d) 75 m/s
Explanation:
1. First ball falls for 18 s: \( \frac{1}{2}(10)(18)^2=1620 \) m.
2. Second ball moves for 12 s: \( 12v+\frac{1}{2}(10)(12)^2=12v+720 \).
3. Equate: \( 12v+720=1620 \), so v = 75 m/s.
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