Download JEE MCQs for JEE Physics: Centre of Mass
Review structured MCQ sets for JEE Physics Centre of Mass. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.
Chapter-wise Objective Questions: Centre of Mass
Navigate directly to the 50 objective questions for Centre of Mass using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.
Multiple Choice Questions
(a) A bangle
(b) A pencil
(c) A shot put
(d) A dice
Show Answer & Explanation
Answer: (a) A bangle
Explanation:
1. The centre of mass of a symmetric body lies at its geometric centre.
2. For a ring-shaped bangle the geometric centre is the empty point at the middle of the ring, which is not part of the material.
Teacher's Note:
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107. .
(a) \( L \)
(b) \( \dfrac{3L}{4} \)
(c) \( \dfrac L4 \)
(d) \( \dfrac L3 \)
Show Answer & Explanation
Answer: (c) \( \dfrac L4 \)
Explanation:
1. There is no external horizontal force, so the centre of mass does not move: \( Mx=\dfrac M3(L-x) \).
2. \( 3x=L-x \), so \( x=L/4 \).
Teacher's Note:
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160. .
(a) 1 m
(b) 2 m
(c) 3 m
(d) 4 m
Show Answer & Explanation
Answer: (b) 2 m
Explanation:
1. Acceleration of the centre of mass \( =F_{ext}/M_{total}=30/30=1 \) m/s\(^2\).
2. \( s=\tfrac12at^2=\tfrac12(1)(4)=2 \) m.
Teacher's Note:
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133. .
(a) \( \dfrac{3}{35}\ \text{cm (on the side opposite to the hole)} \)
(b) \( \dfrac{1}{35}\ \text{cm (towards the hole)} \)
(c) \( \dfrac{3}{10}\ \text{cm (towards the hole)} \)
(d) \( \dfrac{3}{36}\ \text{cm (on the side opposite to the hole)} \)
Show Answer & Explanation
Answer: (a) \( \dfrac{3}{35}\ \text{cm (on the side opposite to the hole)} \)
Explanation:
1. Treat the removed part as negative mass. Masses are proportional to areas: \( m_{hole}\propto\pi(1)^2 \), \( M_{disc}\propto\pi(6)^2=36\pi \).
2. \( x_{cm}=\dfrac{-m_{hole}\,d}{M-m_{hole}}=\dfrac{-1\times3}{36-1}=-\dfrac{3}{35} \) cm.
3. The negative sign means the centre of mass shifts away from the hole.
Teacher's Note:
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116. .
(a) \( \dfrac{2L}{3} \)
(b) \( \dfrac L2 \)
(c) \( \dfrac L3 \)
(d) \( \dfrac{3L}{4} \)
Show Answer & Explanation
Answer: (a) \( \dfrac{2L}{3} \)
Explanation:
1. \( x_{cm}=\dfrac{\int x\lambda\,dx}{\int\lambda\,dx}=\dfrac{\int_0^L x^2dx}{\int_0^L x\,dx}=\dfrac{L^3/3}{L^2/2} \).
2. \( x_{cm}=\dfrac{2L}{3} \).
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126. .
(a) 120 N
(b) 200 N
(c) 180 N
(d) 160 N
Show Answer & Explanation
Answer: (b) 200 N
Explanation:
1. For the climbing boy, \( T-mg=ma \), so the force the boy exerts on the rope is \( 10(10+2)=120 \) N.
2. For the boy descending with uniform velocity, force \( =8\times10=80 \) N.
3. Total pull on the support \( =120+80=200 \) N.
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137. .
(a) \( v \)
(b) \( \dfrac{4v}{3} \)
(c) \( 2v \)
(d) \( \dfrac{3v}{4} \)
Show Answer & Explanation
Answer: (b) \( \dfrac{4v}{3} \)
Explanation:
1. Momentum is conserved: \( Mv=\dfrac M4(0)+\dfrac{3M}{4}v' \).
2. \( v'=\dfrac{4v}{3} \).
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111. .
(a) momentum changes by 2Mv
(b) momentum does not change
(c) kinetic energy changes by \( Mv^2 \)
(d) kinetic energy changes by \( 2Mv^2 \)
Show Answer & Explanation
Answer: (a) momentum changes by 2Mv
Explanation:
1. The velocity reverses, so \( \Delta\vec p=-Mv-Mv \), a magnitude of \( 2Mv \).
2. The speed is constant so the kinetic energy does not change.
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124. .
(a) 2.24 \( \mathring{A} \)
(b) 0.24 \( \mathring{A} \)
(c) 1.24 \( \mathring{A} \)
(d) 3.24 \( \mathring{A} \)
Show Answer & Explanation
Answer: (c) 1.24 \( \mathring{A} \)
Explanation:
1. Take H at the origin and Cl at 1.27 \( \mathring{A} \): \( x_{cm}=\dfrac{1(0)+35.5(1.27)}{1+35.5} \).
2. \( x_{cm}=\dfrac{45.08}{36.5}\approx1.24\ \mathring{A} \).
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126. .
(a) the momentum of the meteorite remains constant
(b) the principle of conservation of momentum applies to the system of meteorite, earth and air molecules
(c) the energy of the meteorite remains constant
(d) the momentum principle applies to the meteorite alone
Show Answer & Explanation
Answer: (b) the principle of conservation of momentum applies to the system of meteorite, earth and air molecules
Explanation:
1. Air molecules exert external forces on the meteorite, so its momentum alone changes.
2. The momentum is conserved only when the meteorite, the air and the earth are all taken in the system.
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121. .
(a) \( \dfrac{3(2+b)}{4(3+b)} \)
(b) \( \dfrac{4(2+b)}{3(3+b)} \)
(c) \( \dfrac{3(2+b)}{4(2+b)} \)
(d) \( \dfrac{4(3+b)}{3(2+b)} \)
Show Answer & Explanation
Answer: (a) \( \dfrac{3(2+b)}{4(3+b)} \)
Explanation:
1. \( \int_0^1 x(1+bx^2)dx=\dfrac12+\dfrac b4=\dfrac{2+b}{4} \).
2. \( \int_0^1(1+bx^2)dx=1+\dfrac b3=\dfrac{3+b}{3} \).
3. \( x_{cm}=\dfrac{(2+b)/4}{(3+b)/3}=\dfrac{3(2+b)}{4(3+b)} \).
Teacher's Note:
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42. o
43. d
44.
45. m
46. u
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48. t
49.
50. g
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53. e
54.
55. 1
56. /
57. 2
58. ,
59.
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63.
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65. h
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67.
68. f
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71. m
72. u
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81.
82. 6
83. /
84. 1
85. 2
86. .
(a) circle
(b) rectangular hyperbola
(c) straight line
(d) parabola
Show Answer & Explanation
Answer: (b) rectangular hyperbola
Explanation:
1. \( p=mv \) is constant, so \( m=p/v \).
2. The product of the two plotted quantities is constant, which is the equation of a rectangular hyperbola.
Teacher's Note:
1. A
2.
3. g
4. r
5. a
6. p
7. h
8.
9. i
10. n
11.
12. w
13. h
14. i
15. c
16. h
17.
18. t
19. h
20. e
21.
22. p
23. r
24. o
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26. u
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28. t
29.
30. o
31. f
32.
33. t
34. h
35. e
36.
37. v
38. a
39. r
40. i
41. a
42. b
43. l
44. e
45. s
46.
47. i
48. s
49.
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51. o
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53. s
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55. a
56. n
57. t
58.
59. i
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61.
62. a
63. l
64. w
65. a
66. y
67. s
68.
69. a
70.
71. r
72. e
73. c
74. t
75. a
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77. g
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80. a
81. r
82.
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84. y
85. p
86. e
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88. b
89. o
90. l
91. a
92. ,
93.
94. a
95. s
96.
97. i
98. n
99.
100. B
101. o
102. y
103. l
104. e
105. '
106. s
107.
108. l
109. a
110. w
111. .
(a) \( \dfrac{mnv}{60} \)
(b) \( 60\,mnv \)
(c) \( \dfrac{60v}{mn} \)
(d) \( \dfrac{mv}{60n} \)
Show Answer & Explanation
Answer: (a) \( \dfrac{mnv}{60} \)
Explanation:
1. Momentum delivered per second \( =\dfrac{n}{60}\times mv \).
2. Force = rate of change of momentum \( =\dfrac{mnv}{60} \).
Teacher's Note:
1. T
2. h
3. e
4.
5. r
6. a
7. t
8. e
9.
10. o
11. f
12.
13. f
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16. i
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29. s
30.
31. p
32. e
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34.
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41.
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50. i
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52. e
53. d
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57.
58. 6
59. 0
60. .
61.
62. M
63. u
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66. i
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68. l
69. y
70.
71. b
72. y
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97. l
98. e
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100. .
(a) momentum
(b) acceleration
(c) velocity
(d) kinetic energy
Show Answer & Explanation
Answer: (b) acceleration
Explanation:
1. Equilibrium means zero net force, so the acceleration must be zero.
2. A body in equilibrium may still move at constant velocity, and so can have momentum and kinetic energy.
Teacher's Note:
1. E
2. q
3. u
4. i
5. l
6. i
7. b
8. r
9. i
10. u
11. m
12.
13. m
14. e
15. a
16. n
17. s
18.
19. z
20. e
21. r
22. o
23.
24. a
25. c
26. c
27. e
28. l
29. e
30. r
31. a
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33. i
34. o
35. n
36. ,
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41.
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43. e
44. r
45. o
46.
47. v
48. e
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54. y
55. .
56.
57. A
58.
59. c
60. a
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62.
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65. v
66. i
67. n
68. g
69.
70. a
71. t
72.
73. c
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77. t
78. a
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80. t
81.
82. s
83. p
84. e
85. e
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90.
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94. t
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118. i
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120. r
121. i
122. u
123. m
124. .
(a) \( 4.8\times10^4\ \text{J} \)
(b) \( 2.5\times10^4\ \text{J} \)
(c) \( 3.5\times10^4\ \text{J} \)
(d) \( 5.2\times10^4\ \text{J} \)
Show Answer & Explanation
Answer: (a) \( 4.8\times10^4\ \text{J} \)
Explanation:
1. Momentum conservation: \( m_1v_1=m_2v_2 \), so \( v_1:v_2=3:1 \).
2. \( KE\propto mv^2 \): \( KE_1:KE_2=1(9):3(1)=3:1 \).
3. \( KE_1=\tfrac34\times6.4\times10^4=4.8\times10^4 \) J.
Teacher's Note:
1. W
2. i
3. t
4. h
5.
6. e
7. q
8. u
9. a
10. l
11.
12. m
13. o
14. m
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18. a
19. ,
20.
21. t
22. h
23. e
24.
25. k
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27. n
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29. t
30. i
31. c
32.
33. e
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35. e
36. r
37. g
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41. s
42.
43. i
44. n
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46. e
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48. s
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54. r
55. o
56. p
57. o
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60. i
61. o
62. n
63. a
64. l
65.
66. t
67. o
68.
69. m
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72. s
73. .
74.
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76. h
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79. l
80. i
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84. e
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88. r
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95.
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99. r
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114.
115. s
116. h
117. a
118. r
119. e
120. .
(a) \( 0^\circ \)
(b) \( 180^\circ \)
(c) \( 30^\circ \)
(d) \( 90^\circ \)
Show Answer & Explanation
Answer: (d) \( 90^\circ \)
Explanation:
1. \( \vec F=d\vec p/dt=(-2\sin t,\ 2\cos t) \).
2. \( \vec F\cdot\vec p=-4\sin t\cos t+4\sin t\cos t=0 \), so they are perpendicular.
Teacher's Note:
1. T
2. h
3. e
4.
5. m
6. o
7. m
8. e
9. n
10. t
11. u
12. m
13.
14. h
15. a
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17.
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30. n
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122. f
123. o
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125. c
126. e
127. .
(a) both the momenta have equal magnitude and the mechanical energy is conserved
(b) the momentum of each block remains constant separately
(c) the mechanical energy of the system decreases
(d) the blocks have equal speeds at all times
Show Answer & Explanation
Answer: (a) both the momenta have equal magnitude and the mechanical energy is conserved
Explanation:
1. There is no external force, so the total momentum remains zero: \( m_1v_1=m_2v_2 \) in magnitude.
2. The surface is smooth and the spring is ideal, so the total mechanical energy is conserved.
Teacher's Note:
1. I
2. n
3. t
4. e
5. r
6. n
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137. .
(a) \( y=-5\ \text{cm} \)
(b) \( y=+5\ \text{cm} \)
(c) \( y=+20\ \text{cm} \)
(d) \( y=-20\ \text{cm} \)
Show Answer & Explanation
Answer: (a) \( y=-5\ \text{cm} \)
Explanation:
1. The centre of mass continues along the x-axis, so \( y_{cm}=0 \).
2. \( \dfrac m4(15)+\dfrac{3m}{4}y=0 \), so \( y=-5 \) cm.
Teacher's Note:
1. I
2. n
3. t
4. e
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130. .
(a) vertically downward with velocity \( v_0 \)
(b) vertically upward with velocity \( v_0 \)
(c) in an arbitrary direction with velocity \( v_0 \)
(d) horizontally with velocity \( v_0 \)
Show Answer & Explanation
Answer: (a) vertically downward with velocity \( v_0 \)
Explanation:
1. The total momentum immediately after the explosion is zero (neglecting gravity during the short explosion).
2. \( m v_0+mv'=0 \), so \( v'=-v_0 \), which is downward.
Teacher's Note:
1. E
2. q
3. u
4. a
5. l
6.
7. m
8. a
9. s
10. s
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92. .
(a) A bullet striking a bag of sand
(b) Striking of two glass balls
(c) An electron captured by a proton
(d) A man jumping onto a moving car
Show Answer & Explanation
Answer: (b) Striking of two glass balls
Explanation:
1. In a perfectly inelastic collision the bodies stick together and move with a common velocity.
2. Glass balls rebound with nearly the same speed, so the collision is nearly elastic.
Teacher's Note:
1. A
2. s
3. k
4.
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117. .
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