JEE Physics Centre of Mass MCQs Set 13

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

Question 1: Which of the following has its centre of mass outside the body?
(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:
1. T
2. h
3. e
4.
5. c
6. e
7. n
8. t
9. r
10. e
11.
12. o
13. f
14.
15. m
16. a
17. s
18. s
19.
20. n
21. e
22. e
23. d
24.
25. n
26. o
27. t
28.
29. b
30. e
31.
32. o
33. n
34.
35. t
36. h
37. e
38.
39. o
40. b
41. j
42. e
43. c
44. t
45. .
46.
47. A
48.
49. r
50. i
51. n
52. g
53. ,
54.
55. a
56.
57. h
58. o
59. l
60. l
61. o
62. w
63.
64. s
65. h
66. e
67. l
68. l
69.
70. a
71. n
72. d
73.
74. a
75.
76. b
77. o
78. o
79. m
80. e
81. r
82. a
83. n
84. g
85.
86. a
87. r
88. e
89.
90. s
91. t
92. a
93. n
94. d
95. a
96. r
97. d
98.
99. e
100. x
101. a
102. m
103. p
104. l
105. e
106. s
107. .

Question 2: A man of mass M stands at one end of a plank of length L and mass M/3, which is at rest on a frictionless horizontal surface. The man walks to the other end of the plank. The distance the man moves relative to the ground is
(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:
1. W
2. h
3. e
4. n
5.
6. t
7. h
8. e
9. r
10. e
11.
12. i
13. s
14.
15. n
16. o
17.
18. e
19. x
20. t
21. e
22. r
23. n
24. a
25. l
26.
27. h
28. o
29. r
30. i
31. z
32. o
33. n
34. t
35. a
36. l
37.
38. f
39. o
40. r
41. c
42. e
43. ,
44.
45. t
46. h
47. e
48.
49. c
50. e
51. n
52. t
53. r
54. e
55.
56. o
57. f
58.
59. m
60. a
61. s
62. s
63.
64. s
65. t
66. a
67. y
68. s
69.
70. a
71. t
72.
73. r
74. e
75. s
76. t
77. .
78.
79. T
80. h
81. e
82.
83. m
84. a
85. n
86.
87. a
88. n
89. d
90.
91. p
92. l
93. a
94. n
95. k
96.
97. m
98. o
99. v
100. e
101.
102. i
103. n
104.
105. o
106. p
107. p
108. o
109. s
110. i
111. t
112. e
113.
114. d
115. i
116. r
117. e
118. c
119. t
120. i
121. o
122. n
123. s
124. ,
125.
126. a
127. n
128. d
129.
130. t
131. h
132. e
133.
134. r
135. e
136. l
137. a
138. t
139. i
140. v
141. e
142.
143. d
144. i
145. s
146. p
147. l
148. a
149. c
150. e
151. m
152. e
153. n
154. t
155.
156. i
157. s
158.
159. L
160. .

Question 3: Two blocks A (10 kg) and B (20 kg) are connected by a massless string on a smooth surface. A force of 30 N is applied on B along the line of the string. The distance travelled by the centre of mass in 2 s, starting from rest, is
(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:
1. T
2. h
3. e
4.
5. c
6. e
7. n
8. t
9. r
10. e
11.
12. o
13. f
14.
15. m
16. a
17. s
18. s
19.
20. a
21. c
22. c
23. e
24. l
25. e
26. r
27. a
28. t
29. e
30. s
31.
32. a
33. s
34.
35. i
36. f
37.
38. a
39. l
40. l
41.
42. t
43. h
44. e
45.
46. m
47. a
48. s
49. s
50.
51. w
52. e
53. r
54. e
55.
56. a
57. t
58.
59. o
60. n
61. e
62.
63. p
64. o
65. i
66. n
67. t
68.
69. u
70. n
71. d
72. e
73. r
74.
75. t
76. h
77. e
78.
79. n
80. e
81. t
82.
83. e
84. x
85. t
86. e
87. r
88. n
89. a
90. l
91.
92. f
93. o
94. r
95. c
96. e
97. .
98.
99. T
100. h
101. e
102.
103. i
104. n
105. t
106. e
107. r
108. n
109. a
110. l
111.
112. t
113. e
114. n
115. s
116. i
117. o
118. n
119.
120. i
121. s
122.
123. i
124. r
125. r
126. e
127. l
128. e
129. v
130. a
131. n
132. t
133. .

Question 4: A circular hole of radius 1 cm is cut from a disc of radius 6 cm. The centre of the hole is 3 cm from the centre of the disc. The centre of mass of the remaining disc is at a distance from the centre of the disc equal to
(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:
1. S
2. u
3. b
4. t
5. r
6. a
7. c
8. t
9.
10. t
11. h
12. e
13.
14. h
15. o
16. l
17. e
18.
19. a
20. s
21.
22. a
23.
24. n
25. e
26. g
27. a
28. t
29. i
30. v
31. e
32.
33. m
34. a
35. s
36. s
37.
38. a
39. t
40.
41. i
42. t
43. s
44.
45. o
46. w
47. n
48.
49. c
50. e
51. n
52. t
53. r
54. e
55. .
56.
57. T
58. h
59. e
60.
61. a
62. r
63. e
64. a
65.
66. o
67. f
68.
69. t
70. h
71. e
72.
73. r
74. e
75. m
76. a
77. i
78. n
79. i
80. n
81. g
82.
83. d
84. i
85. s
86. c
87.
88. i
89. s
90.
91. 3
92. 6
93.
94. m
95. i
96. n
97. u
98. s
99.
100. 1
101.
102. i
103. n
104.
105. u
106. n
107. i
108. t
109. s
110.
111. o
112. f
113.
114. p
115. i
116. .

Question 5: A non-uniform thin rod of length L lies along the x-axis with one end at the origin. Its linear mass density is \( \lambda=\lambda_0x \). The centre of mass is at a distance from the origin equal to
(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. \( x_{cm}=\dfrac{2L}{3} \).

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

Question 6: Two boys of masses 10 kg and 8 kg move along a vertical rope over a fixed support: the first climbs up with an acceleration of 2 m/s\(^2\) and the second descends with uniform velocity. The tension in the rope at the fixed support is (g = 10 m/s\(^2\))
(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.

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

Question 7: A body of mass M moving with velocity v suddenly breaks into two pieces. One part of mass M/4 remains stationary. The velocity of the other part is
(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. \( v'=\dfrac{4v}{3} \).

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

Question 8: A particle of mass M moves in a horizontal circle of radius R with uniform speed v. When it moves from one point to the diametrically opposite point, its
(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.

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

Question 9: In the HCl molecule the separation between the nuclei is about 1.27 \( \mathring{A} \). A chlorine atom is about 35.5 times as massive as a hydrogen atom, and nearly all the mass of an atom is in its nucleus. The centre of mass of the molecule, measured from the hydrogen nucleus, is at about
(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. \( x_{cm}=\dfrac{45.08}{36.5}\approx1.24\ \mathring{A} \).

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

Question 10: When a meteorite burns in the atmosphere,
(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.

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

Question 11: The density of a non-uniform rod of length 1 m is \( \rho(x)=a(1+bx^2) \), where a and b are constants and \( 0\le x\le1 \). The centre of mass of the rod is at
(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:
1. C
2. h
3. e
4. c
5. k
6.
7. t
8. h
9. e
10.
11. s
12. p
13. e
14. c
15. i
16. a
17. l
18.
19. c
20. a
21. s
22. e
23.
24. b
25.
26. =
27.
28. 0
29. :
30.
31. a
32.
33. u
34. n
35. i
36. f
37. o
38. r
39. m
40.
41. r
42. o
43. d
44.
45. m
46. u
47. s
48. t
49.
50. g
51. i
52. v
53. e
54.
55. 1
56. /
57. 2
58. ,
59.
60. a
61. n
62. d
63.
64. t
65. h
66. e
67.
68. f
69. o
70. r
71. m
72. u
73. l
74. a
75.
76. g
77. i
78. v
79. e
80. s
81.
82. 6
83. /
84. 1
85. 2
86. .

Question 12: If the momentum of a body remains constant, the graph of its mass against speed is a
(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
25. d
26. u
27. c
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.
50. c
51. o
52. n
53. s
54. t
55. a
56. n
57. t
58.
59. i
60. s
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
76. n
77. g
78. u
79. l
80. a
81. r
82.
83. h
84. y
85. p
86. e
87. r
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. .

Question 13: A machine gun fires n bullets per minute, each of mass m, into a stationary target which stops them. Each bullet arrives with velocity v. The average force on the target is
(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
14. i
15. r
16. i
17. n
18. g
19.
20. i
21. n
22.
23. b
24. u
25. l
26. l
27. e
28. t
29. s
30.
31. p
32. e
33. r
34.
35. s
36. e
37. c
38. o
39. n
40. d
41.
42. i
43. s
44.
45. n
46.
47. d
48. i
49. v
50. i
51. d
52. e
53. d
54.
55. b
56. y
57.
58. 6
59. 0
60. .
61.
62. M
63. u
64. l
65. t
66. i
67. p
68. l
69. y
70.
71. b
72. y
73.
74. t
75. h
76. e
77.
78. m
79. o
80. m
81. e
82. n
83. t
84. u
85. m
86.
87. o
88. f
89.
90. o
91. n
92. e
93.
94. b
95. u
96. l
97. l
98. e
99. t
100. .

Question 14: A body in equilibrium may not have
(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
32. t
33. i
34. o
35. n
36. ,
37.
38. n
39. o
40. t
41.
42. z
43. e
44. r
45. o
46.
47. v
48. e
49. l
50. o
51. c
52. i
53. t
54. y
55. .
56.
57. A
58.
59. c
60. a
61. r
62.
63. m
64. o
65. v
66. i
67. n
68. g
69.
70. a
71. t
72.
73. c
74. o
75. n
76. s
77. t
78. a
79. n
80. t
81.
82. s
83. p
84. e
85. e
86. d
87.
88. o
89. n
90.
91. a
92.
93. s
94. t
95. r
96. a
97. i
98. g
99. h
100. t
101.
102. r
103. o
104. a
105. d
106.
107. i
108. s
109.
110. i
111. n
112.
113. e
114. q
115. u
116. i
117. l
118. i
119. b
120. r
121. i
122. u
123. m
124. .

Question 15: A bomb at rest explodes into two fragments of masses 1 g and 3 g. The total kinetic energy of the fragments is \( 6.4\times10^4 \) J. The kinetic energy of the smaller fragment is
(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
15. e
16. n
17. t
18. a
19. ,
20.
21. t
22. h
23. e
24.
25. k
26. i
27. n
28. e
29. t
30. i
31. c
32.
33. e
34. n
35. e
36. r
37. g
38. y
39.
40. i
41. s
42.
43. i
44. n
45. v
46. e
47. r
48. s
49. e
50. l
51. y
52.
53. p
54. r
55. o
56. p
57. o
58. r
59. t
60. i
61. o
62. n
63. a
64. l
65.
66. t
67. o
68.
69. m
70. a
71. s
72. s
73. .
74.
75. T
76. h
77. e
78.
79. l
80. i
81. g
82. h
83. t
84. e
85. r
86.
87. f
88. r
89. a
90. g
91. m
92. e
93. n
94. t
95.
96. c
97. a
98. r
99. r
100. i
101. e
102. s
103.
104. t
105. h
106. e
107.
108. l
109. a
110. r
111. g
112. e
113. r
114.
115. s
116. h
117. a
118. r
119. e
120. .

Question 16: A particle moves in the xy-plane so that its linear momentum at any time is \( p_x=2\cos t \), \( p_y=2\sin t \). The angle between the force F and the momentum p is
(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
16. s
17.
18. c
19. o
20. n
21. s
22. t
23. a
24. n
25. t
26.
27. m
28. a
29. g
30. n
31. i
32. t
33. u
34. d
35. e
36.
37. a
38. n
39. d
40.
41. r
42. o
43. t
44. a
45. t
46. i
47. n
48. g
49.
50. d
51. i
52. r
53. e
54. c
55. t
56. i
57. o
58. n
59. ,
60.
61. s
62. o
63.
64. t
65. h
66. e
67.
68. f
69. o
70. r
71. c
72. e
73.
74. i
75. s
76.
77. a
78. l
79. w
80. a
81. y
82. s
83.
84. p
85. e
86. r
87. p
88. e
89. n
90. d
91. i
92. c
93. u
94. l
95. a
96. r
97.
98. t
99. o
100.
101. i
102. t
103. ,
104.
105. l
106. i
107. k
108. e
109.
110. c
111. e
112. n
113. t
114. r
115. i
116. p
117. e
118. t
119. a
120. l
121.
122. f
123. o
124. r
125. c
126. e
127. .

Question 17: Two blocks of masses \( m_1 \) and \( m_2 \) joined by a massless spring on a smooth surface are stretched and released. Then
(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
7. a
8. l
9.
10. f
11. o
12. r
13. c
14. e
15. s
16.
17. c
18. h
19. a
20. n
21. g
22. e
23.
24. t
25. h
26. e
27.
28. m
29. o
30. m
31. e
32. n
33. t
34. u
35. m
36.
37. o
38. f
39.
40. e
41. a
42. c
43. h
44.
45. b
46. l
47. o
48. c
49. k
50.
51. b
52. u
53. t
54.
55. n
56. o
57. t
58.
59. t
60. h
61. e
62.
63. t
64. o
65. t
66. a
67. l
68. .
69.
70. W
71. i
72. t
73. h
74.
75. n
76. o
77.
78. f
79. r
80. i
81. c
82. t
83. i
84. o
85. n
86.
87. t
88. h
89. e
90.
91. e
92. n
93. e
94. r
95. g
96. y
97.
98. s
99. w
100. a
101. p
102. s
103.
104. b
105. e
106. t
107. w
108. e
109. e
110. n
111.
112. s
113. p
114. r
115. i
116. n
117. g
118.
119. a
120. n
121. d
122.
123. k
124. i
125. n
126. e
127. t
128. i
129. c
130.
131. e
132. n
133. e
134. r
135. g
136. y
137. .

Question 18: An isolated particle of mass m moves along the x-axis at some height above the ground. It suddenly explodes into two fragments of masses m/4 and 3m/4. An instant later the smaller fragment is at y = +15 cm. The larger fragment at this instant is at
(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
5. r
6. n
7. a
8. l
9.
10. e
11. x
12. p
13. l
14. o
15. s
16. i
17. o
18. n
19.
20. f
21. o
22. r
23. c
24. e
25. s
26.
27. d
28. o
29.
30. n
31. o
32. t
33.
34. c
35. h
36. a
37. n
38. g
39. e
40.
41. t
42. h
43. e
44.
45. p
46. a
47. t
48. h
49.
50. o
51. f
52.
53. t
54. h
55. e
56.
57. c
58. e
59. n
60. t
61. r
62. e
63.
64. o
65. f
66.
67. m
68. a
69. s
70. s
71. .
72.
73. U
74. s
75. e
76.
77. t
78. h
79. e
80.
81. m
82. a
83. s
84. s
85. -
86. w
87. e
88. i
89. g
90. h
91. t
92. e
93. d
94.
95. p
96. o
97. s
98. i
99. t
100. i
101. o
102. n
103. s
104.
105. t
106. o
107.
108. b
109. a
110. l
111. a
112. n
113. c
114. e
115.
116. a
117. b
118. o
119. u
120. t
121.
122. t
123. h
124. e
125.
126. a
127. x
128. i
129. s
130. .

Question 19: A bomb at rest explodes in air into two equal fragments. If one fragment moves vertically upward with velocity \( v_0 \), the other fragment will move
(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
11. e
12. s
13.
14. f
15. l
16. y
17.
18. a
19. p
20. a
21. r
22. t
23.
24. w
25. i
26. t
27. h
28.
29. e
30. q
31. u
32. a
33. l
34.
35. a
36. n
37. d
38.
39. o
40. p
41. p
42. o
43. s
44. i
45. t
46. e
47.
48. v
49. e
50. l
51. o
52. c
53. i
54. t
55. i
56. e
57. s
58.
59. w
60. h
61. e
62. n
63.
64. t
65. h
66. e
67.
68. i
69. n
70. i
71. t
72. i
73. a
74. l
75.
76. m
77. o
78. m
79. e
80. n
81. t
82. u
83. m
84.
85. i
86. s
87.
88. z
89. e
90. r
91. o
92. .

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

Practice MCQs for JEE Physics Centre of Mass

About Centre of Mass MCQs for JEE Physics

Explore reliable practice questions for Centre of Mass tailored for JEE Physics learners. Use these multiple-choice formats to evaluate preparedness and strengthen problem-solving skills.

How to Verify Your MCQ Answers

Each question includes structured solution keys mapped directly to standard JEE textbooks, helping students evaluate their reasoning and correct mistakes early in their revision.

Enhance Speed with Online MCQ Tests

Follow up your worksheet practice by attempting the interactive online Physics MCQ test for this chapter to evaluate your execution speed. All platform resources are free to access.

FAQs

Where can I access latest JEE Physics Centre of Mass MCQs Set 13?

You can get most exhaustive JEE Physics Centre of Mass MCQs Set 13 for free on StudiesToday.com. These MCQs for JEE Physics are updated for the 2026-27 academic session as per JEE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Physics JEE material?

Yes, our JEE Physics Centre of Mass MCQs Set 13 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.

How do practicing Physics MCQs help in scoring full marks in JEE exams?

By solving our JEE Physics Centre of Mass MCQs Set 13, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Physics.

Do you provide answers and explanations for JEE Physics Centre of Mass MCQs Set 13?

Yes, Physics MCQs for JEE have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused JEE exams.

Can I practice these Physics JEE MCQs online?

Yes, you can also access online interactive tests for JEE Physics Centre of Mass MCQs Set 13 on StudiesToday.com as they provide instant answers and score to help you track your progress in Physics.