Practice MCQs for JEE Physics Circular Motion
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Multiple Choice Questions
(a) \( \dfrac{\pi\sqrt2}{30}\ \text{cm/s} \)
(b) zero
(c) \( \dfrac{\pi}{30}\ \text{cm/s} \)
(d) \( \dfrac{\pi}{30\sqrt2}\ \text{cm/s} \)
Show Answer & Explanation
Answer: (a) \( \dfrac{\pi\sqrt2}{30}\ \text{cm/s} \)
Explanation:
1. \( \omega=2\pi/60=\pi/30 \) rad/s, so the speed of the tip is \( v=\omega r=\pi/30 \) cm/s.
2. In 15 s the hand turns through 90 degrees, so the two velocities are perpendicular.
3. \( |\Delta\vec v|=\sqrt{v^2+v^2}=\sqrt2\,v=\dfrac{\pi\sqrt2}{30} \) cm/s.
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166. .
(a) \( 10\pi \) rad
(b) \( \pi \) rad
(c) \( 5\pi \) rad
(d) \( 20\pi \) rad
Show Answer & Explanation
Answer: (a) \( 10\pi \) rad
Explanation:
1. 300 rpm = 5 revolutions per second.
2. Angle in one second \( =5\times2\pi=10\pi \) rad.
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(a) mg
(b) 3 mg
(c) 2 mg
(d) 5 mg
Show Answer & Explanation
Answer: (b) 3 mg
Explanation:
1. Speed at the bottom: \( v^2=2gl \) (energy conservation from a height \( l \)).
2. Tension at the bottom: \( T=mg+\dfrac{mv^2}{l}=mg+2mg=3mg \).
Teacher's Note:
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138. .
(a) 60.125 kg-wt
(b) 60.175 kg-wt
(c) 60.075 kg-wt
(d) 60.225 kg-wt
Show Answer & Explanation
Answer: (c) 60.075 kg-wt
Explanation:
1. The balance measures the resultant of weight and the centripetal force \( mv^2/r \), which are perpendicular.
2. \( v^2/r=49/100=0.49 \) m/s\(^2\), so the reading is \( 60\sqrt{1+(0.49/9.8)^2}=60\sqrt{1.0025}\approx60.075 \) kg-wt.
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148. .
(a) 41 : 39
(b) 39 : 41
(c) 11 : 12
(d) 12 : 11
Show Answer & Explanation
Answer: (b) 39 : 41
Explanation:
1. \( v^2/r=1600/4=400 \) N per kg.
2. \( T_{top}=m(v^2/r-g)=390 \) N and \( T_{bottom}=m(v^2/r+g)=410 \) N.
3. Ratio \( =390:410=39:41 \).
Teacher's Note:
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155. .
(a) 4 rad/s
(b) 16 rad/s
(c) 21 rad/s
(d) 2 rad/s
Show Answer & Explanation
Answer: (a) 4 rad/s
Explanation:
1. The tension is greatest at the lowest point: \( T=mg+m\omega^2r \).
2. \( 37=0.5(10)+0.5\omega^2(4)=5+2\omega^2 \).
3. \( \omega^2=16 \), so \( \omega=4 \) rad/s.
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132. .
(a) at the top of the circle
(b) halfway down
(c) at the bottom of the circle
(d) None of these
Show Answer & Explanation
Answer: (c) at the bottom of the circle
Explanation:
1. \( mv^2/r=2(16)/1=32 \) N and \( mg=20 \) N.
2. At the bottom, \( T=32+20=52 \) N; at the top, \( T=32-20=12 \) N.
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106. .
(a) 3
(b) 1
(c) 2
(d) 4
Show Answer & Explanation
Answer: (a) 3
Explanation:
1. \( T_{max}=m(v^2/r+g) \) and \( T_{min}=m(v^2/r-g) \).
2. \( v^2/r+g=2(v^2/r-g) \) gives \( v^2/r=3g \).
3. So \( v^2/rg=3 \).
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123. .
(a) \( 30^\circ \)
(b) \( 45^\circ \)
(c) \( 60^\circ \)
(d) \( 90^\circ \)
Show Answer & Explanation
Answer: (d) \( 90^\circ \)
Explanation:
1. At the mean position \( T-mg=mv^2/l \), so \( 2mg=mv^2/l \) and \( v^2=2gl \).
2. If the bob rose from angle \( \theta \), \( v^2=2gl(1-\cos\theta) \).
3. Comparing, \( \cos\theta=0 \), so \( \theta=90^\circ \).
Teacher's Note:
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131. .
(a) \( \dfrac r6 \)
(b) \( \dfrac r3 \)
(c) \( \dfrac r4 \)
(d) \( \dfrac r2 \)
Show Answer & Explanation
Answer: (b) \( \dfrac r3 \)
Explanation:
1. Speed after falling h: \( v^2=2gh \).
2. It leaves when the normal reaction vanishes: \( mg\cos\theta=mv^2/r \) with \( \cos\theta=(r-h)/r \).
3. \( g(r-h)=2gh \), so \( h=r/3 \).
Teacher's Note:
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108. .
(a) \( \sqrt{2\left(u^2-gL\right)} \)
(b) \( \sqrt{u^2-2gL} \)
(c) \( \sqrt{2gL} \)
(d) \( u^2-gL \)
Show Answer & Explanation
Answer: (a) \( \sqrt{2\left(u^2-gL\right)} \)
Explanation:
1. At the horizontal position the speed is \( v \) with \( v^2=u^2-2gL \), directed vertically.
2. The initial velocity is horizontal, so \( |\Delta\vec v|=\sqrt{u^2+v^2} \).
3. \( =\sqrt{2u^2-2gL}=\sqrt{2(u^2-gL)} \).
Teacher's Note:
1. T
2. h
3. e
4.
5. v
6. e
7. l
8. o
9. c
10. i
11. t
12. y
13.
14. a
15. t
16.
17. t
18. h
19. e
20.
21. l
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23. w
24. e
25. s
26. t
27.
28. p
29. o
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32. t
33.
34. i
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36.
37. h
38. o
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40. i
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42. o
43. n
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45. a
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47.
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49. n
50. d
51.
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53. t
54.
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56. h
57. e
58.
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60. o
61. r
62. i
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72. r
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76.
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100. o
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118. i
119. c
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124. .
125.
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127. o
128.
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132.
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135. b
136. t
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141.
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149. e
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152. .
(a) \( T_1>T_2 \)
(b) \( T_1=T_2 \)
(c) \( T_2>T_1 \)
(d) the tension stays the same
Show Answer & Explanation
Answer: (a) \( T_1>T_2 \)
Explanation:
1. \( T=\dfrac{mv^2}{r}+mg\cos\theta \).
2. As \( \theta \) increases, the speed \( v \) decreases (the particle rises) and \( \cos\theta \) decreases, so both terms drop.
3. Hence the tension at 30 degrees is greater than at 60 degrees.
Teacher's Note:
1. T
2. e
3. n
4. s
5. i
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7. n
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70.
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78.
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80. o
81. t
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83.
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85. h
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87.
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89. p
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91. e
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93.
94. t
95. e
96. r
97. m
98.
99. a
100. n
101. d
102.
103. t
104. h
105. e
106.
107. g
108. r
109. a
110. v
111. i
112. t
113. y
114.
115. t
116. e
117. r
118. m
119.
120. f
121. a
122. l
123. l
124.
125. a
126. s
127.
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131.
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140. o
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143. .
(a) 10 J
(b) 30 J
(c) 20 J
(d) 50 J
Show Answer & Explanation
Answer: (c) 20 J
Explanation:
1. The height difference between top and bottom is \( 2r=2 \) m.
2. By energy conservation, \( \Delta KE=mg(2r)=1\times10\times2=20 \) J.
Teacher's Note:
1. T
2. h
3. e
4.
5. d
6. i
7. f
8. f
9. e
10. r
11. e
12. n
13. c
14. e
15.
16. i
17. n
18.
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20. i
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61. i
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103. h
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105.
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107. a
108. i
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110. e
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112. .
113.
114. I
115. t
116.
117. d
118. o
119. e
120. s
121.
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125.
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146. .
(a) \( 10\pi\ \text{rad/s} \)
(b) \( 20\pi\ \text{rad/s} \)
(c) \( 40\pi\ \text{rad/s} \)
(d) \( 60\pi\ \text{rad/s} \)
Show Answer & Explanation
Answer: (b) \( 20\pi\ \text{rad/s} \)
Explanation:
1. 600 rpm = 10 rev/s and 1200 rpm = 20 rev/s.
2. Increase \( =10\text{ rev/s}\times2\pi=20\pi \) rad/s.
Teacher's Note:
1. C
2. o
3. n
4. v
5. e
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9. e
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11. c
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122. .
(a) 1
(b) 4
(c) 3
(d) 2
Show Answer & Explanation
Answer: (a) 1
Explanation:
1. Friction supplies the centripetal force: \( m\omega^2r\le\mu mg \).
2. \( \omega\le\sqrt{\mu g/r}=\sqrt{0.5\times10/5}=1 \) rad/s.
Teacher's Note:
1. T
2. h
3. e
4.
5. l
6. i
7. m
8. i
9. t
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56. u
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60. a
61. l
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65.
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72. a
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105. .
(a) 108 km/h
(b) 30 km/h
(c) 81 km/h
(d) 162 km/h
Show Answer & Explanation
Answer: (a) 108 km/h
Explanation:
1. \( v=\sqrt{\mu rg}=\sqrt{0.3\times300\times10}=30 \) m/s.
2. \( 30\times3.6=108 \) km/h.
Teacher's Note:
1. R
2. e
3. m
4. e
5. m
6. b
7. e
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9.
10. t
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120. .
(a) 12 m/s
(b) 27 m/s
(c) 18 m/s
(d) 22 m/s
Show Answer & Explanation
Answer: (b) 27 m/s
Explanation:
1. To avoid overturning, the torque about the outer rail must balance: \( v^2/(rg)\le x/h \), where \( x=0.75 \) m is half the track width.
2. \( v=\sqrt{rgx/h}=\sqrt{100\times10\times0.75/1}=\sqrt{750}\approx27 \) m/s.
Teacher's Note:
1. T
2. o
3. p
4. p
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128. .
(a) \( 10^4 \) N
(b) \( 10^3 \) N
(c) \( 10^5 \) N
(d) \( 10^2 \) N
Show Answer & Explanation
Answer: (a) \( 10^4 \) N
Explanation:
1. Friction provides the centripetal force: \( f=mv^2/r=2000\times100/20 \).
2. \( f=10^4 \) N.
Teacher's Note:
1. O
2. n
3.
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90. .
(a) 9.9 m/s\(^2\) along the radius
(b) 9.9 m/s\(^2\) along the tangent
(c) 7.9 m/s\(^2\) along the radius
(d) 1.3 m/s\(^2\) along the radius
Show Answer & Explanation
Answer: (a) 9.9 m/s\(^2\) along the radius
Explanation:
1. \( \omega=2\pi\times14/25=3.52 \) rad/s.
2. \( a=\omega^2r=(3.52)^2\times0.8\approx9.9 \) m/s\(^2\).
3. In uniform circular motion the acceleration is directed towards the centre, along the radius.
Teacher's Note:
1. A
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112. .
(a) 6
(b) 7
(c) 8
(d) 5
Show Answer & Explanation
Answer: (a) 6
Explanation:
1. 900 km/h = 250 m/s.
2. \( a=v^2/r=62500/1000=62.5 \) m/s\(^2\).
3. \( a/g=62.5/9.8\approx6.4 \), so about 6.
Teacher's Note:
1. C
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Free study material for Physics
Circular Motion Objective Questions & Solutions for JEE Physics
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FAQs
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