JEE Physics Circular Motion MCQs Set 20

Practice MCQs for JEE Physics Circular Motion

Explore reliable objective questions for Circular Motion tailored for JEE learners. Utilizing these Physics multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

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Multiple Choice Questions

Question 1: The length of the second's hand of a watch is 1 cm. The magnitude of the change in velocity of its tip in 15 s is
(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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Question 2: A wheel rotates with a constant angular velocity of 300 rpm. The angle through which it rotates in one second is
(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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Question 3: A simple pendulum is displaced through 90\(^\circ\) from the vertical and released. The minimum strength of the string needed to withstand the tension as it passes the mean position is
(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: \( T=mg+\dfrac{mv^2}{l}=mg+2mg=3mg \).

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Question 4: An object is weighed on a spring balance moving around a curve of radius 100 m at a speed of 7 m/s. The object weighs 60 kg-wt at rest. The reading of the balance is
(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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Question 5: A stone of mass 1 kg is tied to a string 4 m long and rotated in a vertical circle at a constant speed of 40 m/s. The ratio of the tension at the top to that at the bottom is
(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 \( =390:410=39:41 \).

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Question 6: A weightless thread can bear a tension up to 3.7 kg-wt. A stone of mass 500 g is tied to it and revolved in a vertical circle of radius 4 m. The maximum angular velocity of the stone is (g = 10 m/s\(^2\))
(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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Question 7: A 2 kg stone at the end of a 1 m long string is whirled in a vertical circle at a constant speed of 4 m/s. The tension in the string is 52 N when the stone is (g = 10 m/s\(^2\))
(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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Question 8: An object tied to a string is rotated in a vertical circle of radius r at a constant speed. If \( T_{max}/T_{min}=2 \), then \( v^2/rg \) equals
(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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Question 9: A simple pendulum oscillates in a vertical plane. When it passes through the mean position the tension in the string is 3 times the weight of the bob. The maximum angular displacement is
(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 \( \theta=90^\circ \).

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Question 10: A particle is placed at the highest point of a smooth sphere of radius r and given a slight push. It leaves the sphere at a depth h below the top, where h equals
(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 \( h=r/3 \).

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Question 11: A stone tied to a string of length L is whirled in a vertical circle. At its lowest position its speed is u. The magnitude of the change in velocity as it reaches the position where the string is horizontal is
(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:
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Question 12: A particle moves in a vertical circle. The tensions in the string at positions 30\(^\circ\) and 60\(^\circ\) from the vertical (lowest position) are \( T_1 \) and \( T_2 \) respectively. Then
(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:
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Question 13: A body of mass 1 kg is rotating in a vertical circle of radius 1 m. The difference between its kinetic energies at the bottom and at the top is (g = 10 m/s\(^2\))
(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:
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Question 14: A fan makes 600 revolutions per minute. After some time it makes 1200 revolutions per minute. The increase in its angular velocity is
(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:
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Question 15: A body moves along a circular path of radius 5 m. The coefficient of friction between the path and the body is 0.5. The maximum angular velocity, in rad/s, with which it can move without leaving the path is (g = 10 m/s\(^2\))
(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.

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Question 16: A car moves on a circular level road of radius 300 m. The coefficient of friction is 0.3 and g = 10 m/s\(^2\). The maximum speed of the car is
(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.

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Question 17: A railway carriage has its centre of gravity 1 m above the rails, which are 1.5 m apart. The maximum safe speed on an unbanked curve of radius 100 m is about (g = 10 m/s\(^2\))
(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.

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Question 18: A car of mass 2000 kg moves at 10 m/s on a circular path of radius 20 m on a level road. The frictional force needed so that the car does not slip is
(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.

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Question 19: A stone tied to the end of a string 80 cm long is whirled in a horizontal circle at constant speed. It makes 14 revolutions in 25 s. The magnitude and direction of its acceleration are
(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.

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Question 20: An aircraft executes a horizontal loop of radius 1 km at a speed of 900 km/h. The ratio of its centripetal acceleration to g is about (take g = 9.8 m/s\(^2\))
(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.

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Circular Motion Objective Questions & Solutions for JEE Physics

JEE Physics Circular Motion Objective Test Questions

Review structured objective questions for JEE Physics Circular Motion. Built according to official JEE guidelines, these MCQ sets support daily revision and core concept reinforcement.

NCERT-Aligned Objective Questions and Solutions

Cross-reference your completed choices with comprehensive NCERT solutions for JEE Physics to ensure absolute clarity across all sub-topics in this chapter.

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FAQs

Where can I access latest JEE Physics Circular Motion MCQs Set 20?

You can get most exhaustive JEE Physics Circular Motion MCQs Set 20 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 Circular Motion MCQs Set 20 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 Circular Motion MCQs Set 20, 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 Circular Motion MCQs Set 20?

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 Circular Motion MCQs Set 20 on StudiesToday.com as they provide instant answers and score to help you track your progress in Physics.