JEE Physics Circular Motion MCQs Set 16

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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Access the complete set of multiple-choice questions for Circular Motion below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: A particle P of mass m is attached to a vertical axis by two strings AP and BP of length \( \ell \) each, the separation AB being \( \ell \). P rotates about the axis with angular velocity ω and the tensions in the strings are \( T_1 \) (AP) and \( T_2 \) (BP), with A above B. Which of the following are correct? (1) \( T_1+T_2 = m\omega^2\ell \) (2) \( T_1-T_2 = 2mg \) (3) BP remains taut only if \( \omega\ge\sqrt{2g/\ell} \) (4) \( T_1 = T_2 \)
(a) Only (1) and (4)
(b) Only (1), (2) and (3)
(c) Only (2) and (4)
(d) Only (3) and (4)
Show Answer & Explanation

Answer: (b) Only (1), (2) and (3)

Explanation:
1. The triangle is equilateral, so each string makes 60° with the axis. Vertical: \( (T_1-T_2)\cos60^\circ = mg \Rightarrow T_1-T_2 = 2mg \). Horizontal: \( (T_1+T_2)\sin60^\circ = m\omega^2\ell\sin60^\circ \Rightarrow T_1+T_2 = m\omega^2\ell \).
2. \( T_2 = \frac{m\omega^2\ell-2mg}{2}\ge0 \Rightarrow \omega\ge\sqrt{2g/\ell} \).

Teacher's Note:
1. Resolve vertically and horizontally.
2. T₂ ≥ 0 gives the minimum ω.

Question 2: A cart moves with constant speed along a horizontal circular path. A particle is thrown up vertically from the cart. Which of the following are correct? (1) The particle lands somewhere on the circular path. (2) The particle lands outside the circular path. (3) The particle follows an elliptical path. (4) The particle follows a parabolic path.
(a) Only (1) and (3)
(b) Only (1) and (4)
(c) Only (2) and (3)
(d) Only (2) and (4)
Show Answer & Explanation

Answer: (d) Only (2) and (4)

Explanation:
1. The particle keeps the cart's tangential velocity in the ground frame, so it moves like a projectile (a parabola) in a straight line horizontally.
2. The cart curves away inward, so the particle lands outside the circular path.

Teacher's Note:
1. No horizontal force acts on the particle after release.
2. The cart turns while the particle goes straight.

Question 3: A particle moves in a circle of radius 1.0 cm at a speed v = 2.0t, where v is in cm/s and t in seconds. The radial acceleration of the particle at t = 1 s is
(a) 2.0 cm/s²
(b) 1.0 cm/s²
(c) 4.0 cm/s²
(d) 8.0 cm/s²
Show Answer & Explanation

Answer: (c) 4.0 cm/s²

Explanation:
1. At t = 1 s, v = 2 cm/s, so \( a_c = \frac{v^2}{R} = \frac{4}{1} = 4 \) cm/s².

Teacher's Note:
1. Radial acceleration = v²/R.
2. Evaluate v at the given time.

Question 4: A particle moves in a circle of radius 1.0 cm at a speed v = 2.0t cm/s. The magnitude of its total acceleration at t = 1 s is
(a) \( \sqrt{12} \) cm/s²
(b) 6 cm/s²
(c) \( \sqrt{20} \) cm/s²
(d) 4 cm/s²
Show Answer & Explanation

Answer: (c) \( \sqrt{20} \) cm/s²

Explanation:
1. \( a_t = \frac{dv}{dt} = 2 \) cm/s², \( a_c = 4 \) cm/s².
2. \( a = \sqrt{2^2+4^2} = \sqrt{20} \) cm/s².

Teacher's Note:
1. Tangential = dv/dt.
2. Combine the two components in quadrature.

Question 5: A particle moves in a circle of radius 1 m starting from rest with constant tangential acceleration \( \frac\pi2 \) m/s². Its speed after it has gone halfway around the circle is
(a) \( \frac\pi2 \) m/s
(b) 2 m/s
(c) \( 2\pi \) m/s
(d) \( \pi \) m/s
Show Answer & Explanation

Answer: (d) \( \pi \) m/s

Explanation:
1. The time for half a circle is 2 s, so \( v = a_tt = \frac\pi2\times2 = \pi \) m/s (about 3.14 m/s).

Teacher's Note:
1. v = at from rest.
2. Time was found from s = πR.

Question 6: A solid body rotates about a stationary axis so that its angular velocity depends on the rotation angle φ as \( \omega = \omega_0 - k\phi \), where \( \omega_0 \) and k are positive constants. At t = 0 the angle φ = 0. The dependence of φ on time is
(a) \( \phi = \omega_0t-\frac{kt^2}{2} \)
(b) \( \phi = \frac{\omega_0}{k}\left(1-e^{-kt}\right) \)
(c) \( \phi = \frac{\omega_0}{k}e^{-kt} \)
(d) \( \phi = \omega_0 e^{-kt} \)
Show Answer & Explanation

Answer: (b) \( \phi = \frac{\omega_0}{k}\left(1-e^{-kt}\right) \)

Explanation:
1. \( \frac{d\phi}{dt} = \omega_0-k\phi \Rightarrow \int_0^\phi\frac{d\phi}{\omega_0-k\phi} = \int_0^tdt \).
2. \( -\frac1k\ln\frac{\omega_0-k\phi}{\omega_0} = t \Rightarrow \phi = \frac{\omega_0}{k}(1-e^{-kt}) \).

Teacher's Note:
1. Separate variables and integrate.
2. φ approaches ω₀/k at large t.

Question 7: Two bodies A and B, diametrically opposite on a circle of radius R, move counterclockwise with uniform speed v. A is given a constant tangential acceleration \( a = \frac{72v^2}{25\pi R} \) at t = 0 and collides with B after time \( t = \frac{5\pi R}{6v} \). The angle covered by A in this time is
(a) \( \frac{5\pi}{6} \)
(b) \( \pi \)
(c) \( \frac{11\pi}{6} \)
(d) \( \frac{13\pi}{6} \)
Show Answer & Explanation

Answer: (c) \( \frac{11\pi}{6} \)

Explanation:
1. \( \theta_A = \frac vRt+\pi = \frac{5\pi}{6}+\pi = \frac{11\pi}{6} \) (B has covered \( \frac{5\pi}{6} \), so A gained π).

Teacher's Note:
1. Resolve forces along and perpendicular to the motion carefully.
2. Check units and the special angle values before finalising.

Question 8: For the same two bodies (radius R, initial speed v, A given tangential acceleration \( \frac{72v^2}{25\pi R} \) and colliding after \( t = \frac{5\pi R}{6v} \)), the angular velocity of A at the instant of collision is
(a) \( \frac{17v}{5R} \)
(b) \( \frac{12v}{5R} \)
(c) \( \frac{v}{R} \)
(d) \( \frac{7v}{5R} \)
Show Answer & Explanation

Answer: (a) \( \frac{17v}{5R} \)

Explanation:
1. \( \omega = \frac vR+\frac aRt = \frac vR+\frac{72v^2}{25\pi R^2}\cdot\frac{5\pi R}{6v} = \frac vR+\frac{12v}{5R} = \frac{17v}{5R} \).

Teacher's Note:
1. Resolve forces along and perpendicular to the motion carefully.
2. Check units and the special angle values before finalising.

Question 9: A particle moves in a circle of radius R with constant speed v. The magnitude of its average acceleration during a time interval \( \frac{\pi R}{2v} \) is
(a) \( \frac{2v^2}{\pi R} \)
(b) \( \frac{2\sqrt2\,v^2}{\pi R} \)
(c) \( \frac{v^2}{R} \)
(d) \( \frac{\sqrt2\,v^2}{\pi R} \)
Show Answer & Explanation

Answer: (b) \( \frac{2\sqrt2\,v^2}{\pi R} \)

Explanation:
1. In time \( \frac{\pi R}{2v} \) the particle covers a quarter circle, so velocity turns through 90°: \( |\Delta v| = \sqrt2v \). Average acceleration \( = \frac{\sqrt2v}{\pi R/2v} = \frac{2\sqrt2v^2}{\pi R} \).

Teacher's Note:
1. Resolve forces along and perpendicular to the motion carefully.
2. Check units and the special angle values before finalising.

Question 10: A point moves along a circle of radius 20 cm with a constant tangential acceleration of 5 cm/s², starting from rest. The time after the start of motion when the normal acceleration equals the tangential acceleration is
(a) 1 s
(b) 4 s
(c) 5 s
(d) 2 s
Show Answer & Explanation

Answer: (d) 2 s

Explanation:
1. \( a_c = \frac{(a_tt)^2}{R} = a_t \Rightarrow t^2 = \frac{R}{a_t} = \frac{20}{5} = 4 \), so \( t = 2 \) s.

Teacher's Note:
1. Resolve forces along and perpendicular to the motion carefully.
2. Check units and the special angle values before finalising.

Question 11: Two strings, each 5 m long, tie a 4 kg block to a vertical rod at points 8 m apart (upper string above, lower string below). When the system rotates about the rod both strings are taut. The number of revolutions per minute needed for the tension in the upper string to be 20 kgf (take 1 kgf = 10 N, g = 10 m/s²) is
(a) \( \frac{30}{\pi}\sqrt{\frac{35}{4}} \)
(b) \( \frac{60}{\pi}\sqrt{\frac{35}{2}} \)
(c) \( \frac{30}{\pi}\sqrt{\frac{35}{2}} \)
(d) \( \frac{15}{\pi}\sqrt{\frac{35}{2}} \)
Show Answer & Explanation

Answer: (c) \( \frac{30}{\pi}\sqrt{\frac{35}{2}} \)

Explanation:
1. Each string makes angle φ with the horizontal where \( \sin\phi = 0.8 \), \( \cos\phi = 0.6 \), radius = 3 m. Vertical: \( (T_1-T_2)(0.8) = 40 \Rightarrow T_1-T_2 = 50 \) N, so with \( T_1 = 200 \) N, \( T_2 = 150 \) N. Horizontal: \( (T_1+T_2)(0.6) = m\omega^2(3) \Rightarrow \omega^2 = 17.5 \). Then \( N = \frac{60\omega}{2\pi} = \frac{30}{\pi}\sqrt{\frac{35}{2}} \) rpm.

Teacher's Note:
1. Resolve forces along and perpendicular to the motion carefully.
2. Check units and the special angle values before finalising.

Question 12: In the arrangement of the previous question (4 kg block, two 5 m strings tied 8 m apart on the rod), the upper string has tension 20 kgf. The tension in the lower string is (g = 10 m/s²)
(a) 50 N
(b) 200 N
(c) 150 N
(d) 100 N
Show Answer & Explanation

Answer: (c) 150 N

Explanation:
1. Vertical balance: \( (T_1-T_2)\sin\phi = mg \) with \( \sin\phi = 0.8 \) gives \( T_1-T_2 = 50 \) N. With \( T_1 = 200 \) N, \( T_2 = 150 \) N.

Teacher's Note:
1. Resolve forces along and perpendicular to the motion carefully.
2. Check units and the special angle values before finalising.

Question 13: A circular turntable of radius 0.5 m has a smooth groove along a radius with a spring of constant \( 10^2 \) N/cm fixed at its outer end. A ball of mass 90 g in the groove rests at 0.1 m from the centre with the spring unstretched when the table is at rest. The table is rotated with \( 10^2 \) rad/s. The ball moves away from its initial position, compressing the spring, by a distance nearly equal to
(a) \( 10^{-1} \) m
(b) \( 10^{-3} \) m
(c) 1 m
(d) \( 10^{-2} \) m
Show Answer & Explanation

Answer: (d) \( 10^{-2} \) m

Explanation:
1. \( k = 10^4 \) N/m and \( m\omega^2 = 0.09\times10^4 = 900 \) N/m. Equilibrium in the rotating frame: \( kx = m\omega^2(0.1+x) \Rightarrow x = \frac{90}{9100}\approx10^{-2} \) m.

Teacher's Note:
1. Resolve forces along and perpendicular to the motion carefully.
2. Check units and the special angle values before finalising.

Question 14: A particle is projected with speed 100 m/s at 60° with the horizontal. At time t the velocity vector becomes perpendicular to the direction of the initial velocity. The tangential acceleration of the particle at that time is (g = 10 m/s²)
(a) 10 m/s²
(b) 5 m/s²
(c) \( 5\sqrt3 \) m/s²
(d) 2.5 m/s²
Show Answer & Explanation

Answer: (b) 5 m/s²

Explanation:
1. The velocity is perpendicular to the initial one, so it is directed 30° below the horizontal. The component of g along it is \( g\sin30^\circ = 5 \) m/s².

Teacher's Note:
1. Resolve forces along and perpendicular to the motion carefully.
2. Check units and the special angle values before finalising.

Question 15: The minimum radius of a circle along which a cyclist can ride at 18 km/h if the coefficient of friction between tyres and road is 0.5 (g = 10 m/s²) is
(a) 2.5 m
(b) 10 m
(c) 5 m
(d) 25 m
Show Answer & Explanation

Answer: (c) 5 m

Explanation:
1. \( v = 5 \) m/s. \( r_{min} = \frac{v^2}{\mu g} = \frac{25}{5} = 5 \) m.

Teacher's Note:
1. Resolve forces along and perpendicular to the motion carefully.
2. Check units and the special angle values before finalising.

Question 16: A smooth wire is bent into a vertical circle of radius a and rotated about its vertical diameter AB with angular speed ω. A bead P is at rest relative to the ring at a point whose horizontal distance from the axis is a/2 (below the centre). Then \( \omega^2 \) equals
(a) \( \frac{2g}{a\sqrt3} \)
(b) \( \frac{g}{a} \)
(c) \( \frac{2g}{a} \)
(d) \( \frac{g\sqrt3}{2a} \)
Show Answer & Explanation

Answer: (a) \( \frac{2g}{a\sqrt3} \)

Explanation:
1. Here \( \sin\theta = \frac12 \), \( \theta = 30^\circ \) from the vertical. \( N\cos\theta = mg \), \( N\sin\theta = m\omega^2a\sin\theta \Rightarrow \omega^2 = \frac{g}{a\cos\theta} = \frac{2g}{a\sqrt3} \).

Teacher's Note:
1. Resolve forces along and perpendicular to the motion carefully.
2. Check units and the special angle values before finalising.

Question 17: A point P moves counter-clockwise on a circular path of radius 20 m, sweeping out a length s = t³ + 5 (s in metres, t in seconds). The acceleration of P at t = 2 s is nearly
(a) 13 m/s²
(b) 14 m/s²
(c) 12 m/s²
(d) 7.2 m/s²
Show Answer & Explanation

Answer: (b) 14 m/s²

Explanation:
1. \( v = 3t^2 = 12 \) m/s, \( a_t = 6t = 12 \) m/s², \( a_c = \frac{144}{20} = 7.2 \) m/s².
2. \( a = \sqrt{12^2+7.2^2} = 14 \) m/s².

Teacher's Note:
1. Find both components at t = 2 s.
2. Combine in quadrature.

Question 18: For a particle in uniform circular motion with speed v on a circle of radius R, the acceleration at the point P(R, θ) (θ measured from the x-axis) is
(a) \( -\frac{v^2}{R}\cos\theta\,\hat i+\frac{v^2}{R}\sin\theta\,\hat j \)
(b) \( \frac{v^2}{R}\hat i+\frac{v^2}{R}\hat j \)
(c) \( \frac{v^2}{R}\cos\theta\,\hat i+\frac{v^2}{R}\sin\theta\,\hat j \)
(d) \( -\frac{v^2}{R}\cos\theta\,\hat i-\frac{v^2}{R}\sin\theta\,\hat j \)
Show Answer & Explanation

Answer: (d) \( -\frac{v^2}{R}\cos\theta\,\hat i-\frac{v^2}{R}\sin\theta\,\hat j \)

Explanation:
1. Acceleration is centripetal, of magnitude \( \frac{v^2}{R} \), directed from P towards the centre, i.e. along \( -(\cos\theta\,\hat i+\sin\theta\,\hat j) \).

Teacher's Note:
1. The unit vector of the position is (cos θ, sin θ).
2. Centripetal acceleration points opposite to the position vector.

Question 19: Two cars of masses m₁ and m₂ move in circles of radii r₁ and r₂ and complete their circles in the same time t. The ratio of their centripetal accelerations is
(a) \( r_1 : r_2 \)
(b) \( m_1r_1 : m_2r_2 \)
(c) \( m_1 : m_2 \)
(d) 1 : 1
Show Answer & Explanation

Answer: (a) \( r_1 : r_2 \)

Explanation:
1. Same period means same ω, so \( a_c = \omega^2r\propto r \).

Teacher's Note:
1. Acceleration does not depend on mass.
2. Same T gives same ω.

Question 20: Two bodies A and B rotate about an axis such that the angle \( \theta_A \) covered by A is proportional to the square of time, while \( \theta_B \) varies linearly with time. Both start at \( \theta = 0 \). A completes its first revolution in \( \sqrt\pi \) s and B needs \( 4\pi \) s to complete half a revolution. The ratio \( \omega_A:\omega_B \) at t = 5 s is
(a) 40 : 1
(b) 20 : 1
(c) 80 : 1
(d) 160 : 1
Show Answer & Explanation

Answer: (c) 80 : 1

Explanation:
1. \( \theta_A = kt^2 \) with \( 2\pi = k\pi \Rightarrow k = 2 \), so \( \omega_A = 4t = 20 \) rad/s at t = 5 s.
2. \( \theta_B = ct \) with \( \pi = c(4\pi) \Rightarrow c = \frac14 \), so \( \omega_B = \frac14 \) rad/s. Ratio \( = 80:1 \).

Teacher's Note:
1. Use the given time for one revolution to find k.
2. ω = dθ/dt.

Question 21: A particle moves along a circle of radius \( \frac{20}{\pi} \) m with constant tangential acceleration. If its velocity is 80 m/s at the end of the second revolution after motion has begun, the tangential acceleration is
(a) \( 20\ \text{m/s}^2 \)
(b) \( 80\ \text{m/s}^2 \)
(c) \( 40\ \text{m/s}^2 \)
(d) \( 160\ \text{m/s}^2 \)
Show Answer & Explanation

Answer: (c) \( 40\ \text{m/s}^2 \)

Explanation:
1. Distance in two revolutions \( = 2\times2\pi R = 80 \) m.
2. \( v^2 = 2a_ts \Rightarrow a_t = \frac{6400}{160} = 40 \) m/s².

Teacher's Note:
1. Use v² = 2as for the tangential motion.
2. Two revolutions cover 4πR.

Question 22: Two blocks of masses \( m_1 = 10 \) kg and \( m_2 = 5 \) kg are joined by a light string of length 0.3 m placed along a diameter of a turntable, \( m_1 \) at 0.124 m from the centre O and \( m_2 \) on the other side. The coefficient of friction between the table and \( m_1 \) is 0.5; there is no friction for \( m_2 \). The table rotates at 10 rad/s with the masses at rest relative to it. The frictional force on \( m_1 \) is
(a) 36 N
(b) 88 N
(c) 124 N
(d) 50 N
Show Answer & Explanation

Answer: (a) 36 N

Explanation:
1. \( m_2 \): \( T = m_2\omega^2r_2 = 5\times100\times0.176 = 88 \) N. \( m_1 \): \( T+f = m_1\omega^2r_1 = 124 \) N.
2. So \( f = 36 \) N.

Teacher's Note:
1. The string tension is the same on both blocks.
2. Friction makes up the difference in the centripetal force.

Question 23: For the same two blocks (10 kg at 0.124 m, 5 kg on the other side at 0.176 m, μ = 0.5 for the 10 kg block only, g = 10 m/s²), the minimum angular speed at which the masses will slip is
(a) 10 rad/s
(b) 15 rad/s
(c) 5 rad/s
(d) \( \frac{25\sqrt2}{3} \) rad/s
Show Answer & Explanation

Answer: (d) \( \frac{25\sqrt2}{3} \) rad/s

Explanation:
1. At the limit \( f = \mu m_1g = 50 \) N: \( 50 = (m_1r_1-m_2r_2)\omega^2 = (1.24-0.88)\omega^2 \).
2. \( \omega^2 = \frac{50}{0.36} \Rightarrow \omega = \frac{25\sqrt2}{3}\approx11.8 \) rad/s.

Teacher's Note:
1. Use the maximum static friction.
2. T cancels between the two blocks.

Question 24: For the same two blocks on the turntable (10 kg and 5 kg, string 0.3 m, no friction on the 5 kg), the distances from the centre at which the masses should be placed (with the string taut) so that no frictional force acts on the 10 kg block are
(a) 0.2 m for the 10 kg block and 0.1 m for the 5 kg block
(b) 0.1 m for the 10 kg block and 0.2 m for the 5 kg block
(c) 0.15 m for each
(d) 0.124 m for the 10 kg block and 0.176 m for the 5 kg block
Show Answer & Explanation

Answer: (b) 0.1 m for the 10 kg block and 0.2 m for the 5 kg block

Explanation:
1. No friction means \( T = m_1\omega^2r_1 = m_2\omega^2r_2 \), so \( 10r_1 = 5r_2 \).
2. With \( r_1+r_2 = 0.3 \): \( r_1 = 0.1 \) m, \( r_2 = 0.2 \) m.

Teacher's Note:
1. Centripetal forces must equal the same tension.
2. Heavier block sits closer to the axis.

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Multiple Choice Questions (MCQs) for JEE Physics Circular Motion

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