Kinematics of Circular Motion — JEE Main Physics MCQs with Solutions
Free JEE Main Physics Kinematics of Circular Motion MCQs with step-by-step solutions (8 questions). Part of Circular Motion. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Kinematics of Circular Motion · easy · theory
In non-uniform circular motion, the total acceleration of a particle is:
A. Only a_c
B. a_c − a_t
C. a_c + a_t
D. √(a_c² + a_t²), since the centripetal and tangential accelerations are perpendicular ✓ Correct
Solution: a_c (radial, v²/r) ⊥ a_t (tangential, αr); they add vectorially: a = √(a_c² + a_t²). Trap: forgetting a_t in non-uniform motion.
Q2 — Kinematics of Circular Motion · medium · theory
A particle moves in a circle with uniformly INCREASING speed. The angle between its total acceleration and its velocity is:
A. Exactly 90°
B. Zero
C. Acute (between 0° and 90°) ✓ Correct
D. Obtuse
Solution: a_t is along v (speeding up) and a_c is ⊥ v; the resultant leans forward of the radius — an acute angle with v. (Obtuse if slowing down; 90° only if uniform.)
Q3 — Kinematics of Circular Motion · easy · numerical
A wheel turns at 240 rpm. Its angular velocity in rad/s is:
A. 2π rad/s
B. 4π rad/s
C. 8π ≈ 25.1 rad/s ✓ Correct
D. 240 rad/s
Solution: ω = 2πN/60 = 2π×240/60 = 8π ≈ 25.1 rad/s.
Q4 — Kinematics of Circular Motion · medium · numerical
A particle moves in a circle of radius 2 m. At an instant its speed is 4 m/s and its speed is increasing at 3 m/s². The magnitude of its total acceleration is:
A. 3 m/s²
B. 11 m/s²
C. √73 ≈ 8.5 m/s² ✓ Correct
D. 8 m/s²
Solution: a_c = v²/r = 8 m/s², a_t = 3 m/s²; a = √(64 + 9) = √73 ≈ 8.5 m/s². Trap: not 8 + 3.
Q5 — Kinematics of Circular Motion · hard · numerical
A particle starts from rest on a circle of radius 1 m with constant angular acceleration α = 2 rad/s². At t = 1 s the magnitude of its TOTAL acceleration is:
A. 2 m/s²
B. 6 m/s²
C. 4 m/s²
D. √20 ≈ 4.5 m/s² ✓ Correct
Solution: ω = αt = 2 rad/s ⇒ a_c = ω²r = 4 m/s²; a_t = αr = 2 m/s²; a = √(16+4) = √20 ≈ 4.5 m/s².
Q6 — Kinematics of Circular Motion · medium · numerical
A wheel accelerates uniformly from rest to 20 rad/s in 4 s. The number of revolutions completed in this time is:
A. 20/π ≈ 6.4 ✓ Correct
B. 10
C. 80/π
D. 40
Solution: θ = ½αt² with α = 5 rad/s²: θ = ½×5×16 = 40 rad ⇒ n = 40/2π = 20/π ≈ 6.4 rev.
Q7 — Kinematics of Circular Motion · medium · numerical
A particle moves on a circle of radius 20/π m with speed v = 2t (m/s). At t = 2 s, its centripetal acceleration is:
A. 2 m/s²
B. 4 m/s²
C. 0.8π m/s² ✓ Correct
D. 1.6π m/s²
Solution: v(2) = 4 m/s; a_c = v²/r = 16π/20 = 0.8π ≈ 2.5 m/s².
Q8 — Kinematics of Circular Motion · hard · numerical
A particle in a circle of radius 1 m has speed v = t² (m/s). At t = 1 s, the angle between its total acceleration and the radius is:
A. 45°
B. tan⁻¹(2) ≈ 63.4° ✓ Correct
C. 90°
D. tan⁻¹(1/2) ≈ 26.6°
Solution: a_t = dv/dt = 2t = 2; a_c = v²/r = 1. tanφ = a_t/a_c = 2 ⇒ φ = tan⁻¹2 with respect to the radial direction.