Motion on Flat Circular Tracks & Friction — JEE Main Physics MCQs with Solutions
Free JEE Main Physics Motion on Flat Circular Tracks & Friction 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 — Motion on Flat Circular Tracks & Friction · easy · theory
On a flat (unbanked) road, the centripetal force for a turning car is provided by:
A. The engine thrust directly
B. Static friction between the tyres and the road ✓ Correct
C. The car's weight
D. The normal reaction
Solution: Friction is the only horizontal force available on level ground; v_max = √(μrg).
Q2 — Motion on Flat Circular Tracks & Friction · medium · theory
The maximum safe speed on a flat circular road is independent of:
A. The mass of the vehicle ✓ Correct
B. The value of g
C. The coefficient of friction
D. The radius of the turn
Solution: μmg = mv²/r ⇒ v_max = √(μrg) — mass cancels. Trap: heavier cars can NOT corner faster on this account.
Q3 — Motion on Flat Circular Tracks & Friction · easy · numerical
A car turns on a flat road of radius 45 m with μ = 0.5. The maximum safe speed is:
A. 9 m/s
B. 30 m/s
C. 15 m/s ✓ Correct
D. 22.5 m/s
Solution: v = √(μrg) = √(0.5×45×10) = √225 = 15 m/s.
Q4 — Motion on Flat Circular Tracks & Friction · medium · numerical
A coin sits 0.2 m from the centre of a turntable (μ = 0.5). The maximum angular speed before it slips is:
A. 10 rad/s
B. 25 rad/s
C. 5 rad/s ✓ Correct
D. 2.5 rad/s
Solution: μg = ω²r ⇒ ω = √(μg/r) = √(0.5×10/0.2) = √25 = 5 rad/s.
Q5 — Motion on Flat Circular Tracks & Friction · hard · numerical
A car circles a flat track at the maximum safe speed for radius r. To take a turn of radius r/4 on the same surface, its maximum speed must be:
A. Half the original ✓ Correct
B. Twice the original
C. A quarter of the original
D. Unchanged
Solution: v_max ∝ √r ⇒ √(1/4) = 1/2 — tighter turns demand much lower speeds.
Q6 — Motion on Flat Circular Tracks & Friction · medium · numerical
A cyclist at 10 m/s brakes while turning on a flat road of radius 20 m (μ = 0.5, g = 10). The friction needed just for the turn (per kg) is:
A. 5 m/s² — exactly the friction limit, so no braking margin remains ✓ Correct
B. 10 m/s²
C. 1 m/s²
D. 2.5 m/s²
Solution: a_c = v²/r = 100/20 = 5 m/s² = μg — friction is fully used by the turn; any braking causes a skid. This is why you slow BEFORE a corner.
Q7 — Motion on Flat Circular Tracks & Friction · hard · numerical
Two coins sit at radii 4 cm and 9 cm on a turntable. As the speed slowly increases, the outer coin slips first. The ratio of the angular speeds at which the two coins (outer : inner) would slip is:
A. 2 : 3 ✓ Correct
B. 4 : 9
C. 3 : 2
D. 9 : 4
Solution: ω_slip = √(μg/r) ∝ 1/√r ⇒ ω_out : ω_in = 1/√9 : 1/√4 = 2 : 3 — confirming the outer coin (smaller ω_slip) goes first.
Q8 — Motion on Flat Circular Tracks & Friction · medium · numerical
A car (μ = 0.5) moves on a flat road with speed v = 5t (m/s) on a curve of radius 62.5 m. The time at which it starts to skid (friction fully used by the centripetal demand alone) is:
A. ≈ 3.5 s ✓ Correct
B. 5 s
C. 2.5 s
D. 12.5 s
Solution: Skid when v² = μrg = 312.5 ⇒ v ≈ 17.7 m/s ⇒ t = v/5 ≈ 3.5 s. (Strictly the tangential demand advances this slightly — the trap the options test.)