Banking of Roads & Well of Death — JEE Main Physics MCQs with Solutions
Free JEE Main Physics Banking of Roads & Well of Death 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 — Banking of Roads & Well of Death · easy · theory
Roads are banked at curves so that:
A. The weight of the car decreases
B. Friction increases
C. Cars can stop faster
D. A component of the normal reaction provides the centripetal force, reducing reliance on friction ✓ Correct
Solution: At the design speed v₀ = √(rg tanθ), N sinθ alone supplies mv²/r — no friction needed.
Q2 — Banking of Roads & Well of Death · medium · theory
At the optimum (design) speed on a banked road, the friction force on the tyres is:
A. Maximum down the slope
B. Maximum up the slope
C. Equal to μN
D. Zero ✓ Correct
Solution: The horizontal component of N exactly meets the centripetal demand, so no frictional force is called upon — minimum tyre wear.
Q3 — Banking of Roads & Well of Death · easy · numerical
A curve of radius 40 m is banked at 45°. The optimum speed is:
A. 20 m/s ✓ Correct
B. 40 m/s
C. 10 m/s
D. 14.1 m/s
Solution: v₀ = √(rg tan45°) = √400 = 20 m/s.
Q4 — Banking of Roads & Well of Death · medium · numerical
A road of radius 50 m is designed for 15 m/s. The required banking angle satisfies tanθ =
A. 0.45 ✓ Correct
B. 0.90
C. 0.30
D. 0.15
Solution: tanθ = v²/rg = 225/500 = 0.45 ⇒ θ ≈ 24°.
Q5 — Banking of Roads & Well of Death · hard · numerical
In a well of death of radius 6.4 m with μ = 0.4 between tyres and wall, the minimum speed of the rider is:
A. ≈ 8 m/s
B. ≈ 12.6 m/s ✓ Correct
C. ≈ 25.6 m/s
D. ≈ 16 m/s
Solution: N = mv²/r supplies the wall reaction; friction μN ≥ mg ⇒ v_min = √(rg/μ) = √(6.4×10/0.4) = √160 ≈ 12.6 m/s.
Q6 — Banking of Roads & Well of Death · medium · numerical
A railway track of radius 200 m is banked so that a train at 20 m/s exerts no side thrust on the rails. tanθ equals:
A. 0.05
B. 0.2 ✓ Correct
C. 0.1
D. 0.4
Solution: tanθ = v²/rg = 400/2000 = 0.2.
Q7 — Banking of Roads & Well of Death · hard · numerical
A curve of radius 90 m is banked at tanθ = 0.5 with μ = 0.5 between tyres and road. The MAXIMUM safe speed is:
A. √300 m/s
B. √1200 ≈ 34.6 m/s ✓ Correct
C. √450 ≈ 21.2 m/s
D. 30 m/s
Solution: v_max = √[rg(tanθ + μ)/(1 − μ tanθ)] = √[900×1/0.75] = √1200 ≈ 34.6 m/s.
Q8 — Banking of Roads & Well of Death · medium · numerical
For the same banked curve (r = 90 m, tanθ = 0.5, μ = 0.5), the MINIMUM speed without slipping down is:
A. √300 m/s
B. 10 m/s
C. 0 (the road is not steep enough to need one, since tanθ ≤ μ) ✓ Correct
D. √450 m/s
Solution: v_min² = rg(tanθ − μ)/(1 + μtanθ) = 0 when tanθ = μ — the car can even stand still. Trap: v_min exists only when tanθ > μ.