Banking of Roads & Well of Death — IISER Physics MCQs with Solutions
Free IISER 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. Zero ✓ Correct
B. Maximum up the slope
C. Equal to μN
D. Maximum down the slope
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. 10 m/s
C. 40 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. 0.90
B. 0.15
C. 0.45 ✓ Correct
D. 0.30
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. ≈ 12.6 m/s ✓ Correct
B. ≈ 16 m/s
C. ≈ 8 m/s
D. ≈ 25.6 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. 0.2 ✓ Correct
B. 0.05
C. 0.4
D. 0.1
Solution: tanθ = v²/rg = 400/2000 = 0.2.
Q7 — Banking of Roads & Well of Death · hard · theory
In the well of death, the rider does NOT fall because:
A. Gravity vanishes at speed
B. The normal force balances the weight
C. The centrifugal force presses him up
D. The wall's normal reaction provides the centripetal force while friction balances the weight ✓ Correct
Solution: N = mv²/r is horizontal (centripetal); vertical equilibrium needs f = μN ≥ mg — which sets the minimum speed √(rg/μ). Trap: N does NOT balance gravity; friction does.
Q8 — Banking of Roads & Well of Death · hard · numerical
A rider circles a death well at the minimum speed. If he doubles his speed, the friction force actually needed to hold him up becomes:
A. One quarter
B. Unchanged — still exactly his weight mg ✓ Correct
C. Half
D. Four times larger
Solution: The vertical friction requirement is always exactly mg. Doubling v quadruples N = mv²/r, so the AVAILABLE friction μN quadruples — bigger safety margin, same requirement. A favourite conceptual trap.