A vehicle negotiating a curved road needs centripetal force directed toward the centre of the curve. On a level road, static friction alone supplies this force, capping the safe speed at v_max = √(μ_s g r). Banking the road — tilting the surface inward at angle θ — changes the game. The normal force now has a horizontal component that contributes to centripetal acceleration, reducing the burden on friction.
The core trap: treating centripetal force as a separate force on the free-body diagram. On a banked road, the real forces are weight (mg downward), the normal reaction N (perpendicular to the banked surface), and friction f (along the surface, directed up or down the incline depending on whether the vehicle is slow or fast). There is no additional "centripetal force" arrow. The net inward component of these real forces equals mv²/r.
For the frictionless ideal bank (μ_s = 0), setting N sin θ = mv²/r and N cos θ = mg gives the optimum speed: v₀ = √(gr tan θ). At this speed alone, friction is unnecessary.
With friction, the maximum safe speed becomes:
v_max = √[gr(μ_s + tan θ) / (1 − μ_s tan θ)]
This is the NCERT banked-road formula (Class 11 Physics Chapter 4, page 63). Notice: setting θ = 0 recovers the level-road formula v_max = √(μ_s g r). Setting μ_s = 0 recovers the optimum speed v₀ = √(gr tan θ).
Watch out: the formula assumes 1 − μ_s tan θ > 0. For very steep banks with high friction, this denominator approaches zero and the formula breaks down — but NEET problems stay within the valid regime.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
On a banked road with no friction, the centripetal force required for circular motion is provided by which component of the normal reaction?
Answer: B. On a frictionless banked road, the normal reaction N acts perpendicular to the road surface. Its horizontal component N sin θ points toward the centre of the curve and provides the centripetal force. The vertical component N cos θ balances weight (NCERT Class 11 Physics Chapter 4, page 63).
Why A is wrong: A is wrong because the vertical component N cos θ balances the weight mg — it acts vertically, not toward the centre of the curve. (trap: confusing vertical equilibrium with centripetal direction)
Why C is wrong: C is wrong because the normal force is perpendicular to the surface by definition; it has no tangential component along the surface. Friction acts along the surface, not the normal force.
Why D is wrong: D is wrong because only the horizontal component of N provides centripetal force. Using the full magnitude would overestimate the centripetal force and violate the vertical equilibrium condition N cos θ = mg.
What is the optimum speed for a vehicle on a frictionless banked road of radius r banked at angle θ?
Answer: D. Setting N sin θ = mv²/r and N cos θ = mg, dividing eliminates N and m, giving tan θ = v²/(gr), so v = √(gr tan θ) (NCERT Class 11 Physics Chapter 4, page 63).
Why A is wrong: A is wrong because using sin θ instead of tan θ means the vertical equilibrium (N cos θ = mg) has not been properly used to eliminate N. The correct derivation divides the radial equation by the vertical one, yielding tan θ, not sin θ.
Why B is wrong: B is wrong because the correct formula is v = √(gr tan θ), not v = √(gr / tan θ). Inverting the tangent function gives the wrong dimensional relationship and yields incorrect values for all angles except 45°.
Why C is wrong: C is wrong because v = √(μ_s gr) is the maximum speed on a level (unbanked) road where friction alone provides centripetal force. On a frictionless banked road, μ_s plays no role.
In the banked-road formula v_max = √[gr(μ_s + tan θ)/(1 − μ_s tan θ)], what does setting θ = 0 reduce the formula to?
Answer: C. When θ = 0, tan θ = 0. The formula becomes v_max = √[gr(μ_s + 0)/(1 − 0)] = √(μ_s gr), which is the level-road maximum speed formula (NCERT Class 11 Physics Chapter 4, page 63).
Why A is wrong: A is wrong because v = √(gr tan θ) is the frictionless banked-road optimum speed, not the level-road formula. Setting θ = 0 makes tan θ = 0, which would give v = 0 in this expression — clearly not the level-road answer.
Why B is wrong: B is wrong because v_max = √(gr / μ_s) has μ_s in the denominator. Substituting θ = 0 into the banked-road formula places μ_s in the numerator under the square root, not in the denominator.
Why D is wrong: D is wrong because a level road with friction can still support circular motion. Setting θ = 0 does not eliminate centripetal force — friction provides it, giving v_max = √(μ_s gr).
A car travels on a frictionless banked road of radius 50 m banked at 30°. What is the safe speed? (Take g = 10 m/s², tan 30° = 1/√3.)
Answer: A. v = √(gr tan θ) = √(10 × 50 × 1/√3) = √(500/√3) = √(288.7) ≈ 17.0 m/s (NCERT Class 11 Physics Chapter 4, page 63).
Why B is wrong: B is wrong because 29.4 m/s = √(gr/tan θ) = √(500√3) divides by tan θ instead of multiplying by it. (trap: inverting the angular factor)
Why C is wrong: C is wrong because √(500) ≈ 22.4 m/s omits the tan θ factor entirely, effectively treating the bank angle as 45° where tan 45° = 1. The correct calculation must include tan 30° = 1/√3.
Why D is wrong: D is wrong because 15.8 m/s = √(gr sin θ) = √250 uses sin 30° = 0.5 instead of tan 30° = 1/√3. The derivation yields tan θ, not sin θ, from dividing the two equilibrium equations.
A banked road has radius 100 m, banking angle 45°, and coefficient of static friction μ_s = 0.5. What is the maximum safe speed? (Take g = 10 m/s², tan 45° = 1.)
Answer: A. v_max = √[gr(μ_s + tan θ)/(1 − μ_s tan θ)] = √[10 × 100 × (0.5 + 1)/(1 − 0.5 × 1)] = √[1000 × 1.5/0.5] = √(3000) ≈ 54.8 m/s (NCERT Class 11 Physics Chapter 4, page 63).
Why B is wrong: B is wrong because √(1000) = √(gr × tan θ) is the frictionless optimum speed. This ignores friction entirely, which significantly raises the maximum safe speed when μ_s = 0.5. (mistake: treating centripetal force as only the normal force component, ignoring friction's contribution)
Why C is wrong: C is wrong because √(500) = √(μ_s gr) is the level-road formula with θ = 0. On a banked road at 45° with friction, both the banking and friction contribute to centripetal force, giving a much higher limit.
Why D is wrong: D is wrong because this value results from using (μ_s + tan θ) in the numerator but forgetting to divide by (1 − μ_s tan θ) in the denominator — effectively setting the denominator to 1. The denominator correction is essential when both friction and banking act together.
A highway curve of radius 200 m is designed so that a car travelling at 20 m/s needs no friction. What banking angle θ is required? (Take g = 10 m/s².)
Answer: D. For the frictionless optimum: tan θ = v²/(gr) = (20)²/(10 × 200) = 400/2000 = 0.2, so θ = arctan(0.2) ≈ 11.3° (NCERT Class 11 Physics Chapter 4, page 63).
Why A is wrong: A is wrong because tan θ = 0.1 would correspond to v² = 0.1 × gr = 0.1 × 2000 = 200, giving v ≈ 14.1 m/s, not 20 m/s. It comes from a stray factor 2, v²/(2gr) = 400/4000, carried over from v² = 2gh.
Why B is wrong: B is wrong because tan θ = 2.0 comes from dropping g: v²/r = 400/200 = 2.0 is the centripetal acceleration in m/s², not the dimensionless tan θ. The resulting 63.4° would be impractical for a highway.
Why C is wrong: C is wrong because tan θ = 0.5 would require v² = 0.5 × 2000 = 1000, giving v ≈ 31.6 m/s. This could result from using r = 80 m instead of 200 m, or a computational slip.
A student draws a free-body diagram for a car on a banked road and includes four arrows: weight mg downward, normal reaction N perpendicular to the surface, friction f along the surface, and a separate "centripetal force" arrow pointing toward the centre. What is wrong with this diagram?
Answer: C. Centripetal force is not a new fundamental force. It is the label for the net inward radial force, which on a banked road is provided by the horizontal components of the normal reaction and friction. Adding a separate centripetal force arrow double-counts the inward force (NCERT Class 11 Physics Chapter 4, page 61).
Why A is wrong: A is wrong because friction CAN act along the surface — its direction depends on whether the car is above or below the optimum speed. The direction along the surface is correct; the issue is the fictitious centripetal force arrow, not friction's direction.
Why B is wrong: B is wrong because the normal reaction is always perpendicular to the contact surface by definition. On a banked road, this means N is tilted from the vertical, which is correct. Forcing N to be vertical would violate the definition of normal force.
Why D is wrong: D is wrong because weight always acts vertically downward toward Earth's centre, regardless of the surface orientation. Gravitational force does not depend on the tilt of the road.
A car rounds a banked curve of radius 50 m at 20 m/s. The road is banked at angle θ where tan θ = 0.5. If the car does not skid, what is the minimum coefficient of static friction required? (Take g = 10 m/s².)
Answer: B. B is correct. The car exceeds the optimum speed v₀ = √(gr tan θ) = √(10 × 50 × 0.5) = √250 ≈ 15.8 m/s, so friction must act inward (up the incline). At the friction limit: v² = gr(μ_s + tan θ)/(1 − μ_s tan θ). Substituting: 400 = 500(μ_s + 0.5)/(1 − 0.5μ_s). Cross-multiplying: 400(1 − 0.5μ_s) = 500(μ_s + 0.5), so 400 − 200μ_s = 500μ_s + 250, giving 150 = 700μ_s and μ_s = 150/700 = 0.214 ≈ 0.21. The denominator (1 − μ_s tan θ) is the part that is easy to drop, and dropping it changes the answer by 40% (NCERT Class 11 Physics Chapter 4, page 63).
Why A is wrong: A is wrong because μ_s = 0.5 far exceeds the required minimum of 0.214. This value likely comes from reading tan θ = 0.5 and assuming the friction coefficient must equal the tangent of the banking angle, which has no physical basis.
Why C is wrong: C is wrong because μ_s = 1.0 is unrealistically high for a tyre-road interface and far exceeds the required 0.214. This could result from setting v² = gr·μ_s (the level-road formula) and solving μ_s = v²/(gr) = 400/500 = 0.8, then rounding up. (mistake: ignoring the banking contribution)
Why D is wrong: D is wrong, and it is the trap worth knowing: 0.30 is what the truncated formula μ_s = v²/(gr) − tan θ = 0.8 − 0.5 gives. That form drops the (1 − μ_s tan θ) denominator, which is only valid when μ_s tan θ is negligible. Here it is not, and the true answer 0.214 is 40% lower. This lesson's own MCQ 3 quotes the full formula.
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Given
A circular highway curve has radius r = 100 m and is banked at θ = 30° (tan 30° = 1/√3 ≈ 0.577). The coefficient of static friction between tyres and road is μ_s = 0.3. Take g = 10 m/s².
Required
Find the maximum safe speed v_max for a car on this curve.
Concept
On a banked road with friction, both the horizontal component of the normal force and the friction force (directed up the incline at maximum speed) provide centripetal acceleration. The maximum speed is reached when static friction is at its limit.
Formula
$$v_\max = \sqrt{\frac{gr(\mu_s + \tan\theta)}{1 - \mu_s \tan\theta}}$$
Substitution
$$v_\max = \sqrt{\frac{10 \times 100 \times (0.3 + 0.577)}{1 - 0.3 \times 0.577}}$$
Calculation
Numerator inside square root: 1000 × 0.877 = 877.
Denominator inside square root: 1 − 0.1732 = 0.8268.
Ratio: 877 / 0.8268 = 1060.7.
v_max = √(1060.7) ≈ 32.6 m/s.
Note on exact constants: g = 10 m/s² is a problem-defined exact value and does not limit significant figures. The radius r = 100 m is given as exact. The significant-figure count is governed by μ_s = 0.3 (1 sig fig) and tan 30° = 0.577 (3 sig fig). Reporting to 3 significant figures: v_max ≈ 32.6 m/s.
Final answer
v_max ≈ 32.6 m/s ≈ 117 km/h.
Common trap
Students who treat centripetal force as a separate force on the FBD will double-count the inward force and get an incorrect v_max. The centripetal force is the NET result of N sin θ + f cos θ (horizontal components), not an additional arrow.
Similar NEET-style question
A car travels on a banked curve of radius 50 m at angle θ = 45° with μ_s = 0.2. Find v_max. (Answer: apply the same formula with tan 45° = 1.)
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On a road banked at angle θ with friction coefficient μ_s, the maximum safe speed is v_max = √[g r (μ_s + tan θ) / (1 − μ_s tan θ)]. The optimum (no-friction-needed) speed is v_o = √(g r tan θ). Banking allows higher safe speeds than a level road.
-- NCERT Class 11 Physics, Ch. 4, p. 64On a road banked at angle theta from horizontal with tyre-road friction coefficient mu_s, this is the maximum speed for safe negotiation of a curve of radius r.
| Symbol | Quantity | SI Unit |
|---|---|---|
| v_max | Maximum safe speed | m/s |
| g | Gravitational acceleration | m/s^2 |
| r | Radius of the curve | m |
| mu_s | Coefficient of static friction | (dimensionless) |
| theta | Banking angle | rad/deg |
More in Laws of Motion: 16 exam traps and mistakes · 8 formulas · 10 question patterns from its other lessons.
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