Maximum safe speed on a level circular road of radius r without skidding is v_max = √(μ_s g r). The friction force must provide the centripetal force; speeds higher than this exceed available friction and the vehicle skids outward.
-- NCERT Class 11 Physics, Ch. 4, p. 63Vehicle Level Road
8 MCQs6 revision cards9-step worked example
Source: NCERT Laws of MotionOfficial key: NTA-verifiedLast updated: 26 Sep 2026
Vehicle Level Road, explained for NEET
When a car rounds a flat, unbanked curve, the only horizontal force available to bend its path inward is static friction between the tyres and the road. No banking angle assists; no "centripetal force" magically appears on the free-body diagram. Static friction is the centripetal force here.
The key relationship. For a vehicle of mass m moving at speed v around a level curve of radius r:
- The normal force is N = mg (level road, no vertical acceleration).
- The required centripetal force is mv²/r, directed toward the centre of the curve.
- Static friction supplies this: f_s = mv²/r.
- The ceiling on static friction is f_{s,max} = μ_s N = μ_s mg.
Setting mv²/r = μ_s mg, the mass cancels and the maximum safe speed is:
v_max = √(μ_s g r)
Exceed this and the tyres lose grip — the vehicle skids outward.
High-frequency trap — friction force vs friction-limited acceleration. NEET questions sometimes ask for the maximum acceleration a vehicle can have so that an object resting on it doesn't slide. Students compute μmg (a force in newtons) when the answer is μg (acceleration in m/s²). The mass cancels. Read the question unit carefully: newtons → force; m/s² → acceleration.
Concept trap — centripetal force is not a separate force. On the free-body diagram of the car, draw only real forces: weight mg downward, normal N upward, static friction f_s inward. Their net radial component must equal mv²/r. Never draw an additional arrow labelled "centripetal force" — that double-counts the inward force already provided by friction (NCERT Class 11 Physics, Chapter 4, page 63).
Watch-out. The formula v_max = √(μ_s g r) uses static friction (tyres are not sliding on the road). If a question specifies kinetic friction, the vehicle is already skidding — different analysis applies.
Can you answer these Vehicle Level Road MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Direct ApplicationPractice
A car moves on a level circular road of radius 50 m. If the coefficient of static friction between the tyres and road is 0.4, what is the maximum speed the car can maintain without skidding? (Take g = 10 m/s²)
Show answer and why every option is right or wrong
Answer: B. v_max = √(μ_s g r) = √(0.4 × 10 × 50) = √200 ≈ 14.1 m/s. Option B (14 m/s) is the closest correct value. (NCERT Class 11 Physics, Chapter 4, page 63.)
Why A is wrong: A (10 m/s): This results from computing √(μ_s × r) = √(0.4 × 50) = √20 ≈ 4.5, then rounding incorrectly, or from forgetting to include g in the formula.
Why C is wrong: C (20 m/s): This comes from computing μ_s × g × r = 0.4 × 10 × 50 = 200 and then taking the answer as 20 by mistakenly extracting √(400) instead of √(200), or from confusing v_max² with v_max.
Why D is wrong: D (200 m/s): This is μ_s × g × r = 200 without taking the square root — a common trap of forgetting the √ in the formula.
MCQ 2Easy RecallPractice
On a level road, the maximum safe speed for a vehicle on a circular curve depends on:
Show answer and why every option is right or wrong
Answer: C. v_max = √(μ_s g r). The formula contains only μ_s, g, and r. Mass cancels out. (NCERT Class 11 Physics, Chapter 4, page 63.)
Why A is wrong: A is wrong because mass does not appear in v_max = √(μ_s g r) — it cancels when setting μ_s mg = mv²/r.
Why B is wrong: B is wrong on two counts: the relevant friction is static (tyres are not sliding), and mass does not appear in the formula.
Why D is wrong: D is wrong because mass cancels out of the derivation. v_max is independent of the vehicle's mass.
MCQ 3Easy RecallPractice
Which force provides the centripetal acceleration for a car taking a turn on a level road?
Show answer and why every option is right or wrong
Answer: A. On a level unbanked road, the only horizontal force that can act toward the centre of the curve is static friction. It provides the required centripetal acceleration. (NCERT Class 11 Physics, Chapter 4, page 63.)
Why B is wrong: B is wrong because weight acts vertically downward and is balanced by the normal force on a level road; it has no horizontal component.
Why C is wrong: C is wrong because the engine's driving force acts tangentially (along the road) to maintain speed, not radially inward toward the centre of the curve.
Why D is wrong: D is wrong because centripetal force is not a separate fundamental force. It is the name for the net radial inward force — here provided entirely by static friction. Drawing it as an extra force double-counts the inward force (trap: treating centripetal force as a new force).
MCQ 4Direct ApplicationPractice
A box rests on the floor of a truck moving on a level circular road of radius 100 m. The coefficient of static friction between the box and the truck floor is 0.5. What is the maximum speed of the truck so the box does not slide? (g = 10 m/s²)
Show answer and why every option is right or wrong
Answer: A. The box stays on the truck as long as static friction can supply the centripetal force: μ_s mg ≥ mv²/r. Maximum speed v_max = √(μ_s g r) = √(0.5 × 10 × 100) = √500 ≈ 22.4 m/s. (NCERT Class 11 Physics, Chapter 4, page 63.)
Why B is wrong: B (500 m/s): This is μ_s × g × r = 500 without the square root — forgetting to take √ in v_max = √(μ_s g r).
Why C is wrong: C (5 m/s): Likely computed as μ_s × g = 0.5 × 10 = 5, confusing the friction-limited acceleration (μ_s g) with the speed (trap: friction force vs friction-limited acceleration).
Why D is wrong: D (50 m/s): This is μ_s × r = 0.5 × 100 = 50, dropping both g and the square root; v_max = √(μ_s g r) = √500 ≈ 22.4 m/s.
MCQ 5Direct ApplicationPYQ Pattern
A body of mass 5 kg rests on the floor of a vehicle moving on a level road. If μ_s = 0.6 and g = 10 m/s², what is the maximum acceleration of the vehicle so the body does not slide?
Show answer and why every option is right or wrong
Answer: B. The maximum acceleration before the body slides is a_max = μ_s g = 0.6 × 10 = 6 m/s². The mass cancels — the question asks for acceleration, not force. (Anchored to PYQ pattern: body-on-vehicle friction limit.)
Why A is wrong: A (30 N): This is the maximum friction force (μ_s × m × g = 0.6 × 5 × 10 = 30 N), not the maximum acceleration. The question asks for acceleration in m/s², not force in newtons (trap: friction force vs friction-limited acceleration).
Why C is wrong: C (3 m/s²): This results from dividing μ_s g by 2, likely from a misremembering of the formula or an erroneous extra factor.
Why D is wrong: D (0.6 m/s²): This is just the coefficient of friction μ_s itself, not μ_s × g. The coefficient is dimensionless and cannot be an acceleration.
MCQ 6Concept TrapPractice
A student draws the free-body diagram of a car on a level circular road. They draw four forces: weight downward, normal force upward, static friction inward, and "centripetal force" inward. What is the error?
Show answer and why every option is right or wrong
Answer: D. Centripetal force is the net inward radial force, not an independent force. On a level road, static friction alone provides this inward force. Drawing both "static friction inward" and "centripetal force inward" double-counts the same physical effect. (Mistake anchor: treating centripetal force as a new force.)
Why A is wrong: A is wrong because weight MUST be drawn on a free-body diagram regardless of balance. Being balanced by the normal force doesn't mean it vanishes — both exist and both must appear.
Why B is wrong: B is wrong because static friction does point inward (toward the centre) on a level road. 'Centrifugal force' is a pseudo-force appearing only in a rotating reference frame, not on a free-body diagram in an inertial frame.
Why C is wrong: C is wrong because the normal force on a level road acts perpendicular to the road surface, which is vertically upward. It has no horizontal component.
MCQ 7Direct ApplicationPractice
If the radius of a level circular road is doubled while the coefficient of static friction remains the same, by what factor does the maximum safe speed change?
Show answer and why every option is right or wrong
Answer: B. v_max = √(μ_s g r). If r → 2r, then v_max → √(μ_s g × 2r) = √2 × √(μ_s g r) = √2 × v_max(original). The maximum speed increases by a factor of √2. (NCERT Class 11 Physics, Chapter 4, page 63.)
Why A is wrong: A (doubles): This would require v_max to be proportional to r, but v_max ∝ √r. Doubling r gives √2 times the original speed, not 2 times.
Why C is wrong: C (remains the same): v_max depends on r through √r. Changing the radius changes the maximum safe speed.
Why D is wrong: D (increases by factor of 4): This confuses v_max with v_max². Since v_max ∝ √r, doubling r quadruples v_max² but only multiplies v_max by √2.
MCQ 8Concept TrapPractice
A box of mass 2 kg sits on the floor of a truck. The truck is on a level road. A small horizontal force of 3 N is applied to the box by the truck's acceleration. If μ_s = 0.5 and g = 10 m/s², what is the actual static friction acting on the box?
Show answer and why every option is right or wrong
Answer: C. The maximum static friction is μ_s mg = 0.5 × 2 × 10 = 10 N. But the applied tangential load is only 3 N, which is well below the ceiling. Static friction is self-adjusting: it exactly matches the applied force to prevent sliding. So the actual static friction is 3 N, not 10 N. (Mistake anchor: assuming static friction always equals its maximum value.)
Why A is wrong: A (10 N): This is the MAXIMUM static friction (μ_s mg = 10 N), not the actual friction. Static friction self-adjusts to match the applied force when below the limit. Using μ_s mg always over-estimates friction in non-limiting cases (trap: static friction always at maximum).
Why B is wrong: B (0 N): There is a net horizontal force on the box from the truck's acceleration. Static friction must act to keep the box from sliding, so it cannot be zero.
Why D is wrong: D (5 N): This has no physical basis — it is neither the applied force (3 N) nor the maximum friction (10 N). Likely an arithmetic error or an incorrect averaging.
Free NEET study resources
Get a structured 30-day Mechanics plan and a complete formula booklet — delivered to your inbox instantly.
Vehicle Level Road: quick recall before you leave
How do you solve a Vehicle Level Road question? A worked example
- 1
Given
A car negotiates a level circular curve of radius r = 80 m. The coefficient of static friction between the tyres and road is μ_s = 0.50. Take g = 10 m/s² (exact, problem-defined).
- 2
Required
Maximum speed v_max at which the car can round the curve without skidding.
- 3
Concept
On a level road, static friction is the sole provider of centripetal force. The car skids when the required centripetal force exceeds the maximum static friction.
- 4
Formula
Setting centripetal force equal to maximum static friction:
mv²/r = μ_s mg → v_max = √(μ_s g r) - 5
Substitution
v_max = √(0.50 × 10 × 80) = √(400)
- 6
Calculation
v_max = √400 = 20 m/s
Note: g = 10 m/s² is an exact problem-defined constant, and r = 80 m and μ_s = 0.50 are given values. The integer result 20 m/s is exact under these given values. - 7
Final answer
v_max = 20 m/s
- 8
Common trap
A student who computes μ_s × m × g × r (forgetting the square root and including mass) gets a number in newton-metres — dimensionally wrong for speed. Another common error: computing μ_s g = 5 m/s² (the friction-limited acceleration) and reporting it as the speed. Always check units: speed is in m/s.
- 9
Similar NEET-style question
A box sits on the floor of a truck rounding a level curve of radius 200 m. If μ_s = 0.40, find the maximum speed of the truck so the box does not slide. (Answer: v_max = √(0.40 × 10 × 200) = √800 ≈ 28.3 m/s.)
---
What to remember before solving Vehicle Level Road questions
Which Vehicle Level Road formulas do you need for NEET?
1 formula — click to collapse
Maximum safe speed on a level circular road
v_\max = \sqrt{\mu_s g r}
Static friction is the only force available to provide centripetal acceleration on a level road. Setting mu_s*N = m*v^2/r and N = m*g gives this maximum-safe speed bound.
| Symbol | Quantity | SI Unit |
|---|---|---|
| v_max | Maximum safe speed (no skid) | m/s |
| mu_s | Coefficient of static friction (tyre vs road) | (dimensionless) |
| g | Gravitational acceleration | m/s^2 |
| r | Radius of the circular path | m |
Valid when
- Road is level (no banking)
- Tyres do not slide (static friction regime)
- Driver maintains uniform speed on the curve
Do NOT use when
- Banked road (use the banked-road formula)
- Slippery / wet road where mu_s is reduced
More in Laws of Motion: 16 exam traps and mistakes · 8 formulas · 10 question patterns from its other lessons.
Vehicle Level Road questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
Sources
NCERT refs: Class 11 Physics Chapter 4, p.63
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
Test yourself on this topic with real past-paper questions:
Practice this topic →Prefer to watch this?
8 real NEET Laws of Motion questions (2022–2026) with mechanics you can watch move (22 min)
Watch on YouTube