For a body of mass m moving in a circle of radius r at uniform speed v, the centripetal force required (directed toward the centre) is F_c = m v² / r = m ω² r. This is the net INWARD radial force; it is provided by some real force (tension, friction, gravity, etc.).
-- NCERT Class 11 Physics, Ch. 4, p. 63Centripetal Force
8 MCQs5 revision cards9-step worked example
Source: NCERT Laws of MotionOfficial key: NTA-verifiedLast updated: 26 Sep 2026
Centripetal Force, explained for NEET
The single most common conceptual error in circular motion questions is treating centripetal force as a new, separate force that you add to the free-body diagram. It is not. Centripetal force is the label for whichever real force — tension, friction, gravity component, normal reaction — provides the inward radial acceleration. Drawing it as an extra arrow on top of the real forces double-counts the inward force and wrecks every calculation that follows.
The core formula. For a body of mass m moving at constant speed v in a circle of radius r, the net inward radial force must equal:
F_c = mv²/r = mω²r
(NCERT Class 11 Physics Chapter 4, page 63.)
This is Newton's second law applied radially: the real forces must supply exactly mv²/r inward. No more, no less.
Identifying the real force. On a level road, static friction between tyre and road is the centripetal force. The maximum safe speed before skidding is:
v_max = √(μ_s · g · r)
(NCERT Class 11 Physics Chapter 4, page 63.)
On a banked road at angle θ with friction coefficient μ_s:
v_max = √[gr(μ_s + tan θ) / (1 − μ_s tan θ)]
(NCERT Class 11 Physics Chapter 4, page 63.)
The frictionless banked case (μ_s = 0) reduces to v₀ = √(gr tan θ) — a common special case in NEET.
Watch-out. When you draw a free-body diagram for circular motion, draw only the real forces (weight, normal, friction, tension). Then set their net inward component equal to mv²/r. If you ever write "centripetal force" as a separate arrow pointing inward alongside tension or friction, you have made the classic error — stop and redraw.
Can you answer these Centripetal Force 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 stone of mass 0.5 kg is tied to a string and whirled in a horizontal circle of radius 1.0 m at constant speed 4.0 m/s. What is the centripetal force acting on the stone?
Show answer and why every option is right or wrong
Answer: C. F_c = mv²/r = 0.5 × (4.0)² / 1.0 = 0.5 × 16 / 1.0 = 8.0 N (NCERT Class 11 Physics Chapter 4, page 63).
Why A is wrong: A: This uses F = mv/r instead of mv²/r — forgetting to square the speed.
Why B is wrong: B: This uses ½mv²/r = 4.0 N, carrying the ½ of kinetic energy into the force formula; the centripetal force is mv²/r, with no ½.
Why D is wrong: D: This doubles the correct answer, likely by using m = 1.0 kg instead of 0.5 kg.
MCQ 2Easy RecallPractice
In uniform circular motion, centripetal force is:
Show answer and why every option is right or wrong
Answer: C. Centripetal force is not a new force — it is the net inward radial component provided by real forces such as tension, friction, or gravity (NCERT Class 11 Physics Chapter 4, page 63).
Why A is wrong: A: This is the classic misconception — centripetal force is not a new fundamental force; it is a label for the net radial component of existing real forces (trap: treating centripetal force as an additional force on the FBD).
Why B is wrong: B: Centripetal force is always perpendicular to velocity in UCM, so it does zero work (W = F·d·cos 90° = 0).
Why D is wrong: D: In uniform circular motion speed is constant; centripetal force changes direction, not speed.
MCQ 3Direct ApplicationPractice
A car negotiates a level circular road of radius 50 m. The coefficient of static friction between the tyres and the road is 0.4. Taking g = 10 m/s², the maximum speed at which the car can take the turn without skidding is:
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 ≈ 14 m/s (NCERT Class 11 Physics Chapter 4, page 63).
Why A is wrong: A: This computes √(μ_s · g · r) = √(0.4 × 10 × 50) incorrectly as √100 — likely dropping the factor of 2 inside 200.
Why C is wrong: C: This uses √(2μ_s · g · r) = √400 = 20 m/s, carrying a factor 2 over from v² = 2gh; the friction limit μ_s·mg = mv²/r has no 2.
Why D is wrong: D: This computes μ_s · g · r = 200 and reports that as the speed in m/s — forgetting the square root entirely.
MCQ 4Concept TrapPractice
A student draws a free-body diagram for a ball on a string in vertical circular motion at the top of the circle. The student draws three forces: weight (down), tension (down), and centripetal force (down). What is wrong with this diagram?
Show answer and why every option is right or wrong
Answer: D. Centripetal force is not a separate force — at the top of the circle, the net downward force (weight + tension) provides the centripetal acceleration. Drawing centripetal force as a third arrow double-counts the inward force (NCERT Class 11 Physics Chapter 4, page 63).
Why A is wrong: A: Weight always points toward the Earth's centre (downward), regardless of the ball's position on the circle.
Why B is wrong: B: At the top of a vertical circle, tension in the string does point toward the centre — which is downward. Tension direction is correct.
Why C is wrong: C: The diagram IS wrong — centripetal force is not a separate force to draw alongside the real forces (trap: adding centripetal force as a new force on the FBD).
MCQ 5Easy RecallPractice
On a frictionless banked road at angle θ, the speed at which a vehicle can take the curve without any tendency to slide is:
Show answer and why every option is right or wrong
Answer: A. For a frictionless banked road, setting the horizontal component of the normal force equal to mv²/r gives v = √(gr tan θ) (NCERT Class 11 Physics Chapter 4, page 63, with μ_s = 0).
Why B is wrong: B: This uses cos θ instead of tan θ — confusing the vertical and horizontal resolution of the normal force.
Why C is wrong: C: This uses sin θ instead of tan θ — the centripetal component comes from N sin θ, but the vertical balance gives N cos θ = mg, so the ratio yields tan θ, not sin θ.
Why D is wrong: D: This inverts tan θ, giving cot θ instead — the formula has tan θ in the numerator under the square root, not in the denominator.
MCQ 6Direct ApplicationPractice
A body moves in a circle of radius 2.0 m at a constant speed of 6.0 m/s. The magnitude of centripetal acceleration is:
Show answer and why every option is right or wrong
Answer: A. a_c = v²/r = (6.0)²/2.0 = 36/2 = 18 m/s² (NCERT Class 11 Physics Chapter 4, page 63).
Why B is wrong: B: This computes 2v²/r or v²/(r×1.5) — an arithmetic error; the correct calculation is 36/2 = 18.
Why C is wrong: C: This computes v/r = 6/2 = 3 — using v/r instead of v²/r.
Why D is wrong: D: This gives v² = 36 but forgets to divide by r = 2.0 m.
MCQ 7CalculationPractice
A car moves on a banked road of radius 100 m at angle θ = 30° with coefficient of static friction μ_s = 0.2. Using g = 10 m/s² and tan 30° ≈ 0.577, the maximum safe speed is closest to:
Show answer and why every option is right or wrong
Answer: D. v_max = √[gr(μ_s + tan θ)/(1 − μ_s tan θ)] = √[10 × 100 × (0.2 + 0.577)/(1 − 0.2 × 0.577)] = √[1000 × 0.777/0.8846] = √[1000 × 0.878] = √878 ≈ 29.6 m/s ≈ 30 m/s. (NCERT Class 11 Physics Chapter 4, page 63.)
Why A is wrong: A: This likely uses the frictionless formula v = √(gr tan θ) = √(1000 × 0.577) = √577 ≈ 24 m/s — ignoring the friction contribution that raises the safe speed.
Why B is wrong: B: This is the minimum safe speed, √[gr(tan θ − μ_s)/(1 + μ_s tan θ)] = √338 ≈ 18 m/s, where friction acts up the slope; at the maximum speed friction acts down the slope and adds to the banking.
Why C is wrong: C: This computes (μ_s + tan θ) correctly but drops the denominator (1 − μ_s tan θ): √(1000 × 0.777) = √777 ≈ 28 m/s, a little below the true maximum.
MCQ 8Easy RecallPractice
Which of the following statements about centripetal force is correct?
Show answer and why every option is right or wrong
Answer: B. In orbital motion (e.g. a satellite), gravitational attraction toward the centre provides the entire centripetal force — no string or friction needed (NCERT Class 11 Physics Chapter 4, page 63).
Why A is wrong: A: Centripetal force is perpendicular to velocity at every instant in UCM, so it does zero work (cos 90° = 0).
Why C is wrong: C: Centripetal force equals mv²/r, which depends on speed and radius. It equals weight only in specific cases (e.g. satellite orbit where mg = mv²/r at a particular altitude).
Why D is wrong: D: Centripetal force points radially inward (toward the centre), not tangentially. A tangential force would change speed, not direction.
Free NEET study resources
Get a structured 30-day Mechanics plan and a complete formula booklet — delivered to your inbox instantly.
Centripetal Force: quick recall before you leave
How do you solve a Centripetal Force question? A worked example
Pattern: Banked road with friction — maximum safe speed (based on the banked-road formula, the highest-relevance in-scope topic-specific application).
- 1
Given
A car takes a banked curve of radius r = 50.0 m, banking angle θ = 45°, coefficient of static friction μ_s = 0.3. Take g = 10 m/s² (exact, problem-defined). tan 45° = 1 (exact).
- 2
Required
Maximum speed v_max at which the car can negotiate the curve without skidding.
- 3
Concept
On a banked road with friction, both the horizontal component of the normal force and the friction force contribute to centripetal acceleration. We resolve forces along radial and vertical directions and derive the maximum speed condition.
- 4
Formula
v_max = √[gr(μ_s + tan θ)/(1 − μ_s tan θ)]
(NCERT Class 11 Physics Chapter 4, page 63.) - 5
Substitution
v_max = √[10 × 50.0 × (0.3 + 1)/(1 − 0.3 × 1)]
- 6
Calculation
Numerator inside square root: 10 × 50.0 × 1.3 = 650
Denominator: 1 − 0.3 = 0.7
Ratio: 650/0.7 = 928.57...
v_max = √928.57 ≈ 30.5 m/s
Note on exact constants: g = 10 m/s² and tan 45° = 1 are exact values defined in the problem statement — they do not limit significant figures. The precision is governed by the given values of r (3 sig figs), μ_s (1 sig fig), and θ (exact angle). - 7
Final answer
v_max ≈ 30 m/s (to 1 significant figure, limited by μ_s = 0.3). In scientific notation: 3 × 10¹ m/s.
- 8
Common trap
If you set μ_s = 0 (frictionless), you get v₀ = √(gr tan 45°) = √(500) ≈ 22.4 m/s. With friction assisting, the maximum speed is higher (≈ 30 m/s). A common error is to use the frictionless formula even when friction is given, losing roughly 8 m/s of safe margin.
Another trap: if you forget that centripetal force is supplied by the combination of normal force and friction (not by adding a separate "centripetal force" to the diagram), you double-count forces and get an incorrect result. - 9
Similar NEET-style question
A car moves on a banked road of radius 200 m at an angle of 30° with μ_s = 0.25. Taking g = 10 m/s², find the maximum safe speed. [Answer: Apply the same formula with the new values.]
---
What to remember before solving Centripetal Force questions
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. 63Formula
Vehicle on a banked road
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. 64Which Centripetal Force formulas do you need for NEET?
3 formulas — click to collapse
Maximum safe speed on a banked road (with friction)
v_\max = \sqrt{\tfrac{gr(\mu_s + \tan\theta)}{1 - \mu_s \tan\theta}}
On 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 |
Valid when
- Banked turn at angle theta (theta = 0 reduces to level-road formula)
- 1 - mu_s*tan_theta > 0 (formula breaks down for very steep banks at high friction)
- Optimum/no-friction speed v_o = sqrt(g*r*tan_theta) is a SPECIAL CASE
Do NOT use when
- Banked angle so steep that 1 - mu_s*tan_theta <= 0 (use centripetal limit form)
- Friction direction reversed (very low speed on a steep bank — vehicle slides inward)
Centripetal force in uniform circular motion
The net inward force required to keep a body of mass m moving in a circle of radius r at speed v is m*v^2/r. This 'centripetal' force is NOT a new fundamental force — it is whichever real force (tension, friction, gravity, etc.) provides the inward acceleration.
| Symbol | Quantity | SI Unit |
|---|---|---|
| F_c | Centripetal force | N |
| m | Mass of the body | kg |
| v | Tangential speed | m/s |
| r | Radius of the circle | m |
| omega | Angular speed | rad/s |
Valid when
- Speed is uniform (a_t = 0, only radial acceleration matters)
- Identify the real force that provides F_c (tension, friction, normal component, etc.)
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
Where do students lose marks on Centripetal Force?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
1 item — click to collapse
Root cause: concept gap
Correction
Centripetal force is NOT a new fundamental force. It is the NET inward radial component of the real forces (tension, friction, normal, gravity, etc.). On a free-body diagram, draw only the real forces; their net inward component must equal m*v^2/r.
Wrong option pattern
Distractor sums tension + 'centripetal force' as separate inward forces.
More in Laws of Motion: 15 exam traps and mistakes · 6 formulas · 10 question patterns from its other lessons.
Centripetal Force 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