An object moving in a circle of radius r with constant speed v has acceleration of magnitude a_c = v² / r directed toward the centre of the circle (centripetal). In terms of angular speed ω = v/r, a_c = ω² r. The acceleration changes direction continuously even though the speed is constant.
-- NCERT Class 11 Physics, Ch. 3, p. 17Dyn Ucm
8 MCQs3 revision cards9-step worked example
Source: NCERT Laws of MotionPYQ coverage: NEET 2020, 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026
Lesson
The highest-yield trap in dynamics of uniform circular motion is treating centripetal force as a separate, additional force on the free-body diagram. It is not. Centripetal force is the label for the net inward radial component of whatever real forces act on the body — tension, friction, normal reaction, gravity, or some combination. Drawing a separate "centripetal force" arrow alongside real forces double-counts the inward force and wrecks the entire calculation.
In uniform circular motion, an object moves at constant speed v along a circle of radius r. Because the direction of velocity changes continuously, the object accelerates. This acceleration is centripetal — directed toward the centre — with magnitude a_c = v²/r (NCERT Class 11 Physics, Chapter 3, page 17). By Newton's second law, a net inward force F_c = mv²/r must exist to produce this acceleration (NCERT Class 11 Physics, Chapter 4, page 13).
The procedure is always the same: draw only real forces on the free-body diagram, resolve them along the radial direction, and set the net inward component equal to mv²/r. On a level road, static friction alone provides centripetal force, giving v_max = √(μ_s·g·r). On a banked road, the normal force's horizontal component contributes, leading to the banked-road formula.
A common confusion in NEET: the question asks which quantity is constant in UCM. Speed and kinetic energy are constant. Velocity is not — its direction changes every instant. Acceleration magnitude (v²/r) is constant, but acceleration direction keeps rotating inward. NEET distractors exploit the speed-versus-velocity conflation.
Watch out: if you ever draw "centripetal force" as a separate arrow on an FBD, you have invented a force that does not exist. Identify the real provider — then set it equal to mv²/r.
Practice MCQs
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In uniform circular motion, which of the following quantities remains constant?
MCQ 2Direct ApplicationPractice
A stone of mass 0.50 kg is tied to a string and whirled in a horizontal circle of radius 1.0 m at a constant speed of 4.0 m/s. What is the centripetal force acting on the stone?
MCQ 3Direct ApplicationPractice
A car moves along a level circular road of radius 50 m. If the coefficient of static friction between the tyres and the road is 0.40 and g = 10 m/s², the maximum safe speed of the car is:
MCQ 4Easy RecallPractice
A student draws the free-body diagram of a ball on a string in horizontal circular motion. The student draws three arrows: tension (inward), weight (downward), and a separate "centripetal force" arrow (inward). What is wrong with this diagram?
MCQ 5Easy RecallPractice
In uniform circular motion at constant speed v, the centripetal acceleration has:
MCQ 6Direct ApplicationPractice
A body of mass 2.0 kg moves in a circle of radius 0.50 m. The centripetal force on it is 16 N. What is the speed of the body?
MCQ 7Concept TrapPractice
A car negotiates a frictionless banked curve. The banking angle is θ. Which force provides the centripetal acceleration?
MCQ 8Easy RecallPractice
A particle completes one full revolution in uniform circular motion of radius r at speed v. The magnitude of the displacement and the distance covered after one full revolution are, respectively:
Quick recall before you leave
Worked Example
Pattern: UCM properties at an instant (anchored to PYQ 2024 Q3 Q17)
- 1
Given
- Mass m = 0.20 kg - Radius r = 0.80 m - Frequency f = 5.0 rev/s (2 sig figs)
- 2
Required
(a) Centripetal acceleration a_c (b) Tension T in the string
- 3
Concept
In UCM, centripetal acceleration = ω²r, directed toward the centre. The string tension is the real force providing centripetal force: T = ma_c = mω²r.
- 4
Formula
- ω = 2πf - a_c = ω²r - T = ma_c = mω²r
- 5
Substitution
- ω = 2π × 5.0 = 10π rad/s - a_c = (10π)² × 0.80 = 100π² × 0.80 - T = 0.20 × a_c
- 6
Calculation
- ω = 31.4 rad/s (3 sig figs for intermediate) - a_c = 100 × 9.870 × 0.80 = 789.6 m/s² → 7.9 × 10² m/s² (2 sig figs, limited by f = 5.0) - T = 0.20 × 789.6 = 157.9 N → 1.6 × 10² N (2 sig figs) Note on exact values: π is an exact mathematical constant and the integer 2 in ω = 2πf is exact — neither limits the significant-figure count. The sig-fig count is governed by the measured values (m, r, f), all given to 2 significant figures.
- 7
Final answer
(a) a_c ≈ 7.9 × 10² m/s² (b) T ≈ 1.6 × 10² N
- 8
Common trap
A student might add a separate "centripetal force" on top of the string tension. That would double the inward force and yield T = mv²/r − F_centripetal, a meaningless equation. The tension IS the centripetal force here — no additional force exists.
- 9
Similar NEET-style question
A ball of mass 0.10 kg is attached to a string of length 0.50 m and whirled in a horizontal circle at 4.0 rev/s. Find the tension in the string. (Answer: T = mω²r = 0.10 × (8π)² × 0.50 ≈ 32 N.) ---
Before solving, remember these
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. 13Formulas
9 formulas — click to collapse
Centripetal acceleration in uniform circular motion
An object moving in a circle of radius r at constant speed v has acceleration of magnitude v^2/r (or equivalently omega^2 * r) directed toward the centre. This is centripetal (radially inward), not tangential.
| Symbol | Quantity | SI Unit |
|---|---|---|
| a_c | Centripetal acceleration | m/s^2 |
| v | Tangential speed | m/s |
| r | Radius of circle | m |
| omega | Angular speed | rad/s |
Valid when
- Speed v is constant (uniform circular motion)
- r and the centre are well-defined (instantaneous radius of curvature for general curved motion)
Do NOT use when
- Non-uniform circular motion (then there is also a tangential acceleration component)
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.)
Conservation of linear momentum
If the net external force on a system of particles is zero, the total linear momentum of the system is conserved (vector equality of total p before and after any internal interaction). Basis of all collision and recoil analysis.
| Symbol | Quantity | SI Unit |
|---|---|---|
| F_ext | Net external force on system | N |
| p_total | Sum of m_i*v_i over all particles | kg*m/s |
Valid when
- Net EXTERNAL force is zero (internal forces always cancel by Newton's 3rd law)
- Conservation is vector — apply componentwise (x and y separately)
- Holds independent of whether collisions are elastic or inelastic
Do NOT use when
- External impulses present (gravity over a long time, friction)
Impulse of a force
Impulse equals the product of force and the time interval over which it acts. By Newton's Second Law, impulse equals the change in linear momentum during that interval.
| Symbol | Quantity | SI Unit |
|---|---|---|
| J | Impulse (vector) | N*s = kg*m/s |
| F | Force (treated as average) | N |
| Delta_t | Time interval | s |
| Delta_p | Change in momentum | kg*m/s |
Valid when
- Useful when forces are large but act briefly (collisions, bat-on-ball, kicks)
- Direction of impulse is the direction of average force
Kinetic friction
When two surfaces slide relative to each other, kinetic friction opposes the motion with magnitude proportional to the normal force.
| Symbol | Quantity | SI Unit |
|---|---|---|
| f_k | Kinetic friction force | N |
| mu_k | Coefficient of kinetic friction | (dimensionless) |
| N | Normal force | N |
Valid when
- Surfaces ARE sliding
- Direction: opposite to instantaneous relative velocity of one surface vs the other
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
Newton's Second Law of Motion
The net external force on a body equals the rate of change of its linear momentum. For a body of constant mass, this reduces to F = m*a — net force equals mass times acceleration. Both F and a are vectors; the acceleration is in the direction of the net force.
| Symbol | Quantity | SI Unit |
|---|---|---|
| F | Net external (vector) force | N |
| m | Mass of the body | kg |
| a | Acceleration (vector) | m/s^2 |
| p | Linear momentum (= m*v) | kg*m/s |
| t | Time | s |
Valid when
- F is the resultant (net) of all external forces, not any single force
- Mass is constant for the form F = m*a (use F = dp/dt for variable mass)
- Inertial reference frame (no pseudo-forces); add inertial corrections in non-inertial frames
Do NOT use when
- Frame is non-inertial (need pseudo-forces)
- Mass is varying significantly (use F = dp/dt)
- Quantum / relativistic regimes (Newtonian mechanics breaks down)
Maximum static friction
The maximum value of static friction between two surfaces in contact equals the coefficient of static friction times the normal force. Below f_s_max, static friction self-adjusts to whatever value is needed to prevent relative motion.
| Symbol | Quantity | SI Unit |
|---|---|---|
| f_s_max | Maximum static friction | N |
| mu_s | Coefficient of static friction | (dimensionless) |
| N | Normal force | N |
Valid when
- Surfaces in contact, no relative motion (impending motion limit)
- Below f_s_max, actual static friction = applied tangential load (self-adjusting)
- f_s_max is independent of the apparent area of contact (Coulomb-Amontons assumption)
Do NOT use when
- Surfaces are sliding (use kinetic friction f_k = mu_k * N instead)
- Lubricated / fluid-friction conditions
Exam Traps & Common Mistakes
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
16 items — click to collapse
Category: Similar Terms
Student claims velocity is constant in uniform circular motion (it's not — direction changes).
When it triggers
Question asks 'in uniform circular motion at constant speed, which is also constant?'
How to avoid
In UCM: SPEED constant; KE constant. VELOCITY (vector) NOT constant. ACCELERATION (centripetal, magnitude v²/r) constant in MAGNITUDE but NOT in direction.
Category: Sign Convention
Student picks the direction of MOTION as the direction of the net force, instead of the direction of the change in momentum (Δp). When velocity changes direction at constant speed, the force is perpendicular to BOTH the initial and final velocity vectors (in the limit) — it's the direction of Δv.
When it triggers
Question describes a body changing direction (e.g. turning) and asks for the force direction or the direction of Δp.
How to avoid
F ∝ Δp = m Δv = m (v_f − v_i). Always draw v_i and v_f as vectors and subtract; the result (v_f − v_i) is the direction of net force.
Category: Unit Conversion
Student computes μmg (friction FORCE in Newtons) when asked for the friction-limited ACCELERATION. The two differ by a factor of m: a = μg, F = μmg.
When it triggers
Question gives μ, g, and a body's mass and asks for the maximum acceleration of the supporting surface OR the friction force on the body.
How to avoid
Read carefully: 'maximum acceleration of the vehicle so the body stays still' = μg (no mass). 'Friction force on the body' = μmg (mass present). Units expose the error: N for force, m/s² for acceleration.
Category: Sign Convention
Student forgets that velocity DIRECTION reverses on bounce; computes |v₁ - v₂| instead of |v₁ + v₂|.
When it triggers
Question describes ball dropped from height h₁, rebounding to height h₂; asks for impulse on ball.
How to avoid
Impulse J = Δp = m(v_after - v_before). Take down as positive: v_before = +v₁, v_after = -v₂. So J = m(-v₂ - v₁) = -m(v₁ + v₂); magnitude = m(v₁ + v₂).
Category: Sign Convention
Student writes a = g sin θ for a rough incline (which is the smooth-incline answer); forgets to subtract μ g cos θ.
When it triggers
Question contrasts rough vs smooth inclines, or asks for acceleration on a rough incline.
How to avoid
On a rough incline (block sliding down): a = g(sin θ - μ_k cos θ). On a rough incline (block sliding up): a = -g(sin θ + μ_k cos θ). Smooth case (μ = 0): just g sin θ.
Category: Overthinking
Student takes torque about the centre of mass (introducing all 4 forces with non-zero moment arms) instead of about the floor contact (where 2 forces have zero moment arm and the equation simplifies).
When it triggers
Question gives a uniform rod or ladder leaning against a wall; asks for friction coefficient or limiting condition.
How to avoid
Pick the pivot to ELIMINATE unknown forces from the torque equation. Floor-contact pivot: normal force and friction at floor contribute zero torque; only weight (mid-length) and wall normal (top) appear. Result: μ_min = 1 / (2 tan θ).
Category: Sign Convention
Student adds magnitudes of momenta of fragments instead of vectors; ignores cancellation when fragments fly perpendicular or in opposite directions.
When it triggers
Question describes a body at rest exploding into multiple fragments with given mass ratios and partial velocity info.
How to avoid
Total momentum is a VECTOR. Initial p = 0; therefore Σ m_i v_i = 0 as a vector equation. Decompose along chosen axes (often natural symmetry axes); sum = 0 in each.
Category: Overthinking
Student tries to apply F=ma to the system as a whole (using net force = (m1-m2)g and total mass m1+m2) but loses track of the tension. The correct approach writes Newton's Second Law SEPARATELY for each mass and treats T as an unknown in two simultaneous equations.
When it triggers
Question contains a frictionless pulley with two unequal masses tied to a string. Asks for tension T or acceleration a (or both).
How to avoid
Draw a free-body diagram for EACH mass. Write F=ma per body, treating T as the same magnitude on both sides of the string. Solve simultaneously: a = (m1 - m2) g / (m1 + m2); T = 2 m1 m2 g / (m1 + m2).
Category: Negative Marking
Multi-mass pulley problem requires computing acceleration first, then tension. T = 2 m1 m2 g/(m1+m2). Sign errors in m1−m2 propagate.
When it triggers
Atwood machine or pulley system with multiple masses.
How to avoid
Compute a = (m1-m2)g/(m1+m2) first with m1 the heavier mass and downward as positive. Then T from F=ma on either mass. Always re-check by plugging back into both Newton's 2nd Law equations.
Category: Overthinking
Student applies the full external force F to a single block instead of recognising the system needs to be analysed for the contact (internal) force.
When it triggers
Question gives horizontal force F on block A which pushes block B; asks for contact force between A and B or acceleration.
How to avoid
System acceleration: a = F / (m_A + m_B). Contact force on B from A = m_B × a. The full F acts on the system, not on each block independently.
Category: Overthinking
Student computes P = Mgv (just lifting against gravity) and ignores the friction-opposing-motion term.
When it triggers
Question describes a lift moving at constant speed with explicit friction force on cable or guides.
How to avoid
At constant speed, net force = 0, so cable tension T = Mg + f_friction. Power = T × v = (Mg + f) × v. Always add friction when stated.
Root cause: concept gap
Correction
Action-reaction pairs ALWAYS act on DIFFERENT bodies. The pair to the book's weight is the gravitational pull the book exerts on the Earth. The pair to the normal force from table on book is the force the book exerts on the table. Equal-and-opposite forces on the SAME body are an equilibrium statement, not third-law statement.
Wrong option pattern
Distractor labels two forces on the same body as a Newton's-third-law pair.
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.
Root cause: unit error
Correction
Impulse J has dimensions of momentum (kg*m/s) and equals F*Delta_t. Force has dimensions kg*m/s^2. Confusing them inflates or deflates an answer by a factor of seconds. Always check units before declaring an answer.
Wrong option pattern
Distractor reports an answer in newtons where the correct answer is in N*s (or vice versa).
Root cause: concept gap
Correction
Linear momentum is conserved only when the net EXTERNAL force is zero. Gravity over a finite time changes momentum. For collision problems we use conservation because the collision happens over a brief time interval where external impulses are negligible compared to internal collision forces.
Wrong option pattern
Distractor sets initial momentum = final momentum for a free-fall problem where gravity has acted for several seconds.
Root cause: formula misuse
Correction
Static friction is SELF-ADJUSTING: f_s exactly cancels the applied tangential force up to a ceiling f_s_max = mu_s * N. Below the ceiling, f_s = applied force. At the ceiling, motion is impending. Substituting mu_s * N too early over-estimates the friction.
Wrong option pattern
Distractor uses mu_s * N for the static friction force in a no-slip scenario where the applied force is well below threshold.
Past Year Questions
10 questions from NEET 2020, 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys. — click to collapse
NEET 2025
10.75
20.25
30.40
40.5
NTA Answer: Option 1(final)
NEET 2025
184 NS
221 NS
37 NS
40
NTA Answer: Option 2(final)
NEET 2025
1200 3N
2100 N
3100 3N
4200 N
NTA Answer: Option 3(final)
NEET 2024Revised key
1Zero
24 N
36 N
410 N
NTA Answer: Option 3(revised_final)
NEET 2024Revised key
A particle moving with uniform speed in a circular path maintains:
1Constant velocity
2Constant acceleration
3Constant velocity but varying acceleration
4Varying velocity and varying acceleration
NTA Answer: Option 4(revised_final)
NEET 2023
1Along northward
2Along north-east
3Along south-west
4Along eastward
NTA Answer: Option 2(final)
NEET 2023
1150 m s–2
21.5 m s–2
350 m s–2
41.2 m s–2
NTA Answer: Option 2(final)
NEET 2022
1- 5 - 3 2 v
2v
32 v
42 2v
NTA Answer: Option 4(final)
NEET 2021
11.4 kg m/s
20 kg m/s
34.2 kg m/s
42.1 kg m/s
NTA Answer: Option 3(final)
NEET 2020
1g/10
2g
3g/2
4g/5
NTA Answer: Option 4(final)
How NEET usually asks this
10 recurring patterns from past papers — click to collapse
Particle in uniform circular motion at constant speed; identify which of speed / velocity / acceleration / kinetic energy is constant. Speed and KE constant; velocity (vector) and acceleration (centripetal) are NOT constant — direction changes.
InterpretationEasy
Common distractors
treats velocity as scalar
Conflates speed (scalar) with velocity (vector)
A body at rest explodes into 3 fragments; mass ratios given; some fragments' velocities given; find velocity of remaining fragment. Apply vector momentum conservation: Σ m_i v_i = 0 (since initial p = 0). Common shape: 2:2:1 mass ratio, two equal-mass fragments fly perpendicular with given speed; third fragment recoils.
Multi StepMedium
Common distractors
scalar sum of momenta
Treating momentum as scalar; ignores vector cancellation
A body moving in one direction suddenly changes velocity direction (same or different speed); find the direction of the net force. Force direction = direction of momentum CHANGE (Δp = p_f − p_i), NOT direction of motion. Common shape: 'moving south, suddenly turning east at same speed' → Δp vector points north-east.
InterpretationEasy
Common distractors
force along final velocity
Default to thinking force points in direction of motion
force along initial velocity
Newton-1 misread: object 'wants' to keep moving in original direction
Ball of mass m dropped from height h₁; rebounds to height h₂ (≤ h₁); find impulse on ball from ground. Use v² = 2gh to get speeds at impact and rebound; impulse J = m(v₂ - (-v₁)) = m(v₁ + v₂) where v₁ = √(2g h₁) and v₂ = √(2g h₂). Sign of velocity flips on rebound.
Multi StepEasy
Common distractors
subtracts v2 from v1 instead of adding
Forgetting velocity DIRECTION reversal at bounce
Two inclines of equal length L and same angle θ (e.g. 45°); one rough (with μ), one smooth. Compare time-of-descent or final velocity at bottom. Smooth: a = g sin θ. Rough: a = g(sin θ - μ cos θ). Use L = ½ a t² or v² = 2aL.
Multi StepMedium
Common distractors
ignores mu cos theta term
Forgetting the friction-along-incline component
Uniform rod / ladder of mass M and length L leans against a smooth vertical wall at angle θ; floor is rough with friction coefficient μ. Find limiting condition for static equilibrium. Apply ΣF = 0 (3 equations: horizontal, vertical, torque about base). Wall provides only horizontal reaction (smooth); floor provides normal + friction. μ_min = 1 / (2 tan θ).
Multi StepHard
Common distractors
torque about wrong pivot
Default pivot at center of mass adds complexity
Atwood-style pulley with two unequal masses connected by an inextensible massless string over a frictionless pulley. Apply F = ma to each mass separately along the string direction. The tension is the same throughout the string; the magnitudes of acceleration are equal but oriented oppositely. Solve simultaneous equations for tension T and acceleration a. Common shape: given two masses m1, m2 and asked for a or T, with options testing common confusions (g vs a in equations, treating the system as one body).
Multi StepMedium
Common distractors
uses g where a belongs
Forgetting that the system accelerates, so weight is balanced by net force minus T
confuses tension with weight
Treating T = m·g for one of the masses (which is true only when a=0)
A body rests on the floor of an accelerating vehicle; find the maximum vehicle acceleration before the body slides. Static friction provides the horizontal force on the body; max accel = μ_s g. Above this, body slides backward relative to the vehicle. Common shape: μ_s, g given; find a_max.
Direct ApplicationEasy
Common distractors
uses mu times g times mass
Confusing force (μ·N = μmg) with acceleration
Horizontal force F applied to block A (mass m_A); A pushes B (mass m_B) in front. Find acceleration of system AND contact force between A and B. System: a = F/(m_A + m_B); contact force on B from A = m_B × a = m_B F / (m_A + m_B).
Multi StepEasy
Common distractors
treats each block with full F
Forgetting Newton's 3rd law internal force decomposition
Lift moving up at constant speed v with total mass M; friction force f opposes motion. Power required from cable = (Mg + f) × v. Common shape: M = 2000 kg, v = 1.5 m/s, f given; find motor power.
Multi StepEasy
Common distractors
forgets friction term
Ignoring opposing force
Sources
NCERT refs: Class 11 Physics Chapter 3, p.17 | Class 11 Physics Chapter 4, p.13
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