Velocity Time Graph

8 MCQs3 revision cards9-step worked example

Source: NCERT KinematicsPYQ coverage: NEET 2020, 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

The velocity-time (v-t) graph is one of the most information-dense representations in kinematics. NEET questions built on it test whether you can extract three quantities correctly: velocity at an instant (read the y-value), acceleration (compute the slope), and displacement (compute the area under the curve). The common trap is treating the graph like a position-time graph and misreading what slope and area represent.

What the v-t graph encodes. For straight-line motion under uniform acceleration, v = v₀ + at (NCERT Class 11 Physics Chapter 2, page 5). On the v-t graph this is a straight line whose slope equals the constant acceleration a, and whose y-intercept is the initial velocity v₀. A horizontal line means zero acceleration (uniform motion). A line sloping downward means deceleration (negative a if positive direction is upward).

Displacement as area. The displacement between times t₁ and t₂ equals the area under the v-t curve between those times. For constant acceleration, the shape is a trapezoid (or triangle if v₀ = 0). The second kinematic equation s = v₀t + ½at² is exactly this area. A high-frequency NEET trap: for free fall from rest, successive-second distances follow the odd-number ratio 1 : 3 : 5 : 7, not the linear ratio 1 : 2 : 3 : 4. The quadratic dependence of displacement on time (s = ½gt²) makes equal time intervals yield unequal distances.

When the graph is NOT a straight line. If the v-t curve is nonlinear, acceleration is not constant. You cannot use v = v₀ + at or v² = v₀² + 2as in that region — those three kinematic equations require constant acceleration. For a curved v-t graph, acceleration at any instant is the tangent's slope, and displacement is still the area (computed by integration or geometric estimation).

Watch-out for NEET. When a problem states an object is "thrown downward" with initial speed u, the v-t graph starts at v₀ = u (not zero). Using v² = 2gh instead of v² = u² + 2gh costs the full 4 marks plus the 1-mark negative penalty.

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

The slope of a velocity-time graph for a body moving in a straight line represents:

MCQ 2Easy RecallPractice

The area under a velocity-time graph between two instants gives the:

MCQ 3Easy RecallPractice

A velocity-time graph is a horizontal straight line at v = 5 m/s. Which statement is correct?

MCQ 4Direct ApplicationPractice

A body starts from rest and accelerates uniformly. The distances covered in the 1st, 2nd, and 3rd seconds of its motion are in the ratio:

MCQ 5Direct ApplicationPractice

A ball is thrown vertically downward from a tower with an initial speed of 10 m/s. It strikes the ground with a speed of 30 m/s. Taking g = 10 m/s², the height of the tower is:

MCQ 6Direct ApplicationPractice

The v-t graph of a body is a straight line passing through the origin with a slope of 3 m/s². The displacement of the body in the first 4 seconds is:

MCQ 7Concept TrapPractice

The velocity-time graph of a particle is a curve (not a straight line) between t = 0 and t = 5 s. A student uses v = v₀ + at to find the velocity at t = 5 s. This approach is:

MCQ 8CalculationPractice

A particle starts from rest and moves with constant acceleration. If it covers 10 m in the first 2 seconds, what distance does it cover in the next 2 seconds (from t = 2 s to t = 4 s)?

Quick recall before you leave

Worked Example

1
Given
- Initial velocity: v₀ = 0 (released from rest) - Acceleration: a = g = 10 m/s² (constant, downward) - Required: ratio of distances in the 1st, 2nd, 3rd, and 4th individual seconds
2
Required
Distance covered in each successive 1-second interval: d₁, d₂, d₃, d₄ and their ratio.
3
Concept
Under constant acceleration from rest, cumulative distance grows as s = ½gt². The distance in the nth second equals sₙ − sₙ₋₁ = ½g(n² − (n−1)²) = ½g(2n − 1). The factor (2n − 1) produces the odd-number sequence 1, 3, 5, 7, ...
4
Formula
Distance in nth second: dₙ = ½g(2n − 1) This is derived from the second kinematic equation s = ½at² by subtraction of consecutive cumulative distances.
5
Substitution
- d₁ = ½(10)(2×1 − 1) = 5 × 1 = 5 m - d₂ = ½(10)(2×2 − 1) = 5 × 3 = 15 m - d₃ = ½(10)(2×3 − 1) = 5 × 5 = 25 m - d₄ = ½(10)(2×4 − 1) = 5 × 7 = 35 m
6
Calculation
Ratio d₁ : d₂ : d₃ : d₄ = 5 : 15 : 25 : 35 = 1 : 3 : 5 : 7 Note on exact constants: g = 10 m/s² is a problem-defined exact value, and the integers 1, 2, 3, 4 are counting numbers. These do not limit significant figures in the ratio.
7
Final answer
The distances in the 1st, 2nd, 3rd, and 4th seconds are in the ratio **1 : 3 : 5 : 7**.
8
Common trap
Students often answer 1 : 2 : 3 : 4, assuming distance grows linearly with time. This is wrong because distance grows as t² (quadratic), making successive-interval distances follow the odd-number rule, not a linear sequence. On the v-t graph, the area of each successive strip (between t = n−1 and t = n) is a trapezoid whose area grows as (2n − 1), not as n.
9
Similar NEET-style question
A stone is dropped from a tall building. If it covers 20 m in the first 2 seconds, what distance does it cover in the 3rd and 4th seconds combined? (Use the odd-number ratio for 2-second intervals, or compute directly from s = ½gt².) ---

Before solving, remember these

Key Fact

Area-under-v-t-curve = displacement

For motion in a straight line, the area between the velocity–time curve and the time axis, between two instants, equals the displacement of the object over that interval. This geometric interpretation is the basis for graphical derivations of the kinematic equations.

-- NCERT Class 11 Physics, Ch. 2, p. 5

Formulas

6 formulas — click to collapse

First kinematic equation (uniform acceleration)

v = v_{0} + a t

Final velocity equals initial velocity plus acceleration times the time elapsed, for motion under constant acceleration.

Symbol	Quantity	SI Unit
v	Final velocity	m/s
v0	Initial velocity	m/s
a	Constant (uniform) acceleration	m/s^2
t	Elapsed time	s

Valid when

Acceleration a is CONSTANT (uniform) in both magnitude and direction
All quantities measured in the same inertial reference frame
Motion is along a straight line; signs encode direction along chosen axis

Do NOT use when

Acceleration changes in magnitude or direction (use a(t) integration)
Motion is uniformly circular at constant speed (a is centripetal, not tangential)

Second kinematic equation (displacement under uniform acceleration)

x - x_{0} = v_{0} t + \frac{1}{2} a t^{2}

Displacement equals initial-velocity-times-time plus half of acceleration-times-time-squared. The (1/2) factor is the area of the triangle on the v-t graph.

Symbol	Quantity	SI Unit
x	Final position	m
x0	Initial position	m
v0	Initial velocity	m/s
a	Constant acceleration	m/s^2
t	Time elapsed	s

Valid when

Acceleration constant (magnitude and direction)
Sign convention consistent across x, v, a (one chosen positive direction)

Third kinematic equation (velocity-squared)

v^{2} = v_{0}^{2} + 2 a (x - x_{0})

Relates final velocity to initial velocity, displacement, and acceleration without using time. Most useful when t is unknown or unwanted.

Symbol	Quantity	SI Unit
v	Final velocity	m/s
v0	Initial velocity	m/s
a	Constant acceleration	m/s^2
x - x0	Displacement	m

Valid when

Constant acceleration
Use signed values for v, v0, a, and (x - x0) consistently

Do NOT use when

Time-dependent acceleration
Curvilinear motion where acceleration is not parallel to displacement

Projectile maximum height

H = \frac{v _{0}^{2} s i n ^{2} ( θ _{0} )}{2 g}

Maximum height attained by a projectile launched at speed v0 and angle theta0 above the horizontal, measured above the launch level.

Symbol	Quantity	SI Unit
H	Maximum height (above launch)	m
v0	Launch speed	m/s
theta0	Launch angle	rad/deg
g	Gravitational acceleration	m/s^2

Valid when

Air resistance neglected
Constant g over trajectory

Projectile horizontal range

R = \frac{v _{0}^{2} s i n ( 2 θ _{0} )}{g}

For a projectile launched from and returning to the same horizontal level with initial speed v0 at angle theta0 above the horizontal, the horizontal range R is given by this formula. R is maximised at theta0 = 45 deg.

Symbol	Quantity	SI Unit
R	Horizontal range	m
v0	Launch speed	m/s
theta0	Launch angle above horizontal	rad (or deg with sin in deg)
g	Gravitational acceleration	m/s^2

Valid when

Launch and landing are at the same vertical height
Air resistance neglected
g treated as constant over the trajectory

Do NOT use when

Launch and landing heights differ (use full kinematics)
Significant air drag (e.g. table-tennis ball, badminton shuttle)
Variation of g (ballistic trajectories spanning large altitude changes)

Centripetal acceleration in uniform circular motion

a_{c} = \frac{v ^{2}}{r} = ω^{2} r

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)

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.

14 items — click to collapse

Trap

Free-fall: linear-in-time vs quadratic distance progression

Category: Similar Terms

Student answers 1:2:3:4 for distances in successive 1-second intervals (linear) instead of 1:3:5:7 (Galileo's odd numbers).

When it triggers

Question asks about ratios of distances traversed in successive 1-s intervals during free fall from rest.

How to avoid

Distance grows quadratically (y = ½ g t²); successive interval distances are y_n - y_{n-1} = ½ g (t_n² - t_{n-1}²) = ½ g (2n-1) seconds. The factor (2n-1) gives 1, 3, 5, 7, ...

Trap

Position-time graph slope: tan vs sin/cos

Category: Graph Interpretation

Student uses sin or cos of the angle the line makes with the time axis, instead of tan, to extract velocity.

When it triggers

Question gives an angle the x-t line makes with the t-axis (often 30°, 45°, 60°) and asks for velocity or its ratio.

How to avoid

Velocity = dx/dt = slope of x-t line = tan(angle), where the angle is measured from the time axis. Always tan, not sin or cos.

Trap

Implicit-function kinematics differentiation

Category: Overthinking

Student attempts to invert t(x) algebraically before differentiating, getting tangled in messy algebra; misses chain rule.

When it triggers

Question gives t as function of x (instead of x as function of t), e.g. t = x² + x.

How to avoid

Differentiate the given relation directly: dt/dx = (function of x). Then v = dx/dt = 1/(dt/dx). For acceleration use chain rule: a = dv/dt = (dv/dx)(dx/dt) = v dv/dx.

Trap

Initial velocity dropped from kinematic equation

Category: Sign Convention

Student treats a 'thrown vertically downward' problem as if the object were dropped (u = 0). The result is wrong by an additive u² term in v² = u² + 2gh. When the question explicitly states a launch speed, that speed is non-zero and CANNOT be ignored.

When it triggers

Question phrases: 'thrown vertically downward', 'projected with initial velocity', 'launched with speed u'. If u is given numerically, it MUST appear in the equation.

How to avoid

Always parse the launch verbal cue and write down u with its sign before reaching for v² = 2gh. Use the full v² = u² + 2gh (or u² - 2gh for upward motion).

Trap

Projectile max-height formula confused with range formula

Category: Similar Terms

Student plugs into v₀² sin(2θ)/g (range) when asked for maximum height, or vice versa. The two share v₀ and θ but have different sin-vs-sin² and 2g-vs-g terms.

When it triggers

Question mentions launch speed and angle and asks for max height (H) or range (R). Distractors include the wrong formula's answer.

How to avoid

Memorise BOTH formulas explicitly: H = v₀² sin² θ / (2g) (note sin²); R = v₀² sin(2θ) / g (note sin of doubled angle). Check by setting θ = 45°: max range, half max height.

Trap

Projectile angle: from horizontal vs from vertical

Category: Sign Convention

Student plugs angle θ into v cos θ when the question states 'angle with the vertical' (which makes the horizontal component v sin θ).

When it triggers

Question phrases like 'thrown at angle θ with the vertical direction' or 'with horizontal'.

How to avoid

Always identify reference axis explicitly. From horizontal: vx = v cos θ, vy = v sin θ. From vertical: vx = v sin θ, vy = v cos θ. The two are complementary (θ_h + θ_v = 90°).

Trap

Relative-velocity direction sign

Category: Sign Convention

Student fails to distinguish between same-direction and opposite-direction relative velocities, treating both as magnitudes.

When it triggers

Question describes two objects moving on the same line; observer somewhere between or alongside.

How to avoid

Relative velocity is a VECTOR. Same direction: v_rel = v_a - v_b (smaller magnitude). Opposite direction: v_rel = v_a + v_b (larger magnitude). Use sign convention consistently along chosen axis.

Trap

Linear speed-distance ratio under uniform deceleration

Category: Overthinking

Student assumes proportionality of speed to remaining distance under uniform deceleration. In fact, KE drops linearly with distance (v² is the linear quantity, not v): v² = u² - 2as. Speed-vs-distance is a sqrt-curve, not a line.

When it triggers

Question describes a body decelerating through stages with given speed at one stage; asks for distance to stop or speed at another stage.

How to avoid

Always work with v², not v, when uniform deceleration is in play. The work-energy theorem gives the same answer faster: ½ m v² = work done against constant force over distance.

Trap

UCM-to-projectile speed bridge

Category: Overthinking

Student uses the radius R as the projectile launch height or fails to compute the UCM speed from period.

When it triggers

Question describes a particle in UCM with given (R, T) then says 'now launched vertically up with same speed; find max height'.

How to avoid

Step 1: speed v = 2πR/T (from UCM). Step 2: max projectile height H = v²/(2g) = (2πR/T)² / (2g). Don't shortcut by setting H = R.

Trap

Uniform circular motion: velocity vs speed constancy

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.

Mistake

Student treats average velocity as the arithmetic mean of two given speeds, or treats |average velocity| = average speed.

Root cause: concept gap

Correction

Average velocity = total displacement / total time (vector). Average speed = total path length / total time (scalar, always >= |average velocity|). For round-trip motion, average velocity is zero; average speed is not.

Wrong option pattern

Distractor offers (v1+v2)/2 instead of total-distance/total-time.

Mistake

Student applies v = v0 + a*t (or other constant-acceleration kinematic equations) to a problem where acceleration is not constant.

Root cause: formula misuse

Correction

The three kinematic equations require CONSTANT acceleration. For variable acceleration, use a = dv/dt and integrate, or use v dv = a dx for position-dependent acceleration. Verify constant-a before applying these formulas.

Wrong option pattern

Distractor uses constant-acceleration kinematic equations on a problem where the question explicitly says acceleration changes with time or position.

Mistake

Student forgets that projectile-motion formulas assume air resistance is negligible, or applies them to non-symmetric (different launch and landing height) trajectories without correction.

Root cause: concept gap

Correction

Standard range R = v0^2 sin(2*theta)/g and H = v0^2 sin^2(theta)/(2g) assume (i) launch and landing at the same height, (ii) negligible air drag, and (iii) constant g. For asymmetric trajectories, use the full kinematic decomposition along x and y.

Wrong option pattern

Distractor applies R = v0^2 sin(2*theta)/g to a projectile launched from a cliff.

Mistake

Student claims a body moving in a circle at constant speed has zero acceleration (because speed is constant).

Root cause: concept gap

Correction

Acceleration is the rate of change of VELOCITY (a vector), not speed. In uniform circular motion, the speed is constant but the velocity direction changes continuously, giving a centripetal acceleration of magnitude v^2/r toward the centre.

Wrong option pattern

Distractor option says 'a = 0 because speed is constant'.

Past Year Questions

10 questions from NEET 2020, 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys. — click to collapse

NEET 2025

In some appropriate units, time (t) and position (x) relation of a moving particle is given by t = x2 + x. The acceleration of the particle is 2 2

2− 2x+1 (x+2)3 2 2

3−

4+ (2x+1)3 (x+1)3

NTA Answer: Option 3(final)

NEET 2025

Two cities X and Y are connected by a regular bus service with a bus leaving in either direction every T min. A girl is driving scooty with a speed of 60 km/h in the direction X to Y notices that a bus goes past her every 30 minutes in the direction of her motion, and every 10 minutes in the opposite direction. Choose the correct option for the period T of the bus service and the speed (assumed constant) of the buses.

115 min, 120 km/h

29 min, 40 km/h

325 min, 100 km/h

410 min, 90 km/h

NTA Answer: Option 1(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

A horizontal bridge is built across a river. A student standing on the bridge throws a small ball vertically upwards with a velocity 4 m s–1. The ball strikes the water surface after 4 s. The height of bridge above water surface is (Take g = 10 m s–2)

160 m

264 m

368 m

456 m

NTA Answer: Option 2(final)

NEET 2023

A bullet from a gun is fired on a rectangular wooden block with velocity u. When bullet travels 24 cm through u the block along its length horizontally, velocity of bullet becomes . Then it further penetrates into the block 3 in the same direction before coming to rest exactly at the other end of the block. The total length of the block is

124 cm

228 cm

330 cm

427 cm

NTA Answer: Option 4(final)

NEET 2022

The ratio of the distances travelled by a freely falling body in the 1st, 2nd, 3rd and 4th second

11 : 1 : 1 : 1

21 : 2 : 3 : 4

31 : 4 : 9 : 16

41 : 3 : 5 : 7

NTA Answer: Option 4(final)

NEET 2022

The displacement-time graphs of two moving particles make angles of 30° and 45° with the x-axis as shown in the figure. The ratio of their respective velocity is

1- 6 - 1 : 3

23 : 1

31 : 1

41 : 2

NTA Answer: Option 1(final)

NEET 2022

A ball is projected with a velocity, 10 ms–1, at an angle of 60° with the vertical direction. Its speed at the highest point of its trajectory will be

110 ms–1

2Zero

35 3 m s − 1

45 ms–1

NTA Answer: Option 3(final)

NEET 2021

A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution. If this particle were projected with the 'θ' same speed at an angle to the horizontal, the maximum height attained by it equals 4R. The θ, angle of projection, is then given by : 1 2gT 2 2 θ=sin−1 

1π2  R  1  2  2 gT θ=cos−1

2  π2 R 1 π2  2 R θ=cos−1 

32 gT  1 π2  2 R θ=sin−1 

42 gT 

NTA Answer: Option 1(final)

NEET 2020

U U 20 m/s fl fl U – U 12. A ball is thrown vertically downward with a U 80 m/s fl U U – U ° velocity of 20 m/s from the top of a tower. It hits the ground after some time with a velocity of — (g=10 m/s2) 80 m/s. The height of the tower is : (g=10 m/s2)

1300 m

2360 m

3340 m

4320 m

NTA Answer: Option 1(final)

How NEET usually asks this

10 recurring patterns from past papers — click to collapse

For an object dropped from rest, distances traversed in successive equal time intervals are in the ratio 1:3:5:7:9:... (odd numbers). Derivable from y = ½ g t². Common shape: 'find ratio of distance covered in 1st, 2nd, 3rd, 4th seconds of free fall'.

Direct ApplicationEasy

Common distractors

uses arithmetic progression 1 2 3 4

Linear-time intuition

Time given as implicit function of position, e.g. t = x² + x. To find acceleration, differentiate twice: dt/dx = 2x + 1, so v = dx/dt = 1/(2x+1); a = dv/dt = -2v²·v = -2v³ (chain rule). Multi-step calculus.

Multi StepMedium

Common distractors

treats t x as explicit

Default to t-as-input thinking

A projectile (typically a bullet) penetrates a uniform medium with constant decelerating force; given initial speed and speed after a known distance, find the total stopping distance. Apply v² = u² + 2as to each segment, noting that 'a' is the same throughout. Common shape: bullet hits block at u, slows to u/k after distance d₁; how much further to stop?

Multi StepMedium

Common distractors

treats speed ratio as distance ratio

Linear thinking: 'speed went from u to u/3, so distance traversed should be 3× the original'

Two particles' position-time graphs are straight lines with given angles vs the time axis (e.g. 30° and 45°); find ratio of velocities. Velocity = slope = tan(angle). Common shape: ratio = tan(30°)/tan(45°) = 1/√3.

InterpretationEasy

Common distractors

uses sin or cos instead of tan

Trig confusion under time pressure

A particle in uniform circular motion (period T, radius R) is subsequently launched vertically upward with the SAME speed (v = 2πR/T); find max height. H = v²/(2g) = (2πR/T)²/(2g).

Multi StepMedium

Common distractors

uses radius as height

Surface confusion of geometric R and projectile H

Given launch speed v₀ and angle θ above horizontal, find maximum height. H = v₀² sin² θ / (2g). Common values plug in cleanly when sin θ ∈ {0.5, 0.707, 0.866}. Distractors test (i) using sin instead of sin², (ii) forgetting the factor of 2 in denominator, (iii) using g = 10 vs g = 9.8 inconsistently.

Direct ApplicationEasy

Common distractors

uses sin instead of sin squared

Confusing the height formula H = v²sin²θ/(2g) with the range 2v sin θ / g

Projectile launched at angle θ; find SPEED at highest point. The vertical component vanishes at the apex; horizontal component v₀ cos θ is conserved. Common gotcha: 'angle with the vertical' vs 'angle with the horizontal'.

Direct ApplicationEasy

Common distractors

uses sin instead of cos

Sign-of-angle-axis confusion

speed zero at top

True only for vertical motion (no horizontal v₀)

Buses leave both ends of a route every T minutes; observer travels between them at constant speed; given separate periods at which buses pass observer in same and opposite directions, find bus speed or T. Use relative-velocity equations: same direction T_same = L / (v_bus - v_obs), opposite T_opp = L / (v_bus + v_obs).

Multi StepMedium

Common distractors

ignores direction relative to observer

Treating bus-passing time as absolute

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)

Object given a non-zero downward initial velocity from elevation; asked for the height fallen, time of impact, or final speed. Apply v² = u² + 2gh (or analogous) with proper signs. Common shape: a ball thrown vertically downward from a tower with initial speed u, hitting the ground at speed v; find the tower height. Distractors test (i) sign-of-u confusion, (ii) using v² = 2gh forgetting u², (iii) wrong g unit.

Multi StepEasy

Common distractors

drops initial velocity term

Student conflates 'thrown' with 'dropped' and uses v² = 2gh

Sources

NCERT refs: Class 11 Physics Chapter 2, p.5

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Velocity Time Graph

Lesson

Practice MCQs

Quick recall before you leave

First kinematic equation

Second kinematic equation (displacement)

Third kinematic equation (velocity-squared)

Worked Example

Before solving, remember these

Formulas

Valid when

Do NOT use when

Valid when

Valid when

Do NOT use when

Valid when

Valid when

Do NOT use when

Valid when

Do NOT use when

Exam Traps & Common Mistakes

When it triggers

How to avoid

When it triggers

How to avoid

When it triggers

How to avoid

When it triggers

How to avoid

When it triggers

How to avoid

When it triggers

How to avoid

When it triggers

How to avoid

When it triggers

How to avoid

When it triggers

How to avoid

When it triggers

How to avoid

Correction

Wrong option pattern

Correction

Wrong option pattern

Correction

Wrong option pattern

Correction

Wrong option pattern

Past Year Questions

How NEET usually asks this

Common distractors

Common distractors

Common distractors

Common distractors

Common distractors

Common distractors

Common distractors

Common distractors

Common distractors

Common distractors

Sources

Free NEET study resources