NEET PhysicsUpdated 19 Oct 2026

Motion in a Plane: NEET PYQs and weightage

Motion in a Plane carries 2.4 of 100 Physics marks in NEET by past-paper share: 18 of the 758 tagged NEET Physics PYQs in the ViJEEta bank (NEET 2013–2026). It ranks 24 of 30 Physics chapters.

Take the 25-question Motion in a Plane test Formula sheet ↓All NEET Physics chapters

2.4/100 Weight in Physics

Exam importance (2/5, by past-paper share)

18 NEET PYQs

363 Practice questions

334 With measured difficulty

How the weight is computed: this chapter's tagged NEET PYQs divided by all tagged NEET Physics PYQs in the bank (18 of 758), times 100, with repeated questions counted once. The bank does not store the year of each NEET question, so none is printed. Computed from the bank at build time. Updated 19 Oct 2026.

Motion in a Plane: formula sheet for NEET

14 formulas, each with when to use it. Copy it onto one page of your notebook: that page is this chapter's handbook.

Horizontal range
$R = \frac{v _{0}^{2} sin 2 θ _{0}}{g}$
Use it when: the projectile lands back at the same level it was launched from
Time of flight
$T_{f} = \frac{2 v _{0} sin θ _{0}}{g}$
Use it when: you need the total time in air for a same-level landing (it equals $2 \times$ time to reach the top)
Maximum height
$h_{m} = \frac{( v _{0} sin θ _{0} ) ^{2}}{2 g}$
Use it when: only the vertical part of the launch speed matters for how high it rises
Time to reach the top
$t_{m} = \frac{v _{0} sin θ _{0}}{g}$
Use it when: the question asks when vertical velocity becomes zero
Maximum range
$R_{m a x} = \frac{v _{0}^{2}}{g} (at θ_{0} = 4 5^{\circ})$
$R (4 5^{\circ} + α) = R (4 5^{\circ} - α)$
Angles $(4 5^{\circ} + α)$ and $(4 5^{\circ} - α)$ give equal range.
Use it when: a question asks for the best angle, or two launch angles give the same range
Velocity at time t
$v_{x} = v_{0} cos θ_{0} (constant)$
$v_{y} = v_{0} sin θ_{0} - g t$
Use it when: you need the speed or direction at any instant: $v = v_{x}^{2} + v_{y}^{2}$ , $tan θ = v_{y} / v_{x}$
Position at time t
$x = (v_{0} cos θ_{0}) t$
$y = (v_{0} sin θ_{0}) t - \frac{1}{2} g t^{2}$
Use it when: you need where the projectile is, including horizontal throws from a height (put $θ_{0} = 0$ , so $y = - \frac{1}{2} g t^{2}$ )
Equation of trajectory
$y = x tan θ_{0} - \frac{g x ^{2}}{2 ( v _{0} cos θ _{0} ) ^{2}}$
Use it when: $x$ and $y$ are given with no time, or you must check that the path is a parabola
Resultant of two vectors
$R^{2} = A^{2} + B^{2} + 2 A B cos θ$
$tan α = \frac{B sin θ}{A + B cos θ}$
Use it when: two vectors at angle $θ$ are added ( $θ$ is the angle between them, tail to tail)
Components of a vector
$A_{x} = A cos θ$
$A_{y} = A sin θ$
$A = A_{x}^{2} + A_{y}^{2}$
$tan θ = \frac{A _{y}}{A _{x}}$
Use it when: breaking a velocity or force along $x$ and $y$ , or rebuilding it from components
Constant acceleration in a plane
$v = v_{0} + a t$
$r = r_{0} + v_{0} t + \frac{1}{2} a t^{2}$
Use it when: acceleration is a fixed vector; apply each equation separately along $x$ and along $y$
Velocity and acceleration from position
$v = \frac{d r}{d t}$
$a = \frac{d v}{d t} (component by component)$
Use it when: $r (t)$ is given as a function of time with $\hat{i}, \hat{j}, \hat{k}$
Centripetal acceleration
$a_{c} = \frac{v ^{2}}{R} = ω^{2} R = 4 π^{2} ν^{2} R$
Directed towards the centre.
Use it when: the speed is constant on a circle; magnitude stays fixed but the direction keeps changing
Angular speed
$v = ω R$
$ω = \frac{2 π}{T} = 2 π ν (ω in rad/s)$
Use it when: revolutions per time are given and you need $ω$ or the linear speed

Trap

at the highest point the velocity is NOT zero. Only v_y becomes zero, and the projectile still moves at v₀ cos θ₀. Also, R and T_f above are valid only when it lands back at the launch level.

How does NEET ask Motion in a Plane?

7 real previous-year questions from this chapter, as printed in the paper, each with its worked solution. Answer the first one here with no sign-in.

Q1PhysicsNEET UG PYQUniform Circular Motion

Try right here · No sign-in

A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8 × 10−4J by the end of the second revolution after the beginning of the motion?

Answer key (A) & worked solution

Correct answer: Option A

\frac{1}{2} m v^{2} = E \Rightarrow \frac{1}{2} (\frac{10}{1000}) v^{2} = 8 \times 1 0^{- 4}

⇒ v

^{2}

= 16 × 10−2 ⇒ v = 4 × 10−1 = 0.4 m/s Now, V

^{2}

= u

^{2}

+ 2a

_{t}

s (s = 4π R)

\Rightarrow \frac{16}{100} = 0^{2} + 2 a_{t} (4 \times \frac{22}{7} \times \frac{6.4}{100}) \Rightarrow a_{t} = \frac{16}{100} \times \frac{7 \times 100}{8 \times 22 \times 6.4} = 0.1 m / s^{2}

Q2PhysicsNEET UG PYQProjectile Motion

A bullet is fired from a gun at the speed of

280 m s^{- 1}

in the direction

3 0^{\circ}

above the horizontal. The maximum height attained by the bullet is

(g = 9.8 m s^{- 2}, sin 3 0^{\circ} = 0.5) :

A2000 m

B1000 m

C3000 m

D2800 m

Answer key (B) & worked solution

Correct answer: Option B

H_{m a x} = \frac{u ^{2} s i n ^{2} θ}{2 g}

= \frac{( 280 ) ^{2} ( s i n 3 0 ^{\circ} ) ^{2}}{2 ( 9.8 )}

= 1000 m

Q3PhysicsNEET UG PYQPosition Vector, Displacement, And Motion In Two Dimensions

The x and y coordinates of the particle at any time are x = 5t – 2t

^{2}

and y = 10t respectively, where x and y are in meter and t in second. The acceleration of the particle at t = 2 s is

A– 4 m/s

^{2}

B8 m/s

^{2}

D5 m/s

^{2}

Answer key (A) & worked solution

Correct answer: Option A

x = 5 t - 2 t^{2}; y = 10 t

\frac{d x}{d t} = 5 - 4 t; \frac{d y}{d t} = 10

v_{x} = 5 - 4 t; v_{y} = 10

\frac{d v _{x}}{d t} = - 4; \frac{d v _{y}}{d t} = 0

a_{x} = - 4; a_{y} = 0

Acceleration of particle at

t = 2 s

= - 4 m / s^{2}

Q4PhysicsNEET UG PYQRelative Motion in Two Dimensions

A monoatomic gas at a pressure P, having a volume V expands isothermally to a volume 2V and then adiabatically to a volume 16V. The final pressure of the gas is:

(γ = \frac{5}{3})

\frac{P}{64}

B16P

C64P

D32P

Answer key (A) & worked solution

Correct answer: Option A

In isothermal process

P_{1} V_{1} = P_{2} V_{2}

P V = P_{2} 2 V

P_{2} = \frac{P}{2}

Now in adiabatic process

P V^{γ} = co n s t

\frac{p}{2} (2 V)^{γ} = P_{f} (16 V)^{γ}

P_{f} = \frac{P}{2} (\frac{2 V}{16 V})^{γ}

γ = \frac{5}{3}

\frac{P _{f}}{P} = \frac{1}{2 ^{3 γ + 1}} = \frac{1}{2 ^{6}} = \frac{1}{64}

Q5PhysicsNEET UG PYQUniform Circular Motion

A particle moving with uniform speed in a circular path maintains:

AConstant acceleration

BConstant velocity but varying acceleration

CVarying velocity and varying acceleration

DConstant velocity.

Answer key (C) & worked solution

Correct answer: Option C

Direction of both velocity and acceleration are changing.

Q6PhysicsNEET UG PYQPosition Vector, Displacement, And Motion In Two Dimensions

A particle moves so that its position vector is given by

r = cos ω t \overset{x}{^} + sin ω r \overset{y}{^}

. Where ω is a constant. Which of the following is true?

AVelocity and acceleration both are perpendicular to

r

BVelocity and acceleration both are parallel to

r

CVelocity is perpendicular to

r

and acceleration is directed towards the origin

DVelocity is perpendicular to

r

and acceleration is directed away from the origin

Answer key (C) & worked solution

Correct answer: Option C

r = cos ω t \overset{x}{^} + sin ω t \overset{y}{^} v = - ω sin ω t \overset{x}{^} + ω cos ω t \overset{y}{^} a = - ω^{2} cos ω t \overset{x}{^} - ω sin ω t \overset{y}{^} = - ω^{2} r

r . v = 0

hence

r ⊥ v

a

is directed towards the origin.

Q7PhysicsNEET UG PYQProjectile Motion

A projectile is fired from the surface of the earth with a velocity of 5 ms

^{- 1}

and angle θ with the horizontal. Another projectile fired from another planet with a velocity of 3 ms

^{- 1}

at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is (in ms

^{- 2}

) : (given g = 9.8 ms

^{- 2}

)

A16.3

B110.8

C3.5

D5.9

Answer key (C) & worked solution

Correct answer: Option C

y = x tan θ - \frac{1}{2} g \frac{x}{u ^{2} cos ^{2} θ} [Trajectory equation] \Rightarrow \frac{g _{1}}{( u _{1} ) ^{2}} = \frac{g _{2}}{( u _{2} ) ^{2}} ∴ \frac{g _{2}}{g _{1}} = (\frac{u _{2}}{u _{1}})^{2} = (\frac{3}{5})^{2} = \frac{9}{25} ∴ g_{2} = \frac{9}{25} \times 9.8 = 3.5 m / s^{2}

Which topics in Motion in a Plane carry the most questions?

All 5 topics, by tagged PYQs, then practice questions. Counted from the bank. Updated 19 Oct 2026.

Topic	PYQs	Practice questions
Projectile Motion	6	191
Uniform Circular Motion	6	98
Relative Motion in Two Dimensions	4	51
Position Vector, Displacement, And Motion In Two Dimensions	2	22
Miscellaneous Examples	0	1

Why does Motion in a Plane show "–" until you answer 3 questions?

Unknown is not zero. Until you answer at least 3 graded questions in Motion in a Plane, ViJEEta shows – instead of a zero or a 50% default. Book exercises can raise chapter mastery up to 0.60, adaptive practice up to 0.75, and only a timed chapter test or mock can certify 0.80 and above.

How ViJEEta scores you

What happens when you test Motion in a Plane?

Every chapter test is 25 questions. It starts at moderate difficulty and adapts after each saved answer. If you open a solution before answering, that attempt earns no credit, so your mastery stays honest.

Start the Motion in a Plane test

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Motion in a Plane: NEET PYQs and weightage

Motion in a Plane: formula sheet for NEET

Horizontal range

Time of flight

Maximum height

Time to reach the top

Maximum range

Velocity at time t

Position at time t

Equation of trajectory

Resultant of two vectors

Components of a vector

Constant acceleration in a plane

Velocity and acceleration from position

Centripetal acceleration

Angular speed

How does NEET ask Motion in a Plane?

Which topics in Motion in a Plane carry the most questions?

Why does Motion in a Plane show "–" until you answer 3 questions?

What happens when you test Motion in a Plane?

Chapters either side of Motion in a Plane by weight