JEE Physics Motion In A Plane MCQs Set 29

Multiple Choice Questions (MCQs) for JEE Physics: Motion In A Plane

Explore reliable objective questions for Motion In A Plane tailored for JEE learners. Utilizing these Physics multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

Practice Motion In A Plane MCQs for JEE Physics

Access the complete set of multiple-choice questions for Motion In A Plane below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: A stone of mass m is tied to a string of length l and rotated in a circle with constant speed v. If the string is released, the stone flies
(a) radially outward
(b) radially inward
(c) tangentially outward
(d) with an acceleration mv²/l
Show Answer & Explanation

Answer: (c) tangentially outward

Explanation:
1. At release no force acts horizontally, so the stone moves along the instantaneous velocity.
2. That direction is the tangent to the circle.

Teacher's Note:
1. Velocity is always tangential to the circle.

Question 2: For a particle in uniform circular motion with speed v on a circle of radius R, the acceleration at the point P(R, θ) on the circle (θ is measured from the x-axis) is
(a) \( -\frac{v^{2}}{R}\cos\theta\,\hat i+\frac{v^{2}}{R}\sin\theta\,\hat j \)
(b) \( -\frac{v^{2}}{R}\sin\theta\,\hat i+\frac{v^{2}}{R}\cos\theta\,\hat j \)
(c) \( -\frac{v^{2}}{R}\cos\theta\,\hat i-\frac{v^{2}}{R}\sin\theta\,\hat j \)
(d) \( \frac{v^{2}}{R}\hat i+\frac{v^{2}}{R}\hat j \)
Show Answer & Explanation

Answer: (c) \( -\frac{v^{2}}{R}\cos\theta\,\hat i-\frac{v^{2}}{R}\sin\theta\,\hat j \)

Explanation:
1. The acceleration is directed towards the centre with magnitude v²/R.
2. The unit vector from P to the centre is −(cos θ i + sin θ j).

Teacher's Note:
1. The acceleration points opposite to the position vector.

Question 3: A ball is thrown upwards and returns to the ground describing a parabolic path. Which of the following has the same value at the time of throw and at the time of return? I. Kinetic energy. II. Speed. III. Vertical component of velocity.
(a) I and II
(b) II and III
(c) III only
(d) I, II and III
Show Answer & Explanation

Answer: (a) I and II

Explanation:
1. At the same level the speed is the same, so the kinetic energy is the same.
2. The vertical component reverses direction, so it is not the same.

Teacher's Note:
1. Speed is a magnitude; velocity components have signs.

Question 4: A particle moves in a circular orbit under the action of a central attractive force inversely proportional to the distance r. The speed of the particle is
(a) \( \propto r^{2} \)
(b) independent of r
(c) \( \propto r \)
(d) \( \propto 1/r \)
Show Answer & Explanation

Answer: (b) independent of r

Explanation:
1. k/r = mv²/r gives v² = k/m.
2. The speed does not depend on r.

Teacher's Note:
1. Equate force with the centripetal requirement.

Question 5:
Assertion (A): Two projectiles having the same range must have the same time of flight.
Reason (R): The horizontal component of velocity is constant in projectile motion under gravity.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (d) A is false, but R is true.

Explanation:
1. Equal ranges occur for θ and 90° − θ, which have different times of flight, so the Assertion is false.
2. The Reason is a true statement.

Teacher's Note:
1. Complementary angles give the same range but different flight times.

Question 6: For motion in two or three dimensions, the angle between velocity and acceleration is
(a) 0°
(b) 90°
(c) 180°
(d) any angle between 0° and 180°
Show Answer & Explanation

Answer: (d) any angle between 0° and 180°

Explanation:
1. The acceleration can have any orientation relative to the velocity.
2. Parallel, perpendicular and oblique cases all occur, so any angle from 0° to 180° is possible.

Teacher's Note:
1. General motion has no fixed angle.

Question 7: If t_m is the time taken by a projectile to reach the maximum height, the total time of flight T_f is related to t_m as
(a) t_m = 2T_f
(b) T_f = t_m
(c) T_f = 2t_m
(d) none of these
Show Answer & Explanation

Answer: (c) T_f = 2t_m

Explanation:
1. Time of ascent = u sin θ/g and time of flight = 2u sin θ/g.
2. So T_f = 2t_m.

Teacher's Note:
1. Ascent and descent take equal time on level ground.

Question 8: The position vector of a particle is \( \vec r=(a\cos\omega t)\hat i+(a\sin\omega t)\hat j \). The velocity of the particle is
(a) directed towards the origin
(b) directed away from the origin
(c) parallel to the position vector
(d) perpendicular to the position vector
Show Answer & Explanation

Answer: (d) perpendicular to the position vector

Explanation:
1. v = dr/dt = −aω sin ωt i + aω cos ωt j.
2. Then r·v = 0, so the velocity is perpendicular to the position vector (circular motion).

Teacher's Note:
1. Zero dot product means the vectors are perpendicular.

Question 9: Vector A has components A_x = 2, A_y = 3 and vector B has components B_x = 4, B_y = 5. Match the columns.
Column IColumn II
(A) y-component of A + B(1) 8
(B) Magnitude of A + B(2) \( -2 \)
(C) x-component of A − B(3) \( 2\sqrt2 \)
(D) Magnitude of A − B(4) 10

(a) A-1, B-4, C-2, D-3
(b) A-1, B-2, C-4, D-3
(c) A-4, B-1, C-2, D-3
(d) A-3, B-4, C-2, D-1
Show Answer & Explanation

Answer: (a) A-1, B-4, C-2, D-3

Explanation:
1. A + B = (6, 8), so its y-component is 8 and its magnitude is 10.
2. A − B = (−2, −2), so its x-component is −2 and its magnitude is 2√2.

Teacher's Note:
1. Add or subtract the components, then use the Pythagoras theorem.

Question 10: A body projected at an angle with the horizontal has a range of 300 m. If the time of flight is 6 s, the horizontal component of velocity is
(a) 30 m/s
(b) 50 m/s
(c) 40 m/s
(d) 45 m/s
Show Answer & Explanation

Answer: (b) 50 m/s

Explanation:
1. Range = horizontal velocity × time of flight.
2. ux = 300/6 = 50 m/s.

Teacher's Note:
1. The horizontal velocity is constant.

Question 11: A vector can be written in terms of components as A = Ax i + Ay j + Az k. When will |A| be zero?
(a) Ax = Ay = 0 and Az ≠ 0
(b) Ax = Ay = Az ≠ 0
(c) Ax = Ay = Az = 0
(d) |A| can never be zero
Show Answer & Explanation

Answer: (c) Ax = Ay = Az = 0

Explanation:
1. |A| = √(Ax² + Ay² + Az²) is the sum of squares.
2. It is zero only when every component is zero.

Teacher's Note:
1. A sum of squares vanishes only if every term vanishes.

Question 12: A boat which has a speed of 5 km/h in still water crosses a river of width 1 km along the shortest possible path in 15 minutes. The velocity of the river water is
(a) 1 km/h
(b) 3 km/h
(c) 4 km/h
(d) √41 km/h
Show Answer & Explanation

Answer: (b) 3 km/h

Explanation:
1. The shortest path means the boat heads upstream so that the drift cancels.
2. Crossing speed = 1 km/0.25 h = 4 km/h.
3. River speed = √(5² − 4²) = 3 km/h.

Teacher's Note:
1. Net speed across and the speed in still water give the river speed.

Question 13: If Vr is the velocity of rain falling vertically and Vm is the velocity of a man walking on a level road, the magnitude of the velocity of the rain relative to the man is
(a) \( \sqrt{V_r^{2}+V_m^{2}+2V_rV_m\cos\theta} \)
(b) \( \sqrt{V_r^{2}+V_m^{2}-2V_rV_m\cos\theta} \)
(c) \( \sqrt{V_r^{2}+V_m^{2}} \)
(d) \( \sqrt{V_r^{2}-V_m^{2}} \)
Show Answer & Explanation

Answer: (c) \( \sqrt{V_r^{2}+V_m^{2}} \)

Explanation:
1. The rain velocity is vertical and the man's velocity is horizontal, so the two are perpendicular.
2. Relative speed = √(Vr² + Vm²).

Teacher's Note:
1. Perpendicular components combine by Pythagoras.

Question 14: The range of a projectile is R when the angle of projection is 40°. For the same velocity of projection and range, the other possible angle of projection is
(a) 45°
(b) 50°
(c) 60°
(d) 40°
Show Answer & Explanation

Answer: (b) 50°

Explanation:
1. Equal ranges occur for θ and 90° − θ.
2. The other angle is 90° − 40° = 50°.

Teacher's Note:
1. Complementary angles give equal range.

Question 15: The position of a particle is \( \vec r=2t^{2}\hat i+3t\hat j+4\hat k \) (SI units). The acceleration at t = 1 s is
(a) 4 m/s² along the y-direction
(b) 3 m/s² along the x-direction
(c) 4 m/s² along the x-direction
(d) 2 m/s² along the z-direction
Show Answer & Explanation

Answer: (c) 4 m/s² along the x-direction

Explanation:
1. a = d²r/dt² = 4i.
2. It is constant, so at t = 1 s it is 4 m/s² along the x-direction.

Teacher's Note:
1. Differentiate the position vector twice.

Question 16: If the angles of projection of a projectile with the same initial velocity exceed or fall short of 45° by equal amounts, the ratio of the horizontal ranges is
(a) 1 : 2
(b) 1 : 3
(c) 1 : 4
(d) 1 : 1
Show Answer & Explanation

Answer: (d) 1 : 1

Explanation:
1. The angles are 45° + α and 45° − α, which add up to 90°.
2. They are complementary, so the ranges are equal and the ratio is 1 : 1.

Teacher's Note:
1. Symmetry about 45° gives equal range.

Question 17: ABCDEF is a regular hexagon with centre O. The value of \( \vec{AB}+\vec{AC}+\vec{AD}+\vec{AE}+\vec{AF} \) is
(a) \( 2\vec{AO} \)
(b) \( 4\vec{AO} \)
(c) \( 6\vec{AO} \)
(d) zero
Show Answer & Explanation

Answer: (c) \( 6\vec{AO} \)

Explanation:
1. AB + AE... pair up: AB + AF = AO·2... precisely, AB + AF = AC·… so use the centre: AC + AE = 2AO + ...
2. A cleaner route is to put O as origin: AB + AC + AD + AE + AF = (B + C + D + E + F) − 5A, and B + C + D + E + F = −A since the six vertices sum to zero.
3. The sum is −A − 5A = −6A = 6 AO.

Teacher's Note:
1. Vertices of a regular polygon sum to zero about its centre.

Question 18:
Assertion (A): The horizontal range is the same when the angle of projection is greater than 45° by a certain value and less than 45° by the same value.
Reason (R): If θ = 45° + α, then R1 = u² cos 2α/g, and if θ = 45° − α, then R2 = u² cos 2α/g.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. sin 2(45° + α) = sin(90° + 2α) = cos 2α and sin 2(45° − α) = cos 2α.
2. So the ranges are equal, and the Reason shows why.

Teacher's Note:
1. Angles symmetric about 45° give equal ranges.

Question 19: Given A = i + j and B = i − j, match the columns.
Column IColumn II
(A) (A + B)/2(1) \( \hat i \)
(B) (A − B)/2(2) \( \hat j \)
(C) (A · B)/2(3) \( -\hat k \)
(D) (A × B)/2(4) 0

(a) A-2, B-1, C-4, D-3
(b) A-1, B-3, C-4, D-2
(c) A-3, B-2, C-4, D-1
(d) A-1, B-2, C-4, D-3
Show Answer & Explanation

Answer: (d) A-1, B-2, C-4, D-3

Explanation:
1. (A + B)/2 = i and (A − B)/2 = j.
2. A·B = 1 − 1 = 0.
3. A × B = (i + j) × (i − j) = −2k, so (A × B)/2 = −k.

Teacher's Note:
1. Compute the sum, difference, dot and cross products component-wise.

Question 20: Which of the following statements are incorrect? I. A scalar quantity is one that is conserved in a process. II. A scalar quantity is one that can never take negative values. III. A scalar quantity has the same value for observers with different orientations of the axes.
(a) I and III
(b) II only
(c) II and III
(d) I and II
Show Answer & Explanation

Answer: (d) I and II

Explanation:
1. Conservation has nothing to do with being a scalar, so I is incorrect.
2. Scalars such as charge and temperature can be negative, so II is incorrect.
3. III is the correct definition.

Teacher's Note:
1. Check each statement against the definition of a scalar.

Question 21: At the highest point on the trajectory of a projectile, its
(a) potential energy is minimum
(b) kinetic energy is maximum
(c) total energy is maximum
(d) kinetic energy is minimum
Show Answer & Explanation

Answer: (d) kinetic energy is minimum

Explanation:
1. At the top the vertical velocity is zero, so only the horizontal velocity remains.
2. The speed is least there, so kinetic energy is minimum while potential energy is maximum; the total energy is constant.

Teacher's Note:
1. KE is least and PE is greatest at the top.

Question 22:
Assertion (A): τ = r × F and τ ≠ F × r.
Reason (R): The cross product of vectors is commutative.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (c) A is true, but R is false.

Explanation:
1. F × r = −(r × F) = −τ, so τ ≠ F × r and the Assertion is true.
2. The cross product is anticommutative, so the Reason is false.

Teacher's Note:
1. A × B = −B × A.

Question 23: For two vectors A and B, |A + B| = |A − B| is always true when
(a) |A| = |B| ≠ 0
(b) A ⟂ B
(c) |A| = |B| ≠ 0 and A and B are parallel or antiparallel
(d) none of these
Show Answer & Explanation

Answer: (b) A ⟂ B

Explanation:
1. Squaring gives A² + B² + 2A·B = A² + B² − 2A·B, so A·B = 0.
2. The vectors must be perpendicular.

Teacher's Note:
1. Equal sum and difference magnitudes mean a zero dot product.

Question 24: Which of the following is an essential condition for the horizontal component of velocity of a projectile to remain constant?
(a) Acceleration due to gravity should be exactly constant
(b) Angle of projection should be 45°
(c) There should be no air resistance
(d) All of these
Show Answer & Explanation

Answer: (c) There should be no air resistance

Explanation:
1. With no horizontal force the horizontal acceleration is zero and the horizontal velocity stays constant.
2. Air resistance would give a horizontal force, so it must be absent.
3. The angle of projection and the constancy of g do not affect the horizontal component.

Teacher's Note:
1. Constant horizontal velocity needs zero horizontal force.

Question 25: The shape of the trajectory of the motion of an object is determined by
(a) acceleration
(b) initial position
(c) initial velocity
(d) all of these
Show Answer & Explanation

Answer: (d) all of these

Explanation:
1. The path depends on the initial position and initial velocity as well as the acceleration acting.
2. Changing any one of them changes the trajectory.

Teacher's Note:
1. The path follows from the equations of motion with all three inputs.

Free study material for Physics

Download Chapter MCQs: JEE Physics Motion In A Plane

JEE Physics Motion In A Plane Objective Test Questions

Explore reliable practice questions for Motion In A Plane tailored for JEE Physics learners. Use these multiple-choice formats to evaluate preparedness and strengthen problem-solving skills.

NCERT-Aligned Objective Questions and Solutions

Each question includes structured solution keys mapped directly to standard JEE textbooks, helping students evaluate their reasoning and correct mistakes early in their revision.

Next Steps in Your Exam Preparation

Explore our broader library of printable assignments, chapter notes, and mock tests designed to support continuous revision and secure higher marks in JEE assessments.

FAQs

Where can I access latest JEE Physics Motion In A Plane MCQs Set 29?

You can get most exhaustive JEE Physics Motion In A Plane MCQs Set 29 for free on StudiesToday.com. These MCQs for JEE Physics are updated for the 2026-27 academic session as per JEE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Physics JEE material?

Yes, our JEE Physics Motion In A Plane MCQs Set 29 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.

How do practicing Physics MCQs help in scoring full marks in JEE exams?

By solving our JEE Physics Motion In A Plane MCQs Set 29, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Physics.

Do you provide answers and explanations for JEE Physics Motion In A Plane MCQs Set 29?

Yes, Physics MCQs for JEE have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused JEE exams.

Can I practice these Physics JEE MCQs online?

Yes, you can also access online interactive tests for JEE Physics Motion In A Plane MCQs Set 29 on StudiesToday.com as they provide instant answers and score to help you track your progress in Physics.