
Motion in a Plane is a core chapter in Class 11 Physics, introducing two-dimensional kinematics and vector algebra. As outlined in the Karnataka 1st PUC syllabus, the chapter covers vectors, projectile motion, and uniform circular motion. Board exams frequently test these topics through derivations, conceptual questions, and vector operations.
Practising Karnataka 1st PUC Physics Chapter 3 Important Questions helps students master projectile trajectories and circular paths. It also builds clarity on vector resolution and the formulas for range and height, while improving speed and accuracy in annual exams.
Students can practice these curated questions to master key derivations and formulas. Solving them ensures strong command over vector applications and kinematics.
1. For the resultant of two vectors to be maximum, what must be the angle between them?
(A) 0° (B) 60° (C) 90° (D) 180°
Answer: (A) 0°
Solution: R = √(A² + B² + 2AB cos θ). R is maximum when cos θ = 1, so θ = 0°.
2. If A = 4î − 3ĵ and B = 6î + 8ĵ, find the magnitude and direction (with the positive X-axis) of A + B.
(A) 5, tan⁻¹(3/4) (B) 5√5, tan⁻¹(1/2) (C) 25, tan⁻¹(3/4) (D) 10, tan⁻¹(5)
Answer: (B) 5√5, tan⁻¹(1/2)
Solution: A + B = 10î + 5ĵ. Magnitude = √(10² + 5²) = 5√5. Direction: tan θ = 5/10 = 1/2.
3. The maximum and minimum values of the resultant of two forces acting at a point are 7 N and 3 N, respectively. The smaller force is
(A) 2 N (B) 4 N (C) 5 N (D) 1 N
Answer: (A) 2 N
Solution: Maximum = P + Q = 7 and minimum = P − Q = 3. Subtracting gives 2Q = 4, so the smaller force Q = 2 N.
4. Among the given pairs of vectors, the resultant can never be 3 units for
(A) 1 unit and 2 units (B) 2 units and 5 units (C) 3 units and 6 units (D) 4 units and 8 units
Answer: (D) 4 units and 8 units
Solution: The resultant of two vectors lies between |A − B| and A + B. For 4 and 8, the range is 4 to 12 units, so 3 units is not possible.
5. The range of a projectile fired at an angle of 15° is 50 m. If it is fired with the same speed at an angle of 45°, its range will be
(A) 25 m (B) 37 m (C) 50 m (D) 100 m
Answer: (D) 100 m
Solution: R = v₀² sin 2θ / g. At 15°, v₀²/g × sin 30° = 50, so v₀²/g = 100 m. At 45°, R = (v₀²/g) × sin 90° = 100 m.
6. A projectile is projected at an angle of 60° to the horizontal. If its range is 30√3 m, the maximum height reached is
(A) 15√3 m (B) 60√3 m (C) 50 m (D) 22.5 m
Answer: (D) 22.5 m
Solution: R = (v₀²/g) sin 120° = 30√3, which gives v₀²/g = 60. H = (v₀²/g) × sin²60° / 2 = 60 × (3/4) / 2 = 22.5 m.
7. In the projectile motion of a particle on level ground, which of the following remains constant with respect to time and position?
(A) Vertical component of velocity
(B) Average velocity between any two points on the path
(C) Horizontal component of velocity
(D) Angle between the instantaneous velocity and the horizontal
Answer: (C) Horizontal component of velocity
Solution: No force acts horizontally (neglecting air resistance), so vₓ = v₀ cos θ₀ stays constant. Only the vertical component changes due to gravity.
8. A particle under uniform circular motion makes 600 rpm. In how much time will it complete one revolution?
(A) 0.2 s (B) 0.1 s (C) 0.4 s (D) 0.3 s
Answer: (B) 0.1 s
Solution: 600 rpm = 600/60 = 10 revolutions per second. Time period T = 1/10 = 0.1 s.
9. A particle moves in a circle of radius 5 cm with constant speed and a time period of 0.2π s. The acceleration of the particle is
(A) 15 m/s² (B) 25 m/s² (C) 36 m/s² (D) 5 m/s²
Answer: (D) 5 m/s²
Solution: ω = 2π/T = 2π/0.2π = 10 rad/s. Centripetal acceleration a = ω²r = (10)² × 0.05 = 5 m/s².
10. A particle is projected so that its horizontal range is twice the greatest height attained. If the initial speed is v, the horizontal range is
(A) v²/g (B) v²/2g (C) 2v²/3g (D) 4v²/5g
Answer: (D) 4v²/5g
Solution: Setting R = 2H gives (v² · 2 sin θ cos θ)/g = (v² sin²θ)/g, so tan θ = 2. Then sin 2θ = 2 tan θ/(1 + tan²θ) = 4/5, and R = (v²/g) × (4/5) = 4v²/5g.
Students can download the complete chapter question bank for offline practice. Access the 1st PUC Physics Chapter 3 Important Questions PDF through the link below.
Download Here
Using organised solutions helps students plan their study time effectively. Begin by reviewing the Karnataka Class 11 Physics Syllabus to understand the weightage of Motion in a Plane, then follow these steps to prepare for your Physics exam:
Master Core Vector Concepts: Study vector addition laws and resolution before starting problems. Strong vector fundamentals make two-dimensional kinematics easier to understand.
Practice Standard Derivations: Write down derivations for projectile range, time of flight, and centripetal acceleration. Regular writing practice ensures step-by-step accuracy, and checking the Karnataka 1st PUC Exam Pattern helps you see how marks are distributed across long-answer and short-answer questions.
Solve Motion in a Plane Numericals: Practice numerical problems to build calculation speed. Apply kinematic equations separately along horizontal and vertical directions, and review the Karnataka 1st PUC Mid Term Question Papers to see how such problems are framed and solved.
Review Motion in a Plane Questions and Answers: Check standard questions and answers to verify your problem-solving steps and correct mistakes early. Comparing your solutions with the answer key helps you spot errors in method and calculation.
Attempt Motion in a Plane Important Questions: Solve these questions under timed conditions, then take a full test using the Karnataka 1st PUC Physics Model Papers to improve exam confidence and time management.
Mastering Motion in a Plane requires consistent practice of vector operations, projectile motion, and circular motion problems. Regularly solving these important questions will strengthen your concepts and help you score confidently in your Karnataka 1st PUC Physics exam.
Centripetal acceleration is the inward acceleration directed toward the centre of a circular path. Its magnitude is
ac=v2R

