Important Questions Class 11 Physics Chapter 4: Motion in a Plane

Motion in a plane describes movement using two perpendicular directions instead of one straight line. Vectors help express displacement, velocity and acceleration when direction changes during motion.

Motion in a plane extends one-dimensional kinematics to two-dimensional paths using vectors, components and relative velocity. Important Questions Class 11 Physics Chapter 4 help students practise scalar and vector quantities, vector addition, resolution of vectors, projectile motion, relative velocity in two dimensions and uniform circular motion. The CBSE 2026 chapter builds the foundation for mechanics, projectile numericals, circular motion, relative motion and vector-based problem solving.

Key Takeaways

• Vector Quantity: A vector has magnitude and direction, such as displacement or velocity.
• Projectile Path: Projectile motion under gravity follows a parabolic path.
• Relative Velocity: Velocity of A relative to B is vAB = vA − vB.
• Circular Motion: Uniform circular motion has centripetal acceleration v²/r towards the centre.

Important Questions Class 11 Physics Chapter 4 Structure 2026

Important Questions Class 11 Physics Chapter 4 with Answers

Motion in a Plane combines vector algebra with two-dimensional motion.
Students should resolve every vector into components before using equations.
These motion in a plane class 11 important questions follow the NCERT 2026 flow.

1. What does Important Questions Class 11 Physics Chapter 4 mainly test?

Important Questions Class 11 Physics Chapter 4 mainly test vectors, components, projectile motion, relative velocity and uniform circular motion. The chapter connects direction-based quantities with kinematics.

1. Vector Skill: Add, subtract and resolve vectors.
2. Projectile Skill: Use horizontal and vertical motion separately.
3. Relative Velocity Skill: Use vAB = vA − vB.
4. Circular Motion Skill: Use ac = v²/r.
5. Final Result: The chapter tests two-dimensional motion through vectors.

2. Why are vectors needed in Motion in a Plane?

Vectors are needed because motion in a plane has magnitude and direction. Displacement, velocity and acceleration can point in different directions.

1. One-Dimensional Motion: Direction stays along one line.
2. Plane Motion: Direction can change continuously.
3. Vector Use: Direction and magnitude stay included.
4. Final Result: Vectors describe two-dimensional motion correctly.

3. What is the main difference between motion in a straight line and motion in a plane?

Motion in a straight line needs one coordinate, while motion in a plane needs two coordinates. Plane motion uses x and y components.

1. Straight Line: One axis is enough.
2. Plane Motion: Two perpendicular axes are needed.
3. Example: Projectile motion needs horizontal and vertical analysis.
4. Final Result: Motion in a plane is two-dimensional motion.

Vectors Class 11 Physics Questions

Vectors describe quantities that need both magnitude and direction.
They follow special addition rules instead of ordinary arithmetic addition.
These vectors class 11 physics questions cover basics required for Motion in a Plane.

4. What is a vector quantity?

A vector quantity has both magnitude and direction. It also follows vector addition laws.

1. Magnitude: Numerical size.
2. Direction: Orientation in space.
3. Examples: Displacement, velocity, acceleration and force.
4. Final Result: A vector needs magnitude and direction.

5. What is a scalar quantity?

A scalar quantity has only magnitude and no direction. It follows ordinary algebra.

1. Magnitude: Numerical value.
2. Direction: Not required.
3. Examples: Mass, time, distance and speed.
4. Final Result: A scalar has magnitude only.

6. Is displacement a scalar or vector?

Displacement is a vector quantity. It gives the shortest change in position with direction.

1. Initial Position: Starting point.
2. Final Position: Ending point.
3. Direction: From initial to final point.
4. Final Result: Displacement is a vector.

7. Is speed a scalar or vector?

Speed is a scalar quantity. It gives how fast an object moves without direction.

1. Speed: Distance travelled per unit time.
2. Direction: Not included.
3. Related Vector: Velocity includes direction.
4. Final Result: Speed is scalar, while velocity is vector.

Scalar and Vector Quantities Class 11 Questions

Students often lose marks by mixing scalar and vector rules.
Vector quantities require direction, while scalar quantities need only magnitude.
These scalar and vector quantities class 11 questions clarify common NCERT examples.

8. What is the difference between distance and displacement?

Distance is total path length, while displacement is shortest directed separation. Distance is scalar, and displacement is vector.

1. Distance: Depends on path taken.
2. Displacement: Depends on initial and final positions.
3. Closed Path: Distance is non-zero, displacement is zero.
4. Final Result: Distance and displacement are different physical quantities.

9. What is the difference between speed and velocity?

Speed is distance per unit time, while velocity is displacement per unit time. Speed is scalar, and velocity is vector.

1. Speed Formula: Distance/time.
2. Velocity Formula: Displacement/time.
3. Direction: Velocity needs direction.
4. Final Result: Velocity is directional speed.

10. Can a vector have zero magnitude?

Yes, a vector can have zero magnitude. Such a vector is called a zero vector.

1. Magnitude: Zero.
2. Direction: Not definite.
3. Example: Displacement after a complete round trip.
4. Final Result: A zero vector has magnitude zero.

11. Can two vectors of different magnitudes be equal?

No, two vectors are equal only when their magnitudes and directions are the same. Their initial positions need not be same.

1. Condition 1: Same magnitude.
2. Condition 2: Same direction.
3. Position: Can be shifted parallel to itself.
4. Final Result: Equal vectors have equal magnitude and direction.

Vector Addition Class 11 Questions

Vector addition depends on both magnitudes and directions.
Graphical and analytical methods give the same resultant when used correctly.
These vector addition class 11 questions cover triangle law, parallelogram law and components.

12. State the triangle law of vector addition.

Triangle law states that two vectors can be placed head-to-tail to find their resultant. The third side gives the resultant.

1. First Vector: Draw from starting point.
2. Second Vector: Draw from head of first vector.
3. Resultant: Join starting point to final head.
4. Final Result: Triangle law gives resultant vector.

13. State the parallelogram law of vector addition.

Parallelogram law states that two vectors drawn from one point form adjacent sides of a parallelogram. The diagonal gives the resultant.

1. Two Vectors: Draw from common point.
2. Parallelogram: Complete the figure.
3. Diagonal: Represents resultant.
4. Final Result: The diagonal gives vector sum.

14. What is resultant of two perpendicular vectors A and B?

The resultant magnitude is √(A² + B²). Its direction is tan θ = B/A from vector A.

1. Given Vectors: A and B are perpendicular.
2. Magnitude Formula: R = √(A² + B²).
3. Direction Formula: tan θ = B/A.
4. Final Result: Perpendicular vectors add by Pythagoras theorem.

15. Find resultant of 3 m east and 4 m north displacement.

The resultant displacement is 5 m at tan^-1(4/3) north of east. Use perpendicular vector addition.

1. East Component: 3 m.
2. North Component: 4 m.
3. Magnitude:
   R = √(3² + 4²)
   R = 5 m
4. Direction: tan θ = 4/3.
5. Final Result: Resultant = 5 m north of east.

16. When is magnitude of vector sum maximum?

Magnitude of vector sum is maximum when vectors act in the same direction. The maximum value is A + B.

1. Angle: θ = 0°.
2. Formula: R = √(A² + B² + 2AB cos θ).
3. At θ = 0°: R = A + B.
4. Final Result: Maximum resultant = A + B.

17. When is magnitude of vector sum minimum?

Magnitude of vector sum is minimum when vectors act in opposite directions. The minimum value is |A − B|.

1. Angle: θ = 180°.
2. Formula: R = √(A² + B² + 2AB cos θ).
3. At θ = 180°: R = |A − B|.
4. Final Result: Minimum resultant = |A − B|.

Resolution of Vectors Class 11 Questions

Resolution means splitting a vector into components along chosen directions.
Rectangular components simplify motion in two perpendicular directions.
These resolution of vectors class 11 questions support projectile and circular motion numericals.

18. What is resolution of a vector?

Resolution of a vector means splitting it into components along chosen axes. The vector equals the sum of its components.

1. Original Vector: A.
2. x-component: Ax.
3. y-component: Ay.
4. Final Result: A = Ax i + Ay j.

19. What are rectangular components of a vector A at angle θ with x-axis?

The rectangular components are Ax = A cos θ and Ay = A sin θ. They lie along x and y axes.

1. Vector Magnitude: A.
2. Angle with x-axis: θ.
3. Horizontal Component: Ax = A cos θ.
4. Vertical Component: Ay = A sin θ.
5. Final Result: A = A cos θ i + A sin θ j.

20. How do you find magnitude from components?

Magnitude is found using A = √(Ax² + Ay²). This follows from Pythagoras theorem.

1. x-component: Ax.
2. y-component: Ay.
3. Formula Used: A = √(Ax² + Ay²).
4. Final Result: Magnitude comes from perpendicular components.

21. How do you find direction from components?

Direction is found using tan θ = Ay/Ax. The quadrant decides the final angle.

1. Horizontal Component: Ax.
2. Vertical Component: Ay.
3. Formula Used: tan θ = Ay/Ax.
4. Final Result: Direction depends on component signs.

22. Resolve 10 m/s velocity at 30° with horizontal.

The components are 8.66 m/s and 5 m/s. Use cos 30° and sin 30°.

1. Given Data:
   v = 10 m/s
   θ = 30°
2. Horizontal Component:
   vx = 10 cos 30° = 8.66 m/s
3. Vertical Component:
   vy = 10 sin 30° = 5 m/s
4. Final Result: vx = 8.66 m/s and vy = 5 m/s.

Motion in a Plane Class 11 Questions

Motion in a plane uses position, velocity and acceleration vectors.
The x and y components can be treated independently when axes are perpendicular.
These motion in a plane class 11 questions cover vector kinematics and components.

23. What is position vector in a plane?

Position vector gives the location of a particle from the origin. In a plane, it is written as r = xi + yj.

1. x-coordinate: x.
2. y-coordinate: y.
3. Position Vector: r = xi + yj.
4. Final Result: Position vector locates a particle in a plane.

24. What is displacement vector in a plane?

Displacement vector is the change in position vector. It is Δr = r2 − r1.

1. Initial Position: r1.
2. Final Position: r2.
3. Formula Used: Δr = r2 − r1.
4. Final Result: Displacement is change in position vector.

25. What is average velocity in a plane?

Average velocity is displacement divided by time interval. It is a vector quantity.

1. Displacement: Δr.
2. Time Interval: Δt.
3. Formula Used: vavg = Δr/Δt.
4. Final Result: Average velocity points along displacement.

26. What is instantaneous velocity in a plane?

Instantaneous velocity is the limiting value of average velocity as time interval tends to zero. It is tangent to the path.

1. Average Velocity: Δr/Δt.
2. Limit: Δt → 0.
3. Direction: Tangent to path.
4. Final Result: Instantaneous velocity is tangent to trajectory.

27. What is acceleration in a plane?

Acceleration is the rate of change of velocity vector with time. It may change speed, direction or both.

1. Velocity Change: Δv.
2. Time Interval: Δt.
3. Formula Used: a = Δv/Δt.
4. Final Result: Acceleration measures change in velocity vector.

Motion in a Plane with Constant Acceleration Questions

Constant acceleration in a plane can be solved by components.
Each component follows the same equations used for straight-line motion.
These questions connect Class 11 kinematics with two-dimensional motion.

28. What are equations of motion in a plane for constant acceleration?

The equations are vector forms of straight-line kinematics. They apply when acceleration remains constant.

1. Velocity Equation: v = u + at.
2. Position Equation: r = r0 + ut + 1/2 at².
3. Velocity-Displacement Equation: v² relation applies component-wise.
4. Final Result: Constant-acceleration equations work in vector form.

29. How are x and y motions treated in two-dimensional kinematics?

The x and y motions are treated independently. Time remains common for both directions.

1. Horizontal Equation: Use x-components.
2. Vertical Equation: Use y-components.
3. Common Quantity: Same time t.
4. Final Result: Component motions stay independent.

30. A particle has ux = 3 m/s, uy = 4 m/s, ax = 2 m/s² and ay = 0. Find velocity after 2 s.

The velocity after 2 s is 7i + 4j m/s. Use component equations.

1. Given Data:
   ux = 3 m/s
   uy = 4 m/s
   ax = 2 m/s²
   ay = 0
   t = 2 s
2. x-component:
   vx = ux + axt = 3 + 2 × 2 = 7 m/s
3. y-component:
   vy = uy + ayt = 4 + 0 = 4 m/s
4. Final Result: v = 7i + 4j m/s.

31. For the same particle, find displacement after 2 s.

The displacement is 10i + 8j m. Use s = ut + 1/2at² in components.

1. x-displacement:
   sx = 3 × 2 + 1/2 × 2 × 2²
   sx = 10 m
2. y-displacement:
   sy = 4 × 2 + 0
   sy = 8 m
3. Vector Form: s = 10i + 8j.
4. Final Result: Displacement = 10i + 8j m.

Projectile Motion Class 11 Questions

Projectile motion is two-dimensional motion under gravity alone.
Horizontal velocity stays constant, while vertical velocity changes due to gravity.
These projectile motion class 11 questions cover trajectory, time, height and range.

32. What is projectile motion?

Projectile motion is motion of an object thrown into air under gravity. Air resistance is neglected in basic Class 11 problems.

1. Initial Velocity: Has horizontal and vertical components.
2. Acceleration: g acts downward.
3. Path: Parabolic.
4. Final Result: Projectile motion is two-dimensional motion under gravity.

33. Why is projectile path parabolic?

Projectile path is parabolic because horizontal motion is uniform and vertical motion is accelerated. Eliminating time gives a quadratic equation.

1. Horizontal Motion: x = u cos θ × t.
2. Vertical Motion: y = u sin θ × t − 1/2 gt².
3. Eliminate t: y becomes a quadratic function of x.
4. Final Result: Projectile trajectory is a parabola.

34. What are horizontal and vertical components of projectile velocity?

The components are ux = u cos θ and uy = u sin θ. Gravity affects only the vertical component.

1. Initial Speed: u.
2. Projection Angle: θ.
3. Horizontal Component: u cos θ.
4. Vertical Component: u sin θ.
5. Final Result: Projectile velocity is resolved into u cos θ and u sin θ.

35. What is time of flight of a projectile?

The time of flight is T = 2u sin θ/g for projection and landing at the same level. It depends on vertical motion.

1. Initial Vertical Speed: u sin θ.
2. At Highest Point: Vertical speed becomes zero.
3. Total Time: Twice time to highest point.
4. Final Result: T = 2u sin θ/g.

36. What is maximum height of a projectile?

The maximum height is H = u² sin²θ/(2g). It occurs when vertical velocity becomes zero.

1. Initial Vertical Speed: u sin θ.
2. At Top: vy = 0.
3. Formula Used: H = u² sin²θ/(2g).
4. Final Result: Maximum height depends on vertical speed squared.

37. What is horizontal range of a projectile?

The horizontal range is R = u² sin 2θ/g for same-level landing. It is maximum at 45°.

1. Horizontal Speed: u cos θ.
2. Time of Flight: 2u sin θ/g.
3. Range: R = u cos θ × T.
4. Final Result: R = u² sin 2θ/g.

38. At what angle is projectile range maximum?

Projectile range is maximum at 45° for fixed speed and same-level landing. This follows from sin 2θ.

1. Range Formula: R = u² sin 2θ/g.
2. Maximum Condition: sin 2θ = 1.
3. Angle: 2θ = 90°, so θ = 45°.
4. Final Result: Maximum range occurs at 45°.

Projectile Motion Numericals Class 11

Projectile numericals need separate horizontal and vertical equations.
Use g = 9.8 m/s² unless the question gives another value.
These projectile motion numericals class 11 questions support CBSE 2026 practice.

39. A projectile is thrown at 20 m/s at 30°. Find time of flight using g = 10 m/s².

The time of flight is 2 s. Use T = 2u sin θ/g.

1. Given Data:
   u = 20 m/s
   θ = 30°
   g = 10 m/s²
2. Formula Used: T = 2u sin θ/g.
3. Calculation:
   T = 2 × 20 × 1/2 / 10
   T = 2 s
4. Final Result: Time of flight = 2 s.

40. For the same projectile, find maximum height.

The maximum height is 5 m. Use H = u² sin²θ/(2g).

1. Given Data:
   u = 20 m/s
   θ = 30°
   g = 10 m/s²
2. Formula Used: H = u² sin²θ/(2g).
3. Calculation:
   H = 20² × (1/2)² / 20
   H = 5 m
4. Final Result: Maximum height = 5 m.

41. For the same projectile, find horizontal range.

The horizontal range is 20√3 m. Use R = u² sin 2θ/g.

1. Given Data:
   u = 20 m/s
   θ = 30°
   g = 10 m/s²
2. Formula Used: R = u² sin 2θ/g.
3. Calculation:
   R = 400 × sin 60° / 10
   R = 40 × √3/2
   R = 20√3 m
4. Final Result: Range = 20√3 m.

42. A ball is projected horizontally at 15 m/s from height 20 m. Find time to hit ground using g = 10 m/s².

The time to hit the ground is 2 s. Vertical motion decides the time.

1. Given Data:
   h = 20 m
   uy = 0
   g = 10 m/s²
2. Formula Used: h = 1/2gt².
3. Calculation:
   20 = 1/2 × 10 × t²
   t² = 4
   t = 2 s
4. Final Result: Time = 2 s.

43. For the same ball, find horizontal distance travelled.

The horizontal distance is 30 m. Horizontal velocity remains constant.

1. Given Data:
   ux = 15 m/s
   t = 2 s
2. Formula Used: x = uxt.
3. Calculation:
   x = 15 × 2
   x = 30 m
4. Final Result: Horizontal distance = 30 m.

Relative Velocity in Two Dimensions Class 11 Questions

Relative velocity compares motion of one object with respect to another.
In two dimensions, velocities must be subtracted as vectors.
These relative velocity in two dimensions class 11 questions cover rain, river and moving observers.

44. What is relative velocity in two dimensions?

Relative velocity of A with respect to B is the velocity of A seen from B. It is given by vAB = vA − vB.

1. Velocity of A: vA.
2. Velocity of B: vB.
3. Formula Used: vAB = vA − vB.
4. Final Result: Relative velocity is vector difference.

45. What is velocity of B relative to A?

Velocity of B relative to A is vBA = vB − vA. It is the negative of vAB.

1. Formula 1: vAB = vA − vB.
2. Formula 2: vBA = vB − vA.
3. Relation: vBA = −vAB.
4. Final Result: Relative velocities are equal and opposite.

46. Two cars move east at 20 m/s and 12 m/s. Find relative velocity of first car with respect to second.

The relative velocity is 8 m/s east. Subtract velocities along the same direction.

1. First Car: vA = 20 m/s east.
2. Second Car: vB = 12 m/s east.
3. Formula Used: vAB = vA − vB.
4. Calculation: vAB = 20 − 12 = 8 m/s east.
5. Final Result: Relative velocity = 8 m/s east.

47. A river flows east at 3 m/s, and a swimmer swims north at 4 m/s relative to water. Find velocity relative to ground.

The ground velocity is 5 m/s at tan^-1(3/4) east of north. Add perpendicular velocities.

1. Swimmer Relative to Water: 4 m/s north.
2. River Velocity: 3 m/s east.
3. Magnitude:
   v = √(4² + 3²)
   v = 5 m/s
4. Final Result: Velocity = 5 m/s east of north.

48. Rain falls vertically at 12 m/s, and a man runs east at 5 m/s. Find rain velocity relative to man.

The rain appears to fall with speed 13 m/s slanting westward. Use vector subtraction.

1. Rain Velocity: 12 m/s downward.
2. Man Velocity: 5 m/s east.
3. Relative Velocity Components: 5 m/s west and 12 m/s downward.
4. Magnitude: √(5² + 12²) = 13 m/s.
5. Final Result: Rain appears at 13 m/s downward-west.

Uniform Circular Motion Class 11 Questions

Uniform circular motion has constant speed but changing velocity direction.
The acceleration always points towards the centre of the circle.
These uniform circular motion class 11 questions cover angular speed, period and centripetal acceleration.

49. What is uniform circular motion?

Uniform circular motion is motion in a circle with constant speed. Velocity changes because direction changes continuously.

1. Path: Circle.
2. Speed: Constant.
3. Velocity: Changes direction.
4. Final Result: Uniform circular motion is accelerated motion.

50. Why is uniform circular motion accelerated?

Uniform circular motion is accelerated because velocity direction changes at every point. Acceleration points towards the centre.

1. Speed: Constant.
2. Velocity Direction: Changes continuously.
3. Acceleration Direction: Towards centre.
4. Final Result: Direction change causes acceleration.

51. What is centripetal acceleration?

Centripetal acceleration is acceleration directed towards the centre of circular motion. Its magnitude is ac = v²/r.

1. Speed: v.
2. Radius: r.
3. Formula Used: ac = v²/r.
4. Final Result: Centripetal acceleration points inward.

52. What is angular velocity in circular motion?

Angular velocity is rate of change of angular displacement. Its formula is ω = 2π/T.

1. Angular Displacement: θ.
2. Time: t.
3. Formula: ω = θ/t.
4. For One Revolution: ω = 2π/T.
5. Final Result: Angular velocity measures rotation rate.

53. What is relation between linear speed and angular speed?

The relation is v = rω. Linear speed increases with radius for the same angular speed.

1. Radius: r.
2. Angular Speed: ω.
3. Formula Used: v = rω.
4. Final Result: Linear speed equals radius times angular speed.

54. Find centripetal acceleration for speed 10 m/s and radius 5 m.

The centripetal acceleration is 20 m/s². Use ac = v²/r.

1. Given Data:
   v = 10 m/s
   r = 5 m
2. Formula Used: ac = v²/r.
3. Calculation:
   ac = 10²/5
   ac = 20 m/s²
4. Final Result: ac = 20 m/s².

55. A body completes 10 revolutions in 5 s on a circle of radius 2 m. Find angular speed.

The angular speed is 4π rad/s. Use ω = 2πf.

1. Revolutions: 10.
2. Time: 5 s.
3. Frequency: f = 10/5 = 2 Hz.
4. Angular Speed: ω = 2π × 2 = 4π rad/s.
5. Final Result: ω = 4π rad/s.

56. For the same body, find linear speed.

The linear speed is 8π m/s. Use v = rω.

1. Given Data:
   r = 2 m
   ω = 4π rad/s
2. Formula Used: v = rω.
3. Calculation:
   v = 2 × 4π
   v = 8π m/s
4. Final Result: v = 8π m/s.

NCERT Class 11 Physics Chapter 4 Questions

NCERT questions often combine vectors, projectile motion and circular motion.
Students should draw axes and resolve every vector before substituting values.
These NCERT Class 11 Physics Chapter 4 questions follow the 2026 exercise style.

57. Can average velocity be zero when average speed is not zero?

Yes, average velocity can be zero when average speed is not zero. This happens in a complete round trip.

1. Round Trip: Initial and final positions match.
2. Displacement: Zero.
3. Distance: Non-zero.
4. Final Result: Average velocity can be zero while average speed is positive.

58. Can speed remain constant while acceleration is non-zero?

Yes, speed can remain constant while acceleration is non-zero. Uniform circular motion is the standard example.

1. Speed: Constant.
2. Velocity Direction: Changes continuously.
3. Acceleration: Directed towards centre.
4. Final Result: Constant speed can still mean accelerated motion.

59. Why is acceleration in projectile motion constant?

Acceleration in projectile motion is constant because only gravity acts. It acts vertically downward with magnitude g.

1. Horizontal Acceleration: Zero.
2. Vertical Acceleration: −g.
3. Assumption: Air resistance neglected.
4. Final Result: Projectile acceleration is constant downward.

60. Why is horizontal velocity constant in projectile motion?

Horizontal velocity is constant because horizontal acceleration is zero. Gravity acts only vertically.

1. Horizontal Force: Zero in basic projectile motion.
2. Horizontal Acceleration: ax = 0.
3. Result: vx remains constant.
4. Final Result: Horizontal component of velocity stays constant.

61. Why are two projection angles possible for the same range?

Two projection angles are possible because sin 2θ has the same value for complementary angles. The angles are θ and 90° − θ.

1. Range Formula: R = u² sin 2θ/g.
2. Complementary Angles: θ and 90° − θ.
3. Same Value: sin 2θ remains same.
4. Final Result: Complementary projection angles give equal range.

62. Why does a projectile have minimum speed at highest point?

A projectile has minimum speed at the highest point because vertical velocity becomes zero. Horizontal velocity remains unchanged.

1. At Highest Point: vy = 0.
2. Horizontal Velocity: vx = u cos θ.
3. Total Speed: v = vx.
4. Final Result: Minimum speed equals horizontal velocity.

Class 11 Physics Chapter 4 Numericals

Numericals in this chapter become easier after component separation.
Projectile and circular motion formulas must use SI units throughout.
These Class 11 Physics Chapter 4 numericals support exam-style practice.

63. A vector has components 6 and 8. Find its magnitude.

The magnitude is 10. Use A = √(Ax² + Ay²).

1. Given Data:
   Ax = 6
   Ay = 8
2. Formula Used: A = √(Ax² + Ay²).
3. Calculation:
   A = √(6² + 8²)
   A = √100
   A = 10
4. Final Result: Magnitude = 10.

64. A vector of magnitude 20 makes 60° with x-axis. Find its components.

The components are 10 and 10√3. Use Ax = A cos θ and Ay = A sin θ.

1. Given Data:
   A = 20
   θ = 60°
2. x-component:
   Ax = 20 cos 60° = 10
3. y-component:
   Ay = 20 sin 60° = 10√3
4. Final Result: Components are 10 and 10√3.

65. A projectile has time of flight 4 s and range 80 m. Find horizontal velocity.

The horizontal velocity is 20 m/s. Use range = horizontal velocity × time.

1. Given Data:
   R = 80 m
   T = 4 s
2. Formula Used: R = uxT.
3. Calculation:
   ux = R/T
   ux = 80/4
   ux = 20 m/s
4. Final Result: Horizontal velocity = 20 m/s.

66. A body moves in a circle of radius 0.5 m with angular speed 6 rad/s. Find centripetal acceleration.

The centripetal acceleration is 18 m/s². Use ac = ω²r.

1. Given Data:
   r = 0.5 m
   ω = 6 rad/s
2. Formula Used: ac = ω²r.
3. Calculation:
   ac = 6² × 0.5
   ac = 18 m/s²
4. Final Result: ac = 18 m/s².

CBSE Class 11 Physics Chapter-Wise Important Questions