Unit 4: Motion in a Plane — Online MCQ Test
PHYSICS · CLASS 11 INTERMEDIATE 1 YEAR · Telangana State Board
Practice Unit 4: Motion in a Plane with a free chapter-wise online MCQ test.
This chapter covers: This chapter covers scalar and vector quantities vector addition and resolution relative velocity in two dimensions uniform circular motion and projectile motion parameters..
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Unit 4: Motion in a Plane — Important Questions & Answers
Which of the following is a vector quantity?
- A. Mass
- B. Speed
- C. Displacement
- D. Time
Answer: C. Displacement
Displacement has both magnitude and direction, so it is a vector quantity. Mass, speed, and time are scalars.
Displacement has both magnitude and direction, so it is a vector quantity. Mass, speed, and time are scalars.
The magnitude of the resultant of two vectors is maximum when the angle between them is:
- A. 0°
- B. 45°
- C. 90°
- D. 180°
Answer: A. 0°
The resultant is maximum when the two vectors act in the same direction, i.e., the angle between them is 0°.
The resultant is maximum when the two vectors act in the same direction, i.e., the angle between them is 0°.
A person walking 3 km east and then 4 km north has a displacement of:
- A. 5 km
- B. 7 km
- C. 1 km
- D. 12 km
Answer: A. 5 km
Displacement is the straight-line distance from start to end. Using Pythagoras, it is √(3² + 4²) = 5 km.
Displacement is the straight-line distance from start to end. Using Pythagoras, it is √(3² + 4²) = 5 km.
A projectile is thrown with speed u at an angle θ to the horizontal. Its time of flight is:
- A. u sinθ / g
- B. 2u sinθ / g
- C. u cosθ / g
- D. u / g
Answer: B. 2u sinθ / g
The total time of flight for a projectile launched and landing at the same level is T = 2u sinθ / g.
The total time of flight for a projectile launched and landing at the same level is T = 2u sinθ / g.
A body is projected with speed u at angle θ. Which of the following is NOT a correct expression for the maximum height reached?
- A. u² sin²θ / 2g
- B. H = (u sinθ)² / 2g
- C. u² sin2θ / 2g
- D. v_y² / 2g, where v_y = u sinθ
Answer: C. u² sin2θ / 2g
The correct maximum height is based on the vertical component of velocity: H = u² sin²θ / 2g. The expression with sin2θ is not correct.
The correct maximum height is based on the vertical component of velocity: H = u² sin²θ / 2g. The expression with sin2θ is not correct.