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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 for Telangana State Board CLASS 11 INTERMEDIATE 1 YEAR PHYSICS. 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.. AI-generated questions from basic to board-exam level, with instant results and explanations.

10
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20m
Time Limit
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  • Questions you've seen before won't repeat until the pool resets
  • You have 20 minutes — exam auto-submits when time is up
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  • Results and explanations shown immediately after submission
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Unit 4: Motion in a Plane — Important Questions & Answers (FAQ)

Frequently asked questions from Telangana State Board CLASS 11 INTERMEDIATE 1 YEAR PHYSICS — Unit 4: Motion in a Plane, with answers and explanations. These are sample questions; the exam has its own separate question set.

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.
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°.
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.
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.
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.

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