Class 11 Physics, Chapter 3 (NCERT). 15 questions by concept, 5 with pictures. Open “Show answer” for the answer and why.
How to teach this chapter · Common misconceptions & worked visual for this chapter
Two free displacement vectors both have magnitude 5 m. A points south and B points north. Are A and B equal?
(b) No; their directions differ.
Vector equality requires equal magnitude and direction. Translation of a free vector is allowed, but changing direction is not.
The diagram shows vectors A and B with labelled endpoints. Which vector expression represents the arrow from Q to P?
(b) A − B
Write QP = OP − OQ. Therefore QP = A − B; joining common-tail tips gives a difference.
For the two given vectors in the diagram, which construction produces A + B?
(c) Translate B so its tail is at P; join O to its new tip.
In head-to-tail addition, the second vector is translated without changing direction. The sum runs from the first tail to the final head.
A skater travels along a semicircular edge of radius 1000 m from one end of a diameter to the other. What is the displacement magnitude?
(c) 2000 m
Displacement joins the start and finish by a straight vector. The endpoints are separated by a diameter, so its magnitude is twice the radius.
Which classification of electric current and impulse, respectively, is correct?
(a) Scalar; vector.
A directional context does not by itself define a vector. The quantity electric current is scalar, whereas impulse has vector addition and direction.
A displacement has magnitude 20 m and makes 30 degrees with the positive x-axis in the first quadrant. What is its x-component?
(b) 10√3 m
The x-component is adjacent to the stated angle: Ax = A cos(theta). Substituting gives 10√3 m.
A projectile has initial upward velocity 14 m/s and returns to its launch height. For g = 9.8 m/s² and negligible drag, what is its total flight time?
(c) 20/7 s
The apex time is uy/g. The equal-height descent takes the same time, so total time is 2uy/g = 20/7 s.
A particle starts at x = 0 with vx = 5 m/s and constant ax = 3 m/s². What is x after 6 s?
(a) 84 m
Use x = ux t + ax t²/2 for constant acceleration. The acceleration-generated speed increases during the interval, requiring the factor one half.
A particle has vx = 3 + (2)t m/s and vy = 10 m/s, with t in seconds. What is its acceleration vector at t = 3 s?
(c) (2, 0) m/s²
Acceleration is the derivative of velocity. The constant velocity terms contribute zero; only the coefficient of t remains.
In metres, A = (5, 12) and B = (-2, -7), where entries are signed Cartesian components. What are the components of A + B?
(a) (3, 5) m
Add x-components and y-components separately. Keep the negative signs; opposite components partially cancel.
From height 20 m, ball A is dropped and ball B is launched horizontally at 12 m/s at the same instant. Neglect air resistance. Which reaches level ground first?
(b) They reach the ground together.
Both balls start at height 20 m with zero vertical velocity.
With air resistance neglected, their vertical motion obeys the same equation h = gt²/2.
Ball B’s horizontal speed 12 m/s does not change its vertical fall time, so they reach the level ground together.
Therefore, the requested answer is They reach the ground together.
A ball is launched at 28 m/s at 60 degrees above horizontal. Ignoring air resistance, what is its acceleration at maximum height?
(a) g vertically downward.
Only the vertical velocity is instantaneously zero at the apex. Gravity still gives acceleration g downward.
A ball is thrown from the floor at speed 40 m/s in a hall of height 25 m. Let g = 10 m/s² and neglect drag. Which condition must be checked before using 45 degrees as the best launch angle?
(b) The maximum height u² sin²(theta)/(2g) must not exceed the ceiling.
For launch speed u at angle θ, maximum height is u²sin²θ/(2g).
Before choosing 45°, check that u²/(4g) ≤ h so that the trajectory stays below the ceiling.
Use u = 40 m/s, g = 10 m/s² and h = 25 m in this check. The usual unconstrained 45° result is valid only when the height condition is met.
Therefore, the requested answer is The maximum height u² sin²(theta)/(2g) must not exceed the ceiling.
For a floor launch at speed 40 m/s below a 25 m ceiling, take g = 10 m/s² and no drag. What is the largest allowed initial upward velocity component?
(c) 10√5 m/s
Require uy²/(2g) <= 25 m. Thus uy <= √(2gH), giving 10√5 m/s; the unconstrained 45-degree choice would hit the ceiling.
An object completes one uniform circular revolution in 40 s. Which comparison of instantaneous acceleration and average acceleration over that revolution is correct?
(a) Instantaneous acceleration is nonzero; the full-turn average is zero.
In uniform circular motion, speed stays constant but velocity changes direction. Instantaneous acceleration points toward the centre.
After a complete revolution in 40 s, final velocity equals initial velocity.
Average acceleration is (v_final − v_initial)/40 = 0, even though instantaneous acceleration is nonzero.
Therefore, the requested answer is Instantaneous acceleration is nonzero; the full-turn average is zero.
Wrong
An upward-moving projectile accelerates upward.
Why it feels right. Its motion and acceleration seem to point the same way.
Right
With air resistance neglected, gravity accelerates the projectile downward throughout flight.
Worked example
Successive vertical velocities can be +20, +10, 0, −10 m/s at equal time intervals in an illustrative model with g = 10 m/s².
