Vector Addition — NEET Physics MCQs with Solutions
Free NEET Physics Vector Addition MCQs with step-by-step solutions (16 questions). Part of Vectors. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Vector Addition · medium · theory
Two equal forces of $10$ N each act at a point at right angles to each other. Their resultant is
A. $0$
B. $10\sqrt{2}$ N at $45^\circ$ to each force ✓ Correct
C. $10\sqrt{2}$ N along one of the forces
D. $20$ N at $45^\circ$ to each force
Solution: $R = \sqrt{10^2 + 10^2} = 10\sqrt{2}$ N, and for equal forces the resultant bisects the $90^\circ$ angle, i.e. $45^\circ$ from each.
Q2 — Vector Addition · medium · theory
Five equal forces act on a particle at the centre of a regular pentagon, each directed towards a vertex. The resultant of the five forces is
A. $\sqrt{5}$ times one force
B. Zero ✓ Correct
C. $5$ times one force
D. Equal to one force
Solution: By symmetry the five equal vectors separated by $72^\circ$ form a closed polygon — their components cancel in every direction, so the resultant is zero.
Q3 — Vector Addition · medium · theory
The vectors $3\hat{i} - 4\hat{j}$ and $-3\hat{i} + 4\hat{j}$ are
A. Equal in magnitude and opposite in direction ✓ Correct
B. Equal vectors
C. At $45^\circ$ to each other
D. Perpendicular to each other
Solution: The second is $-1$ times the first: same magnitude ($5$), exactly opposite direction. Their sum is the zero vector.
Q4 — Vector Addition · medium · theory
The resultant of two vectors of equal magnitude $a$ can have a magnitude SMALLER than $a$ when the angle between them is
A. Greater than $120^\circ$ ✓ Correct
B. Exactly $90^\circ$
C. Any angle
D. Less than $60^\circ$
Solution: $R = 2a\cos(\theta/2)$. $R < a$ requires $\cos(\theta/2) < \tfrac{1}{2}$, i.e. $\theta/2 > 60^\circ \Rightarrow \theta > 120^\circ$.
Q5 — Vector Addition · medium · theory
If $\vec A = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec B = \hat{i} - 2\hat{j} + 2\hat{k}$, then $|\vec A + \vec B|$ is
A. $11$
B. $\sqrt{14}$
C. $\sqrt{11}$ ✓ Correct
D. $3\sqrt{2}$
Solution: $\vec A + \vec B = 3\hat{i} + \hat{j} + \hat{k}$, so the magnitude is $\sqrt{9+1+1} = \sqrt{11}$.
Q6 — Vector Addition · medium · numerical
Two forces of 3 N and 4 N act at right angles. Their resultant is:
A. 1 N
B. 5 N ✓ Correct
C. 12 N
D. 7 N
Solution: R = √(3²+4²) = 5 N.
Q7 — Vector Addition · hard · numerical
Two equal vectors of magnitude 10 units act at 120° to each other. Their resultant is:
A. 10√3 units
B. 10 units ✓ Correct
C. 0
D. 20 units
Solution: R = √(A²+A²+2A²cos120°) = √(100+100−100) = 10 units.
Q8 — Vector Addition · medium · numerical
Two vectors of magnitudes 8 and 6 give a maximum resultant of:
A. 2
B. 14 ✓ Correct
C. 10
D. 48
Solution: Maximum when parallel: 8 + 6 = 14.
Q9 — Vector Addition · medium · numerical
Two forces of 5 N and 12 N act at right angles. The resultant magnitude is:
A. 7 N
B. 60 N
C. 13 N ✓ Correct
D. 17 N
Solution: R = √(5²+12²) = 13 N.
Q10 — Vector Addition · hard · numerical
Two vectors of 10 units each act at 60°. Their resultant is:
A. 10 units
B. 0
C. 10√3 units ✓ Correct
D. 20 units
Solution: R = √(100+100+2(100)(0.5)) = √300 = 10√3 units.
Q11 — Vector Addition · medium · numerical
The minimum resultant of two vectors of magnitude 7 and 4 is:
A. 28
B. 3 ✓ Correct
C. 0
D. 11
Solution: Minimum when antiparallel: 7 − 4 = 3.
Q12 — Vector Addition · hard · numerical
The resultant of two vectors P and Q is R. When Q is doubled, the new resultant is perpendicular to P. Then R equals:
A. P + Q
B. Q ✓ Correct
C. 2Q
D. P
Solution: Perpendicularity gives P + 2Q cosθ = 0 ⇒ cosθ = −P/2Q. Then R² = P² + Q² + 2PQcosθ = P² + Q² − P² = Q² ⇒ R = Q — a classic identity.
Q13 — Vector Addition · hard · numerical
Two forces P + Q and P − Q act at right angles to each other. The magnitude of their resultant is:
A. 2P
B. √(P² + Q²)
C. P + Q
D. √(2P² + 2Q²) ✓ Correct
Solution: R² = (P+Q)² + (P−Q)² = 2P² + 2Q².
Q14 — Vector Addition · hard · numerical
The resultant of two equal vectors of magnitude A is also of magnitude A. The angle between the two vectors is:
A. 120° ✓ Correct
B. 45°
C. 90°
D. 60°
Solution: A² = A² + A² + 2A²cosθ ⇒ cosθ = −1/2 ⇒ θ = 120°.
Q15 — Vector Addition · hard · numerical
Vectors A and B satisfy |A + B| = |A − B|. The angle between A and B is:
A. 0°
B. 45°
C. 180°
D. 90° ✓ Correct
Solution: Squaring both: 2A·B = −2A·B ⇒ A·B = 0 ⇒ perpendicular.
Q16 — Vector Addition · hard · numerical
The maximum and minimum resultants of two forces are in the ratio 3 : 1. The forces are in the ratio:
A. 9 : 1
B. 3 : 1
C. 2 : 1 ✓ Correct
D. 4 : 3
Solution: (P+Q)/(P−Q) = 3 ⇒ P + Q = 3P − 3Q ⇒ 4Q = 2P ⇒ P : Q = 2 : 1.