CBSE Class 11 Physics Vectors Competitive Exam MCQs Set 02

Answer: (a) 0 N

Explanation:
1. Equal angles between six coplanar forces means each angle is 360°/6 = 60°.
2. Each force has an equal and opposite partner (at 180°), so they cancel in pairs.
3. Resultant = 0 N

Teacher's Note:
1. Any number of equal forces spread evenly around a point (like spokes) have zero resultant.
2. Drawn head to tail, the six forces form a closed regular hexagon.

Answer: (d) none of above

Explanation:
1. Take east = \( \hat{i} \), north = \( \hat{j} \), vertically up = \( \hat{k} \). Then \( \vec{A} \) is along \( -\hat{k} \) and \( \vec{B} \) along \( \hat{i} \).
2. \( (-\hat{k}) \times \hat{i} = -(\hat{k} \times \hat{i}) = -\hat{j} \)
3. So \( \vec{A} \times \vec{B} \) is along south, which is none of the given directions.

Teacher's Note:
1. The cross product is perpendicular to both vectors, so it can never be along east here.
2. Use the cyclic order î → ĵ → k̂ → î to read off products like k̂ × î = ĵ.

Answer: (c) A is true, but R is false.

Explanation:
1. Statement-1 is the definition of a vector, so it is true.
2. Statement-2 is false: e.g. two equal vectors at 180° give zero resultant, which is less than either.
3. Hence Statement-1 is true and Statement-2 is false.

Teacher's Note:
1. The resultant of two vectors can be anything from |A − B| to A + B, so it can be smaller than each.
2. Having magnitude and direction is not enough; the quantity must also follow the vector law of addition.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Statement-1: \( F = \sqrt{8^2 + 6^2} = 10 \) N, true.
2. Statement-2: \( |\vec{A} \times \vec{B}|^2 + |\vec{A} \cdot \vec{B}|^2 = A^2B^2(\sin^2\theta + \cos^2\theta) = 1 \), true.
3. Statement-2 does not explain Statement-1, so both are true but Statement-2 is not the explanation.

Teacher's Note:
1. Rectangular components are perpendicular, so the magnitude follows from Pythagoras.
2. For any two vectors, |A × B|² + (A · B)² = A²B².

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Zero dot products mean \( \vec{A} \perp \vec{B} \) and \( \vec{A} \perp \vec{C} \), so \( \vec{A} \) is perpendicular to the plane of \( \vec{B} \) and \( \vec{C} \) (Statement-2 true).
2. \( \vec{B} \times \vec{C} \) is also perpendicular to that plane, so \( \vec{A} \parallel \vec{B} \times \vec{C} \) (Statement-1 true).
3. Statement-2 is the reason for Statement-1.

Teacher's Note:
1. Two directions perpendicular to the same plane must be parallel (or antiparallel).
2. The cross product of two vectors is always normal to the plane containing them.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Two vectors of unequal magnitude can never cancel, since |A − B| > 0.
2. Three unequal vectors forming a closed triangle taken in order give zero resultant (Statement-2 true).
3. So the minimum number is three (Statement-1 true), and Statement-2 explains it.

Teacher's Note:
1. Two vectors cancel only if they are equal in magnitude and opposite in direction.
2. A closed triangle of vectors is the simplest closed polygon, hence three is the minimum.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Statement-2 is the standard formula from \( \vec{A} \cdot \vec{B} = AB\cos\theta \): true.
2. \( (\hat{i} + \hat{j}) \cdot \hat{k} = 0 \) ⇒ cos θ = 0 ⇒ θ = π/2: Statement-1 true.
3. Statement-1 follows directly from Statement-2.

Teacher's Note:
1. A zero dot product between non-zero vectors always means they are perpendicular.
2. Any vector in the xy-plane is perpendicular to k̂.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Distance has only magnitude, so it is a scalar: Statement-1 true.
2. Distance is defined as the length of the actual path covered: Statement-2 true.
3. Statement-2 only defines distance; it does not by itself explain why it is a scalar.

Teacher's Note:
1. Distance (path length) is a scalar while displacement (shortest straight line from start to end) is a vector.
2. A quantity is scalar because it needs no direction for its full description.

Answer: (c) 2

Explanation:
1. R² = P² + Q² + 2PQ cos 60° ⇒ 7Q² = P² + Q² + PQ
2. ⇒ P² + PQ − 6Q² = 0 ⇒ (P + 3Q)(P − 2Q) = 0
3. Since P > 0, P = 2Q, i.e. P/Q = 2

Teacher's Note:
1. Dividing the equation by Q² turns it into a quadratic in P/Q, which is quick to solve.
2. Reject the negative root because magnitudes are positive.

Answer: (a) 2(F₁² + F₂²)

Explanation:
1. P² = F₁² + F₂² + 2F₁F₂ cos θ
2. Q² = F₁² + F₂² − 2F₁F₂ cos θ (reversing F₂ changes θ to 180° − θ)
3. Adding: P² + Q² = 2(F₁² + F₂²)

Teacher's Note:
1. P and Q are the two diagonals of the parallelogram formed by F₁ and F₂.
2. Sum of squares of the diagonals = sum of squares of all four sides of a parallelogram.

Answer: (b) 0 m

Explanation:
1. Take east = +x, north = +y. Displacements: (0, 3), (4, 0) and 5 m at 37° south of west = (−5 cos 37°, −5 sin 37°) = (−4, −3).
2. Sum: x = 0 + 4 − 4 = 0, y = 3 + 0 − 3 = 0
3. Net displacement = 0 m

Teacher's Note:
1. The first two moves give 5 m at 37° north of east; the third move of 5 m is exactly opposite to it.
2. Use cos 37° = 4/5 and sin 37° = 3/5 for such problems.

Answer: (d) \( \sqrt{3} \) : 2 : 1

Explanation:
1. The angle between R and P = 360° − 150° − 120° = 90°.
2. By Lami's theorem, each force is proportional to the sine of the angle between the other two: P : Q : R = sin 120° : sin 90° : sin 150°
3. = \( \frac{\sqrt{3}}{2} : 1 : \frac{1}{2} = \sqrt{3} : 2 : 1 \)

Teacher's Note:
1. For three forces in equilibrium, P/sin α = Q/sin β = R/sin γ, where each angle is the one between the other two forces.
2. The three angles between the forces must add up to 360°.

Answer: (a) 9 km/hr

Explanation:
1. North-west is 45° north of west; the shoreline is 15° south of west.
2. Angle between the velocity and the shoreline = 45° + 15° = 60°
3. Component along the shoreline = 18 cos 60° = 9 km/hr

Teacher's Note:
1. The component of a vector along a line is (magnitude) × cos(angle with that line).
2. Measure both directions from the same reference (west here) before finding the angle between them.

Answer: (d) \( 6\hat{i} + 8\hat{j} \)

Explanation:
1. Displacement = \( (4 - 1)\hat{i} + (2 + 2)\hat{j} + (3 - 3)\hat{k} = 3\hat{i} + 4\hat{j} \), magnitude 5
2. Unit vector along the motion = \( 0.6\hat{i} + 0.8\hat{j} \)
3. Velocity = 10 × \( (0.6\hat{i} + 0.8\hat{j}) = 6\hat{i} + 8\hat{j} \) m/s

Teacher's Note:
1. Velocity vector = speed × unit vector in the direction of motion.
2. Option (c) is only the unit vector; its magnitude is 1, not 10.

Answer: (d) 120°

Explanation:
1. Let the forces be F and 2F at angle θ. Resultant ⊥ F means its component along F is zero.
2. F + 2F cos θ = 0 ⇒ cos θ = −½
3. θ = 120°

Teacher's Note:
1. 'Resultant perpendicular to a force' means the total component along that force is zero.
2. Equivalently, tan α = (2F sin θ)/(F + 2F cos θ) becomes infinite, so the denominator is zero.

Answer: (b) 1

Explanation:
1. \( |\hat{a} - \hat{b}|^2 = 1 + 1 - 2 \times 1 \times 1 \times \cos 60^\circ = 2 - 1 = 1 \)
2. \( |\hat{a} - \hat{b}| = 1 \)

Teacher's Note:
1. For unit vectors, |â − b̂| = 2 sin(θ/2), so at 60° it equals 2 sin 30° = 1.
2. The two unit vectors and their difference form an equilateral triangle here.

Answer: (a) 10 dynes

Explanation:
1. One force points away from the vertex along the line; the other points towards the vertex along the line inclined at 60°.
2. Placing them tail to tail, the angle between the forces is 180° − 60° = 120°.
3. R = 2F cos(120°/2) = 2 × 10 × ½ = 10 dynes

Teacher's Note:
1. Always check the arrow directions: the angle between vectors is measured with both tails at the same point.
2. Two equal forces at 120° have a resultant equal to either force.

Answer: (c) π/2

Explanation:
1. A² + B² + 2AB cos θ = A² + B² − 2AB cos θ
2. ⇒ 4AB cos θ = 0 ⇒ cos θ = 0
3. θ = π/2

Teacher's Note:
1. The sum and difference are the diagonals of the parallelogram; equal diagonals mean a rectangle.
2. Equivalently, |A + B|² − |A − B|² = 4A · B, so A · B = 0.

Answer: (a) 2r sin (θ/2)

Explanation:
1. The initial and final position vectors from the centre both have magnitude r with angle θ between them.
2. Displacement = \( |\vec{r}_2 - \vec{r}_1| = \sqrt{r^2 + r^2 - 2r^2\cos\theta} = \sqrt{2r^2(1 - \cos\theta)} \)
3. \( = \sqrt{4r^2\sin^2(\theta/2)} = 2r\sin(\theta/2) \)

Teacher's Note:
1. Linear displacement is the chord joining the two positions, not the arc length rθ.
2. Use 1 − cos θ = 2 sin²(θ/2) to simplify.

Answer: (a) P + Q

Explanation:
1. A direction making 120° with the x-axis makes 120° − 90° = 30° with the y-axis.
2. So \( \vec{Q} \) (at 30° with the y-axis, on the same side) is along the same direction as \( \vec{P} \).
3. Parallel vectors add directly: resultant = P + Q

Teacher's Note:
1. Convert both directions to angles from the same axis before comparing them.
2. Q could also be read as 30° on the other side of the y-axis, but then the resultant √(P² + Q² + PQ) is not among the options.

Answer: (b) 5 miles

Explanation:
1. Net east = 1 + 2 = 3 miles; net north = 9 − 5 = 4 miles
2. Distance from start = \( \sqrt{3^2 + 4^2} = 5 \) miles

Teacher's Note:
1. Add all east–west moves and all north–south moves separately, then combine with Pythagoras.
2. 'How far from the starting point' asks for the magnitude of displacement, not the total path (17 miles).

Answer: (b) tan⁻¹(2/3)

Explanation:
1. Angle with the y-axis: tan β = (component perpendicular to y-axis)/(component along y-axis)
2. tan β = A_x/A_y = 2/3
3. β = tan⁻¹(2/3)

Teacher's Note:
1. tan⁻¹(3/2) is the angle with the x-axis; the angle with the y-axis is its complement.
2. cos⁻¹(3/2) is impossible because cosine cannot exceed 1.

Answer: (b) −7

Explanation:
1. Perpendicular vectors have zero dot product: 3 × 2 + 2 × x + 8 × 1 = 0
2. 6 + 2x + 8 = 0 ⇒ 2x = −14
3. x = −7

Teacher's Note:
1. 'At right angles' is the cue to set the dot product equal to zero.
2. Multiply matching components and add; do not mix î with ĵ terms.

Answer: (a) a₁ = 0.6, a₂ = 0.8

Explanation:
1. Perpendicular: 4a₁ − 3a₂ = 0 ⇒ a₁ : a₂ = 3 : 4
2. Unit vector: a₁² + a₂² = 1 ⇒ a₁ = 3/5 = 0.6, a₂ = 4/5 = 0.8
3. Check: 4(0.6) − 3(0.8) = 0 ✓

Teacher's Note:
1. Both conditions are needed: zero dot product and magnitude 1.
2. Option (b) is perpendicular but has magnitude 5, so it is not a unit vector.

Answer: (b) \( x\vec{a} \) is a vector collinear to \( \vec{a} \)

Explanation:
1. For x > 0, \( x\vec{a} \) points along \( \vec{a} \); for x < 0 it points opposite to \( \vec{a} \).
2. In both cases it lies along the same line as \( \vec{a} \).
3. So \( x\vec{a} \) is always collinear to \( \vec{a} \); option (a) fails for negative x.

Teacher's Note:
1. Multiplying by a scalar can stretch, shrink or reverse a vector, but never turns it off its line.
2. Collinear includes both 'same direction' and 'opposite direction'.