JEE Physics Vectors and Calculus MCQs Set 02

Answer: (b) \( 0^\circ \) or \( 180^\circ \)

Explanation:
1. With \( |\vec a| = |\vec b| = a \): \( |\vec a + \vec b| = 2a\cos\frac\theta2 \) and \( |\vec a - \vec b| = 2a\sin\frac\theta2 \).
2. The condition gives \( \cos\frac\theta2 + \sin\frac\theta2 = 1 \), which holds for \( \frac\theta2 = 0 \) or \( \frac\pi2 \).
3. So \( \theta = 0^\circ \) or \( 180^\circ \).

Teacher's Note:
1. For equal vectors the sum and difference are the diagonals of a rhombus.
2. The sum of the cosine and sine of a half angle equals 1 only at 0 and 90 degrees.

Answer: (d) \( \cos\theta = \frac{1}{\sqrt3} \)

Explanation:
1. The direction cosines satisfy \( \cos^2\theta + \cos^2\theta + \cos^2\theta = 1 \).
2. So \( \cos\theta = \frac{1}{\sqrt3} \) and \( \sin\theta = \sqrt{2/3} \). The values \( 2/\sqrt3 \) exceed 1 and are impossible.

Teacher's Note:
1. Direction cosines of any vector obey l² + m² + n² = 1.
2. Equal inclination means l = m = n.

Answer: (a) 2 or 1/2

Explanation:
1. \( \frac{A + B}{|A - B|} = 3 \) gives \( A + B = 3|A - B| \).
2. If \( A > B \): \( A = 2B \). If \( B > A \): \( B = 2A \).
3. So the ratio is 2 or 1/2.

Teacher's Note:
1. Since the minimum magnitude is |A − B|, either vector may be the larger one.
2. Both ratios give identical maximum to minimum ratios.

Answer: (c) Both (1) and (2)

Explanation:
1. Let the wind speed be w at 45°. Relative velocity \( = (w/\sqrt2 - 8,\ w/\sqrt2) \) must point 120° from east, so \( \tan120^\circ = -\sqrt3 = \frac{w/\sqrt2}{w/\sqrt2 - 8} \).
2. This gives \( \frac{w}{\sqrt2}(1+\sqrt3) = 8\sqrt3 \), so \( w = \frac{8\sqrt6}{1+\sqrt3} \).
3. Relative speed \( = \frac{w/\sqrt2}{\sin120^\circ} = \frac{16}{1+\sqrt3} \) km/h.

Teacher's Note:
1. Relative velocity is the wind velocity minus the man's velocity.
2. Resolve the relative velocity along the stated direction to find w.

Answer: (c) less than AB, equal to AB or zero, but never greater than AB

Explanation:
1. \( |\vec A\times\vec B| = AB\sin\theta \) and \( 0 \le \sin\theta \le 1 \).
2. So the magnitude ranges from 0 to AB.

Teacher's Note:
1. The sine of an angle never exceeds 1.
2. Zero occurs for parallel or antiparallel vectors; AB occurs for perpendicular ones.

Answer: (a) (1), (2) and (3) only

Explanation:
1. \( C^2 = A^2 + B^2 + 2AB\cos\theta \).
2. If \( C^2 = A^2 + B^2 \) then \( \cos\theta = 0 \). If \( C^2 < A^2 + B^2 \) then \( \cos\theta < 0 \) so \( \theta > 90^\circ \). If \( C^2 > A^2 + B^2 \) then \( \cos\theta > 0 \) so \( \theta < 90^\circ \).
3. So (1), (2) and (3) are correct and (4) is wrong.

Teacher's Note:
1. The sign of the term 2AB cos θ decides the angle.
2. Compare C² with A² + B² to classify the angle as acute, right or obtuse.

Answer: (d) equals the sum of their x-components, can be smaller than the sum of the magnitudes and can equal the sum of the magnitudes

Explanation:
1. Components add algebraically, so the x-component of the resultant is the sum of the individual x-components.
2. Each x-component is at most the magnitude of its vector, so the sum cannot exceed the sum of magnitudes. It equals it when every vector lies along +x.

Teacher's Note:
1. Resolve each vector, then add the components along each axis separately.
2. The sum of magnitudes is only reached for parallel vectors.

Answer: (b) 2 and \( -\tfrac23 \)

Explanation:
1. Squaring: \( 4(10 + x^2) = 16 + (4x-2)^2 + 4 \).
2. \( 40 + 4x^2 = 24 + 16x^2 - 16x \) gives \( 3x^2 - 4x - 4 = 0 \).
3. \( (3x + 2)(x - 2) = 0 \), so \( x = 2 \) or \( x = -\tfrac23 \).

Teacher's Note:
1. Square both sides to remove the roots.
2. A quadratic gives two roots, and both satisfy the given condition.

Answer: (b) Any of (−4, 9), (−4, −3), (12, 9) or (12, −3)

Explanation:
1. The x-component has magnitude \( \sqrt{10^2 - 6^2} = 8 \) m.
2. The signs of both components are free, so final x is \( 4 \pm 8 = 12 \) or \( -4 \) and final y is \( 3 \pm 6 = 9 \) or \( -3 \).
3. All four combinations are possible.

Teacher's Note:
1. Magnitude of a component fixes it only up to sign.
2. With no direction given, every sign choice is allowed.

Answer: (c) \( 3\overrightarrow{AD} \)

Explanation:
1. The diagonal AD passes through O, so \( \overrightarrow{AD} = 2\overrightarrow{AO} \).
2. \( \overrightarrow{AB} + \overrightarrow{AF} = \overrightarrow{AO} \) (diagonal of the rhombus ABOF) and \( \overrightarrow{AC} + \overrightarrow{AE} = 3\overrightarrow{AO} \) (triangle ACE has centroid O).
3. Total \( = \overrightarrow{AO} + 3\overrightarrow{AO} + 2\overrightarrow{AO} = 6\overrightarrow{AO} = 3\overrightarrow{AD} \).

Teacher's Note:
1. Use symmetry about the centre of the hexagon.
2. For an equilateral triangle, the sum of the two side vectors from one vertex is three times the vector to the centroid.

Answer: (a) \( 2\overrightarrow{AF} \)

Explanation:
1. \( \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD} + \overrightarrow{DE} + \overrightarrow{EF} = \overrightarrow{AF} \) by the polygon law.
2. Adding the last term \( \overrightarrow{AF} \) gives \( 2\overrightarrow{AF} \).

Teacher's Note:
1. Notice that the last vector is AF, not FA, so the polygon does not close.
2. A closed set would end with FA and sum to zero.

Answer: (d) \( \vec 0 \)

Explanation:
1. Opposite vertices give \( \overrightarrow{AO} + \overrightarrow{DO} = \vec 0 \), \( \overrightarrow{BO} + \overrightarrow{EO} = \vec 0 \) and \( \overrightarrow{CO} + \overrightarrow{FO} = \vec 0 \).
2. So the total is the null vector.

Teacher's Note:
1. Vectors from opposite vertices to the centre are equal and opposite.
2. The centre is the centroid, so the position vectors about it sum to zero.

Answer: (b) 8 units

Explanation:
1. Displacement \( \vec d = (3-2)\hat i + (2+1)\hat j + (-2-1)\hat k = \hat i + 3\hat j - 3\hat k \).
2. \( W = \vec F\cdot\vec d = 2(1) + 1(3) + (-1)(-3) = 8 \).

Teacher's Note:
1. Work is the dot product of force and displacement.
2. Displacement is final position minus initial position.

Answer: (d) \( 2(F_1^2 + F_2^2) \)

Explanation:
1. \( P^2 = F_1^2 + F_2^2 + 2F_1F_2\cos\theta \).
2. Reversing F\(_2\) changes cos θ to −cos θ, so \( Q^2 = F_1^2 + F_2^2 - 2F_1F_2\cos\theta \).
3. Adding: \( P^2 + Q^2 = 2(F_1^2 + F_2^2) \).

Teacher's Note:
1. Adding the two expressions removes the cosine term.
2. Subtracting them would instead give 4F₁F₂ cos θ.

Answer: (a) zero

Explanation:
1. The last leg has components \( 5\cos37^\circ = 4 \) m west and \( 5\sin37^\circ = 3 \) m south.
2. North: +3 − 3 = 0. East: +4 − 4 = 0.
3. The man returns to the starting point.

Teacher's Note:
1. 3-4-5 legs form a closed triangle.
2. Displacement is zero whenever the vectors close to form a polygon.

Answer: (c) a = −7, b = −3, c = −4

Explanation:
1. Equilibrium needs each component of the resultant to vanish: \( 7 + b + c = 0 \), \( a + c + 11 = 0 \), \( -10 - a - b = 0 \).
2. Adding gives \( a + b + c = -14 \), so \( a = -7 \), \( b = -3 \), \( c = -4 \).

Teacher's Note:
1. Set the sum of x, y and z components separately to zero.
2. Three equations, three unknowns, solved by adding all three first.

Answer: (c) 2

Explanation:
1. \( 7Q^2 = P^2 + Q^2 + 2PQ\cos60^\circ = P^2 + Q^2 + PQ \).
2. \( P^2 + PQ - 6Q^2 = 0 \), so \( (P + 3Q)(P - 2Q) = 0 \) and \( P/Q = 2 \) (the negative root is rejected).

Teacher's Note:
1. Divide the quadratic by Q² and solve for the ratio P/Q.
2. Magnitudes are positive, so only the positive root is allowed.

Answer: (a) \( \sqrt3 : 2 : 1 \)

Explanation:
1. The angle between R and P is \( 360^\circ - 150^\circ - 120^\circ = 90^\circ \).
2. By Lami's theorem, \( \frac{P}{\sin120^\circ} = \frac{Q}{\sin90^\circ} = \frac{R}{\sin150^\circ} \).
3. So \( P : Q : R = \frac{\sqrt3}{2} : 1 : \frac12 = \sqrt3 : 2 : 1 \).

Teacher's Note:
1. Lami's theorem links each force to the sine of the angle between the other two.
2. Find the third angle from the fact that the three angles add up to 360°.

Answer: (d) 9 km/h

Explanation:
1. North-west is 45° north of west, and the shoreline is 15° south of west, so the angle between them is \( 45^\circ + 15^\circ = 60^\circ \).
2. Component \( = 18\cos60^\circ = 9 \) km/h.

Teacher's Note:
1. The component along a line is v cos of the angle between v and that line.
2. North-west and the shoreline lie on opposite sides of west, so the angles add.

Answer: (b) \( 6\hat i + 8\hat j \) m/s

Explanation:
1. Displacement \( = (3, 4, 0) \), with magnitude 5.
2. The direction is the unit vector \( 0.6\hat i + 0.8\hat j \), so velocity \( = 10(0.6\hat i + 0.8\hat j) = 6\hat i + 8\hat j \).

Teacher's Note:
1. Velocity vector = speed × unit vector along the motion.
2. The unit vector alone would have no unit and magnitude 1.

Answer: (b) \( 120^\circ \)

Explanation:
1. Let the forces be F and 2F. The resultant is perpendicular to F, so its component along F is zero: \( F + 2F\cos\theta = 0 \).
2. So \( \cos\theta = -\tfrac12 \) and \( \theta = 120^\circ \).

Teacher's Note:
1. Set the component of the resultant along the smaller force to zero.
2. A negative cosine means the angle is obtuse.

Answer: (c) 1

Explanation:
1. \( |\hat a - \hat b|^2 = 1 + 1 - 2\cos60^\circ = 1 \).
2. So the magnitude is 1.

Teacher's Note:
1. Two unit vectors at 60° form an equilateral triangle with their difference.
2. The same formula gives 2 sin(θ/2) = 1.

Answer: (a) \( 10\sqrt3 \) dyne

Explanation:
1. \( R = 2F\cos\frac\theta2 = 2(10)\cos30^\circ = 10\sqrt3 \) dyne.

Teacher's Note:
1. For equal vectors, the resultant is 2F cos(θ/2).
2. It lies along the bisector of the angle.

Answer: (d) 5 miles

Explanation:
1. East: \( 1 + 2 = 3 \) miles. North: \( 9 - 5 = 4 \) miles.
2. Distance \( = \sqrt{3^2 + 4^2} = 5 \) miles.

Teacher's Note:
1. Add all east-west legs and all north-south legs separately.
2. The 3-4-5 triangle gives the straight-line distance.