JEE Physics Vectors and Calculus MCQs Set 01

Answer: (b) 30 m east

Explanation:
1. Take east as x and north as y. The displacement is the vector sum of the three legs.
2. North and south legs: \( +40 - 40 = 0 \) along y. East leg: \( +30 \) along x.
3. Displacement = 30 m towards east.

Teacher's Note:
1. Displacement depends only on the initial and final positions, not on the path length.
2. Opposite legs of equal length cancel.

Answer: (d) \( \sqrt{F_1^2 + F_2^2} \)

Explanation:
1. \( R^2 = F_1^2 + F_2^2 + 2F_1F_2\cos 90^\circ \) and \( \cos 90^\circ = 0 \).
2. So \( R = \sqrt{F_1^2 + F_2^2} \).

Teacher's Note:
1. The cross term vanishes for perpendicular vectors, which gives the Pythagoras result.
2. This is the quickest way to combine two perpendicular components.

Answer: (a) 50 and 53° with east, towards north

Explanation:
1. \( R = \sqrt{30^2 + 40^2} = 50 \).
2. \( \tan\theta = \frac{40}{30} = \frac43 \), so \( \theta = 53^\circ \) from the east axis towards north.

Teacher's Note:
1. Memorise the 3-4-5 triangle: angles near 37° and 53°.
2. The angle is always measured from the vector that is taken as the reference axis.

Answer: (c) \( 250\sqrt5 \) N, \( \tan^{-1}(2) \) west of north

Explanation:
1. Take east as x and north as y: \( \vec F_2 - \vec F_1 = -500\hat i + 250\hat j \).
2. Magnitude \( = \sqrt{500^2 + 250^2} = 250\sqrt5 \) N.
3. The vector points to the north and west. The angle from north is \( \tan^{-1}\frac{500}{250} = \tan^{-1} 2 \), towards west.

Teacher's Note:
1. Subtraction is addition of the reversed vector, so the x-component changes sign.
2. Always check the quadrant of the final vector from the signs of its components.

Answer: (c) cannot be zero

Explanation:
1. \( \vec A + \vec B \) lies in the plane, while \( \vec C \) has a component perpendicular to it.
2. That perpendicular component cannot be cancelled by anything in the plane, so the resultant is never zero.

Teacher's Note:
1. Three vectors can sum to zero only if they are coplanar.
2. A component outside a plane can only be cancelled by another component outside it.

Answer: (a) 8 N

Explanation:
1. The resultant lies between \( 10 - 6 = 4 \) N and \( 10 + 6 = 16 \) N.
2. Only 8 N falls inside this range.

Teacher's Note:
1. Minimum resultant is the difference of magnitudes and maximum is their sum.
2. Any value in between is possible for a suitable angle.

Answer: (d) null vector

Explanation:
1. In the polygon law, the resultant is the line joining the start of the first to the end of the last vector.
2. For a closed polygon these points coincide, so the resultant is a null (zero) vector.

Teacher's Note:
1. A closed vector polygon means that the vectors are in equilibrium.
2. A null vector has zero magnitude and no definite direction.

Answer: (b) \( \pi \)

Explanation:
1. \( R^2 = P^2 + Q^2 + 2PQ\cos\theta \) is least when \( \cos\theta = -1 \).
2. So \( \theta = \pi \), and \( R_{\min} = |P - Q| \).

Teacher's Note:
1. Opposite vectors subtract fully, giving the minimum resultant.
2. Parallel vectors (angle 0) give the maximum resultant, the sum of magnitudes.

Answer: (b) \( \pi/2 \)

Explanation:
1. \( 12^2 + 5^2 = 144 + 25 = 169 = 13^2 \).
2. So \( C^2 = A^2 + B^2 \), which means \( 2AB\cos\theta = 0 \) and \( \theta = 90^\circ \).

Teacher's Note:
1. Pythagorean triples such as 5, 12, 13 signal perpendicular vectors.
2. Compare C² with A² + B² to decide whether the angle is acute, right or obtuse.

Answer: (c) \( \theta = 90^\circ \)

Explanation:
1. Squaring: \( P^2 + Q^2 + 2PQ\cos\theta = P^2 + Q^2 - 2PQ\cos\theta \).
2. So \( 4PQ\cos\theta = 0 \), which for non-zero vectors gives \( \theta = 90^\circ \).

Teacher's Note:
1. The sum and difference vectors have equal length only for perpendicular vectors (the diagonals of a rectangle).
2. Zero vectors are excluded because the angle is then undefined.

Answer: (a) \( -25\cos30^\circ \) and \( +25\sin30^\circ \)

Explanation:
1. A clockwise angle of 210° is the same as \( -210^\circ \), that is \( 150^\circ \) measured anticlockwise.
2. \( x = 25\cos150^\circ = -25\cos30^\circ \) and \( y = 25\sin150^\circ = +25\sin30^\circ \).

Teacher's Note:
1. Convert a clockwise angle to the anticlockwise one by changing its sign.
2. Check the quadrant: 150° lies in the second quadrant, where x is negative and y is positive.

Answer: (d) \( 30\sqrt3 \) km h\(^{-1}\)

Explanation:
1. \( v_y = \sqrt{60^2 - 30^2} = \sqrt{2700} = 30\sqrt3 \) km h\(^{-1}\).

Teacher's Note:
1. The components and the vector form a right triangle.
2. Here the 30-60-90 pattern gives sides in the ratio 1 : √3 : 2.

Answer: (b) 3 and \( -\hat i + 2\hat j - \hat k \)

Explanation:
1. Dot product: \( 1\cdot2 + 1\cdot1 + 1\cdot0 = 3 \).
2. Cross product determinant: \( \hat i(1\cdot0 - 1\cdot1) - \hat j(1\cdot0 - 1\cdot2) + \hat k(1\cdot1 - 1\cdot2) = -\hat i + 2\hat j - \hat k \).

Teacher's Note:
1. The scalar product multiplies like components and adds; the vector product uses the 3 × 3 determinant.
2. Check with A·(A×B), which must be zero: -1 + 2 - 1 = 0.

Answer: (d) 6 and \( 6\sqrt3 \)

Explanation:
1. \( \vec A\cdot\vec B = 4\cdot3\cos60^\circ = 6 \).
2. \( |\vec A\times\vec B| = 4\cdot3\sin60^\circ = 6\sqrt3 \).

Teacher's Note:
1. Dot product uses cosine and cross product magnitude uses sine.
2. At 60°, cos = 1/2 and sin = √3/2.

Answer: (a) \( \pm\sqrt{0.11} \)

Explanation:
1. A unit vector has magnitude 1: \( 0.25 + 0.64 + c^2 = 1 \).
2. So \( c^2 = 0.11 \) and \( c = \pm\sqrt{0.11} \).

Teacher's Note:
1. The squares of the direction components of a unit vector add up to 1.
2. Both signs of c satisfy the condition.

Answer: (c) \( 15^\circ \)

Explanation:
1. The first vector makes \( \tan^{-1}(2/2) = 45^\circ \) with the x-axis.
2. The second makes \( \tan^{-1}\sqrt3 = 60^\circ \).
3. The angle between them is \( 60^\circ - 45^\circ = 15^\circ \).

Teacher's Note:
1. Find each vector's angle with the axis and subtract.
2. You can confirm using cos θ = (2 + 2√3)/(2√2 · 2) = 0.966, i.e. 15°.

Answer: (c) \( \vec B \times \vec C \)

Explanation:
1. \( \vec A \) is perpendicular to both \( \vec B \) and \( \vec C \).
2. \( \vec B\times\vec C \) is perpendicular to both of them as well, so \( \vec A \) can be parallel to it.
3. \( \vec B\cdot\vec C \) is a scalar, so it has no direction.

Teacher's Note:
1. A vector perpendicular to two non-parallel vectors lies along their cross product.
2. A scalar can never be parallel to a vector.

Answer: (a) \( 60^\circ \)

Explanation:
1. \( \tan\theta = \frac{AB\sin\theta}{AB\cos\theta} = \frac{8\sqrt3}{8} = \sqrt3 \).
2. The scalar product is positive, so the angle is acute: \( \theta = 60^\circ \).

Teacher's Note:
1. Dividing the cross-product magnitude by the dot product gives tan θ directly.
2. Note that the problem says the magnitude of the scalar product; if the dot product were negative the angle would be 120°.

Answer: (d) \( \cos^{-1}\frac{5}{13} \)

Explanation:
1. \( R^2 = 180 = 1 + 169 + 2(1)(13)\cos\theta \).
2. So \( 26\cos\theta = 10 \) and \( \cos\theta = \frac{5}{13} \).

Teacher's Note:
1. Keep the 2PQ cos θ term when solving for the angle.
2. An inverse cosine of a number greater than 1 (such as 13/5) is impossible, which quickly rules out some options.

Answer: (b) \( 2|\vec b| \)

Explanation:
1. \( \frac{a + b}{a - b} = 3 \Rightarrow a + b = 3a - 3b \).
2. So \( 2a = 4b \) and \( a = 2b \).

Teacher's Note:
1. Maximum resultant is a + b and minimum is |a − b|.
2. Assume a is larger so that the minimum is a − b.

Answer: (b) \( \tan^{-1}(2/3) \)

Explanation:
1. The angle with the y-axis satisfies \( \cos\alpha = \frac{3}{\sqrt{13}} \) and \( \sin\alpha = \frac{2}{\sqrt{13}} \).
2. So \( \tan\alpha = \frac{2}{3} \).

Teacher's Note:
1. The angle with the y-axis uses the y-component as the adjacent side.
2. tan⁻¹(3/2) would be the angle with the x-axis.

Answer: (c) \( \hat i - 4\hat j + 5\hat k \)

Explanation:
1. \( \overrightarrow{AB} = \vec r_B - \vec r_A = (2-1)\hat i + (-3-1)\hat j + (4+1)\hat k \).
2. \( = \hat i - 4\hat j + 5\hat k \).

Teacher's Note:
1. Always subtract the tail (starting point) from the head (end point).
2. Reversing the order gives the opposite vector.

Answer: (a) \( \sqrt{58} \) units

Explanation:
1. \( \vec A + \vec B = 7\hat i - 3\hat j \).
2. Magnitude \( = \sqrt{49 + 9} = \sqrt{58} \).

Teacher's Note:
1. Add components first, then find the magnitude.
2. Never add magnitudes of vectors unless they are parallel.

Answer: (d) \( \frac{3\hat i + \hat k}{\sqrt{10}} \)

Explanation:
1. \( \vec A - \vec B = 3\hat i + 0\hat j + \hat k \), with magnitude \( \sqrt{10} \).
2. The unit vector is the vector divided by its magnitude.

Teacher's Note:
1. A unit vector is the vector divided by its own magnitude.
2. The negative option points the opposite way (along B − A).