JEE Physics Vectors and Calculus MCQs Set 09

Answer: (a) \( 5\sqrt2 \)

Explanation:
1. \( |\vec v|=\sqrt{3^2+4^2+5^2}=\sqrt{50} \).
2. \( \sqrt{50}=5\sqrt2 \).

Teacher's Note:
1. The magnitude is the square root of the sum of the squares of the components.

Answer: (a) \( -2\hat j+3\hat k \)

Explanation:
1. \( \vec A+\vec B=\hat i \).
2. \( \vec B=\hat i-\vec A=\hat i-\hat i-2\hat j+3\hat k=-2\hat j+3\hat k \).

Teacher's Note:
1. Subtract the given vector from the target vector component by component.

Answer: (b) 5

Explanation:
1. \( B_x=11-7=4,\ B_y=9-6=3 \).
2. \( |\vec B|=\sqrt{16+9}=5 \).

Teacher's Note:
1. Subtract the components of A from those of A + B, then use Pythagoras.

Answer: (c) \( 30\sqrt3\ \text{km/h} \)

Explanation:
1. \( v_2=\sqrt{60^2-30^2}=\sqrt{2700} \).
2. \( v_2=30\sqrt3 \) km/h.

Teacher's Note:
1. The squares of the two rectangular components add up to the square of the magnitude.

Answer: (b) \( 2P\cos\dfrac\theta2 \)

Explanation:
1. \( R^2=P^2+P^2+2P^2\cos\theta=2P^2(1+\cos\theta)=4P^2\cos^2\tfrac\theta2 \).
2. \( R=2P\cos\tfrac\theta2 \).

Teacher's Note:
1. Use 1 plus cos theta equals 2 cos squared of half theta to simplify the parallelogram law.

Answer: (a) 45°

Explanation:
1. \( 10A^2=4A^2+2A^2+2(2A)(\sqrt2A)\cos\theta \).
2. \( 4=4\sqrt2\cos\theta \), so \( \cos\theta=1/\sqrt2 \).
3. \( \theta=45^\circ \).

Teacher's Note:
1. Put the given magnitudes in the cosine rule for the resultant and solve for the angle.

Answer: (b) 0.33

Explanation:
1. \( 0.25+0.64+c^2=1 \), so \( c^2=0.11 \).
2. \( c=\sqrt{0.11}=0.33 \).

Teacher's Note:
1. A unit vector has magnitude one, so the squares of the components add to one.

Answer: (a) \( \cos^{-1}\left(\dfrac{5}{\sqrt{50}}\right) \)

Explanation:
1. \( \vec P+\vec Q=5\hat i-4\hat j+3\hat k \), magnitude \( \sqrt{25+16+9}=\sqrt{50} \).
2. \( \cos\alpha=5/\sqrt{50} \).

Teacher's Note:
1. The cosine of the angle with an axis is the component along that axis divided by the magnitude.

Answer: (d) 120°

Explanation:
1. \( C^2=A^2+B^2+2AB\cos\theta \): \( 9=9+9+18\cos\theta \).
2. \( \cos\theta=-\tfrac12 \), so \( \theta=120^\circ \).

Teacher's Note:
1. Three equal vectors that sum to give a vector of the same size form an equilateral arrangement, giving 120 degrees between the first two.

Answer: (a) 90°

Explanation:
1. \( \vec A\cdot\vec B=1-1=0 \).
2. The dot product is zero, so the angle is 90°.

Teacher's Note:
1. A zero dot product of two non-zero vectors means they are perpendicular.

Answer: (b) 180°

Explanation:
1. \( |\vec A+\vec B|=A-B \) requires the minimum possible resultant.
2. This happens only when the vectors are antiparallel, so \( \theta=180^\circ \).

Teacher's Note:
1. The smallest resultant of two vectors is the difference of their magnitudes, obtained when they point in opposite directions.

Answer: (c) 6 units and 10 units

Explanation:
1. Let the smaller be A and the larger B: \( A+B=16 \). Since R is perpendicular to A, \( B^2=A^2+R^2 \), so \( B^2-A^2=64 \).
2. \( B-A=64/16=4 \).
3. \( B=10 \) and \( A=6 \).

Teacher's Note:
1. A right-angled triangle is formed by the smaller vector, the resultant and the larger vector.

Answer: (b) \( \cos^{-1}\left[-\dfrac{A^2+B^2}{2(A^2-B^2)}\right] \)

Explanation:
1. The two forces have magnitudes \( A+B \) and \( A-B \), so the sum of their squares is \( 2(A^2+B^2) \) and the product of the magnitudes is \( A^2-B^2 \).
2. The condition \( R^2=2A^2+2B^2+2(A^2-B^2)\cos\theta=A^2+B^2 \).
3. \( \cos\theta=-\dfrac{A^2+B^2}{2(A^2-B^2)} \).

Teacher's Note:
1. Apply the parallelogram law with the two magnitudes treated as A plus B and A minus B.

Answer: (a) \( \sqrt{90} \)

Explanation:
1. \( 2\vec A=2\hat i+2\hat j-4\hat k \), \( 3\vec B=6\hat i-3\hat j+3\hat k \).
2. \( 2\vec A-3\vec B=-4\hat i+5\hat j-7\hat k \).
3. Magnitude \( =\sqrt{16+25+49}=\sqrt{90} \).

Teacher's Note:
1. Scale each vector, subtract component-wise, and take the root of the sum of squares.

Answer: (b) 3 N and 4 N

Explanation:
1. Right angle: \( F_1^2+F_2^2=25 \). At 120°: \( F_1^2+F_2^2-F_1F_2=13 \).
2. So \( F_1F_2=12 \).
3. \( F_1,F_2=3,4 \) N.

Teacher's Note:
1. Use the cosine rule twice.
2. The product of the forces follows from the difference of the two equations.

Answer: (a) \( -2\hat i-\hat k \)

Explanation:
1. Sum of the first two: \( 2\hat i+\hat j+\hat k \).
2. \( \vec C=\hat j-(2\hat i+\hat j+\hat k)=-2\hat i-\hat k \).

Teacher's Note:
1. The third vector equals the required resultant minus the sum of the known vectors.

Answer: (a) \( \hat j-\hat k \)

Explanation:
1. \( \vec Q=\hat i-\vec P=\hat i-\hat i+\hat j-\hat k \).
2. \( \vec Q=\hat j-\hat k \).

Teacher's Note:
1. Subtract P from the unit vector along x.

Answer: (b) \( -5\hat i+3\hat j \)

Explanation:
1. Sum \( =5\hat i-3\hat j+\hat k \).
2. Needed vector \( =\hat k-(5\hat i-3\hat j+\hat k)=-5\hat i+3\hat j \).

Teacher's Note:
1. Required result minus the existing sum gives the vector to be added.

Answer: (c) \( \vec A \) is parallel to \( \vec B \)

Explanation:
1. \( \vec B=3\vec A \), so the vectors are parallel.
2. \( \vec A\cdot\vec B=3+3+12=18\ne0 \), so they are not perpendicular, and \( |\vec B|=3|\vec A| \).

Teacher's Note:
1. One vector being a scalar multiple of the other means they are parallel.

Answer: (a) 20

Explanation:
1. \( \vec p\cdot\vec Q=3(0)+0(3)\ldots \) taking components: \( (0)(2)+(3)(0)+(4)(5) \).
2. \( \vec p\cdot\vec Q=20 \).

Teacher's Note:
1. Multiply matching components and add.
2. Only the z-components are non-zero in both vectors.