JEE Physics Vectors and Calculus MCQs Set 08

Answer: (d) Potential energy

Explanation:
1. Weight and momentum have direction, and nuclear spin is an angular momentum.
2. Potential energy is a scalar.

Teacher's Note:
1. Scalars have only magnitude.
2. Energy, work and power are scalars.

Answer: (c) \( \dfrac{5}{\sqrt2} \)

Explanation:
1. Component \( = \dfrac{\vec A\cdot\vec B}{|\vec B|} = \dfrac{2+3}{\sqrt2} = \dfrac{5}{\sqrt2} \).

Teacher's Note:
1. Component of A along B is \( \vec A\cdot\hat B \).
2. Divide by the magnitude of B.

Answer: (c) 87 N

Explanation:
1. \( F\cos60^\circ = 50 \Rightarrow F = 100 \) N.
2. Vertical component \( = F\sin60^\circ = 100\times0.866 \approx 87 \) N.

Teacher's Note:
1. Horizontal component is \( F\cos\theta \), vertical is \( F\sin\theta \).
2. Equivalently \( F_y = F_x\tan\theta \).

Answer: (b) \( 45^\circ \)

Explanation:
1. Magnitude \( = \sqrt{1+1+2} = 2 \).
2. \( \cos\theta = \dfrac{\sqrt2}{2} \Rightarrow \theta = 45^\circ \).

Teacher's Note:
1. The direction cosine with the Z-axis is the k-component divided by the magnitude.

Answer: (c) \( (a_x - a_y)/\sqrt2 \)

Explanation:
1. Unit vector along \( (\hat{i}-\hat{j}) \) is \( \dfrac{\hat{i}-\hat{j}}{\sqrt2} \).
2. Component \( = \vec A\cdot\hat n = \dfrac{a_x-a_y}{\sqrt2} \).

Teacher's Note:
1. Always divide by the magnitude of the direction vector.

Answer: (c) \( F_1 = 3 \) N, \( F_2 = 5 \) N, \( F_3 = 6 \) N

Explanation:
1. The resultant of 3 N and 5 N lies between 2 N and 8 N.
2. For equilibrium the third force must lie in this range: only 6 N does.

Teacher's Note:
1. Three forces balance if they can form a triangle.
2. Triangle rule: \( |F_1-F_2|\le F_3\le F_1+F_2 \).

Answer: (a) \( 10t \)

Explanation:
1. \( v_x = 8t,\ v_y = 6t \).
2. Speed \( = \sqrt{64t^2+36t^2} = 10t \).

Teacher's Note:
1. Differentiate each coordinate to get velocity components.
2. Combine with the Pythagoras rule.

Answer: (b) 3

Explanation:
1. Two vectors of different magnitude cannot cancel.
2. Three vectors can, if their magnitudes satisfy the triangle condition. So the minimum is 3.

Teacher's Note:
1. Vectors adding to zero form a closed polygon.
2. A triangle is the simplest polygon.

Answer: (d) \( \dfrac{\pi}{3} \)

Explanation:
1. \( R^2 = 2F^2(1+\cos\theta) = 3F^2 \).
2. \( \cos\theta = \dfrac12 \Rightarrow \theta = \dfrac{\pi}{3} \).

Teacher's Note:
1. Use \( R^2 = P^2+Q^2+2PQ\cos\theta \).
2. Equal forces give \( R = 2F\cos(\theta/2) \).

Answer: (a) \( \cos^{-1}(3/5) \)

Explanation:
1. \( 5^2 = 4^2+3^2 \), so \( \vec B \perp \vec C \) and the vectors form a right triangle with hypotenuse A.
2. The angle between A and C has \( \cos\theta = \dfrac{C}{A} = \dfrac35 \).

Teacher's Note:
1. Check whether the magnitudes form a Pythagorean triple.
2. The hypotenuse is the resultant.

Answer: (a) \( \cos^{-1}\left[-\dfrac{x^2+y^2}{2(x^2-y^2)}\right] \)

Explanation:
1. \( R^2 = (x+y)^2+(x-y)^2+2(x^2-y^2)\cos\theta = x^2+y^2 \).
2. \( 2x^2+2y^2 + 2(x^2-y^2)\cos\theta = x^2+y^2 \Rightarrow \cos\theta = -\dfrac{x^2+y^2}{2(x^2-y^2)} \).

Teacher's Note:
1. Expand carefully: \( (x+y)^2+(x-y)^2 = 2(x^2+y^2) \).

Answer: (c) \( (12.0\ \text m)\hat{i}-(16.0\ \text m)\hat{k} \)

Explanation:
1. Displacement = final minus initial.
2. \( x: 9-(-3) = 12;\ y: 2-2 = 0;\ z: -8-8 = -16 \).

Teacher's Note:
1. Subtract component by component.

Answer: (b) \( C < |\vec A-\vec B| \)

Explanation:
1. \( C^2 = A^2+B^2+2AB\cos120^\circ = A^2+B^2-AB \).
2. \( |\vec A-\vec B|^2 = A^2+B^2+AB \), which is larger.
3. So \( C < |\vec A-\vec B| \).

Teacher's Note:
1. For an obtuse angle the sum is smaller than the difference.
2. (The option C = |A+B| is trivially true by definition but is not the intended relation; the strict inequality is the one asked.)

Answer: (a) \( \cos^{-1}(5/13) \)

Explanation:
1. \( 13^2 = 12^2+5^2 \): B and C are perpendicular.
2. \( \cos\theta = \dfrac{C}{A} = \dfrac{5}{13} \).

Teacher's Note:
1. Pythagorean triple 5-12-13.
2. \( \cos^{-1}(13/12) \) is impossible since cosine cannot exceed 1.

Answer: (c) cannot keep the body in equilibrium

Explanation:
1. Equilibrium would need \( 1+2 = 3 \) with all three along one line, which is excluded because the forces act along different directions.
2. 2 N and 3 N at right angle give \( \sqrt{13}\ne1 \); 1 N and 2 N give \( \sqrt5\ne3 \); 1 N and 3 N at an acute angle give a resultant above 2 N.
3. So equilibrium is not possible.

Teacher's Note:
1. The extreme case 1 + 2 = 3 needs collinear forces.
2. Use the resultant formula for each option.

Answer: (b) \( Q \)

Explanation:
1. New resultant \( \vec P+2\vec Q \perp \vec P \): \( P^2+2\vec P\cdot\vec Q = 0 \).
2. \( R^2 = P^2+Q^2+2\vec P\cdot\vec Q = P^2+Q^2-P^2 = Q^2 \).
3. So \( R = Q \).

Teacher's Note:
1. Perpendicular means dot product zero.
2. Substitute \( 2\vec P\cdot\vec Q = -P^2 \).

Answer: (c) \( P \)

Explanation:
1. \( R = 2P\cos(120^\circ/2) = 2P\times\tfrac12 = P \).

Teacher's Note:
1. For equal vectors \( R = 2P\cos(\theta/2) \).
2. At \( 120^\circ \) the resultant equals each vector.

Answer: (c) \( 12\hat{i}+16\hat{j}-3\hat{k} \)

Explanation:
1. \( \vec C = \vec A+3\vec B \).
2. \( = (3+9)\hat{i}+(1+15)\hat{j}+(3-6)\hat{k} = 12\hat{i}+16\hat{j}-3\hat{k} \).

Teacher's Note:
1. Add the components after multiplying B by 3.

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

Explanation:
1. \( R = 2A\cos(\theta/2) = A \Rightarrow \cos(\theta/2) = \tfrac12 \).
2. \( \theta/2 = 60^\circ \Rightarrow \theta = 120^\circ \).

Teacher's Note:
1. Same result as \( R^2 = 2A^2(1+\cos\theta) = A^2 \).

Answer: (d) \( \sqrt{202} \)

Explanation:
1. Sum \( = 9\hat{i}+11\hat{j} \).
2. Magnitude \( = \sqrt{81+121} = \sqrt{202} \).

Teacher's Note:
1. Add components, then take the root of the sum of squares.

Answer: (c) \( -2\hat{i}+3\hat{j}+5\hat{k} \)

Explanation:
1. Displacement \( = \) B minus A \( = (-2-0, 6-3, 4+1) \).
2. \( = -2\hat{i}+3\hat{j}+5\hat{k} \).

Teacher's Note:
1. Final position minus initial position.

Answer: (a) \( 8\hat{i}+10\hat{j}+12\hat{k} \)

Explanation:
1. \( \vec\tau = \vec r\times\vec F \), with rows (1, -2, 1) and (5, 2, -5).
2. \( i: (-2)(-5)-(1)(2) = 8;\ j: -[(1)(-5)-(1)(5)] = 10;\ k: (1)(2)-(-2)(5) = 12 \).

Teacher's Note:
1. Torque is \( \vec r\times\vec F \), in that order.
2. Expand the determinant with alternating signs.

Answer: (b) \( -\hat{i}+\hat{j}+3\hat{k} \)

Explanation:
1. \( \vec L = \vec r\times\vec p \) with rows (1, -2, 1) and (2, -1, 1).
2. \( i: (-2)(1)-(1)(-1) = -1;\ j: -[(1)(1)-(1)(2)] = 1;\ k: (1)(-1)-(-2)(2) = 3 \).

Teacher's Note:
1. Angular momentum is \( \vec r\times\vec p \).

Answer: (a) 1 : 1

Explanation:
1. \( \vec A\cdot\vec C = 2-3-8 = -9 \), \( \vec B\cdot\vec C = 2+3+4 = 9 \).
2. \( \vec X\cdot\vec C = -9\alpha+9\beta = 0 \Rightarrow \alpha = \beta \).

Teacher's Note:
1. Perpendicular means the dot product is zero.