Multiple Choice Questions, Vectors and Equilibrium
Choose the best possible answer;
- Two vectors lie with their tails at the same point. When the angle between them is increased by 20° their scalar product has the same magnitude but changes from positive to negative. The original angle between them was:
(A) 0 (B) 60° (C) 70° (D)80°
- The minimum number of vectors of unequal magnitudes required to produce zero resultant is
(A) 2 (B) 3 (C) 4 (D)5
- If the resultant of two vectors, each of magnitude A is also a vector of magnitude A, the angle between the two vectors will be
(A) 30° (B) 45° (C) 60° (D)120°
- The magnitude of vector A = 2î + ĵ + 2k̂ is
(A) 9 (B) 5 3 (D) 1
- When Fx = 3 N and F = 5 N then Fy =
(A) 6 N (B) 4 N (C) 2 N (D) 0 N
- A meter stick is supported by a knife edge at the 50 cm mark. Asif hangs masses of 0.4 kg and 0.6 kg from the 20 cm and 80 cm marks respectively. Where should Asif hang a third mass of 0.30 kg to keep the stick balanced?
20 cm (B) 70 cm (C)30 cm (D)25 cm
- Ax = 1.5 cm, Ay = -1.0 cm, into which quadrant do the vector A point?
(A) I (B) II (C) III IV
- A.(A × B) = …
(A) 0 (B) 1 (C) AB (D) A2
- Two forces of magnitudes 20 N and 50 N act simultaneously on a body. Which one of the following cannot be the resultant of the two forces?
(A) 20 N (B) 30 N 40 N (D) 70 N
- If the dot product of two non-zero vectors A and B is zero, then the magnitude of their cross product is
(A) 0 (B) 1 (C) AB (D)-AB
- 11. The sum of magnitudes of two forces is 16 N. If the resultant force is 8 N and its direction is perpendicular to minimum force the force are
(A) 6 N and 10 N (B) 8 N and 8 N (C) 4N and 12N (D) 2 N and 14 N
We know that for a right angled triangle, (Hypotenuse)2 = (base)2 + (perpendicular)2.
Given in the problem is
Put this value of x2 in equation (1)
Put x1 = 6 in equation (2), we have x2 = 16 – 6 = 10 N
- Find the mass of the uneven rod shown in the figure. If its center of gravity is 14 cm from end A.
(A) 100 g (B) 150 g (C) 80 g (D) 5 g
SOLUTION: Clockwise torque produced by 80 g (= 0.08 kg) mass to the right = 20 × 9.8 × 0.08 = 15.68 N m.
Let m be the mass of the rod, then the anti-clockwise torque produced by the weight mg to the left = 9.8m × 16 = 156.8m N m. Since the rod is in equilibrium, both torques are equal. 156.8m = 15.68 OR m = 15.68/156.8 = 0.1 kg = 100 g.
- The following diagrams show a uniform rod with its midpoint on the pivot. Two equal forces F are applied on the rod as shown in the figure. Which diagram shows the rod in equilibrium?
Solution: Please note a body is in equilibrium when the resultant of all forces acting on it is zero and the sum of all torques acting on the body is zero.
(a) Anti-clockwise torque (b) clockwise torque (c) no torque (both torques cancel each other) (d) Anti-clockwise torque.
Hence (c) is the correct option.
- For which angle the equation
(A) 30° (B) 45° (C) 60° (D) 90°
- What is the torque acting on the wheel of radius 2 m?
(A)10 N clockwise (B) 10 N m anticlockwise (C) 10 N m clockwise (D) 5 N m clockwise
Solution Hint: The force 10 N is producing a clockwise torque. Take it as negative. The 5 N force is producing anti-clockwise torque. Take it positive. Apply the formula and find both torques. Then add them. Remember the anti-clockwise torque is positive. Θ in both cases is 90o.