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Wednesday, 13 November 2013

F.Sc Notes: Physics XI: Chapter 02 (11-21) Vectors and Equilibrium Exercise Short Questions:

FSc Notes: Physics XI: Chapter 02 (11-21) Vectors and Equilibrium Exercise SQ:

Majority of these question need diagram to explain but i have not drawn the diagrams.

Question 2.11 Two vectors have unequal magnitudes. Can their sum be zero? Explain.
Answer 2.11 No
. The sum of two unequal vectors cannot be zero. For the sum of vectors to be zero, the  vectors must have equal magnitude with opposite directions.

Question 2.12 Show that the sum and differences of two perpendicular vectors of equal lengths are also perpendicular and of the same length.
Answer 2.12
Vectors A and B are perpendicular to each other having equal lengths we have:
(A + B) is ⊥ to (A - B) i.e. sum and difference of the vectors are perpendicular to each other.

Question 2.13 How would the two vectors of the same magnitude have to be oriented, if they were to be combined to give a resultant equal to a vector of the same magnitude?
Answer 2.13
When the angle between two vectors of same magnitude is 120 degree, the magnitudes of the resultant will be same.

Question 2.14  The two vectors to be combined have magnitude 60N and 35N. Pick the correct answer from those given and tell why it is only one of the three that is correct. (i) 100N (ii) 70N (iii) 20N.
Answer 2.14
  A1 = 60N and A2 = 35N
Answer (ii) 70N is correct.
For maximum value, both vectors in same direction,
A1 + A2 = 60 + 35 = 95 → cannot be (i) 100N.
For minimum value, both vectors having opposite direction,
A1 - A2 = 60 + 35 = 25 → cannot be (iii) 20N.

Question 2.15 Suppose the sides of a closed polygon represent vector arranged head to tail .What is the sum of these vectors?
Answer 2.15
The sum will be zero. Because the tail of first vector meets the head of last vector. Whenever vector representation makes a closed shape its resultant vector is a null vector.

Question 2.16  Identify the correct answer.
  1. Two ships X and Y are traveling in different directions at equal speeds. The actual direction of motion of X is due north but to an observer on Y, the apparent direction of motion of X is north-east. The actual direction of motion of Y as observed from shore will be (A) East (B) West (C) south-east (D) south-west.
  2. A horizontal force F is applied to a small object P of mass m at rest on a smooth plane inclined at an angle θ to the horizontal as shown in fig The magnitude of the resultant force acting up and along the surface of the plane, on the object is (a) F cos θ -mg sin θ  (b) F sin θ -mg cos θ (c) F cos θ +mg cos θ (d) Fsin θ+mg sin θ (e) mgtan θ.
Answer 2.16 (1) The correct answer is (B) West.
(2)  The correct answer is (a) F cos θ - mg sin θ the resultant force acting Up and along the surface of the plane is = F cos θ - mg sin θ.

Question 2.17   If all the components of the vectors, A1 and A2 were reversed, how would this alter A1 x A2?
Answer 2.17
    For A1 = - A1& A2= - A2
 -A1 x -A2 = A1 x A2  
There will be no effect, if all the components of the vectors A1 & A2 are reversed.

Question 2.18   Name the three different conditions that could make A1 x A2 = 0.
Answer 2.18
   A1 x A2 could be zero, if
  1. A1 is null vector; 0 x A2 = 0
  2. A2 is null vector; A1 x 0 = 0      
  3. A1 x A2 are parallel or anti-parallel, i.e. A1 x A2 = A1 A2 sin θ = A1 A2 sin 180 degree = 0

Question 2.19   Identify true or false statement and explain the reason.
  1. A body in equilibrium implies that it is neither moving nor rotating.
  2. If co planar forces acting on a body from a closed polygon, then the body is said to be in equilibrium.
Answer 2.19 a) It is false. A body moving with constant velocity can also be in equilibrium.
b) It is true. The vector sum will be zero, for the coplanar forces forming a closed polygon, fulfills the 1st condition of equilibrium.

Question 2.20   A picture is suspended from a wall by two strings. Show by diagram the configuration of the strings  for which the tension in the strings will be minimum.
Answer 2.20
    For T minimum, θ = 90 degree
Σ Fy = 0                 Ty + Ty - w = 0                    2Ty - w = 0
2 T sin θ = w                                           T = w / 2 sinθ
For minimum T, θ = 90 degree  i.e. T = w / 2 sin 90 degree 
T = w / 2.

Question 2.21 Can a body rotate about its center of gravity under the action of its weight?
Answer 2.21 No
. A body cannot rotate about its center of gravity under the action of its weight. Because moment arm will be zero, so torque or turning effect will be zero.  

Written By: Asad Hussain

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