Q # 1. Define the terms (i) Unit Vector (ii) Position Vector (iii) Component of a Vector.
Unit Vector
A vector having the unit magnitude is called the unit vector. It is used to indicate the direction of any vector. The unit vector in the direction of vector A is expressed as:
\[\widehat{A} = \frac{\mathbf{A}}{|A|}\]
where\(\widehat{A}\) is the unit vector in the direction of vector A and \(|A|\) is its magnitude.
Position Vector
The position vector describes the location of a point with respect to origin. In two dimension, the position vector `\(\overrightarrow{\mathbf{r}}\)' of point \(P\ (a,\ b)\) is describe as
\[\overrightarrow{\mathbf{r}}\mathbf{=}a\widehat{i} + b\widehat{j}\]
In three dimensional Cartesian coordinate system, the position vector `\(\overrightarrow{\mathbf{r}}\)' of point \(P\ (a,\ b,\ c)\) is describe as
\[\overrightarrow{\mathbf{r}}\mathbf{=}a\widehat{i} + b\widehat{j} + c\widehat{k}\]
Component of a Vector
A component of a vector is its effective value in a specific direction.
Q # 2. The vector sum of three vectors gives a zero resultant. What can be the orientation of the vectors?
Ans. If the three vectors are represented by the sides of a triangle taken in cyclic order, then the vector sum of three vectors will be zero.
Let three vectors \(\overrightarrow{\mathbf{A}}\mathbf{,\ }\overrightarrow{\mathbf{B}}\) and \(\overrightarrow{\mathbf{C}}\)are the three vectors acting along the sides of triangle \(PQR\) as shown in the figure. As the head of \(\overrightarrow{\mathbf{C}}\)coincides with the tail of \(\overrightarrow{\mathbf{A}}\), so by head to tail rule, the resultant of these three vectors will be zero.
Q # 3. Vector \(\mathbf{A}\) lies in xy plane. For what orientation will both of its rectangular components be negative? For what components will its components have opposite signs?
Ans. Figure shows sign with rectangular components of an arbitrary vector \(\overrightarrow{\mathbf{A}}\) in different quadrants.
It is clear from figure:
- When the vector lies in 3rd quadrant, then both of its rectangular components of vector will negative.
- The components of a vector have opposite sign when the vector lies in 2nd or 4th quadrant.
Q # 4. If one of the rectangular components of a vector is not zero, can its magnitude be zero? Explain.
Ans. If \(A_{x}\) and \(A_{y}\) are the rectangular components of vector \(\overrightarrow{\mathbf{A}}\), then magnitude of vector is described by formula:
\[A = \left| \overrightarrow{\mathbf{A}} \right| = \ \sqrt{{A_{x}}^{2} + {A_{y}}^{2}}\]
It is clear from the expression that the magnitude of a vector will be zero only if all of its rectangular components are zero.
Therefore if one of the components is not zero, then the magnitude of vector can't be zero.
Q # 5. Can a vector have a component greater than the vector's magnitude?
Ans. No, the component of a vector can never be greater than the vector's magnitude because the component of a vector is its effective value in a specific direction.
If \(A_{x}\) and \(A_{y}\) are the rectangular components of vector \(\overrightarrow{\mathbf{A}}\), then magnitude of vector is described by formula:
\[\left| \overrightarrow{\mathbf{A}} \right| = \ \sqrt{{A_{x}}^{2} + {A_{y}}^{2}}\]
It is clear from the expression that maximum magnitude of component can be equal to the magnitude of the vector.
Q # 6. Can the magnitude of a vector have a negative value?
Ans. No, the magnitude of a vector cannot be negative, because the magnitude of vector \(\overrightarrow{\mathbf{A}}\) can be described by the formula:
\[\left| \overrightarrow{\mathbf{A}} \right| = \ \sqrt{{A_{x}}^{2} + {A_{y}}^{2}}\]
Where \(A_{x}\) and \(A_{y}\) are the rectangular components of \(\overrightarrow{\mathbf{A}}\).
As the squares of real quantities always gives the positive values. Therefore, the magnitude of a vector will always be positive.
Q # 7.If \(\mathbf{A\ + \ B = 0}\), what can you say about the components of the two vectors.
Ans. Given that:
\[\overrightarrow{\mathbf{A}}\mathbf{\ + \ }\overrightarrow{\mathbf{B}}\mathbf{=}0\]
\[\mathbf{\Longrightarrow}\overrightarrow{\mathbf{A}}\mathbf{\ \ = - \ \ }\overrightarrow{\mathbf{B}}\]
These vectors can be expressed in terms of rectangular components,
\[A_{x}\widehat{i} + A_{y}\widehat{j} = - \left( B_{x}\widehat{i} + B_{y}\widehat{j} \right)\]
\[\mathbf{\Longrightarrow}A_{x}\widehat{i} + A_{y}\widehat{j} = - B_{x}\widehat{i} - B_{y}\widehat{j}\]
Comparing the coefficients of unit vectors \(\widehat{i}\)and \(\widehat{j}\), we get:
\[A_{x} = - B_{x}\]
and
\[A_{y} = - B_{y}\]
Hence the components of both vectors are equal in magnitude but opposite in direction.
Q # 8. Under what circumstances would a vector have components that are equal in magnitude?
Ans. If \(A_{x}\)\&\(A_{y}\) are rectangular components of vector \(\overrightarrow{\mathbf{A}}\), then according to given condition:
\[A_{x} = A_{y}\]
\[\Longrightarrow A\cos\theta = A\sin\theta \Longrightarrow \sin\theta = \cos\theta \Longrightarrow \frac{\sin\theta}{\cos\theta} = 1\]
\[\Longrightarrow \tan\theta = 1 \Longrightarrow \theta = \tan^{- 1}(1) = 45{^\circ}\]
So the components of a vector will have equal magnitude when vector makes an angle of 45˚ with horizontal.
Q # 9. Is it possible to add a vector quantity to a scalar quantity?
Ans. No it is not possible to add a vector quantity to a scalar quantity because only the physical quantities of same nature can be added. Vectors and scalars are different physical quantities.
Q # 10. Can you add zero to a null vector?
Ans. No, zero can't be added to a null vector because zero is a scalar and scalars can't be added to vectors. Only the physical quantities of same nature can be added.
Q # 11. Two vectors have unequal magnitudes. Can their sum be zero? Explain.
Ans. No, the sum of two vectors having unequal magnitudes can't be zero. The sum of two vectors will be zero only when their magnitudes are equal and they act in opposite direction.
Q # 12. Show that the sum and the difference of two perpendicular vectors of equal lengths are also perpendicular and of same length.
Ans. Consider two vectors \(\overrightarrow{\mathbf{A}}\) and \(\overrightarrow{\mathbf{B}}\) of equal magnitude which are perpendicular to each other. The sum and the difference of both vectors gives the resultant \(\overrightarrow{\mathbf{R}}\)and\(\overrightarrow{\mathbf{R'}}\),respectively, and are described below:
\[\overrightarrow{\mathbf{R}}\mathbf{=}\overrightarrow{\mathbf{A}}\mathbf{+}\overrightarrow{\mathbf{B}}\mathbf{=}A\ \widehat{i}\mathbf{+}B\widehat{j}\]
\[\overrightarrow{\mathbf{R'}}\mathbf{=}\overrightarrow{\mathbf{A}}\mathbf{-}\overrightarrow{\mathbf{B}}\mathbf{=}A\ \widehat{i}\mathbf{-}B\widehat{j}\]
Magnitude of\(\overrightarrow{\mathbf{R}}\mathbf{=}R = \sqrt{A^{2} + B^{2}}\)
Magnitude of\(\overrightarrow{\mathbf{R'}}\mathbf{=}R' = \sqrt{A^{2} + B^{2}}\)
So, it is clear that the sum and the difference of two perpendicular vectors of equal magnitude have the same lengths.
Now taking dot product of \(\mathbf{R}\)and \(\mathbf{R'}\), we get:
\[\overrightarrow{\mathbf{R}}\mathbf{\ .\ }\overrightarrow{\mathbf{R'}}\mathbf{=}\left( A\ \widehat{i}\mathbf{+}B\widehat{j} \right).\left( A\ \widehat{i}\mathbf{-}B\widehat{j} \right) = A^{2} - B^{2}\]
\(\because\left| \overrightarrow{\mathbf{A}} \right|\mathbf{=}\left| \overrightarrow{\mathbf{B}} \right|\mathbf{\ \Longrightarrow}A = B\)
\[\overrightarrow{\mathbf{R}}\mathbf{\ .\ }\overrightarrow{\mathbf{R'}} = A^{2} - A^{2} = 0\]
As \(\overrightarrow{\mathbf{R}}\mathbf{\ .\ }\overrightarrow{\mathbf{R'}} = 0\), therefore, the sum and the difference of two perpendicular vectors of equal magnitude are perpendicular to each other.
Q # 13. How would the two vector same magnitude have to be oriented, if they were to be combined to give a resultant equal to a vector of same magnitude?
Ans. The two vectors of equal magnitudes are combined to give a resultant vector of same magnitude when they act along the sides of equilateral triangle.
Consider two vectors \(\overrightarrow{\mathbf{A}}\) and \(\overrightarrow{\mathbf{B}}\)of equal magnitude which act along the sides of equilateral triangle such that they make an angle of \(120˚\) with each other. Both vectors are added by head to tail rule to give resultant \(\overrightarrow{\mathbf{R}}\), which is along the third side of equilateral triangle as shown in the figure below:
From figure it is clear that \(\overrightarrow{\mathbf{R}}\mathbf{=}\overrightarrow{\mathbf{A}}\mathbf{+}\overrightarrow{\mathbf{B}}\) and \(\left| \overrightarrow{\mathbf{R}} \right|\mathbf{=}\left| \overrightarrow{\mathbf{A}} \right|\mathbf{=}\left| \overrightarrow{\mathbf{B}} \right|\), as all the sides of equilateral triangle have equal lengths.
Q # 14. The two vectors to be combined have magnitude 60 N and 35 N. Pick the correct answer from those given below and tell why is it the only one of the three that is correct. (i) 100 N (ii) 70 N (iii) 20 N
Ans. The correct answer is 70 N.
- The resultant of two vectors has maximum magnitude when they act in same direction. Thus if both vectors are parallel, then the magnitude of resultant will be: \(60\ N + 35\ N = 95\ N\).
- The resultant of two vectors has minimum magnitude when they act in opposite direction. Thus if both vectors are anti-parallel, then the magnitude of resultant is \(60\ N - \ 35\ N = 25\ N\).
Hence the sum can't be less than \(25\ N\) and more than \(95\ N\). Therefore, the only possible value for correct answer is \(70\ N\).
Q # 15. Suppose the sides of a closed polygon represent vector arranged head to tail. What is the sum of these vectors?
Ans.If there are five vectors \(\overrightarrow{\mathbf{A}}\mathbf{,\ \ }\overrightarrow{\mathbf{B}}\mathbf{,\ \ }\overrightarrow{\mathbf{C}}\mathbf{,\ \ }\overrightarrow{\mathbf{D}}\) and \(\overrightarrow{\mathbf{E}}\) which are acting along the sides of close polygon as shown in the figure:
As the tail of the first vector meets with the head of last vector, so by head to tail rule:
\[\overrightarrow{\mathbf{A}}\mathbf{+}\overrightarrow{\mathbf{B}}\mathbf{+}\overrightarrow{\mathbf{C}}\mathbf{+}\overrightarrow{\mathbf{D}}\mathbf{+}\overrightarrow{\mathbf{E}}\mathbf{=}0\]
Hence the sum of vectors arranged along the sides of polygon will be zero.
Q # 16. Identify the correct answer.
Two ships X and Y are travelling in different direction at equal speeds. The actual direction of X is due to north but to an observer on Y, the apparent direction of motion X is north-east. The actual direction of motion of Y as observed from the shore will be (A) East (B) West (C) South-east (D) South-West
Ans. The correct answer is (B) West
The horizontal force F is applied to a small object P of mass m at rest on a smooth plane inclined at an angle \(\mathbf{\theta}\) \textbf{to the horizontal as shown in the figure below. The magnitude of the resultant force acting up and along the surface of the plane, on the object is
\[\mathbf{\ F\ cos\ \theta\ - \ mg\ sin\ \theta}\]
\[\mathbf{F\ sin\ \theta\ - \ mg\ cos\ \theta}\]
\[\mathbf{F\ cos\ \theta + mg\ sin\ \theta}\]
\[\mathbf{\ F\ sin\ \theta + mg\ cos\ \theta}\]
\[\mathbf{ \ mg\ tan\theta\ }\]
Ans.The forces acting up and along the surface of plane is\(\mathbf{F\ cos\ \theta\ - \ mg\ sin\ \theta}\), therefore the correct option is (a)
Q # 17.If all the components of the vectors, \({\overrightarrow{\mathbf{A}}}_{\mathbf{1}}\) and \({\overrightarrow{\mathbf{A}}}_{\mathbf{2}}\) were reversed, how would this alter \({\overrightarrow{\mathbf{A}}}_{\mathbf{1}}\mathbf{\times} {\overrightarrow{\mathbf{A}}}_{\mathbf{2}}\).
Ans. If all the components of the vectors \({\overrightarrow{\mathbf{A}}}_{\mathbf{1}}\) and \({\overrightarrow{\mathbf{A}}}_{\mathbf{2}}\)are reversed, then both vectors will be represented as \(- {\overrightarrow{\mathbf{A}}}_{\mathbf{1}}\)and \(- {\overrightarrow{\mathbf{A}}}_{\mathbf{2}}\), respectively.
As
\[\mathbf{-}{\overrightarrow{\mathbf{A}}}_{\mathbf{1}}\mathbf{\times} - {\overrightarrow{\mathbf{A}}}_{\mathbf{2}}\mathbf{=}{\overrightarrow{\mathbf{A}}}_{\mathbf{1}}\mathbf{\times}{\overrightarrow{\mathbf{A}}}_{\mathbf{2}}\]
So the vector product of two vectors will remain unchanged even when the components of the vectors are reversed.
Q # 18.Name the three different conditions that could makes \({\overrightarrow{\mathbf{A}}}_{\mathbf{1}}\mathbf{\times}{\overrightarrow{\mathbf{A}}}_{\mathbf{2}}\mathbf{=}\overrightarrow{\mathbf{O}}\)
Ans. The conditions that could make the \({\overrightarrow{\mathbf{A}}}_{\mathbf{1}}\mathbf{\times}{\overrightarrow{\mathbf{A}}}_{\mathbf{2}}\mathbf{=}O\) are as follows:
- If \({\overrightarrow{\mathbf{A}}}_{\mathbf{1}}\) or \({\overrightarrow{\mathbf{A}}}_{\mathbf{2}}\)is the null vector
- If both \({\overrightarrow{\mathbf{A}}}_{\mathbf{1}}\) and \({\overrightarrow{\mathbf{A}}}_{\mathbf{2}}\)are the null vectors
- If the vectors \({\overrightarrow{\mathbf{A}}}_{\mathbf{1}}\)and \({\overrightarrow{\mathbf{A}}}_{\mathbf{2}}\)are parallel or anti-parallel with each other.
Q # 19. Identify true or false statements and explain the reason.
(a) A body in equilibrium implies that it is neither moving nor rotating.
(b) If the coplanar forces acting on a body form a close polygon, then the body is said to be in equilibrium.
Ans.
(i) Statement (a) is false. Because a body may be in equilibrium if it is moving or rotating with uniform velocity.
(ii) Statement (b) is correct. Since the vector sum of all the forces acting on the body along close polygon is zero, then the first condition of equilibrium will be satisfied and the body will be in state of equilibrium.
Q # 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 string is minimum.
Ans. Consider a picture of weight \(\mathbf{W}\) is suspended by two strings as shown in the figure.
From figure,
\[2T\sin\theta = W\]
\[\Longrightarrow T = \frac{W}{2\sin\theta}\]
It is clear from the expression that the tension in the string will be minimum when the factor \(\sin\theta\) will be maximum (equal to 1). This condition can be obtained for \(\theta = 90{^\circ}\).
Q # 21. Can a body rotate about its center of gravity under the action of its weight?
Ans. No a body can't rotate about the center of gravity under the action of its weight.
The whole weight of the body acts on the center of gravity. Therefore, the perpendicular distance between line of action of weight and axis of rotation, called moment arm, is zero.
As
\[Torque = (Force)(Moment\ Arm)\]
So the torque due to weight will be zero because the moment arm is zero. Hence, a body cannot rotate about center of gravity under the action of its weight.
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