Class 11 Physics

Vector Multiple Choice Questions And Answers

Question 1. What is the condition for \(\vec{A}+\vec{B}=\vec{A}-\vec{B}\) to be valid?

1. \(\vec{A}=0\)
2. \(\vec{B}=0\)
3. \(\vec{A}=\vec{B}\)
4. \(\vec{A}=-\vec{B}\)

Answer: 2. \(\vec{B}=0\)

Question 2. The magnitudes of the sum and difference of the two vectors are equal. The angle between the vectors is

1. 0°
2. 90°
3. 120°
4. 60°

Answer: 2. 90°

Read And Learn More WBCHSE Class 11 Physics MCQs

Question 3. The magnitude of each of the two vectors is P. The Magnitude of the resultant of the two is also P. The Angle between the vectors is

1. 0°
2. 60°
3. 120°
4. 90°

Answer: 3. 120°

Question 4.Two forces of equal magnitude act simultaneously on a particle. If the resultant of the forces is equal to the magnitude of each of them then the angle between the forces is

1. An acute angle
2. An obtuse angle
3. A right angle
4. Of any value

Answer: 2. An obtuse angle

Question 5. A man travels 30 m to the north, then 20 m to the east, and after that 30√2 m to the south-west. His displacement from the starting point is

1. 15 m to the east
2. 28 m to the south
3. 10 m to the west
4. 15 m to the southwest

Answer: 3. 10 m to the west

WBBSE Class 11 Vector MCQs

Question 6. \(0.2 \hat{i}+0.6 \hat{j}+a \hat{k}\) is a unit vector. The value of a should be

1. 0.3
2. 0.4
3. 0.6
4. 0.8

Answer: 3. 0.6

Question 7. \(\vec{R}\) is the resultant of the vectors \(\vec{A}\) and \(\vec{B}\). If R = \(\frac{B}{\sqrt{2}}\), then the angle θ is

1. 30°
2. 45°
3. 60°
4. 75°

Answer: 2. 45°

Question 8. The position vector of a particle is related to time t as \(\vec{r}=\left(t^2-1\right) \hat{i}+2 t \hat{j}\). The locus of the particle on the x-y plane is

1. Parabolic
2. Circular
3. Straight line
4. Elliptical

Answer: 1. Parabolic

Question 9. Two forces of the same magnitude F are at right angles to each other. The magnitude of the net force (total force) acting on the object is

1. F
2. 2 F
3. Between F and 2F
4. More than 2 F

Answer: 3. Between F and 2 F

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Question 10. If a and b are two unit vectors inclined at an angle of 60° to each other, then

1. \(|a+b|>1\)
2. \(|a+b|<1\)
3. \(|a-b|>1\)
4. \(|a-b|<1\)

Answer: 1. \(|a+b|>1\)

Question 11. The condition (a-b)² = a²b² is satisfied when

1. a is parallel to b
2. a ≠ b
3. a-b = 1
4. a ⊥ b

Answer: 3. a-b = 1

Vector Operations and Their Applications MCQs

Question 12. If \(\vec{P}+\vec{Q}=\vec{R} \text { and }|\vec{P}|=|\vec{Q}|=\sqrt{3} \text { and }|\vec{R}|=3\), then the angle between \(\vec{P}\) and \(\vec{Q}\) is

1. \(\frac{\pi}{4}\)
2. \(\frac{\pi}{6}\)
3. \(\frac{\pi}{3}\)
4. \(\frac{\pi}{2}\)

Answer: 3. \(\frac{\pi}{3}\)

Question 13. The angles which the vector \(\vec{A}=3 \hat{i}+6 \hat{j}+2 \hat{k}\) makes with the coordinate axes are

1. \(\cos ^{-1} \frac{3}{7}, \cos ^{-1} \frac{6}{7}\) and \(\cos ^{-1} \frac{2}{7}\)
2. \(\cos ^{-1} \frac{4}{7}, \cos ^{-1} \frac{5}{7} \)and \(\cos ^{-1} \frac{3}{7}\)
3. \(\cos ^{-1} \frac{3}{7}, \cos ^{-1} \frac{4}{7}\) and \(\cos ^{-1} \frac{1}{7}\)
4. None of the above

Answer: 1. \(\cos ^{-1} \frac{3}{7}, \cos ^{-1} \frac{6}{7}\) and \(\cos ^{-1} \frac{2}{7}\)

Question 14. If  \(\vec{a}+\vec{b}=\vec{c}\), then the angle between \(\vec{a}\) and \(\vec{b}\) is

1. 90°
2. 180°
3. 120°
4. Zero

Answer: 4. Zero

Question 15. If at and are two non-collinear unit vectors and if \(\left|a_1+a_2\right|=\sqrt{3}\) then the value of \(\left(a_1-a_2\right) \cdot\left(2 a_1+a_2\right)\) is

1. 2
2. 3/2
3. 1/2
4. 1

Answer: 3. 1/2

Question 16. Which of the following is a vector quantity?

1. Temperature
2. Impulse
3. Gravitational potential
4. Power

Answer: 2. Impulse

Question 17. The resultant of two forces of magnitude (x+y) and (x-y) is \(\sqrt{x^2+y^2}\). The angle between them is

1. \(\cos ^{-1}\left[-\frac{\left(x^2+y^2\right)}{2\left(x^2-y^2\right)}\right]\)
2. \(\cos ^{-1}\left[-\frac{2\left(x^2-y^2\right)}{x^2+y^2}\right]\)
3. \(\cos ^{-1}\left[-\frac{x^2+y^2}{x^2-y^2}\right]\)
4. \(\cos ^{-1}\left[-\frac{x^2-y^2}{x^2+y^2}\right]\)

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

Question 18. ABC is an equilateral triangle and O is its centre of mass. Find n if \(\overrightarrow{A B}+\overrightarrow{A C}=n \overrightarrow{A O}\)

1. 1
2. 2
3. 3
4. 4

Hint: \(\overrightarrow{A B}+\overrightarrow{A C}=(\overrightarrow{A O}+\overrightarrow{O B})+(\overrightarrow{A O}+\overrightarrow{O C})\)

= \(2 \overrightarrow{A O}+(\overrightarrow{O B}+\overrightarrow{O C})=2 \overrightarrow{A O}+\overrightarrow{A O}\)

= \(3 \overrightarrow{A O}\)

Answer: 3. 3

Short Notes on Vector Properties with MCQs

Question 19. The sum of the magnitudes of two forces acting at a point is 16 N. The resultant has a magnitude of 8N and is perpendicular to the force of lower magnitude. The two forces are

1. 6N and 10N
2. 8N and 8N
3. 4N and 12N
4. 2N and 14N

Answer: 1. 6N and 10N

Question 20. Two men have velocities of 4 m/s towards the east and 3 m/s towards west, respectively. The velocity of the first man relative to the second is

1. \((4 \hat{i}+3 \hat{j})\)
2. \((3 \hat{i}+4 \hat{j})\)
3. \((4 \hat{i}-3 \hat{j})\)
4. \((3 \hat{i}-4 \hat{j})\)

Answer: 1. \((4 \hat{i}+3 \hat{j})\)

Question 21. Two forces in the ratio 1:2 act simultaneously on a particle. The result of these forces is three times the first force. The angle between them is

1. 0°
2. 60°
3. 90°
4. 45°

Answer: 1. 0°

Question 22. Given \(\vec{A}=2 \hat{i}+3 \hat{j} \text { and } \vec{B}=\hat{i}+\hat{j}\). The component of vector \(\vec{A}\) along vector \(\vec{B}\) is

1. \(\frac{1}{\sqrt{2}}\)
2. \(\frac{3}{\sqrt{2}}\)
3. \(\frac{5}{\sqrt{2}}\)
4. \(\frac{7}{\sqrt{2}}\)

Answer: 3. \(\frac{5}{\sqrt{2}}\)

Question 23. One of the components of a velocity vector of magnitude 50 m · s^-1 Is 30 m · s^-1, Its other orthogonal component Is

1. 15 m · s^-1
2. 20 m · s^-1
3. 25 m · s^-1
4. 40 m · s^-1

Answer: 4. 40 m · s^-1

Question 24. The times of flight of the two projectiles are t₁ and t₂ If R is the horizontal range of each of them, then

1. \(t_1 t_2 \propto R\)
2. \(t_1 t_2 \propto R^2\)
3. \(t_1 t_2 \propto R^3\)
4. \(t_1 t_2 \propto R^{\frac{1}{2}}\)

Hint: The angles of projection are \(\theta\) and \(\left(90^{\circ}-\theta\right)\).

∴ \(t_1 t_2=\frac{2 u \sin \theta}{g} \cdot \frac{2 u \sin \left(90^{\circ}-\theta\right)}{g}\)

= \(\frac{4 u^2}{g^2} \sin \theta \cos \theta=\frac{2}{g} \cdot \frac{u^2 \sin 2 \theta}{g}=\frac{2}{g} R\)

∴ \(t_1 t_2\propto R\)

Answer: 1. \(t_1 t_2 \propto R\)

Question 25. For angles θ and (90° – θ) of projection, a projectile has the same horizontal range R. The maximum heights attained are H₁ and H₂ respectively. Then the relation among R, H_1, and H₂ is

1. \(R=\sqrt{H_1 H_2}\)
2. \(R=\sqrt{H_1^2+H_2^2}\)
3. \(R=H_1+H_2\)
4. \(R=4 \sqrt{H_1 H_2}\)

Answer: 4. \(R=4 \sqrt{H_1 H_2}\)

Question 26. An airplane is flying with a velocity of 216 km/h at an altitude of 1960 m relative to the ground. It drops a bomb when it is just above point A on the ground. The bomb hits the ground at B. The distance AB is

1. 1.2 km
2. 0.33 km
3. 3.33 km
4. 33 km

Answer: 1. 1.2 km

Real-Life Examples of Vectors in Motion

Question 27. The initial velocity and the acceleration of a particle are \(\vec{u}\) = 3\(\hat{i}\)+4\(\hat{j}\) and \(\vec{a}\) = 0.3 \(\hat{i}\)+0.4\(\hat{j}\). The magnitude of its velocity after 10 s is

1. 10 units
2. 8.5 units
3. 77√2 units
4. 7 units

Answer: 1. 10 units

Question 28. The equations of motion of a projectile are x = 36t and 2y = 96t- 9.8 t². The angle of projection is

1. \(\sin ^{-1}\left(\frac{4}{5}\right)\)
2. \(\sin ^{-1}\left(\frac{3}{5}\right)\)
3. \(\sin ^{-1}\left(\frac{3}{4}\right)\)
4. \(\sin ^{-1}\left(\frac{4}{3}\right)\)

Hint: x= 36t and y = 48t – \(\frac{1}{2}\) x 9.8t²—they suggest that the horizontal and vertical components of initial velocity are 36 and 48 units, respectively.

∴ \(\tan \theta=\frac{48}{36}=\frac{4}{3} \text { or, } \sin \theta=\frac{4}{5}\)

Answer: 1. \(\sin ^{-1}\left(\frac{4}{5}\right)\)

Question 29. Two projectiles, projected with angles (45°-θ) and (45° +θ) respectively, have their horizontal ranges in the ratio

1. 2:1
2. 1:1
3. 2:3
4. 1:2

Answer: 2. 1:1

Question 30. For a projectile, (horizontal range)² = 48 x (maximum height)². The angle of projection is

1. 45°
2. 60°
3. 75°
4. 30°

Answer: 4. 30°

Question 31. Two railway tracks are parallel to the west-east direction. Along one track train A moves with a speed of 30 m · s^-1 from west to east, while along the second track, train B moves with a speed of 48 m · s^-1 from east to west. The relative speed of B with respect to A is

1. 48 m · s^-1
2. -78 m · s^-1
3. 30 m · s^-1
4. Zero

Answer: 2. 48 m · s^-1

Question 32. Angle between the vectors \(\hat{i}+\hat{j} \text { and } \hat{i}-\hat{k}\) is

1. 60°
2. 30°
3. 45°
4. 90°

Answer: 1. 60°

Question 33. A vector is multiplied by (-2). As a result

1. The magnitude of the vector is doubled and the direction is unaltered
2. The magnitude of the vector remains the same and the direction is reversed
3. The magnitude of the vector is doubled and the direction is reversed
4. No change in the magnitude or direction of the vector

Answer: 3. The Magnitude of the vector is doubled and the direction is reversed

Question 34. In a clockwise system

1. \(\hat{j} \times \hat{j}=1\)
2. \(\hat{k} \cdot \hat{i}=1\)
3. \(\hat{j} \times \hat{k}=\hat{i}\)
4. \(\hat{i} \cdot \hat{i}=0\)

Answer: 3. \(\hat{j} \times \hat{k}=\hat{i}\)

Question 35. A vector \(\vec{P}=3 \hat{i}-2 \hat{j}+a \hat{k}\) is perpendicular to the vector \(\vec{Q}=2 \hat{i}+\hat{j}-\hat{k}\). The value of a is

1. 2
2. 1
3. 4
4. 3

Answer: 3. 4

Step-by-Step Solutions to Vector MCQs

Question 36. For any two vectors \(\vec{A} \text { and } \vec{B} \text {, if } \vec{A} \cdot \vec{B}=|\vec{A} \times \vec{B}|\), the magnitude of \(\vec{C}=\vec{A}+\vec{B}\) is equal to

1. \(\sqrt{A^2+B^2}\)
2. \(A+B\)
3. \(\sqrt{A^2+B^2+\frac{A B}{\sqrt{2}}}\)
4. \(\sqrt{A^2+B^2+\sqrt{2} A B}\)

Answer: 4. \(\sqrt{A^2+B^2+\sqrt{2} A B}\)

Question 38. A vector perpendicular to both of \((3 \hat{i}+\hat{j}+2 \hat{k})\) and \((2 \hat{i}- 2 \hat{j}+4 \hat{k})\) is

1. \(\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}-\hat{k})\)
2. \(\hat{i}-\hat{j}-\hat{k}\)
3. \(\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})\)
4. \((\sqrt{3} \hat{i}-\hat{j}-\hat{k})\)

Answer: 2. \(\hat{i}-\hat{j}-\hat{k}\)

Question 39. A vector normal to \(a \cos \theta \hat{i}+b \sin \theta \hat{j}\) is

1. \(b \sin \theta \hat{i}-a \cos \theta \hat{j}\)
2. \(\frac{1}{a} \sin \theta \hat{i}-\frac{1}{b} \cos \theta \hat{j}\)
3. \(5 \hat{k}\)
4. All of the above

Answer: 4. All of the above

Question 40. A vector \(\vec{A}\) of magnitude 2 units is inclined at angles 30° and 60° with positive x- and y-axes, respectively. Another vector of magnitude 5 units is aligned along the positive x-axis. Then \(\vec{A}\) • \(\vec{B}\) is

1. 5√3
2. 3√5
3. 2√3
4. 3√2

Answer: 1. 5√3

Question 41. For what values of a and b, the vector \((a \hat{i}+b \hat{j})\) will be a unit vector perpendicular to the vector \((\hat{i}+\hat{j})\)?

1. 1,0
2. 0,1
3. \(\frac{1}{\sqrt{3}},-\frac{2}{\sqrt{3}}\)
4. \(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\)

Answer: 4. \(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\)

Question 42. Two billiard balls, starting from the same point, have velocities \((\hat{i}+\sqrt{3} \hat{j}) \text { and }(2 \hat{i}+2 \hat{j})\), respectively. The angle between them is

1. 60°
2. 15°
3. 45°
4. 30°

Answer: 2. 15°

Question 43. If \(|\vec{A} \times \vec{B}|=\sqrt{3} \vec{A} \cdot \vec{B}\), then \(|\vec{A}+\vec{B}|=\)?

1. \(\left(A^2+B^2+A B\right)^{\frac{1}{2}}\)
2. (\(\left(A^2+B^2+\frac{A B}{\sqrt{3}}\right)^{\frac{1}{2}}\)
3. A+B
4. \(\left(A^2+B^2+\sqrt{3} A B\right)^{\frac{1}{2}}\)

Answer: 1. \(\left(A^2+B^2+A B\right)^{\frac{1}{2}}\)

Question 44. The vector product of \(\vec{A}\) and \(\vec{B}\) is zero. The scalar product of \(\vec{A}\) and \((\vec{A}+\vec{B})\) is

1. 0
2. \(A^2\)
3. \(A B\)
4. \(A^2+A B\)

Answer: 4. \(A^2+A B\)

Question 45. If \(\vec{A} \cdot \vec{B}=\vec{A} \cdot \vec{C}=0\), then the vector parallel to \(\vec{A}\) would be

1. \(\vec{C}\)
2. \(\vec{B}\)
3. \(\vec{B} \times \vec{C}\)
4. \(\vec{A} \times(\vec{B} \times \vec{C})\)

Answer: 3. \(\vec{B} \times \vec{C}\)

In this type of question, more than one option is correct.

Question 46. In the following figure which of the statements is correct?

1. The sign of x -component of \(\vec{l}_1\) and \(\vec{l}_2\) is negative
2. The signs of the y -component of \(\vec{l}_1 \text { and } \vec{l}_2\) are positive and negative respectively
3. The signs of x and y-components of \(\vec{l}_1+\vec{l}_2\) are both positive
4. None of these

Answer:

1. The sign of x -component of \(\vec{l}_1\) and \(\vec{l}_2\) is negative

3. The signs of x and y-components of \(\vec{l}_1+\vec{l}_2\) are both positive

Common Vector MCQs for Class 11 –

Question 47. Two particles are projected in the air with speed v₀ at angles θ₁ and θ₂ (both acute) to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then which are the correct choices?

1. The angle of projection:  θ₁ > θ₂
2. Time of flight: T₁ > T₂
3. Horizontal range: R₁ > R₂
4. Total energy: U₁ > U₂

Answer:

Question 48. For two vectors \(\vec{A} \text { and } \vec{B},|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|\) is always true when

1. \(|\vec{A}|=|\vec{B}| \neq 0\)
2. \(\vec{A} \perp \vec{B}\)
3. \(|\vec{A}|=|\vec{B}| \neq 0\) and \(\vec{A}\) and \(\vec{B}\) are parallel or antiparallel
4. When either \(|\vec{A}|\) or \(|\vec{B}|\) is zero

Answer:

2. \(\vec{A} \perp \vec{B}\)

4. When either \(|\vec{A}|\) or \(|\vec{B}|\) is zero

Question 49. Given \(\vec{a}+\vec{b}+\vec{c}+\vec{d}=\overrightarrow{0}\). Which of the following statements is correct?

1. \(\vec{a}, \vec{b}, \vec{c}\) and \(\vec{d}\) must each be a null vector
2. The magnitude of \((\vec{a}+\vec{c})\) equals the magnitude of \((\vec{b}+\vec{d})\)
3. The magnitude of \(\vec{a}\) can never be greater than the sum of the magnitudes of \(\vec{b}, \vec{c}\), and \(\vec{d}\)
4. \((\vec{b}+\vec{c})\) must lie on the plane of \(\vec{a}\) and \(\vec{d}\) if \(\vec{a}\) and \(\vec{d}\) are not collinear and in the line of \(\vec{a}\) and \(\vec{d}\), if they are collinear

Answer:

2. The magnitude of \((\vec{a}+\vec{c})\) equals the magnitude of \((\vec{b}+\vec{d})\)

3. The magnitude of \(\vec{a}\) can never be greater than the sum of the magnitudes of \(\vec{b}, \vec{c}\) and \(\vec{d}\)

4. \((\vec{b}+\vec{c})\) must lie on the plane of \(\vec{a}\) and \(\vec{d}\) if \(\vec{a}\) and \(\vec{d}\) are not collinear and in the line of \(\vec{a}\) and \(\vec{d}\), if they are collinear

Question 50. State which of the following statements are false. A scalar quantity is one that

1. Is conserved in a process
2. Can never take negative values
3. Must be dimensionless
4. Has the same value for observers with different orientations of axes

Answer:

1. Is conserved in a process
2. Can never take negative values
3. Must be dimensionless

Question 51. The resultant of \(\vec{P}\) and \(\vec{Q}\) is \(\vec{R}\). The angle between \(\vec{P}\) and \(\vec{Q}\) is

1. 90°, if R² = P² + Q²
2. Less than 90°, if R2 > (P² + Q²)
3. Greater than 90°, if R² < (P2 + Q²)
4. Greater than 90°, if R² > (P² + Q²)

Answer:

1. 90°, if R² = P² + Q²
2. Less than 90°, if R2 > (P² + Q²)
3. Greater than 90°, if R² < (P2 + Q²)

Question 52. The magnitude of the vector product of A and B may be

1. Greater than AB
2. Equal to AB
3. Less than AB
4. Zero

Answer:

2. Equal to AB

3. Less than AB

4. Zero

Question 53. If \(\vec{A}=5 \hat{i}+6 \hat{j}+3 \hat{k} \text { and } \vec{B}=6 \hat{i}-2 \hat{j}-6 \hat{k}\), then

1. \(\vec{A}\) and \(\vec{B}\) are perpendicular
2. \(\vec{A} \times \vec{B}=\vec{B} \times \vec{A}\)
3. \(\vec{A}\) and \(\vec{B}\) have the same magnitude
4. \(\vec{A} \cdot \vec{B}=0\)

Answer:

1. \(\vec{A}\) and \(\vec{B}\) are perpendicular

4. \(\vec{A} \cdot \vec{B}=0\)

Question 54. Two projectiles, thrown from the same point with the same magnitude of velocity, have angles 60° and 30° of their projection. Then their

1. Maximum heights attained are the same
2. Horizontal ranges are the same
3. Magnitudes of velocity at the instants of hitting the ground are the same
4. Times of flight are the same

Answer:

2. Horizontal ranges are the same

3. Magnitudes of velocity at the instants of hitting the ground are the same

Question 55. A body is projected with a velocity u at an angle θ with the horizontal. At t = 2s, the body makes an angle of 30° with the horizontal. 1 s later, it attains its maximum height. Then

1. u = 20√3 m/s
2. θ = 60°
3. θ = 45
4. u = \(\frac{20}{\sqrt{3}} \mathrm{~m} / \mathrm{s}\)

Answer:

1. u = 20√3 m/s
2. θ = 60°

Question 56. \(\vec{A} \perp \vec{B}\); \(\vec{C}\) is coplanar with \(\vec{A}\) and \(\vec{B}\). Therefore,

1. \(\vec{A}=x \vec{B}+y \vec{C}\), where x and y are scalars
2. \(\vec{A} \cdot(\vec{B} \times \vec{C})=0\)
3. \(|(\vec{A} \times \vec{B}) \times \vec{C}|=A B C\)
4. \(\vec{A} \cdot \vec{B}=0\)

Answer:

1. \(\vec{A}=x \vec{B}+y \vec{C}\), where x and y are scalars
2. \(\vec{A} \cdot(\vec{B} \times \vec{C})=0\)
3. \(|(\vec{A} \times \vec{B}) \times \vec{C}|=A B C\)
4. \(\vec{A} \cdot \vec{B}=0\)

Product Of A Scalar And A Vector

WBBSE Class 11 Scalar and Vector Product Notes

A scalar has magnitude but no direction. On the other hand, a vector has both magnitude and direction. Therefore, when a scalar is multiplied by a vector, the product has both magnitude and direction. Thus, the product is a vector.

1. When a vector is multiplied by a scalar, the product is also a vector. If the scalar is positive then the direction of the vector remains unaltered but, if the scalar is negative, the direction of the vector becomes opposite to that of the original vector.

Example: 10 eastwards x 5 = 50 eastwards but, 10 eastwards x (-5) =(-50) eastwards = 50 westwards

2. The product of a scalar quantity with a vector quantity produces, in general, some other vector quantity.

3. If the vector is a unit vector, then the product’s magnitude is the same as the scalar’s. For example, if \(\vec{n}\) is a unit vector due east, then \(\vec{n}\) x 5 =5 due east or, \(\vec{n}\) x 5 m · s^-1 speed = 5 m · s^-1 velocity eastwards.

Product Of Two Vectors

Two vectors, when multiplied, may produce either a scalar or a vector. Accordingly, they are called scalar or dot product and vector or cross product.

Scalar Or Dot Product

Two vectors may give a scalar product like: force • displacement = work

Here, although force and displacement are both vector quantities, their product, work, is a scalar quantity. Such type of product of two vectors is called a scalar product or dot product. A dot (⋅) symbol is used between the two vectors to represent such a product mathematically.

Scalar Or Dot Product Definition: The scalar product of two vectors \(\vec{A}\) and \(\vec{B}\) is defined as \(\vec{A}\)–\(\vec{B}\) = AB cosθ, where θ is the angle between A and B.

Some Properties Of Scalar Products:

1. \(\vec{A}\) · \(\vec{A}\) = A², i.e., the scalar product of a vector with itself is the square of its magnitude.

If the same vector is taken twice, then the angle between them =0°.

∴\(\vec{A}\) ⋅ \(\vec{A}\) = AAcosθ = A²

Hence, A = \(\sqrt{\vec{A} \cdot \vec{A}}\)

This rule is often used for determining the magnitude of any vector.

Understanding Scalar Multiplication of Vectors

2. \(\vec{A}\) ⋅ \(\vec{B}\)= \(\vec{B}\) ⋅ \(\vec{A}\), i.e., scalar products are commutative.

If the angle from \(\vec{A}\) to \(\vec{B}\) is θ, the angle from \(\vec{B}\) to \(\vec{A}\) is -θ.

∴ \(\vec{A}\) ⋅ \(\vec{B}\) = ABcosθ

and \(\vec{B}\)–\(\vec{A}\) = BAcos(-θ) = ABcosθ

Hence, \(\vec{A}\) ⋅ \(\vec{B}\) = \(\vec{B}\) ⋅ \(\vec{A}\)

3. The scalar product of two orthogonal vectors, i.e., the vectors that are mutually at right angles, is zero.

If \(\vec{A}\) ⊥\(\vec{B}\), then the angle between them, θ = 90°. Hence, cosθ = 0°

∴ \(\vec{A}\) ⋅ \(\vec{B}\) = ABcosθ = 0

Alternatively, if \(\vec{A}\) ⋅ \(\vec{B}\) = 0 for two non-zero vectors \(\vec{A}\) and \(\vec{B}\), then \(\vec{A}\) ⊥ \(\vec{B}\).

4. If θ is the angle between two vectors \(\vec{A}\) and \(\vec{B}\), then \(\vec{A}\) ⋅ \(\vec{B}\) = ABcosθ or, \(\frac{\vec{A} \cdot \vec{B}}{A B}=\cos \theta\)

or, \(\theta=\cos ^{-1}\left(\frac{\vec{A} \cdot \vec{B}}{A B}\right)\)

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

5. Scalar Products Of Mutually Perpendicular Unit Vectors: This rule is often used to determine the angle between two vectors, \(\hat{i}, \hat{j}, \hat{k}\) are three unit vectors along the directions of positive x, y, z axes respectively. Each has a value of 1 and they are mutually perpendicular.

If we take the same vector twice, the angle between them will be 0°; for example, the angle between \(\hat{i} \text { and } \hat{i}\) is 0°. If we take two different vectors, then the angle between them will be 90°; for example, the angle between \(\hat{i} \text { and } \hat{J}\)is 90°.

Therefore, \(\hat{i} \cdot \hat{i}=(1)(1) \cos 0^{\circ}=1\) and \(\hat{i} \cdot \hat{j}=(1)(1) \cos 90^{\circ}=0\)

So, the general rule for the scalar product of these unit vectors \(\hat{i} \cdot \hat{i}=1, \quad \hat{j} \cdot \hat{j}=1, \quad \hat{k} \cdot \hat{k}=1\)

and \(\hat{i} \cdot \hat{j}=\hat{j} \cdot \hat{i}=\hat{j} \cdot \hat{k}=\hat{k} \cdot \hat{j}=\hat{k} \cdot \hat{i}=\hat{i} \cdot \hat{k}=0\)

Effects of Scalar Multiplication on Vector Magnitude

6. Scalar Product Of Two Vectors Using The Position Coordinates: Let \(\vec{A}=A_x \hat{i}+A_y \hat{j}+A_z \hat{k}\) and \(\vec{B}=B_x \hat{i}+B_y \hat{j}+B_z \hat{k}\) i.e., components of \(\vec{A}\) and \(\vec{B}\) along x, y and z axes are \(A_x\), \(A_y, A_z\) and \(B_x, B_y, B_z\) respectively.

Now, \(\vec{A} \cdot \vec{B}=\left(A_x \hat{i}+A_y \hat{j}+A_z \hat{k}\right) \cdot\left(B_x \hat{i}+B_y \hat{j}+B_z \hat{k}\right)\)

Following the rules of simple multiplication, and omitting the 6 terms containing \(\hat{i} \cdot \hat{j}, \hat{i} \cdot \hat{k}\), etc. that are equal to zero, \(vec{A} \cdot \vec{B}=A_x B_x+A_y B_y+A_z B_z\)

6. Dot product of a vector with \(\hat{i}, \hat{j}, \hat{\boldsymbol{k}}\) Let a vector be \(\vec{A}=A_x \hat{i}+A_y \hat{j}+A_z \hat{k}\).

So, \(\hat{i} \cdot \vec{A}=\hat{i} \cdot\left(A_x \hat{i}+A_y \hat{j}+A_z \hat{k}\right)\)

= \(A_x \text { [using, } \hat{i} \cdot \hat{i}=\cdots=1 \text {, and } \hat{i} \cdot \hat{j}=\cdots=0 \text { ] }\)

Similarly, \(\hat{j} \cdot \vec{A}=A_y\), and \(\hat{k} \cdot \vec{A}=A_z\).

Thus, the dot product of any vector with \(\hat{i}, \hat{j}, \hat{k}\) denotes the magnitudes of the x, y, and z components respectively of the vector itself. This property stands for any unit vector \(\hat{n}\):

∴ \(\hat{n} \cdot \vec{A}=\) magnitude of the component of \(\vec{A}\) along the direction of the unit vector \(\hat{n}\). i.e., the projection of \(\vec{A}\) along the direction of \(\hat{n}\).

Product Of Two Vectors Numerical Examples

Short Answer Questions on Scalar Multiplication

Example 1. Find the magnitude of the vectors \(3 \hat{i}-4 \hat{j}+12 \hat{k}\).
Solution:

⇒ \(\vec{A}=(3 \hat{i}-4 \hat{j}+12 \hat{k})\)

∴ \(\vec{A} \cdot \vec{A}=(3 \hat{i}-4 \hat{j}+12 \hat{k}) \cdot(3 \hat{i}-4 \hat{j}+12 \hat{k})\)

= \(3^2+(-4)^2+12^2=169\)

Now \(\vec{A} \cdot \vec{A}=A^2\); therefore \(A^2=169\)

∴ A = \(\sqrt{169}=13 \text { units. }\)

Example 2. Find the angle between the vectors \(\hat{i}+\hat{j} \text { and } \hat{i}-\hat{k}\)
Solution:

Let us consider, \(\vec{A}=(\hat{i}+\hat{j})\) and \(\vec{B}=(\hat{i}-\hat{k})\)

∴ \(\vec{A} \cdot \vec{A}=(\hat{i}+\hat{j}) \cdot(\hat{i}+\hat{j})=1^2+1^2=2=A^2\)

∴ A = \(sqrt{2} \text { units }\)

and \(\vec{B} \cdot \vec{B}=(\hat{i}-\hat{k}) \cdot(\hat{i}-\hat{k})=1^2+(-1)^2=2=B^2\)

∴ B = \(\sqrt{2} \text { units }\)

⇒ \(\vec{A} \cdot \vec{B}=(\hat{i}+\hat{j}) \cdot(\hat{i}-\hat{k})=\hat{i} \cdot \hat{i}-\hat{i} \cdot \hat{k}+\hat{j} \cdot \hat{i}-\hat{j} \cdot \hat{k}\)

= 1-0+0-0 = 1

If the angle between the two vectors is θ, \(\vec{A} \cdot \vec{B}=A B \cos \theta\)

or, \(\cos \theta =\frac{\vec{A} \cdot \vec{B}}{A B}=\frac{1}{\sqrt{2} \cdot \sqrt{2}}=\frac{1}{2}=\cos 60^{\circ}\)

∴ θ = 60°

Example 3. Find the angle between the vectors \(\vec{A}=2 \hat{i}+3 \hat{j}\) and \(\vec{B}=-3 \hat{i}+2 \hat{j}\)
Solution:

Let the angle between the two vectors \(\vec{A}\) and \(\vec{B}\) be θ.

Now, \(\vec{A} \cdot \vec{B}=(2 \hat{i}+3 \hat{j}) \cdot(-3 \hat{i}+2 \hat{j})=-6+6=0\)

As this dot product is zero, \(\vec{A}\) and \(\vec{B}\) are perpendicular to each other. So, the angle between them = 90°.

Example 4. Prove that die diagonals of a rhombus are perpendicular to each other.
Solution:

ABCD is a rhombus. \(\overrightarrow{A B}=\overrightarrow{D C}=\vec{a}\)

and \(\overrightarrow{A D}=\overrightarrow{B C}=\vec{b}\) the values of \(\vec{a}\) and \(\vec{b}\) are the same for a rhombus.

Let a = b = m

⇒ \(\overrightarrow{A C}=\overrightarrow{A B}+\overrightarrow{B C}=\vec{a}+\vec{b}\)

and \(\overrightarrow{B D}=\overrightarrow{B A}+\overrightarrow{A D}=-\vec{a}+\vec{b}\)

∴ \(\overrightarrow{A C} \cdot \overrightarrow{B D}=(\vec{a}+\vec{b}) \cdot(-\vec{a}+\vec{b})\)

= \(-\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}-\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}\)

= \(-a^2+\vec{a} \cdot \vec{b}-\vec{a} \cdot \vec{b}+b^2=-m^2+m^2=0\)

As the dot product is 0, the angle between the vectors = 90°

i.e., \(\overrightarrow{A C}=\overrightarrow{B D}\) are perpendicular to each other.

Example 5. \(\vec{a}, \vec{b}\) and \(\vec{c}\) are three unit vectors. Show that, \(|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2\) ≤ 9
Solution:

⇒ \(|\vec{a}+\vec{b}+\vec{c}|^2=(\vec{a}+\vec{b}+\vec{c}) \cdot(\vec{a}+\vec{b}+\vec{c})\)

= \(\vec{a} \cdot \vec{a}+\vec{b} \cdot \vec{b}+\vec{c} \cdot \vec{c}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})\)

= \(3+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})\)

Now, \(|\vec{a}+\vec{b}+\vec{c}|^2\) ≥ 0

∴ \(2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})\) ≥ -3……(1)

Also, \(|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2\)

= \(2[\vec{a} \cdot \vec{a}+\vec{b} \cdot \vec{b}+\vec{c} \cdot \vec{c}-(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})]\)

or, \(2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})\)

= \(6-\left[|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2\right]\)

From (1) and (2), we get, \(6-\left[|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2\right]\) ≥-3

or, \(\left[|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2\right]\) ≤ 9

Example 6. Find the pojetion of vector \(\vec{P}=2 \hat{i}-3 \hat{j}+6 \hat{k}\) on the vector = \(\vec{Q}=\hat{i}+2 \hat{j}+2 \hat{k}\)
Solution:

The projection of \(\vec{P}\) on \(\vec{Q}\) is

P cosθ = \(\frac{\vec{P} \cdot \vec{Q}}{|\vec{Q}|}=\frac{P_x Q_x+P_y Q_y+P_z Q_z}{\sqrt{Q_x^2+Q_y^2+Q_z^2}}\)

(where θ is the angle between \(\vec{P}\) and \(\vec{Q}\))

= \(\frac{(2 \cdot 1)+[(-3) \cdot 2]+(6 \cdot 2)}{\sqrt{1^2+2^2+2^2}}=\frac{2-6+12}{\sqrt{9}}=\frac{8}{3}\)

Example 7. A particle is moving in a curvilinear path defined by the equations x = 2t², y = t²-4t, and z = 3t-5. Find out the magnitudes of the components of velocity and acceleration along (\(\hat{i}-3 \hat{j}+2 \hat{k})\)) at time t = 1
Solution:

Velocity, \(\vec{v}=\frac{d \vec{r}}{d t}=\frac{d}{d t}(x \hat{i}+y \hat{j}+z \hat{k})\)

= \(\frac{d x}{d t} \hat{i}+\frac{d y}{d t} \hat{j}+\frac{d z}{d t} \hat{k}\)

= \(4 t \hat{i}+(2 t-4) \hat{j}+3 \hat{k}\)

∴ At t=1, \(\vec{v}=4 \hat{i}-2 \hat{j}+3 \hat{k}\)……(1)

Acceleration, \(\vec{a}=\frac{d \vec{v}}{d t}=4 \hat{i}+2 \hat{j}=\) constant ….(2)

Unit vector along \(\vec{A}=\hat{i}-3 \hat{j}+2 \hat{k}\), \(\hat{n}=\frac{\vec{A}}{A}=\frac{\hat{i}-3 \hat{j}+2 \hat{k}}{\sqrt{1^2+(-3)^2+2^2}}=\frac{1}{\sqrt{14}}(\hat{i}-3 \hat{j}+2 \hat{k})\)….(3)

∴ At t=1, the component of \(\vec{v}\) along \(\vec{A}\) is, from (1) and (3),

∴ \(\nu_A =\hat{n} \cdot \vec{v}=\frac{1}{\sqrt{14}}(\hat{i}-3 \hat{j}+2 \hat{k}) \cdot(4 \hat{i}-2 \hat{j}+3 \hat{k})\)

= \(\frac{1}{\sqrt{14}} \times 16=\frac{8 \sqrt{14}}{7}\)

The component of \(\vec{a}\) along \(\vec{A}\) is, from (2) and (3), \(a_A=\hat{n} \cdot \vec{a}=\frac{1}{\sqrt{14}}(\hat{i}-3 \hat{j}+2 \hat{k}) \cdot(4 \hat{i}+2 \hat{j})\)

= \(\frac{1}{\sqrt{14}} \times(-2)=-\frac{\sqrt{14}}{7}\)

∴ \(\left|a_A\right|=\frac{\sqrt{14}}{7}\)

Example 8. Prove that a right-angled triangle can be formed using the vectors \(\vec{A}=\hat{i}-3 \hat{j}+5 \hat{k}, \quad \vec{B}=2 \hat{i}+\hat{j}-4 \hat{k}\) and \(\vec{C}=3 \hat{i}-2 \hat{j}+\hat{k}\)
Solution:

∴ \(\vec{A}+\vec{B}=(\hat{i}-3 \hat{j}+5 \hat{k})+(2 \hat{i}+\hat{j}-4 \hat{k})\)

= \(3 \hat{i}-2 \hat{j}+\hat{k}=\vec{C}\)

i.e., \(\vec{A}+\vec{B}=\vec{C}\)

So, \(\vec{A}, \vec{B} and \vec{C}\) can form a triangle.

Again, \(\vec{B} \cdot \vec{C}=(2 \hat{i}+\hat{j}-4 \hat{k}) \cdot(3 \hat{i}-2 \hat{j}+\hat{k})=6-2-4=0\)

∴ In the triangle formed by \(\vec{A}, \vec{B}\) and \(\vec{C}\), the angle between \(\vec{B}\) and \(\vec{C}\) is \(90^{\circ}\). Thus, it is a right-angled triangle.

[The dot product \(\vec{B} \cdot \vec{C}\) has been calculated, because \(\vec{A}\) forms the largest side of the triangle. An angle of \(90^{\circ}\) is always opposite to the largest side. In this example, \(A=\sqrt{35}, B=\sqrt{21}\) and \(C=\sqrt{14}\), i.e., A>B, C]

Applications of Scalar and Vector Multiplication

Vector Or Cross Product: The product of two vectors can be a vector. For example, angular velocity x position vector = linear velocity

Here, both angular velocity and position vector are vector quantities, and their product, linear velocity, is also a vector quantity. Such type of product of two vectors is called a vector product or cross product and is represented by putting a cross (x) sign between the two vectors.

Vector Or Cross Product Definition: The vector product or cross product of two vectors \(\vec{a}\) and \(\vec{b}\) is defined as \(\vec{a}\) x \(\vec{b}\)= ab sin\(\theta \hat{n}\), where d is the angle from \(\vec{a}\) to \(\vec{b}\), and \(\hat{n}\) is a unit vector. The direction of \(\hat{n}\) is the direction of advance of a right-handed screw when it is rotated from \(\vec{a}\) and \(\vec{b}\).

Here, \(\hat{n} \perp \vec{a} \text { and } \hat{n} \perp \vec{b}\). So the product \(\vec{a} \times \vec{b}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\). The unit vector \(\hat{n}\) has a magnitude, [latx]|\hat{n}|[/latex] = 1.

So the magnitude of the vector product is \(|\vec{a} \times \vec{b}|=a b \sin \theta\)

Some Properties Of Vector Products:

1. \(\vec{A}\) x \(\vec{A}\) = \(\vec{0}\), i.e., vector product of a vector with itself is a null vector.

The angle between the same vector taken twice is zero.

∴ The magnitude of \(\vec{A}\)x \(\vec{A}\) = \(|\vec{A} \times \vec{A}|=A A \sin 0^{\circ}=0\)

Hence, \(\vec{A}\) x \(\vec{A}\) = \(\vec{0}\)

2. \(\vec{A}\) x \(\vec{B}\) = –\(\vec{B}\) x \(\vec{A}\), i.e., vector product is not commutative.

If the angle from \(\vec{A}\) to \(\vec{B}\) is θ, the angle from \(\vec{B}\) to \(\vec{A}\) is -θ. As sin(-θ) = -sinθ, we get, \(\vec{A}\) x \(\vec{B}\) = –\(\vec{B}\) x \(\vec{A}\).

3. Vector Product Of Two Mutually Perpendicular Vectors: If \(\vec{A}\) ⊥ \(\vec{B}\), then, the magnitude of \(\vec{A}\) x \(\vec{B}\) = \(|\vec{A} \times \vec{B}|\) = ABsin90° = AB, and the direction of \(\vec{A}\) x \(\vec{B}\) is perpendicular to both of \(\vec{A}\) and \(\vec{B}\).

Hence, \(\vec{A}\), \(\vec{B}\) and (\(\vec{A} \times \vec{B}\))—all these three vectors are mutually perpendicular, i.e., if \(\vec{A}\) and \(\vec{B}\) are along x and y axes respectively, then their product \(\vec{A} \times \vec{B}\) will be along z -axis.

4. Vector Products Of Mutually Perpendicular Unit Vectors: \(\hat{i}, \hat{j}, \hat{k}\) are the unit vectors along positive x, y, and z axes respectively. The magnitude of each is 1.

As \(\vec{A}\) x \(\vec{A}\) = 0, we get, \(\hat{i} \times \hat{i}=\hat{j} \times \hat{j}=\hat{k} \times \hat{k}=0\)

Again, the magnitude of \(\hat{i} \times \hat{j}=|\hat{i} \times \hat{j}|=(1)(1) \sin 90^{\circ}=1\)

According to the right-handed corkscrew rule, the direction of \(\hat{i}\) x \(\hat{j}\) is along the z-axis. Again, the unit vector along the z-axis is \(\hat{k}\). Hence, \(\hat{i}\) x \(\hat{j}\) = \(\hat{k}\).

Similarly, \(\hat{i} \times \hat{j}=\hat{k}, \hat{j} \times \hat{k}=\hat{i}, \hat{k} \times \hat{i}=\hat{j}\)

Again, since \(\vec{A} \times \vec{B}=-\vec{B} \times \vec{A}, \hat{j} \times \hat{i}=-\hat{k}, \hat{k} \times \hat{j}=-\hat{i}\), \(\hat{i} \times \hat{k}=-\hat{j}\)

Here, the cyclic order of \(\hat{i}, \hat{j}, \hat{k}\) is important. If this cyclic order is maintained, the product of the first two vectors leads to the third with a positive sign, whereas if the cyclic order is not maintained, the product becomes the third vector with a negative sign.

Example: \(\hat{i} \times \hat{j}=\hat{k}, \hat{j} \times \hat{k}=\hat{i}, \hat{k} \times \hat{i}=\hat{j}, \hat{i} \times \hat{k}=-\hat{j}, \hat{k} \times \hat{j}=-\hat{i}, \hat{j} \times \hat{i}=-\hat{k}\)

5. Vector Product In Terms Of Positional Coordinates: Let \(\vec{A}=A_x \hat{i}+A_y \hat{j}+A_z \hat{k}\) and \(\vec{B}=B_x \hat{i}+B_y \hat{j}+B_z \hat{k}\).

∴ \(\vec{A} \times \vec{B}=\left(A_x \hat{i}+A_y \hat{j}+A_z \hat{k}\right) \times\left(B_x \hat{i}+B_y \hat{j}+B_z \hat{k}\right)\)

Out of the 9 terms of this product, the terms containing \(\hat{i} \times \hat{i}, \hat{j} \times \hat{j}\) and \(\hat{k} \times \hat{k}\) are zero. Writing the remaining 6 terms using the relations like \(\hat{i} \times \hat{j}=\hat{k}\),

we get, \(\vec{A} \times \vec{B}=\hat{i}\left(A_y B_z-A_z B_y\right)+\hat{j}\left(A_z B_x-A_x B_z\right)\) + \(\hat{k}\left(A_x B_y-A_y B_x\right)\)

It is convenient to write this product in a determinant form:

⇒ \(\vec{A} \times \vec{B}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
A_x & A_y & A_z \\
B_x & B_y & B_z
\end{array}\right|\)

Area Of Triangle: If two vectors \(\vec{a}\) and \(\vec{b}\) represent the two sides of a triangle then half of the magnitude of their cross-product will give the area of the triangle.

In ΔABC \(\overrightarrow{C B}=\vec{a}, \overrightarrow{C A}=\vec{b}\) and the angle between them = C.

Height of the triangle, AD = \(A C \cdot \frac{A D}{A C}=b \sin C\)

∴ Area of \(\triangle A B C=\frac{1}{2} \times(\text { base }) \times(\text { height })\)

= \(\frac{1}{2} \times(B C) \times(A D)=\frac{1}{2} a b \sin C\)

= \(\frac{1}{2}|\vec{a} \times \vec{b}|\)

Area Of Parallelogram: ABCD is a parallelogram. Let us denote the two adjacent sides BC and BA by two vectors, \(\vec{a}=\overrightarrow{B C}\) and \(\vec{b}=\overrightarrow{B A}\); θ = angle between the vectors \(\vec{a}\) and \(\vec{b}\).

Now, we draw AD ⊥ BC.

From AD = AB \(\frac{A D }{A B}\) = b sinθ. So, the area of the parallelogram, S = base x height = (BC)(AD) = \(a b \sin \theta=|\vec{a} \times \vec{b}|\)

This means that the magnitude of the cross product of two vectors is geometrically represented by the area of a parallelogram whose adjacent sides stand for the two given vectors.

Surace Of A Vector: The preceding relation S = \(|\vec{a} \times \vec{b}|\) hints at the possibility of writing S as a vector product, \(\vec{S}\) = \(\vec{a}\) x \(\vec{b}\)

• But it would mean that the area S is a vector quantity. Now, we have to check whether it can be true. For this, let us consider a plane mirror in a room. To describe its effect precisely, we have to specify not only its surface area but also its orientation, i.e., how it is placed in the room.
• A statement like a mirror of area 600 cm² is not sufficient; it should be stated like ‘a mirror of area 600 cm” facing north’.
• This confirms that a plane surface should indeed be treated as a vector quantity; its magnitude equals the area of the surface, and its direction is perpendicular to the surface. A curved surface does not have a definite direction, and cannot be treated as a vector.
• However, an infinitesimally small area ds on this surface is effectively plane, and may be treated as a vector \(\overrightarrow{d s}\). This concept is widely used in different branches of physics.
• In this context, the relation \(\vec{S}\) = \(\vec{a}\) x \(\vec{b}\) is valid for a parallelogram. Consequently, the cross product of two vectors is geometrically represented by the area vector of a parallelogram whose adjacent sides stand for the two given vectors.

Vector Or Cross Product Numerical Examples

Example 1. \(\vec{A}=2 \hat{i}+3 \hat{j}+4 \hat{k}\) and \(\vec{B}=\hat{i}-\hat{j}+\hat{k}\) are two vectors. Find \(\vec{A} \times \vec{B}\).
Solution:

⇒ \(\vec{A} \times \vec{B}=\hat{i}\left(A_y B_z-A_z B_y\right)+\hat{j}\left(A_z B_x-A_x B_z\right)\)

+ \(\hat{k}\left(A_x B_y-A_y B_x\right)\)

= \(\hat{i}\{3 \cdot 1-4 \cdot(-1)\}+\hat{j}(4 \cdot 1-2 \cdot 1)\) + \(\hat{k} \hat{i}+2 \hat{j}-5 \hat{k}\)

Alternative method:

⇒ \(\vec{A} \times \vec{B} =\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 3 & 4 \\
1 & -1 & 1
\end{array}\right|=\hat{i}(3+4)-\hat{j}(2-4)+\hat{k}(-2-3)\)

= \(7 \hat{i}+2 \hat{j}-5 \hat{k}\)

Example 2. Using the vector method in a triangle, prove that,

1. \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\) and
2. \(\cos A=\frac{b^2+c^2-a^2}{2 b c}\).

Solution: According to \(\vec{a}+\vec{c}=\vec{b}\)

1. \(\vec{a} \times \vec{b}=\vec{a} \times(\vec{a}+\vec{c})=\vec{a} \times \vec{c}\) (because \(\vec{a} \times \vec{a}=0\))

∴ \(|\vec{a} \times \vec{b}|=|\vec{a} \times \vec{c}|\)

or, \(a b \sin C=a c \sin \theta\)

(where \(\theta\) is the angle between \(\vec{a}\) and \(\vec{c}\))

or, \(b \sin C=c \sin \left(180^{\circ}-B\right)\)

or, \(b \sin C=c \sin B or, \frac{b}{\sin B}=\frac{c}{\sin C}\)

Proceeding in the same way, we get \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\)

2. \(b^2=\vec{b} \cdot \vec{b}=(\vec{a}+\vec{c}) \cdot(\vec{a}+\vec{c})=\vec{a} \cdot \vec{a}+\vec{c} \cdot \vec{c}+2 \vec{a} \cdot \vec{c}\)

= \(a^2+c^2+2 a c \cos \theta\)

= \(a^2+c^2+2 a c \cos \left(180^{\circ}-B\right)\)

= \(a^2+c^2-2 a c \cos B\)

Similarly, \(a^2=b^2+c^2-2 b c \cos A\)

or, \(\cos A=\frac{b^2+c^2-a^2}{2 b c}\).

Example 3. If the two diagonals of a parallelogram are given by \(\vec{R}_1=3 \hat{i}-2 \hat{j}+7 \hat{k}\) and \(\vec{R}_2=5 \hat{i}+6 \hat{j}-3 \hat{k}\), find out the area of this parallelogram.
Solution:

Let, the two adjacent sides of the parallelogram \(\vec{A}\) and \(\vec{B}\).

Then, \(\vec{R}_1=\vec{A}+\vec{B} \text { and } \overrightarrow{R_2}=\vec{A}-\vec{B}\)

Now, \((\vec{A}+\vec{B}) \times(\vec{A}-\vec{B})=\vec{A} \times \vec{A}-\vec{A} \times \vec{B}+\vec{B} \times \vec{A}-\vec{B} \times \vec{B}\)

= \(0-\vec{A} \times \vec{B}-\vec{A} \times \vec{B}-0=-2(\vec{A} \times \vec{B})\)

From this given data, \((\vec{A}+\vec{B}) \times(\vec{A}-\vec{B})=\vec{R}_1 \times \vec{R}_2\)

= \((3 \hat{i}-2 \hat{j}+7 \hat{k}) \times(5 \hat{i}+6 \hat{j}-3 \hat{k})\)

= \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & -2 & 7 \\
5 & 6 & -3
\end{array}\right|=\hat{i}[(-2)(-3)-7 \times 6]+\hat{j}[7 \times 5-3(-3)]\) + \(\hat{k}[3 \times 6-5(-2)]\)

= \(-36 \hat{i}+44 \hat{j}+28 \hat{k}\)

On comparison, \(-2(\vec{A} \times \vec{B})=-36 \hat{i}+44 \hat{j}+28 \hat{k}\)

or, \(\vec{A} \times \vec{B}=18 \hat{i}-22 \hat{j}-14 \hat{k}\)

Here, \(\vec{A} \times \vec{B}=\vec{S}=\) area vector for the parallelogram.

∴ Its area \(|\vec{S}|=|\vec{A} \times \vec{B}|=\sqrt{18^2+(-22)^2+(-14)^2}\)

=31.7 square units.

Example 4. Find the angle betwen force \(\vec{F}=(3 \hat{i}+4 \hat{j}-5 \hat{k})\) and displacement \(\vec{d}=(5 \hat{i}+4 \hat{j}-3 \hat{k})\). Also find the projection of \(\vec{F}\) on \(\vec{d}\).
Solution:

Let \(\theta\) be the angle between the vectors \(\vec{F}\) and \(\vec{d}\). Then by the definition of the scalar product of two vectors, we nave \(\vec{F} \cdot \vec{d}=F d \cos \theta \quad \text { or, } \cos \theta=\frac{\vec{F} \cdot \vec{d}}{F d}\)

Now, \(|\vec{F}|=\sqrt{3^2+4^2+(-5)^2}=\sqrt{50}\)

and \(\vec{d}=\sqrt{5^2+4^2+(-3)^2}=\sqrt{50}\)

∴ \(\vec{F} \cdot \vec{d}=(3 \hat{i}+4 \hat{j}-5 \hat{k}) \cdot(5 \hat{i}+4 \hat{j}-3 \hat{k})\)

= \((3 \cdot 5)+(4 \cdot 4)+(-5) \cdot(-3)\)

= \(15+16+15=46\)

∴ \(\cos \theta=\frac{46}{\sqrt{50} \times \sqrt{50}}=\frac{46}{50}=0.92\)

or, \(\theta=\cos ^{-1} 0.92 \approx 23^{\circ}\)

Hence, the projection of \(\vec{F}\) on \(\vec{d}\) is \(F \cos \theta=\sqrt{50} \times(0.92)=7.07 \times 0.92=6.50 \text { units. }\)

Example 5. Prove that \(|\vec{P} \times \vec{Q}|^2=|\vec{P}|^2|\vec{Q}|^2-|\vec{P} \cdot \vec{Q}|^2\).
Solution:

∴ \(|\vec{P} \times \vec{Q}|^2+|\vec{P} \cdot \vec{Q}|^2=(P Q \sin \theta)^2+(P Q \cos \theta)^2\)

(θ= angle from \(\vec{P} \text { to } \vec{Q}\))

= \(P^2 Q^2\left(\sin ^2 \theta+\cos ^2 \theta\right)=P^2 Q^2\)

= \(|\vec{P}|^2|\vec{Q}|^2\)

∴ \(|\vec{P} \times \vec{Q}|^2=|\vec{P}|^2|\vec{Q}|^2-|\vec{P} \cdot \vec{Q}|^2\).

Example 6. The resultant of two vectors A and B acting through a point O Is R. A certain straight line Intersects the lines representing the vectors \(\vec{A}\), \(\vec{B}\) and \(\vec{R}\) at points P, Q, and S, respectively. Prove that \(\frac{A}{O P}+\frac{B}{O Q}=\frac{R}{O S}\)
Solution:

Let, the unit vectors along \(\vec{A}, \vec{B}\) and \(\vec{R}\) be \(\hat{a}\), and \(\hat{c}\) respectively.

Then \(\vec{A}=A \hat{a}, \vec{B}=B \hat{b}, \vec{R}=R \hat{c}\)

and \(A \hat{a}+B \hat{b}=R \hat{c}\)

or, \(A \frac{\overrightarrow{O P}}{O P}+B \frac{\overrightarrow{O Q}}{O Q}=R \frac{\overrightarrow{O S}}{O S}\)

Graphical Representation of Scalar Multiplication

If \(\hat{n}\) is the unit vector along the line C D, \(\overrightarrow{O P} \times \hat{n}=O P \sin \alpha \hat{m}=O N \hat{m}\): \(\overrightarrow{O Q} \times \hat{n}=O N \hat{m} ; \overrightarrow{O S} \times \hat{n}=O N \hat{m}\)

where \(\hat{m}\) is a unit vector perpendicular to the plane of the vectors.

Then, \(\frac{A}{O P} \overrightarrow{O P} \times \hat{n}+\frac{B}{O Q} \overrightarrow{O Q} \times \hat{n}=\frac{R}{O S} \overrightarrow{O S} \times \hat{n}\)

or, \(\frac{A}{O P} O N \hat{m}+\frac{B}{O Q} O N \hat{m}=\frac{R}{O S} O N \hat{m}\)

or, \(\frac{A}{O P}+\frac{B}{O Q}=\frac{R}{O S}\).

Relative Velocity Of Rain With Respect To A Moving Observer

WBBSE Class 11 Relative Velocity of Rain Notes

In the absence of wind, rain falls vertically. But when a man moves forward during a rainfall, he has to slant his umbrella in front of him. This is due to the fact that the relative velocity of the falling rain with respect to the man makes an angle with the vertical.

Let the actual downward velocity of rain = \(\vec{v}\); the horizontal velocity of the man = \(\vec{u}\).

Angle of Rain with Respect to a Moving Observer

Hence, the relative velocity of rain with respect to the man, \(\vec{w}\) = \(\vec{v}\) – \(\vec{u}\)

∴ w = \(\sqrt{v^2+u^2}\)….(1)

If the relative velocity makes an angle θ with the vertical, then, tanθ = 2……(2)

From equation (1) the magnitude of the relative velocity of rain and from equation (2) its direction can be determined. As the relative velocity of rain makes an angle θ with the vertical, it seems that rain comes down at an angle when we move. This is why, while riding a cycle, one has to hold an umbrella along CO to save himself from getting wet.

For the same reason, a speeding vehicle receives more rain on the front windscreen than on the rear one. Similarly, a person sitting inside a moving train feels that the raindrops follow a slanted path.

Read and Learn More: Class 11 Physics Notes

Relative Velocity Of Rain With Respect To A Moving Observer Numerical Examples

Short Answer Questions on Relative Velocity of Rain

Example 1. A man is walking on a horizontal road at 3 km · h^-1 t while rain is falling vertically with a velocity of 4 km · h^-1. Find the magnitude and direction of the i velocity of rain with respect to the man.
Solution:

Velocity of the man, u = 3 km· h^-1; velocity of rainfall, v = 4 km · h^-1

Since, ∠AOB = 90°, the magnitude of the relative velocity of rain with respect to the man,

w = \(\sqrt{u^2+v^2}=\sqrt{3^2+4^2}=5 \mathrm{~km} \cdot \mathrm{h}^{-1}\)

If \(\angle B O C=\theta, \tan \theta=\frac{u}{v}\) or, \(\theta=\tan ^{-1} \frac{u}{v}\)

∴ \(\theta=\tan ^{-1} \frac{3}{4}=36.9^{\circ}\)

Hence, the velocity of rain with respect to the man makes an angle of 36.9° with the vertical.

Example 2. To a man, walking on a horizontal path at 2 km · h^-1, rain appears to fall vertically at 2 km · h^-1. Find the magnitude and the direction of the actual velocity of rainfall.
Solution:

⇒ \(\vec{u}\) = velocity of the man, \(\vec{v}\)= actual velocity of the raindrops ; so, the relative velocity of the rain¬drops with respect to the man

⇒ \(\vec{w}=\vec{v}-\vec{u} \quad \text { or, } \vec{v}=\vec{u}+\vec{w}\)

Therefore, the resultant of the velocity of the man (\(\vec{u}\)) and the relative velocity of rain (\(\vec{w}\)) will give the real velocity of rain (\(\vec{v}\)).

From the figure, \(v=\sqrt{2^2+2^2}=2 \sqrt{2} \mathrm{~km} \cdot \mathrm{h}^{-1}\)

If the actual direction of rain is inclined at an angle θ with the vertical, then

tanθ = \(\frac{u}{w}\) = \(\frac{2}{2}\) = 1

∴ θ = 45°.

Example 3. To a car driver moving at 40 km · h^-1 towards the south, the wind appears to blow towards the east. When the speed of the car is reduced to 20km· h^-1 wind appears to blow from the north-west. Find the magnitude and direction of the actual velocity of the wind.
Solution:

Let us choose: the x-axis along the east and the y-axis along the north.

Initial velocity of the car, \(\vec{u}_1=-40 \hat{j} \mathrm{~km} \cdot \mathrm{h}^{-1}\)

Final velocity of the car, \(\vec{u}_2=-20 \hat{j} \mathrm{~km} \cdot \mathrm{h}^{-1}\)

Let, \(\vec{v}\) actual velocity of the wind.

Then, the relative velocity of the wind with respect to the car, \(\vec{w}=\vec{v}-\vec{u} \quad \text { or, } \vec{v}=\vec{u}+\vec{w}\)

Initially, \(\vec{w}_1=w_1 \hat{i}\); so \(\vec{v}=\vec{u}_1+\vec{w}_1=w_1 \hat{i}-40 \hat{j}\)…..(1)

Finally, \(\vec{w}_2=w_2 \cos 45^{\circ} \hat{i}-w_2 \sin 45^{\circ} \hat{j}\) (as it is towards south-east)

= \(\frac{w_2}{\sqrt{2}} \hat{i}-\frac{w_2}{\sqrt{2}} \hat{j}\)

∴ \(\vec{v}=\vec{u}_2+\vec{w}_2=\frac{w_2}{\sqrt{2}} \hat{i}-\left(\frac{w_2}{\sqrt{2}}+20\right) \hat{j}\)

Comparing the coefficients of \(\hat{j}\) in (1) and (2), \(\frac{w_2}{\sqrt{2}}+20=40 \text { or, } \frac{w_2}{\sqrt{2}}=20\)

Then from (2), \(\vec{v}=20 \hat{i}-(20+20) \hat{j}=(20 \hat{i}-40 \hat{j}) \mathrm{km} \cdot \mathrm{h}^{-1}\) (between east and south)

Magnitude of \(\vec{v}=v=\sqrt{(20)^2+(40)^2}=20 \sqrt{1+4}\)

= \(20 \sqrt{5} \mathrm{~km} \cdot \mathrm{h}^{-1}\)

If \(\vec{v}\) is inclined at an angle θ with east, then \(\tan \theta=\frac{-40}{20}=-2=\tan \left(-63.4^{\circ}\right) \quad \text { or, } \theta=-63.4^{\circ}\)

So, the wind velocity is at an angle of \(63.4^{\circ}\) south of east.

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Application Of Relative Velocity In One An Two Dimensional Motion

Applications of Relative Velocity in Rain Problems

Motion Of A Boat On A River

Let \(\vec{v}\) be the speed of a boat in still water. If the river current has a velocity of u along its length, the boat is subjected to this velocity too. Therefore, it will move with a resultant velocity, \(\vec{w}\) = \(\vec{v}\) + \(\vec{u}\), and the motion of the boat will, in general, be oblique.

Shortest Distance Traversed In Crossing The River: To find the direction of \(\vec{v}\) such that the distance traveled by the boat is the least, let the angle made by \(\vec{v}\) with the shortest distance of crossing be θ. Resolving the components of \(\vec{v}\) in two mutually perpendicular directions

1. Against the current \(\vec{u}\) and
2. In the direction of the shortest distance AB, we get v sinθ and v cosθ respectively. The component v sinθ balances the effect of the current u.

Therefore, \(u=v \sin \theta \quad \text { or, } \theta=\sin ^{-1} \frac{u}{v}\)

Understanding Relative Velocity of Rain

Component v cosθ enables the boat to cross the river along AB and the time t required for that is

∴ t = \(\frac{l}{v \cos \theta}\)……(1)

The vector vcosθ is actually \(\vec{v}\) + \(\vec{u}\) = \(\vec{w}\) = the resultant velocity. So, to cross along the shortest distance, the boat should be steered in such a way that the resultant velocity is directly across the river.

w = \(\sqrt{v^2-u^2}\), and the angle of steering is, \(\theta=\sin ^{-1} \frac{u}{v}\). If l be the width of the river, the time taken to cross it is,

t = \(\frac{l}{w}=\frac{l}{\sqrt{v^2-u^2}}\)….(2)

Crossing The River In Minimum Time: In equation (1). t will be minimum when the value of cos# is maximum, i.e., cosθ = 1 or, θ = 0°. Hence, to cross the river in minimum time, the boat should be driven along the width of the river.

In that case, the situation is as shown. The boat crosses obliquely along AB with the resultant velocity w. The time required to cross the river is

t’ = \(\frac{A B}{w}=\frac{A C}{v}=\frac{l}{v}\)…..(3)

This is exactly the time required when no current is present. The longitudinal displacement of the boat due to oblique crossing is, \(C B=A C \tan \alpha=\frac{l u}{v}\)….(4)

If there is no current, u=0. Then from equations (2), (3), and (4), we get, t = \(t^{\prime}=\frac{l}{v} \quad \text { and } C B=0\)

Motion Of A Boat On A River Numerical Examples

Example 1. The velocity of a boat in still water is 5 km · h^-1. It takes 15 min to cross a river along the width. The river is 1 km wide. Find the velocity of the current.
Solution:

The boat crosses a river of width 1 km in 15 min or in \(\frac{1}{4}\) h.

The resultant of the velocity of the boat and that of the current is \(\vec{w}\)(say) and

∴ w = \(\frac{1}{1 / 4}=4 \mathrm{~km} \cdot \mathrm{h}^{-1} \text {. }\)

Let the velocity of the boat in still water be \(\vec{v}\) and the velocity of the current be \(\vec{u}\).

Frome figure, \(v^2=u^2+w^2\)

or, \(u^2=v^2-w^2\)

or, \(u=\sqrt{v^2-w^2}=\sqrt{5^2-4^2}=3 \mathrm{~km} \cdot \mathrm{h}^{-1}\)

Example 2. A man can reach the point directly opposite on the other bank of a river by swimming across the river in time t₁ and crossing the same distance in time t₂ while swimming along the current. If the velocity of the man in still water is v and the velocity of the water current is u, find the ratio between t₁ and t₂.
Solution:

Let the resultant velocity with which the man swims across the river be w. Hence, w = \(\sqrt{v^2-u^2}\) and therefore, time required to cross the river,

∴ \(t_1=\frac{l}{\sqrt{v^2-u^2}}\)…..(1)

Where l is the width of the river

When the man swims in the direction of the current, the resultant velocity, w’ = v+u, and time required to cross the same distance,

∴ \(t_2=\frac{l}{v+u}\)…….(2)

From (1) and (2), \(\frac{t_1}{t_2}=\frac{v+u}{\sqrt{v^2-u^2}}=\frac{v+u}{\sqrt{v-u} \cdot \sqrt{v+u}}=\sqrt{\frac{v+u}{v-u}}\)

Real-Life Examples of Rain and Relative Velocity

Example 3. Two boats, each with a velocity 8 km · h^-1, attempt to cross a river of width 800 m. The velocity of river current is 5 km · h^-1. One of the boats crosses the river following the shortest path and the other follows the route in which the time taken is minimum. If they start simultaneously, what would be the time difference between their arrivals at the other bank?
Solution:

In order to follow the shortest path, the boat should set itself at a particular angle with the current such that the boat’s resultant velocity is perpendicular to the direction of the current. Hence, the magnitude of the resultant velocity of the boat,

w = \(\sqrt{v^2-u^2}\)

and the time required to cross the river along the shortest path, \(t_1=\frac{x}{w}=\frac{x}{\sqrt{v^2-u^2}}=\frac{0.8}{\sqrt{8^2-5^2}}=0.128 \mathrm{~h}=7.69 \mathrm{~min}\)

[x = 800 m = 0.8 km, v = 8 km · h^-1 and u = 5 km · h^-1]

To cross in minimum time, the boat should travel at right angles to the current and the time required is

Required time difference, t₁ – t₂ = 7.69 – 6 = 1.69 min

Example 4. A person can swim at 4 km · h^-1 in still water. At what angle should he set himself to cross the river in a direction perpendicular to the river current of velocity 2 km · h^-1?
Solution:

Let the velocity of the person be \(\vec{v}\), velocity of the current be \(\vec{u}\) and the resultant of \(\vec{u}\) and \(\vec{v}\) be \(\vec{w}\). The person crosses the river along its width from one bank to the other.

Hence, \(\vec{w}\) and \(\vec{u}\) are perpendicular to each other. If the angle between \(\vec{v}\) and \(\vec{w}\) is θ, then \(\sin \theta=\frac{u}{v}\)= \(\frac{2}{4}=\frac{1}{2}\) or, \(\theta=30^{\circ} \text {. }\)

Hence, the person has to swim at an inclination of 30° with the width of the river, or at an angle of (90° + 30°) or, 120° with the direction of the current as shown.

Example 5. The width of a river is D. A man can cross the river in time t₁ in the absence of any river current. But in the presence of a certain river current, the man takes a time t₂ to cross the river directly. Show that the velocity of the current is v = \(D \sqrt{\frac{1}{t_1^2}}-{\frac{1}{t_2^2}}\)
Solution:

Let u = velocity of the man on the river. \(t_1=\frac{D}{u} \quad \text { or, } u=\frac{D}{t_1}\)….(1)

In the presence of a river current of velocity v, the effective velocity of the man will be the resultant w of u and v. \(\vec{w} = \vec{u}+\vec{v}\)

To cross the river directly, the man has to swim in such a direction that the resultant w is across the width of the river.

Clearly, w = \(\sqrt{u^2-v^2}\)

and the time taken to cross the river is \(t_2=\frac{D}{w}=\frac{D}{\sqrt{u^2-v^2}}\).

∴ \(t_2^2=\frac{D^2}{u^2-v^2}\) or, \(u^2=v^2+\frac{D^2}{t_2^2}\)……(2)

From (1) and (2), \(\frac{D^2}{t_1^2}=v^2+\frac{D^2}{t_2^2}\) or, \(v^2=D^2\left(\frac{1}{t_1^2}-\frac{1}{t_2^2}\right)\)

∴ v = \(D \sqrt{\frac{1}{t_1^2-\frac{1}{t_2}}}\)

Relative Velocity And Relative Acceleration

WBBSE Class 11 Relative Velocity Notes

Relative Velocity: When we consider the motion of a body either on the earth’s surface or close to it, it is assumed that the earth is a stationary frame of reference. The velocity of a body with respect to a stationary observer on the earth’s surface is considered as the actual velocity.

• Often we consider the motion of a body with respect to another one.
• A passenger in a moving train observes the surrounding objects in apparent motion even if they actually are stationary. When both of two trains are moving side by side, an observer in one of them would not be able to perceive the actual motion of the other—only an apparent motion relative to himself would be observed.
• In such cases, it is necessary to consider the idea of relative motion or relative velocity between two bodies where the reference point may not be at rest.
• We already know that there is nothing like absolute rest or absolute motion. So there should be nothing like absolute velocity—all observed velocities are effectively relative.

Relative Velocity Definition: Relative velocity is defined as the apparent velocity of a body with respect to another body that may be in motion.

Calculation Of Relative Velocity:

Case 1: The observer and the object move in the same direction: Let us assume that the velocity of the observer is u and that of the object is v. If they move in the same direction, the object would seem to be slower to the observer.

The relative velocity of the object with respect to the observer is, w = actual velocity of the object – actual velocity of the observer

= V— u…..(1)

Case 2: The observer and the object move in opposite directions: Let the velocity of the observer be u and that of the object moving in the opposite direction be -v.

Read and Learn More: Class 11 Physics Notes

Here again, the relative velocity of the object with respect to the observer is,

w = actual velocity of the object – actual velocity of the observer

= – v- u = -(v+ u)….(2)

The magnitude of the relative velocity of the object, (u + v) will always be greater than the actual velocities (u or v).

Case 3: The observer and the object move in different directions: If the object and the observer are moving in different directions that are not in the same line or plane, then again equation (1) is used to find the relative velocity. However, the subtraction has to be done using vector algebra.

In general, if \(\vec{u}\) = velocity of an object,

and \(\vec{v}\) = velocity of an observer, then, the relative velocity of the object with respect to the observer is, \(\vec{w}=\vec{v}-\vec{u}\)….(3)

We may rewrite equation (3) as \(\vec{w}=\vec{v}-\vec{u}=\vec{v}+(-\vec{u})\) = the resultant of the vectors \(\vec{v}\) and —\(\vec{u}\).

If we interchange the observer and the object keeping the velocities unchanged, then the relative velocity is given by, \(\vec{w}^{\prime}=\vec{u}-\vec{v}=-(\vec{v}-\vec{u})=-\vec{w}\)

As \(\vec{w}^{\prime}=-\vec{w}\), these two relative velocities Eire eqeal but opposite. Thus, out of two trains running on parallel tracks if the second train moves forward relative to the first one, then the first train will be observed to move backward relative to the second.

Since velocity is a vector quantity, the relative velocity of a body involves a vector subtraction and can be found out by the vector addition of \(\vec{v}\) and \(-\vec{u}\).

For example, a train moving due east with a velocity u sees another train moving due west with the same velocity. Then, relative velocity = \(\vec{u}-(-\vec{u})=2 \vec{u}\)and they will appear to approach each other at twice the speed.

If both the trains are moving in the same direction, relative velocity = \((\vec{u}-\vec{u})\) = 0. The trains will appear to be at rest with respect to each other.

Understanding Relative Velocity in Physics

Magnitude And Direction Of Relative Velocity: Let us consider that a body A (for example, an observer) is moving with velocity \(\vec{u}\) along OX and B, and another body is moving with velocity \(\vec{v}\) along OY as shown. Taking O as origin, \(\overrightarrow{O A}=\vec{u}\), \(\overrightarrow{O B}=\vec{v}\) and the angle between them is ∠AOB = α.

To compute the relative velocity of B with respect to A, OA’ is drawn so that AO = OA’. Then \(\overrightarrow{O A^{\prime}}\) is the opposite vector of \(\overrightarrow{O A}\), i.e., \(\overrightarrow{O A^{\prime}}\) = –\(\vec{u}\). Completing the parallelogram OA’CB, the diagonal \(\overrightarrow{O C}\) is drawn. OC represents the resultant of \(\vec{v}\) and –\(\vec{u}\), i.e., the relative velocity of B with respect to A.

∴ \(\overrightarrow{O A}\) = \(\vec{w}\)

And OC  = \(|\vec{w}|=\sqrt{u^2+v^2+2 u v \cos \left(180^{\circ}-\alpha\right)}\)

= \(\sqrt{u^2+v^2-2 u v \cos \alpha}\)…..(4)

If the relative velocity \(\overrightarrow{O C}\) is inclined to \(\overrightarrow{O B}\) at an angle θ then,

tanθ \(=\frac{u \sin \left(180^{\circ}-\alpha\right)}{v+u \cos \left(180^{\circ}-\alpha\right)}=\frac{u \sin \alpha}{\nu-u \cos \alpha}\)….(5)

Also, since, \(\vec{w}^{\prime}=-\vec{w}\), the relative velocity of A with respect to B is represented by the vector \(\overrightarrow{C P}\).

Relative Acceleration Explained

Relative Acceleration: By similar reasoning, relative acceleration is defined as the apparent acceleration of a body with respect to another body that may be in an accelerated motion.

Let A and B be two moving objects. The rate of change of velocity of A with respect to B is called the relative acceler¬ation of A with respect to B. Actual acceleration of A is obtained when B is at rest or in uniform motion.

If the actual accelerations of A and B are \(\vec{a}_A\) and \(\vec{a}_B\) respectively, then the relative acceleration of B with respect to A

∴ \(\vec{a}_{B A}=\vec{a}_B-\vec{a}_A\)…..(1)

and relative acceleration of A with respect to B

∴ \(\vec{a}_{A B}=\vec{a}_A-\vec{a}_B\)….(2)

Relative Acceleration Example:

1. The actual acceleration of a freely falling body is the downward acceleration due to gravity (\(\vec{g}\)). Hence, two bodies, falling freely, will have a relative acceleration \(\vec{g}\)–\(\vec{g}\) = 0. They are, therefore, stationary or moving with uniform velocity with respect to each other.
2. Let a body A fall vertically downwards with an acceleration \(\vec{g}\) due to gravity. Another body B is moving horizontally with a uniform acceleration \(\vec{a}\).

Relative Velocity Formula and Examples

Then the relative acceleration of A with respect to B is, \(\vec{g}^{\prime}=\vec{g}-\vec{a}=\vec{g}+(-\vec{a})\) = resultant of \(\vec{g}\) and –\(\vec{a}\).

This resultant \(\vec{g}^{\prime}\) is shown.

Its magnitude is, g’ = \(\sqrt{g^2+a^2} \text { (obviously, } g^{\prime}>g \text { ) }\)

g’ is inclined with the vertical at an angle, \(\theta=\tan ^{-1} \frac{a}{g}\)

For example, when a car or a train accelerates suddenly, any object hanging from its roof is tilted away from the vertical due to the inclination of the relative acceleration \(\vec{g}^{\prime}\). For any random motion of the car or of the train, the relative acceleration also changes randomly, and angle θ does not have a steady value. That is why hanging objects are often observed to be oscillating.

Relative Acceleration Numerical Examples

Short Answer Questions on Relative Motion

Example 1. A car is moving at 80 km · h^-1 towards the north. Another car is moving at 80^2 km · h^-1 towards the northwest. Find the relative velocity of the second car with respect to the first.
Solution:

The velocity of the first car, \(\vec{u}\) = 80 km • h^-1 towards north = \(\overrightarrow{A B}\) the velocity of the 2nd car, \(\vec{v}\) = 80√2 km · h^-1 towards north-west = \(\overrightarrow{A C}\).

Hence, relative velocity of the second car with respect to the \(\vec{w}=\vec{v}-\vec{u}=\vec{v}+(-\vec{u})=\overrightarrow{A C}+\overrightarrow{B A}=\overrightarrow{B C}\) (from triangle law of vector addition)

As ∠BAC = \(45^{\circ}\), we get from \(\triangle A B C\), \(w^2=u^2+v^2-2 u v \cos 45^{\circ}\)

= \((80)^2+(80 \sqrt{2})^2-2.80 \cdot 80 \sqrt{2} \cdot \frac{1}{\sqrt{2}}\)

∴ \(w^2=80^2\left[1+(\sqrt{2})^2-2\right]=80^2\)

∴ w = 80 km · h^-1

Since, w = u, from ΔABC,

∠ACB = ∠BAC = 45°

∴ ∠ABC = 90°

Hence, the vector \(\overrightarrow{B C}=\vec{w}\) is directed towards the west.

The relative velocity of 1st car with respect to the second will also have the same magnitude but will be directed towards the east.

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Example 2. Two bodies are moving such that the velocity of one is twice that of the other and they make an angle of 60° with each other. Find the relative velocity of one with respect to the other.
Solution:

Representing the velocities in the vector diagram, we get, \(\overrightarrow{A B}=\vec{u}\) = velocity of one body, \(\overrightarrow{A B}=\vec{2 u}\) = velocity of the other.

The relative velocity of the second with respect to the first, \(\vec{w}=\overrightarrow{A C}-\overrightarrow{A B}=\overrightarrow{A C}+\overrightarrow{B A}=\overrightarrow{B C}\) [from triangle law of vector addition]

From ΔABC, \(w^2=(-u)^2+(2 u)^2+2 \cdot(-u) \cdot 2 u \cos 60^{\circ}\)

= \(5 u^2-2 u^2=3 u^2\)

∴ \(w=\sqrt{3} u\)

Also from the trigonometric rule, \(\frac{2 u}{\sin \angle A B C}=\frac{w}{\sin 60^{\circ}}=\frac{\sqrt{3} u \times 2}{\sqrt{3}}\)

or, \(\sin \angle A B C=1=\sin 90^{\circ}\)

∴ \(\angle A B C=90^{\circ} \text {. }\)

Hence \(\vec{w}\), i.e., \(\overrightarrow{B C}\) is perpendicular to \(\vec{u}\), i.e., \(\overrightarrow{A B}\).

Similarly, the relative velocity of the first with respect to the second can be found. In this case, the relative velocity will be –\(\vec{w}\) as the direction will be opposite.

Example 3. At any instant of time, two ships A and B are 70 km apart along a line AB which is directed from north to south. A starts moving towards west at 25 km · h^-1 and at the same time B starts moving towards the north at 25 km · h^-1. Find the distance of closest approach between the two ships and the time required for this.
Solution:

Solution: Let us choose the origin at point A; x-axis along east; y-axis along north.

Initial position of ship A=0 and that of ship B=-70\(\hat{j}\) km

Velocity of ship A = \(-25 \hat{i} \mathrm{~km} \cdot \mathrm{h}^{-1}\) and that of ship \(B=25 \hat{j} \mathrm{~km} \cdot \mathrm{h}^{-1}\).

After a time of t hours, the positions of the ships are,

⇒ \(\vec{r}_A=0+(-25 \hat{i}) t=-25 t \hat{i} \mathrm{~km}\)

⇒ \(\vec{r}_B=-70 \hat{j}+(25 \hat{j}) t=(25 t-70) \hat{j} \mathrm{~km}\)

Position of ship B relative to ship A, \(\vec{r}=\vec{r}_B-\vec{r}_A=25 t \hat{i}+(25 t-70) \hat{j}\)

The distance between them is r = \(|\vec{r}|\).

So, \(r^2 =(25 t)^2+(25 t-70)^2\)

= \(625 t^2+625 t^2-3500 t+4900\)

= \(50\left(25 t^2-70 t+98\right)\)

= \(50\left[(5 t-7)^2-7^2+98\right]\)

= \(50\left[(5 t-7)^2+49\right]\)

When the distance r between the ships is minimum, \(r^2\) is also minimum.

This happens when \((5 t-7)^2\), a squared quantity, is minimum, i.e., zero.

∴ 5t – 7 = 0

or, t = \(\frac{7}{5} \mathrm{~h}=84 \mathrm{~min}=1 \mathrm{~h} 24 \mathrm{~min}\)

At this instant of time, \(r^2=50[0+49]=50 \times 49=25 \times 49 \times 2\)

∴ The distance of the closest approach between the two ships, \(r_{\min }=\sqrt{25 \times 49 \times 2}=35 \sqrt{2} \mathrm{~km}\)

Example 4. A ship is moving towards the east at 10 km · h^-1. A boat is moving north of east making an angle of 30° with the north. What should be the velocity of the boat so that the boat always appears, from the ship, to move towards the north?
Solution:

Let the velocity of the ship be \(\vec{u}=\overrightarrow{O A}\) and velocity of the boat be \(\vec{v}=\overrightarrow{O B}\)

Hence, the velocity of the boat with respect to the ship, \(\vec{w}=\vec{v}-\vec{u}=\overrightarrow{A B}\)

From the figure, \(\sin 30^{\circ}=\frac{u}{v} \text { or, } \frac{1}{2}=\frac{u}{v}\)

or, \(v =2 u=2 \times 10 \mathrm{~km} \cdot \mathrm{h}^{-1} \text { [Given, } u=10 \mathrm{~km} \cdot \mathrm{h}^{-1} \text { ] }\)

= \(20 \mathrm{~km} \cdot \mathrm{h}^{-1}\)

Graphical Representation of Relative Velocity

Example 5. A man is in a car moving with an acceleration of 5 m · s^-2. Find the apparent value of the acceleration due to gravity and the direction of pull of the earth with respect to him.
Solution:

Let the acceleration of the man in the car be \(\vec{a}\) and acceleration due to gravity be \(\vec{g}\). Hence acceleration due to gravity’s relation to the man is, \(\overrightarrow{g^{\prime}}=\vec{g}-\vec{a}=\vec{g}+(-\vec{a})\)

g’ = \(\sqrt{g^2+a^2}=\sqrt{9.8^2+5^2}\)

= \(\sqrt{121.04}=11 \mathrm{~m} \cdot \mathrm{s}^{-2}\)

The angle θ, which \(\vec{g}^{\prime}\) makes with \(\vec{g}\) is given by \(\tan \theta=\frac{a}{g}=\frac{5}{9.8}=0.51=\tan 27^{\circ}\)

∴ \(\theta=27^{\circ}\)

Hence, with respect to the man the earth’s pull is acting downward at an angle 27° with the vertical in the side opposite to his direction of motion.

Example 6. A lift is moving up with a constant acceleration a. A man standing on the lift, throws a ball vertically upwards with a velocity v, which returns to the thrower after a time t. Show that v = (a + g)\(\frac{t}{2}\) where g is the acceleration due to gravity.
Solution:

The downward acceleration of the ball with respect to the lift = g-(-a) = g+ a.

The initial velocity of the ball is v upward. As the ball returns to the thrower, its relative displacement is zero.

Hence, from the equation h = ut- \(\frac{1}{2}\)gt², we get,

0 = \(v t-\frac{1}{2}(g+a) t^2 \text { or, } 0=t\left\{v-\frac{1}{2}(g+a) t\right\}\)

Since, t ≠ 0,

∴ \(v-\frac{1}{2}(g+a) t=0 \quad \text { or, } v=(a+g) \frac{t}{2}\)

Example 7. A lift is moving up with an acceleration of 2 m · s^-2. A nail gets dislodged from the roof of the lift when its speed reaches 8 m · s^-1. If the height of the lift cage is 3 m, find the time taken by the nail to touch the floor of the lift.
Solution:

Relative to the lift, the initial velocity of the nail is u = 0 and it falls through a height of h = 3 m. Inside the lift, the relative acceleration due to gravity acting on the nail downwards is

g’ = \(g-(-a)=9.8-(-2)=11.8 \mathrm{~m} \cdot \mathrm{s}^{-2}\)

∴ h = \(u t+\frac{1}{2} g^{\prime} t^2 \text { or, } 3=0 \cdot t+\frac{1}{2} \times 11.8 \times t^2\)

or, \(t^2=\frac{2 \times 3}{11.8} \text { or, } t=\sqrt{\frac{6}{11.8}}=0.713 \mathrm{~s}\)

Example 8. A simple pendulum is suspended from the roof of a car moving horizontally with an acceleration of 10 m · s^-2. What will be the angle made by the pendulum in its equilibrium position with the vertical? [g = 10 m · s^-2]
Solution:

The acceleration of the car, a = 10 m · s^-2, and acceleration due to gravity g = 10 m · s^-2. The pendulum will be inclined along the direction of the apparent pull of the earth.

Let the angle made by the pendulum with the vertical be θ.

∴ tanθ = \(\frac{a}{g}\) = \(\frac{10}{10}\) = 1 or, θ = 45

Example 9. Two parallel rail lines are directed as north-south. A train X runs towards north with a speed of 15 m · s^-1 and another train Y runs towards south with a speed of 25 m · s^-1. Find

1. The velocity of Y relative to X,
2. The velocity of ground with respect to Y,
3. The velocity of a monkey running on the roof of X against its motion with a velocity of 5 m · s^-1 relative to X, as observed by a man standing on the ground.

Solution:

Let us assume the velocity from south to north is positive.

Then, the velocity of X, v_x = +15 m · s^-1

and velocity of Y, u_y = -25 m · s^-1 m

The velocity of train Y relative to train X is, v_YX = v_Y – v_X=-25-15 = -40 m · s^-1

The relative velocity of ground with respect to train Y is, v_GY = v_G – V_Y = o – (-25) = 25 m · s^-1

The relative velocity of the monkey with respect to training X is, VMX= v_M-u_X=-5m · s^-1

(where is the velocity of the monkey with respect to the ground)

∴ v_M = v_X + v_MX = 15 + (-5) = 10 m · s^-1

Example 10. A steamer is moving towards the east with a velocity u. A second steamer is moving with a velocity 2u at angle θ north of east The motion of the second steamer relative to the first is along the northeast. Show that, cosθ-sinθ = \(\frac{1}{2}\)
Solution:

Let us choose, the x-axis along the east and the y-axis along the north. Then the velocities are of the first steamer, \(\vec{v}_1=u \hat{i}\); of the second steamer, \(\overrightarrow{v_2}=2 u \cos \theta \hat{i}+2 u \sin \theta \hat{j}\)

∴ Velocity of the second steamer relative to the first is, \(\vec{w}=\vec{v}_2-\vec{v}_1=u(2 \cos \theta-1) \hat{i}+2 u \sin \theta \hat{j}\)

As \(\vec{w}\) is along the north-east, its x- and y-components are equal.

∴ u(2cosθ-1) = 2usinθ or, 2cosθ-2sinθ = 1

or, cosθ – sinθ = \(\frac{1}{2}\)

Example 11. A stone is dropped from a tower 400 m high. Simultaneously, another stone is thrown upwards from the earth’s surface with a velocity of 100 m/s. When and where would these two stones meet? (g = 9.8 m/s²)
Solution:

Both the stones move with a uniform acceleration g, the acceleration due to gravity, acting downwards.

∴ The acceleration of one relative to the other = g- g = 0.

At time t = 0, the downward velocities of the two stones are, respectively, 0 and -100m/s. So, relative velocity = 0 – (-100) s 100 m/s. The stones move with this relative velocity, which is uniform in the absence of any relative acceleration.

The initial distance between the stones = 400 m.

Thus, they will meet after a time t = \(\frac{400}{100}\) = 4 s.

The height through which the first stone falls in time t = 4 s is, h = \(\frac{1}{2}\)gt² = \(\frac{1}{2}\)x 9.8 x 42 = 78.4 m

∴ The stones meet after 4 s at a height of (400 – 78.4), or 321.6 m.

Example 12. A rubber bail Is thrown downwards from the top of a lower with a velocity of 14 m/s. A second ball Is dropped from the same place 1 s later. The first ball reaches the ground in 2 seconds and rebounds with the same magnitude of velocity. How much later would the two balls collide with each other?
Solution:

Height of the tower, h = Distance travelled by the first ball in 2 s (t = 0 to t = 2s)

= 14 x 2 + \(\frac{1}{2}\) x 9.8 x 2² = 47.6 m

The second ball starts at t = 1 s. From t = Is to t = 2s, this ball comes down through a height, h’ = \(\frac{1}{2}\) x 9.8 x (2 – 1)² = 4.9 m

∴ At t = 2 s , the distance between the two balls, H = h-h’ = 47.6-4.9 = 42.7 m

For the first ball, velocity at t = 2 s is, v₁ = 14 + 9.8 x 2 = 33.6 m/s

As it rebounds with the same magnitude of velocity, its upward velocity at the same instant, i.e., at t = 2 s is v’₁= -33.6 m/s.

For the second belli, at t = 1 s, the velocity u = 0.

So, at t = 2 s, the velocity, v₂ = 0 + 9.8 x (2 – 1) = 9.8 m/s

∴ Relative velocity of the second ball with respect to the first at t = 2 s is,

V = v₂-v’₁ = 9.8-(-33.6) = 43.4 m/s

Relative acceleration =g-g = 0. So, for relative motion, the velocity V is uniform. Thus, time required to travel the height of H = 42.7 m is,

t = \(\frac{H}{V}\) = \(\frac{42.7}{43.4}\) = 0.98 s-1

Therefore, the balls will collide at T = 2 + t = 2.98 s.

Example 13. An object falling freely from a height H hits an inclined plane at a height h in its trajectory. At the instant of collision, the velocity of the object changes to become horizontal. What is the value of \(\frac{h}{H}\) for which it spends maximum time to reach the ground?

Solution:

The trajectory of the object is ABC

Let AB is a free fall through a height (H-h) in time t₁.

∴ (H- h) = \(0+\frac{1}{2} g t_1^2 \quad \text { or, } t_1=\sqrt{\frac{2(H-h)}{g}}\)

BC is a projectile motion for which the vertical component of velocity initially (at B) is zero. Also for the path BC, the vertical drop is through a height h. If the time taken is t₂

h = \(0+\frac{1}{2} g t_2^2 \quad \text { or, } t_2=\sqrt{\frac{2 h}{g}}\)

Total time, t = \(t_1+t_2=\sqrt{\frac{2(H-h)}{g}}+\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2}{g}}(\sqrt{H-h}+\sqrt{h}]\)

∴ \(t^2=\frac{2}{g}[(H-h)+h+2 \sqrt{h(H-h)}]\)

= \(\frac{2}{g}[H+\sqrt{4 h(H-h)}]=\frac{2}{g}\left[H+\sqrt{H^2-(2 h-H)^2}\right]\)

For maximum t, i.e., for maximum t², the term (2h- H)² should be minimum. This minimum value is zero, as it is a square.

∴ (2h— H)² = 0 or, 2h = H or, \(\frac{h}{H}\) = \(\frac{1}{2}\)

Composition Of Several Vectors By Resolution

WBBSE Class 11 Vector Resolution Notes

Three or more vectors can be added by the law of polygon of vectors. This method is geometrically convenient but is inconvenient mathematically. However, when vectors lie on a plane, an elegant method is to resolve each of the vectors into two mutually perpendicular components while performing addition.

Composition Of Several Vectors By Resolution Process:

The vectors are drawn on a plane from a common initial point. Two mutually perpendicular axes X and Y are drawn from that point and each vector is resolved into components along these two axes.

Next, algebraic sums of the x-components and of the y -y-components of the vectors with proper sign (+ or -) are determined separately. Now, vector addition of the resultant x and y components gives the resultant of all the vectors.

Composition Of Several Vectors By Resolution Calculation:

Let \(\vec{P}, \vec{Q}, \vec{R}\) be three coplanar vectors having a common initial point O. To find the resultant, axes OX and OY are drawn, coplanar with the vectors.

Let  \(\vec{P}, \vec{Q}, \vec{R}\) make angles α, β, γ respectively with the x-axis. Thus their respective components along the x-axis are \(P_x=P \cos \alpha, Q_x=Q \cos \beta, and R_x=R \cos \gamma\).

Similarly \(P_y=P \sin \alpha, Q_y=Q \sin \beta and R_y=R \sin \gamma\).

Composition of Vectors by Resolution Explained

Hence, the resultants (sums) F_x and F_y along x and y axes are \(F_x=P \cos \alpha+Q \cos \beta+R \cos \gamma\)….(1)

and \(F_y=P \sin \alpha+Q \sin \beta+R \sin \gamma\)….(2)

Read and Learn More: Class 11 Physics Notes

For performing the summation, positive or negative values of sine and cosine functions for the angles α, β, γ should be taken into consideration. \(\vec{F}\) represents the resultant of F_x and F_y, according to the triangle or the parallelogram law of vector addition.

F = \(\sqrt{F_x^2+F_y^2}\)…(3)

Key Concepts in Vector Addition and Resolution

and the inclination θ of \(\vec{F}\) with the x-axis is given by \(\tan \theta=\frac{F_y}{F_x}\)…(4)

Equations (3) and (4) give the magnitude and direction respectively of the resultant of the three vectors \(\vec{P}, \vec{Q}, \vec{R}\). The resultant of any number of coplanar vectors can be found using this method.

Composition Of Several Vectors By Resolution Numerical Examples

Examples of Vector Composition Using Resolution

Example 1. Compute the resultant of three coplanar vectors P, 2P and 3P inclined at 120° with one another.
Solution:

Computing The resultant of three coplanar vectors P, 2P and 3P inclined at 120° with one another

Let the x-axis be taken along vector P. Then the vectors 2P and 3P will make angles 120° and 240° respectively with the x-axis. Resolving the vectors into components along the x and y axes and adding, we get the stuns R_x and R_y respectively along the x and y axes.

⇒ \(R_x=P \cos 0^{\circ}+2 P \cos 120^{\circ}+3 P \cos 240^{\circ}\)

∴ \(R_x=P+2 P\left(-\frac{1}{2}\right)+3 P\left(-\frac{1}{2}\right)=P-P-\frac{3}{2} P=-\frac{3}{2} P\)

⇒ \(R_y=P \sin 0^{\circ}+2 P \sin 120^{\circ}+3 P \sin 240^{\circ}\)

= \(0+2 P \times \frac{\sqrt{3}}{2}+3 P\left(-\frac{\sqrt{3}}{2}\right)=(2-3) \frac{\sqrt{3}}{2} P=-\frac{\sqrt{3}}{2} P\)

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Hence, the value of the resultant \(\vec{R}\) is, R = \(\sqrt{R_x^2+R_y^2}=\sqrt{\frac{9}{4} P^2+\frac{3}{4} P^2}=\sqrt{3} P\)

∴ \(\vec{R}\) makes an angle θ with the x-axis such that,\(\tan \theta=\frac{R_y}{R_x}=\frac{-\frac{\sqrt{3}}{2} P}{-\frac{3}{2} P}=\frac{1}{\sqrt{3}}=\tan 30^{\circ} \text { or, } \tan 210^{\circ}\)

As \(R_x\) and \(R_y\) are both negative, R must be in the third

As R_x and R_y are both negative, R must be in the third quadrant, and θ = 210°.

Short Answer Questions on Vector Resolution

Example 2. Five coplanar forces, each of magnitude F, are acting on a particle. Each force is inclined at an angle of 30° with the previous one. Find out the magnitude and direction of the resultant force on the particle.
Solution:

Given

Five coplanar forces, each of magnitude F, are acting on a particle. Each force is inclined at an angle of 30° with the previous one.

Let us choose the x-axis along the first force, and y-axis along the perpendicular direction on the plane of the forces

The angles made by the five forces with the x-axis are 0°, 30°, 60°, 90° and 120°.

∴ Component of resultant force along x-axis, F_x = Fcos0° + Fcos30° + Fcos60° + Fcos90° + Fcos120°

= \(F\left[1+\frac{\sqrt{3}}{2}+\frac{1}{2}+0-\frac{1}{2}\right]\)

= \(F\left(1+\frac{\sqrt{3}}{2}\right)\)

and component of resultant force along y-axis, Fy = Fsin0° + Fsin30° + Fsin60° + Fsin90° + Fsinl20°

= \(F\left[0+\frac{1}{2}+\frac{\sqrt{3}}{2}+1+\frac{\sqrt{3}}{2}\right]=F\left(\frac{3}{2}+\sqrt{3}\right)\)

= \(\sqrt{3} F\left(1+\frac{\sqrt{3}}{2}\right)\)

So, the magnitude of the resultant force, \(F^{\prime}=\sqrt{F_x^2+F_y^2}=F\left(1+\frac{\sqrt{3}}{2}\right) \sqrt{1^2+(\sqrt{3})^2}\)

= \(2 F\left(1+\frac{\sqrt{3}}{2}\right)=(2+\sqrt{3}) F\)

The angle of inclination of F’ with the x-axis, \(\theta=\tan ^{-1} \frac{F_y}{F_x}=\tan ^{-1} \sqrt{3}=60^{\circ} \text { (along the third force) }\)

Three Dimensional Resolution Of Vectors Algebraic Representation

Let the magnitude and direction of a vector \(\vec{r}\), in space, be represented by \(\overrightarrow{O P}\). Our aim here is to find an elegant representation of \(\vec{r}\)

O is taken as the origin of the 3-dimensional space and three mutually perpendicular axes x, y, and z are drawn. A special rule is followed to indicate the directions of the three axes.

If we hold the first 3 fingers of our right hand at right angles to one another, then the forefinger points toward the x-axis, the middle finger toward the y-axis, and the thumb towards the z-axis. This is called the right-handed cartesian coordinate system.

Here, the origin O (0,0,0) is the initial point, and P (x, y, z) is the endpoint of the vector \(\overrightarrow{O P}\). Taking the line segment OP as a diagonal, the cuboid ADPEOBFC is drawn.

Understanding Three Dimensional Vector Components

The projections of the vector \(\vec{r}\) along the three axes are, from OA = x, OB = AD = y, and OC = DP = z. These x, y, and z are three mutually perpendicular components of the vector \(\vec{r}\).

When the line segments OA, AD, and DP are taken as vectors, the three-dimensional polygon OADP gives, as per the law of polygon of vector addition, \(\overrightarrow{O A}+\overrightarrow{A D}+\overrightarrow{D P}=\overrightarrow{O P}\)….(1)

If \(\hat{i}, \hat{j}, \hat{k}\) are the unit vectors along the x, y and z axes respectively in the positive direction,

∴ \(\overrightarrow{O A}=x \hat{i}, \overrightarrow{A D}=y \hat{j} \quad \text { and } \overrightarrow{D P}=z \hat{k}\)

Since, \(\vec{r}=\overrightarrow{O P}\), from equation (1)

∴ \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\)………..(2)

⇒ \(\vec{r}\) is the position vector of the point P w.r.t origin O.

Equation (2) shows the algebraic representation of a position vector with initial point (0,0,0) and final point (x, y, Z).

Determination Of The Magnitude Of The Vector: From the geometrical property of a cuboid, (diagonal)² = (length)³ + (breadth)2 + (height)²

or, OP² = OA² + OB² + OC² or, r² = x² + y² + z²

or, \(r=\sqrt{x^2+y^2+z^2}\)….(3)

Hence, the magnitude or value of a vector with its initial point as the origin can be determined easily with the help of the coordinates of its terminal point.

Direction Cosine: Let \(\overrightarrow{O P}\) be inclined at an angle a with the x-axis, i.e., ∠POA = α. As per the property of the cuboid, ΔOAP is a right-angled triangle with ∠OAP = 90°.

Hence, \(\cos \alpha=\frac{O A}{O P}=\frac{x}{r}=l\) (say)….(4)

Similarly, if \(\overrightarrow{O P}\) makes angles β and γ with y and z axes respectively,

⇒ \(\left.\begin{array}{rl}\cos \beta =\frac{y}{r}=m \\ \text { and } \cos \gamma =\frac{z}{r}=n\end{array}\right\}\)

From (4) and (5), \(l^2+m^2+n^2=\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=\frac{x^2}{r^2}+\frac{y^2}{r^2}+\frac{z^2}{r^2}\)

= \(\frac{x^2+y^2+z^2}{r^2}=\frac{r^2}{r^2}=1\)…..(6) [with the help of equation (3)]

Hence, to know the direction of the vector, the angles α, β,γ can be determined using equations (4) and (5). cos α, cosβ, cosγ or l, m, n is called the direction cosines of vector r with respect to the axes x, y, and z respectively. Equation (6) indicates the relationship among the direction cosines.

 

WBBSE Class 11 Three Dimensional Vector Resolution Notes

Sum And Difference Of Two Vectors: Let \(\overrightarrow{r_1}=x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k} \text { and } \vec{r}_2=x_2 \hat{i}+y_2 \hat{j}+z_2 \hat{k}\) be two vectors.

Sum or resultant of these two vectors is \(\vec{r}_1+\vec{r}_2=\left(x_1+x_2\right) \hat{i}+\left(y_1+y_2\right) \hat{j}+\left(z_1+z_2\right) \hat{k}\)

and its magnitude is, \(\left|\vec{r}_1+\vec{r}_2\right|=\sqrt{\left(x_1+x_2\right)^2+\left(y_1+y_2\right)^2+\left(z_1+z_2\right)^2}\)

The direction of the resultant is specified by the direction cosines. They are \(\cos \alpha=\frac{x_1+x_2}{\left|\vec{r}_1+\vec{r}_2\right|}, \cos \beta=\frac{y_1+y_2}{\left|\vec{r}_1+\vec{r}_2\right|} \text { and } \cos \gamma=\frac{z_1+z_2}{\left|\vec{r}_1+\vec{r}_2\right|}\)

Similarly, the difference between the two vectors, \(\overrightarrow{r_1} \text { and } \overrightarrow{r_2}\) can also be found out. The resultant in this case will be \(\vec{r}_2-\vec{r}_1=\left(x_2-x_1\right) \hat{i}+\left(y_2-y_1\right) \hat{j}+\left(z_2-z_1\right) \hat{k}\)

The magnitude of the resultant will be, \(\left|\vec{r}_2-\vec{r}_1\right|=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}\)

The corresponding direction cosines are \(\cos \alpha=\frac{x_2-x_1}{\left|\overrightarrow{r_2}-\vec{r}_1\right|}, \cos \beta=\frac{y_2-y_1}{\left|\vec{r}_2-\vec{r}_1\right|} \text { and } \cos \gamma=\frac{z_2-z_1}{\left|\vec{r}_2-\vec{r}_1\right|}\)

These relations are applied to determine the sum or difference of any number of vectors.

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Sum And Difference Of Two Vectors Discussions:

1. Equation (2) represents the 3-dimensional cartesian form of any vector \(\vec{r}\). However, the symbol \(\vec{r}\) is usually reserved for the position vector of a particle.

Any vector \(\vec{A}\) is represented in the system as \(\vec{A}=A_x \hat{i}+A_y \hat{j}+A_z \hat{k}\)….(7)

where A_x, A_y, and A_z stand for the x, y, and z components respectively of the vector \(\vec{A}\). Then, the magnitude of \(\vec{A}\) is

A= \(\sqrt{A_x^2+A_y^2+A_z^2}\)….(8)

The direction cosines, respectively, are l = \(\frac{A_x}{A}, m=\frac{A_y}{A} \quad \text { and } n=\frac{A_z}{A}\)…..(9)

with \(l^2+m^2+n^2=1\)….(10)

2. This is an example of the resolution of a vector \(\vec{A}\) into three components \(A_x \hat{i}, A_y \hat{j} \text { and } A_z \hat{k}\).

We dealt with coplanar vectors, and it was sufficient to resolve a vector into two components only. But, for a system of non-coplanar vectors, it is absolutely necessary to resolve each vector into three mutually perpendicular components.

3. Once every vector under consideration in a problem can be represented in the form of equation (7), vector geometry essentially transforms into vector algebra, and geometrical figures and rules are no longer necessary. Whereas vector geometry is practicable only for a few vectors, there is no limit on the number of vectors that can be handled with the help of vector algebra.

Algebraic Representation Numerical Examples

Short Answer Questions on 3D Vector Components

Example 1. The coordinates of the endpoint of a vector \(\overrightarrow{O P}\) is (4,3, -5). Express the vector in terms of its coordinates and find its absolute value.
Solution:

If the coordinates of P is (x, y, z) then, \(\overrightarrow{O P}=x \hat{i}+y \hat{j}+z \hat{k}\)

Here, x=4, y=3 and z=-5

∴ \(\overrightarrow{O P}=4 \hat{i}+3 \hat{j}-5 \hat{k}\)

and the absolute value of the vector,

OP = \(|\overrightarrow{O P}|=\sqrt{x^2+y^2+z^2}\)

= \(\sqrt{4^2+3^2+(-5)^2}\)

= \(\sqrt{16+9+25}=\sqrt{50}=5 \sqrt{2}\)

Example 2. Find the magnitude of the vector \(\vec{A}=\hat{i}-2 \hat{j}+3 \hat{k}\). Also, find the unit vector in the direction of \(\vec{A}\)
Solution:

Magnitude of \(\vec{A}\), A = \(\sqrt{1^2+(-2)^2+3^2}=\sqrt{14}\)

Unit vector in the direction of \(\vec{A}\) is \(\hat{n}=\frac{\vec{A}}{\vec{A}}=\frac{1}{\sqrt{14}} \hat{i}-\frac{2}{\sqrt{14}} \hat{j}+\frac{3}{\sqrt{14}} \hat{k}\)

Example 3. Vectors \(\vec{A}\) and \(\vec{B}\) can be expressed as \(\vec{A}=10 \hat{i}-12 \hat{j}+5 \hat{k} \text { and } \vec{B}=7 \hat{i}+8 \hat{j}-12 \hat{k}\), where \(\hat{i}\), \(\hat{j}\), k\(\hat{k}\) are unit vectors along x, y, z axes respectively. Find the resultant of the two vectors and its magnitude.
Solution:

The resultant vector \(\vec{C}=\vec{A}+\vec{B}=(10+7) \hat{i}+(-12+8) \hat{j}+(5-12) \hat{k}\)

or, \(\vec{C}=17 \hat{i}-4 \hat{j}-7 \hat{k}\)

Magnitude of the resultant, C = \(\sqrt{17^2+(-4)^2+(-7)^2}=\sqrt{354}=18.81 .\)

Example 4. Position coordinates of A and B are (-1,5,7) and (3,2,-5) respectively. Express \(\overrightarrow{A B}\) in terms of position coordinates.
Solution:

Let O be the origin, \(\overrightarrow{O A}=\vec{a}, \overrightarrow{O B}=\vec{b}\) and \(\overrightarrow{A B}=\vec{c}=\overrightarrow{O B}-\overrightarrow{O A}=\vec{b}-\vec{a}\)

Here, \(\vec{a}=-\hat{i}+5 \hat{j}+7 \hat{k} and \vec{b}=3 \hat{i}+2 \hat{j}+(-5) \hat{k}\)

∴ \(\overrightarrow{A B}=\vec{c}=\vec{b}-\vec{a}\)

= \((3 \hat{i}+2 \hat{j}-5 \hat{k})-(-\hat{i}+5 \hat{j}+7 \hat{k})\)

= \(4 \hat{i}-3 \hat{j}-12 \hat{k}\)

Example 5. \(3 \hat{i}+4 \hat{j}+12 \hat{k}\) is a vector. Find the magnitude of the vector and the angles it makes with the x, y, and z axes.
Solution:

If \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\), then the magnitude \(|\vec{r}|=\sqrt{x^2+y^2+z^2}\)

∴ Magnitude of the vector \(3 \hat{i}+4 \hat{j}+12 \hat{k}\) is,

a = \(\sqrt{3^2+4^2+12^2}=\sqrt{169}=13\)

Let the corresponding angles the vector makes with the x, y, and z axes be α, β, and γ respectively.

From definitions, \(\cos \alpha=\frac{x}{a}=\frac{3}{13} \text { or, } \alpha=\cos ^{-1} \frac{3}{13}\)

and \(\cos \beta=\frac{y}{a}=\frac{4}{13} \text { or, } \beta=\cos ^{-1} \frac{4}{13}\)

and \(\cos \gamma=\frac{z}{a}=\frac{12}{13} \text { or, } \gamma=\cos ^{-1} \frac{12}{13}\)

Mathematical Formulation for 3D Vector Resolution

Example 6. Two vectors \(\vec{A}\) and \(\vec{B}\) are \(\vec{A}=5 \hat{i}+3 \hat{j}-4 \hat{k}\) and \(\vec{B}=5 \hat{i}+2 \hat{j}+4 \hat{k}\). Find the unit vectors along \(\vec{A}+\vec{B}\) and \(\vec{A}-\vec{B}\).
Solution:

⇒ \(\vec{A}=5 \hat{i}+3 \hat{j}-4 \hat{k} and \vec{B}=5 \hat{i}+2 \hat{j}+4 \hat{k}\)

∴ \(\vec{A}+\vec{B}=(5+5) \hat{i}+(3+2) \hat{j}+(-4+4) \hat{k}=10 \hat{i}+5 \hat{j}\)

and \(|\vec{A}+\vec{B}|=\sqrt{10^2+5^2}=5 \sqrt{5}\)

Hence, unit vector along \(\vec{A}+\vec{B}=\frac{\vec{A}+\vec{B}}{|\vec{A}+\vec{B}|}=\frac{2}{\sqrt{5}} \hat{i}+\frac{1}{\sqrt{5}} \hat{j}\)

Similarly, \(\vec{A}-\vec{B}=\hat{j}-8 \hat{k}\)

and \(|\vec{A}-\vec{B}|=\sqrt{1^2+(-8)^2}=\sqrt{65}\)

and unit vector along \(\vec{A}-\vec{B}=\frac{\vec{A}-\vec{B}}{|\vec{A}-\vec{B}|}=\frac{1}{\sqrt{65}} \hat{j}-\frac{8}{\sqrt{65}} \hat{k}\).

Example 7. Two velocities \(\vec{v}_1\) and \(\vec{v}_2\) are 3 m/s towards north and 4 m/s towards east, respectively. Find \(\vec{v}_1-\vec{v}_2\).
Solution:

Let us choose, the x-axis along the east and the y-axis along the north.

∴ \(\vec{v}_1=3 \hat{j} \mathrm{~m} / \mathrm{s}\) and \(\vec{v}_2=4 \hat{i} \mathrm{~m} / \mathrm{s}\)

∴ \(\vec{v}_1-\vec{v}_2=-4 \hat{i}+3 \hat{j}\) (between west and north)

∴ \(\left|\vec{v}_1-\vec{v}_2\right|=\sqrt{(-4)^2+3^2}=5 \mathrm{~m} / \mathrm{s}\)

If \(\vec{\nu}_1-\vec{v}_2\) makes an angle θ with east, then \(\tan \theta=\frac{3}{-4}=\tan 143.1^{\circ}=\tan \left(180^{\circ}-36.9^{\circ}\right)\)

∴ \(\vec{v}_1-\vec{v}_2\) is inclined at an angle of 143.1^{\circ} north of east, i.e., \(36.9^{\circ}\) north of west.

Example 8. Find out the resultant of the following three displacement vectors: \(\vec{A}\) = 10 m, along north-west; \(\vec{B}\) = 20 m, 30° north of east; \(\vec{C}\) = 35 m, along the south.
Solution:

Let us choose, x-axis along the east and the y-axis along the north. Taking into account the x- and y-components of the given vectors, they can be written as (in meter),

⇒ \(\vec{A}=\hat{i}\left(-10 \cos 45^{\circ}\right)+\hat{j} 10 \sin 45^{\circ}=-5 \sqrt{2} \hat{i}+5 \sqrt{2} \hat{j}\)

⇒ \(\vec{B}=\hat{i} 20 \cos 30^{\circ}+\hat{j} 20 \sin 30^{\circ}=10 \sqrt{3} \hat{i}+10 \hat{j}\)

⇒ \(\vec{C}=-35 \hat{j}\)

∴ Resultant, \(\vec{R}=\vec{A}+\vec{B}+\vec{C}\)

= \(\hat{i}(-5 \sqrt{2}+10 \sqrt{3})+\hat{j}(5 \sqrt{2}+10-35)\)

= \(\hat{i} 5(2 \sqrt{3}-\sqrt{2})-\hat{j} 5(5-\sqrt{2})\) [between east and south]

∴ R = \(|\vec{R}|=\sqrt{\{5(2 \sqrt{3}-\sqrt{2})\}^2+\{-5(5-\sqrt{2})\}^2}\)

= \(5 \sqrt{(2 \sqrt{3}-\sqrt{2})^2+(5-\sqrt{2})^2}=20.65 \mathrm{~m}\)

If \(\vec{R}\) makes an angle θ with the x-axis, then \(\tan \theta=\frac{R_y}{R_x}=\frac{-5(5-\sqrt{2})}{5(2 \sqrt{3}-\sqrt{2})}=-1.75=\tan \left(-60.2^{\circ}\right)\)

So, \(\vec{R}\) is inclined at \(60.2^{\circ}\) south of east.

Example 9. The characteristic equations of a particle moving in a curved path are x = e^-t, y = 2 cos 3t, and z = 2sin3t, where t stands for time. Find out

1. velocity and acceleration at any instant,
2. velocity and acceleration at t = 0.

Solution:

Position vector of the particle at any instant, \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}=e^{-t} \hat{i}+2 \cos 3 \hat{t}+2 \sin 3 t \hat{k}\)

1. Velocity, \(\vec{v}=\frac{d \vec{r}}{d t}=-e^{-t} \hat{i}-6 \sin 3 t \hat{j}+6 \cos 3 t \hat{k}\)

= \(-x \hat{i}-3 z \hat{j}+3 y \hat{k}\)

Acceleration, \(\vec{a}=\frac{d \vec{v}}{d t}=e^{-t} \hat{i}-18 \cos 3 t \hat{j}-18 \sin 3 t \hat{k}\)

= \(x \hat{i}-9 y \hat{j}-9 z \hat{k}\)

2. At t = 0, \(\vec{v}=-e^0 \hat{i}-6 \sin (3 \times 0) \hat{j}+6 \cos (3 \times 0) \hat{k}\)

= \(-\hat{i}-6 \times 0 \hat{j}+6 \times 1 \hat{k}\)

= \(-\hat{i}+6 \hat{k}\)

v = \(|\vec{v}|=\sqrt{(-1)^2+6^2}=\sqrt{37}\)

At t = \(0, \vec{a}=e^0 \hat{i}-18 \cos (3 \times 0) \hat{j}-18 \sin (3 \times 0) \hat{k}\)

= \(\hat{i}-(18 \times 1) \hat{j}-(18 \times 0) \hat{k}\)

= \(\hat{i}-18 \hat{j}\)

a = \(|\vec{a}|=\sqrt{1^2+(-18)^2}=\sqrt{325}=5 \sqrt{13}\)

Real-Life Examples of Three Dimensional Vectors

Example 10. The position vector \(\vec{r}\) of a particle with respect to the origin changes with time t as \(\vec{r}=A t \hat{i}-B t^2 \hat{j}\), where A and B are positive constants. Determine

1. The locus of the particle,
2. The nature of variation with time of the velocity and acceleration vectors, and also the moduli of them.

Solution:

1. \(\vec{r}=A t \hat{i}-B t^2 \hat{j}=x i+y \hat{j}\)

x = \(A t \quad \text { and } y=-B t^2\)

Then, \(x^2=A^2 t^2 or, t^2=\frac{x^2}{A^2}\)

also, \(t^2=-\frac{y}{B}\)

∴ \(\frac{x^2}{A^2}=-\frac{y}{B}\) or, \(x^2=-\frac{A^2}{B} y\)

This is the locus of the particle, which is a parabola.

2. Velocity, \(\vec{v}=\frac{d \vec{r}}{d t}=A \hat{i}-2 B t \hat{j}\)

Its modulus, \(|\vec{v}|=\sqrt{A^2+(-2 B t)^2}=\sqrt{A^2+4 B^2 t^2}\)

Acceleration, \(\vec{a}=\frac{d \vec{v}}{d t}=-2 B \hat{j}\)

Its modulus, \(|\vec{a}|=\sqrt{(-2 B)^2}=2 B\)

Example 11. The position vector of a particle is, \(\vec{r}=3 t \hat{i}-2 t^2 \hat{j}+4 \hat{k}\). Find out

1. Its velocity \(\vec{v}\) and acceleration \(\vec{a}\),
2. The magnitude and direction of its velocity at t = 2 s.

Solution:

∴ \(\vec{r}=3 t \hat{i}-2 t^2 \hat{j}+4 \hat{k}\)

1. \(\vec{v}=\frac{d \vec{r}}{d t}=3 \hat{i}-4 t \hat{j}\)

∴ \(\vec{a}=\frac{d \vec{v}}{d t}=-4 \hat{j}\)

2. At t = \(2 \mathrm{~s} \quad \vec{v}=3 \hat{i}-(4 \times 2) \hat{j}=3 \hat{i}-8 \hat{j}\) (between the x-axis and negative y-axis)

∴ \(|\vec{v}|=\sqrt{3^2+(-8)^2}=\sqrt{73} \text { unit }\)

Inclination of \(\vec{v}\) with the x-axis, \(\theta=\tan ^{-1}\left(\frac{-8}{3}\right)=-69.4^{\circ}\)

Subtraction Of Two Vectors

The general rule for subtraction in algebra can be represented in two ways: 5-3 = 2 and 5 + (-3) = 2. This implies that the result of subtraction of a positive quantity from a number is the same as the result of adding a negative quantity of the same value to the number. This rule is followed in vector subtraction too.

If the difference between two vectors \(\vec{a}\) and \(\vec{b}\) is \(\vec{d}\), then \(\vec{a}\) – \(\vec{b}\) = \(\vec{d}\) and also \(\vec{a}\) + (-\(\vec{b}\)) = \(\vec{d}\), where –\(\vec{b}\) is the opposite vector of \(\vec{b}\)

Line segments OA and OB represent the magnitudes and directions of the vectors \(\vec{a}\) and \(\vec{b}\) respectively. BO is extended up to D, so that OD = BO.

This makes \(\overrightarrow{O D}\) and \(\overrightarrow{O B}\) two opposite vectors i.e., \(\overrightarrow{O D}\)  = –\(\overrightarrow{O B}\) = -b .

WBBSE Class 11 Vector Subtraction Notes

In the parallelogram OAED, the diagonal \(\overrightarrow{O E}\) gives the resultant of \(\vec{a}\) and –\(\vec{b}\) which is the difference of \(\vec{a}\) and \(\vec{b}\).

Hence, \(\overrightarrow{O A}+\overrightarrow{O D}=\overrightarrow{O E}\)

or, \(\vec{a}+(-\vec{b})=\vec{d}\)

or, \(\vec{a}-\vec{b}=\vec{d}\)

Read and Learn More: Class 11 Physics Notes

Relation Between The Resultant And The Difference Of Two Vectors: The diagonal \(\overrightarrow{O F}\) of the parallelogram OAFB represents the resultant of two vectors \(\vec{a}\) and \(\vec{a}\).

Now parallelograms OAFB and OAED are congruent. Therefore, it is evident that \(\overrightarrow{O E}\) and \(\overrightarrow{O F}\) can be represented by the diagonals (\(\overrightarrow{B A}\) and \(\overrightarrow{O F}\)) of the same parallelogram.

Hence, we may conclude that, if a diagonal of a parallelogram represents the resultant of two vectors, the other diagonal would represent the difference between them, with proper assignment of directions.

 

Relation Between The Resultant And The Difference Of Two Vectors Special Cases:

1. When the angle between the two vectors \(\vec{a}\) and \(\vec{b}\) is 90° i.e., they are a pair of orthogonal vectors, the parallelograms change into rectangles, for which the diagonals are equal and make equal angles with the base vector \(\vec{a}\).

Hence, the resultant and the difference of these two vectors are equal in magnitude and they are inclined at the same angle as the base vector but in opposite directions.

Analytically, \(\left.\begin{array}{l}
c^2=a^2+b^2 \text { or, } c=\sqrt{a^2+b^2} \\
d^2=a^2+b^2 \text { or, } d=\sqrt{a^2+b^2}
\end{array}\right\}\)….(1)

⇒ \(\left.\begin{array}{r}
\tan \theta=\frac{b}{a} \\
\tan \theta^{\prime}=-\frac{b}{a}
\end{array}\right\}\)…..(2)

∴ θ = θ’

2. For two orthogonal vectors of equal value, i.e., when a = b, the parallelogram changes into a square for which the diagonals are of equal length and are mutu¬ally perpendicular. In this case, the resultant and the difference of the two vectors are of equal magnitude; they are mutually orthogonal as well.

Putting a = b in equation (2), tanθ =1 or, θ = 45° and tanθ’ = -1 or, θ = -45°

Hence, the angle between the resultant and the difference of the two vectors =45° + 45° = 90°.

Applications of Vector Subtraction in Physics

Zero Of Null Vector: A vector having a magnitude zero (0) and no fixed direction is a null vector. If the initial and the terminal points of a line segment representing a vector coincide, then the vector is called a zero or null vector.

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Indispensability Of The Use Of A Null Vector: if a particle starting from a point returns to the same point after some time, its displacement, which is a vector quantity, is a zero vector. Also, two equal and opposite forces acting on a body have a zero resultant, which is a zero vector \(\vec{0}\), i.e., \(\vec{F}-\vec{F}=\overrightarrow{0}\). Hence the difference of two equal vectors is a zero vector.

Again, the acceleration of a particle moving with uniform velocity is zero. This is also another example of a zero or null vector, as acceleration is a vector quantity. Hence, the use of zero or null vectors is indispensable in vector algebra.

The following two operations give rise to a zero vector.

1. When a vector is multiplied by zero, the result is a zero vector. Thus, \(0(\vec{A})=\overrightarrow{0}.\)
2. When the negative of a vector is added to the vector itself, the result is a zero vector, thus \(\vec{A}+(-\vec{A})=\overrightarrow{0}\).

Properties Of Zero Or Null Vector:

1. The addition or subtraction of a zero vector from a vector results in the same vector. Thus \(\vec{A} \pm \overrightarrow{0}=\vec{A}.\)
2. The multiplication of a non-zero real number with a zero vector is again a zero vector. If n is a non-zero real number, then \(n(\overrightarrow{0})=\overrightarrow{0}.\)
3. If n₁ and n₂ are two different non-zero real numbers, then the relation \(n_1 \vec{A}=n_2 \vec{B}\) can hold only if both \(\vec{A}\) and \(\vec{B}\) are zero vector.

Position And Displacement Vectors: In the chapter One-dimensional Motion, we defined the position and the displacement vectors as two important properties related to the motion of a particle. Here, we shall discuss some important points about them vectorically.

Position vector Definition: A vector used to specify the position of a point with reference to the origin of the coordinate system is called a position vector.

Consider a point P having coordinates (x, y, z). If O is the origin then \(\overrightarrow{O P}\) is called the position vector, \(\vec{r}\). The distance between the origin and the point gives the magnitude of the position vector. Here, \(\overrightarrow{O P}=\vec{r}\) (position vector).

Thus, \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\)]

Hence, magnitude of \(\vec{r}\) is given by \(|\vec{r}|=\sqrt{x^2+y^2+z^2}\)

Displacement Definition: The vector connecting the initial and final positions of a particle is called the displacement vector.

Suppose a particle is moved from point A(x₁, y₁, z₁) to another point B(x₂, y₂, z₂).

The shortest distance between the points A and B is the displacement. So \(\overrightarrow{A B}\) represents the displacement vector, \(\vec{r}\) as shown.

Graphical Method for Vector Subtraction

Applying triangle law of vectors to the ΔOAB we get, \(\overrightarrow{O A}+\overrightarrow{A B}=\overrightarrow{O B} \quad \text { or, } \overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=\vec{r}_2-\vec{r}_1=\vec{s}\)

The length of the straight line AB gives the magnitude of the displacement vector. The direction of the vector is along the direction of motion of the particle.

With respect to the origin O, \(\vec{r}_1=x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k}\) = initial position vector

and \(\vec{r}_2=x_2 \hat{i}+y_2 \hat{j}+z_2 \hat{k}\) = final position vector

∴ \(\overrightarrow{A B}=\vec{s}=\left(\vec{r}_2-\vec{r}_1\right)=\left(x_2-x_1\right) \hat{i}+\left(y_2-y_1\right) \hat{j}+\left(z_2-z_1\right) \hat{k}\)

The magnitude of the displacement vector is independent of the choice of origin of the cartesian coordinate system.

∴ \(|\overrightarrow{A B}|=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}\)

The points to be noted here are:

1. The displacement of a particle is the vector difference between its final and initial position vectors.
2. The direction of \(\overrightarrow{A B}\) is different from that of \(\overrightarrow{O A}\) and \(\overrightarrow{O B}\). So, in general, the displacement vector is directed neither along the initial nor along the final position vector of a particle. In other words, generally ‘position’ and ‘change of position’ are vectors that are not in the same direction.
3. Velocity is defined as the rate of displacement with time, i. e., the rate of change of position with time. There is nothing like a ‘change of displacement’; hence, the velocity vector always lies along the displacement vector. The average velocity of the particle during the time spent while going from A to B, is directed along \(\overrightarrow{A B}\).

Position And Displacement Vectors Numerical Examples

Mathematical Formulation of Vector Subtraction

Example 1. A car is traveling towards the east at 10 m · s^-1. It takes 10 seconds to change its direction of motion to the north and continues with the same magnitude of velocity. Find the magnitude and direction of the average acceleration of the car.
Solution:

Change in the velocity of the car

= final velocity – initial velocity

= (10 m • s^-1 towards north) – (10 m • s^-1 towards east)

= \(\overrightarrow{A B}-\overrightarrow{O A}=\overrightarrow{A B}+\overrightarrow{A O}=\overrightarrow{A C}\) [from parallelogram law of vector addition]

∴ AC² =AB² + BC² =AB² + AO²

= 10² + 10² = 100 + 100 = 200

∴ AC = 10√2 m · s^-1

Average acceleration = \(\frac{\text { change in velocity }}{\text { time }}\)

= \(\frac{10 \sqrt{2}}{10}=\sqrt{2} \mathrm{~m} \cdot \mathrm{s}^{-2}\)

It is directed along \(\overrightarrow{A C}\) that makes angle 6 with \(\overrightarrow{A B}\), and \(\tan \theta=\frac{B C}{A B}=\frac{A O}{A B}=\frac{10}{10}=1=\tan 45^{\circ}\)

∴ \(\theta=45^{\circ}\)

Hence, average acceleration is directed towards the north-west.

Real-Life Examples of Vector Subtraction

Example 2. A boy runs 210 m along the corridor of his school and turns; right at the end of the corridor runs 180 m to the end of the building and then turns right and runs 30 m.

1. Construct a vector diagram that represents this motion. Indicate your choice of unit vectors,
2. What is the direction and magnitude of the straight line between start and finish?

Solution:

1. We choose, \(\hat{i}\) = unit vector along x-axis, \(\hat{j}\) = unit vector along y-axis,

The consecutive displacements are, \(\overrightarrow{A B}=210 \hat{i} \mathrm{~m}, \overrightarrow{B C}=-180 \hat{j} \mathrm{~m}, \overrightarrow{C D}=-30 \hat{i} \mathrm{~m}\)

2. Resultant, \(\overrightarrow{A D}=\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C D}=(180 \hat{i}-180 \hat{j}) \mathrm{m}\)

∴ \(\tan \theta=\frac{E D}{A E}=\frac{B C}{A E}=\frac{-180}{180}=-1\)

or, θ = 45°

So \(\overrightarrow{A D}\) is inclined at 45° with the x-axis in the negative direction.

∴ \(|\overrightarrow{A D}|=\sqrt{(180)^2+(-180)^2}=180 \sqrt{2} \mathrm{~m}\)

Short Answer Questions on Vector Subtraction

Example 3. A particle is moving in a circular path with a uniform speed v. Show that, when the particle traverses through an angle of 120°, the change in its velocity is √3v.
Solution:

The initial velocity \(\overrightarrow{v_1}\) and the final velocity \(\overrightarrow{v_2}\) are shown.

They are drawn from the same initial point.

Here, \(\left|\overrightarrow{v_1}\right|=\left|\vec{v}_2\right|=v\)

Change in velocity = \(\vec{v}_2-\vec{v}_1\)

∴ \(\left|\vec{v}_2-\vec{v}_1\right|=\sqrt{v_1^2+v_2^2-2 v_1 v_2 \cos 120^{\circ}}\)

= \(\sqrt{v^2+v^2-2 v \cdot v \cdot\left(-\frac{1}{2}\right)}=\sqrt{3 v^2}=\sqrt{3} \nu\)

WBCHSE Class 11 Physics Transmission Of Heat

Unit 7 Properties Of Bulk Matter Chapter 9 Transmission Of Heat Short Answer Type Questions

Question 1. At what temperature does a body stop radiating?
Answer:

A body stop radiating at absolute zero or 0 K temperature.

Question 2. The ratio of thermal conductivity of two rods is 4:3. The radii and thermal resistances of the two rods are equal. Calculate the ratio of their length.
Answer:

The ratio of thermal conductivity of two rods is 4:3. The radii and thermal resistances of the two rods are equal.

Thermal resistance = \(\frac{1}{k A} \frac{l}{k}=\frac{1}{\pi r^2}\)

The radii and thermal resistance of the two rods are equal.

∴ \(\frac{1}{k_1} \frac{l_1}{\pi r^2}=\frac{1}{k_2} \frac{l_2}{\pi r^2} \quad \text { or, } \frac{l_1}{l_2}=\frac{k_1}{k_2}=\frac{4}{3}\)

Hence, ratio of their lengths is 4 : 3.

Question 3. The thermal conductivity of aluminum is 0.5 cal • s^-1 • cm^-1 • °C^-1. Express it in SI system.
Answer:

The thermal conductivity of aluminum is 0.5 cal • s^-1 • cm^-1 • °C^-1.

Thermal conductivity of aluminium, k = 0.5 cal • s^-1 • cm^-1 • °C^-1 [In CGS system]

In SI, k = 0.5 x 4.2 x 10² J • m^-1 • K^-1 • s^-1

= 210 J • m^-1 • K^-1 • s^-1

WBCHSE Physics Chapter 9 Solutions 

Question 4. Consider a blackbody radiation In a cubical box at absolute temperature T. If the length of each side of the box is doubled and the temperature of the walls of the box and that of the radiation is halved, then the total energy

1. Halves
2. Doubles
3. Quadruples
4. Remains the same

Answer:

According to Stefan’s law, \(E \propto A \sigma T^4\)

∴ \(\frac{E_1}{E_2}=\frac{A_1 T_1^4}{A_2 T_2^4}=\frac{A_1 T_1^4}{\left(4 A_1\right)\left(\frac{1}{2} T_1^4\right)}\)

because \(A_2=4 A_1 \text { and } T_2=\frac{1}{2} T_1\)

or, \(\frac{E_1}{E_2}=4: 1\)

The option 3 is correct.

WBBSE Class 11 Transmission of Heat Short Answer Questions

Question 5. The same quantity of ice is filled in each of the two metal containers P and Q having the same size, shape, and wall thickness but made of different materials. The containers are kept in identical surroundings. The ice in P melts completely in time t₁ whereas that in Q takes a time t₂. The ratio of thermal conductivities of the materials of P and Q is

1. \(t_2: t_1\)
2. \(t_1: t_2\)
3. \(t_1^2: t_2^2\)
4. \(t_2^2: t_1^2\)

Answer:

⇒ \(\left(k A \frac{d T}{d x}\right) t=m L\),

so, \(k \propto \frac{1}{t}\).

Hence, \(\frac{k_1}{k_2}=\frac{t_2}{t_1}\)

The option 1 is correct.

WBCHSE Class 11 Physics Transmission of Heat 

Question 6. A solid maintained at t₁°C Is kept In an evacuated chamber at temperature t°C(t₂>>t₁). The rate of boat absorbed by the body is proportional to

1. \(t_2^4-t_1^4\)
2. \(\left(t_2^4+273\right)-\left(t_1^4+273\right)\)
3. \(t_2-t_1\)
4. \(t_2^2-t_1^2\)

Answer:

None of the options Is correct.

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Question 7. If the temperature of the sun gets doubled, the rate of energy received on the earth will increase by a factor of

1. 2
2. 4
3. 8
4. 16

Answer:

The rate of energy received on the earth,

E ∝ T⁴ [T = temperature of the sun]

∴ Now, if the temperature of the sun gets doubled, then the rate of energy received on the earth will increase 2⁴ = 16 times.

The option 4 is correct.

Question 8. The temperature of the water of a pond is 0°C while that of the surrounding atmosphere is -20 °C. If the density of ice is ρ, coefficient of thermal conductivity is k and the latent heat of melting is L then the thickness Z of the ice layer formed increases as a function of time as,

1. \(Z^2=\frac{60 k}{\rho L} t\)
2. \(Z=\sqrt{\frac{40 k}{\rho L}} t\)
3. \(Z^2=\frac{40 k}{\rho L} \sqrt{t}\)
4. \(Z^2=\frac{40 k}{\rho L} t\)

Answer:

If the thickness of the ice layer increases by Z in time t, then

t = \(\frac{\rho L}{2 k \theta} Z^2\)

Now, \(\theta =[0-(-20)]=20^{\circ} \mathrm{C}\)

∴ t = \(\frac{\rho L}{2 k \cdot 20} Z^2 \quad \text { or, } Z^2=\frac{40 k t}{\rho L}\)

The option D is correct.

Understanding Heat Transfer Short Answer Questions

Question 9. The temperature of a blackbody radiation enclosed in a container of volume V in increased from 100°C to 1000°C. The heat required in the process is

1. 4.79 x 10^-4 cal
2. 9.21 x 10^-5 cal
3. 2.17 x 10^-4 cal
4. 7.54 x 10^-4 cal

Answer: Data insufficient.

WBCHSE Class 11 Physics Transmission of Heat 

Question 10. Three rods of copper, brass, and steel are welded together to form a Y -shaped structure. Aren of a cross-section of each rod Is 4 cm². The end of the copper rod Is maintained at 100 °C whereas ends of brass and stool are kept at 0 °C. Lengths of the copper, brass, and steel rods are 46,13 and 12 cm respectively. The rods are thermally Insulated from surroundings except at the ends. Thermal conductivities of copper, brass, and steel are 0.92, 0.26, and 0.12 in CGS units, respectively. Rate of heat flow through copper rod is

1. 1.2 cal/s
2. 2.4 calls
3. 4.0 cal/s
4. 6.0 cal/s

Answer:

⇒ \(Q=Q_1+Q_2\)

or, \(\frac{0.92 \times 4(100-T)}{46}\)

= \(\frac{0.26 \times 4 \times(T-0)}{13}+\frac{0.12 \times 4 \times T}{12} \)

or, \(200-2 T=2 T+T\)

or, \(T=40^{\circ} \mathrm{C}\)

Q = \(\frac{0.92 \times 4 \times 60}{46}=4.8 \mathrm{cal} / \mathrm{s}\)

The option 3 is correct.

WBCHSE Physics Chapter 9 Solutions 

Question 11. Consider a spherical shell of radius R at temperature 7. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume u = \(\frac{U}{V} \propto T^4\) and pressure \(P=\frac{1}{3}\left(\frac{U}{V}\right)\). If the shell now undergoes an adiabatic expansion the relationship between T and R is

1. \(T \propto e^{-R}\)
2. \(T \propto e^{-3 R}\)
3. \(T \propto \frac{1}{R}\)
4. \(T \propto \frac{1}{R^3}\)

Answer:

For 1 mol ideal gas, \(P V=R_0 T\) [where R₀ is the universal gas constant]

or, \(P=\frac{R_0 T}{V}=\frac{1}{3}\left(\frac{U}{V}\right) \propto T^4\)

So, \(\frac{1}{V} \propto T^3\)

or, \(\frac{1}{\frac{4}{3} \pi R^3} \propto T^3\)

or, \(T^3 \propto \frac{1}{R^3}\)

∴ \(T \propto \frac{1}{R}\)

The option 3 is correct.

Key Concepts in Heat Transfer Short Answers

Question 12. A certain quantity of water cools from 70 °C to 60 °C in the first 5 minutes and to 54 °C in the next 5 minutes. The temperature of the surroundings is

1. 45°C
2. 20°C
3. 42°C
4. 10°C

Answer:

From Newton’s law of cooling, \(\frac{(70+273)-\theta_0}{(60+273)-\theta_0}=\frac{(60+273)-\theta_0}{(54+273)-\theta_0}\)

[θ₀ = temperature of the surrounding]

or, θ₀ = 318 K = 45°C

The option 1 is correct

Question 13. The two ends of a metal rod are maintained at temperatures 100 °C and 110 °C. The rate of heat flow in the rod is found to be 4.0 J/s. If the ends are maintained at temperatures 200 °C and 210 °C, the rate of heat flow will be

1. 44.0 J/s
2. 16.8 J/s
3. 8.0 J/s
4. 4.0 J/s

Answer:

Rate of heat flow is proportional to the difference in temperatures of the two ends.

∴ \(\frac{\left(\frac{d \theta}{d t}\right)_{1 \text { st case }}}{\left(\frac{d \theta}{d t}\right)_{2 \mathrm{nd} \text { case }}}=\frac{\left(\theta_1-\theta_2\right)_{1 \text { st case }}}{\left(\theta_1-\theta_2\right)_{2 \mathrm{nd} \text { case }}}\)

or, \(\left(\frac{d \theta}{d t}\right)_{2 \text { nd case }}=\left(\frac{d \theta}{d t}\right)_{1 \text { st case }} \times \frac{\left(\theta_1-\theta_2\right)_{2 \text { nd case }}}{\left(\theta_1-\theta_2\right){ }_{1 \text { st case }}}\)

= \(4 \times \frac{10}{10}=4 \mathrm{~J} / \mathrm{s}\)

= 4 x 10/10 = 4 J/s

The option 4 is correct.

WBCHSE Physics Chapter 9 Solutions 

Question 14. A black body is at a temperature of 5760 K. The energy of radiation emitted by the body at wavelength 250 nm is U₁, at wavelength 500 nm is U_2, and that at 1000 nm is U₃. Wien’s constant b = 2.88 x 10⁶ nm • K. Which of the following is correct?

1. U₃ = 0
2. U₁ >U₂
3. U₂>U₁
4. U₁ = 0

Answer:

According to the Wien’s displacement law, \(\lambda_m T =b\)

or, \(\lambda_m=\frac{b}{T}=\frac{2.88 \times 10^6}{5760}=500 \mathrm{~nm}\)

Now, the maximum amount of emitted radiation corresponding to λ_m is U₂.

From graph, U₁ < U₂ > U₃

∴ U₂ > U₁

The option 3 is correct.

Fourier’s Law of Heat Conduction Short Answers

Class 11 Physics Heat Transfer Questions 

Question 15. In a certain planetary system, it is observed that one of the celestial bodies having a surface temperature of 200 K, emits radiation of maximum intensity near the wavelength 12μm. The surface temperature of a nearby star which emits light of maximum intensity at a wavelength λ = 4800Å, is

1. 7500 K
2. 5000 K
3. 2500 K
4. 10000 K

Answer:

According to the Wien’s displacement law, \(\lambda_m T=\text { constant }\)

∴ \(\lambda_{m_1} T_1=\lambda_{m_2} T_2\)

or, \(T_2=T_1 \frac{\lambda_{m_1}}{\lambda_{m_2}}=200 \times \frac{12 \times 10^{-6}}{4800 \times 10^{-10}}=5000 \mathrm{~K}\)

The option 2 is correct.

Transmission of Heat Class 11 Notes 

Question 16. A wall consists of alternating blocks of length ‘ d ’ and coefficient of thermal conductivity K₁ and K₂ respectively as shown. The cross-sectional area of the blocks are the same. The equivalent coefficient of thermal conductivity of the wall between left and right is

1. \(\frac{K_1+K_2}{2}\)
2. \(\frac{2 K_1 K_2}{K_1+K_2}\)
3. \(\frac{K_1+K_2}{3}\)
4. \(\frac{3 K_1 K_2}{K_1+K_2}\)

Answer:

Thermal resistance of the blocks, \(R_1=R_3=R_5=\frac{1}{K_1} \frac{d}{A} \text { and } R_2=R_4=R_6=\frac{1}{K_2} \frac{d}{A}\)

If R be the equivalent thermal resistance of the parallel combination, then

⇒ \(\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\cdots=3 \frac{K_1 A}{d}+3 \frac{K_2 A}{d}=\frac{3 A}{d}\left(K_1+K_2\right)\)

If K be the equivalent thermal conductivity of the wall, then

⇒ \(\frac{1}{R}=\frac{6 A}{d} K\)

∴ \(\frac{6 A}{d} K=\frac{3 A}{d}\left(K_1+K_2\right)\)

or, \(K=\frac{K_1+K_2}{2}\)

The option 1 is correct

Question 17. The power radiated by a black body is P and it radiates maximum energy at wavelength, \(\lambda_0\). If the temperature of the black body is now changed so that it radiates maximum energy at wavelength \(\frac{3}{4} \lambda_0\), the power radiated by it becomes nP. The value of n is

1. \(\frac{256}{81}\)
2. \(\frac{4}{3}\)
3. \(\frac{3}{4}\)
4. \(\frac{81}{256}\)

Answer:

From Wien’s displacement law

∴ \(\lambda_{\max } \cdot T=\text { constant }\)

∴ \(\lambda_1 T_1=\lambda_2 T_2 \quad \text { or, } \lambda_0 T_1=\frac{3}{4} \lambda_0 T_2 \quad \text { or, } \frac{T_2}{T_1}=\frac{4}{3}\)

∴ \(\frac{P_2}{P_1}=\left(\frac{T_2}{T_1}\right)^4 \quad \text { or, } \frac{n P}{P}=\left(\frac{4}{3}\right)^4\)

or, \(n=\frac{256}{81}\)

The option 1 is correct.

Heat Transfer MCQs for Class 11 

Real-Life Examples of Heat Transmission

Question 18. Using it find the time taken by a hot food in a pan to cool from 71 °C to 69°C If the room temperature is 20°C. The food cools from 94°C to 86°C in 2 minutes.
Answer:

The time taken by a hot food in a pan to cool from 71 °C to 69°C If the room temperature is 20°C. The food cools from 94°C to 86°C in 2 minutes

Let θ₀ = room temperature; t = time taken by the hot food to cool from θ₁ °C to θ₂ °C Then, from

Newton’s law of cooling, \(k t=\log \frac{\theta_1-\theta_0}{\theta_2-\theta_0}, where k=\mathrm{a} constant\)

In this problem, \(\theta_0=20^{\circ} \mathrm{C}\).

In the hotter region, \(\theta_1=94^{\circ} \mathrm{C}, \theta_2=86^{\circ} \mathrm{C} and t=2 \mathrm{~min}=120 \mathrm{~s}\).

In the colder region, \(\theta_1=71^{\circ} \mathrm{C} and \theta_2=69^{\circ} \mathrm{C}\)

So, \(120 k=\log \frac{94-20}{86-20}=\log \frac{74}{66}=0.0497\)

and \(k t=\log \frac{71-20}{69-20}=\log \frac{51}{49}=0.0174\)

Then, \(\frac{120}{t}=\frac{0.0497}{0.0174}, \quad or, t=\frac{120 \times 0.0174}{0.0497}=42 \mathrm{~s}\)

WBCHSE Physics Chapter 9 Solutions 

Question 19. Show graphically the temperature variation with time associated with a cooling hot body. Why burns from steam are more serious than those from boiling water?
Answer:

Shows the exponentially decreasing nature of temperature with the passage of time.

Every 1 g of steam at 100 °C contains a latent heat excess of about 540 cal, compared to 1 g of water at 100°C.

So, steam releases a greater amount of energy, than that released by boiling water, to a cold body brought to its contact. That is why steam burns are more serious.

Heat Transfer MCQs for Class 11 

Question 20. What is thermal conductivity of perfect heat conductors?
Answer:

Thermal conductivity of perfect heat conductors

Thermal conductivity of a perfect heat conductor is infinity.

Unit 7 Properties Of Bulk Matter Chapter 9 Transmission Of Heat Long Answer Type Questions

Question 1. Two rods A and B are of equal length. Each rod has the ends at temperature T₁ and T₂ respectively, what is the condition that will ensure equal rates of flow of heat through the rods A and B?
Answer:

Given:

Two rods A and B are of equal length. Each rod has the ends at temperature T₁ and T₂ respectively

Let the cross-section and conductivity of rod A be α₁ and k₁ and those of rod B be α₂ and k₂ respectively.

Also, let the length of each rod be l.

Since, heat current in both rods are the same, \(\frac{Q}{t}=\frac{k_1 \alpha_1\left(T_1-T_2\right)}{l}\)

= \(\frac{k_2 \alpha_2\left(T_1-T_2\right)}{l}\)

or, \(k_1\alpha_1=k_2 \alpha_2\)

This is a required condition.

Read and Learn More Class 11 Physics Long Answer Questions

Question 2. At what temperature will a wooden and a metal block appear to be equally Hot or cold upon touching them?
Answer:

If the temperatures of the wooden block and of the metal block are the same as our body temperature, then they would appear to be equally hot or cold on touch.

Question 3. If a piece of paper Is wrapped around a wooden rod and held over a lighted candle the paper burns to ashes almost immediately. On the other hand, If the piece of paper Is wrapped on a metal rod and similarly held over the lighted candle then the paper takes some time to burn. Explain.
Answer:

Given:

If a piece of paper Is wrapped around a wooden rod and held over a lighted candle the paper burns to ashes almost immediately. On the other hand, If the piece of paper Is wrapped on a metal rod and similarly held over the lighted candle then the paper takes some time to burn.

Wood is a bad conductor of heat Obviously, the heat the wood receives from the candle is not conducted through it. So, the region that is heated retains the heat, as a result of which the paper reaches its ignition temperature and burns to ashes almost immediately.

If the wood is replaced by a metal rod, the heat is conducted through the latter, as the metal is a good conductor of heat. So, it takes some time for the paper to reach its ignition point and to start burning.

Question 4. Thermal conductivity of air is less than that of felt, still felt is more widely used as heat insulator. Why?
Answer:

Given:

Thermal conductivity of air is less than that of felt, still felt is more widely used as heat insulator.

Air has a thermal conductivity less than felt. Yet when a hot body is kept in air, a convection current develops. So the body starts losing heat. On the other hand, this is not the case if the body is covered with felt.

• This is because felt itself is a bad conductor of heat and the narrow perforations within the felt fibres are filled up with air. Air being a bad conductor of heat, no heat loss takes place due to conduction.
• Again, the body being covered with felt, it is not open to air. Hence, no heat loss takes place due to convec¬tion as well. So, the hot body can retain the heat. This is why felt is more widely used as a heat insulator.

Question 5. Why snow is a better insulator than ice?
Answer:

Snow is a better insulator than ice:

Ice being solid and having no pores do not enclose air. Snow being porous encloses air in it. Air is a bad conductor of heat hence snow is a better insulator.

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Question 6. We feel wanner when the sky is cloudy—explain.
Answer:

Given:

We feel wanner when the sky is cloudy

Heat waves cannot penetrate cloud. Earth’s surface receives heat during daytime and radiates that out during night hours, thereby cooling the surface, Cloud acta m a reflector of this radiated float and sends It back to earth, So the cooling effect Is less and the nights are warm.

Question 7. A metal piece cooled In liquid air (temperature -180°C) produces a burning sensation when It is touched, why?
Answer:

Given:

A metal piece cooled In liquid air (temperature -180°C) produces a burning sensation when It is touched

The temperature of liquid air is -180°C l.e., much less than our body temperature. Hence, the temperature difference between the metal piece and our body is very large.

Metal Is a good conductor of heat, So, when the metal piece Is touched, heat conducts very quickly from the body to the metal piece. This very rapid loss of heat will produce a burning sensation, as our body cannot withstand It.

Question 8. Three specimens made of the same material are kept at 200°C In a room. They arc respectively a sphere, a cube, and a circular lamina of the same mass. Compare the rates of their cooling.
Answer:

Given:

Three specimens made of the same material are kept at 200°C In a room. They arc respectively a sphere, a cube, and a circular lamina of the same mass.

The mass, the temperature, and the material of the three specimens being the same, the rate of radiation from the surfaces is directly proportional to the surface areas. Hence, the circular lamina, having the largest surface area among the three, cools the fastest. The sphere, being of the smallest surface area, will have the slowest rate of cooling.

Fourier’s Law of Heat Conduction Long Answer Questions

Question 9. How does a heater coll attain a steady temperature when current is on, after an Initial steady rise In temperature?
Answer:

Given:

Current through the coil of the heater produces heat and the coil shows a steady rise in temperature initially. With the increase in temperature rate of radiation from the coil also increases.

At a certain temperature, the rate of heating due to current equals the rate of radiation, i.e., the heater cannot retain any heat for a further rise in temperature. So, the coil maintains its constant temperature.

Question 10. Explain why the climate of a harbor town is more temperate than that of a town in a desert at the same latitude.
Answer:

Sea water has a high specific beat So the temperature of a harbour town does not increase very much due to heat absorption in the daytime, also if does not decrease very much due to the radiation of heat in bight time.

On the other hand desert sand has a low specific heat so due to heat absorption and radiation the temperature variation is much Higher, That is why (he climate of a harbor town is more temperate.

Question 11. A solid copper sphere of radius ft and a hollow sphere of the same material of external and internal radii R and r are raised to the same temperature and are allowed to cool under similar conditions. Which one of the two will have a faster rate of cooling?
Answer:

Given:

A solid copper sphere of radius ft and a hollow sphere of the same material of external and internal radii R and r are raised to the same temperature and are allowed to cool under similar conditions.

The external radius of the hollow sphere and the radius pf the solid sphere being the same, the outer surface areas of the two spheres are equal. If they are heated to the same temperature and are exposed in the same surroundings, the rate of loss of heat will be the same as their outer surface areas and Initial temperatures are equal.

But, for the same rate of heat loss, the temperature of the hollow sphere will drop at a faster rate as Its mass is less. So the hollow sphere will have a faster rate of cooling.

Question 12. Bulbs of two thermometers are coated with lampblack and sliver. Compare their readings when they are

1. Kept immersed In water inside a dark room,
2. In open daylight and
3. In a clear night.

Answer:

1. insulation is produced because of air being trapped between the bulb and the coating of lampblack. This will delay the thermometer to attain the temperature of water, Silver being a good conductor, the thermometer with the silver-coated bulb will attain thermal equilibrium with water instantaneously.

2. Lamp black is a good absorber and a bad reflector of heat. On the other hand silver is a good reflector and a bad absorber of heat. Hence the bulb coated with lamp black will absorb larger amount of heat in the sun¬light. On the other hand, the bulb coated with silver will reflect most of the heat. So, the first thermometer will show a higher reading.

3. In a clear cloudless night, the bulb coated with lamp black will radiate faster as it is a good absorber and hence a good radiator of heat. Hence, its temperature reading will be lower.

Real-Life Examples of Heat Transmission

Question 13. Two spheres of the same material and of radii 1 m and 4 m are kept at 4000 K and 2000 K respectively, Showing that the heat radiated from them per second Is the same.
Answer:

We know that the relations among the rate of radiation (E), surface area (A), and absolute temperature (T) are \(E \propto A \text { and } E \propto T^4 \quad \text { or, } E \propto A T^4\)

∴ In case of the two spheres in question, \(\frac{E_1}{E_2}=\frac{A_1 T_1^4}{A_2 T_2^4}=\left(\frac{1}{4}\right)^2 \times\left(\frac{4000}{2000}\right)^4=\frac{1}{16} \times 16=1 \text { or, } E_1=E_2\)

Question 14. Water boils faster in a metal vessel with a rough and black bottom than in a metal vessel with a smooth bottom. Explain.
Answer:

Given:

Water boils faster in a metal vessel with a rough and black bottom than in a metal vessel with a smooth bottom.

The rough and dark-bottomed surface has a greater ability to absorb heat than the smooth-bottomed metal vessel. The smooth surface of the metal vessel reflects a reasonable part of the incident heat. Obviously, the water in the metal vessel with a rough and black bottom will boil faster.

Question 15. The bottom surface of a cooking utensil is made j rough and black, whereas a calorimeter surface is kept smooth and shiny. Explain.
Answer:

Given:

The bottom surface of a cooking utensil is made j rough and black, whereas a calorimeter surface is kept smooth and shiny.

The basic calorimetric principle (heat lost = heat gained) is easily applicable when no heat is exchanged between the calorimeter and its surroundings. The calorimeter surface is made bright and shiny to reduce heat exchange by radiation. But a rough and black-bottomed surface of a cooking utensil absorbs heat faster from the source.

Question 16. When we cover the bulb of a thermometer with a piece of quilt why is the reading of the instrument comparatively less? What changes in the reading will be recorded if the quilt is soaked?

1. In either or
2. In water?

Answer:

A quilt is a bad conductor of heat because of the wool and the air trapped within it. So the temperature of the surroundings cannot be attained by the bulb covered with a piece of quilt. Therefore, the reading of the thermometer is comparatively less.

1. If the quilt is soaked in ether, which is a highly volatile substance, the ether will evaporate quickly and take its latent heat of vaporization from the bulb of the thermometer. So, the reading in the thermometer will fall quickly.

2. If water is used instead of ether, it is less volatile, the reading in the thermometer will fall slowly.

Question 17. Two friends, waiting for their third companion in a restaurant, ordered two cups of tea. While one of them poured milk into his cup of tea, the other waited till their friend arrived and then mfwH the cold milk with the tea. Whose cup of tea would be comparatively warmer?
Answer:

Given:

Two friends, waiting for their third companion in a restaurant, ordered two cups of tea. While one of them poured milk into his cup of tea, the other waited till their friend arrived and then mfwH the cold milk with the tea.

The friend who mixed milk with the tea earlier would enjoy a warmer cup of tea. We know from Newton’s law of cooling, the more the difference in the temperature of the substance with its surroundings, more is the rate of heat radiation from it.

Since the first friend’s cup of tea became comparatively cooler on mixing the colder milk, the rate of heat radiation diminished in this case. Again, the second friend poured the cold milk afterward so, the rate of heat radiation from his tea was faster and hence it became colder

Thermal Conductivity and Its Importance

Question 18. What will be the nature of v_m T graph for a per fleetly black body?
Answer:

We know that for a perfectly black body

⇒ \(\lambda_m T=\text { constant }\)

or, \(\frac{T}{\nu_m}=\text { constant }\)

∴ \(\nu_m \propto T\)

So, v_m -T graph will be a straight line passing through the origin.

Question 19. Five rods of the same dimensions are arranged as shown. They have thermal conductivities k₁, k₂, k₃, k₄ and k₅. When points A and B are maintained at different temperatures, under what condition no heat flow through the central rod?

Answer:

The arrangement of the rods is similar to a bal¬anced Wheatstone Bridge.

No heat flows through the central rod if we have,

∴ \(\frac{k_1}{k_2}=\frac{k_3}{k_4} \text { or, } k_1 k_4=k_2 k_3\)

Question 20. The plots of intensity (I) versus wavelength (λ) for three black bodies at temperatures T₁, T₂, and T₃ respectively are as shown. What will be the relation of these temperatures?
Answer:

Given:

The plots of intensity (I) versus wavelength (λ) for three black bodies at temperatures T₁, T₂, and T₃ respectively are as shown.

According to Wein’s displacement law for blackbody radiation,

λ_m = constant

Therefore, we can conclude that the higher the value of λ_m, the smaller is the value of temperature.

According to the figure, λ₂ > λ₃ > λ₁

Therefore, T₁ > T₃ > T₂.

Factors Affecting Heat Transfer Rates

Question 21. A planet is situated at a mean distance d from the sun and its average surface temperature is T. Suppose that the planet receives energy only from the sun and loses energy from its surface by the process of radiation. Ignore all atmospheric actions. If \(T \propto d^{-n}\) then show that n = \(\frac{1}{2} \text {. }\)
Answer:

Given:

A planet is situated at a mean distance d from the sun and its average surface temperature is T. Suppose that the planet receives energy only from the sun and loses energy from its surface by the process of radiation.

Suppose, the power of radiation by the sun = P, radius of the planet = R

So, energy received by the planet = \(\frac{P}{4 \pi d^2} \times \pi R^2\)

Energy radiated by the planet = 4πR² • σT⁴

Therefore, for thermal equilibrium, \(\frac{P}{4 \pi d^2} \times \pi R^2=4 \pi R^2 \sigma T^4\)

or, \(T^4=\frac{P}{16 \pi \sigma d^2}\)

∴ \(T^4 \propto \frac{1}{d^2} \quad \text { or, } T \propto \frac{1}{d^{1 / 2}}\)

or, \(T \propto d^{-1 / 2}\)

Comparing with the given equation \(T \propto d^n \text {, we get } n=\frac{1}{2}\)

Unit 7 Properties Of Bulk Matter Chapter 9 Transmission Of Heat Multiple Choice Questions And Answers

Question 1. Through a solid medium heat

1. Cannot be conducted but convection occurs
2. Cannot be transmitted by convection
3. Cannot be transmitted by conduction
4. Can be transmitted by both conduction and convection

Answer: 2. Cannot be transmitted by convection

Question 2. Through a solid or a liquid medium heat

1. Cannot be conducted but convection occurs
2. Cannot be transmitted by convection
3. Cannot be transmitted by conduction
4. Can be transmitted by both conduction and convection

Answer: 4. Can be transmitted by both conduction and convection

Read And Learn More WBCHSE Class 11 Physics MCQs

Question 3. During the conduction of heat

1. The molecules of the substance remain static
2. The molecules of the substance move in the direction of heat conduction
3. The molecules of the substance move randomly in different directions
4. The molecules vibrate around their equilibrium positions

Answer: 4. The molecules vibrate around their equilibrium positions

Question 4. An example of a liquid which is a good conductor of heat is

1. Water
2. Alcohol
3. Mercury
4. Aqueous solution of an acid or a base or a salt

Answer: 3. Aqueous solution of an acid or a base or a salt

Question 5. If the thickness of a plate is d, the cross-sectional area is A and the difference in temperature of its two ends is T then the amount of heat conducted through it in unit time is

1. \(Q \propto d A T\)
2. \(Q \propto \frac{d T}{A}\)
3. \(Q \propto \frac{A T}{d}\)
4. \(Q \propto \frac{T}{d A}\)

Answer: 3. \(Q \propto \frac{T}{d A}\)

WBBSE Class 11 Heat Transfer MCQs

Question 6. The SI unit of the coefficient of thermal conductivity is

1. J · m^-1 K^-1
2. W · m^-1 · K^-1
3. J · m^-2 · K^-1
4. J · m^-2 · K^-2

Answer: 2. W · m^-1 · K^-1

Question 7. The coefficient of thermal conductivity of copper is nine times that of steel in the composite cylindrical bar shown in the figure. What will be the temperature at the 18cm 6cm junction of copper and steel?

1. 33°C
2. 90°C
3. 75°C
4. 65°C

Answer: 3. 75°C

Question 8. The area of the cross-section of a conductor of thickness d is A, the coefficient of thermal conductivity of its material is k and the difference in temperature of its two ends is T. If Q amount of heat is conducted through it in time t then the thermal resistance of the conductor is

1. \(\frac{Q}{t}\)
2. \(\frac{A T}{d}\)
3. \(\frac{k A}{d}\)
4. \(\frac{d}{k A}\)

Answer: 4. \(\frac{d}{k A}\)

Question 9. The ratio of the lengths, cross-sectional areas and the differences in temperature of the two ends of two conducting rods are 2:3 each. If the rate of heat conduction through them be equal, then the ratio of the coefficients of thermal conductivity of their materials is

1. 2:3
2. 3:2
3. 4:9
4. 9:4

Answer: 2. 3:2

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Question 10. Thermal resistance of a substance denotes

1. The coefficient of thermal conductivity
2. The rate of heat conduction
3. The reciprocal of the coefficient of thermal conductivity
4. The reciprocal of the rate of heat conduction

Answer: 3. The reciprocal of the coefficient of thermal conductivity

Question 11. The rate of heat conduction through the window of a room is 273 J • s^-1 when the difference in temperatures of air inside and outside the room is 20°C. If this difference in temperature be 20 K, then the rate of heat conduction will be

1. 293 J · s^-1
2. 273 J · s^-1
3. 253 J · s^-1
4. 263 J · s^-1

Answer: 2. 273 J · s^-1

Thermal Properties and Heat Transfer MCQs

Question 12. The thicknesses of two metallic plates of a combined strip are equal. The coefficients of their thermal conductivities are k₁ and k₂ and the temperatures of their hot and cold ends are T₁ and T₂ respectively. The temperature of the junction of the combined strip will be

1. \(\frac{k_1 T_1+k_2 T_2}{k_1+k_2}\)
2. \(\frac{k_1 T_1-k_2 T_2}{k_1-k_2}\)
3. \(\frac{k_2 T_1+k_1 T_2}{k_1+k_2}\)
4. \(\frac{k_2 T_1-k_1 T_2}{k_2-k_1}\)

Answer: 1. \(\frac{k_1 T_1+k_2 T_2}{k_1+k_2}\)

Question 13. The areas of a cross-section of two metallic rods of equal lengths, having coefficients of thermal conductivity k₁ and k₂, are A₁ and A₂ respectively. The rate of heat conduction in them will be the same if

1. \(k_1 A_1^2=k_2 A_2^2\)
2. \(k_1 A_2=k_2 A_1\)
3. \(k_1 A_1=k_2 A_2\)
4. \(k_1^2 A_1=k_2^2 A_2\)

Answer: 3. \(k_1 A_2=k_2 A_1\)

Question 14. The temperatures of the two ends A and B of a rod AB of length 20 cm are 100°C and 0°C respectively. In thermal equilibrium the temperature of the rod at a distance 6 cm from the end A will be

1. 50°C
2. 80°C
3. 70°C
4. 60°C

Answer: 3. 70°C

Question 15. Three rods made of the same material and having the same cross-section have been joined as shown. Each rod is of the same length. The left and right ends are kept at 0°C and 90°C respectively. The temperature of the junction of the three rods will be

1. 45°C
2. 60°C
3. 30°C
4. 20°C

Answer: 2. 60°C

Question 16. Two metallic rods of lengths x₁ and x₂ have equal cross-sectional area. Their thermal conductivities are k₁ and k₂ respectively. If they are connected in series, then the equivalent thermal conductivity of the combination will be

1. \(\frac{\frac{x_1}{k_1}+\frac{x_2}{k_2}}{k_1+k_2}\)
2. \(\frac{x_1+x_2}{\frac{x_1}{k_1}+\frac{x_2}{k_2}}\)
3. \(\frac{k_1+k_2}{\frac{x_1}{k_1}+\frac{x_2}{k_2}}\)
4. \(\frac{\frac{x_1}{k_1}+\frac{x_2}{k_2}}{x_1+x_2}\)

Answer: 2. \(\frac{x_1+x_2}{\frac{x_1}{k_1}+\frac{x_2}{k_2}}\)

Question 17. The dimension of thermal resistance is

1. \(\mathrm{ML}^2 \mathrm{~T}^{-2} \Theta\)
2. \(\mathrm{M}^{-2} \mathrm{~L}^{-2} \mathrm{~T}^2 \Theta^3\)
3. \(\mathrm{ML}^2 \mathrm{~T}^{-3} \Theta\)
4. \(\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^3 \Theta\)

Answer: 4. \(\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^3 \Theta\)

Short Notes on Modes of Heat Transfer with MCQs

Question 18. The process by virtue of which water in a pond gets heated at noon in summer up to a certain depth is

1. Conduction
2. Convection
3. Conduction as well as convection
4. Convection as well as radiation

Answer: 1. Conduction

Question 19. If the velocity of light in a vacuum be c then the velocity of radiant heat in any other medium is

1. Always equal to c
2. Always greater than c
3. Always less than c
4. Equal to or less than c

Answer: 3. Always less than c

Question 20. We can protect our body from the heat of the sun by means of an umbrella. From this, it can be said that

1. Radiant heat travels in a straight line
2. Velocity of radiant heat is equal to that of light
3. Radiant heat does not warm the medium
4. Radiant heat spreads in the form of waves

Answer: 1. Radiant heat travels in a straight line

Question 21. With the increase in temperature of a radiator

1. Intensity of radiant heat increases
2. The wavelength of radiant heat increases
3. The absorptive power of the radiator increases
4. The emissive power of the radiator increases.

Answer: 1. Intensity of radiant heat increases

Question 22. The specific design, for which heat lost by radiation from a thermos flask is the least, is

1. Glass vessel with stopper made of cork
2. Shiny wall of the glass vessel
3. Vacuous space in between two walls of the glass vessel
4. Both the characteristics 1 and 3

Answer: 2. Shiny wall of the glass vessel

Question 23. Felt is widely used as an insulator of heat instead of air, because

1. The coefficient of thermal conductivity of air is more than that of felt
2. Convection of heat occurs in air but not in felt
3. No air remains trapped inside the fibres of felt
4. Felt is a better insulator of heat than air

Answer: 2. Convection of heat occurs in air but not in felt

Step-by-Step Solutions to Heat Transfer MCQs

Question 24. If the temperature of the sun gets doubled, the rate of energy received on earth will be

1. Doubled
2. Quadrupled
3. 16 times
4. 8 times

Answer: 3. 16 times

Question 25. The coefficients of radiation of two bodies of equal shapes and volumes are 0.2 and 0.8 respectively. If they radiate heat at the same rate then the ratio of their temperatures will be

1. √3:1
2. √2:1
3. 1:√5
4. 1:√5

Answer: 2. √2:1

Question 26. If the temperature of a black body rises from T to 2 T, how many times will its rate of radiation be?

1. 4
2. 2
3. 16
4. 8

Answer: 3. 16

Question 27. The radii of two spheres made of the same material are 1 m and 4 m respectively. If their temperatures are 4000 K and 2000 K respectively, then the ratio of the amount of heat radiated per second is

1. 1:1
2. 16:1
3. 4:1
4. 1:9

Answer: 1. 16:1

Question 28. Which law is used to determine the temperature of stars?

1. Stefan’s law
2. Wien’s law
3. Kirchhoff’s law
4. Planck’s law

Answer: 1. Stefan’s law

Question 29. When the body has the same temperature as that of its surroundings, it

1. Does not radiate heat
2. Radiates same quantity of heat as it absorbs
3. Radiates less quantity of heat as it receives from surroundings
4. Radiates more quantity of heat as it receives from surroundings

Answer: 2. Radiates same quantity of heat as it absorbs

Question 30. A piece of iron appears colder in winter than a piece of wood when touched because

1. The temperature of iron is less than that of wood
2. The temperature of iron is less than our body temperature but that of wood is higher on
3. Wood being an insulator of heat, no heat from it can be transmitted to our body
4. Iron being a good conductor, heat can be transmitted from our body to the piece of iron

Answer: 4. The temperature of iron is less than that of wood

Question 31. The specific property that a cooking utensil must possess, is

1. High absorptive power
2. High thermal conductivity
3. Low specific heat
4. All of the above three characteristics

Answer: 4. All of the above three characteristics

Question 32. Woolen clothes are used in winter because wool is

1. An insulator of heat
2. A material having higher specific heat.
3. A material haring lower specific heat
4. A good conductor of heat

Answer: 1. An insulator of heat

Question 33. The mode in which a cup of hot tea placed on a metallic table loses heat is

1. Conduction
2. Convection
3. Radiation
4. All of them

Answer: 4. All of them

Real-Life Examples of Heat Transfer Applications

Question 34. Which mode of transmission of heat depends on the force of gravity?

1. Conduction
2. Convection
3. Radiation
4. None of 1, 2 and 3

Answer: 2. Convection

Question 35. The depletion in the ozone layer is mainly caused by

1. Nitrous oxide
2. Methane
3. Sulphur dioxide
4. Clilorofluro carbon

Answer: 4. Clilorofluro carbon

In this type of question, more than one options are correct.

Question 36. A hollow and a solid sphere of same material and identical outer surface are heated to the same temperature.

1. In the beginning, both will emit equal amount of radiation per unit of time
2. In the beginning, both will absorb an equal amount of radiation per unit of time
3. Both spheres will have the same rate of fall of temperature \(\left(\frac{d t}{d t}\right)\)
4. Both spheres will have equal temperatures at any moment

Answer:

1. In the beginning, both will emit equal amount of radiation per unit of time
2. In the beginning, both will absorb equal amount of radiation per unit time
3. Both spheres will have same rate of fall of temperature \(\left(\frac{d t}{d t}\right)\)

Question 37. A composite block is made of slabs A, B, C, D, and E of different thermal conductivities (given in terms of a constant k) and size (given in terms of length, L) as shown. All slabs are of same width. Heat Q flows only from left to right through the blocks. Then in a steady state

1. Heat flow through slabs A and E are same
2. Heat flow through slab E is maximum
3. The temperature difference across slab E is the smallest
4. Heat flow through C = heat flow through B + heat flowthrough D

Answer:

1. Heat flow through slabs A and E are the same

4. Heat flow through C = heat flow through B + heat flowthrough D

Question 38. The two ends of a uniform rod of thermal conductivity k are maintained at different but constant temperatures. The temperature gradient at any point on the rod is \(\frac{d \theta}{d l}\) (equal to the difference in temperature per unit length). The heat flow per unit time per unit cross-section of the rod is l.

1. \(\frac{d \theta}{d l}\) is the same for all points on rod
2. l will decrease as we move from higher to lower temperature
3. q = \(k \cdot \frac{d \theta}{d l}\)
4. None of these

Answer:

1. \(\frac{d \theta}{d l}\) is the same for all points on rod

3. q = \(k \cdot \frac{d \theta}{d l}\)

Question 39. A heated body emits radiation which has maximum intensity at frequency v_m. If the temperature of the body is doubled

1. The maximum intensity of radiation will be at frequency 2v_m
2. The maximum intensity of radiation will be at frequency
3. The total emitted energy will increase by a factor 16
4. The total emitted energy will increase by a factor 2

Answer:

1. The maximum intensity of radiation will be at frequency 2v_m

3. The total emitted energy will increase by a factor 16

Question 40. The energy radiated by a body depends on

1. Area of body
2. Nature of surface
3. Mass of body
4. Temperature of body

Answer:

1. Area of body

2. Nature of surface

4. Temperature of body