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LIGHT WAVES
OBJECTIVE-II
1. A light wave can travel
(a) in vacuum
(b) in vacuum only
(c) in a material medium
(d) in a material medium only.
ANSWER: (a), (c)
EXPLANATION: The light is an electromagnetic wave that can travel in a vacuum as well as in a transparent material medium.
1. A light wave can travel
(a) in vacuum
(b) in vacuum only
(c) in a material medium
(d) in a material medium only.
ANSWER: (a), (c)
EXPLANATION: The light is an electromagnetic wave that can travel in a vacuum as well as in a transparent material medium.
(a) in vacuum
(b) in vacuum only
(c) in a material medium
(d) in a material medium only.
ANSWER: (a), (c)
EXPLANATION: The light is an electromagnetic wave that can travel in a vacuum as well as in a transparent material medium.
2. Which of the following properties of light conclusively support wave theory of light?
(a) Light obeys laws of reflection.
(b) The speed of light in water is smaller than the speed in a vacuum.
(c) Light shows interference.
(d) Light shows the photoelectric effect.
ANSWER: (b), (c)
EXPLANATION: The reflection can happen in both waves and particles. The photoelectric effect shows the particle nature of light. Hence (a) and (d) are not true.
The smaller speed of light in water than the speed in the vacuum cannot be explained by the particle nature of light. But it can be explained by the wave theory of light. Hence the option (b) is true.
The phenomenon of interference can occur only in waves. (c) is true.
2. Which of the following properties of light conclusively support wave theory of light?
(a) Light obeys laws of reflection.
(b) The speed of light in water is smaller than the speed in a vacuum.
(c) Light shows interference.
(d) Light shows the photoelectric effect.
ANSWER: (b), (c)
EXPLANATION: The reflection can happen in both waves and particles. The photoelectric effect shows the particle nature of light. Hence (a) and (d) are not true.
The smaller speed of light in water than the speed in the vacuum cannot be explained by the particle nature of light. But it can be explained by the wave theory of light. Hence the option (b) is true.
The phenomenon of interference can occur only in waves. (c) is true.
(a) Light obeys laws of reflection.
(b) The speed of light in water is smaller than the speed in a vacuum.
(c) Light shows interference.
(d) Light shows the photoelectric effect.
ANSWER: (b), (c)
EXPLANATION: The reflection can happen in both waves and particles. The photoelectric effect shows the particle nature of light. Hence (a) and (d) are not true.
The smaller speed of light in water than the speed in the vacuum cannot be explained by the particle nature of light. But it can be explained by the wave theory of light. Hence the option (b) is true.
The phenomenon of interference can occur only in waves. (c) is true.
3. When light propagates in the vacuum there is an electric field and a magnetic field. These fields
(a) are constant in time
(b) have zero average value
(c) are perpendicular to the direction of propagation of light
(d) are mutually perpendicular.
ANSWER: (b), (c), (d)
EXPLANATION: The electric and magnetic fields vary with time and position, hence (a) is not true.
Like in sinewave where the average value of displacement within a wavelength is zero, the average value of electric and magnetic fields are zero in the light waves. (b) is true.
The direction of electric and magnetic fields are perpendicular to the direction of propagation of light. Option (c) is true.
The electric and the magnetic fields are also mutually perpendicular to each other. i.e. if light travels in x-direction then electric and magnetic fields will be either in y and z-direction or in z and y directions. Option (d) is true.
3. When light propagates in the vacuum there is an electric field and a magnetic field. These fields
(a) are constant in time
(b) have zero average value
(c) are perpendicular to the direction of propagation of light
(d) are mutually perpendicular.
ANSWER: (b), (c), (d)
EXPLANATION: The electric and magnetic fields vary with time and position, hence (a) is not true.
Like in sinewave where the average value of displacement within a wavelength is zero, the average value of electric and magnetic fields are zero in the light waves. (b) is true.
The direction of electric and magnetic fields are perpendicular to the direction of propagation of light. Option (c) is true.
The electric and the magnetic fields are also mutually perpendicular to each other. i.e. if light travels in x-direction then electric and magnetic fields will be either in y and z-direction or in z and y directions. Option (d) is true.
(a) are constant in time
(b) have zero average value
(c) are perpendicular to the direction of propagation of light
(d) are mutually perpendicular.
ANSWER: (b), (c), (d)
EXPLANATION: The electric and magnetic fields vary with time and position, hence (a) is not true.
Like in sinewave where the average value of displacement within a wavelength is zero, the average value of electric and magnetic fields are zero in the light waves. (b) is true.
The direction of electric and magnetic fields are perpendicular to the direction of propagation of light. Option (c) is true.
The electric and the magnetic fields are also mutually perpendicular to each other. i.e. if light travels in x-direction then electric and magnetic fields will be either in y and z-direction or in z and y directions. Option (d) is true.
4. Huygens' principle of secondary wavelets may be used to
(a) find the velocity of light in vacuum.
(b) explain the particle behavior of light.
(c) find the new position of a wavefront.
(d) explain the Snell's law.
ANSWER: (c), (d)
EXPLANATION: The Huygen's principle of secondary wavelets says that the points of a wavefront act as secondary sources of light emitting secondary wavelets. The line joining the points on the secondary wavelets that are in the same phase gives the new position of a wavefront. Hence the option (c) is true.
It can also be used to explain the Snell's law. Hence the option (d) is true.
Options (a) and (b) are not true.
4. Huygens' principle of secondary wavelets may be used to
(a) find the velocity of light in vacuum.
(b) explain the particle behavior of light.
(c) find the new position of a wavefront.
(d) explain the Snell's law.
ANSWER: (c), (d)
EXPLANATION: The Huygen's principle of secondary wavelets says that the points of a wavefront act as secondary sources of light emitting secondary wavelets. The line joining the points on the secondary wavelets that are in the same phase gives the new position of a wavefront. Hence the option (c) is true.
It can also be used to explain the Snell's law. Hence the option (d) is true.
Options (a) and (b) are not true.
(a) find the velocity of light in vacuum.
(b) explain the particle behavior of light.
(c) find the new position of a wavefront.
(d) explain the Snell's law.
ANSWER: (c), (d)
EXPLANATION: The Huygen's principle of secondary wavelets says that the points of a wavefront act as secondary sources of light emitting secondary wavelets. The line joining the points on the secondary wavelets that are in the same phase gives the new position of a wavefront. Hence the option (c) is true.
It can also be used to explain the Snell's law. Hence the option (d) is true.
Options (a) and (b) are not true.
5. Three observers A, B, and C measure the speed of light coming from a source to be vₐ, vᵦ and vᵨ. The observer A moves towards the source and C moves away from the source at the same speed. The observer B stays stationary. The surrounding space is vacuum everywhere.
(a) vₐ > vᵦ > vᵨ
(b) vₐ < vᵦ < vᵨ
(c) vₐ = vᵦ = vᵨ
(d) vᵦ =½(vₐ + vᵨ)
ANSWER: (c), (d)
EXPLANATION: The speed of light in the vacuum is a universal constant denoted by c and it does not depend on the relative motion of the observer. It is the basis of the special theory of relativity. Hence vₐ = vᵦ = vᵨ, thus option (c) is true and the options (a) and (b) are false.
Since vₐ = vᵦ = vᵨ, Hence ½(vₐ + vᵨ) = vᵦ. Thus option (d) is also true.
5. Three observers A, B, and C measure the speed of light coming from a source to be vₐ, vᵦ and vᵨ. The observer A moves towards the source and C moves away from the source at the same speed. The observer B stays stationary. The surrounding space is vacuum everywhere.
(a) vₐ > vᵦ > vᵨ
(b) vₐ < vᵦ < vᵨ
(c) vₐ = vᵦ = vᵨ
(d) vᵦ =½(vₐ + vᵨ)
ANSWER: (c), (d)
EXPLANATION: The speed of light in the vacuum is a universal constant denoted by c and it does not depend on the relative motion of the observer. It is the basis of the special theory of relativity. Hence vₐ = vᵦ = vᵨ, thus option (c) is true and the options (a) and (b) are false.
Since vₐ = vᵦ = vᵨ, Hence ½(vₐ + vᵨ) = vᵦ. Thus option (d) is also true.
(a) vₐ > vᵦ > vᵨ
(b) vₐ < vᵦ < vᵨ
(c) vₐ = vᵦ = vᵨ
(d) vᵦ =½(vₐ + vᵨ)
ANSWER: (c), (d)
EXPLANATION: The speed of light in the vacuum is a universal constant denoted by c and it does not depend on the relative motion of the observer. It is the basis of the special theory of relativity. Hence vₐ = vᵦ = vᵨ, thus option (c) is true and the options (a) and (b) are false.
Since vₐ = vᵦ = vᵨ, Hence ½(vₐ + vᵨ) = vᵦ. Thus option (d) is also true.
6. Suppose the medium in the previous problem is water. Select the correct option(s) from the list given in that question.
ANSWER: (a), (d)
EXPLANATION: The speed of light c is constant for any observer in the vacuum only. When the light is traveling in a medium its speed reduces to c/µ, where µ is the refractive index of the medium. In a medium, the speed of light measured by an observer depends on the relative motion of the observer. Hence the observer A will measure the speed vₐ > vᵦ and the observer C will measure vᵨ < vᵦ. Hence the option (a) is true.
Since both A and C move with the same speed in opposite directions, they will find the change equally in the speed. Hence ½(vₐ + vᵨ) = vᵦ. Thus option (d) is also true.
6. Suppose the medium in the previous problem is water. Select the correct option(s) from the list given in that question.
ANSWER: (a), (d)
EXPLANATION: The speed of light c is constant for any observer in the vacuum only. When the light is traveling in a medium its speed reduces to c/µ, where µ is the refractive index of the medium. In a medium, the speed of light measured by an observer depends on the relative motion of the observer. Hence the observer A will measure the speed vₐ > vᵦ and the observer C will measure vᵨ < vᵦ. Hence the option (a) is true.
Since both A and C move with the same speed in opposite directions, they will find the change equally in the speed. Hence ½(vₐ + vᵨ) = vᵦ. Thus option (d) is also true.
ANSWER: (a), (d)
EXPLANATION: The speed of light c is constant for any observer in the vacuum only. When the light is traveling in a medium its speed reduces to c/µ, where µ is the refractive index of the medium. In a medium, the speed of light measured by an observer depends on the relative motion of the observer. Hence the observer A will measure the speed vₐ > vᵦ and the observer C will measure vᵨ < vᵦ. Hence the option (a) is true.
Since both A and C move with the same speed in opposite directions, they will find the change equally in the speed. Hence ½(vₐ + vᵨ) = vᵦ. Thus option (d) is also true.
7. Light waves travel in vacuum along the X-axis. Which of the following may represent the wavefronts?
(a) x = c
(b) y = c
(c) z = c
(d) x + y + z = c.
ANSWER: (a)
EXPLANATION: The wavefronts are perpendicular to the direction of the light. Since here the light is traveling along X-axis, the wavefront will be a plane perpendicular to the X-axis. Out of the four options only (a) represents a plane perpendicular to X-axis.
7. Light waves travel in vacuum along the X-axis. Which of the following may represent the wavefronts?
(a) x = c
(b) y = c
(c) z = c
(d) x + y + z = c.
ANSWER: (a)
EXPLANATION: The wavefronts are perpendicular to the direction of the light. Since here the light is traveling along X-axis, the wavefront will be a plane perpendicular to the X-axis. Out of the four options only (a) represents a plane perpendicular to X-axis.
(a) x = c
(b) y = c
(c) z = c
(d) x + y + z = c.
ANSWER: (a)
EXPLANATION: The wavefronts are perpendicular to the direction of the light. Since here the light is traveling along X-axis, the wavefront will be a plane perpendicular to the X-axis. Out of the four options only (a) represents a plane perpendicular to X-axis.
8. If the source of light used in Young's double slit experiment is changed from red to violet,
(a) the fringes will become brighter
(b) consecutive fringes will come closer
(c) the intensity of minima will increase
(d) the central bright fringe will become a dark fringe.
ANSWER: (b)
EXPLANATION: The options (a), (c) and (d) are related to the intensity changes. Since the intensity of light incident on the slits is not changed hence there will be no change in the intensity of the fringes. These options are not true.
The fringe width is given as
w = D𝝺/d.
In this experiment D and d are not changed only 𝛌 is changed. Since the violet has smaller 𝛌 than the red light, so the fringe width will decrease. Hence the consecutive fringes will come closer. Option (b) is true.
8. If the source of light used in Young's double slit experiment is changed from red to violet,
(a) the fringes will become brighter
(b) consecutive fringes will come closer
(c) the intensity of minima will increase
(d) the central bright fringe will become a dark fringe.
ANSWER: (b)
EXPLANATION: The options (a), (c) and (d) are related to the intensity changes. Since the intensity of light incident on the slits is not changed hence there will be no change in the intensity of the fringes. These options are not true.
The fringe width is given as
w = D𝝺/d.
In this experiment D and d are not changed only 𝛌 is changed. Since the violet has smaller 𝛌 than the red light, so the fringe width will decrease. Hence the consecutive fringes will come closer. Option (b) is true.
(a) the fringes will become brighter
(b) consecutive fringes will come closer
(c) the intensity of minima will increase
(d) the central bright fringe will become a dark fringe.
ANSWER: (b)
EXPLANATION: The options (a), (c) and (d) are related to the intensity changes. Since the intensity of light incident on the slits is not changed hence there will be no change in the intensity of the fringes. These options are not true.
The fringe width is given as
w = D𝝺/d.
In this experiment D and d are not changed only 𝛌 is changed. Since the violet has smaller 𝛌 than the red light, so the fringe width will decrease. Hence the consecutive fringes will come closer. Option (b) is true.
9. A Young's double slit experiment is performed with white light.
(a) The central fringe will be white.
(b) There will not be a completely dark fringe.
(c) The fringe next to central will be red.
(d) The fringe next to central will be violet.
ANSWER: (a), (b), (c)
EXPLANATION: The central fringe will be white because all the component colors band will superimpose here. The option (a) is true.
Since the fringes of different wavelengths of light have different fringe-widths, there will not be a completely dark fringe. Option (b) is true.
Since the wavelength of the violet light is nearly half of the red light, the fringe-width of the violet light is half of the red light. Since the central bright violet fringe combines with other fringes to form the white fringe, the next to it will be a minima for violet light. So violet will not be next to the central white fringe. Next will be a very thin band of white light minus violet light which will give a shade of off white. In this way, the components of indigo, blue, green, yellow and orange colors will be withdrawn respectively in the very thin consecutive bands. These will effect in different off white shades which cannot be clearly differentiated from the white band. The first color away from the center will be recognized as red when in these very thin bands yellow+orange+red, orange+red or red remains. hence the option (c) is true.
Young's Double slit experiment with white light
9. A Young's double slit experiment is performed with white light.
(a) The central fringe will be white.
(b) There will not be a completely dark fringe.
(c) The fringe next to central will be red.
(d) The fringe next to central will be violet.
ANSWER: (a), (b), (c)
EXPLANATION: The central fringe will be white because all the component colors band will superimpose here. The option (a) is true.
Since the fringes of different wavelengths of light have different fringe-widths, there will not be a completely dark fringe. Option (b) is true.
Since the wavelength of the violet light is nearly half of the red light, the fringe-width of the violet light is half of the red light. Since the central bright violet fringe combines with other fringes to form the white fringe, the next to it will be a minima for violet light. So violet will not be next to the central white fringe. Next will be a very thin band of white light minus violet light which will give a shade of off white. In this way, the components of indigo, blue, green, yellow and orange colors will be withdrawn respectively in the very thin consecutive bands. These will effect in different off white shades which cannot be clearly differentiated from the white band. The first color away from the center will be recognized as red when in these very thin bands yellow+orange+red, orange+red or red remains. hence the option (c) is true.
(a) The central fringe will be white.
(b) There will not be a completely dark fringe.
(c) The fringe next to central will be red.
(d) The fringe next to central will be violet.
ANSWER: (a), (b), (c)
EXPLANATION: The central fringe will be white because all the component colors band will superimpose here. The option (a) is true.
Since the fringes of different wavelengths of light have different fringe-widths, there will not be a completely dark fringe. Option (b) is true.
Since the wavelength of the violet light is nearly half of the red light, the fringe-width of the violet light is half of the red light. Since the central bright violet fringe combines with other fringes to form the white fringe, the next to it will be a minima for violet light. So violet will not be next to the central white fringe. Next will be a very thin band of white light minus violet light which will give a shade of off white. In this way, the components of indigo, blue, green, yellow and orange colors will be withdrawn respectively in the very thin consecutive bands. These will effect in different off white shades which cannot be clearly differentiated from the white band. The first color away from the center will be recognized as red when in these very thin bands yellow+orange+red, orange+red or red remains. hence the option (c) is true.
 |
| Young's Double slit experiment with white light |
10. Four light waves are represented by
(i) y = a₁ sin ⍵t
(ii) y = a₂ sin (⍵t+ε)
(iii) y = a₁ sin 2⍵t
(iv) y = a₂ sin 2(⍵t+ε)
Interference fringes may be observed due to the superposition of
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iv)
(d) (iii) and (iv).
ANSWER: (a), (d)
EXPLANATION: The interference fringes can be observed with coherent sources of light which have a constant initial phase difference and same frequency. Only the waves of option (a) and (d) satisfy this.
10. Four light waves are represented by
(i) y = a₁ sin ⍵t
(ii) y = a₂ sin (⍵t+ε)
(iii) y = a₁ sin 2⍵t
(iv) y = a₂ sin 2(⍵t+ε)
Interference fringes may be observed due to the superposition of
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iv)
(d) (iii) and (iv).
ANSWER: (a), (d)
EXPLANATION: The interference fringes can be observed with coherent sources of light which have a constant initial phase difference and same frequency. Only the waves of option (a) and (d) satisfy this.
===<<<O>>>===
Links to the Chapters
===<<<O>>>===
Links to the Chapters
CHAPTER- 16 - Sound Waves
OBJECTIVE-I
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES -Q-21 TO Q-30
EXERCISES -Q-31 TO Q-40
EXERCISES -Q-41 TO Q-50
EXERCISES -Q-51 TO Q-60
EXERCISES -Q-61 TO Q-70
EXERCISES - Q-71 TO Q-80
EXERCISES - Q-81 TO Q-89
CHAPTER- 15 - Wave Motion and Waves on a String
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES - Q-21 TO Q-30
EXERCISES - Q-31 TO Q-40
EXERCISES - Q-41 TO Q-50
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q1 TO Q10
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
EXERCISES- Q51 TO Q58 (2-Extra Questions)
CHAPTER- 11 - Gravitation
EXERCISES -Q 31 TO 39
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 16 - Sound Waves
OBJECTIVE-I
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES -Q-21 TO Q-30
EXERCISES -Q-31 TO Q-40
EXERCISES -Q-41 TO Q-50
EXERCISES -Q-51 TO Q-60
EXERCISES -Q-61 TO Q-70
EXERCISES - Q-71 TO Q-80
EXERCISES - Q-81 TO Q-89
CHAPTER- 15 - Wave Motion and Waves on a String
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES - Q-21 TO Q-30
EXERCISES - Q-31 TO Q-40
EXERCISES - Q-41 TO Q-50
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
EXERCISES- Q51 TO Q58 (2-Extra Questions)
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 8 - Work and Energy
Click here for → Question for Short Answers
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → Exercises (1-10)
Click here for → Question for Short Answers
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → Exercises (1-10)
Click here for → Exercises (11-20)
Click here for → Exercises (21-30)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
Click here for → Exercises (21-30)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
CHAPTER- 7 - Circular Motion
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Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
CHAPTER- 6 - Friction
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Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
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Click here for → EXERCISES (11-20)
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CHAPTER- 6 - Friction
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Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
For more practice on problems on friction solve these- "New Questions on Friction".
Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
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CHAPTER- 5 - Newton's Laws of Motion
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)
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CHAPTER- 4 - The Forces
The Forces-
"Questions for short Answers"
Click here for "The Forces" - OBJECTIVE-I
Click here for "The Forces" - OBJECTIVE-II
Click here for "The Forces" - Exercises
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CHAPTER- 3 - Kinematics - Rest and Motion
Click here for "Questions for short Answers"
Click here for "OBJECTIVE-I"
Click here for EXERCISES (Question number 1 to 10)
Click here for EXERCISES (Question number 11 to 20)
Click here for EXERCISES (Question number 21 to 30)
Click here for EXERCISES (Question number 31 to 40)
Click here for EXERCISES (Question number 41 to 52)
CHAPTER- 2 - "Vector related Problems"
CHAPTER- 2 - "Vector related Problems"
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