51. An observer moves towards a stationary source of sound with a speed 1/5th of the speed of sound. The wavelength and frequency of the sound emitted are λ and ƒ respectively. The apparent frequency and wavelength recorded by the observer are respectively.

1. 1. 0.8ƒ, 0.8λ

   
   
   

   2. 1.2ƒ, 1.2λ

   
   
   

   3. 1.2ƒ, λ

   
   
   

   4. ƒ, 1.2λ

   
   
   

---

1. View Hint View Answer Discuss in Forum

   ƒapparent =			u + u/5		ƒ =	6	ƒ = 1.2 ƒ
   			u			5

   
   Wavelength remains constant (unchanged) in this case.

   Correct Option: C

   ƒapparent =			u + u/5		ƒ =	6	ƒ = 1.2 ƒ
   			u			5

   
   Wavelength remains constant (unchanged) in this case.

---

52. A whistle of frequency 385 Hz rotates in a horizontal circle of radius 50 cm at an angular speed of 20 radians s^–1. The lowest frequency heard by a listener a long distance away at rest with respect to the
    centre of the circle, given velocity of sound equal to 340 ms^–1, is

1. 1. 396 Hz

   
   
   

   2. 363 Hz

   
   
   

   3. 374 Hz

   
   
   

   4. 385 Hz

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Velocity of source
   vs = rω = 0.50 × 20 = 10 ms^-1
   

   n' =	v	n =	340 × 385	= 374 Hz
   	v - vs		340 + 10

   Correct Option: C

   Velocity of source
   vs = rω = 0.50 × 20 = 10 ms^-1
   

   n' =	v	n =	340 × 385	= 374 Hz
   	v - vs		340 + 10

---

---

53. A vehicle, with a horn of frequency n is moving with a velocity of 30 m/s in a direction perpendicular to the straight line joining the observer and the vehicle. The observer perceives the sound to have a frequency n + n₁. Then (if the sound velocity in air is 300 m/s)

1. 1. n₁ = 10n

   
   
   

   2. n₁ = 0

   
   
   

   3. n₁ = 0.1n

   
   
   

   4. n₁ = – 0.1n

   
   
   

---

1. View Hint View Answer Discuss in Forum

   As the source is not moving towards or away from the observer in a straight line, so the Doppler’s effect will not be observed by the observer.

   Correct Option: B

   As the source is not moving towards or away from the observer in a straight line, so the Doppler’s effect will not be observed by the observer.

---

54. A star, which is emitting radiation at a wavelength of 5000 Å, is approaching the earth with a velocity of 1.50 × 10⁶ m/s. The change in wavelength of the radiation as received on the earth is

1. 1. 0.25 Å

   
   
   

   2. 2.5 Å

   
   
   

   3. 25 Å

   
   
   

   4. 250 Å

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Given : Wavelength (λ) = 5000 Å
   velocity of star (v) = 1.5 × 10⁶ m/s.
   We know that wavelength of the approaching star (λ')
   

   (λ') = λ	c - v
   	c

   
   

   or,	λ'	=	c - v	= 1 -	v
   	λ		c		c

   
   

   or,	v	= 1 -	λ'	=	λ - λ'	=	Δλ
   	c		λ		λ		λ

   
   Therefore,
   

   Δλ = λ ×	v	= 5000 ×	1.5 × 106	= 25 Å
   	c		3 × 10 × 108

   
   [where ∆λ = Change in the wavelength]

   Correct Option: C

   Given : Wavelength (λ) = 5000 Å
   velocity of star (v) = 1.5 × 10⁶ m/s.
   We know that wavelength of the approaching star (λ')
   

   (λ') = λ	c - v
   	c

   
   

   or,	λ'	=	c - v	= 1 -	v
   	λ		c		c

   
   

   or,	v	= 1 -	λ'	=	λ - λ'	=	Δλ
   	c		λ		λ		λ

   
   Therefore,
   

   Δλ = λ ×	v	= 5000 ×	1.5 × 106	= 25 Å
   	c		3 × 10 × 108

   
   [where ∆λ = Change in the wavelength]

---

---

55. A cylindrical resonance tube open at both ends, has a fundamental frequency, ƒ, in air. If half of the length is dipped vertically in water, the fundamental frequency of the air column will be

1. 1. 2ƒ

   
   
   

   2. ​3ƒ/2

   
   
   

   3. ƒ

   
   
   

   4. ƒ/2

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Fundamental frequency of open pipe,
   

   ƒ =	v
   	2l

   
   When half of tube is filled with water, then the length of air column becomes half
   

   l' =			l
   			2

   
   and the pipe becomes closed.
   So, new fundamental frequency
   

   ƒ' =	v	=	v					=	v
   	4l'		4			l			2l
   						2

   
   Clearly ƒ' = ƒ.

   Correct Option: C

   Fundamental frequency of open pipe,
   

   ƒ =	v
   	2l

   
   When half of tube is filled with water, then the length of air column becomes half
   

   l' =			l
   			2

   
   and the pipe becomes closed.
   So, new fundamental frequency
   

   ƒ' =	v	=	v					=	v
   	4l'		4			l			2l
   						2

   
   Clearly ƒ' = ƒ.

---