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Q1. A cylindrical tube containing air is open at both ends. If the shortest length of the tube for resonance with a given fork is 2 cm, the next shortest length for resonance with the same fork will be
  •  80 cm
  •  90 cm
  •  60 cm
  •  40 cm
Solution
As the tube is open at both ends, therefore, next shortest length for resonance =2×20=40 cm

Q2.On which principle does sonometer works?
  •  Hooke’s law
  •  Resonance
  •  Newton’s law
  •  Elasticity
Solution
Sonometer works on the principle of resonance. At resonance the wire of sonometer vibrate with maximum amplitude.

Q3.  The intensity ratio of two waves is 1:9. The ratio of their amplitudes, is
  •   3:1
  •  9:1
  •  1:9
  •  1:3
Solution


Q4. The driver of a car travelling with speed 30 metres per second towards a hill sounds a horn of frequency 600 Hz. If the velocity of sound in air is 330 metres per second, the frequency of the reflected sound as heard by the driver is
  •  500 Hz
  •  550 Hz
  •  555.5 Hz
  •  720 Hz
Solution

Q5.The musical interval between two tones of frequencies 320 Hz and 240 Hz is
  •  (4/3)
  •  560
  •  320×240
  •  80
Solution
 Musical interval is the ratio of frequencies =320/240=4/3

Q6. Two tuning fork, A and B produce notes of frequencies 258 Hz and 262 Hz. An unknown note sounded with a produces certain beats. When the same note is sounded with B, the beat frequency gets doubled, the unknown frequency is
  •  280 Hz
  •  254 Hz
  • 256 Hz
  •  300 Hz
Solution

Q7.The amplitude of two waves are in ratio 5:2. If all other conditions for the two waves Are same, then what is the ratio of their energy densities?
  •  5:2
  •  5:4
  •  25:4
  •  4:5
Solution

Q8.v_1 and v_2 are the velocities of sound at the same temperature in two monoatomic gases of densities ρ_1 and ρ_2 respectively. If ρ_1/ρ_2=1/4 then the ratio of velocities v_1 and v_2 will be
  •  1∶2
  •  2∶1
  •  1∶4
  •  4∶1
Solution



Q9.The equation y=A cos^2⁡〖(2π nt-2π x/λ)〗 represents a wave with
  •  Amplitude A, frequency n and wavelength λ
  •  Amplitude A, frequency 2n and wavelength 2λ
  •  Amplitude A/2, frequency 2n and wavelength λ
  •  Amplitude A/2, frequency 2n and wavelength λ/2
Solution



Q10. A wavelength 0.60 cm is produced in air and it travels at a speed of 300 ms^(-1). It will be an
  •  Infrasonic wave
  •  Ultrasonic wave
  •  Audible wave
  • None of the above
Solution
n=v/λ=300/(0.6×〖10〗^(-2) ) Hz=3/6×〖10〗^4 Hz=50,000Hz ⇒ Wave is ultrasonic

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