A sound wave of wavelength `90 cm` in glass is reflected into air. If the speed of sound in glass is `5400 m//s`, the wavelength of wave in air (speed of sound in air `= 330 m//s`) is :
A. 55 cm
B. 5.5 cm
C. 55 m
D. 5.5 m
A. 55 cm
B. 5.5 cm
C. 55 m
D. 5.5 m
Correct Answer – B
In glass, `lamda_g = 0.9 m, v_g = 5400 m//s`
`v = (v_g)/(lamda_g) = (5400 m//s)/(0.9 m) = 6000 Hz`
In air, `v = 6000 Hz` (as v does not change with change of medium), `v_a = 330 m//s`
`lamda_a = (v a)/(v) = (330 m//s)/(6000 Hz) = 0.055 m = 5.5 cm`
Aliter. As `v_g = v lamda_g and v_a = v lamda_a`,
`(v g)/(v a) = (lamda_g)/(lamda_a)` or `lamda_a = lamda_g xx (v a)/(v g)` or `lamda_a = (0.9 m) ((330 m//s)/(5400 m//s)) = 0.055 m = 5.5 cm`.