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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 cmB. 5.5 cmC. 55 mD. 5.5 m |
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Answer» 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`. |
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