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An organ pipe of cross-sectional area 100 `cm^(2)` resonates with a tuning fork of frequency 1000 Hz in fundamental tone. The minimum volume of water to be drained so the pipe again resonates with the same tuning fork is (Take velocity of wave = 320 m `s^(-1)`)A. `800 cm^(3)`B. `1200 cm^(3)`C. `1600 cm^(3)`D. `2000 cm^(3)` |
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Answer» Correct Answer - C Here, `upsilon=1000" Hz"` `1000=(v)/(4l_(1))=(3v)/(4l_(2))` Using `v= 320" m "s^(-1)`, we get, `l_(1)`= 8 cm and `l_(2)` = 24 cm `therefore" "` Minimum volume `=16xx100=1600" cm"^(3)`. |
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