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 waterto be drained so the pipe again resonates with the same tuning fork is (Take velocity of wave = 320 m s^(-1))

An organ pipe of cross-sectional area 100 cm^(2) resonates with a tuni

1.	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 waterto be drained so the pipe again resonates with the same tuning fork is (Take velocity of wave = 320 m s^(-1))
Answer» `800 cm^(3)` `1200 cm^(3)` `1600 cm^(3)` `2000 cm^(3)` SOLUTION :Here, `upsilon=1000" Hz"` `1000=(V)/(4l_(1))=(3v)/(4l_(2))` Using `v= 320" m "s^(-1)`, we get, `l_(1)`= 8 cmand `l_(2)` = 24 cm `therefore""` Minimum volume `=16xx100=1600" cm"^(3)`.