Correct Answer - C::D
For a closed tube
`f_(n) = ( nv)/( 4 L)`
`L = 1.1 m , v = 330 m//s`
`f_(n) = ( n xx 330)/( 4 xx 1.1) = 500 Hz rArr n = 6.66`
highest frequency ,
`f_(h) = ( 6 xx 330)/( 4 xx 1.1) = 450 Hz`
Lowest frequency ,
`f_(1) = ( 1 xx 330)/(4.4) = 75 Hz`