A glass tube of `1.0 m` length is filled with water . The water can be drained out slowly at the bottom of the tube . If a vibrating tuning fork of frequency `500 Hz` is brought at the upper end of the tube and the velocity of sound is `300 m//s`, then the total number of resonances obtained will be

Answer

2

Explanation

`v=500Hz`
`V=300m//sec`
`lamda=(V)/(v)=(300)/(500)=(3)/(5)m`
Resonating length `l=((2N-1)xxV)/(4V)=((2N-1).300)/(4xx500)`
`implies (3(2N-1))/(20) le 1m" "therefore 2N-1=(20)/(3)implies 2N=(20)/(3)+1`
`2N=(23)/(3) implies N=(23)/(6)=4`
Since only odd harmonics are possible, there will be only two resonance lengths.