In a resonance tube apparatus, the first and second resonance lengths are `l_(1)` and `l_(2)` respectively. If `v` is the velocity of wave. Then

Options

A. Frequency is , `u=(V)/(2(l_(2)-l_(1))`
B. End correction , is `e=(l_(2)-3l_(1))/(2)`
C. End correction is, `e=(l_(2)-3l_(1))/(2)`
D. Frequency is , `u=(V)/(4(l_(2)-l_(1))`

Answer

A, B

Explanation

For end correction `e`, `l_(1)+e=(lambda)/(4)` and `l_(2)+e=(3lambda)/(4)`
`:. L_(2)-l_(1)=(lambda)/(2)=(v)/(2v)`, `u=(v)/(2(l_(2)-l_(1))`
So, choice (`a` ) is correct and choice (`d`) is wrong
Put`lambda=4(l_(1)+e)` and `l_(2)+e=(3lambda)/(4)`
Then `L_(2)+e=3(l_(1)+e)`
`l_(2)-3l_(1)=2e` `:. e=(l_(2)-3l_(1))/(2)`
So, choice (`b`) is correct and choice (`c`) is wrong