The the resonance tube experiment first resonant length is `l_(1)` and the second resonant length is `l_(2)`, then the third resonant length will be ?

Options

A. `5l`
B. `2(l_(2)-l_(1))`
C. `2l_(2)-l_(1)`
D. `3l_(2)-2l_(1)`

Answer

C

Explanation

For first resonance
`l_(1) + epsilon = (V)/(4 f_(0))`
for second resonance
`l_(2) = epsilon = (3v)/(4f_(0))`
for the third resonance
`l_(3) + epsilon = (5v)/(4f_(0))`
Solving get `l_(3) = 2l_(2) - l_(1)`.