In a resonance pipe the first and second resonances are obtained at depths $22.7 cm$ and $70.2 cm$ respectively. What will be the end correction ..... $cm$
Diffcult
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(a) For end correction $ x, \frac{{{l_2} + x}}{{{l_1} + x}} = \frac{{3\lambda /4}}{{\lambda /4}} = 3$
$x = \frac{{{l_2} - 3{l_1}}}{2} = \frac{{70.2 - 3 \times 22.7}}{2} = 1.05\,cm$
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