Question

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 ?

• A. $1.05\,cm$
• B. $115.5\,cm$
• C. $92.5\,cm$
• D. $113.5\,cm$

Answer 1

Answer: (A)

$1.05\,cm$

Answer 2

For end correction $x$,
$\frac{l_{2}+x}{l_{1}+x} =\frac{3 \lambda / 4}{\lambda / 4}=3$
$\Rightarrow x =\frac{l_{2}-3 l_{1}}{2}$
$=\frac{70.2-3 \times 22.7}{2}=1.05 cm$