Question

In a resonance pipe the first and second resonances are obtained at depths $22.7cm$ and $70.2cm$ respectively. What will be the end correction?

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

Answer 1

The correct option is A $1.05cm$

For end correction $x$,
$\frac{l_2+x}{l_1+x}=\frac{3\lambda /4}{\lambda /4}=3$
$\Rightarrow x=\frac{l_2-3l_1}{2}$
$\Rightarrow =\frac{70.2-3\times 22.7}{2}=1.05$cm