Question

In a resonance tube experiment the first resonance is obtained for $10cm$ of air column and the second for $32cm$. The end correction for this apparatus is equal to

• A. $0.5cm$
• B. $1.0cm$
• C. $1.5cm$
• D. $2cm$

Answer 1

Answer: (B)

$1.0cm$

Here, $l_1+e=\frac{\lambda }{4}$ and $l_2+e=\frac{3\lambda }{4}$
So, $3l_1+3e=l_2+e$
$2e=l_2-3l_1$
$e=\frac{1}{2}(l_2-3l_1)=\frac{1}{2}(32-3(10\left)\right)=1.0cm$