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

The correct option is (A): 1.05 cm

Explanation:
Let the depth of first resonance of the pipe be l1 = 22.7 cm
Let the depth of second resonance of the pipe be l2 = 70.2 cm

For the end correction,
$\rightarrow \frac{l_{2}+x}{l_{1}+x}=\frac{3\frac{\pi}{4}}{\frac{\pi}{4}}$
$\rightarrow \frac{l_{2}+x}{l_{1}+x}=3$
$\rightarrow x=\frac{l_{2}-3l_{1}}{2}$

After substituting the values, we get:
$\rightarrow x=\frac{70.2−3 \times 22.7}{2}$
$\rightarrow x=\frac{70.2−68.1}{2}$
$\rightarrow x=\frac{2.1}{2}$
$\rightarrow x=1.05\text{ cm}$

Therefore, the end correction is 1.05 cm.