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

The Correct Option is A

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.