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(A) Let $l_1$ and $l_2$ be the lengths of the air column for the first and second resonance,and $e$ be the end correction.For the first resonance: $l_

Vedclass · Accepted Answer

$1.05$

Vedclass · Answer

(A) Let $l_1$ and $l_2$ be the lengths of the air column for the first and second resonance,and $e$ be the end correction.For the first resonance: $l_1 + e = \frac{\lambda}{4}$ (Equation $1$)For the second resonance: $l_2 + e = \frac{3\lambda}{4}$ (Equation $2$)Dividing Equation $2$ by Equation $1$:$\frac{l_2 + e}{l_1 + e} = 3$$l_2 + e = 3l_1 + 3e$$2e = l_2 - 3l_1$$e = \frac{l_2 - 3l_1}{2}$Substituting the given values $l_1 = 22.7 \ cm$ and $l_2 = 70.2 \ cm$:$e = \frac{70.2 - 3(22.7)}{2} = \frac{70.2 - 68.1}{2} = \frac{2.1}{2} = 1.05 \ cm$.

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

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