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(A) Let the end correction be $e$. The resonance condition for a tube closed at one end is given by $l_n + e = (2n-1) \frac{\lambda}{4}$.For the first

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(A) Let the end correction be $e$. The resonance condition for a tube closed at one end is given by $l_n + e = (2n-1) \frac{\lambda}{4}$.For the first resonance $(n=1)$: $24.1 + e = \frac{\lambda}{4} \quad ---(1)$For the second resonance $(n=2)$: $74.1 + e = \frac{3\lambda}{4} \quad ---(2)$Subtracting equation $(1)$ from equation $(2)$:$(74.1 + e) - (24.1 + e) = \frac{3\lambda}{4} - \frac{\lambda}{4}$$50 = \frac{2\lambda}{4} = \frac{\lambda}{2}$$\lambda = 100 \ cm$Substituting $\lambda$ in equation $(1)$:$24.1 + e = \frac{100}{4} = 25$$e = 25 - 24.1 = 0.9 \ cm$The end correction $e$ is related to the diameter $D$ by the formula $e = 0.3D$.$0.9 = 0.3D$$D = \frac{0.9}{0.3} = 3 \ cm$.

In a resonance tube experiment,the first and second resonance are heard when the water level is $24.1 \ cm$ and $74.1 \ cm$ respectively,below the open end of the tube. The inner diameter of the tube is (in $cm$)

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