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(B) In a resonance column experiment,the first resonance length $l_1$ is given by the formula: $\frac{\lambda}{4} = l_1 + e$,where $e$ is the end corr

Vedclass · Accepted Answer

$15.2$

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(B) In a resonance column experiment,the first resonance length $l_1$ is given by the formula: $\frac{\lambda}{4} = l_1 + e$,where $e$ is the end correction.The end correction $e$ for a tube of diameter $d$ is given by $e = 0.3d$.The wavelength $\lambda$ is calculated as $\lambda = \frac{v}{f}$,where $v = 336 \ m/s$ and $f = 512 \ Hz$.$\lambda = \frac{336}{512} \approx 0.65625 \ m = 65.625 \ cm$.Therefore,$\frac{\lambda}{4} = \frac{65.625}{4} = 16.40625 \ cm$.Given the diameter $d = 4 \ cm$,the end correction $e = 0.3 	imes 4 = 1.2 \ cm$.Substituting these values into the resonance formula: $l_1 = \frac{\lambda}{4} - e = 16.40625 - 1.2 = 15.20625 \ cm$.Rounding to one decimal place,the reading is $15.2 \ cm$.

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