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(D) Given: Diameter $d = 6 \, cm = 0.06 \, m$,Frequency $f = 504 \, Hz$,Speed of sound $v = 336 \, m/s$.The end correction $e$ for a resonance tube is

Vedclass · Accepted Answer

$14.8$

Vedclass · Answer

(D) Given: Diameter $d = 6 \, cm = 0.06 \, m$,Frequency $f = 504 \, Hz$,Speed of sound $v = 336 \, m/s$.The end correction $e$ for a resonance tube is given by $e = 0.3 	imes d = 0.3 	imes 6 \, cm = 1.8 \, cm$.For the first resonance,the length of the air column $l$ satisfies the condition $l + e = \frac{\lambda}{4}$.We know that $\lambda = \frac{v}{f} = \frac{336}{504} \, m = \frac{2}{3} \, m = \frac{200}{3} \, cm \approx 66.67 \, cm$.Thus,$l + e = \frac{66.67}{4} = 16.67 \, cm$.Substituting the value of $e$,we get $l + 1.8 = 16.67 \, cm$.Therefore,$l = 16.67 - 1.8 = 14.87 \, cm$.Rounding to the nearest provided option,the reading is $14.8 \, cm$.

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