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Find out the coefficient of linear expansion of the material of an open pipeso that the frequency of any tone emitted from itdoes not very with temperature .

Answer» <html><body><p></p>Solution :The fundamental <a href="https://interviewquestions.tuteehub.com/tag/frequencies-999868" style="font-weight:bold;" target="_blank" title="Click to know more about FREQUENCIES">FREQUENCIES</a> at ` 0^(@)` C and` t^(@)` C are respectively ,<br/>`n_(1) = (V_(0))/(2l_(0)) and n_(2) (V_(t))/(2l_(t))`<br/>Here,`V_(0)`= velocity of <a href="https://interviewquestions.tuteehub.com/tag/sound-648690" style="font-weight:bold;" target="_blank" title="Click to know more about SOUND">SOUND</a> at ` 0^(@) C l_(0)`= length of the pipeat ` 0^(@) C , V_(t)` = velocity of sound at ` t^(@) C , l_(t)` = length of the pipe at ` t^(@) C ` .<br/>According to the question, ` n_(1) = n_(2)`<br/>so , ` (V_(0))/(2l_(0)) = (V_(t))/(2l_(t))`<br/>or , `V_(0) l_(t) = V_(t) l_(0) or V_(0) l_(0) (1 + alpha t) = V_(0) ( 1 + 0 .00183 t) l_(0)`<br/>[`alpha = ` coefficient of <a href="https://interviewquestions.tuteehub.com/tag/linear-1074458" style="font-weight:bold;" target="_blank" title="Click to know more about LINEAR">LINEAR</a> expansion]<br/>or, ` alpha = 0 . 00183^(@) C^(-1)` .</body></html>


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