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The internal radius of a 1m long resonance tube is measured as 3.0 cm. A tuning fork of frequency 2000 Hz is used. The first resonating length is measured as 4.5 cm and the second resonating length is measured as 14.0 cm. We shall calculate the following.

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :(i) Maximum percentage error in measurement of as, given by Reyleigh'sformula <br/> (Given error in measurement of radius is 0.1 m) `Deltae= 0.6 Delta R = 0.6 xx 0.1 = 0.06 cm` <br/> Percentage error is `(Delta e)/(e) xx 100 = (0.06)/(0.6 xx <a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>) xx 100 = 3.33 %` <br/> (<a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>) Speed of sound at the room temperature <br/> `l_(1)=4.6m,l_(2)=14.0cm,lamda = 2(l_(2)-l_(1))(14.0-4.6)=18.8cm,<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a> =flamda = 2000 xx (18.8)/(100) = 376 m//s` <br/> (ii) End correction obtained in the experiment, `e = (l_(2)-3l_(1))/(2)=(14.0-3xx4.6)/(2)=0.1cm` <br/> (<a href="https://interviewquestions.tuteehub.com/tag/iv-501699" style="font-weight:bold;" target="_blank" title="Click to know more about IV">IV</a>) Percentage error in the calculation of a with respect to theoretical valuess <br/> Percentage error `= (0.6xx3-0.1)/(0.6xx3)xx100=94.44%`</body></html>


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