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A mass of 2kg is suspendedfrom a fixed point by a wire of length 3m and diameter 0.5 mm. Initially the wire is just unstretched, the mass resting on a fixed support. By hoH' much must the temperature fall if the mass is to be entirely supported by the wire? Given that Y of the wire = 206 G Pa, alpha = 11 xx 10^(-6)//^(0) C .

Answer» <html><body><p></p>Solution :When the temperature <a href="https://interviewquestions.tuteehub.com/tag/decreases-946143" style="font-weight:bold;" target="_blank" title="Click to know more about DECREASES">DECREASES</a> the wire contracts in length and <a href="https://interviewquestions.tuteehub.com/tag/lifts-1073381" style="font-weight:bold;" target="_blank" title="Click to know more about LIFTS">LIFTS</a> the mass up. But the weight begins to produce a stress and counteracts the contraction <a href="https://interviewquestions.tuteehub.com/tag/produced-592947" style="font-weight:bold;" target="_blank" title="Click to know more about PRODUCED">PRODUCED</a>. Hence the contraction due to cooling is equal to the stretching produced by the weight W.<br/> `therefore <a href="https://interviewquestions.tuteehub.com/tag/delta-947703" style="font-weight:bold;" target="_blank" title="Click to know more about DELTA">DELTA</a> L = (WL)/(AY) = (2 xx 9.8 xx 3)/(pi (0.25)^(2) xx 10^(-6) xx 206 xx 10^(9))` <br/> Now the contraction due to cooling <br/>` = L alpha d theta = 3 xx 11 xx 10^(-6) xx d theta` <br/> Since the contraction due to cooling = extension produced by the weight. On solving `d theta = 44^(@)`C</body></html>


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