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The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law : R=R_(0)[1+alpha(T-T_(0))] The resitance is 101.6" "Omega at the triple-point of water 273.16 K, and 165.5" "Omega at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4" "Omega ?

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> :`R=R_(0)[1+alpha(T-T_(0))]` <br/> `:.R-R_(0)=R_(0)xx5xx10^(-3)(T-T_(0))` <br/> `:.T-T_(0)=(R-R_(0))/(<a href="https://interviewquestions.tuteehub.com/tag/5xx10-1901302" style="font-weight:bold;" target="_blank" title="Click to know more about 5XX10">5XX10</a>^(-3)R_(0))` <br/> At triple <a href="https://interviewquestions.tuteehub.com/tag/point-1157106" style="font-weight:bold;" target="_blank" title="Click to know more about POINT">POINT</a> of water : <br/> `T_(1)-T_(0)=(R_(1)-R_(0))/(5xx10^(-3)R_(0))` . . .(1) <br/> and at triple point of <a href="https://interviewquestions.tuteehub.com/tag/leat-2785238" style="font-weight:bold;" target="_blank" title="Click to know more about LEAT">LEAT</a> : <br/> `T_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)-T_(0)=(R_(2)-R_(0))/(5xx10^(-3)R_(0))`. . . (2) <br/> By taking ratio of eauation (2) to equation (1), <br/> `(T_(2)-T_(0))/(T_(1)-T_(0))=(R_(2)-R_(0))/(R_(1)-R_(0))` <br/> Here `R_(0)=101.6Omega to` resistance at temp. of triple point <br/> `T_(0)=273.16K` <br/> `R_(1)=165.5Omega to` temp. `T_(1)=600.5K` <br/> `R_(2)=123.4Omega to T_(2)=?` <br/> `:.(T_(2)-273.16)/(600.5-273.16)=(123.4-101.6)/(165.5-101.6)` <br/> `:.(T_(2)-273.16)/(327.34)=(21.8)/(63.9)` <br/> `:.T_(2)~~385K`</body></html>


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