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V - I graph of a conductor at temperature T_(1) and T_(2) are shown in the figure (T_(2)-T_(1)) is proportional to |
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Answer» Solution :Slope of LINE gives RESISTANCE So, `R_(1)=tan theta=R_(0)(1+alphaT_(1))` `R_(2)=tan (90-theta)=cot theta=R_(0)(1+alphaT_(2))`  `cot theta-tan theta=R_(0)alpha(T_(2)-T_(1))` `or (cos theta)/(sin theta)-(sin theta)/(cos theta)=R_(0)alpha(T_(2)-T_(1))` `R_(0)alpha(T_(2)-T_(1))=(cos 2 theta)/(((sin 2 theta))/(2)) or T_(2)-T_(1) alpha cot 2 theta` |
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