• # question_answer sThe current I vs voltage V graphs for a given metallic wire at two different temperatures ${{T}_{1}}$ and ${{T}_{2}}$ are shown in the figure. It is concluded that A)  ${{T}_{1}}>{{T}_{2}}$                   B)  ${{T}_{1}}<{{T}_{2}}$C)  ${{T}_{1}}={{T}_{2}}$              D)  ${{T}_{1}}=2{{T}_{2}}$

R increases with increasing temp: $V=IR$ Slope of graph $=\frac{1}{V}=\frac{1}{R};$. Slope of ${{T}_{1}}$is more i.e., $\frac{1}{{{R}_{1}}}$ is more, hence ${{R}_{1}}$ is less. This concludes that ${{T}_{1}}$ will be less than ${{T}_{2}}$ as ${{R}_{1}}$is less than ${{R}_{2}}$.