The temperature gradient in a rod of $0.5 m$ long is ${80^o}C/m$. If the temperature of hotter end of the rod is ${30^o}C$, then the temperature of the cooler end is ...... $^oC$
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(b) $\frac{{{\theta _1} - {\theta _2}}}{l} = 80 \Rightarrow \frac{{30 - {\theta _2}}}{{0.5}} = 80 \Rightarrow {\theta _2} = - {10^o}C$
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