The temperature gradient across a copper rod of thickness 0.02m, maintained at two temperature junctions of 20º and 80ºC respectively is?
3.0 x 10\(^{2}\)km\(^{-1}\)
3.0 x 10\(^{3}\)km\(^{-1}\)
5.0 x 10\(^{3}\)km\(^{-1}\)
3.0 x 10\(^{4}\)km\(^{-1}\)
Correct answer is B
\(\Delta\) = \(\frac{\bigtriangleup\theta}{\delta} = \frac{80-20}{0.02}\)
= 3.0 x 10\(^{3}\)km\(^{-1}\)