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\(^{-1}\)
The correct answer is: B
Explanation
\(\Delta\) = \(\frac{\bigtriangleup\theta}{\delta} = \frac{80-20}{0.02}\)
= 3.0 x 10\(^{3}\)km\(^{-1}\)
There is an explanation video available .