A)
B)
C)
D)
Correct Answer: C
Solution :
\[J=A{{T}^{2}}{{e}^{-b/T}}\] Þ \[\frac{J}{{{T}^{2}}}\propto {{e}^{-b/T}}\] i.e. \[\frac{J}{{{T}^{2}}}\] will vary exponentially with \[\frac{1}{T}\], having negative slope.You need to login to perform this action.
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