A) \[({{\theta }_{i}}-{{\theta }_{0}}){{e}^{-kt}}\]
B) \[({{\theta }_{i}}-{{\theta }_{0}})\,\,In\,\,(kt)\]
C) \[{{\theta }_{0}}+{{e}^{-kt/C}}\]
D) \[{{\theta }_{i}}{{e}^{-kt}}-{{\theta }_{0}}\]
Correct Answer: C
Solution :
[c] \[-\frac{\theta t}{dt}=K(\theta -{{\theta }_{0}})\] \[\frac{dt}{\theta -{{\theta }_{0}}}=-Kdt\] \[\log \,\,(\theta -{{\theta }_{0}})=-Kt+C\] \[\theta -{{\theta }_{0}}={{e}^{-Kt/C}}\] \[\theta ={{\theta }_{0}}={{e}^{-Kt/C}}\]You need to login to perform this action.
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