A) \[100{}^\circ C\]
B) \[200{}^\circ C\]
C) \[300{}^\circ C\]
D) \[400{}^\circ C\]
Correct Answer: D
Solution :
Comparing the given equation with standard equation \[E=\alpha t+\frac{1}{2}\beta {{t}^{2}}\] \[\alpha =40\] and \[\frac{1}{2}\beta =-\frac{1}{20}\] Þ \[\beta =-\frac{1}{10}\] Hence neutral temperature \[{{t}_{n}}=-\frac{\alpha }{\beta }=\frac{-\,40}{-1/10}\] Þ \[{{t}_{n}}={{400}^{o}}C\]
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