A) \[350{}^\circ C\]
B) 350 K
C) \[560{}^\circ C\]
D) 560 K
Correct Answer: A
Solution :
Comparing the given equation with standard equation \[E=\alpha t+\frac{1}{2}\beta \,{{t}^{2}},\] we get \[\alpha =14\] and \[\frac{1}{2}\beta =-\,0.02\] Þ b = ? 0.04 Hence neutral temperature \[{{t}_{n}}=\,-\,\frac{\alpha }{\beta }\]\[=\,-\,\frac{14}{-\,0.04}=350{}^\circ C\]
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