MCQ
If $E=a t+b t^2$, what is the temperature of inversion
- ✓$-\frac{a}{2 b}$
- B$+\frac{a}{2 b}$
- C$-\frac{a}{b}$
- D$+\frac{a}{b}$
✓
Answer
Correct option: A.
$-\frac{a}{2 b}$
(a) $E=a t+b t^2$ at inversion temperature $E$ will be minimumThus $\frac{d E}{d t}=0 \Rightarrow \frac{d}{d t}\left[a t+b t^2\right]=0$$\Rightarrow a+2 b t=0 \Rightarrow t=-\frac{a}{2 b}$
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