JEE MAIN - Mathematics (2024 - 31st January Evening Shift - No. 12)
The temperature $$T(t)$$ of a body at time $$t=0$$ is $$160^{\circ} \mathrm{F}$$ and it decreases continuously as per the differential equation $$\frac{d T}{d t}=-K(T-80)$$, where $$K$$ is a positive constant. If $$T(15)=120^{\circ} \mathrm{F}$$, then $$T(45)$$ is equal to
90$$^\circ$$ F
85$$^\circ$$ F
80$$^\circ$$ F
95$$^\circ$$ F
Explanation
$$\begin{aligned}
& \frac{\mathrm{dT}}{\mathrm{dt}}=-\mathrm{k}(\mathrm{T}-80) \\
& \int_\limits{160}^{\mathrm{T}} \frac{\mathrm{dT}}{(\mathrm{T}-80)}=\int_\limits0^{\mathrm{t}}-\mathrm{Kdt} \\
& {[\ln |\mathrm{T}-80|]_{160}^{\mathrm{T}}=-\mathrm{kt}} \\
& \ln |\mathrm{T}-80|-\ln 80=-\mathrm{kt} \\
& \ln \left|\frac{\mathrm{T}-80}{80}\right|=-\mathrm{kt} \\
& \mathrm{T}=80+80 \mathrm{e}^{-\mathrm{kt}} \\
& 120=80+80 \mathrm{e}^{-\mathrm{k} .15} \\
& \frac{40}{80}=\mathrm{e}^{-\mathrm{k} 15}=\frac{1}{2} \\
& \therefore \mathrm{T}(45)=80+80 \mathrm{e}^{-\mathrm{k} .45} \\
& =80+80\left(\mathrm{e}^{-\mathrm{k} .15}\right)^3 \\
& =80+80 \times \frac{1}{8} \\
& =90
\end{aligned}$$
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