JEE Main 2024 (Online) 31st January Evening ShiftQuestion 1
The temperature $T(t)$ of a body at time $t=0$ is $160^{\circ}$F and it decreases continuously as per the differential equation $\frac{dT}{dt}=-K(T-80)$, where $K$ is a positive constant. If $T(15)=120^{\circ}$F, then $T(45)$ is equal to
Differential Equation
Solution
Concept. $T-80$ decays exponentially. Formulas used. (1) $\dfrac{dT}{dt}=-K(T-80)\Rightarrow T-80=(T_0-80)e^{-Kt}$Step 1 - Solve. By (1), $T-80=80e^{-Kt}$. Step 2 - Use $T(15)=120$. $40=80e^{-15K}$, so $e^{-15K}=\frac12$. Step 3 - At $t=45$. $T-80=80\left(\frac12\right)^3=10$, so $T(45)=90^\circ$F. Answer: (C)