The temperature $\mathrm{T}(\mathrm{t})$ of a body at time $\mathrm{t}=0$ is $160^{\circ}$ F and it decreases continuously as per the differential equation $\frac{\mathrm{dT}}{\mathrm{dt}}=-\mathrm{K}(\mathrm{T}-80)$, where K is positive constant. If $\mathrm{T}(15)=120^{\circ}$ F, then $\mathrm{T}(45)$ is equal to

JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Differential Equations

The temperature

T (t)

of a body at time

t = 0

16 0^{\circ}

F and it decreases continuously as per the differential equation

\frac{dT}{dt} = - K (T - 80)

, where K is positive constant. If

T (15) = 12 0^{\circ}

F, then

T (45)

is equal to

Answer: (C)

\frac{d T}{d t} = - K (T - 80)

, so

T - 80 = (T (0) - 80) e^{- K t} = 80 e^{- K t}

.

Using

T (15) = 120

40 = 80 e^{- 15 K} \Rightarrow e^{- 15 K} = \frac{1}{2}

T (45) - 80 = 80 e^{- 45 K} = 80 (e^{- 15 K})^{3} = 80 (\frac{1}{2})^{3} = \frac{80}{8} = 10

T (45) = 9 0^{\circ}

JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Single correct option

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