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Mathematics · Differential Equations

JEE Main 2024 — 31 January, Shift 2 — Question 8

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

  1. Option A:

    85∘F85^{\circ} \mathrm{F}

  2. Option B:

    95∘F95^{\circ} \mathrm{F}

  3. Option C:

    90∘F90^{\circ} \mathrm{F}

    Correct
  4. Option D:

    80∘F80^{\circ} \mathrm{F}

Answer: C

Step-by-step solution

dTdt=−k( T−80)\frac{\mathrm{dT}}{\mathrm{dt}}=-\mathrm{k}(\mathrm{~T}-80) ∫160TdT(T−80)=∫0t−Kdt\int_{160}^{\mathrm{T}} \frac{\mathrm{dT}}{(\mathrm{T}-80)}=\int_{0}^{\mathrm{t}}-\mathrm{Kdt}

[ln⁡∣T−80∣]160T=−kt[\ln |\mathrm{T}-80|]_{160}^{\mathrm{T}}=-\mathrm{kt}

ln⁡∣T−80∣−ln⁡80=−kt\ln |\mathrm{T}-80|-\ln 80=-\mathrm{kt}

ln⁡∣T−8080∣=−kt\ln \left|\frac{\mathrm{T}-80}{80}\right|=-\mathrm{kt}

T=80+80e−kt\mathrm{T}=80+80 \mathrm{e}^{-\mathrm{kt}}

120=80+80e−k.15120=80+80 \mathrm{e}^{-\mathrm{k} .15}

4080=e−k15=12\frac{40}{80}=\mathrm{e}^{-\mathrm{k} 15}=\frac{1}{2}

∴T(45)=80+80e−k. 45\therefore \mathrm{T}(45)=80+80 \mathrm{e}^{- \text {k. } 45}

=80+80(e−k.15)3=80+80\left(\mathrm{e}^{-\mathrm{k} .15}\right)^{3}

=80+80×18=80+80 \times \frac{1}{8}

=90=90

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
31 January, Shift 2
Subject
Mathematics
Chapter
Differential Equations
Topic
Applications of Differential Equations

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