ExamBro
ExamBro
JEE Mains · Maths · STD 12 - 9. differential equations

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

  1. A \(85^{\circ} \mathrm{F}\)
  2. B \(95^{\circ} \mathrm{F}\)
  3. C \(90^{\circ} \mathrm{F}\)
  4. D \(80^{\circ} \mathrm{F}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(90^{\circ} \mathrm{F}\)

Step-by-step Solution

Detailed explanation

\( \frac{\mathrm{dT}}{\mathrm{dt}}=-\mathrm{k}(\mathrm{T}-80) \) \( \int_{160}^{\mathrm{T}} \frac{\mathrm{dT}}{(\mathrm{T}-80)}=\int_0^{\mathrm{t}}-\mathrm{Kdt} \) \( {[\ln |\mathrm{T}-80|]_{160}^{\mathrm{T}}=-\mathrm{kt}} \) \( \ln |\mathrm{T}-80|-\ln 80=-\mathrm{kt}\)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app