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

एक वस्तु का तापमान \(\mathrm{T}(\mathrm{t})\), समय \(\mathrm{t}=0\) पर \(160^{\circ} \mathrm{F}\) है तथा यह अवकल समीकरण \(\frac{\mathrm{dT}}{\mathrm{dt}}=-\mathrm{K}(\mathrm{T}-80)\) के अनुसार लगातार कम हो रहा है, जहाँ \(\mathrm{K}\) धनात्मक अचर है। यदि \(\mathrm{T}(15)=120^{\circ} \mathrm{F}\) है, तो \(\mathrm{T}(45)\) = ........... है।

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

Correct Answer

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

Step-by-step Solution

Detailed explanation

\( \frac{\mathrm{dT}}{\mathrm{dt}}=-\mathrm{k}(\mathrm{T}-80) \) \( \int_{160}^{\mathrm{T}} \frac{\mathrm{dT}}{(\mathrm{T}-80)}=\int_0^{\mathrm{t}}-\mathrm{Kdt} \) \( {[\ln |\mathrm{T}-80|]_{160}^{\mathrm{T}}=-\mathrm{kt}} \) \( \ln |\mathrm{T}-80|-\ln 80=-\mathrm{kt}\)…
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