MHT CET · Maths · Application of Derivatives
If surrounding air is kept at \(20^{\circ} \mathrm{C}\) and body cools from \(80^{\circ} \mathrm{C}\) to \(70^{\circ} \mathrm{C}\) in 5 minutes, then the temperature of the body after 15 minute will be
- A \(52.7^{\circ} \mathrm{C}\)
- B \(51.7^{\circ} \mathrm{C}\)
- C \(54.7^{\circ} \mathrm{C}\)
- D \(50.7^{\circ} \mathrm{C}\)
Answer & Solution
Correct Answer
(C) \(54.7^{\circ} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
\(\frac{d T}{d t}=-k(T-20) \Rightarrow T-20=e^{-k t+c} \Rightarrow T=\) \(20+e^c e^{-k t}\)
for \(t=0, T=80 \Rightarrow e^c=60\)
i.e. \(T=20+60 \cdot e^{-k t}\)
For \(t=5, T=70 \Rightarrow 70=20+60 e \Rightarrow-5 k=\log \frac{5}{6}\)
i.e. \(T=20+60 e^{\frac{1}{5} \log \frac{5}{6} \times t}\)
Now, for \(t=15\)
\(T=20+60 e^{15 \times \frac{1}{5} \log \frac{5}{6}}=20+60 \times\left(\frac{5}{6}\right)^3=54.7\)
for \(t=0, T=80 \Rightarrow e^c=60\)
i.e. \(T=20+60 \cdot e^{-k t}\)
For \(t=5, T=70 \Rightarrow 70=20+60 e \Rightarrow-5 k=\log \frac{5}{6}\)
i.e. \(T=20+60 e^{\frac{1}{5} \log \frac{5}{6} \times t}\)
Now, for \(t=15\)
\(T=20+60 e^{15 \times \frac{1}{5} \log \frac{5}{6}}=20+60 \times\left(\frac{5}{6}\right)^3=54.7\)
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