A body takes $10$ minutes to cool down from $62^o C$ to $50^o C$. If the temperature of surrounding is $26^o C$ then in the next $10$ minutes temperature of the body will be ......... $^oC$
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From Newton's cooling law
$\frac{\theta_{1}-\theta_{2}}{t}=K\left(\frac{\theta_{1}+\theta_{2}}{2}-26\right)$
$Case-I:$ $\frac{12}{10}=\mathrm{K} \times 30$
$\mathrm{K}=\frac{12}{10 \times 30}=\frac{1}{25}$
$Case-II$ $: \frac{50-\theta_{2}}{10}=\mathrm{K}\left(\frac{50+\theta_{2}}{2}-26\right)$
$\frac{50-\theta_{2}}{10}=\frac{1}{25}\left(\frac{\theta_{2}-2}{2}\right)$
$250-5 \theta_{2}=\theta_{2}-2$
$6 \theta_{2}=252$
$\theta_{2}=\frac{252}{6}=42^{\circ} \mathrm{C}$
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