The temperature of a body falls from $62^oC\, to\, 50^oC$ in $10$ minutes. If the temperature of the surroundings is $26^oC$, the temperature in next $10$ minutes will become ...... $^oC$
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(a) $\frac{{62 - 50}}{{10}} = K\left( {\frac{{62 + 50}}{2} - 26} \right)$==> $\frac{6}{5} = K \times 30$
==> $K = \frac{1}{{25}}$
And, $\frac{{50 - \theta }}{{10}} = \frac{1}{{25}}\left( {\frac{{50 + \theta }}{2} - 26} \right)$
==> $\theta = 42^\circ C$.
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