A) \[36{}^\circ C\]
B) \[36.4{}^\circ C\]
C) \[37{}^\circ C\]
D) \[37.5{}^\circ C\]
Correct Answer: B
Solution :
[b] We have\[\theta -{{\theta }_{s}}=({{\theta }_{0}}-{{\theta }_{s}}){{e}^{-kt}}\] ....(l) where \[{{\theta }_{0}}\] = initial temperature of body \[=40{}^\circ C\] \[\theta =\] temperature of body after time t \[{{\theta }_{s}}=\]temperature of surrounding Since body cools from \[40{}^\circ C to 38{}^\circ C\]in 10 min, we have \[38{}^\circ -30{}^\circ = \left( 40{}^\circ -30{}^\circ \right) {{e}^{-10k}}\] ....(2) Let after 10 min, the body temp. be\[\theta \]. \[\theta -30{}^\circ =\left( 38{}^\circ -30{}^\circ \right){{e}^{-10k}}\] .....(3) Dividing equ. (2) by equ. (3) gives, \[\frac{{{8}^{\operatorname{o}}}}{\theta -{{30}^{\operatorname{o}}}}=\frac{{{10}^{\operatorname{o}}}}{{{8}^{\operatorname{o}}}}\Rightarrow \theta -{{30}^{\operatorname{o}}}={{6.4}^{\operatorname{o}}}\Rightarrow \theta =36.{{4}^{\operatorname{o}}}\]
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