A) 50 sec
B) 90 sec
C) 60 sec
D) 48 sec
Correct Answer: D
Solution :
According to Newtons cooling law. \[\frac{{{\theta }_{1}}-{{\theta }_{2}}}{t}=K\left( \frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}-\theta \right)\] \[\frac{{{80}^{o}}-{{60}^{o}}}{1}=K\left( \frac{{{80}^{o}}+{{60}^{o}}}{2}-{{30}^{o}} \right)\] \[{{20}^{o}}=K\times {{40}^{o}}\] \[\Rightarrow \] \[K=\frac{1}{2}\] For 2nd case \[\frac{{{60}^{o}}-{{50}^{o}}}{t}=K\left( \frac{{{60}^{o}}+{{50}^{o}}}{2}-{{30}^{o}} \right)\] \[\frac{{{10}^{o}}}{t}=\frac{1}{2}({{55}^{o}}-{{30}^{o}})\] \[\frac{10}{t}=\frac{1}{2}\times 25\] \[t=\frac{10\times 2}{25}\min .\] \[=\frac{20}{25}\times 60=48\sec \]
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