A) 4
B) 5
C) 6
D) 7
Correct Answer: A
Solution :
Accord to Newtons law of cooling Rate of cooling \[\propto \] excess of temperature or rate of cooling \[=K\times \]excess of temperature. If \[{{\theta }_{1}}\] is initial temperature, \[{{\theta }_{2}}\] is the final temperature and \[{{\theta }_{0}}\] is the room temperature, then \[\frac{{{\theta }_{1}}-{{\theta }_{2}}}{t}=K\left( \frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}-{{\theta }_{0}} \right)\] Since, body cools from, \[{{75}^{o}}\]to \[{{65}^{o}}\]in 2 min and surrounding temperature is \[{{30}^{o}}.\] \[\frac{75-65}{2}=K\left( \frac{75+65}{2}-30 \right)\] or \[5=K\times 40\Rightarrow K=\frac{5}{40}=\frac{1}{8}\] When body cools from \[55{{\,}^{o}}C\]to \[45{{\,}^{o}}C,\]we have \[\frac{55-45}{{{t}_{2}}}=\frac{1}{8}\left( \frac{55+45}{2}-30 \right)\] \[\frac{10}{{{t}_{2}}}=\frac{1}{8}\times 20\] or \[{{t}_{2}}=\frac{10\times 8}{20}=4\min \]
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