A) \[{{20}^{o}}C\]
B) \[{{30}^{o}}C\]
C) \[{{15}^{o}}C\]
D) \[{{10}^{o}}C\]
Correct Answer: D
Solution :
According to Newton's law of cooling \[\frac{{{\theta }_{2}}-{{\theta }_{1}}}{t}=K\left[ \frac{{{\theta }_{1}}-{{\theta }_{2}}}{2}-{{\theta }_{s}} \right]\] where, \[{{\theta }_{s}}\] is the temperature of the surroundings. \[\frac{60-50}{10}=K\left[ \frac{60+50}{2}-{{\theta }_{s}} \right]\] Similarly, \[\frac{50-42}{10}=K\,(46-{{\theta }_{s}})\] \[\frac{8}{10}=K\,(46-{{\theta }_{s}})\] ... (ii) Dividing Eq. (i) by Eq. (ii), we get \[\frac{10}{8}=\frac{K(55-{{\theta }_{s}})}{K(46-{{\theta }_{s}})}\] \[\Rightarrow \] \[{{\theta }_{s}}={{10}^{o}}C\]
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