A) \[24{}^\circ C\]
B) \[20{}^\circ C\]
C) \[14{}^\circ C\]
D) \[28{}^\circ C\]
Correct Answer: D
Solution :
Let after next 7 min, its temperature be\[\theta \]. From Newton's law of cooling. \[\frac{{{\theta }_{1}}-{{\theta }_{2}}}{t}\propto \left( \frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}-{{\theta }_{0}} \right)\] where\[{{\theta }_{0}}=\]temperature of surrounding. \[\therefore \] \[\frac{60-40}{7}\propto \left( \frac{60+40}{2}-10 \right)\] ??(i) and \[\frac{40-\theta }{7}\propto \left( \frac{40+\theta }{2}-10 \right)\] ...(ii) Dividing Eq. (i) by Eq. (ii), we obtain \[\frac{20}{7}\times \frac{7}{(40-\theta )}=\frac{40}{(20+\theta )/2}\] \[\Rightarrow \] \[\frac{20}{40-\theta }=\frac{40\times 2}{20+\theta }\] \[\Rightarrow \] \[20+\theta =160-4\theta \] \[\Rightarrow \] \[5\theta =160-20=140\] \[\therefore \] \[\theta =\frac{140}{5}={{28}^{o}}C\]
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