A) \[{{45}^{o}}C\]
B) \[{{42.85}^{o}}C\]
C) \[{{40}^{o}}C\]
D) \[{{38.5}^{o}}C\]
Correct Answer: B
Solution :
Using the relation we have \[\frac{ms({{60}^{o}}-{{50}^{o}})}{10}=K\left( \frac{{{60}^{o}}+{{50}^{o}}}{2}-{{25}^{o}} \right)\] \[\frac{10ms}{10}=K({{55}^{o}}-{{25}^{o}})\] ... (1) Suppose the required temperature is\[T\] Then \[\frac{ms({{50}^{o}}-T)}{100}=K\left( \frac{{{50}^{o}}+T}{2}-{{25}^{o}} \right)\] \[=K\left( \frac{{{50}^{o}}+T-{{50}^{o}}}{2} \right)\] \[\frac{m({{50}^{o}}-T)}{10}=\frac{KT}{2}\] ? (2) Dividing equation (2) by (1) we get \[\frac{{{50}^{o}}-T}{10}=\frac{\frac{T}{2}}{{{30}^{o}}}=\frac{T}{{{60}^{o}}}\] \[{{50}^{o}}-T=\frac{T}{6}\] or \[{{300}^{o}}-6T=T\] So, \[T={{300}^{o}}\] or \[T=\frac{{{300}^{o}}}{7}={{42.85}^{o}}C\]
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