A) \[\frac{{{s}_{1}}{{\theta }_{1}}+{{s}_{2}}{{\theta }_{2}}+{{s}_{3}}{{\theta }_{3}}}{{{m}_{1}}{{s}_{1}},\,\,{{m}_{2}}{{s}_{2}},\,\,{{m}_{3}}{{s}_{3}}}\]
B) \[\frac{{{m}_{1}}{{s}_{1}}{{\theta }_{1}}+{{m}_{2}}{{s}_{2}}{{\theta }_{2}}+{{m}_{3}}{{s}_{3}}{{\theta }_{3}}}{{{m}_{1}}{{s}_{1}},\,\,{{m}_{2}}{{s}_{2}},\,\,{{m}_{3}}{{s}_{3}}}\]
C) \[\frac{{{m}_{1}}{{s}_{1}}{{\theta }_{1}}+{{m}_{2}}{{s}_{2}}{{\theta }_{2}}+{{m}_{3}}{{s}_{3}}{{\theta }_{3}}}{{{m}_{1}}{{\theta }_{1}},\,\,{{m}_{2}}{{\theta }_{2}},\,\,{{m}_{3}}{{\theta }_{3}}}\]
D) \[\frac{{{m}_{1}}{{\theta }_{1}}+{{m}_{2}}{{\theta }_{2}}+{{m}_{3}}{{\theta }_{3}}}{{{\operatorname{s}}_{1}}{{\theta }_{1}},\,\,{{\operatorname{s}}_{2}}{{\theta }_{2}},\,\,{{\operatorname{s}}_{3}}{{\theta }_{3}}}\]
Correct Answer: B
Solution :
[b] Let lost temperature is \[T{}^\circ C\]then \[{{m}_{1}}{{s}_{1}}{{T}_{1}}+{{m}_{2}}{{s}_{2}}{{T}_{2}}+{{m}_{3}}{{s}_{3}}{{T}_{3}}\] = released energy =\[{{m}_{1}}{{s}_{1}}{{T}_{1}}+{{m}_{2}}{{s}_{2}}{{T}_{2}}+{{m}_{3}}{{s}_{3}}{{T}_{3}}\]= energy taken \[\Rightarrow T=\frac{{{m}_{1}}{{s}_{1}}{{T}_{1}}+{{m}_{2}}{{s}_{2}}{{T}_{2}}+{{m}_{3}}{{s}_{3}}{{T}_{3}}}{{{m}_{1}}{{s}_{1}}+{{m}_{2}}{{s}_{2}}+{{m}_{3}}{{s}_{3}}}\]
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