A) \[{{23.33}^{o}}C\] and \[A\]
B) \[{{83.33}^{o}}C\] and \[{{20}^{o}}C\]
C) \[{{50}^{o}}C\] and \[{{30}^{o}}C\]
D) \[{{30}^{o}}C\] and \[{{50}^{o}}C\]
Correct Answer: B
Solution :
If suppose \[{{K}_{Ni}}=K\Rightarrow {{K}_{Al}}=3K\] and \[{{K}_{Cu}}=6K.\] Since all metal bars are connected in series So \[{{\left( \frac{Q}{t} \right)}_{Combination}}={{\left( \frac{Q}{t} \right)}_{Cu}}={{\left( \frac{Q}{t} \right)}_{Al}}={{\left( \frac{Q}{t} \right)}_{Ni}}\] and \[\frac{3}{{{K}_{eq}}}=\frac{1}{{{K}_{Cu}}}+\frac{1}{{{K}_{Al}}}+\frac{1}{{{K}_{Ni}}}=\frac{1}{6K}+\frac{1}{3K}+\frac{1}{K}=\frac{9}{6K}\] Þ \[{{K}_{eq}}=2K\] Hence, if \[{{\left( \frac{Q}{t} \right)}_{Combination}}={{\left( \frac{Q}{t} \right)}_{Cu}}\] Þ \[\frac{{{K}_{eq}}\,A(100-0)}{{{l}_{Combination}}}=\frac{{{K}_{Cu}}A(100-{{\theta }_{1}})}{{{l}_{Cu}}}\] Þ \[\frac{2K\,A\,(100-0)}{(25+10+15)}=\frac{6K\,A\,(100-{{\theta }_{1}})}{25}\] Þ \[{{\theta }_{1}}=83.33{}^\circ C\] Similar if \[{{\left( \frac{Q}{t} \right)}_{Combination}}={{\left( \frac{Q}{t} \right)}_{Al}}\] Þ \[\frac{2K\,A(100-0)}{50}=\frac{3K\,A({{\theta }_{2}}-0)}{15}\]Þ \[{{\theta }_{2}}={{20}^{o}}C\]You need to login to perform this action.
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