A) 1 minute
B) 2 minutes
C) 4 minutes
D) 16 minutes
Correct Answer: A
Solution :
\[T=\frac{{{K}_{1}}{{\theta }_{1}}+{{K}_{2}}{{\theta }_{2}}}{{{K}_{1}}+{{K}_{2}}}\]\[=\frac{300\times 100+200\times 0}{300+200}=60{}^\circ C\] (R = Thermal resistance) Þ \[\left( \frac{Q}{t} \right)=\frac{k\pi {{r}^{2}}({{\theta }_{1}}-{{\theta }_{2}})}{L}\propto \frac{{{r}^{2}}}{L}\] (\[\frac{{{Q}_{1}}}{{{Q}_{2}}}={{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}\left( \frac{{{l}_{2}}}{{{l}_{1}}} \right)={{\left( \frac{1}{2} \right)}^{2}}\times \left( \frac{2}{1} \right)=\frac{1}{2}\] Q and \[{{Q}_{2}}=2{{Q}_{1}}\] are same) Þ \[\frac{Q}{t}=\frac{KA\Delta \theta }{l}\]Þ\[6000=\frac{200\times 0.75\times \Delta \theta }{1}\] (Series resistance \[\Delta \theta =\frac{6000\times 1}{200\times 0.75}=40{}^\circ C\] and parallel resistance \[\frac{{{K}_{A}}A({{\theta }_{1}}-\theta )}{l}=\frac{{{K}_{B}}A(\theta -{{\theta }_{2}})}{l}\])
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