A) \[{{K}_{1}}{{A}_{2}}={{K}_{2}}{{A}_{1}}\]
B) \[{{K}_{1}}{{A}_{1}}={{K}_{2}}{{A}_{2}}\]
C) \[{{K}_{1}}={{K}_{2}}\]
D) \[{{K}_{1}}A_{1}^{2}={{K}_{2}}A_{2}^{2}\]
Correct Answer: B
Solution :
\[{{\left( \frac{Q}{t} \right)}_{1}}=\frac{{{K}_{1}}{{A}_{1}}({{\theta }_{1}}-{{\theta }_{2}})}{l}\] and \[{{\left( \frac{Q}{t} \right)}_{2}}=\frac{{{K}_{2}}{{A}_{2}}({{\theta }_{1}}-{{\theta }_{2}})}{l}\] given \[{{\left( \frac{Q}{t} \right)}_{1}}={{\left( \frac{Q}{t} \right)}_{2}}\]Þ \[{{K}_{1}}{{A}_{1}}={{K}_{2}}{{A}_{2}}\]
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