A) \[{{K}_{1}}={{K}_{4}}\,\text{and}\,\ {{K}_{2}}={{K}_{3}}\]
B) \[{{K}_{1}}{{K}_{4}}={{K}_{2}}{{K}_{3}}\]
C) \[{{K}_{1}}{{K}_{2}}={{K}_{3}}{{K}_{4}}\]
D) \[\frac{{{K}_{1}}}{{{K}_{4}}}=\frac{{{K}_{2}}}{{{K}_{3}}}\]
Correct Answer: B
Solution :
For no current flow between C and D \[{{\left( \frac{Q}{t} \right)}_{AC}}={{\left( \frac{Q}{t} \right)}_{CB}}\]Þ \[\frac{{{K}_{1}}A({{\theta }_{A}}-{{\theta }_{C}})}{l}=\frac{{{K}_{2}}A({{\theta }_{C}}-{{\theta }_{B}})}{l}\] Þ \[\frac{{{\theta }_{A}}-{{\theta }_{C}}}{{{\theta }_{C}}-{{\theta }_{B}}}=\frac{{{K}_{2}}}{{{K}_{1}}}\] ...(i) Also \[{{\left( \frac{Q}{t} \right)}_{AD}}={{\left( \frac{Q}{t} \right)}_{DB}}\]Þ\[\frac{{{K}_{3}}A({{\theta }_{A}}-{{\theta }_{D}})}{l}=\frac{{{K}_{4}}A({{\theta }_{D}}-{{\theta }_{B}})}{l}\] Þ\[\frac{{{\theta }_{A}}-{{\theta }_{D}}}{{{\theta }_{D}}-{{\theta }_{B}}}=\frac{{{K}_{4}}}{{{K}_{3}}}\] ...(ii) It is given that \[{{\theta }_{C}}={{\theta }_{D}},\] hence from equation (i) and (ii) we get \[\frac{{{K}_{2}}}{{{K}_{1}}}=\frac{{{K}_{4}}}{{{K}_{3}}}\] Þ \[{{K}_{1}}{{K}_{4}}={{K}_{2}}{{K}_{3}}\]
You need to login to perform this action.
You will be redirected in
3 sec
