[AIPMT 2014]
A) \[10\Omega \]
B) \[15\Omega \]
C) \[20\Omega \]
D) \[25\Omega \]
Correct Answer: B
Solution :
For first case, \[\frac{5}{{{l}_{1}}}=\frac{R}{(100-{{l}_{1}})}\,\,\,\,\,\,\,\,...(i)\] Now, by shunting resistance R by an equal resistance R, new resistance in that arm become \[\frac{R}{2}\] So \[\frac{5}{1.6{{l}_{1}}}\frac{R/2}{(100-{{l}_{1}})}\,\,\,\,\,\,\,\,\,\,....(ii)\] From Eqs. (i) and (ii) \[\frac{1.6}{1}=\frac{(100-1.6{{l}_{1}})}{100-{{l}_{1}}}\times 2\] Þ \[160-1.6{{l}_{1}}=200-3.2{{l}_{1}}\] \[1.6{{l}_{1}}=40\] \[{{l}_{1}}=\frac{400}{1.6}=25m\] From Eq. (i), \[(i),\frac{5}{25}=\frac{R}{75}\Rightarrow R=15\Omega \]
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