A) 4 : 1
B) 1 : 4
C) 1:1
D) 2 : 1
Correct Answer: A
Solution :
Let\[AC={{l}_{1}},CB={{l}_{2}}\]and r the resistance per unit length of wire AB. As there is no current in galvanometer G, the resistance 60 Q, 15 0, AC and CB form the arms of a balanced Wheatstone's bridge \[\therefore \] \[\frac{P}{Q}=\frac{R}{S}\] \[\Rightarrow \] \[\frac{60}{15}=\frac{r\times {{l}_{1}}}{r\times {{l}_{2}}}\] Or \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{4}{1}\]
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