(4) \(\frac{3l}{(100-l)}\)
Given \(\frac{x}{R}=\frac{l}{100-l}\)
When 5 P is connected is parallel with X, the effective resistance in Gap 1 is \(\frac{x\times5p}{x+5p}\)
When 15 Q is connected in parallel with R, the effective resistance in Gap 2 is \(\frac{R\times15Q}{R+15Q}\)
Since balancing length remains same,
![](https://www.sarthaks.com/?qa=blob&qa_blobid=8305373803370997300)
Alternatively: Since the balacing length is the same, potential drops across the gaps remain the same. Therefore, the ratio of resistances would be the same along individual paths.
![](https://www.sarthaks.com/?qa=blob&qa_blobid=11797461831456496598)