Physics - Current Electricity Question with Solution | TestHub

Let the unknown resistance be$\mathrm{X}$. Then the condition of Wheatstone's bridge gives$\frac{\text{X}}{\text{R}}=\frac{20\text{r}}{\text{80r}}$, where$\mathrm{r}$is the resistance of wire per$\mathrm{cm}$.

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$\therefore \text{X}=\frac{20}{80}\times \text{R}=\frac{1}{4}\times 1=\text{0.25 }\mathrm{\Omega }$