Two wires 'A' and 'B' of equal lengths are connected in left and right gaps, of meter bridge, respectively. The null point is obtained at $40 \mathrm{~cm}$ from left end. Diameters of the wires 'A' and 'B' are in the ratio $3: 1$, the ratio of specific resistance of 'A' to that of 'B' is

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Aicharya · Accepted Answer

$\begin{array}{l}\frac{R_{1}}{R_{2}}=\frac{40}{60}=\frac{2}{3}=\frac{\frac{ho_{1} \ell_{1}}{A_{1}}}{\frac{ho_{2} \ell_{2}}{A_{2}}}=\frac{ho \pi r_{2}^{2}}{ho_{2} \pi r_{1}^{2}}=\frac{ho_{1} r_{2}^{2}}{ho_{2} r_{1}^{2}} \quad 	herefore \ell_{1}=\ell_{2} \ 	herefore \frac{ho_{1}}{ho_{2}}=\frac{r_{1}^{2}}{r_{2}^{2}} 	imes \frac{2}{3}=\frac{6}{1}\end{array} \quad \because \frac{r_{1}}{r_{2}}=\frac{3}{1}$

Two wires 'A' and 'B' of equal lengths are connected in left and right gaps, of meter bridge, respectively.…

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