A wire of length 10 cm and radius 7 × 10 - 4 m connected across the right gap of a meter bridge. When a…

From the null point condition of a Meter Bridge, it follows that,

$$\begin{gathered} \frac{4.5 \mathrm{\Omega }}{60}=\frac{X}{100-60} \\ \Rightarrow X=\frac{40}{60}\times 4.5 \mathrm{\Omega } \\ =3 \mathrm{\Omega } \end{gathered}$$

Also,

$\begin{array}{rcl}X& =& \frac{\rho l}{\mathrm{A}}\\ & =& \frac{\rho l}{\pi r^2} ...\left(1\right)\end{array}$

From equation (1), it follows that

$$\begin{gathered} 3=\rho \times \frac{0.1}{\pi \times {\left(\sqrt{7}\times {10}^{-4}\right)}^2} \\ \Rightarrow \rho =\frac{3\times \pi \times {\left(\sqrt{7}\times {10}^{-4}\right)}^2}{0.1} \\ \approx 66\times {10}^{-7} \mathrm{\Omega } \mathrm{m} \end{gathered}$$

Therefore, $R=66$.