A wire of length 10cm and radius 7 × 10 ^ - 4 m connected across...

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

$\dfrac{4 . 5Ω}{60} = \dfrac{X}{100 - 60} ⇒ X = \dfrac{40}{60} \times 4 . 5Ω = 3Ω$

Also,

$\begin{cases}X & = & \dfrac{ρl}{A} \\ & = & \dfrac{ρl}{\pi{r}^{2}} . . . \left(1\right)\end{cases}$

From equation (1), it follows that

$3 = ρ \times \dfrac{0 . 1}{\pi \times {\left(\sqrt{7} \times {10}^{- 4}\right)}^{2}} ⇒ ρ = \dfrac{3 \times \pi \times {\left(\sqrt{7} \times {10}^{- 4}\right)}^{2}}{0 . 1} \approx 66 \times {10}^{- 7}Ωm$

Therefore, $R = 66$.