A wire of length $$10$$ cm and radius $$\sqrt{7} \times 10^{-4}$$ m connected across the right gap of a meter bridge. When a resistance of $$4.5 \; \Omega$$ is connected on the left gap by using a resistance box, the balance length is found to be at $$60$$ cm from the left end. If the resistivity of the wire is $$R \times 10^{-7} \; \Omega$$ m, then value of $$R$$ is :

In a meter bridge, at balance: $$\frac{R_1}{R_2} = \frac{l}{100 - l}$$

Given: $$R_1 = 4.5 \; \Omega$$ (left gap), balance length $$l = 60$$ cm.

$$\frac{4.5}{R_2} = \frac{60}{40} = \frac{3}{2}$$

$$R_2 = \frac{4.5 \times 2}{3} = 3 \; \Omega$$

The wire in the right gap has length $$l_w = 10$$ cm $$= 0.1$$ m and radius $$r = \sqrt{7} \times 10^{-4}$$ m.

Using $$R_2 = \frac{\rho l_w}{\pi r^2}$$:

$$3 = \frac{\rho \times 0.1}{\pi \times 7 \times 10^{-8}}$$

$$\rho = \frac{3 \times \pi \times 7 \times 10^{-8}}{0.1} = \frac{21\pi \times 10^{-8}}{0.1} = 21\pi \times 10^{-7}$$

$$\rho = 21 \times 3.14159 \times 10^{-7} \approx 65.97 \times 10^{-7} \approx 66 \times 10^{-7} \; \Omega \text{m}$$

So $$R = 66$$.

The answer is $$66$$, which corresponds to Option (3).