NEET 2020 Physics Question 179 Solution

For a metre bridge with unknown resistance $$R$$ in the left gap and known resistance $$S = 10\ \Omega$$ in the right gap, at null deflection: $$\frac{R}{S} = \frac{l_1}{l_2}$$, where $$l_1 : l_2$$ is the ratio of balancing wire lengths.

Given $$l_1 : l_2 = 3 : 2$$: $$\frac{R}{10} = \frac{3}{2} \implies R = 10 \times \frac{3}{2} = 15\ \Omega$$.

The physical wire has total length $$L = 1.5\text{ m}$$ and total resistance $$R = 15\ \Omega$$. Linear resistance density: $$\lambda_{\text{res}} = \frac{R}{L} = \frac{15\ \Omega}{1.5\text{ m}} = 10\ \Omega/\text{m}$$.

$$\text{Length for } 1\ \Omega = \frac{1\ \Omega}{10\ \Omega/\text{m}} = 0.1\text{ m} = 1.0 \times 10^{-1}\text{ m}$$. Matches Option 2.