Current Electricity

Let $R_1$ be the initial resistance of the wire in the left gap.$R_2$ be the resistance in the right gap $l_1$ be the initial balancing length ( 40 cm )Let $l_2$ be the remaining length ( $100-40=60 \text{ cm}$ )The initial ratio is $\; \frac{R_1}{R_2} = \; \frac{40}{60} = \; \frac{2}{3}$When the wire is stretched to double its length, its resistance becomes $4 R_1$Let $l_1$ " be the new balancing length $l_2$ 'be the new remaining length (100- $l_1$ )The new ratio is $\; \frac{4 R_1}{R_2} = \; \frac{l_1'}{100 - l_1'}$$\; 4 \cdot \left( \; \frac{2}{3} \right) = \; \frac{l_1'}{100 - l_1'}$$\; 8 (100 - l_1') = 3 l_1'$$\; 800 - 8 l_1' = 3 l_1'$$\; 11 l_1' = 800$$\; l_1' = \; \frac{800}{11}$The new balancing point is approximately $\left( \; \frac{800}{11} \right) \text{ cm}$ from the left end.