Question

When a conducting wire is connected in the right gap and a known resistance in the left gap, the balancing length is 60cm. If the wire is stretched so that its length increases by $20\%$, then the balancing length becomes

• A. 54 cm
• B. 65.6 cm
• C. 64 cm
• D. 51 cm

Answer 1

The correct option is D 51 cm

$\frac{R_1}{R_2}=\frac{60}{40}=\frac{3}{2}\to \left(1\right)\frac{R_1}{R_2^1}=\frac{l^1}{100-l^1}\to \left(2\right)\frac{R_2^1}{R_2}={\left[\frac{l_2^1}{l_2}\right]}^2={\left[\frac{120}{100}\right]}^2=1.44FromEq\left(2\right)\frac{R_1}{1.44R_2}=\frac{l^1}{100-l^1}\frac{3}{1.44\times 2}=\frac{l^1}{100-l^1}\frac{3}{2.88}=\frac{l^1}{100-l^1}300-3l^1=2.88l^{1}300=5.88l^1{l}^1=\frac{300}{5.88}=51cm$