A wire of resistance 20 ohm, length 10 cm and cross-sectional area is 5 cm2 is melted to form a wire of cross-sectional area 1.5cm2. What is the resistance of new wire?

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Solution

$R_{1} = 20 ohm l_{1} = 10 cm A_{1} = 5 {cm}^{2} A_{2} = 1.5 {cm}^{2} Now, when the wire is melted, its volume will remain unchanged . l_{1} A_{1} = l_{2} A_{2} 10 \times 5 = l_{2} \times 1.5 Or l_{2} = \frac{10 \times 5}{1.5} = 33.3 cm Now, Resistance R = \frac{ρl}{A} So, R_{1} = \frac{{ρl}_{1}}{A_{1}} and R_{2} = \frac{{ρl}_{2}}{A_{2}} Or \frac{R_{1}}{R_{2}} = \frac{{ρl}_{1}}{A_{1}} \times \frac{A_{2}}{{ρl}_{2}} = \frac{l_{1}}{A_{1}} \times \frac{A_{2}}{l_{2}} = \frac{10}{5} \times \frac{1.5}{\frac{10 \times 5}{1.5}} = 2 \times \frac{1.5 \times 1.5}{10 \times 5} \frac{20}{R_{2}} = 2 \times \frac{1.5 \times 1.5}{10 \times 5} Or R_{2} = \frac{20 \times 10 \times 5}{2 \times 1.5 \times 1.5} = \frac{1000}{4.5} = 222.2 ohm$

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