Question

1 kg piece of copper is drawn into a wire of cross-

sectional area 1 mm2, and then the same wire is


remoulded again into a wire of cross-sectional area 2 mm2.

Find the ratio of the resistance of the first wire to the second

wire.

Answer 1

$\large\underline{\bigstar \: \: {\sf Given-}}$

• Cross section area of first wire ${\sf (A_1)}$ = 1mm²
• Cross section of second wire ${\sf (A_2)}$ = 2mm²

$\large\underline{\bigstar \: \: {\sf To \; Find -}}$

• Ratio of Resistance of first wire to the second wire ${\sf \left(\dfrac{R_1}{R_2}\right)}$

$\large\underline{\bigstar \: \: {\sf Formula \: Used -}}$

$\large\implies\underline{\boxed{\sf R=\rho\dfrac{\ell}{A}}}$

$\large\underline{\bigstar \: \: {\sf Solution-}}$

We know that from the given formula -

Resistance is directly proportional to Lenght of wire

$\implies{\sf R\: \: \propto \: \: \ell}$

Resistance is inveresly proportional to Cross-section Area of wire

$\implies{\sf R \: \; \propto \: \; \dfrac{1}{A} }$

➩ Convert Cross-section area from mm² to m²

$\implies{\sf A_1=1mm^2}$

$\implies{\bf A_1=1 \times 10^{-6}m^2}$

$\implies{\sf A_2=2mm^2}$

$\implies{\bf A_2=2 \times 10^{-6}m^2}$

__________________________________

$\implies{\sf \dfrac{R_1}{R_2}=\dfrac{A_2}{A_1}}$

$\implies{\sf \dfrac{R_1}{R_2}=\dfrac{2\times 10^{-6}}{1 \times 10^{-6}}}$

$\implies{\sf \dfrac{R_1}{R_2}=\dfrac{2}{1} }$

$\implies{\bf R_1:R_2=2:1 }$

$\large\underline{\bigstar \: \: {\sf Answer-}}$

Ratio of Resistance for first wire to the second wire is ${\bf R_1:R_2=2:1}$