Question

Calculate the area of cross section of a wire if is length is 1.0m it's resistance is 2.3ohm and the resistivityof the material of the wire is 1.84×10–6ohm m

Answer 1

Given :

• Length of wire, L = 1.0 m
• Resistance of wire, R =2.3 Ω
• Resistivity of wire, ρ = 1.84 × 10⁻⁶  Ω m

To find :

• Area of cross section of wire, A = ?

Formula required :

• Relation between Resistance (R), resistivity (ρ), Length (L) and area of cross section (A) of conductor

$\red{\bigstar}\boxed{\sf{R=\dfrac{\rho\;\; L}{A}}}$

Solution :

Using formula      

$\implies\sf{R=\dfrac{\rho\;\;L}{A}}$

$\implies\sf{2.3=\dfrac{1.84 \times 10^{-6}\times 1.0}{A}}$

$\implies\sf{A=\dfrac{1.84 \times 10^{-6}}{2.3}}$

$\implies\underline{\underline{\red{\sf{A=0.8\times10^{-6}\;\;m^2}}}}$

Therefore,

• Area of cross section of conductor is 0.8 × 10⁻⁶ m².

Answer 2

Given:

• Resistance (R) = 2.3 Ω

• Length of the wire (l) = 1 m

• Resistivity of wire (ρ) = 1.84 × $\sf 10^-6\:$ Ω

To Find :

Cross section area of wire = ?

Solution :

$\huge\bigstar \: \: \boxed{\tt R = \rho \dfrac{l}{A}}\: \: \bigstar$

$: \implies\sf 23 \: = \: 1.84 \: \times \: {10}^{ - 6} \: \times \: \dfrac{1}{A}$

$: \implies\sf A \: = \: \dfrac{1.84 \: \times \: {10}^{ - 6}}{23}$

$: \implies\underline{ \boxed{\sf A \: = \: 0.8 \: \times \: {10}^{ - 6} \:m^{2}}} \: \: \bigstar$

$\underline{\sf Therefore,\:the \: area \: of \: cross \: section \: of \: wire \: is \: \bf{ 0.8 \times 10^{-6}\:m^2}}.$