Question

An Aluminium wire of length 2.0 m and cross-section area of 1.8 × 10^-6 m² has a resistance of 0.03 Ω. Find the resistivity of aluminium.​

Answer 1

Given :

• Length, $\sf \ell$ = 2.0 m
• Area of cross-section = 1.8 × $\sf 10^{-6}m^2$
• Resistance, R = 0.03 Ω

To Find :

• Resistivity of aluminium, ρ = ?

Solution :

As, we have :

• Length, $\sf \ell$ = 2.0 m
• Area of cross-section = 1.8 × $\sf 10^{-6}m^2$
• Resistance, R = 0.03 Ω

And we have to find resistivity of aluminium (ρ) :

We know that :

$\large \blue{ \dag} \underline{ \boxed{ \pink{\bf R = \dfrac{ \rho \ell}{A} }}}$

By substituting values :

$\sf : \implies 0.03 = \dfrac{ \rho \times 2.0}{1.8 \times 10^{ - 6} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf : \implies 0.03 \times 1.8 \times 10^{ - 6} = \rho \times 2.0 \\ \sf : \implies 0.054 \times 10^{ - 6} = \rho \times 2.0 \: \: \: \: \: \: \: \: \: \\ \sf : \implies \dfrac{ 0.054 \times 10^{ - 6}}{2.0} = \rho \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf : \implies 0.027 \times 10^{ - 6} = \rho \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf : \implies 2.7 \times 10^{ - 8} = \rho \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\large \blue{ \dag} \underline{ \boxed{ \pink{\bf \rho \: = 2.7 \times 10^{ - 8}Ωm}}}$

$\underline{\tt Hence, \: value \: of \: resistivity \: of \: aluminium \: is \: \pink{27 \times 10^{ - 8}Ω m.}}$

Answer 2

$\large \pmb{ \bf{GiVEN }} \\ \tt Length, \sf ℓ = 2.0 m \\ \tt Area \: of \: cross-section = 1.8 × \sf 10^{-6}m² \\ \tt \: Resistance, R = 0.03 Ω \\ \\ \\ \large \pmb{ \bf{ To \: FiND}} \\ \tt \: Resistivity \: o f \: aluminium, \: ρ = ? \\ \\ \\ \large \pmb{ \bf{SOLUTiON}} \\ \tt \: We \: know \: that \\ \large \blue \leadsto \underline{ \boxed{ \blue{\bf R = \dfrac{ \rho \ell}{A} }}} \\ \cr \: By \: substituting \: values : \\ \\ \begin{gathered} \sf : \mapsto 0.03 = \dfrac{ \rho \times 2.0}{1.8 \times 10^{ - 6} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf : \mapsto 0.03 \times 1.8 \times 10^{ - 6} = \rho \times 2.0 \\ \\ \sf : \mapsto 0.054 \times 10^{ - 6} = \rho \times 2.0 \: \: \: \: \: \: \: \: \: \\ \\ \sf : \mapsto \dfrac{ 0.054 \times 10^{ - 6}}{2.0} = \rho \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf : \mapsto 0.027 \times 10^{ - 6} = \rho \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf : \mapsto 2.7 \times 10^{ - 8} = \rho \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \end{gathered} \\ \\ \large \blue{ \ddag} \: \: \underline{ \boxed{ \red{\bf \rho \: = 2.7 \times 10^{ - 8}Ωm}}} \: \: \blue{ \ddag} \\ \\ \small{\underline{\tt Hence, \: value \: of \: resistivity \: of \: aluminium \: is \: \pink{27 \times 10^{ - 8}Ω m.}} }$