Question

The resistance of a wire of 80cm and of uniform area of cross section 0.025square is found to be 1.50 ohm .calculate resistivity of the wire

Answer 1

Answer:

The resistivity of the wire is 4.687 × 10^-4 Ωm.

Explanation:

Given :-

• The resistance of a wire of 80 cm and of uniform area of cross section 0.025 cm² is found to be 1.50 Ω.

To find :-

• Resistivity of the wire.

Solution :-

Formula used :

${\boxed{\sf{R=\dfrac{\rho\:l}{A}}}}$

• Terms identification :
• R = Resistance
• $\rho$=Resistivity
• l = Length
• A = cross section area

Given :-

• R = 1.50 Ω

• L = 80 cm

• A = 0.025 cm²

$\mapsto\sf{R=\dfrac{\rho\:l}{A}}$

$\mapsto\sf{1.50=\dfrac{\rho\:\times\:80}{0.025}}$

$\mapsto\sf{80\rho=1.50\times\:0.025}$

$\mapsto\sf{80\rho=0.0375}$

$\mapsto\sf{\rho=\dfrac{0.0375}{80}}$

$\mapsto\sf{\rho=4.687\times\:10^{-4}}$

Therefore the resistivity of the wire is 4.687 × 10^-4 Ωm.

Answer 2

AnsWer :

$\tt {\pink{We \:have}}\begin{cases} \sf{\green{Resistance \: (R)=1.50 \:\Omega}}\\ \sf{\blue{Area\:of\:cross\: section \: (A)=0.025\:cm^2}}\\ \sf{\orange{Length \: of \: the wire \: (l)=80\:cm}}\\ \sf{\red{Resistivity \: ( \rho)=?}}\end{cases}$

$: \implies\sf R = \dfrac{\rho \: l}{A}$

$: \implies\sf \rho = \dfrac{ A R }{l}$

$: \implies\sf \rho = \dfrac{0.025 \times 1.50 }{80}$

$: \implies\sf \rho = \dfrac{0.0375}{80}$

$: \implies \underline{ \boxed{\sf \rho =4.687 \times {10}^{ - 4} \: \Omega \: m}}$

$\therefore\underline{\textsf { Resistivity of the wire is \textbf{4.687 \times 10 ^{-4} \Omega m}}}.$