Question

1. A wire has a circular cross-section of radius 2mm and resistivity of its material is 4x10-6 ohm-m. Find the length of thin wire whose resistance is 60ohm.

Answer 1

Answer:

Resistance = resistivity * length / area

length = resistance * area / resistivity

length = 60*π*(2*10-³)² / (4*10-⁶) = 60π m

Explanation:

i hope this helps

Answer 2

Answer:

$\boxed{\sf Length \ of \ wire = 188.4 \ m \ \ or \ \ 60 \pi \ m}$

Given:

Radius of cross-section (r) = 2 mm = $\sf 2 \times 10^{-3} \ m$

Resistance (R) = 60 ohm

Resistivity $(\rho)$ = $4 \times 10^{-6} \ \Omega -m$

To Find:

Length of wire

Explanation:

$\sf Resistance \propto Length$$\sf Resistance \propto \frac{1}{Area \ of \ cross-section}$

$\sf \implies Resistance \propto \frac{Length}{Area \ of \ cross-section} \\ \\ \sf \implies Resistance = \rho (\frac{Length}{Area \ of \ cross-section} )\\ \\ \sf \implies R = \rho \frac{l}{A} \\ \\ \sf \implies R = \rho \frac{l}{\pi {r}^{2} } \\ \\ \sf \implies \rho \frac{l}{\pi {r}^{2} } = R \\ \\ \sf \implies \rho l = R \times \pi {r}^{2} \\ \\ \sf \implies l = \frac{R \times \pi {r}^{2}}{ \rho}$

$\sf \implies l = \frac{60 \pi \times (2 \times {10}^{ - 3})^{2} }{4 \times {10}^{ - 6} } \\ \\ \sf \implies l = \frac{60 \pi \times \cancel{4 \times {10}^{ -6 }} }{ \cancel{4 \times {10}^{ - 6}} } \\ \\ \sf \implies l = 60 \pi \\ \\ \sf \implies l = 60 \times 3.14 \\ \\ \sf \implies l = 188.4 \: m$