Question

The length of a conducting wire is 50 cm and its radius is 0.5 mm. If its resistance is 30 Ω , what is the Resistivity of it's material?

Answer 1

▪︎■ ANSWER ■▪︎

$\bf\underline{Solution-}$

$L = 50 \: cm = 50 \times 10 {}^{ - 2}m \\ r = 0.5 \: mm = 0.5 \times 10 {}^{ - 3} m \\ = 5 \times 10 {}^{ - 4}m \: and \: r = 30 \: Ω \:$

$Resistivity \: ρ \: = \frac{ra}{l} \\ and \: a = \pi {r}^{2} \\ \therefore \: ρ \: = r \frac{\pi {r}^{2} }{l}$

$= \frac{30 \times 3.14 \times (5 \times {10}^{ - 4 ){}^{2} } }{50 \times 10} \\ \\ = \frac{30 \times3 .14 \times 25 \times {10}^{ - 8} }{50 \times - 10 {}^{ - 2} } \\ \\ = 47.1 \times {10}^{ - 6} \: Ω \: m\\ = 4.71 \times {10}^{ - 5} \: Ω \: m$

$\therefore$ Resistivity of the wire is $\boxed{ 4.71×10^{-5}Ω\:m}$

Answer 2

Answer:

$50cm=50×10−2mr=0.5mm=0.5×10−3m=5×10−4mandr=30Ω</p><p></p><p>\begin{gathered}Resistivity \: ρ \: = \frac{ra}{l} \\ and \: a = \pi {r}^{2} \\ \therefore \: ρ \: = r \frac{\pi {r}^{2} }{l} \end{gathered}</p><p></p><p>$

Step-by-step explanation:

$=50×1030×3.14×(5×10−4)2=50×−10−230×3.14×25×10−8=47.1×10−6Ωm=4.71×10−5Ωm</p><p></p><p>\therefore</p><p></p><p>$

$∴ Resistivity of the wire is \boxed{ 4.71×10^{-5}Ω\:m}4.71×10−5Ωm</p><p>$