Question

The specific resistance of a wire if its length is 100cm and cross sectional area is 0.0020squarecm and resistance is 5 Ohms

Answer 1

Length of the wire (l) = 100 cm = 1 m

Are of cross section (A) = 0.002 cm²

                                       = 0.002 × 10⁻⁴ m²

                                     = 2 × 10⁻³ × 10⁻⁴ m²

                                       = 2 × 10⁻⁷ m²

Resistance of the wire (R) = 5 Ω

Specific resistance of the wire / Resistivity = ρ Ω-m

$\bigstar\ \; \bf \pink{R=\dfrac{\rho l}{A} }$

where ,

• R denotes resistance
• ρ denotes specific resistance/resistivity
• l denotes length
• A denotes cross sectional area

Plugging values we get ,

$\to \sf R=\dfrac{\rho l}{A}$

$\to \sf 5=\dfrac{\rho (1)}{2\times 10^{-7}}$

$\to \sf \rho =10\times 10^{-7}\ \Omega .m\ \; \; \\\\\to \sf \rho=1\times 10^{-6}\ \Omega.m\ \; \orange{\bigstar}$

Answer 2

Answer:

• The specific resistance of a wire is $\sf{ 1 × 10^{-6} \Omega m.}$

Explanation:

Given that,

• Length of wire (l) = 100cm ↪1m.
• Cross sectional area (A) = 0.002 cm² ↪ $\sf{ 2 × 10^{-7} m^2}$
• Resistance (R) = 5 $\Omega$

As we know that,

$\red \bigstar \: { \boxed {\sf{ R = \frac{\rho l}{A} }}}\: \:$

[ Putting values ]

$\sf \: \leadsto \: 5 = \frac{ \rho \times 1}{2 \times 10 {}^{ - 7} } \\ \\ \sf \: \leadsto \rho \: = 5 \times 2 \times 10 {}^{7} \: \Omega m \\ \\ \sf \: \leadsto \rho \: = 10 \times 10 {}^{ - 7} \Omega \: m \\ \\ \leadsto \sf \underbrace{ \rho = 1 \times 10^{-6} \Omega m} \: \: \green \bigstar$