Determine the required length of a copper wire having a cross-sectional area of 0.01 mm^2 to produce a resistance of 1 k . The resistivity of copper is given as 1.7 × 10^ -8 m .

Correct answer: C. 0.59 km The resistance of a wire is given by the formula: R = (l)/(A) Rearranging for length l : l = (R A)/( ) Given values are: R = 1 k = 10^3 A = 0.01 mm^2 = 0.01 × 10^ -6 m^2 = 10^ -8 m^2 = 1.7 × 10^ -8 m Substituting the values into the formula: l = 10^3 × 10^ -8 1.7 × 10^ -8 l = (1000)/(1.7) m l 588.2 m l 0.588 km 0.59 km

Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: Electric Current in Conductors

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Electric Current in Conductors

Determine the required length of a copper wire having a...

Determine the required length of a copper wire having a cross-sectional area of

0.01 \text{ mm}^2

to produce a resistance of

1 \text{ k}\Omega

. The resistivity of copper is given as

1.7 \times 10^{-8} \Omega \text{ m}

Options

A $0.059 \text{ km}$
B $1.7 \text{ km}$
C $0.59 \text{ km}$
D $5.9 \text{ km}$

Correct answer

C. $0.59 \text{ km}$

Step-by-step solution

The resistance of a wire is given by the formula:

R = \rho \dfrac{l}{A}

Rearranging for length

l

l = \dfrac{R A}{\rho}

Given values are:

R = 1 \text{ k}\Omega = 10^3 \Omega

A = 0.01 \text{ mm}^2 = 0.01 \times 10^{-6} \text{ m}^2 = 10^{-8} \text{ m}^2

\rho = 1.7 \times 10^{-8} \Omega \text{ m}

Substituting the values into the formula:

l = \dfrac{10^3 \times 10^{-8}}{1.7 \times 10^{-8}}

l = \dfrac{1000}{1.7} \text{ m}

l \approx 588.2 \text{ m}

l \approx 0.588 \text{ km} \approx 0.59 \text{ km}

Practice Electric Current in Conductors on Quantrex Academy →

More from Electric Current in Conductors