A wire with a length of 1 m and a radius of 0.1 mm possesses a resistance of 100 . Determine the resistivity of the material.

Correct answer: A. π × 10^ -6 m Given, length of the wire, l = 1 m Radius of the wire, r = 0.1 mm = 10^ -4 m Resistance of the wire, R = 100 The cross-sectional area of the wire is A = π r^2 = π (10^ -4 )^2 = π × 10^ -8 m^2 Using the formula for resistance, R = (l)/(A) The resistivity of the material is = (R A)/(l) Substituting the given values, we get = 100 × π × 10^ -8 1 = π × 10^ -6 m

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

A wire with a length of 1 m and a radius of 0.1 mm possesses a...

A wire with a length of

1 \text{ m}

and a radius of

0.1 \text{ mm}

possesses a resistance of

100 \Omega

. Determine the resistivity of the material.

Options

A $\pi \times 10^{-6} \Omega \text{ m}$
B $\pi \times 10^{-4} \Omega \text{ m}$
C $\pi \times 10^{-8} \Omega \text{ m}$
D $\pi \times 10^{-5} \Omega \text{ m}$

Correct answer

A. $\pi \times 10^{-6} \Omega \text{ m}$

Step-by-step solution

Given, length of the wire,

l = 1 \text{ m}

Radius of the wire,

r = 0.1 \text{ mm} = 10^{-4} \text{ m}

Resistance of the wire,

R = 100 \Omega

The cross-sectional area of the wire is

A = \pi r^2 = \pi (10^{-4})^2 = \pi \times 10^{-8} \text{ m}^2

Using the formula for resistance,

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

The resistivity of the material is

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

Substituting the given values, we get

\rho = \dfrac{100 \times \pi \times 10^{-8}}{1} = \pi \times 10^{-6} \Omega \text{ m}

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