Question

The resistance of a wire of length 2m and
area of cross-section 0.5 mm^2 is 2.2 ohm.Calculate the specific resistance of thematerial of the wire.
​

Answer 1

Answer:

length,l=2m

Area, A=0.5mm^2

A=0.5×1/10^6 m^2

A=0.5×10^-6m^2

Resistance, R=2.2 ohm

R=p×l/A

p=R×A/l

p=2.2×0.5×10^-6/2

p=0.55 ×10^-6 ohm-m

p=5.5×10^-7 ohm-m

Answer 2

The specific resistance of the material of the wire is ${5.5 \times10^-^7}$Ωm

Concept:

Resistivity is the resistance of a conductor whose length is 1 m and the area of cross-section is 1 m². It is also known as specific resistance. It is denoted by ρ (rho).

Formula of  resistivity:

Specific resistance/ Resistivity ρ$=R\times \frac{A}{L}$

Where R = resistance, L = length of the conductor, and A = area of cross-section of the conductor.

Given :

Length of a wire, L= 2m

Area of cross-section, $A = 0.5 mm^2$

$A = 0.5 \times \frac{1}{10^6} m^2$

$A=0.5\times10^-^6 m^2$

Resistance, R = 2.2 Ω (ohm)

To find : The specific resistance of the material of the wire .

Solution:

Step 1 : Put the given values in the formula and simplify:

$R = \rho\times \frac{L}{A}$

$\rho = R\times \frac{A}{L}$

$\rho =\frac{2.2\times0.5 \times10^-^6}{2}$

$\rho ={1.1\times0.5 \times10^-^6}$

$\rho ={0.55 \times10^-^6}$ ohm m

$\rho ={5.5 \times10^-^7}$ Ωm

Hence, the specific resistance of the material of the wire is ${5.5 \times10^-^7}$ Ωm.

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