Question

Calculate the resistance of a wire
1.0 km long and 0.50mm diameter if
resistivity of copper is 1.7 × 10^ -8 ohm metre.
Please help me.
Correct answer will be marked as brainliest.

Answer 1

Explanation:

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Answer 2

Given:

Length of wire, l= 1Km= 1000m= 10³m

Diameter of wire, D= 0.5mm= 0.0005 m

Resistivity of wire, ρ= 1.7×10⁻⁸ Ωm

To Find:

Resistance of given wire

Solution:

We know that,

• Expression for resistance R of a conductor is given by

$\pink{\boxed{\bf{R=\frac{\rho l}{A}}}}$

where,

l is the length of the conductor

A is the area of the conductor

ρ is the resistivity of the material if the conductor

• Area of a cross-section of wire is given by

$\purple{\boxed{\bf{A=\pi r^{2}}}}$

where,

r is the radius of wire

• Diameter= 2(radius)

$\rule{190}{2}$

Let the radius, area and resistance of given wire be 'r', 'A' and 'R' respectively

So,

$\longrightarrow\rm{D=2r}$

$\longrightarrow\rm{0.0005=2r}$

$\longrightarrow\rm{2r=0.0005}$

$\longrightarrow\rm{r=\dfrac{0.0005}{2}}$

$\longrightarrow\rm{r=0.00025=2.5\times10^{-4} \:m}$

$\rule{190}{2}$

Also,

$\longrightarrow\rm{A=\pi r^{2}}$

$\longrightarrow\rm{A=\dfrac{22}{7} \times(2.5\times10^{-4})^{2}}$

$\longrightarrow\rm{A=\dfrac{22}{7} \times6.25\times10^{-8}}$

$\longrightarrow\rm{A=\dfrac{137.5}{7}\times10^{-8}\:m^{2}}$

$\rule{190}{2}$

Now,

$\longrightarrow\rm{R=\dfrac{\rho l}{A}}$

$\longrightarrow\rm{R=\dfrac{1.7\times\cancel{10^{-8}}\times10^{3}}{\dfrac{137.5}{7}\times\cancel{10^{-8}}}}$

$\longrightarrow\rm{R=\dfrac{7\times1.7\times10^{3}}{137.5}}$

$\longrightarrow\rm{R=\dfrac{11.9\times10^{3}}{137.5}}$

$\longrightarrow\rm{R=0.08655\times10^{3}}$

$\longrightarrow\rm\green{R=86.55\:\Omega}$

Hence, the resistance of given wire is 86.55 Ω.