(a) Write the relation between resistance and electrical resistivity of the material of a conductor in the shape of a cylinder of length ‘l’ and area of cross-section ‘A’. Hence derive the S.I. unit of electrical resistivity. (b) Resistance of a metal wire of length 5 m is 100Ω. If the area of cross-section of the wire is 3 x 10^–7 m², calculate the resistivity of the metal.

(a) Relation between resistance and electrical resistivity of the material of a conductor  in the shape of a cylinder of length ‘l’ and area of cross-section ‘A’ is:

$R=ρ\frac{l}{A}$

where,

$R-$ Resistance of the conductor.

$ρ(rho)-$ Resistivity (constant).

$l-$ Length of the conductor.

$A-$ Area of a cross-section of the conductor.

Derivation of the S.I. unit of electrical resistivity. 

For resistivity $ρ$, the above expression for resistance can be rearranged as-

$ρ=\frac{R\times A}{l}$

Now, putting the units of $R,\ A$ and $l$ in the expression.

$ρ=\frac{\Omega\ m^2}{m}$

$ρ=\Omega -m$

Thus, the S.I unit of resistivity is Ohm-meter $(\Omega -m)$.

(b) Given:

Resistance, $R=100 \Omega$

Length, $l=5m$

Area, $A=3\times {10^{-7}}m^2$

To find: Resistivity of the metal $ρ$.

Solution:

We know that resistivity, $ρ$ is given as-

$ρ=\frac{R\times A}{l}$

Putting the given values, we get-

$ρ=\frac{100\times 3\times {10^{-7}}}{5}$

$ρ=\frac{300\times {10^{-7}}}{5}$

$ρ=60\times {10^{-7}}\Omega -m$

$ρ=6\times {10}\times {10^{-7}}\Omega -m$

$ρ=6\times {10^{-6}}\Omega -m$

Thus, the resistivity of the metal is $6\times {10^{-6}}\Omega -m$.

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Updated on: 10-Oct-2022

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