Question

Resistivity of iron is $1\times {10}^{-7}ohmm$. The resistance of the given wire of a particular thickness and length is $1ohm$. If the diameter and length of the wire both are doubled find the resistivity and resistance of the wire.

Answer 1

Given: Resistivity of iron is $1\times {10}^{-7}\Omega m$. The resistance of the given wire of a particular thickness and length is $1\Omega$

To find the new resistivity and resistance of the wire, if the diameter and length of the wire both are doubled

Solution:

Resistivity of some material is its intrinsic property and is constant at particular temperature. Resistivity does not depend upon shape.

i.e., the resistivity of a material is independent of physical dimension of the material. Hence the resistivity will remain same, i.e., $1\times {10}^{-7}\Omega m$

According to the question, new diameter, $d_{new}=2d$

and new length, $L_{new}=2L$, $R=1\Omega$, $\rho =1\times {10}^{-7}\Omega m$

Let new resistance be $R_{new}$

Now we know,

$R=\rho \frac{L}{A}$, and $A=\pi r^2=\pi {\left(\frac{d}{2}\right)}^2$

Therefore, new area, $A_{new}=\pi {\left(\frac{d_{new}}{2}\right)}^2$

$A_{new}=\pi {\left(\frac{2d}{2}\right)}^2⟹A_{new}=4\pi {\left(\frac{d}{2}\right)}^2=4A$

Therefore, new resistance,

$R_{new}=\rho \frac{L_{new}}{A_{new}}$

$⟹R_{new}=\rho \frac{2L}{4A}$ by substituting the corresponding values

$R_{new}=\frac{1}{2}\times \rho \frac{L}{A}⟹R_{new}=\frac{1}{2}\times R⟹R_{new}=\frac{1}{2}\times 1=0.5$

Therefore the new resistance is $0.5\Omega$