Question

A wire is stretched to double its length .what will be its new resistivity?

Answer 1

Answer:

The new resistivity of the material of the wire is $\dfrac{\rho R_{2}}{4R_{1}}$.

Explanation:

Given that,

length $l_{2}=2l_{1}$

The resistance of the wire is equal to the product of resistivity of the material and length of the wire divided by the area of cross section.

The resistance of the wire is defined as,

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

Where, $\rho$= resistivity of the material

A = area of cross section

R = resistance of the wire

The volume of both wire is same.

The volume of original wire = volume of new wire

So,$V_{1}=V_{2}$

$A_{1}l_{1}=A_{2}l_{2}$

If a wire is stretched to double its length

$l_{2}=2l_{1}$

Then, $A_{1}l_{1}=A_{2}\times2l_{1}$

$A_{2}=\dfrac{A_{1}}{2}$

The resistivity of the material is ,

$\rho = \dfrac{RA}{l}$

The resistivity of the material of the original wire is

$\rho = \dfrac{R_{1}A_{1}}{l_{1}}$.....(I)

The resistivity of the material of the new wire is

$\rho' = \dfrac{R_{2}A_{2}}{l_{2}}$.....(II)

The ratio of the resistivity of the material of the wire is

$\dfrac{\rho}{\rho'}=\dfrac{ \dfrac{R_{1}A_{1}}{l_{1}}}{ \dfrac{R_{2}A_{2}}{l_{2}}}$

Put the value in the equation

$\dfrac{\rho}{\rho'}=\dfrac{R_{1}2A_{1}2l_{1}}{R_{2}A_{1}l_{1}}$

$\dfrac{\rho}{\rho'}=\dfrac{4R_{1}}{R_{2}}$

$\rho'=\dfrac{\rho R_{2}}{4R_{1}}$

Hence, The new resistivity of the material of the wire is $\dfrac{\rho R_{2}}{4R_{1}}$.

Answer 2

Answer:

4 times

Explanation:

Resistance= Resistivity x length/area

When the wire is stretched to double the length , the area of cross section gets reduced to half.

So

New Resistance = Resistivity x 2 length/area/2

I.e,

New resistance = (Resistivity x length/area)x 4

i.e, New resistance = Resistance x 4

So when the wire is stretched, the resistance multiplies by four times.

Hope you understood. Keep exploring.