Question

If a copper wire is stretched to increase its length by $0.1%$ then percentage increase in its resistance will be
A. $0.2%$
B. $2%%
C. 1%$
D. $0.1%$

Answer 1

Hint:Volume of the wire remains constant on stretching but its length increases and area decreases due to which its resistance changes. Resistivity is a characteristic property and doesn’t change with the shape and size of material. Using these properties of a material, we can find out the change in resistance of copper.

Complete step by step answer:
Let the initial length of the wire be $L$ and area of cross-section be $A$ so volume of the wire, $V = LA$. The Initial resistance of the wire, $R = \rho \dfrac{L}{A}$ , where ρ is the resistivity of the wire. After stretching, the length of the wire becomes,
$L' = L + \dfrac{{0.1}}{{100}}L \\
\Rightarrow L' = L + \dfrac{1}{{1000}}L \\
\Rightarrow L' = \dfrac{{1001}}{{1000}}L \\
\Rightarrow L' = 1.001L\\$
But the volume of the wire remains the same as it doesn’t change on stretching, so
$LA = L'A'$ , here $A'$ is the area of the wire after stretching.
$
A' = \dfrac{{LA}}{{L'}} \\
\Rightarrow A' = \dfrac{{LA}}{{1.001L}} \\
\Rightarrow A' = \dfrac{A}{{1.001}} \\ $
Resistivity is a characteristic property of each material so it is dependent only on the nature of the material of the wire; it doesn’t change with a change in length or area. So the resistivity of wire after stretching remains ρ. But Resistance is directly proportional to the length of wire and inversely proportional to its area of cross-section, so after stretching the wire its resistance changes. So new resistance,
$
R' = \rho \dfrac{{L'}}{{A'}} \\
\Rightarrow R' = \rho \dfrac{{1.001L}}{{\dfrac{A}{{1.001}}}} \\
\Rightarrow R' = \rho {\dfrac{{\left( {1.001} \right)}}{A}^2}L \\
\Rightarrow R' = 1.002 \times \rho \dfrac{L}{A} \\
\Rightarrow R' = 1.002R \\ $
Now change in resistance = New resistance – initial resistance = $R' - R$
Percentage change in resistance $ = \dfrac{{R' - R}}{R} \times 100$
$
\Rightarrow\dfrac{{\left( {1.002} \right)R - R}}{R} \times 100 \\
\Rightarrow\dfrac{{0.002R}}{R} \times 100 \\
\Rightarrow 0.2\% \\ $
$\therefore$ Resistance is increased by 0.2% on stretching the wire.

Hence option A is the correct answer.

Additional information:
On increasing the temperature of a metallic conductor, the free electrons start moving with higher speed and thus collide more frequently which increases the collisions per unit time. Thus resistivity of metallic conductors increases with an increase in temperature while in the case of semiconductors when the temperature is increased the forbidden gap between the conduction band and valence band decreases due to which electrons move easily from valence band to conduction band. This increases the conductivity of the material and thus resistivity of semiconductors decreases with an increase in temperature. Similarly, the resistivity of insulators decreases with an increase in temperature but the increase in temperature needs to be high.

Note:Volume remains constant after stretching because the mass of the wire and its density remains the same. Be careful with the calculations and the student should not get confused between the terms resistance and resistivity. Resistivity changes with change in temperature. We can make a material superconductor by providing it with required temperature conditions.