Wires give shocks, and so do their questions!

Resistance $R$ of a cylindrical wire is directly proportional to its length $l$ and inversely proportional to its cross sectional area $A$ ; that is $R \propto \dfrac{l}{A}$ .

Since volume of the wire $V = Al$ , therefore $A = \dfrac{V}{l}$ , for a constant $V$ , $A \propto \dfrac{1}{l}$ .

Therefore, we have: $R \propto \dfrac{l}{A} \quad \Rightarrow R \propto l^2$ and $\dfrac{R_2}{R_1} = \dfrac{l_2^2}{l_1^2} = 1.08^2 = 1.1664$ or an increase of $\boxed{16.64}$ % in resistance.

We know that for resistance $R$ , resistivity $ρ$ , wire length $l$ and cross section area $A$ , we have:

Clearly,

Then clearly the new wire will have more resistance. Now, let the original wire length be $l$ and the new wire length be $L$ . Let the new wire have variables in capital letters and original wire variables be in small letters. Then,

Now, change in $R$ :

In the above case, $V$ is the volume of the wire. It will be equal in both the cases because increasing the length of a wire by stretching it doesn’t affect its volume. Now, original resistance:

Now, increase in percentage of resistance: