If $U$ = $k x^a y^b z^c$

Then, $\frac{\triangle U}{U}$= $a\frac{\triangle x}{x}$ + $b\frac{\triangle y}{y}$ + $c\frac{\triangle z}{z}$, for small changes.

Resistance of a wire of lenght l and cross-sectional area A, is given by:

$R =$ $\rho \frac{l}{A}$ ...1

When the wire is stretched, both its length (l) as well as area of cross-section (A) changes but the volume of the material of the wire remains same.

$\Rightarrow$ $V$=$l A$ = constant ...2

If change in length is given, we can substitute $A$ from equation 2 to equation 1, to get a relation between $R$ and $l$.

$R =$ $\rho \frac{l^2}{V}$

$\Rightarrow$ $\frac{dR}{R} \times 100$ = $ 2\left( \frac{dl}{l} \times 100 \right)$

$\Rightarrow$ Percentage change in Resistance = $2$ $\times$ Percentage change in length

If change in cross-sectional area is given, we can substitute $l$ from equation 2 to equation 1, to get a relation between $R$ and $A$.

$R =$ $\rho \frac{V}{A^2}$

$\Rightarrow$ $\frac{dR}{R} \times 100$ = $- 2\left( \frac{dA}{A} \times 100 \right)$

$\Rightarrow$ Percentage change in Resistance = $-2$ $\times$ Percentage change in cross-sectional area

Above formula for percentage change are valid only if the change given is very small i.e. < 2%.

Length of a wire is doubled by stretching it. How many times does the resistance increase.

$R_1$ = $\rho \frac{l_1}{A_1}$

$R_2$ = $\rho \frac{l_2}{A_2}$

$l_1 A_1$ = $l_2 A_2$ $\Rightarrow$ $A_2$ = $A_1$ $\frac{l_1}{l_2}$ = $A_1$ $\frac{l_1}{2 l_1}$ = $\frac{A_1}{2}$

$\Rightarrow$ $R_2$ = $\rho \frac{2 l_1}{A_1/2}$ = $\frac{4\rho l_1}{A_1}$ = 4$R_1$

Thus, resistance becomes four times.

Percentage Increase =$\frac{R_2 - R_1}{R_1} \times 100$ = $300$ %.

If we would have used the above formula, i.e. $\frac{dR}{R} \times 100$ = $ 2\left( \frac{dl}{l} \times 100 \right)$ , answer would have been 400%, which is wrong because the change in length is not small.

Length of a wire is increased by $1$% by stretching it. What will be the percentage change in the resistance of the wire?

Here, change in length is very small, so by using formula, $\frac{dR}{R} \times 100$ = $ 2\left( \frac{dl}{l} \times 100 \right)$, the percentage change in Resistance will be $2 \times 1$ = 2%.