The length of a given cylindrical wire is increased to double of its original length. The percentage increase in the resistance of the wire will be ______ %.

Given: The length of a cylindrical wire is increased to double its original length.

$$R = \frac{\rho L}{A}$$

where $$\rho$$ is the resistivity, $$L$$ is the length, and $$A$$ is the cross-sectional area.

When the wire is stretched, its volume remains constant:

$$A_1 L_1 = A_2 L_2$$

If $$L_2 = 2L_1$$, then:

$$A_2 = \frac{A_1 L_1}{2L_1} = \frac{A_1}{2}$$

$$R_1 = \frac{\rho L_1}{A_1}$$

$$R_2 = \frac{\rho L_2}{A_2} = \frac{\rho \times 2L_1}{A_1/2} = \frac{4\rho L_1}{A_1} = 4R_1$$

$$\% \text{ increase} = \frac{R_2 - R_1}{R_1} \times 100 = \frac{4R_1 - R_1}{R_1} \times 100 = 3 \times 100 = 300\%$$

Hence, the percentage increase in resistance is 300%.