The resistance of a wire is 5 $$\Omega$$. Its new resistance in ohm, if stretched to 5 times of its original length will be:

For a wire:

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

When the wire is stretched, its volume remains constant:

$$A_1L_1=A_2L_2$$

Given:

$$L_2=5L_1$$

So,

$$A_2=\frac{A_1}{5}$$

Now new resistance:

$$R_2=\rho\frac{5L_1}{A_1/5}$$

$$R_2=25\rho\frac{L_1}{A_1}$$

$$R_2=25R_1$$

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