If a wire of resistance 1ohm is stretched to double it's length,then resistance will be

The resistance $R$ of a wire is given by:
$R = \rho \frac{L}{A}$
where $\rho$ is the resistivity of the material, $L$ is the length, and $A$ is the cross-sectional area.

The wire is stretched to double its length, so the new length $L'$ is:
$L' = 2L$

When a wire is stretched, its volume remains constant.
Initial volume $V = A \times L$
New volume $V' = A' \times L'$
Since $V = V'$, we have:
$A \times L = A' \times (2L)$
$A = 2A'$
So, the new cross-sectional area $A'$ is:
$A' = \frac{A}{2}$

Substitute the new length $L'$ and new area $A'$ into the resistance formula:
$R' = \rho \frac{L'}{A'}$
$R' = \rho \frac{2L}{A/2}$
$R' = \rho \frac{4L}{A}$
$R' = 4 \left( \rho \frac{L}{A} \right)$
Since $R = \rho \frac{L}{A}$, we get:
$R' = 4R$

Given the initial resistance $R = 1\ \Omega$:
$R' = 4 \times 1\ \Omega$
$R' = 4\ \Omega$

The new resistance will be $4\ \Omega$.