A wire of resistance X ohm is gradually stretched till its length becomes twice its original length. If its new resistance becomes $\displaystyle 40 \Omega$, find the value of X.

$\displaystyle R=n^{2} x$ $\displaystyle 40=(2)^{2} x$ $\displaystyle x=10 \Omega$Volume of wire before and after stretching will remain the same: $\displaystyle A_{1} l_{1}=A_{2} l_{2}$ $\displaystyle A_{2}=\frac{A l}{2 l}=\frac{A}{2}$ $\displaystyle R=\rho \frac{l_{2}}{A_{2}}$ $\displaystyle 40=4 \rho \frac{l}{A}$ $\displaystyle \therefore x=10 \Omega$