A wire of resistance $160 ~\Omega$ is melted and drawn in a wire of one-fourth of its length. The new resistance of the wire will be

Let the original length of the wire be L and its cross-sectional area be A. Then, its resistance R is given by:<br/><br/> $R = \frac{\rho L}{A}$<br/><br/> where $\rho$ is the resistivity of the material of the wire. <br/><br/> When the wire is melted and drawn into a wire of one-fourth of its length, its new length is L/4 and its new cross-sectional area is 4A (since the same amount of material is now spread over a longer length). Therefore, its new resistance R' is given by:<br/><br/> $R' = \frac{\rho (L/4)}{4A} = \frac{R}{16}$ <br/><br/> Substituting the given value of R, we get:<br/><br/> $R' = \frac{160}{16} = 10 ~\Omega$ <br/><br/> Therefore, the new resistance of the wire is $10 ~\Omega$.