Question

A uniform wire of resistance $100\Omega$ is melted and recasted into a wire of length twice that of the original. The resistance of the new wire is _____.

• A. $100\Omega$
• B. $200\Omega$
• C. $300\Omega$
• D. $400\Omega$

Answer 1

The correct option is D $400\Omega$

We know that resistance, $R=\rho \frac{l}{A}$, where $\rho$ is the specific resistance or resistivity of the material, $l$ is the length of the material and $A$ is the area of cross-section.

In both the cases, the material remains the same, $\rho$ will be constant. Also, volume of the wire, $V=Al$ is also constant.
$⟹A=\frac{V}{l}$ and resistance, $R$ can be re-written as $R=\rho \frac{l^2}{V}$
So, $\frac{R_1}{R_2}=\frac{(l_1{)}^2}{(l_2{)}^2}=\frac{l^2}{(2l{)}^2}=\frac{1}{4}$
Given, resistance $R_1=100\Omega$
$⟹R_2=4R_1=4\times 100=400\Omega$
Hence, the new wire will have a resistance of $400\Omega$.