Question

If a resistance R Is melted and recast to half of its length, then the new resistance of the wire will be

• A. $\frac{R}{2}$
• B. $\frac{R}{4}$
• C. $2R$
• D. $R$

Answer 1

The correct option is B

$\frac{R}{4}$

Step 1: Given

Original resistance$=R$

New resistance$=R\text{'}$

Original length$=L$

New length$=L\text{'}=\frac{L}{2}$

Step 2: Determine the resistance

Resistance is given by $R=\frac{\rho L}{A}$

Volume is given by $$\begin{gathered} V=AL \\ \Rightarrow A=\frac{V}{L} \end{gathered}$$

Replacing area,

$$\begin{gathered} R=\frac{\rho L}{{\displaystyle \frac{V}{L}}} \\ =\frac{\rho L^2}{{\displaystyle V}} \end{gathered}$$

Step 3: Determine the new resistance

When the wire is recast, the volume will remain the same.

New resistance,

$$\begin{gathered} R\text{'}=\frac{\rho L{\text{'}}^2}{V} \\ =\frac{\rho {\left({\displaystyle \frac{L}{2}}\right)}^2}{V} \\ =\frac{\rho L^2}{4V} \\ =\frac{R}{4} \end{gathered}$$

Hence, option B is the correct option.