Question

If a resistance $R$ is melted and recasted to half of its length, then the new resistance of the wire will be

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

Answer 1

Answer: (D)

$\frac{R}{4}$

Resistance of wire $R=\frac{\rho L}{A}$

Also volume of wire $V=AL$

$⟹$ $R=\rho \frac{L^2}{V}$

We get $R\propto L^2$ (as volume of the wire remains the same when it is recasted)

Now the wire is melted and recasted to half of its length i.e. $L^{\prime }=\frac{L}{2}$

Thus new resistance of wire $R^{\prime }=R\frac{(L^{\prime }{)}^2}{L^2}$

$\therefore$ $R^{\prime }=R\frac{(L/2{)}^2}{L^2}=\frac{R}{4}$