Question

A wire of length $L$ is drawn such that its diameter is reduced to half of its original diameter. If the initial resistance of the wire were $10\Omega$ , its new resistance would be

• A. $40\Omega$
• B. $80\Omega$
• C. $120\Omega$
• D. $160\Omega$

Answer 1

Answer: (D)

Let the original diameter of the wirc be $D$.
Therefore the new diameter is $D/2.$
Original area of cross-section is $\frac{\pi D^2}{4}$
and the final area of cross-section is $\frac{\pi D^2}{16}$
The new length of the wire is given by
$L\times \frac{\pi D^2}{4}=L^{\prime }\times \frac{\pi D^2}{16}$
$\Rightarrow L^{\prime }=\frac{16}{4}L=4L$
Now, we know that the resistance is given by
$R=\rho \frac{L}{A}$.
$\therefore R^{\prime }=\rho \frac{L^{\prime }}{A^{\prime }}=\rho \frac{4L}{A/4}=16R$
$\left[\because A^{\prime }=\frac{\pi D^2}{16}=\frac{A}{4}\right]$
$\therefore R^{\prime }=16\times 10=160\Omega$