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)

$160\, \Omega$

Answer 2

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 $L \times \frac{\pi D^{2}}{4}=L' \times \frac{\pi D^{2}}{16}$ $\Rightarrow L'=\frac{16}{4} L=4 \,L$ Now, we know that the resistance is given by $R=\rho \frac{L}{A}$. $\therefore R'=\rho \frac{L'}{A'}=\rho \frac{4 L}{A / 4}=16 R$ ${\left[\because A'=\frac{\pi D^{2}}{16}=\frac{A}{4}\right]}$ $\therefore R'=16 \times 10=160 \,\Omega$