A resistance of \( 2\,\Omega \) is to be made from a copper wire (resistivity \( 1.7\times10^{-8}\,\Omega\,\mathrm{m} \)) of length 50 cm. The area of cross-section of the wire must be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 4.25\times10^{-9}\,\mathrm{m^2} \)
The resistance of a wire of length \( l \) and area of cross-section \( A \) is \( R=\rho l/A \), so the area needed for a given resistance is \( A=\rho l/R \).
In SI units the data are:
\( \rho=1.7\times10^{-8}\,\Omega\,\mathrm{m} \)
\( l=50\,\mathrm{cm}=0.5\,\mathrm{m} \)
\( R=2\,\Omega \)
The area of cross-section is:
\( \begin{aligned} A &= \frac{\rho\,l}{R} \\ &= \frac{1.7\times10^{-8}\,\Omega\,\mathrm{m}\times0.5\,\mathrm{m}}{2\,\Omega} \\ &= 4.25\times10^{-9}\,\mathrm{m^2} \end{aligned} \)