Question

Two wires A and B are made up of the same material and have the same mass. Wire A has radius \(2.0 \, \text{mm}\) and wire B has radius \(4.0 \, \text{mm}\). The resistance of wire B is \(2 \, \Omega\). The resistance of wire A is _____ \( \Omega \).

Answer 1

Correct Answer: 32

Given that both wires share identical material composition and mass, their volumes are also equal. Let \(\rho\) denote resistivity, \(A\) represent the cross-sectional area, \(V\) denote volume, and \(\ell\) signify length.

The resistance \(R\) of a wire is defined as: 
\[ R = \frac{\rho \ell}{A} = \frac{\rho V}{A^2}. \] 
As \(V\) remains constant for both wires, the ratio of their resistances can be expressed as: 
\[ \frac{R_A}{R_B} = \frac{A_B^2}{A_A^2} = \frac{r_B^4}{r_A^4}. \] 
Substituting the given values \(R_B = 2 \, \Omega\), \(r_B = 4 \, \text{mm}\), and \(r_A = 2 \, \text{mm}\): 
\[ \frac{R_A}{2} = \left(\frac{4 \times 10^{-3}}{2 \times 10^{-3}}\right)^4. \] 
Simplification yields: 
\[ \frac{R_A}{2} = 16, \] 
resulting in: 
\[ R_A = 32 \, \Omega. \]