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

Since both wires are made of the same material and have the same mass, they also have the same volume. Let \(\rho\) be the resistivity, \(A\) the cross-sectional area, \(V\) the volume, and \(\ell\) the length. 

The resistance \(R\) of a wire is given by: 
\[ R = \frac{\rho \ell}{A} = \frac{\rho V}{A^2}. \] 
Since \(V\) is constant for both wires, we can write: 
\[ \frac{R_A}{R_B} = \frac{A_B^2}{A_A^2} = \frac{r_B^4}{r_A^4}. \] 
Given \(R_B = 2 \, \Omega\), \(r_B = 4 \, \text{mm}\), and \(r_A = 2 \, \text{mm}\), we substitute these values: 
\[ \frac{R_A}{2} = \left(\frac{4 \times 10^{-3}}{2 \times 10^{-3}}\right)^4. \] 
Simplifying this, we get: 
\[ \frac{R_A}{2} = 16, \] 
which gives: 
\[ R_A = 32 \, \Omega. \]

Answer 2

Step 1: Write down given data
For both wires, the material is the same ⇒ resistivity \( \rho \) is the same.
Both wires have the same mass ⇒ volume × density is the same ⇒ their volumes are equal.

Given:
Radius of wire A, \( r_A = 2.0 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)
Radius of wire B, \( r_B = 4.0 \, \text{mm} = 4 \times 10^{-3} \, \text{m} \)
Resistance of wire B, \( R_B = 2 \, \Omega \)
We must find \( R_A \).

Step 2: Recall the resistance formula
Resistance of a wire is given by: \[ R = \rho \frac{L}{A} \] where \(L\) = length of the wire, \(A = \pi r^2\) = cross-sectional area.

Step 3: Use the condition of equal mass (or volume)
For wires of the same material and same mass: \[ \text{Volume} = A \times L = \text{constant}. \] Hence, for wires A and B: \[ A_A L_A = A_B L_B \quad \Rightarrow \quad L_A = L_B \frac{A_B}{A_A} = L_B \left( \frac{r_B^2}{r_A^2} \right). \] So, the thinner wire (smaller radius) is longer.

Step 4: Express ratio of resistances
Since \( R = \rho \frac{L}{A} \), we can write: \[ \frac{R_A}{R_B} = \frac{L_A / A_A}{L_B / A_B} = \frac{L_A A_B}{L_B A_A}. \] Substitute \( L_A = L_B \frac{A_B}{A_A} \): \[ \frac{R_A}{R_B} = \frac{A_B^2}{A_A^2}. \] Now, \( A \propto r^2 \), so: \[ \frac{R_A}{R_B} = \left( \frac{r_B}{r_A} \right)^4. \]

Step 5: Substitute numerical values
\[ \frac{r_B}{r_A} = \frac{4}{2} = 2 \quad \Rightarrow \quad \left( \frac{r_B}{r_A} \right)^4 = 2^4 = 16. \] Thus: \[ R_A = R_B \times 16 = 2 \times 16 = 32 \, \Omega. \]

Step 6: Interpretation
Since both wires have equal mass, the thinner one (smaller radius) must be much longer to have the same volume. The longer and thinner shape increases its resistance dramatically, by a factor of \(16\).

Final answer

32