Calculating Cross-Sectional Area Ratio in Parallel Wires

This problem involves two wires, one made of lead and the other of iron, connected in parallel. We are given information about their specific resistances, the current flowing through them, and their lengths. We need to find the ratio of their cross-sectional areas.

Let's denote the properties of the lead wire with subscript 'L' and the iron wire with subscript 'Fe'.

• Specific resistance ratio: \(\frac{\rho_L}{\rho_{Fe}} = \frac{40}{20} = 2\)
• Current relationship: The lead wire carries 80% more current than the iron wire. So, \(I_L = I_{Fe} + 0.80 I_{Fe} = 1.80 I_{Fe}\). This gives the current ratio \(\frac{I_L}{I_{Fe}} = 1.8\).
• Length relationship: The iron wire is 45% longer than the lead wire. So, \(L_{Fe} = L_L + 0.45 L_L = 1.45 L_L\). This gives the length ratio \(\frac{L_{Fe}}{L_L} = 1.45\).

Since the two wires are connected in parallel, the potential difference (voltage) across them is the same. Let this voltage be \(V\).

According to Ohm's Law, \(V = IR\), where \(I\) is the current and \(R\) is the resistance.

Thus, for the lead wire and the iron wire in parallel, we have:

\(V = I_L R_L = I_{Fe} R_{Fe}\)

The resistance of a wire is given by the formula \(R = \frac{\rho L}{A}\), where \(\rho\) is the specific resistance, \(L\) is the length, and \(A\) is the cross-sectional area.

Substituting the resistance formula into the parallel voltage equation:

\(I_L \left(\frac{\rho_L L_L}{A_L}\right) = I_{Fe} \left(\frac{\rho_{Fe} L_{Fe}}{A_{Fe}}\right)\)

We want to find the ratio of the cross-sectional areas of the latter to the former, which is \(A_{Fe}/A_L\). Let's rearrange the equation to solve for this ratio:

\(\frac{A_{Fe}}{A_L} = \frac{I_{Fe}}{I_L} \frac{\rho_{Fe}}{\rho_L} \frac{L_{Fe}}{L_L}\)

Now, substitute the ratios we found from the given information:

• \(\frac{I_{Fe}}{I_L} = \frac{1}{I_L/I_{Fe}} = \frac{1}{1.8}\)
• \(\frac{\rho_{Fe}}{\rho_L} = \frac{1}{\rho_L/\rho_{Fe}} = \frac{1}{2}\)
• \(\frac{L_{Fe}}{L_L} = 1.45\)

Plugging these values into the equation for the area ratio:

\(\frac{A_{Fe}}{A_L} = \left(\frac{1}{1.8}\right) \times \left(\frac{1}{2}\right) \times (1.45)\)

\(\frac{A_{Fe}}{A_L} = \frac{1.45}{1.8 \times 2}\)

\(\frac{A_{Fe}}{A_L} = \frac{1.45}{3.6}\)

Now, we calculate the numerical value:

\(\frac{A_{Fe}}{A_L} \approx 0.4027...\)

Rounding to one decimal place as suggested by the options, the ratio is approximately 0.4.

The ratio of their cross-sectional areas latter to former (Iron to Lead) is approximately 0.4.