Question

Two cylindrical wires, 1 and 2 , made of the same material, have the same resistance. If the length of wire 2 is twice that of wire \(1,\) what is the ratio of their cross-sectional areas, \(A_{1}\) and \(A_{2} ?\) a) \(A_{1} / A_{2}=2\) c) \(A_{1} / A_{2}=0.5\) b) \(A_{1} / A_{2}=4\) d) \(A_{1} / A_{2}=0.25\)

Short answer

Answer: a) \(A_{1} / A_{2}=2\)

Step-by-step solution

Understanding the resistivity formula

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

Setting up the equation for the wires

Since the two wires have the same resistance and made of the same material, their resistivity (\(\rho\)) will also be the same. Let's denote the resistance of both wires as R, the resistivity as \(\rho\), the length of wire 1 as \(L_1\), the length of wire 2 as \(L_2\), the area of wire 1 as \(A_1\), and the area of wire 2 as \(A_2\). Given that the length of wire 2 is twice that of wire 1, we can write \(L_2 = 2L_1\). Now, we can set up the equations for the resistance of both wires: 1. \(R = \frac{\rho L_1}{A_1}\) 2. \(R = \frac{\rho L_2}{A_2}\)

Eliminate resistance and resistivity

Since the resistance and resistivity are the same for both wires, we can eliminate those terms by equating the two equations from the previous step: \(\frac{L_1}{A_1}=\frac{L_2}{A_2}\) Now, substitute \(L_2\) with \(2L_1\): \(\frac{L_1}{A_1}=\frac{2L_1}{A_2}\)

Solve for the cross-sectional area ratio

To find the ratio of cross-sectional areas, \(\frac{A_1}{A_2}\), we can cross-multiply and solve for the ratio: \(A_1 \cdot 2L_1 = A_2 \cdot L_1\) Now, divide both sides by \(L_1\): \(2A_1 = A_2\) Now, divide both sides by \(A_2\) to find the desired ratio: \(\frac{A_1}{A_2}=\frac{2}{1}\) or just 2. So, the correct answer is: a) \(A_{1} / A_{2}=2\)