Problem 31 - A Complete Resource Book in Physics for JEE Main

Question

A wire \(l=8 \mathrm{~m}\) long of uniform cross-sectional area \(A=8 \mathrm{~mm}^{2}\) has a conductance of \(G=2.45 \Omega^{-1}\). The resistivity of material of the wire will be (A) \(2.1 \times 10^{-7} \Omega \mathrm{m}\) (B) \(3.1 \times 10^{-7} \Omega \mathrm{m}\) (C) \(4.1 \times 10^{-7} \Omega \mathrm{m}\) (D) \(5.1 \times 10^{-7} \Omega \mathrm{m}\)

Step-by-Step Solution

Answer

The resistivity of the material of the wire is approximately \(4.1 \times 10^{-7}~Ωm\). The correct answer is (C) \(4.1 \times 10^{-7} \Omega \mathrm{m}\).

Step 1: Determine the Resistance

First, we need to find the resistance of the wire. We know: \(G = 2.45~Ω^{-1}\) Because conductance is the reciprocal of resistance, we can find the resistance by inverting the given conductance value: \(R = \frac{1}{G}\) Calculate the resistance: \(R = \frac{1}{2.45} \approx 0.4082~Ω\)

Step 2: Use the Resistance formula to find Resistivity

Now that we have the resistance value, we can use the formula \(R = \frac{\rho L}{A}\) to solve for the resistivity. Rearrange this formula to isolate the resistivity term, \(\rho\): \(\rho = \frac{RA}{L}\) Plug in the given values for resistance, length and cross-sectional area, noting that we need to convert the cross-sectional area from square millimeters (\(mm^2\)) to square meters (\(m^2\)): \(R = 0.4082~Ω\) \(L = 8~m\) \(A = 8~mm^2 = 8 \times 10^{-6}~m^2\) Now, calculate the resistivity (\(\rho\)): \(\rho = \frac{(0.4082~Ω)(8 \times 10^{-6}~m^2)}{8~m} \approx 4.1 \times 10^{-7}~Ωm\) The resistivity of the material of the wire is approximately \(4.1 \times 10^{-7}~Ωm\). The correct answer is (C) \(4.1 \times 10^{-7} \Omega \mathrm{m}\).