Question

If the conductivity of the thread results from the aqueous coating only, how does the cross-sectional area A of the coating compare when the thread is 13 mm long versus the starting length of 5 mm? Assume that the resistivity of the coating remains constant and the coating is uniform along the thread. ${\mathbf{A}}_{\mathbf{13}\mathbf{mm}}$is about $\left(a\right)=\frac{\mathbf{1}}{\mathbf{10}}{A}_5;\left(b\right)=\frac{\mathbf{1}}{\mathbf{4}}{A}_5;\left(c\right)=\frac{\mathbf{2}}{\mathbf{5}}{A}_5;$(d) the same as ${\mathbf{A}}_{\mathbf{5}\mathbf{mm}}$.

Short answer

Option b is correct

Step-by-step solution

Resistivity

The resistance offered by a material of length unity and also the unit cross-sectional area is called Resistivity

$\rho =R\frac{A}{I}$

Solution

Using the resistivity formula;

$\rho =R\frac{A}{I}$

Thus,

The required area at the length L=13mm;

$$\begin{gathered} A_{13mm}=\frac{\rho L_{13mm}}{R_{13mm}}=\frac{\left(13\times {10}^{-3}m\right)\rho }{102\times {10}^9\Omega } \\ =0.13\times {10}^{-12}\rho \end{gathered}$$

The required area at the length L=5mm

$$\begin{gathered} A_{5mm}=\frac{\rho L_{5mm}}{R_{5mm}}=\frac{\left(5\times {10}^{-3}m\right)\rho }{9\times {10}^9\Omega } \\ =0.55\times {10}^{-12}\rho \end{gathered}$$

On comparing both the results;

$$\begin{gathered} \frac{A_{13mm}}{A_{5mm}}=\frac{0.13\times {10}^{-12}\rho }{0.55\times {10}^{-12}\rho } \\ \frac{A_{13mm}}{A_{5mm}}=\frac{1}{4} \\ {A}_{13mm}=\frac{1}{4}{A}_{5mm} \end{gathered}$$

Hence, option b is correct.