Question

A cylindrical sample of solid germanium has length 55.0 mm and diameter 5.0 mm. In a test circuit, 0.150 A of current flowed through this sample when the voltage applied between its ends was 17.5 V. What is the electrical conductivity of this sample?

Short answer

The electrical conductivity value is $240\Omega$.

Step-by-step solution

Step 1:The given data

The vessel is a cylindrical vessel. So, its surface area is$\pi r^2\left(ristheradius\right).$

The length of the solid germanium is 55.0 mm.

The diameter of cylindrical germanium rod is 5.0 mm.

The current that is flown through the germanium rod is 0.150 A.

The voltage is 17.5 V.

Step 2:The formula for electrical conductivity

The electrical conductivity is inversely related to resistivity.

The formula for electrical conductivity is as shown below:

$\begin{array}{c}\sigma =\frac{1}{\rho }\\ =\frac{I\times l}{V\times A}\\ =\frac{I\times l}{V\times \pi r^2}\\ =\frac{I\times l}{V\times \pi {\left(\frac{d}{2}\right)}^2}\end{array}$

Calculation

$\begin{array}{c}\sigma =\frac{I\times l}{V\times \pi {\left(\frac{d}{2}\right)}^2}\\ =\frac{0.150A\times 55.0\times {10}^{-3}mm}{17.5V\times \frac{22}{7}\times {\left(\frac{5\times {10}^{-3}}{2}\right)}^{2}mm}\\ =240\Omega \end{array}$