Question

The image contains a hand-drawn circuit diagram and a question. Here is a detailed transcription: Text: At the top left, slightly above the diagram: "A Area" with an arrow pointing to a cylindrical shape representing the cross-sectional area. Inside the diagram, near the top left corner: "Ja". Above the cuboid: "b". Below the cuboid: "Vo . 5amp". Below the voltage source and current indication: "find resistivity of Cuboid".

Diagram Description: The diagram shows a cuboid shape connected to a circuit. The cuboid has a rectangular base with length labeled as 'b' and a height labeled as 'a'. The current is shown flowing horizontally through the cuboid along its length 'b'. The circuit includes a voltage source represented by the symbol for a DC voltage source (a long line and a short parallel line) labeled as 'Vo'. An arrow indicates the direction of current flow, labeled as "5amp", originating from the positive terminal of the voltage source and passing through the cuboid. The current enters the cuboid from the left face (with area a x width, where width is perpendicular to the drawing plane) and exits from the right face. The top of the cuboid is also connected by lines to what appears to be the rest of the circuit (though it's not fully drawn). Above the cuboid, to the right, there is a small sketch showing the cross-sectional area 'A' through which the current flows. This area is perpendicular to the direction of current and would be equal to the height 'a' multiplied by the width of the cuboid (which is not explicitly labeled but can be inferred).

Answer 1

$\rho=\frac{V_o\,A}{5b}$

Answer 2

1. The resistance of the cuboid is given by

   $$R=\frac{V_o}{I}=\frac{V_o}{5}.$$

2. Also, for a uniform material,

   $$R=\rho\,\frac{b}{A},$$

   where $b$ is the length along which the current flows and $A$ is the cross‐sectional area.

3. Equate the two expressions for $R$:

   $$\frac{V_o}{5}=\rho\,\frac{b}{A}.$$

4. Solving for $\rho$:

   $$\rho=\frac{V_o}{5}\,\frac{A}{b}.$$