Q.A metal rod of length $10\ \text{cm}$ and a rectangular cross-section of $1\ \text{cm} \times \tfrac{1}{2}\ \text{cm}$ is connected to a battery across opposite faces. The resistance will be

Concept: Resistance and geometry, $R = \rho\dfrac{L}{A}$. The cuboid $10\ \text{cm}\times1\ \text{cm}\times\tfrac12\ \text{cm}$ has three pairs of opposite faces, each giving a different $L/A$:

• Across the $1\ \text{cm}\times\tfrac12\ \text{cm}$ end faces: $L = 10\ \text{cm}$, $A = 0.5\ \text{cm}^2$ → $R = \rho\dfrac{0.10}{0.5\times10^{-4}} = 2000\,\rho$.
• Across the $10\ \text{cm}\times\tfrac12\ \text{cm}$ faces: $L = 1\ \text{cm}$, $A = 5\ \text{cm}^2$ → $R = 20\,\rho$.
• Across the $10\ \text{cm}\times1\ \text{cm}$ faces: $L = \tfrac12\ \text{cm}$, $A = 10\ \text{cm}^2$ → $R = 5\,\rho$.

The three values are in the ratio $2000:20:5 = 400:4:1$. Resistance is largest when the current runs the long way through the smallest faces (greatest $L$, smallest $A$).

Principle

For a uniform conductor,

where $L$ is the length along the current (between the connected faces) and $A$ is the area of those faces. A cuboid has three distinct pairs of opposite faces, so three possible $(L,A)$ combinations and three resistances.

The three orientations

Sides: $10\ \text{cm} = 0.10\ \text{m}$, $1\ \text{cm} = 0.01\ \text{m}$, $\tfrac12\ \text{cm} = 0.005\ \text{m}$.

1. Across the $1\ \text{cm}\times\tfrac12\ \text{cm}$ end faces — longest path, smallest area:

2. Across the $10\ \text{cm}\times\tfrac12\ \text{cm}$ faces:

3. Across the $10\ \text{cm}\times1\ \text{cm}$ faces — shortest path, largest area:

Comparison

Resistance is greatest when the current is forced through the greatest length and smallest cross-section — the $1\ \text{cm}\times\tfrac12\ \text{cm}$ end faces — and least across the broad $10\ \text{cm}\times1\ \text{cm}$ faces.

Method: Resistance Dependence on Geometry — Choosing the Current-Carrying Faces

For a conductor of given shape, this method finds how resistance changes depending on which pair of faces the current is driven across — apply it whenever a rod, block, or wire is offered in more than one possible orientation between the terminals.

Steps

Step 1: Write the resistance formula in terms of length and area

For a uniform conductor of resistivity $\rho$, length $L$ (measured ALONG the direction of current flow) and cross-sectional area $A$ (the area of the face the current enters/exits through, PERPENDICULAR to the flow):

Step 2: Identify every distinct pair of current-entry faces

For a cuboid (or any solid with more than one pair of parallel faces), each pair of opposite faces defines a different possible current path: the length $L$ for that orientation is the dimension connecting the two chosen faces, and $A$ is the area of one of those faces (the product of the other two dimensions).

Step 3: Compute $R$ for each orientation

Substitute the correct $(L, A)$ pair for each orientation into Step 1's formula, keeping units consistent (convert every length to the same unit, typically metres, before multiplying or dividing).

Step 4: Compare using ratios, not absolute values

Since $\rho$ is the same in every orientation (same material), it cancels out of any comparison: the RATIO of resistances between two orientations is simply the ratio of their respective $L/A$ values. This lets you rank the orientations without needing to know $\rho$ numerically.

Step 5: Applying the geometric rule

Resistance is largest for the orientation with the GREATEST length and SMALLEST cross-section (current forced the "long way" through the "thin way"), and smallest for the orientation with the shortest length and largest cross-section. Identify these directly from the object's dimensions before computing any numbers, then confirm with the Step 3 calculation.