A metel rod of the length 10cm and a rectangular cross-section of 1 cm xx `1//2` cm is connected to a battery across opposite faces. The resistance will be

Options

A. maximum when the battery is connected across `"1 cm"xx(1//2)` cm faces
B. maximum when the battery is connected across `"10 cm"xx(1)` cm faces
C. maximum when the battery is connected across `"1= cm"xx(1//2)` cm faces
D. same irrepective of the three faces

Answer

To find the resistance of a metal rod with given dimensions and connected across different faces, we can use the formula for resistance:

\[ R = \frac{\rho L}{A} \]

where:
- \( R \) is the resistance,
- \( \rho \) is the resistivity of the material,
- \( L \) is the length of the rod,
- \( A \) is the cross-sectional area.

Let's analyze each case step by step.

### Step 1: Identify the dimensions of the rod
The rod has a length of \( L = 10 \, \text{cm} \) and a rectangular cross-section of \( 1 \, \text{cm} \times \frac{1}{2} \, \text{cm} \).

### Step 2: Calculate the cross-sectional area
The area \( A \) of the cross-section is given by:
\[ A = \text{width} \times \text{height} = 1 \, \text{cm} \times \frac{1}{2} \, \text{cm} = \frac{1}{2} \, \text{cm}^2 \]

### Step 3: Case A - Battery connected across 1 cm x ½ cm faces
- Length \( L = 10 \, \text{cm} \)
- Area \( A = \frac{1}{2} \, \text{cm}^2 \)

Using the resistance formula:
\[ R_1 = \frac{\rho L}{A} = \frac{\rho \times 10}{\frac{1}{2}} = 20\rho \, \Omega \]

### Step 4: Case B - Battery connected across 10 cm x 1 cm faces
- Length \( L = \frac{1}{2} \, \text{cm} \)
- Area \( A = 10 \, \text{cm} \times 1 \, \text{cm} = 10 \, \text{cm}^2 \)

Using the resistance formula:
\[ R_2 = \frac{\rho L}{A} = \frac{\rho \times \frac{1}{2}}{10} = \frac{\rho}{20} \, \Omega \]

### Step 5: Case C - Battery connected across 10 cm x ½ cm faces
- Length \( L = 1 \, \text{cm} \)
- Area \( A = 10 \, \text{cm} \times \frac{1}{2} \, \text{cm} = 5 \, \text{cm}^2 \)

Using the resistance formula:
\[ R_3 = \frac{\rho L}{A} = \frac{\rho \times 1}{5} = \frac{\rho}{5} \, \Omega \]

### Step 6: Compare the resistances
Now we have:
- \( R_1 = 20\rho \, \Omega \)
- \( R_2 = \frac{\rho}{20} \, \Omega \)
- \( R_3 = \frac{\rho}{5} \, \Omega \)

Among these, \( R_1 \) is the maximum resistance.

### Conclusion
The resistance will be maximum when the battery is connected across the 1 cm x ½ cm faces, giving us:

\[ \text{Maximum Resistance} = 20\rho \, \Omega \]