Question

A metal rod of length 10cm and a rectangular cross - section of 1cm×\(\frac{1}{2}\)​cm is connected to a battery across opposite faces. The resistance will be

• A. maximum when the battery is connected across 1cm×\(\frac{1}{2}\)​ cm faces
• B. maximum when the battery is connected across 10cm×\(\frac{1}{2}\)​cm faces
• C. maximum when the battery is connected across 10cm× 1 cm faces
• D. same irrespective of the three faces

Answer 1

The Correct Option is A

The resistance \(R\) of a conductor is given by the formula:
\[ R = \rho \cdot \frac{L}{A} \] Where:
\(\rho\) is the resistivity of the material (a constant),
\(L\) is the length of the conductor,
\(A\) is the cross-sectional area through which current flows. For a rectangular cross-section, the area \(A\) is the product of the two dimensions of the cross-section.
1. When the battery is connected across the 1 cm \(\times\) \( \frac{1}{2} \) cm faces, the cross-sectional area \(A = 1 \times \frac{1}{2} = \frac{1}{2} \, \text{cm}^2\).
2. When the battery is connected across the 10 cm \(\times\) \( \frac{1}{2} \) cm faces, the cross-sectional area \(A = 10 \times \frac{1}{2} = 5 \, \text{cm}^2\).
3. When the battery is connected across the 10 cm \(\times\) 1 cm faces, the cross-sectional area \(A = 10 \times 1 = 10 \, \text{cm}^2\).
Since the resistance is inversely proportional to the cross-sectional area, the resistance will be maximum when the cross-sectional area is minimum.

Therefore, the resistance is maximum when the battery is connected across the 1 cm \(\times\) \( \frac{1}{2} \) cm faces.

Answer 2

The resistance \( R \) of a conductor is given by the formula: \[ R = \rho \frac{L}{A} \] where: - \( \rho \) is the resistivity of the material, - \( L \) is the length of the conductor, - \( A \) is the cross-sectional area of the conductor. For a rectangular cross-section, the area \( A \) is given by: \[ A = \text{length} \times \text{width} \] Let’s consider the three different ways the battery can be connected.
Case 1: Battery connected across \( 1 \, \text{cm} \times 1.5 \, \text{cm} \) faces The cross-sectional area \( A = 1 \times 1.5 = 1.5 \, \text{cm}^2 \). The resistance is: \[ R_1 = \rho \frac{10}{1.5} \]
Case 2: Battery connected across \( 10 \, \text{cm} \times 1.5 \, \text{cm} \) faces The cross-sectional area \( A = 10 \times 1.5 = 15 \, \text{cm}^2 \). The resistance is: \[ R_2 = \rho \frac{10}{15} = \rho \frac{2}{3} \]
Case 3: Battery connected across \( 10 \, \text{cm} \times 1 \, \text{cm} \) faces The cross-sectional area \( A = 10 \times 1 = 10 \, \text{cm}^2 \). The resistance is: \[ R_3 = \rho \frac{10}{10} = \rho \] Comparing the three cases, we see that the resistance is maximum when the battery is connected across the \( 1 \, \text{cm} \times 1.5 \, \text{cm} \) faces, as it has the smallest cross-sectional area. Therefore, the correct answer is \( {A} \).