QJEEM23Phc5cb: JEE Main 2023 Physics - Class 12 JEEM23Phc5cb Solution: A rectangular parallelopiped is measured as . If its specific…

The resistance of a material can be calculated using the formula:

$R = \rho \frac{L}{A}$

where

• $R$ is the resistance,
• $\rho$ (rho) is the resistivity or specific resistance of the material,
• $L$ is the length (or distance over which the resistance is being measured), and
• $A$ is the cross-sectional area through which the current flows.

In the context of a rectangular parallelepiped, the "length" and "cross-sectional area" can vary depending on which faces of the shape are considered.

In this particular calculation, the resistance is being measured between the two smaller faces of the parallelepiped, which are squares of side length 1 cm:

The cross-sectional area $A$ should be:

$A = 100 \, \text{cm} \times 1 \, \text{cm} = 100 \, \text{cm}^2$

Converting this to meters gives:

$A = 100 \, \text{cm}^2 = 1 \, \text{m} \times 0.01 \, \text{m} = 0.01 \, \text{m}^2$

So, if we use these values for the length $L$ and cross-sectional area $A$ in the resistance formula, we get:

$R = \rho \frac{L}{A} = 3 \times 10^{-7} \Omega \, m \times \frac{0.01 \, m}{0.01 \, \text{m}^2} = 3 \times 10^{-7} \Omega$

Therefore, the resistance between the two smaller faces of the rectangular parallelepiped is $3 \times 10^{-7} \Omega$ .