Question

The resistance of two conductors of equal cross-sectional area and equal lengths are compared, and are found to be in the ratio 1:2. The resistivities of the materials from which they are constructed must therefore be in what ratio? (A) 1:1 (B) 1:2 (C) 2:1 (D) 4:1

Short answer

(B) 1:2

Step-by-step solution

- Understand the relationship

The resistance (\textbf{R}) of a conductor is given by the formula \[ R = \rho \frac{L}{A} \] where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area of the conductor.

- Set up the proportion

Since the cross-sectional area and lengths are equal, we can write the resistances for the two conductors as: \[ R_1 = \rho_1 \cdot \frac{L}{A} \] and \[ R_2 = \rho_2 \cdot \frac{L}{A} \] Given that \( \frac{R_1}{R_2} = 1:2 \), substitute the resistances to get: \[ \frac{\rho_1 \cdot \frac{L}{A}}{\rho_2 \cdot \frac{L}{A}} = 1:2 \]

- Simplify the equation

Since the lengths \( L \) and cross-sectional areas \( A \) are the same, they cancel out. Therefore, we have: \[ \frac{\rho_1}{\rho_2} = 1:2 \]

- Interpret the result

This shows that \( \rho_1 \) (the resistivity of the first conductor) is half of \( \rho_2 \) (the resistivity of the second conductor). This means the ratio of their resistivities is \( 1:2 \). Therefore, the correct answer is:

Key concepts

resistance formula

The resistance of a conductor is a crucial property that determines how much it opposes the flow of electric current. This resistance, denoted by \( R \), can be described mathematically using the formula \[ R = \rho \frac{L}{A} \]. Here, \( \rho \) stands for resistivity, which is an intrinsic material property. It quantifies how strongly a given material opposes the flow of electric current. \( L \) is the length of the conductor, and \( A \) is its cross-sectional area.

The relationship depicted in this formula shows us how these three factors—resistivity, length, and area—combine to determine the overall resistance.

Understanding this formula helps in deciphering why different materials and different shapes of conductors have varying resistances. It provides the foundation to explore further details about the resistors and their behavior in electrical circuits.

proportional relationships

Proportional relationships are a fundamental concept in many areas of physics, including the study of resistivity in conductors. When we say two quantities are proportional, we mean that as one quantity changes, the other changes in a corresponding manner. In our resistance formula, we see proportional relationships at play.

Given equal lengths and cross-sectional areas, the resistance of two conductors is proportional to their resistivities. This is showcased in the simplified version of our exercise formula: \[ \frac{R_1}{R_2} = \frac{\rho_1}{\rho_2} \].

This proportional relationship allows us to make comparisons easily. For instance, if we know the resistances and lengths of two conductors, we can directly relate their resistivities using proportions. In our exercise, we deduce that \( \rho_1 \) is half of \( \rho_2 \) because their resistance ratio was given as 1:2.

Understanding these proportional relationships aids in solving problems efficiently and provides deeper insights into the nature of conductive materials.

cross-sectional area and length

Cross-sectional area and length are two significant factors that influence the resistance of a conductor. In the formula \[ R = \rho \frac{L}{A} \], \( A \) and \( L \) play pivotal roles in determining the resistance.

**Length (\( L \)):** The longer a conductor, the greater its resistance, provided other factors remain constant. This is because the electrons have to travel a greater distance through the material, facing more opposition.

**Cross-sectional Area (\( A \)):** Conversely, a larger cross-sectional area reduces resistance, assuming other variables stay the same. A wider path allows more electrons to pass through simultaneously, making the flow of current easier.

In practical applications, knowing the interplay between these factors helps in designing efficient electrical components. For example, in electrical wiring, balancing these dimensions ensures minimal resistance without overusing materials.

In our exercise, since both conductors have equal lengths and cross-sectional areas, these variables cancel each other out in the calculations. This simplifies our problem, allowing us to focus solely on comparing resistivities.