Question

Length of a wire of resistance \(R \Omega\) is increased to 10 times, so its resistance becomes \(1000 \Omega\), therefore \(R=\ldots .\) (The volume of the wire remains same during increase in length) (A) \(0.01 \Omega\) (B) \(0.1 \Omega\) (C) \(1 \Omega\) (D) \(10 \Omega\)

Short answer

The initial resistance of the wire is \(10 \ \Omega\), which matches the answer choice (D).

Step-by-step solution

Understand the resistance formula

The resistance (R) of a wire is given by the formula: \[R = \rho \frac{L}{A}\] Where: - \(R\) = resistance (in ohms, Ω) - \(\rho\) = resistivity of the wire material (a constant) - L = length of the wire (in meters, m) - A = cross-sectional area of the wire (in square meters, m²) Since the volume remains constant, we will also use the relationship between length, cross-sectional area, and volume to solve the problem.

Relate volume and cross-sectional area

The volume of a wire is given by the formula: \[V = A \times L\] Since the problem states that the volume remains constant, we can represent the initial volume (V₁) and the final volume (V₂) as follows: \[V_1 = A_1 \times L_1\] and \[V_2 = A_2 \times L_2\] The lengths of the wires are related by the given condition, that is \(L_2 = 10 \times L_1\). Thus, we can write: \[V_1 = A_1 \times L_1\] \[V_2 = A_2 \times (10 \times L_1)\]

Use the resistance formula for both cases and create an equation

Now, we can apply the resistance formula for the initial and final cases: \[R_1 = \rho \frac{L_1}{A_1}\] \[1000 = \rho \frac{10 \times L_1}{A_2}\] We are looking for the initial resistance R₁. Since the resistivity \(\rho\) is constant, we can create an equation involving the areas and lengths of the wire by dividing the second equation by the first equation: \[\frac{1000}{R_1} = \frac{10 \times L_1}{L_1} \times \frac{A_1}{A_2}\]

Use the volume relationship to solve for R₁

Since we know that the volume remains constant: \[V_1 = A_1 \times L_1 = A_2 \times (10 \times L_1)\] \[\frac{A_1}{A_2} = \frac{10 \times L_1}{L_1}\] Now we substitute this relationship into the equation: \[\frac{1000}{R_1} = 10 \times \frac{10 \times L_1}{L_1}\] \[\frac{1000}{R_1} = 100\]

Solve for R₁ and match with given answers

We can now solve for R₁ by dividing both sides by 100: \[R_1 = \frac{1000}{100}\] \[R_1 = 10 \ \Omega\] Thus, the initial resistance of the wire is 10 Ω, which matches the answer choice (D).

Key concepts

Resistivity

When we talk about resistivity, we're dealing with a fundamental property of a material that affects how it opposes current flow. Every material has its own resistivity, denoted by the symbol \( \rho \) (rho). Lower resistivity means better conduction, whereas higher resistivity means the material is more resistant to electric flow. This property is crucial in determining the resistance of a wire when combined with other factors like length and area. Remember, resistivity is a constant specific to the material type and remains unchanged even if the wire's dimensions change.

Cross-sectional Area

The cross-sectional area of a wire significantly influences its resistance. The more area there is for electrons to flow through, the less resistance they encounter, similar to how a wide highway allows more cars to travel smoothly. Mathematically, if the cross-sectional area (\( A \)) increases, the resistance decreases, assuming other factors remain constant. Hence, changes in this area due to stretching or compressing the wire alter the resistance as described by the formula \( R = \rho \frac{L}{A} \). Keeping the area large is favorable for reducing resistance in conductive paths.

Length and Volume Relationship

An interesting aspect of wire manipulation involves understanding how its length and cross-sectional area relate to volume. Volume \(V\) is calculated using the formula \( V = A \times L \), where \(A\) is the area and \(L\) is the length. If the length of the wire increases but the volume remains constant, the cross-sectional area must decrease proportionately.

• Initial: \(V_1 = A_1 \times L_1\)
• Final: \(V_2 = A_2 \times L_2\).

In the exercise, when length becomes 10 times the original, the area must be reduced by a factor of 10. Understanding this relationship helps us determine the changes in resistance during the exercise.

Resistance Calculation

Calculating resistance involves understanding how length, area, and resistivity interplay. The resistance formula \( R = \rho \frac{L}{A} \) links these parameters. When length increases, resistance naturally increases because electrons have more material to traverse. On the other hand, if cross-sectional area decreases (because of length increase and constant volume), this further ups the resistance.
For our specific problem, the resistance increased to 1000 ohms when the length was 10x the original, and the calculation involved balancing these changes using the given formulae.

Electrical Resistivity Formula

The electrical resistivity formula is a critical tool for calculating how a wire's size and material affect its resistance. Given by \( R = \rho \frac{L}{A} \), this formula simplifies to determining the effects of geometry and material on a wire's opposition to current. Resistivity itself is independent of the wire's dimensions, being an intrinsic property of the material. It helps us quantify how changes in length \(L\) and cross-sectional area \(A\) impact resistance, all while maintaining constant resistivity \(\rho\). This formula underpins the reasoning behind why the original resistance \(R_1\) was solved to be 10 ohms, showing the decisive role each component plays in the final outcome.