Question

The masses of three wires of copper are in the ratio of \(1: 3: 5\) and their lengths are in the ratio of \(5: 3: 1\). The ratio of their electrical resistance is: (A) \(1: 1: 1\) (B) \(1: 3: 5\) (C) \(5: 3: 1\) (D) \(125: 15: 1\)

Short answer

The ratio of the electrical resistances of the three copper wires is (D) \(125: 15: 1\).

Step-by-step solution

Remember the electrical resistance formula

For a wire, the electrical resistance (R) is given by the formula: \(R = \rho \frac{L}{A}\) Where: - R = electrical resistance - \(\rho\) = resistivity (a constant for a given material, in this case, copper) - L = length of the wire - A = cross-sectional area of the wire However, the problem gives us the mass ratios of the wires, so we need to find a way to relate the mass to the cross-sectional area.

Relate the mass to the cross-sectional area

The mass (m) of the wire can be written as the volume (V) multiplied by the density (\(\delta\)) of the material: \(m = \delta V\) The volume (V) of the wire can be written as the product of its length (L) and cross-sectional area (A): \(V = A \cdot L\) Now we can rewrite the mass formula in terms of A and L: \(m = \delta (A \cdot L)\) We can then solve for the cross-sectional area (A) in terms of the mass (m) and length (L): \(A = \frac{m}{\delta \cdot L}\)

Substitute the cross-sectional area into the resistance formula

Now we'll substitute the expression for the cross-sectional area (A) from step 2 into the resistance formula from step 1: \(R = \rho \frac{L}{\frac{m}{\delta \cdot L}}\) We can now simplify the formula: \(R = \rho \frac{L^2 \cdot \delta}{m}\)

Apply the given mass and length ratios

According to the problem, the mass ratios of the three wires are 1:3:5, and the length ratios are 5:3:1. We will assign variables to the masses (m1, m2, m3) and lengths (L1, L2, L3) of the wires: \(m1 : m2 : m3 = 1 : 3 : 5\) \(L1 : L2 : L3 = 5 : 3 : 1\) Now we can write the resistances (R1, R2, R3) of the three wires using the simplified formula from step 3: \(R1 = \rho \frac{L1^2 \cdot \delta}{m1}\) \(R2 = \rho \frac{L2^2 \cdot \delta}{m2}\) \(R3 = \rho \frac{L3^2 \cdot \delta}{m3}\)

Find the ratio of the resistances

To find the ratio of the resistances, we will form a proportion between the three resistances: \(\frac{R1}{R2} = \frac{\rho \frac{L1^2 \cdot \delta}{m1}}{\rho \frac{L2^2 \cdot \delta}{m2}}\) and \(\frac{R2}{R3} = \frac{\rho \frac{L2^2 \cdot \delta}{m2}}{\rho \frac{L3^2 \cdot \delta}{m3}}\) Simplifying and substituting the given mass and length ratios: \(\frac{R1}{R2} = \frac{(5)^2}{(3)^2} \cdot \frac{3}{1} = \frac{25}{9} \cdot 3 = \frac{75}{9}\) \(\frac{R2}{R3} = \frac{(3)^2}{(1)^2} \cdot \frac{5}{3} = \frac{9}{1} \cdot \frac{5}{3} = \frac{45}{3}\) Now we can find the ratio of all three resistances: \(R1 : R2 : R3 = \frac{75}{9} : \frac{45}{3} : 1 = 125 : 15 : 1\) Hence, the ratio of their electrical resistances is: (D) \(125: 15: 1\)

Key concepts

Resistivity

Resistivity is a fundamental property of materials that quantifies how strongly a given material opposes the flow of electrical current. It is denoted by the Greek letter \(\rho\) (rho) and is expressed in ohm-meters (Ω⋅m). Resistivity is crucial in determining how much resisting force will act against the flow of electric current in a wire, for a specific material and its dimensions.

• Low resistivity materials, such as copper, are good conductors of electricity, meaning they allow electric current to flow easily.
• High resistivity materials are insulators and they are used to prevent the loss of current from electrical circuits.

The resistivity of a wire influences its electrical resistance when combined with the wire's length and cross-sectional area. A key relationship here is the resistivity formula:\[R = \rho \frac{L}{A}\]where:

• \(R\) is the electrical resistance, measured in ohms (Ω).
• \(L\) is the length of the wire, affecting how much distance the current must travel.
• \(A\) is the cross-sectional area, influencing the ease of electron flow.

This formula indicates that the resistance increases with an increase in resistivity and length, and decreases with an increase in the cross-sectional area.

Cross-sectional Area

The cross-sectional area of a wire is a crucial geometrical parameter that affects the wire's resistance to current flow. The larger the cross-sectional area, the more space there is for current to flow, thus reducing resistance, and vice versa. To determine the cross-sectional area, one can use the following relationship derived from the volume of the wire:\[A = \frac{m}{\delta \cdot L}\]where:

• \(A\) is the cross-sectional area.
• \(m\) represents the mass of the wire.
• \(\delta\) is the material's density.
• \(L\) is the length of the wire.

Thus, for given mass and length, a higher density lowers the cross-sectional area for the same mass. This relationship shows how mass, density, and length come together to influence the wire's cross-sectional area, thereby affecting its overall resistance.

Mass Ratios

Mass ratios are used in this context to compare the relative masses of different wires. Understanding these ratios allows us to determine how the mass of each wire affects its resistance when combined with other parameters like length and density.

• The given ratio of masses \(1:3:5\) signifies that if the first wire has a mass \(m\), the second wire has a mass \(3m\), and the third has a mass \(5m\).
• These mass differences directly affect the wire's cross-sectional area, and thus the resistance.

Since resistance depends inversely on mass (through its influence on cross-sectional area), the ratio of mass influences the calculated resistance for each wire. When substituting mass ratios into our resistance calculation, it is important to consider how these ratios will interact with other factors like length ratios to determine the final resistance values.

Density

Density, denoted by \(\delta\), is defined as mass per unit volume and is a key property that helps relate the mass of a wire to its volume (and hence its cross-sectional area). It is expressed in units such as kilograms per cubic meter (kg/m³).
Density plays a crucial role in calculating electrical resistance because:

• For a constant mass, higher density means a smaller volume—and therefore, a smaller cross-sectional area.
• This smaller area increases resistance, as seen in the resistance formula \(R = \rho \frac{L^2 \cdot \delta}{m}\).

In practical terms, density helps determine how materials can be used effectively in various applications, such as creating thin but strong wires. When solving problems involving resistance and density, considering how density alters the relationship between mass and volume is essential for grasping the full picture of what affects electrical resistance.