Two wires have their lengths, diameters and resistivities, all in the ratio of 1: 2. If the resistance of the thinner wire is 10 ohms, the resistance of the thicker wire is

The resistance of a wire is determined by the formula: Resistance equals Resistivity multiplied by Length divided by Area. The Area is proportional to the square of the Diameter.

Let the properties of the thinner wire be Length (L1), Diameter (D1), and Resistivity (R1). Let the properties of the thicker wire be Length (L2), Diameter (D2), and Resistivity (R2).

The ratio is given as 1 to 2. Therefore: L2 is 2 times L1. D2 is 2 times D1 (making the Area 4 times larger because Area depends on the square of the diameter). R2 is 2 times R1.

Using the formula for the thicker wire: Resistance equals (2 times Resistivity) multiplied by (2 times Length) divided by (4 times Area). This simplifies to (4 divided by 4) times the original resistance.

The calculation shows that the resistance remains the same if the ratio applied to length and resistivity is balanced by the square of the diameter ratio. However, in this specific problem, let us look at the values: Resistance of thick wire = (Resistivity 1 x 2) x (Length 1 x 2) / (Area 1 x 4). The factors 2 x 2 in the numerator and 4 in the denominator cancel each other out (4/4 = 1).

Therefore, the resistance of the thicker wire is the same as the thinner wire, which is 10 ohms.