Two wires of the same metal have same length, but their cross-sections are in the ratio 3:1 . They are joined in series. The resistance of thicker wire is `10 Omega`. The total resistance of the combination will be

Options

A. `2.5Omega`
B. `40//3Omega`
C. `40Omega`
D. `100Omega`

Explanation

To solve the problem step by step, we will analyze the given information and apply the formulas related to resistance in series.

### Step 1: Understand the Given Information
We have two wires made of the same metal, which means they have the same resistivity (ρ). They have the same length (L), but their cross-sectional areas (A) are in the ratio of 3:1. The resistance of the thicker wire (let's call it R1) is given as 10 Ω.

### Step 2: Define the Cross-Sectional Areas
Let the cross-sectional area of the thicker wire (R1) be A1 and the cross-sectional area of the thinner wire (R2) be A2. According to the problem, we have:
- A1 : A2 = 3 : 1
This implies:
- A1 = 3A and A2 = A (for some area A).

### Step 3: Calculate the Resistance of the Thinner Wire
The resistance of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
For the thicker wire (R1):
\[ R1 = \frac{\rho L}{A1} = \frac{\rho L}{3A} \]
For the thinner wire (R2):
\[ R2 = \frac{\rho L}{A2} = \frac{\rho L}{A} \]

### Step 4: Relate the Resistances
We can express the resistance of the thinner wire (R2) in terms of the resistance of the thicker wire (R1):
From the ratio of the areas:
\[ R1 : R2 = A2 : A1 = 1 : 3 \]
Thus, we can write:
\[ R2 = 3R1 \]
Given that R1 = 10 Ω:
\[ R2 = 3 \times 10 = 30 \, \Omega \]

### Step 5: Calculate the Total Resistance in Series
When resistors are connected in series, the total resistance (R_total) is the sum of the individual resistances:
\[ R_{total} = R1 + R2 \]
Substituting the values we found:
\[ R_{total} = 10 \, \Omega + 30 \, \Omega = 40 \, \Omega \]

### Final Answer
The total resistance of the combination of the two wires in series is **40 Ω**.

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