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

Options

A. `40Omega`
B. `(40)/(3)Omega`
C. `(5)/(2)Omega`
D. `100Omega`

Answer

To solve the problem step by step, we will follow these steps:

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

### Step 2: Define the Cross-Sectional Areas
Let the cross-sectional area of wire A (the thicker wire) be 3A and the cross-sectional area of wire B (the thinner wire) be A. Thus, we can express the areas as:
- Area of wire A (thicker) = 3A
- Area of wire B (thinner) = A

### Step 3: Calculate the Resistances
Using the formula for resistance:
\[ R = \frac{\rho L}{A} \]
Since both wires have the same length and resistivity, we can express their resistances as:
- Resistance of wire A (thicker wire):
\[ R_A = \frac{\rho L}{3A} \]
- Resistance of wire B (thinner wire):
\[ R_B = \frac{\rho L}{A} \]

### Step 4: Relate the Resistances
From the above equations, we can see that:
\[ R_B = 3R_A \]
This means that the resistance of wire B is three times that of wire A.

### Step 5: Substitute Known Values
We know that the resistance of wire A (the thicker wire) is given as 10 Ω:
\[ R_A = 10 \, \Omega \]
Thus, we can find the resistance of wire B:
\[ R_B = 3 \times R_A = 3 \times 10 \, \Omega = 30 \, \Omega \]

### Step 6: Calculate Total Resistance in Series
When resistors are connected in series, the total resistance (R_total) is the sum of the individual resistances:
\[ R_{\text{total}} = R_A + R_B \]
Substituting the values we found:
\[ R_{\text{total}} = 10 \, \Omega + 30 \, \Omega = 40 \, \Omega \]

### Final Answer
The total resistance of the combination of the two wires is:
\[ R_{\text{total}} = 40 \, \Omega \]

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