Two wires `A` and `B` made of same material and having their lengths in the ratio `6:1` are connected in series The potential difference across the wire `3V` and `2V` respectively. If `r_A` and `r_B` are the radii of `A` and `B` respectively, then `r_B/r_A` is

Options

A. `1/4`
B. `1/2`
C. 1
D. 2

Explanation

To find the ratio of the radii of two wires A and B connected in series, we will follow these steps:

### Step 1: Understand the relationship between voltage, current, and resistance
When two resistors (or wires in this case) are connected in series, the same current flows through both. The potential difference across each wire can be expressed using Ohm's law:
\[ V = I \cdot R \]
where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance.

### Step 2: Write down the given information
We are given:
- The lengths of the wires A and B are in the ratio \( 6:1 \).
- The potential difference across wire A is \( 3V \) and across wire B is \( 2V \).

### Step 3: Set up the equations for resistance
Let \( R_A \) and \( R_B \) be the resistances of wires A and B, respectively. From Ohm's law, we can write:
\[ I \cdot R_A = 3 \quad \text{(1)} \]
\[ I \cdot R_B = 2 \quad \text{(2)} \]

### Step 4: Find the ratio of the resistances
Dividing equation (1) by equation (2):
\[ \frac{R_A}{R_B} = \frac{3}{2} \]

### Step 5: Relate resistance to dimensions
The resistance \( R \) of a wire can also be expressed as:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area. For a circular wire, the area \( A \) can be expressed in terms of the radius \( r \):
\[ A = \pi r^2 \]
Thus, we can write:
\[ R_A = \frac{\rho L_A}{\pi r_A^2} \]
\[ R_B = \frac{\rho L_B}{\pi r_B^2} \]

### Step 6: Substitute into the resistance ratio
Substituting these expressions into the resistance ratio:
\[ \frac{\frac{\rho L_A}{\pi r_A^2}}{\frac{\rho L_B}{\pi r_B^2}} = \frac{3}{2} \]
The \( \rho \) and \( \pi \) cancel out:
\[ \frac{L_A r_B^2}{L_B r_A^2} = \frac{3}{2} \]

### Step 7: Substitute the length ratio
Since the lengths are in the ratio \( L_A : L_B = 6 : 1 \), we can write:
\[ L_A = 6L_B \]
Substituting this into the equation gives:
\[ \frac{6L_B r_B^2}{L_B r_A^2} = \frac{3}{2} \]
This simplifies to:
\[ \frac{6 r_B^2}{r_A^2} = \frac{3}{2} \]

### Step 8: Solve for the ratio of the radii
Cross-multiplying gives:
\[ 12 r_B^2 = 3 r_A^2 \]
Dividing both sides by 3:
\[ 4 r_B^2 = r_A^2 \]
Taking the square root:
\[ \frac{r_B}{r_A} = \frac{1}{2} \]

### Final Answer
Thus, the ratio of the radii \( \frac{r_B}{r_A} \) is \( \frac{1}{2} \).

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