Two wires each of radius of cross section `r` but of different materials are connected together end to end (in series). If the densities of charge charge carries in the two wires are in the ratio `1:4`, the drift velocity of electrons in the two wires will be in the ratio:

Options

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

Explanation

To solve the problem step by step, let's denote the two wires as Wire 1 and Wire 2.

### Step 1: Understand the relationship between current, charge density, and drift velocity.
The current \( I \) flowing through a wire can be expressed as:
\[ I = n \cdot A \cdot v_d \]
where:
- \( n \) is the density of charge carriers (number of charge carriers per unit volume),
- \( A \) is the cross-sectional area of the wire,
- \( v_d \) is the drift velocity of the charge carriers.

### Step 2: Set up the ratio of charge densities.
Given that the densities of charge carriers in the two wires are in the ratio \( n_1 : n_2 = 1 : 4 \), we can express this as:
\[ n_1 = k \quad \text{and} \quad n_2 = 4k \]
for some constant \( k \).

### Step 3: Use the fact that the wires are in series.
Since the two wires are connected in series, the current \( I \) through both wires must be the same:
\[ I_1 = I_2 \]
Thus, we can write:
\[ n_1 \cdot A \cdot v_{d1} = n_2 \cdot A \cdot v_{d2} \]
Since the cross-sectional area \( A \) is the same for both wires, it cancels out from both sides:
\[ n_1 \cdot v_{d1} = n_2 \cdot v_{d2} \]

### Step 4: Substitute the values of \( n_1 \) and \( n_2 \).
Substituting \( n_1 \) and \( n_2 \) into the equation gives:
\[ k \cdot v_{d1} = 4k \cdot v_{d2} \]
Dividing both sides by \( k \) (assuming \( k \neq 0 \)):
\[ v_{d1} = 4 \cdot v_{d2} \]

### Step 5: Find the ratio of drift velocities.
Now we can express the ratio of the drift velocities:
\[ \frac{v_{d1}}{v_{d2}} = 4 \]
Thus, the ratio of the drift velocities in the two wires is:
\[ \frac{v_{d1}}{v_{d2}} = 4 : 1 \]

### Final Answer:
The drift velocities of the electrons in the two wires will be in the ratio \( 4 : 1 \).

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