Two wires X,Y have the same resistivity, but their cross-sectional areas are in the ratio `2 : 3` and lengths in the ratio `1 : 2`. They are first connected in series and then in parallel to a.d.c. source. Find out the ratio of drift speeds of the electrons in the two wires for the two cases.

Explanation

To solve the problem, we need to find the ratio of drift speeds of electrons in two wires X and Y when they are connected in series and in parallel.

### Given Data:
- Resistivity of both wires (ρ) is the same.
- Cross-sectional areas (A) are in the ratio: \( A_X : A_Y = 2 : 3 \)
- Lengths (L) are in the ratio: \( L_X : L_Y = 1 : 2 \)

### Step 1: Determine the resistance of each wire
The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
For wire X:
\[ R_X = \frac{\rho L_X}{A_X} \]
For wire Y:
\[ R_Y = \frac{\rho L_Y}{A_Y} \]

### Step 2: Substitute the ratios into the resistance formulas
Using the given ratios:
- Let \( L_X = L \) and \( L_Y = 2L \)
- Let \( A_X = 2A \) and \( A_Y = 3A \)

Now substituting these into the resistance formulas:
\[ R_X = \frac{\rho L}{2A} \]
\[ R_Y = \frac{\rho (2L)}{3A} = \frac{2\rho L}{3A} \]

### Step 3: Calculate the total resistance in series
When connected in series, the total resistance \( R_{series} \) is:
\[ R_{series} = R_X + R_Y = \frac{\rho L}{2A} + \frac{2\rho L}{3A} \]
To add these fractions, find a common denominator:
\[ R_{series} = \frac{3\rho L}{6A} + \frac{4\rho L}{6A} = \frac{7\rho L}{6A} \]

### Step 4: Find the drift speed in series
The drift speed \( v_d \) is given by:
\[ v_d = \frac{I}{nAe} \]
In series, the current \( I \) is the same for both wires. Thus, the drift speeds will be proportional to:
\[ v_{dX} \propto \frac{1}{A_X} \quad \text{and} \quad v_{dY} \propto \frac{1}{A_Y} \]
Therefore, the ratio of drift speeds in series is:
\[ \frac{v_{dX}}{v_{dY}} = \frac{A_Y}{A_X} = \frac{3}{2} \]

### Step 5: Calculate the total resistance in parallel
When connected in parallel, the total resistance \( R_{parallel} \) is given by:
\[ \frac{1}{R_{parallel}} = \frac{1}{R_X} + \frac{1}{R_Y} \]
Substituting the values:
\[ \frac{1}{R_{parallel}} = \frac{2A}{\rho L} + \frac{3A}{2\rho L} = \frac{4A}{2\rho L} + \frac{3A}{2\rho L} = \frac{7A}{2\rho L} \]
Thus,
\[ R_{parallel} = \frac{2\rho L}{7A} \]

### Step 6: Find the drift speed in parallel
In parallel, the voltage across both wires is the same. The drift speeds will be proportional to:
\[ v_{dX} \propto \frac{V}{nA_X R_X} \quad \text{and} \quad v_{dY} \propto \frac{V}{nA_Y R_Y} \]
Thus, the ratio of drift speeds in parallel is:
\[ \frac{v_{dX}}{v_{dY}} = \frac{R_Y A_Y}{R_X A_X} \]
Substituting the resistances and areas:
\[ \frac{v_{dX}}{v_{dY}} = \frac{\left(\frac{2\rho L}{3A}\right) \cdot (3A)}{\left(\frac{\rho L}{2A}\right) \cdot (2A)} = \frac{2\rho L}{3A} \cdot \frac{3A}{\rho L} = 2 \]

### Final Ratios:
- In series: \( \frac{v_{dX}}{v_{dY}} = \frac{3}{2} \)
- In parallel: \( \frac{v_{dX}}{v_{dY}} = 2 \)

### Summary:
- The ratio of drift speeds in series is \( \frac{3}{2} \).
- The ratio of drift speeds in parallel is \( 2 \).