A current `I` is passing through a wire having two sections `P` and `O` of uniform diameters `d` and `d//2` respectively. If the mean drift velocity of electrons in section `P` and `Q` is denoted by `v_(P)` and `v_(Q)` respectively, then

Options

A. `v_(P) = v_(Q)`
B. `v_(P) = 1/2v_(Q)`
C. `v_(p) = (1/4)v_(Q)`
D. `v_(P) = 2v_(Q)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between current, drift velocity, and cross-sectional area.
The current \( I \) flowing through a conductor is given by the formula:
\[ I = n e A v_d \]
where:
- \( n \) = number of charge carriers per unit volume,
- \( e \) = charge of an electron,
- \( A \) = cross-sectional area of the wire,
- \( v_d \) = drift velocity of the charge carriers.

### Step 2: Calculate the cross-sectional areas of sections P and Q.
For section P with diameter \( d \):
\[ A_P = \pi \left(\frac{d}{2}\right)^2 = \pi \frac{d^2}{4} \]

For section Q with diameter \( \frac{d}{2} \):
\[ A_Q = \pi \left(\frac{d/2}{2}\right)^2 = \pi \left(\frac{d}{4}\right)^2 = \pi \frac{d^2}{16} \]

### Step 3: Write the expressions for current in both sections.
For section P:
\[ I = n e A_P v_P \implies I = n e \left(\pi \frac{d^2}{4}\right) v_P \]

For section Q:
\[ I = n e A_Q v_Q \implies I = n e \left(\pi \frac{d^2}{16}\right) v_Q \]

### Step 4: Set the two expressions for current equal to each other.
Since the current \( I \) is the same in both sections:
\[ n e \left(\pi \frac{d^2}{4}\right) v_P = n e \left(\pi \frac{d^2}{16}\right) v_Q \]

### Step 5: Cancel out common terms.
We can cancel \( n e \) and \( \pi d^2 \) from both sides:
\[ \frac{d^2}{4} v_P = \frac{d^2}{16} v_Q \]

### Step 6: Simplify the equation.
Dividing both sides by \( d^2 \):
\[ \frac{1}{4} v_P = \frac{1}{16} v_Q \]
Multiplying both sides by 16:
\[ 4 v_P = v_Q \]

### Step 7: Rearranging gives the relationship between \( v_P \) and \( v_Q \).
Thus, we can express \( v_P \) in terms of \( v_Q \):
\[ v_P = \frac{1}{4} v_Q \]
or
\[ \frac{v_P}{v_Q} = \frac{1}{4} \]

### Final Answer:
The relationship between the mean drift velocities is:
\[ \frac{v_P}{v_Q} = \frac{1}{4} \]

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