The drift velocity of free electrons in a conductor of length 6 m is 0.25 `ms^(-1)` under the application of potential difference of 100 V. The mobility of free electrons is

Options

A. `2 * 10^(-3) m^2 V^(-1) s^(-1)`
B. `2.5 * 10^(-2) m^2 V^(-1) s^(-1)`
C. `1.5 * 10^(-2) m^2 V^(-1) s^(-1)`
D. `1.5 * 10^(-3) m^2 V^(-1) s^(-1)`

Answer

To find the mobility of free electrons in the given conductor, we can follow these steps:

### Step 1: Understand the given values
- Length of the conductor (L) = 6 m
- Drift velocity (V_d) = 0.25 m/s
- Potential difference (V) = 100 V

### Step 2: Calculate the Electric Field (E)
The electric field (E) in the conductor can be calculated using the formula:
\[ E = \frac{V}{L} \]
Substituting the given values:
\[ E = \frac{100 \, \text{V}}{6 \, \text{m}} = \frac{100}{6} \, \text{V/m} = \frac{50}{3} \, \text{V/m} \]

### Step 3: Use the formula for Mobility (μ)
The mobility (μ) of free electrons is defined as the ratio of drift velocity (V_d) to the electric field (E):
\[ \mu = \frac{V_d}{E} \]
Substituting the values we have:
\[ \mu = \frac{0.25 \, \text{m/s}}{\frac{50}{3} \, \text{V/m}} \]

### Step 4: Simplify the expression for Mobility
To simplify:
\[ \mu = 0.25 \, \text{m/s} \times \frac{3}{50 \, \text{V/m}} \]
\[ \mu = \frac{0.75}{50} \, \text{m}^2/\text{V}\cdot\text{s} \]
\[ \mu = 0.015 \, \text{m}^2/\text{V}\cdot\text{s} \]
\[ \mu = 1.5 \times 10^{-2} \, \text{m}^2/\text{V}\cdot\text{s} \]

### Conclusion
The mobility of free electrons is:
\[ \mu = 1.5 \times 10^{-2} \, \text{m}^2/\text{V}\cdot\text{s} \]