`w g` of copper is deposited in a copper voltameter when an electric current of 2 ampere is passed for 2 hours. If one ampere of electric current is passed for 4 hours in the same voltameter, copper deposited will be:

Options

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

Answer

To solve the problem, we will use Faraday's laws of electrolysis, which state that the amount of substance deposited or liberated during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.

### Step-by-Step Solution:

1. **Calculate the total charge (Q) for the first scenario:**
- Current (I) = 2 A
- Time (t) = 2 hours = 2 × 3600 seconds = 7200 seconds
- Total charge (Q) = I × t = 2 A × 7200 s = 14400 C

2. **Use Faraday's law to find the weight of copper deposited (W):**
- According to Faraday's first law of electrolysis, the weight (W) of the substance deposited is given by:
\[ W = \frac{Q \cdot M}{F} \]
where:
- \( M \) = molar mass of copper (approximately 63.5 g/mol)
- \( F \) = Faraday's constant (approximately 96500 C/mol)
- For the first case, we can denote the weight of copper deposited as \( W_1 \).

3. **Calculate the total charge (Q) for the second scenario:**
- Current (I) = 1 A
- Time (t) = 4 hours = 4 × 3600 seconds = 14400 seconds
- Total charge (Q) = I × t = 1 A × 14400 s = 14400 C

4. **Use Faraday's law to find the weight of copper deposited in the second scenario (W2):**
- For the second case, we can denote the weight of copper deposited as \( W_2 \).
- Since the total charge (Q) in both scenarios is the same (14400 C), we can conclude:
\[ W_2 = W_1 \]

5. **Conclusion:**
- The weight of copper deposited in the second case will be the same as in the first case, which means \( W_2 = W_1 \).

### Final Answer:
The weight of copper deposited when 1 ampere of electric current is passed for 4 hours in the same voltameter will be the same as the weight deposited when 2 amperes are passed for 2 hours.