The density of eq. wt of a metal are 10.5 g `cm^(-3)` and 100, respectively. The time required for a current of 3 amp to deposit a 0.005 mm thick layer of the same metal on an area of 80 `cm^(2)` is closest to

Options

A. 120s
B. 135s
C. 67.5s
D. 270s

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the Volume of the Metal Layer
The volume (V) of the metal layer can be calculated using the formula:
\[ V = \text{Area} \times \text{Thickness} \]

Given:
- Area = 80 cm²
- Thickness = 0.005 mm = 0.0005 cm (since 1 mm = 0.1 cm)

Now substituting the values:
\[ V = 80 \, \text{cm}^2 \times 0.0005 \, \text{cm} = 0.04 \, \text{cm}^3 \]

### Step 2: Calculate the Mass of the Metal Layer
Using the density (D) to find the mass (m):
\[ m = D \times V \]

Given:
- Density = 10.5 g/cm³

Now substituting the values:
\[ m = 10.5 \, \text{g/cm}^3 \times 0.04 \, \text{cm}^3 = 0.42 \, \text{g} \]

### Step 3: Use Faraday's Law to Find Time
According to Faraday's law, the relationship between mass (m), equivalent weight (Z), current (I), and time (t) is given by:
\[ m = \frac{Z \cdot I \cdot t}{96500} \]

Where:
- \( Z \) (equivalent weight) = 100 g/equiv
- \( I \) (current) = 3 A

Rearranging the formula to solve for time (t):
\[ t = \frac{m \cdot 96500}{Z \cdot I} \]

Substituting the known values:
\[ t = \frac{0.42 \, \text{g} \times 96500}{100 \, \text{g/equiv} \times 3 \, \text{A}} \]

### Step 4: Calculate the Time
Now performing the calculation:
\[ t = \frac{0.42 \times 96500}{100 \times 3} \]
\[ t = \frac{40530}{300} \]
\[ t = 135.1 \, \text{seconds} \]

### Final Answer
The time required for the current to deposit the metal layer is approximately **135 seconds**.

---