With t taken in seconds and I taken in amp, the variation of I follows the equation `t^(2) + I^(2) = 25`
What amount of Ag will be electrodeposited with this current flowing in the interval 0-5 second? (Ag=108)

Options

A. 22 mg
B. 66 mg
C. 77 mg
D. 88 mg

Explanation

To solve the problem of how much silver (Ag) will be electrodeposited when the current varies according to the equation \( t^2 + I^2 = 25 \) over the interval from 0 to 5 seconds, we will follow these steps:

### Step 1: Understand the relationship between current (I) and time (t)
Given the equation:
\[ t^2 + I^2 = 25 \]
We can express the current \( I \) in terms of time \( t \):
\[ I = \sqrt{25 - t^2} \]

### Step 2: Determine the charge (Q) over the time interval
The charge \( Q \) can be calculated using the formula:
\[ Q = \int_0^T I \, dt \]
where \( T = 5 \) seconds. Substituting \( I \) from Step 1:
\[ Q = \int_0^5 \sqrt{25 - t^2} \, dt \]

### Step 3: Evaluate the integral
The integral \( \int \sqrt{25 - t^2} \, dt \) represents the area of a quarter circle with radius 5. The area of a quarter circle is given by:
\[ \text{Area} = \frac{1}{4} \pi r^2 = \frac{1}{4} \pi (5^2) = \frac{25\pi}{4} \]
Thus, the charge \( Q \) over the interval from 0 to 5 seconds is:
\[ Q = \frac{25\pi}{4} \]

### Step 4: Calculate the weight of silver deposited
Using Faraday's law of electrolysis, the weight \( W \) of the substance deposited is given by:
\[ W = \frac{Q \cdot M}{F} \]
where:
- \( M \) is the molar mass of silver (Ag), which is 108 g/mol,
- \( F \) is Faraday's constant, approximately \( 96500 \, C/mol \).

Substituting \( Q \):
\[ W = \frac{\left(\frac{25\pi}{4}\right) \cdot 108}{96500} \]

### Step 5: Calculate the numerical value
Calculating \( W \):
1. Calculate \( \frac{25\pi}{4} \):
\[ \frac{25\pi}{4} \approx \frac{25 \times 3.14}{4} \approx 19.63 \, C \]
2. Substitute into the weight equation:
\[ W = \frac{19.63 \cdot 108}{96500} \approx \frac{2128.04}{96500} \approx 0.022 \, g \]
3. Convert grams to milligrams:
\[ 0.022 \, g = 22 \, mg \]

### Final Answer
The amount of silver (Ag) electrodeposited is **22 mg**.