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

Answer

To solve the problem, we need to find the amount of silver (Ag) that will be deposited when a current varies according to the equation \( t^2 + I^2 = 25 \) over the interval from 0 to 5 seconds.

### Step-by-Step Solution:

1. **Understand the Equation**:
The equation \( t^2 + I^2 = 25 \) describes a relationship between time \( t \) and current \( I \). We can rearrange this to express \( I \) in terms of \( t \):
\[ I = \sqrt{25 - t^2} \]

2. **Determine the Current Over Time**:
We need to calculate the total charge \( Q \) that flows through the circuit from \( t = 0 \) to \( t = 5 \) seconds. The charge can be calculated using the formula:
\[ Q = \int_{0}^{5} I(t) \, dt = \int_{0}^{5} \sqrt{25 - t^2} \, dt \]

3. **Evaluate the Integral**:
The integral \( \int \sqrt{25 - t^2} \, dt \) can be solved using trigonometric substitution. Let \( t = 5 \sin(\theta) \), then \( dt = 5 \cos(\theta) \, d\theta \). The limits change from \( t = 0 \) to \( t = 5 \) which corresponds to \( \theta = 0 \) to \( \theta = \frac{\pi}{2} \).
\[ Q = \int_{0}^{\frac{\pi}{2}} \sqrt{25 - (5 \sin(\theta))^2} \cdot 5 \cos(\theta) \, d\theta \]
Simplifying inside the square root:
\[ Q = \int_{0}^{\frac{\pi}{2}} \sqrt{25(1 - \sin^2(\theta))} \cdot 5 \cos(\theta) \, d\theta = \int_{0}^{\frac{\pi}{2}} 5 \cdot 5 \cos^2(\theta) \, d\theta = 25 \int_{0}^{\frac{\pi}{2}} \cos^2(\theta) \, d\theta \]

4. **Use the Integral of Cosine Squared**:
The integral \( \int \cos^2(\theta) \, d\theta \) can be evaluated using the identity:
\[ \int \cos^2(\theta) \, d\theta = \frac{1}{2} \left( \theta + \frac{1}{2} \sin(2\theta) \right) \]
Evaluating from \( 0 \) to \( \frac{\pi}{2} \):
\[ \int_{0}^{\frac{\pi}{2}} \cos^2(\theta) \, d\theta = \frac{\pi}{4} \]
Therefore,
\[ Q = 25 \cdot \frac{\pi}{4} = \frac{25\pi}{4} \]

5. **Calculate the Amount of Silver Deposited**:
According to Faraday's law, the amount of substance deposited is given by:
\[ m = \frac{Q \cdot M}{F} \]
where \( M \) is the molar mass of silver (Ag = 108 g/mol) and \( F \) is Faraday's constant (\( F = 96500 \, C/mol \)):
\[ m = \frac{\frac{25\pi}{4} \cdot 108}{96500} \]

6. **Final Calculation**:
Calculate \( m \):
\[ m = \frac{25 \cdot 108 \cdot \pi}{4 \cdot 96500} \]
This can be simplified to find the numerical value.