The relation between an ac voltage source and time in SI units is `V=120 sin(100 pi t)cos (100 pi t) V`. The value of peak voltage and frequency will be respectively

Options

A. 120 V and 100 Hz
B. `(120)/(sqrt(2)) V` and 100 Hz
C. 60 and 200 Hz
D. 60 V and 100 Hz

Explanation

To solve the problem, we need to analyze the given AC voltage equation and extract the peak voltage and frequency from it.

### Step-by-Step Solution:

1. **Identify the given voltage equation**:
The voltage equation is given as:
\[ V = 120 \sin(100 \pi t) \cos(100 \pi t) \]

2. **Use the trigonometric identity**:
We can use the trigonometric identity:
\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \]
Here, let \(\theta = 100 \pi t\). Therefore, we can rewrite the equation as:
\[ V = 120 \cdot \frac{1}{2} \sin(200 \pi t) \]
This simplifies to:
\[ V = 60 \sin(200 \pi t) \]

3. **Identify the peak voltage**:
The general form of the AC voltage equation is:
\[ V = V_0 \sin(\omega t) \]
From our equation \(V = 60 \sin(200 \pi t)\), we can see that:
\[ V_0 = 60 \text{ volts} \]
Thus, the peak voltage \(V_0\) is 60 volts.

4. **Identify the angular frequency**:
From the equation \(V = 60 \sin(200 \pi t)\), we have:
\[ \omega = 200 \pi \]

5. **Calculate the frequency**:
The relationship between angular frequency \(\omega\) and frequency \(f\) is given by:
\[ \omega = 2 \pi f \]
Substituting for \(\omega\):
\[ 200 \pi = 2 \pi f \]
Dividing both sides by \(2 \pi\):
\[ f = \frac{200 \pi}{2 \pi} = 100 \text{ Hz} \]

6. **Final results**:
The peak voltage and frequency are:
- Peak Voltage: \(60\) volts
- Frequency: \(100\) Hz

### Conclusion:
The value of peak voltage and frequency will be respectively \(60\) volts and \(100\) Hz.