Statement-1 in a diode AM detector `R=1kOmega` and `C=10pF` circuit is good enough to detect a carrier signal of 100 kHz.
Statement-2: the condition `(1)/(f_(c))ltltRC`

Options

A. Statement-1 is true, statement-2 is true, statement-2 is correct explanation for statement-1
B. statement-1 is true, statement-2 is true, statement-2 is not a correct explanation for statement-1
C. statement-1 is true,statement-2 is false
D. statement-1 is false, statement-2 is true.

Answer

To solve the problem, we need to analyze both statements regarding the diode AM detector circuit.

### Step 1: Calculate the RC time constant
The time constant \( RC \) is calculated using the formula:
\[ RC = R \cdot C \]
Given:
- \( R = 1 \, k\Omega = 1 \times 10^3 \, \Omega \)
- \( C = 10 \, pF = 10 \times 10^{-12} \, F \)

Substituting the values:
\[ RC = (1 \times 10^3) \cdot (10 \times 10^{-12}) = 10 \times 10^{-9} = 10^{-8} \, seconds \]

### Step 2: Calculate the reciprocal of the carrier frequency
The carrier frequency \( f_c \) is given as \( 100 \, kHz \). The reciprocal of the frequency is calculated as:
\[ \frac{1}{f_c} = \frac{1}{100 \times 10^3} = 10^{-5} \, seconds \]

### Step 3: Compare \( \frac{1}{f_c} \) and \( RC \)
Now we need to compare \( \frac{1}{f_c} \) and \( RC \):
- \( \frac{1}{f_c} = 10^{-5} \, seconds \)
- \( RC = 10^{-8} \, seconds \)

We can see that:
\[ 10^{-5} \gg 10^{-8} \]
This means:
\[ \frac{1}{f_c} \gg RC \]

### Step 4: Analyze the statements
- **Statement 1**: The circuit is good enough to detect a carrier signal of \( 100 \, kHz \). This is true since the condition \( \frac{1}{f_c} \gg RC \) is satisfied.
- **Statement 2**: The condition \( \frac{1}{f_c} \ll RC \) is incorrect based on our calculations.

### Conclusion
- **Statement 1** is **True**.
- **Statement 2** is **False**.

Thus, the answer is that Statement 1 is correct and Statement 2 is incorrect.