An electro magnetic wave of wavelength 300 metre can be transmitted by a transmission centre. A condenser of capacity `2.5muF` is available. Calculate the inductance of the required coil for a resonant circuit.Use `pi^(2)=10`.

Answer

To find the inductance of the required coil for a resonant circuit, we will follow these steps:

### Step 1: Calculate the frequency of the electromagnetic wave
The frequency \( f \) of the electromagnetic wave can be calculated using the formula:
\[ f = \frac{c}{\lambda} \]
where:
- \( c \) is the speed of light, approximately \( 3 \times 10^8 \, \text{m/s} \)
- \( \lambda \) is the wavelength, given as \( 300 \, \text{m} \)

Substituting the values:
\[ f = \frac{3 \times 10^8}{300} = 10^6 \, \text{Hz} \]

### Step 2: Use the resonant frequency formula
For a resonant circuit, the frequency is also given by:
\[ f = \frac{1}{2\pi \sqrt{LC}} \]
Squaring both sides gives:
\[ f^2 = \frac{1}{4\pi^2 LC} \]

### Step 3: Rearrange to find inductance \( L \)
Rearranging the equation to solve for \( L \):
\[ L = \frac{1}{4\pi^2 f^2 C} \]

### Step 4: Substitute the values
We know:
- \( f = 10^6 \, \text{Hz} \)
- \( C = 2.5 \, \mu F = 2.5 \times 10^{-6} \, F \)
- Given \( \pi^2 = 10 \)

Substituting these values into the equation for \( L \):
\[ L = \frac{1}{4 \times 10 \times (10^6)^2 \times (2.5 \times 10^{-6})} \]

### Step 5: Calculate \( L \)
Calculating the denominator:
\[ 4 \times 10 \times (10^{12}) \times (2.5 \times 10^{-6}) = 100 \times 10^{6} = 10^8 \]

Thus:
\[ L = \frac{1}{10^8} = 10^{-8} \, H \]

### Final Answer
The inductance \( L \) required for the resonant circuit is:
\[ L = 10^{-8} \, H \]

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