Radiowaves of wavelength `360` m are transmitted from a transmitter. The inductance of the coil which must be connected with capacitor of capacity `3.6 muF` in a resonant citcuit to receive these waves will be appoximately

Options

A. `10^(3)H`
B. `10^(2)H`
C. `10^(-4)H`
D. `10^(-8)H`

Explanation

To find the inductance \( L \) of the coil that must be connected with a capacitor of capacity \( 3.6 \, \mu F \) in a resonant circuit to receive radio waves of wavelength \( 360 \, m \), we can follow these steps:

### Step 1: Calculate the frequency of the radio waves
The frequency \( f \) of the radio waves can be calculated using the formula:
\[ f = \frac{c}{\lambda} \]
where:
- \( c = 3 \times 10^8 \, m/s \) (speed of light)
- \( \lambda = 360 \, m \)

Substituting the values:
\[ f = \frac{3 \times 10^8}{360} \]

### Step 2: Simplify the frequency calculation
Calculating the value:
\[ f = \frac{3 \times 10^8}{360} = 833333.33 \, Hz \approx 8.33 \times 10^5 \, Hz \]

### Step 3: Use the resonant frequency formula
The resonant frequency \( f \) is also given by the formula:
\[ f = \frac{1}{2\pi \sqrt{LC}} \]
Squaring both sides gives:
\[ f^2 = \frac{1}{4\pi^2 LC} \]

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

### Step 5: Substitute the values of \( f \) and \( C \)
We know:
- \( C = 3.6 \, \mu F = 3.6 \times 10^{-6} \, F \)
- \( f = 8.33 \times 10^5 \, Hz \)

Substituting these values into the equation for \( L \):
\[ L = \frac{1}{4\pi^2 (8.33 \times 10^5)^2 (3.6 \times 10^{-6})} \]

### Step 6: Calculate \( L \)
Calculating \( (8.33 \times 10^5)^2 \):
\[ (8.33 \times 10^5)^2 = 6.94 \times 10^{11} \]
Now substituting this back into the equation for \( L \):
\[ L = \frac{1}{4\pi^2 (6.94 \times 10^{11}) (3.6 \times 10^{-6})} \]

Calculating \( 4\pi^2 \):
\[ 4\pi^2 \approx 39.478 \]

Now substituting:
\[ L = \frac{1}{39.478 \times 6.94 \times 10^{11} \times 3.6 \times 10^{-6}} \]

Calculating the denominator:
\[ 39.478 \times 6.94 \approx 274.6 \]
\[ 274.6 \times 3.6 \times 10^{-6} \approx 9.87 \times 10^{-4} \]

Finally, calculating \( L \):
\[ L \approx \frac{1}{9.87 \times 10^{-4}} \approx 1013.1 \, H \]

Thus, the inductance \( L \) is approximately \( 1013.1 \, H \).