The height of a T.V. tower is `100m`. If radius of earth is `6400km,` then what is the maximum distance of transmission from it ?

Options

A. `100km`
B. `64sqrt(10)km`
C. `6.4sqrt(10)km`
D. `8sqrt(20)km.`

Answer

To find the maximum distance of transmission from a TV tower, we can use the formula:

\[ d = \sqrt{2rh} \]

where:
- \(d\) is the maximum distance of transmission,
- \(r\) is the radius of the Earth,
- \(h\) is the height of the tower.

### Step 1: Convert the radius of the Earth to meters
The radius of the Earth is given as \(6400 \, \text{km}\). To convert this to meters, we multiply by \(1000\):

\[ r = 6400 \, \text{km} \times 1000 \, \text{m/km} = 6400000 \, \text{m} \]

### Step 2: Identify the height of the tower
The height of the TV tower is given as:

\[ h = 100 \, \text{m} \]

### Step 3: Substitute the values into the formula
Now we can substitute \(r\) and \(h\) into the formula for \(d\):

\[ d = \sqrt{2 \times 6400000 \, \text{m} \times 100 \, \text{m}} \]

### Step 4: Calculate the value inside the square root
Calculating the multiplication inside the square root:

\[ d = \sqrt{2 \times 6400000 \times 100} = \sqrt{1280000000} \]

### Step 5: Calculate the square root
Now we will calculate the square root:

\[ d = \sqrt{1280000000} = \sqrt{128 \times 10^6} = \sqrt{128} \times 10^3 \]

### Step 6: Simplify \(\sqrt{128}\)
We can simplify \(\sqrt{128}\):

\[ \sqrt{128} = \sqrt{64 \times 2} = 8\sqrt{2} \]

So, we have:

\[ d = 8\sqrt{2} \times 1000 = 8000\sqrt{2} \, \text{m} \]

### Step 7: Convert to kilometers
To convert \(d\) from meters to kilometers, we divide by \(1000\):

\[ d = 8\sqrt{2} \, \text{km} \]

### Step 8: Approximate the value
Using the approximate value of \(\sqrt{2} \approx 1.414\):

\[ d \approx 8 \times 1.414 \approx 11.312 \, \text{km} \]

### Conclusion
The maximum distance of transmission from the tower is approximately \(11.312 \, \text{km}\).