The maximum distance upto which ` TV` transmission from a ` TV` tower of height `h` can be received is proportional to

Options

A. `h^(t//2)`
B. h
C. h
D. `h^2`

Explanation

To solve the problem of determining the maximum distance up to which TV transmission from a TV tower of height \( h \) can be received, we can use the following steps:

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- Consider the Earth as a sphere with radius \( R \).
- Let \( h \) be the height of the TV tower above the surface of the Earth.
- The maximum distance \( d \) from the top of the tower to the horizon can be visualized as a right triangle formed by the radius of the Earth, the height of the tower, and the line of sight to the horizon.

2. **Using the Pythagorean Theorem**:
- In the right triangle formed, we can express the relationship between the radius of the Earth, the height of the tower, and the distance to the horizon:
\[ (R + h)^2 = R^2 + d^2 \]
- Here, \( R + h \) is the distance from the center of the Earth to the top of the tower, and \( d \) is the distance from the tower to the point where the signal can be received.

3. **Expanding the Equation**:
- Expanding the left side:
\[ R^2 + 2Rh + h^2 = R^2 + d^2 \]

4. **Simplifying the Equation**:
- Cancel \( R^2 \) from both sides:
\[ 2Rh + h^2 = d^2 \]

5. **Neglecting \( h^2 \)**:
- Since \( h \) is much smaller than \( R \) (the height of the tower is much smaller than the radius of the Earth), we can neglect \( h^2 \):
\[ d^2 \approx 2Rh \]

6. **Finding the Distance \( d \)**:
- Taking the square root of both sides gives:
\[ d \approx \sqrt{2Rh} \]

7. **Proportionality**:
- From the equation \( d \approx \sqrt{2Rh} \), we see that the maximum distance \( d \) is proportional to the square root of the height of the tower:
\[ d \propto \sqrt{h} \]

### Conclusion:
The maximum distance up to which TV transmission from a TV tower of height \( h \) can be received is proportional to \( \sqrt{h} \).