If the height of TV tower is increased by `21%`, then the transmission range is enhanced by

Options

A. `10%`
B. `5%`
C. `15%`
D. `25%`

Answer

To solve the problem of how much the transmission range is enhanced when the height of a TV tower is increased by 21%, we can follow these steps:

### Step 1: Understand the relationship between height and transmission range
The transmission range \( R \) of a TV tower is given by the formula:
\[ R \propto \sqrt{H} \]
where \( H \) is the height of the tower.

### Step 2: Define the initial height
Let the initial height of the tower be \( H \).

### Step 3: Calculate the new height after the increase
If the height is increased by 21%, the new height \( H' \) can be calculated as:
\[ H' = H + 0.21H = 1.21H \]

### Step 4: Determine the initial and new transmission ranges
Using the proportionality established in Step 1, the initial transmission range \( R \) is:
\[ R = k \sqrt{H} \]
where \( k \) is a constant of proportionality.

The new transmission range \( R' \) with the increased height \( H' \) is:
\[ R' = k \sqrt{H'} = k \sqrt{1.21H} \]

### Step 5: Simplify the new transmission range
Now substituting \( H' \):
\[ R' = k \sqrt{1.21} \sqrt{H} \]
Using the fact that \( \sqrt{1.21} = 1.1 \):
\[ R' = 1.1 k \sqrt{H} = 1.1 R \]

### Step 6: Calculate the percentage increase in transmission range
To find the percentage increase in the transmission range, we calculate:
\[ \text{Percentage Increase} = \left( \frac{R' - R}{R} \right) \times 100 \]
Substituting \( R' = 1.1 R \):
\[ \text{Percentage Increase} = \left( \frac{1.1R - R}{R} \right) \times 100 = \left( \frac{0.1R}{R} \right) \times 100 = 10\% \]

### Final Answer
Thus, the transmission range is enhanced by **10%**.

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