Question

A T.V tower has a height of \(80 \mathrm{~m}\). The maximum distance up to which T.V transmission can be received is equal to (radius of earth $=6.4 \times 10^{6} \mathrm{~m}$ ) (A) \(16 \mathrm{~km}\) (B) \(32 \mathrm{~km}\) (C) \(80 \mathrm{~km}\) (D) \(160 \mathrm{~km}\)

Short answer

The maximum distance up to which T.V transmission can be received is approximately \(32 \mathrm{~km}\) (Option B).

Step-by-step solution

Understanding the Earth's geometry in relation to the T.V tower

In this problem, we will use the Earth's spherical geometry and the height of the T.V tower to find the maximum distance up to which T.V transmission can be received. We can envision this situation as a right triangle, with the Earth's center (O) and the top of the T.V tower (T) connected by a hypotenuse (OT), the Earth's surface as one side of the triangle, and a line segment connecting T to the Earth's surface as the other side. The Earth's radius (r) will be one side of the triangle, and the height of the T.V tower (h) will be added to the Earth's radius.

Apply the Pythagorean theorem

We will use the Pythagorean theorem to find the OT length, which is the Earth's radius plus the height of the tower (r+h). Let's denote the distance up to which T.V transmission can be received as d. Then, we have: \( (OT)^2 = (r + h)^2 \) \( (r + h)^2 = r^2 + d^2 \) Where OT is the Earth's radius plus the tower's height, r is the Earth's radius, and d is the maximum distance up to which T.V transmission can be received.

Calculate the maximum distance

Rearrange the formula obtained in Step 2 to solve for the maximum distance up to which T.V transmission can be received (d). \( (r + h)^2 = r^2 + d^2 \) \( d^2 = (r + h)^2 - r^2 \) Now, plug in the given values for r and h and solve for d. \( d^2 = (6.4 \times 10^{6} + 80)^2 - (6.4 \times 10^{6})^2 \) Solve for d: \( d = \sqrt{(6.4 \times 10^{6} + 80)^2 - (6.4 \times 10^{6})^2} \) \( d \approx 31999.75 \mathrm{~m} \) Since we want the answer in kilometers, we divide by 1000: \( d \approx 31.99975 \mathrm{~km} \)

Compare the result with the given options

Now, we need to compare our result with the given options: (A) \(16 \mathrm{~km}\) (B) \(32 \mathrm{~km}\) (C) \(80 \mathrm{~km}\) (D) \(160 \mathrm{~km}\) Our result of approximately \(32 \mathrm{~km}\) corresponds with option (B). So, the correct answer is: (B) \(32 \mathrm{~km}\)