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) \(32 \mathrm{~km}\)

Key concepts

Understanding Earth's Geometry

When calculating heights and distances related to Earth's features, it's crucial to remember that Earth is not flat. Instead, our planet is a sphere. This fundamental characteristic impacts how we measure distances and heights, particularly in calculations involving lines of sight, like TV tower transmissions. Consider the Earth as a giant sphere, and visualize a right-angled triangle that connects the center of the Earth (O), a point on the Earth's surface directly beneath the tower (C), and the top of the TV tower (T). The hypotenuse will be the line connecting O to T, which consists of the Earth's radius (r) plus the height of the TV tower (h). The leg connecting O to C represents the radius of Earth alone. This geometric relationship forms the basis of calculations involving sight and reach from elevated towers.

Calculation of T.V Tower Height

The height of a TV tower is critical in determining how far its transmission can reach. In this scenario, the height is a straightforward value given as 80 meters. However, this height does not work alone in calculations. Rather, it's the point from which the signal begins, creating a scenario where the tower's height is effectively adding to the Earth’s radius.

• The height contributes as an extension above the Earth, impacting the transmission's line of sight.
• Coupled with Earth's radius, it affects how the Pythagorean theorem applies in calculating the transmission distance.

Once the height is understood, it serves as a basis for further calculations, such as employing geometry or trigonometry to determine reach over the Earth’s curvature.

Transmission Distance Calculation

The key mathematical tool used to calculate the maximum distance up to which TV transmission can be received is the Pythagorean theorem. In this context, the theorem helps us find the distance from the top of the tower to where the signal hits the ground horizontally, essentially the maximum line of sight of the signal.

• This calculation uses the equation \( (OT)^2 = (r + h)^2 \).
• The term \(d^2 = (r + h)^2 - r^2\) solves for the distance \(d\), combining the height of the tower added to the Earth's radius as one side, the Earth radius as another, and the distance \(d\) as the third.

To find \(d\), substitute known values for the radius \(r = 6.4 \times 10^6\) and height \(h = 80\) meters, leading to \[ d = \sqrt{(6.4 \times 10^6 + 80)^2 - (6.4 \times 10^6)^2} \] This will result in an approximate value of \(31.99975\) km; thus, the broadcast's maximum distance is about \(32\) km, fitting within option (B) of the problem's given choices.