/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 The maximum distance upto which ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 22: Problem 8

The maximum distance upto which 'TV transmission from a TV tower of height \(h\) can be received is proportional to (a) \(h^{b}\) (b) \(h\) (c) \(h^{0}\) (d) \(h^{2}\)

Short Answer

Expert verified
The correct option is (a) with \( b = 0.5 \).

Step by step solution

01

Understanding the Problem

The problem asks about the relationship between the height of a TV tower and the maximum distance up to which TV transmission can be received. We need to determine which option correctly describes this proportional relationship.
02

Revisiting the formula for signal range

The maximum distance for TV transmission can be estimated using the line of sight formula:\[ d = \sqrt{2 h R} \]where \( d \) is the distance, \( h \) is the height of the TV tower, and \( R \) is the radius of the Earth. We need to express this in terms of a power of \( h \).
03

Simplifying the formula

Simplify the formula \( d = \sqrt{2 h R} \) to show its dependency on \( h \). Since \( 2 \) and \( R \) are constants, we focus on how \( h \) affects the formula:\[ d \propto \sqrt{h} \] This implies the maximum distance up to which TV transmission can be received is proportional to \( h^{0.5} \).
04

Matching with given options

Compare \( h^{0.5} \) with the given options:- (a) \( h^{b} \): matches if \( b = 0.5 \)- (b) \( h \)- (c) \( h^{0} \)- (d) \( h^{2} \)Option (a) is correct when \( b = 0.5 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Line of Sight Formula
The line of sight formula is essential when determining how far signals, like those from a TV tower, can travel. The core idea is that signals need to travel in a straight path from the source (TV tower) to the receiver (TV antennas), which means they are primarily limited by the line of sight. This is particularly relevant when considering the Earth's curvature, as it limits how far a signal can travel directly without obstruction.

Understanding this formula relies on the basic geometric and physical principle that if the receiver is below the horizon line from the transmitter, the signal can no longer be received. Knowing this principle helps engineers estimate where and how to place transmission towers for optimal coverage.
  • Direct, unobstructed path based on Earth's curvature.
  • Helps in planning tower heights and placements.
  • Ensures signal reaches intended areas effectively.
Height and Signal Range
The height of a TV tower directly influences the range of the TV signal. The taller the tower, the further the signal can potentially travel, due to the line of sight extending over a larger area. This is because taller towers can bypass more obstacles and have a broader horizon.

In simple terms, the TV tower's height determines the maximum reach of the signal on the Earth's curved surface. If a tower is too short, buildings, terrain, and other structures may obstruct the path, dramatically reducing its range. Thus, considering both population density and geography is crucial when setting tower heights.
  • Taller towers extend coverage area.
  • Reduced interference from natural and man-made structures.
  • Critical for widespread broadcasting needs.
Square Root Proportionality
In the context of signal transmission, square root proportionality refers to the mathematical relationship between the signal range and the height of the TV tower. Specifically, the maximum signal distance is proportional to the square root of the tower's height. This is derived from the formula \( d = \sqrt{2 h R} \), where \( d \) is the distance, \( h \) is the height, and \( R \) is the Earth's radius.

This means doubling the height of the tower doesn't double the distance; instead, it increases the range by a factor of \( \sqrt{2} \). Such understanding helps improve the design and planning of broadcast networks, ensuring that network resources are used most efficiently.
  • Proportional to \( h^{0.5} \), not \( h \) or \( h^2 \).
  • Sensitive to height changes.
  • Essential for efficient broadcast network design.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In an electromagnetic wave, the amplitude of electric field is \(1 \mathrm{Vm}^{-1}\). The frequency of wave is \(5 \times 10^{14} \mathrm{~Hz}\). The wave is propagating along z-axis. The average energy density of electrie field in \(\mathrm{J} / \mathrm{m}^{\text {a }}\) will be (a) \(3.2 \times 10^{-12} \mathrm{Jm}^{-1}\) (b) \(2.2 \times 10^{-12} \mathrm{Jm}^{-1}\) (c) \(5.2 \times 10^{-13} \mathrm{Jm}^{-3}\) (d) \(7.2 \times 10^{-12} \mathrm{Jm}^{-1}\)

A plane electromagnetic waves travelling along the \(x\)-direction has a wavelength of \(3 \mathrm{~mm}\). The variation in the electric field occurs in the \(y\)-direction with an amplitude \(66 \mathrm{Vm}^{-1}\). The equation for the electric and magnetic fields as a function of \(x\) and \(t\) are respectively (a) \(E_{y}=33 \cos \pi \times 10^{11}\left(t-\frac{x}{c}\right)\) and \(B_{2}=1.1 \times 10^{-7} \cos \pi \times 10^{11}\left(t-\frac{x}{c}\right)\) (b) \(E_{y}=11 \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right)\) and \(B_{y}=11 \times 10^{-7} \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right)\) (c) \(E_{y}=33 \cos \pi \times 10^{11}\left(t-\frac{x}{c}\right)\) and \(B_{y}=11 \times 10^{-7} \cos \pi \times 10^{11}\left(t-\frac{x}{c}\right)\) (d) \(E_{y}=66 \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right)\) and \(B_{z}=2.2 \times 10^{-7} \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right)\)

The wavelength of X-rays lies between (a) maximum to finite limits (b) minimum to certain limits (c) minimum to infinite limits (d) infinite to finite limits

An electromagnetic wave travels in vacuum along \(z\) direction: \(E=\left(E_{1} \hat{\mathbf{i}}+E_{2} \hat{\mathbf{j}}\right) \cos (k z-\omega t)\). Choose the correct options from the following INCERT Exemplar] (a) The associated magnetic field is given as \(\mathbf{B}=\frac{1}{c}\left(E \hat{\mathbf{i}}+E_{j} \hat{\mathbf{j}}\right) \cos (k z-\omega t)\) (b) The associated magnetic ficld is given as \(\mathbf{B}=\frac{1}{c}(E \hat{\mathbf{i}}-E \hat{\mathbf{j}}) \cos (k z-\omega t)\) (c) The given electromagnetic field is circularly polarised (d) The given electromagnetic wave is plane polarised

Given the wave function (in SI units) for a wave to be \(\psi_{(x, n}=10^{3} \sin \pi\left(3 \times 10^{6} x-9 \times 10^{14} t\right)\). The speed of wave is (a) \(9 \times 10^{14} \mathrm{~ms}^{-1}\) (b) \(3 \times 10^{8} \mathrm{~ms}^{-1}\) [c) \(3 \times 10^{6} \mathrm{~ms}^{-1}\) [d) \(3 \times 10^{7} \mathrm{~ms}^{-1}\)
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Electric Charge Field and Potential

Read Explanation

Radiation

Read Explanation

Kinematics Physics

Read Explanation

Waves Physics

Read Explanation

Fields in Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.