CBSE Class 12 Physics Chapter 28: Communication Systems NCERT Solutions

The correct option is (b) 10 MHz.

For communication that extends beyond the horizon, the radio waves need to travel long distances. This is typically achieved using sky wave propagation, where waves are reflected by the ionosphere.

Let's analyze the given frequencies:

• 10 kHz: These very low frequencies require extremely large antennas for efficient radiation, making them impractical for this purpose.
• 10 MHz: Frequencies in this range are effectively reflected by the ionosphere, making them suitable for beyond-the-horizon communication.
• 1 GHz to 1000 GHz: These are very high frequencies (microwaves and above). Signals in this range tend to penetrate the ionosphere rather than being reflected, making them unsuitable for sky wave communication. They are more suited for line-of-sight communication.

Therefore, 10 MHz is the most suitable frequency for beyond-the-horizon communication using sky waves.

The correct option is (d) Space waves.

Ultra High Frequency (UHF) waves fall within the frequency range of 300 MHz to 3 GHz. At these high frequencies:

• Ground waves (or surface waves) follow the curvature of the Earth but are effective only at lower frequencies (typically up to a few MHz).
• Sky waves rely on reflection from the ionosphere. However, UHF waves are too high in frequency to be reflected by the ionosphere; they tend to penetrate it.
• Space waves propagate in a straight line from the transmitting antenna to the receiving antenna. This is also known as line-of-sight communication. Since UHF waves are not reflected by the ionosphere and do not follow the Earth's curvature effectively, they travel directly to the receiver, provided there is a clear line of sight.

Therefore, UHF frequencies normally propagate by means of space waves.

The correct option is (c) (i), (ii) and (iii) but not (iv).

Let's analyze each statement regarding digital signals:

• (i) Do not provide a continuous set of values: This is true. Digital signals represent information using distinct, separate levels, not a continuous range.
• (ii) Represent values as discrete steps: This is also true. The distinct levels are represented as steps in a sequence.
• (iii) Can utilize binary system: This is true. The most common representation for digital signals is the binary system (0s and 1s), which allows for efficient processing and transmission.
• (iv) Can utilize decimal as well as binary systems: This is false. While digital systems are fundamentally based on discrete states, the standard and most practical representation is binary. Decimal representation is characteristic of analog signals or requires complex encoding/decoding for digital systems, and is not the primary or direct method.

Therefore, statements (i), (ii), and (iii) are true, while statement (iv) is not accurate for standard digital signal representation.

For line-of-sight communication, the primary requirement is that there should be no physical obstruction between the transmitting and receiving antennas. The waves travel in a straight line. Therefore, it is not necessary for the transmitting and receiving antennas to be at the same height. However, the height of the antennas significantly affects the range of communication.

Given:

• Height of the transmitting antenna, $h\; =\; 81\; \backslash text\{\; m\}$
• Radius of the Earth, $R\; =\; 6.4\; \backslash times\; 10^6\; \backslash text\{\; m\}$

The maximum distance (range, $d$) that a signal can travel from a transmitting antenna of height $h$ to a receiving antenna at ground level is given by the formula derived from the tangent to the Earth's curvature:

$$d\; =\; \backslash sqrt\{2Rh\}$$

The service area ($A$) covered by the antenna is the area of a circle with radius $d$:

$$A\; =\; \backslash pi\; d^2\; =\; \backslash pi\; (2Rh)$$

Now, we can calculate the service area:

$A\; =\; \backslash pi\; \backslash times\; (2\; \backslash times\; 6.4\; \backslash times\; 10^6\; \backslash text\{\; m\}\; \backslash times\; 81\; \backslash text\{\; m\})$

$A\; =\; 3.14159\; \backslash times\; (1036.8\; \backslash times\; 10^6\; \backslash text\{\; m\}^2)$

$A\; \backslash approx\; 3257.2\; \backslash times\; 10^6\; \backslash text\{\; m\}^2$

To express this in square kilometers, we divide by $(1000\; \backslash text\{\; m/km\})^2\; =\; 10^6\; \backslash text\{\; m\}^2/\backslash text\{km\}^2\$:$

$A\; \backslash approx\; 3257.2\; \backslash text\{\; km\}^2$

Rounding this value, the service area covered by the antenna is approximately 3256 km².