Match List-I (Event) with List-II (Order of the time interval for the happening of the event) and select the correct option from the options given below the lists.

List-I:                                                             List-II:
(a) The rotation period of earth             (i) 10$$^5$$ s
(b) Revolution period of earth               (ii) 10$$^7$$ s
(c) Period of a light wave                      (iii) 10$$^{-15}$$ s
(d) Period of a sound wave                    (iv) 10$$^{-3}$$ s

We need to match each event from List-I with the appropriate time interval from List-II. Let's recall the approximate values for each event.

Starting with (a) the rotation period of Earth: This is the time for one complete rotation, which is one day. One day has 24 hours. Converting to seconds: 24 hours × 3600 seconds per hour = 86400 seconds. In scientific notation, this is 8.64 × 10$$^4$$ seconds. Since 8.64 is greater than 5, we round up to the next power of 10, giving an order of magnitude of 10$$^5$$ seconds. Therefore, (a) matches with (i).

Next, (b) the revolution period of Earth: This is the time for one orbit around the Sun, which is one year. One year has approximately 365 days. Converting to seconds: 365 days × 24 hours/day × 3600 seconds/hour = 365 × 86400 seconds. Calculating: 365 × 86400 = 31,536,000 seconds. In scientific notation, this is 3.1536 × 10$$^7$$ seconds. The order of magnitude is 10$$^7$$ seconds. Therefore, (b) matches with (ii).

Now, (c) the period of a light wave: Light waves are electromagnetic waves, and for visible light, the frequency is around 10$$^{14}$$ Hz to 10$$^{15}$$ Hz. Taking a typical frequency of 5 × 10$$^{14}$$ Hz, the period T is the reciprocal of frequency: T = 1 / f = 1 / (5 × 10$$^{14}$$) = 2 × 10$$^{-15}$$ seconds. The order of magnitude is 10$$^{-15}$$ seconds. Therefore, (c) matches with (iii).

Finally, (d) the period of a sound wave: Sound waves are audible in the frequency range of 20 Hz to 20,000 Hz. A typical frequency is 1000 Hz (1 kHz). The period T is the reciprocal: T = 1 / 1000 = 0.001 seconds = 10$$^{-3}$$ seconds. Even though periods can range from 5 × 10$$^{-2}$$ seconds (for 20 Hz) to 5 × 10$$^{-5}$$ seconds (for 20,000 Hz), the typical order for common sounds (like 1000 Hz) is 10$$^{-3}$$ seconds. Therefore, (d) matches with (iv).

Summarizing the matches:

• (a) matches with (i) 10$$^5$$ s
• (b) matches with (ii) 10$$^7$$ s
• (c) matches with (iii) 10$$^{-15}$$ s
• (d) matches with (iv) 10$$^{-3}$$ s

Comparing with the options:

• Option A: (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
• Option B: (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
• Option C: (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii)
• Option D: (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)

Our matches correspond to Option A.

Hence, the correct answer is Option A.