Match Column I (Event) with Column II (order of the time interval for happening of the event) and select the correct combination from the options given below.
`{:("Column I","Column II"),("(i) Rotation period of Earth",(p) 10^(5)s),("(ii) Revolution period of Earth",(q) 10^(7)s),("(iii) Period of a light wave",(r) 10^(-15)s),("(iv) Period of a sound wave",(s) 10^(-3)s):}`

Options

A. (i) `rarr` (p), (ii) `rarr` (q), (iii) `rarr` (r ), (iv) `rarr` (s)
B. (i) `rarr` (q), (ii) `rarr` (p), (iii) `rarr` (s), (iv) `rarr` (r )
C. (i) `rarr` (p), (ii) `rarr` (q), (iv) `rarr` (s), (iv) `rarr` (r )
D. (i) `rarr` (q), (ii) `rarr` (p), (v) `rarr` (r ), (iv) `rarr` (s)

Answer

To solve the problem of matching events from Column I with their corresponding time intervals from Column II, we will analyze each event and determine the order of the time interval for each event.

### Step-by-Step Solution:

1. **Rotation Period of Earth**:
- The rotation period of Earth is approximately 24 hours.
- In seconds, this is calculated as:
\[ 24 \text{ hours} = 24 \times 60 \times 60 = 86400 \text{ seconds} \approx 10^5 \text{ seconds} \]
- Therefore, the match is: **(i) Rotation period of Earth → (p) \(10^5\) s**.

2. **Revolution Period of Earth**:
- The revolution period of Earth around the Sun is 1 year, which is approximately 365 days.
- In seconds, this is calculated as:
\[ 365 \text{ days} = 365 \times 24 \times 60 \times 60 \approx 3.15 \times 10^7 \text{ seconds} \approx 10^7 \text{ seconds} \]
- Therefore, the match is: **(ii) Revolution period of Earth → (q) \(10^7\) s**.

3. **Period of a Light Wave**:
- The period of a light wave can be calculated using the speed of light (approximately \(3 \times 10^8\) m/s) and a typical wavelength (e.g., 400 nm or \(4 \times 10^{-7}\) m).
- The frequency \(f\) can be calculated as:
\[ f = \frac{c}{\lambda} \Rightarrow f = \frac{3 \times 10^8}{4 \times 10^{-7}} = 7.5 \times 10^{14} \text{ Hz} \]
- The period \(T\) is the inverse of frequency:
\[ T = \frac{1}{f} \approx \frac{1}{7.5 \times 10^{14}} \approx 1.33 \times 10^{-15} \text{ seconds} \approx 10^{-15} \text{ seconds} \]
- Therefore, the match is: **(iii) Period of a light wave → (r) \(10^{-15}\) s**.

4. **Period of a Sound Wave**:
- The frequency of a typical sound wave that we can hear is around 1000 Hz (1 kHz).
- The period \(T\) can be calculated as:
\[ T = \frac{1}{f} = \frac{1}{1000} = 0.001 \text{ seconds} = 10^{-3} \text{ seconds} \]
- Therefore, the match is: **(iv) Period of a sound wave → (s) \(10^{-3}\) s**.

### Final Matching:
- (i) Rotation period of Earth → (p) \(10^5\) s
- (ii) Revolution period of Earth → (q) \(10^7\) s
- (iii) Period of a light wave → (r) \(10^{-15}\) s
- (iv) Period of a sound wave → (s) \(10^{-3}\) s