The problem asks for a possible difference between the times shown by two digital clocks, given their initial times and how they change over 30 seconds.

Initial Clock Difference Calculation

At the first instant:

• Clock 1 shows: 09h 13m
• Clock 2 shows: 09h 17m
• The initial difference is: $D_0 = 09\text{h } 17\text{m} - 09\text{h } 13\text{m} = 4$ minutes.

Clock Rate Analysis

After 30 seconds of real time:

• Clock 1 shows: 09h 14m
• Clock 2 shows: 09h 17m

Let's determine the rate of each clock relative to real time:

• Clock 1: In 30 seconds of real time, Clock 1 advanced by $09\text{h } 14\text{m} - 09\text{h } 13\text{m} = 1$ minute $= 60$ seconds. Its rate is $\frac{60 \text{ seconds}}{30 \text{ seconds}} = +2$ seconds per second. It runs fast.
• Clock 2: In 30 seconds of real time, Clock 2 advanced by $09\text{h } 17\text{m} - 09\text{h } 17\text{m} = 0$ seconds. Its rate is $\frac{0 \text{ seconds}}{30 \text{ seconds}} = 0$ seconds per second. It runs at the correct speed relative to real time.

Difference Rate Calculation

The difference between the clocks changes based on the difference in their rates:

• Rate of difference change = Rate(Clock 2) - Rate(Clock 1)
• Rate of difference change = $0 \text{ s/s} - 2 \text{ s/s} = -2$ s/s.
• This means the time difference between the clocks decreases by 2 seconds every second.

Calculating Possible Differences

Let $t$ be the time in seconds elapsed since the initial observation.

The difference $D(t)$ at time $t$ can be calculated using the formula:

$D(t) = D_0 + (\text{rate of difference change}) \times t$

Converting the initial difference to seconds: $D_0 = 4 \text{ minutes} = 240$ seconds.

$D(t) = 240 \text{ seconds} + (-2 \text{ s/s}) \times t = 240 - 2t$ seconds.

Now, let's check the given options by finding the time $t$ at which each difference occurs:

• Option 1: 3m 00s = 180s
  $180 = 240 - 2t \implies 2t = 60 \implies t = 30$s. (This is the difference exactly 30 seconds later).
• Option 2: 3m 30s = 210s
  $210 = 240 - 2t \implies 2t = 30 \implies t = 15$s. (This occurs 15 seconds after the initial observation).
• Option 3: 4m 00s = 240s
  $240 = 240 - 2t \implies 2t = 0 \implies t = 0$s. (This is the initial difference).
• Option 4: 4m 30s = 270s
  $270 = 240 - 2t \implies 2t = -30 \implies t = -15$s. (This occurred 15 seconds before the initial observation, assuming constant rates).

Conclusion on Possible Difference

All the options represent differences that are mathematically possible if the clocks maintained constant rates. However, typically, such questions seek a value that occurs within or is representative of the observed period. The difference of 3m 30s occurs exactly at the midpoint ($t=15$s) of the observation interval (0s to 30s).