**Given Information:**

- Two water taps can fill a tank together in 9 and 3/8 hours.
- The tap of the larger diameter takes 10 hours less than the smaller one to fill the tank separately.

**Let's assume:**

- The time taken by the smaller tap to fill the tank alone = x hours
- The time taken by the larger tap to fill the tank alone = (x - 10) hours

**Calculating the Rate:**

- The rate of the smaller tap = 1/x (as it fills 1 tank in x hours)
- The rate of the larger tap = 1/(x - 10) (as it fills 1 tank in x - 10 hours)

**Calculating the Combined Rate:**

- The combined rate of both taps = 1/(9 and 3/8) (as they fill 1 tank in 9 and 3/8 hours)
- We can convert 9 and 3/8 hours into an improper fraction:
- 9 and 3/8 = (8*9 + 3)/8 = 75/8
- Therefore, the combined rate of both taps = 1/(75/8) = 8/75

**Using the Formula:**

- The combined rate of the taps = sum of individual rates
- 8/75 = 1/x + 1/(x - 10)

**Simplifying the Equation:**

- Multiplying the equation by x(x - 10) to eliminate the denominators:
- (8/75)(x)(x - 10) = (1/x)(x)(x - 10) + (1/(x - 10))(x)(x - 10)
- 8(x - 10) = (x - 10) + x
- 8x - 80 = 2x - 10
- 8x - 2x = 80 - 10
- 6x = 70
- x = 70/6
- x = 11 and 2/3

**Calculating Individual Times:**

- The time taken by the smaller tap to fill the tank alone = x = 11 and 2/3 hours = 11 hours and 40 minutes
- The time taken by the larger tap to fill the tank alone = (x - 10) = (11 and 2/3 - 10) = 1 and 2/3 hours = 1 hour and 40 minutes

Therefore, the smaller tap can fill the tank alone in 11 hours and 40 minutes, while the larger tap can fill the tank alone in 1 hour and 40 minutes.