**Given information:**
- Drawdown in an observation well located 10 m away from the pumping well is 3 m after 10 min of pumping.
- The pumping well is drilled in an artesian aquifer.

**Objective:**
To find the time since pumping started for the same drawdown in another observation well located 20 m away from the pumping well.

**Assumptions:**
- The aquifer is homogeneous and isotropic.
- There are no other pumping wells or significant water extraction nearby.
- The pumping rate remains constant throughout the pumping period.
- The observation wells are fully penetrating and have the same characteristics.

**Explanation:**
1. The drawdown in an observation well is a measure of the decrease in water level caused by pumping. It is influenced by various factors including the pumping rate, aquifer properties, and distance from the pumping well.
2. The relationship between drawdown and distance from the pumping well is described by the Theis equation, which is a solution to the groundwater flow equation.
3. According to the Theis equation, the drawdown at a given observation well is proportional to the pumping rate, time since pumping started, and a function of the distance between the pumping well and the observation well.
4. Theis equation can be expressed as:
S = (Q / (4πT)) * W(u)
where:
- S is the drawdown at a given observation well,
- Q is the pumping rate,
- T is the transmissivity of the aquifer,
- W(u) is the well function, which depends on the dimensionless distance u = (r²S) / (4Tt),
- r is the distance between the pumping well and the observation well,
- t is the time since pumping started.
5. In this case, we are given that the drawdown in the observation well located 10 m away is 3 m after 10 min of pumping. Let's assume the pumping rate is Q and the transmissivity is T.
6. Using the given information, we can calculate the dimensionless distance u for the first observation well:
u₁ = (10² * 3) / (4 * T * 10) = 7.5 / T
7. Now, we need to find the time since pumping started for the same drawdown in another observation well located 20 m away. Let's assume this time is t₂.
8. Using the same pumping rate Q and transmissivity T, we can calculate the dimensionless distance u for the second observation well:
u₂ = (20² * 3) / (4 * T * t₂) = 60 / (T * t₂)
9. Since the drawdown in both observation wells is the same, we can equate the dimensionless distances:
u₁ = u₂
7.5 / T = 60 / (T * t₂)
10. Solving for t₂, we get:
t₂ = 60 / (7.5 * T)
11. Therefore, the time since pumping started for the same drawdown in the second observation well located 20 m away is 60 / (7.5 * T) minutes.
12. The value of T is not given in the question, so we cannot determine the exact time. However, we can conclude that the time since