A pool has a uniform circular cross-section of radius 5m and uniform depth 1.4m. It is filled by a pipe which delivers water at the rate of 20 litres per sec. Calculate, in minutes, the time taken to fill the pool. If the pool is emptied in 42min, by another cylindrical pipe through which water flows at 2m per sec, calculate the radius of the pipe in cm.

Answer

To solve the problem step by step, we will break it down into two parts:

1. **Calculating the time taken to fill the pool.**
2. **Calculating the radius of the pipe that empties the pool.**

### Part 1: Time Taken to Fill the Pool

**Step 1: Calculate the Volume of the Pool**
- The pool has a uniform circular cross-section, which means we can use the formula for the volume of a cylinder:
\[ V = \pi r^2 h \]
where \( r \) is the radius and \( h \) is the height (or depth) of the cylinder.

- Given:
- Radius \( r = 5 \) m
- Depth \( h = 1.4 \) m
- Use \( \pi \approx \frac{22}{7} \)

- Substitute the values:
\[ V = \frac{22}{7} \times (5^2) \times 1.4 \]
\[ V = \frac{22}{7} \times 25 \times 1.4 \]
\[ V = \frac{22 \times 25 \times 1.4}{7} \]
\[ V = \frac{770}{7} \approx 110 \text{ m}^3 \]

**Step 2: Convert Volume to Liters**
- Since \( 1 \text{ m}^3 = 1000 \text{ liters} \):
\[ V = 110 \text{ m}^3 \times 1000 = 110000 \text{ liters} \]

**Step 3: Calculate the Time to Fill the Pool**
- The pipe delivers water at a rate of 20 liters per second.
- Time taken to fill the pool:
\[ \text{Time} = \frac{\text{Total Volume}}{\text{Rate}} = \frac{110000 \text{ liters}}{20 \text{ liters/second}} = 5500 \text{ seconds} \]

**Step 4: Convert Time to Minutes**
- To convert seconds to minutes:
\[ \text{Time in minutes} = \frac{5500 \text{ seconds}}{60} \approx 91.67 \text{ minutes} \]
- This can be expressed as \( 91 \frac{2}{3} \) minutes.

### Part 2: Radius of the Pipe that Empties the Pool

**Step 1: Calculate the Volume of Water Emptied per Second**
- The pool is emptied in 42 minutes, which is:
\[ 42 \text{ minutes} = 42 \times 60 = 2520 \text{ seconds} \]
- The volume of the pool is \( 110 \text{ m}^3 \) or \( 110000 \text{ liters} \).
- The rate of emptying:
\[ \text{Rate} = \frac{110000 \text{ liters}}{2520 \text{ seconds}} \approx 43.65 \text{ liters/second} \]

**Step 2: Relate the Flow Rate to the Pipe's Radius**
- The water flows through the pipe at a speed of 2 m/s.
- The volume flow rate can also be expressed as:
\[ \text{Flow Rate} = \text{Area} \times \text{Velocity} = \pi r^2 \times v \]
where \( v = 2 \text{ m/s} \).

- Setting the two expressions for flow rate equal:
\[ \pi r^2 \times 2 = 43.65 \]

**Step 3: Solve for the Radius \( r \)**
- Rearranging gives:
\[ r^2 = \frac{43.65}{2\pi} \]
\[ r^2 = \frac{43.65}{2 \times \frac{22}{7}} = \frac{43.65 \times 7}{44} \approx 6.93 \]
\[ r \approx \sqrt{6.93} \approx 2.63 \text{ m} \]

**Step 4: Convert Radius to Centimeters**
- Since \( 1 \text{ m} = 100 \text{ cm} \):
\[ r \approx 2.63 \times 100 \approx 263 \text{ cm} \]

### Final Answers
- **Time taken to fill the pool:** \( 91 \frac{2}{3} \) minutes or approximately 91.67 minutes.
- **Radius of the pipe:** \( 263 \) cm.