A capillary tube of length 5 cm and diameter 1 mm is connected to a tank horizontally. The rate of flow of water is 10 ce per minute. Calculate the rate of flow of water through another capillary tube of diameter 2 mm and length 50 cm is connected in series with the first capillary?

Explanation

To solve the problem, we need to analyze the flow of water through two capillary tubes connected in series. We will use the principle of conservation of mass, which states that the flow rate (discharge) must remain constant in a closed system.

### Step-by-Step Solution:

1. **Identify Given Data for the First Capillary Tube:**
- Length (L1) = 5 cm = 5 × 10^(-2) m
- Diameter (d1) = 1 mm = 1 × 10^(-3) m
- Rate of flow (Q1) = 10 cc/min = 10 × 10^(-6) m³/min

2. **Convert the Rate of Flow to SI Units:**
- Convert 10 cc/min to m³/s:
\[ Q1 = 10 \, \text{cc/min} = 10 \times 10^{-6} \, \text{m}^3 \times \frac{1 \, \text{min}}{60 \, \text{s}} = \frac{10 \times 10^{-6}}{60} \, \text{m}^3/\text{s} \approx 1.67 \times 10^{-7} \, \text{m}^3/\text{s} \]

3. **Identify Given Data for the Second Capillary Tube:**
- Length (L2) = 50 cm = 50 × 10^(-2) m
- Diameter (d2) = 2 mm = 2 × 10^(-3) m

4. **Apply the Continuity Equation:**
- According to the continuity equation, the flow rate through both capillary tubes must be equal:
\[ Q1 = Q2 \]
- Therefore, the rate of flow through the second capillary tube (Q2) will also be:
\[ Q2 = Q1 = 1.67 \times 10^{-7} \, \text{m}^3/\text{s} \]

5. **Convert Q2 Back to cc/min for Final Answer:**
- Convert Q2 from m³/s back to cc/min:
\[ Q2 = 1.67 \times 10^{-7} \, \text{m}^3/\text{s} \times 60 \, \text{s/min} \times 10^6 \, \text{cc/m}^3 \approx 10 \, \text{cc/min} \]

### Final Answer:
The rate of flow of water through the second capillary tube is **10 cc/min**.