Let's break down the problem: ### Given: - Mass flow rate (\(\dot{m}\)) = 0.05 kg/s - Water density (\(\rho\)) = 996 kg/m³ - Diameter of the hole (d) = 0.02 cm = 0.0002 m ### To Find: **Minimum pressure at which the water supplying pump must operate.** --- #### **Step 1: Find the velocity of water (v) using mass flow rate** \[ \dot{m} = \rho \cdot A \cdot v \] Where: - \(A\) = area of the hole = \(\pi/4 \times d^2\) \[ A = \frac{\pi}{4} \times (0.0002)^2 = 3.14 \times 10^{-8} \text{ m}^2 \] \[ 0.05 = 996 \cdot 3.14 \times 10^{-8} \cdot v \] \[ v = \frac{0.05}{996 \cdot 3.14 \times 10^{-8}} \] \[ v \approx \frac{0.05}{3.129 \times 10^{-5}} \] \[ v \approx 1598.8 \text{ m/s} \] --- #### **Step 2: Find the pressure using Bernoulli’s equation** Neglect losses and potential head differences: \[ P = \frac{1}{2} \rho v^2 \] \[ P = 0.5 \cdot 996 \cdot (1598.8)^2 \] \[ P = 0.5 \cdot 996 \cdot 2,556,166 \] \[ P \approx 0.5 \cdot 2,546,941,736 \] \[ P \approx 1,273,470,868 \text{ N/m}^2 \] \[ P \approx 1,271 \text{ MPa} \] --- ### **Answer:** **The minimum pressure is 1271 MPa.** So, the correct answer is: - **1271 MPa** (Option 1 or 2, both are the same)