Let's break down the problem step-by-step: ### **Problem Restatement** - A jet of water supports a weight of **100 N** at a height of **5 m**. - Water leaves the nozzle with a velocity (**v**) of **20 m/s**. - Find the **diameter (d)** of the nozzle. - Given options: 1. 19.14 mm 2. 5.12 mm 3. 12.32 mm 4. 26.21 mm --- ## **Step 1: Force Balance** The force required to support the weight is equal to the **momentum change per second** of the water jet (Newton's Second Law): \[ F = \text{Rate of change of momentum} = \dot{m} v \] Where: - \( \dot{m} \) = mass flow rate - \( v \) = velocity of jet ### The weight being supported: \[ F = 100\, \text{N} \] --- ## **Step 2: Mass Flow Rate** \[ \dot{m} = \rho Q \] Where: - \( \rho \) = density of water (\(100\, \text{kg/m}^3\)) - \( Q \) = volumetric flow rate (\(Q = A v\)) - \( A \) = area of nozzle (\(A = \frac{\pi d^2}{4}\)) So, \[ \dot{m} = \rho \cdot A \cdot v \] --- ## **Step 3: Substitute Values** Since \( \dot{m} v = 100 \): \[ \rho \cdot A \cdot v^2 = 100 \] \[ 100 \cdot \frac{\pi d^2}{4} \cdot (20)^2 = 100 \] \[ 100 \cdot \frac{\pi d^2}{4} \cdot 400 = 100 \] \[ 100 \cdot 400 \cdot \frac{\pi d^2}{4} = 100 \] \[ 400000 \cdot \frac{\pi d^2}{4} = 100 \] \[ 100000 \cdot \pi d^2 = 100 \] \[ \pi d^2 = \frac{100}{100000} \] \[ \pi d^2 = .001 \] \[ d^2 = \frac{.001}{\pi} \] \[ d^2 \approx \frac{.001}{3.1416} \approx .0003183 \] \[ d = \sqrt{.0003183} \approx .01785\, \text{m} \] \[ d = 17.85\, \text{mm} \] --- ## **Step 4: Closest Option** The closest option to **17.85 mm** is **19.14 mm**. --- ## **Final Answer** ### **The correct option is:** **19.14 mm**