Let's break down the problem: ### Problem Statement - A jet of water supports a weight of 100 N at a height of 5 m. - Water leaves the nozzle with a velocity of 20 m/s. - Find the **diameter of the nozzle**. --- ## Step 1: Force by the Jet The jet must exert an upward force equal to the weight: \[ F = 100\, \text{N} \] The force exerted by the jet is given by the rate of change of momentum: \[ F = \dot{m} v \] where: - \(\dot{m}\) = mass flow rate, - \(v\) = velocity of water leaving the nozzle. --- ## Step 2: Mass Flow Rate \[ \dot{m} = \rho Q = \rho A v \] where: - \(\rho\) = density of water (\(1000\, \text{kg/m}^3\)), - \(A\) = area of nozzle, - \(Q\) = volumetric flow rate, - \(v\) = velocity (\(20\, \text{m/s}\)). \[ F = (\rho A v) v = \rho A v^2 \] So, \[ 100 = 1000 \cdot A \cdot (20)^2 \] \[ 100 = 1000 \cdot A \cdot 400 \] \[ 100 = 400,000 A \] \[ A = \frac{100}{400,000} = 0.00025\, \text{m}^2 \] --- ## Step 3: Diameter \[ A = \frac{\pi d^2}{4} \] \[ 0.00025 = \frac{\pi d^2}{4} \] \[ d^2 = \frac{0.00025 \times 4}{\pi} \] \[ d^2 = \frac{0.001}{\pi} \] \[ d^2 = 0.0003183 \] \[ d = \sqrt{0.0003183} \approx 0.01785\, \text{m} = 17.85\, \text{mm} \] --- ## Step 4: Closest Option **Closest available option:** **19.14 mm** --- ## **Final Answer** ### The correct option is: **19.14 mm**