Question

An ideal water jet with volume flow rate of 0.05 m³/s strikes a flat plate placed normal to its path and exerts a force of 1000 N. Considering the density of water as 1000 Kg/m³, the diameter (in mm) of the water jet is _______

Answer 1

Concept:

Rate of change of momentum in the direction of force is,

\({F_x} = \frac{{mass}}{{time}} = \rho AV\left( {{V_i} - {V_f}} \right)\)

Calculation:

Given: Q = 0.05 m³/s, F_x = 1000 N, ρ = 1000 Kg/m³

Therefore,

1000 = 1000 × AV (V – 0)

\(\Rightarrow 1000 = 1000 \times A \times {\left( {\frac{Q}{A}} \right)^2}\)

⇒ A = Q² = (0.05)²

\(\Rightarrow \frac{\pi }{4}{d^2} = 2.5 \times {10^{ - 3}} \Rightarrow d = 0.05641~m = 56.41~mm\)