Water is flowing continuously from a tap having an internal diameter $$8 \times {10^{ - 3}}m.$$   The water velocity as it leaves the tap is $$0.4\,m{s^{ - 1}}.$$   The diameter of the water stream at a distance $$2 \times {10^{ - 1}}m$$   below the tap is close to:

Solution :
From Bernoulli's theorem,
$$\eqalign{ & {P_0} + \frac{1}{2}\rho v_1^2\rho gh = {P_0} + \frac{1}{2}\rho v_2^2 + 0 \cr & {v_2} = \sqrt {v_1^2 + 2gh} = \sqrt {0.16 + 2 \times 10 \times 0.2} = 2.03\,m/s \cr} $$
From equation of continuity
$$\eqalign{ & {A_2}{v_2} = {A_1}{v_1} \cr & \pi \frac{{D_2^2}}{4} \times {v_2} = \pi \frac{{D_1^2}}{4}{v_1} \cr & \Rightarrow {D_1} = {D_2}\sqrt {\frac{{{v_1}}}{{{v_2}}}} = 3.55 \times {10^{ - 3}}\,m \cr} $$