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

After coming out from up th water will move under gravity.  Let the velocity at a distance h from up be $v_2.$ Using $v_2^2=v_1^2+2gh,$ Here $v_1=0.4m{s}^{-1}$ and h = 0.2 m.

$\Rightarrow v_2=\sqrt{v_1^2+2gh}=\sqrt{{\left(0.4\right)}^2+2\times 10\times 0.2}=2m{s}^{\begin{array}{l} \\ -1\end{array}}$

According to equation of continuity at section (1) and section (1)

$a_1{v}_1=a_2{v}_2\Rightarrow \pi {\left(\frac{d_1}{2}\right)}^2{v}_1=\pi {\left(\frac{d_2}{2}\right)}^2{v}_2$

$Here,d_1=8\times {10}^{-3}m$

$\pi \times {\left(\frac{8\times {10}^{-3}}{2}\right)}^2\times 0.4=\pi \times {\left(\frac{d_2}{2}\right)}^2\times 2$

$\Rightarrow d_2=3.6\times {10}^{-3}m$