The tube of length L is shown in the figure. The radius of cross section at the point (1) is 2 cm and at the point (2) is 1 cm, respectively. If the velocity of water entering at point (1) is 2 m/s, then velocity of water leaving the point (2) will be
\begin{aligned}
& \mathrm{A}_1 \mathrm{~V}_1=\mathrm{A}_2 \mathrm{~V}_2 \Rightarrow 2 \pi(2 \mathrm{R})^2=\mathrm{V}_2 \pi \mathrm{R}^2 \\
& \therefore \mathrm{~V}_2=8 \mathrm{~m} / \mathrm{s}
\end{aligned}
