Question 51 of 192

A 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 $2 \mathrm{~m} / \mathrm{s}$, then velocity of water leaving the point (2) will be

A$4 \mathrm{~m} / \mathrm{s}$B$2 \mathrm{~m} / \mathrm{s}$C$6 \mathrm{~m} / \mathrm{s}$D$8 \mathrm{~m} / \mathrm{s}$

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Answer: (D) $8 \mathrm{~m} / \mathrm{s}$

$\begin{aligned} & \mathrm{A}_1 \mathrm{~V}_1=\mathrm{A}_2 \mathrm{~V}_2 \Rightarrow 2 \pi(2 \mathrm{R})^2=\mathrm{V}_2 \pi \mathrm{R}^2 \\ & \therefore \mathrm{~V}_2=8 \mathrm{~m} / \mathrm{s}\end{aligned}$