JEE MAIN - Physics (2025 - 22nd January Evening Shift - No. 11)
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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
4 m/s
6 m/s
2 m/s
8 m/s
Explanation
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$$\begin{aligned} & \mathrm{A}_1 \mathrm{v}_1=\mathrm{A}_2 \mathrm{~V}_2 \Rightarrow 2 \pi(2 \mathrm{R})^2=\mathrm{V}_2 \pi \mathrm{R}^2 \\ & \therefore \mathrm{~V}_2=8 \mathrm{~m} / \mathrm{s} \end{aligned}$$
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