JEE Mains · Physics · STD 11 - 9.1 fluid mechanics
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 \(8 \mathrm{~m} / \mathrm{s}\)
Answer & Solution
Correct Answer
(D) \(8 \mathrm{~m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{A}_1 \mathrm{~V}_1=\mathrm{A}_2 \mathrm{~V}_2 \Rightarrow 2 \pi(2 \mathrm{R})^2=\mathrm{V}_2 \pi \mathrm{R}^2 \\ & \therefore \mathrm{~V}_2=8 \mathrm{~m} / \mathrm{s}\end{aligned}\)
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