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Question
The maximum average velocity of water in a tube of diameter 2 cm so that the flow becomes laminar is (The viscosity of water is 10−3 Nm−2 s1)
Options
0.1 ms−1
1 ms−1
10 ms−1
100 ms−1
MCQ
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Solution
0.1 ms−1
Explanation:
Vc = `"N" η/"ρD"`
For laminar flow, Reynold's number,
N = 2000, η = 10−3 NM−2 s−1
∴ Vc = `(2000 xx 10^-3)/(10^3 xx 2 xx 10^-2)`
= 0.1 m/s
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Critical Velocity
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