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प्रश्न
The maximum average velocity of water in a tube of diameter 4 cm so that the flow becomes laminar is [Viscosity of water is 10-3 N m-1, take Rn = 200] ____________.
पर्याय
1 m s-1
5 x 10-2 m s-1
10 m s-1
5 x 10-3 m s-1
MCQ
रिकाम्या जागा भरा
उत्तर
The maximum average velocity of water in a tube of diameter 4 cm so that the flow becomes laminar is 5 x 10-3 m s-1.
Explanation:
`"v"_"C" = "R"_"n" eta/(rho"D")`
For laminar flow, Reynold 's number Rn = 200`therefore "v"_"C" = (200 xx 10^-3)/(10^3 xx 4 xx 10^-2)`
`= 5 xx 10^-3 xx "m s"^-1`
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Critical Velocity and Reynolds Number
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