Question

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 R_n = 200] ____________.

Options

• 1 m s^-1

• 5 x 10^-2 m s^-1

• 10 m s^-1

• 5 x 10^-3 m s^-1

Answer 1

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 R_n = 200`therefore "v"_"C" = (200 xx 10^-3)/(10^3 xx 4 xx 10^-2)`

`= 5 xx 10^-3 xx "m s"^-1`