​The diameter of a capillary is 4mm. If Reynolds's number of the tube is 1000 and the co-efficient of viscosity for the water is 0.02 poise, determine the maximum speed of water for the stream line flow in the tube?​

Correct option is C

​$$\text{The Reynolds number is given by:} \\R = \dfrac{\rho v d}{\eta} \\\text{Where:} \\R \text{ is the Reynolds number,} \\\rho \text{ is the density of water (1000 kg/m}^3), \\v \text{ is the average velocity of the flow,} \\d \text{ is the diameter of the capillary (4 mm = 0.004 m),} \\\eta \text{ is the coefficient of viscosity (0.02 poise = 0.002 Pa.s).} \\\text{Calculation:} \\\text{Given:} \\\text{Coefficient of viscosity: } \eta = 0.02 \times 0.1 = 0.002 \, \text{Pa.s,} \\\text{Density of water: } \rho = 1000 \, \text{kg/m}^3, \\\text{Diameter of the capillary: } d = 4 \, \text{mm} = 0.004 \, \text{m,} \\\text{Reynolds number: } R = 1000. \\\text{Using the formula:} \\v = \dfrac{Rn}{\rho d} \\\text{Substitute the given values:} \\v = \dfrac{1000 \times 0.002}{1000 \times 0.004} = 0.5 \, \text{m/s or 50 cm/s} \\\text{Now, the maximum velocity } v_{\text{max}} \text{ is twice the average velocity:} \\v_{\text{max}} = 2v = 2 \times 50 = 100 \, \text{cm/s}$$​​