According to Stoke’s Law for settling velocity of small particles and viscous flow.

Question

TuteeHUB · Accepted Answer

${v_s} = \frac{{18}}{g}.\left( {G - 1} \right).\frac{{{d^2}}}{v}$

TuteeHUB · Answer

${v_s} = \frac{g}{{18}}.\left( {G - 1} \right).\frac{d}{v}$

TuteeHUB · Answer

${v_s} = \frac{g}{{18}}.\left( {G - 1} \right).\frac{{{d^2}}}{v}$

TuteeHUB · Answer

${v_s} = \frac{g}{{18}}.\left( {G + 1} \right).\frac{{{d^2}}}{v}$

1.	According to Stoke’s Law for settling velocity of small particles and viscous flow.
A.	\({v_s} = \frac{g}{{18}}.\left( {G - 1} \right).\frac{d}{v}\)
B.	\({v_s} = \frac{g}{{18}}.\left( {G - 1} \right).\frac{{{d^2}}}{v}\)
C.	\({v_s} = \frac{{18}}{g}.\left( {G - 1} \right).\frac{{{d^2}}}{v}\)
D.	\({v_s} = \frac{g}{{18}}.\left( {G + 1} \right).\frac{{{d^2}}}{v}\)
Answer» C. \({v_s} = \frac{{18}}{g}.\left( {G - 1} \right).\frac{{{d^2}}}{v}\)

Discussion