A) \[4\pi \eta r{{v}^{2}}\]
B) \[4\pi \eta {{r}^{2}}v\]
C) \[2\pi \eta {{r}^{2}}v\]
D) \[6\pi \eta rv\]
Correct Answer: D
Solution :
[d] We can thus say that the viscous force (F) is the function of radius (r), velocity (v) and viscosity (\[\eta \]). \[or\,\,F=f(\eta ,r,v)or\,F=k{{\eta }^{x}}{{r}^{y}}{{v}^{z}}\] ...... (1) Where k is a constant. Now, dimensions of the constituents are \[\therefore \,\,\,\,[ML{{T}^{-2}}]={{[M{{L}^{-1}}{{T}^{-1}}]}^{x}}{{[L]}^{y}}{{[L{{T}^{-1}}]}^{z}}\] \[=[{{M}^{x}}{{L}^{-x+y+z}}{{T}^{-x-z}}]\] Equating the exponents of similar quantities of both sides we get, \[x=1;\text{ }-x+y+z=1\]and \[-x-z=-2\] Solving for\[x,y\And z,\], we get \[x=y=z=1\] Equation (1) becomes \[F=k\eta rv\] Experimentally, it was found that \[k=6\pi \,or\,F=6\pi \eta rv\], which is the famous Stokes' law.
You need to login to perform this action.
You will be redirected in
3 sec
