Physics

# The critical velocity $v$ of a body depends on the coefficient of viscosity $\eta$, the density $d$ and radius of the drop $r$. If $K$ is a dimensionless constant then $v$ is equal to :

$\dfrac { K\eta }{d r }$

##### SOLUTION
Let's consider,
$d=$ density
$r=$ radius of the drop
$v=$ critical velocity o the body
$\eta=$ coefficient of viscosity
$K=$ dimensionless constant called Reynaldo's number
As we know, up to specific flow velocity, the fluid flow remains to streamline and beyond critical velocity, it becomes turbulent.
The critical velocity depends on the factor:-
1) The critical velocity is directly proportional to the coefficient of viscosity.
$v\propto \eta$. . . . . . . . .(1)
2) The critical velocity is inversly proportional to the density of fluid.
$v\propto \dfrac{1}{d}$. . . . . . . . .(2)
3) The critical velocity is inversly proportional to the radius of the drop
$v\propto \dfrac{1}{r}$. . . . . . . . . .(3)
Combining equation (1), (2) and (3), we get,
$v\propto \dfrac{\eta}{dr}$
$v=K\dfrac{\eta}{dr}$
$v=\dfrac{K\eta}{dr}$
The correct option is C.

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Single Correct Medium Published on 18th 08, 2020
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