When a solid sphere moves through a liquid. The liquid opposes the motion with a force. The magnitude of the force depends upon the coefficient of viscosity of the liquid and the radius of the sphere find the expression of force using dimension formula:
$\begin{align}
  & A){{V}_{t}}=\dfrac{2(d-\rho )g{{r}^{2}}}{8\eta } \\
 & B){{V}_{t}}=\dfrac{(d-\rho )g{{r}^{2}}}{6\eta } \\
 & C){{V}_{t}}=\dfrac{2(d-\rho )g{{r}^{2}}}{9\eta } \\
 & D){{V}_{t}}=\dfrac{(d-\rho )g{{r}^{2}}}{9\eta } \\
\end{align}$

Hint: We need to know about all the forces acting on the ball. You should formulate all the forces and find the direction along which it is acting. Equate upward and downward forces for the point when the ball moves with terminal velocity. On solving the equation we get the terminal velocity and the expression of force.

Complete step by step answer:
Let us assign variables to different quantities required
r - Radius of the solid sphere
$\eta -$Coefficient of viscosity of the liquid
 d- Density of the ball
$\rho -$Density of the liquid
g- Acceleration due to gravity
When the ball moves in the liquid we know that it is subjected to three forces namely, weight of the ball, upthrust by the liquid and the viscous force.
Weight of the ball= $mg=\dfrac{4}{3}\pi {{r}^{3}}dg$
Upthrust by the liquid= $v\rho g$=$\dfrac{4}{3}\times \pi {{r}^{3}}\times \rho g$.
To find the viscous force we have to use the stokes law,
Viscous force =$6\pi \eta V$, where V is instantaneous velocity.
When the downward force is greater than the upward force combined, the ball accelerates. The viscous force depends upon velocity. So with increase in velocity, upward force increases. At some point in time, the net force on the ball becomes zero and the velocity becomes constant.
This velocity is known as the terminal velocity ${{V}_{t}}$
Thus when the ball moves at terminal velocity,
Net downward force = Net upward force
$\begin{align}
  & \dfrac{4}{3}\pi {{r}^{3}}dg=\dfrac{4}{3}\times \pi {{r}^{3}}\times \rho g+6\pi \eta {{V}_{t}} \\
 & \Rightarrow 6\pi \eta {{V}_{t}}=\dfrac{4}{3}\pi {{r}^{3}}g(d-\rho ) \\
 & \therefore {{V}_{t}}=\dfrac{2(d-\rho )g{{r}^{2}}}{9\eta } \\
\end{align}$………. Expression of force

So, the correct answer is “Option C”.

Note:
As the ball moves in liquid, depending upon the viscosity of the liquid, liquid friction acts on the ball. More the viscosity of the liquid more is the liquid friction. Liquid friction otherwise known as viscous force is a function of velocity of the ball. So the Stokes law is an expression describing the resisting force on a particle travelling through a viscous fluid. It shows that a maximum velocity is reached in some cases like an object free falling through a fluid.