Question

A container filled with a viscous liquid is moving vertically downwards with a constant speed $3v_0$. At the instant shown, a sphere of radius $r$ is moving vertically downwards (in liquid) has speed $v_0$. There is no relative motion between fluid and container. The coefficient of viscosity is $\eta$. Then at the shown instant, the magnitude of viscous force acting on sphere is

• A. $6\pi \eta r{v}_0$
• B. $12\pi \eta r{v}_0$
• C. $18\pi \eta r{v}_0$
• D. $24\pi \eta r{v}_0$

Answer 1

The correct option is B $12\pi \eta r{v}_0$

As we know the viscous force on the body can be given as,
$F_v=6\pi \eta rv$


If we take the container as our frame of reference then if the sphere is moving down with $v_0$ and container gives its velocity to this sphere then $3v_0$ will act on the sphere in the upward direction as shown in free body diagram of the sphere.
So, the relative velocity of this body is $3v_0-v_0=2v_0$
$F_v=6\pi \eta r\left(2v_0\right)$
$F_v=12\pi \eta r{v}_0$
As the net velocity is in upward direction, then this viscus force will oppose the motion so it will act in downward direction.

For detailed solution watch the next video.