Question

A container filled with viscous liquid is moving vertically downward with constant speed $3v_o$. At the instant shown, a sphere of radius $r$ is moving downward in the liquid with speed $v_o.$ The coefficient of viscosity is $\eta$. There is no relative motion between the liquid and the container. At the shown instant, the magnitude of viscous force on the sphere is

• A. $11\pi \eta r{v}_o$
• B. $12\pi \eta r{v}_o$
• C. $13\pi \eta r{v}_o$
• D. $14\pi \eta r{v}_o$

Answer 1

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

There is no relative motion between liquid and container.
So, velocity of liquid $=$ velocity of container
$\Rightarrow$ velocity of liquid $=3v_o$ (downward)
Relative velocity of sphere with respect to liquid,
$v=v_o-3v_o=-2v_o$
So, the velocity of the sphere is $2v_o$ upward $w.r.t$ liquid.
Now, viscous force on sphere,
$F_v=6\pi r\eta \left(2v_o\right)$
$\Rightarrow F_v=12\pi \eta r{v}_o$

$\begin{array}{c}\text{Why this question?}\\ \text{Drag force on a solid body moving inside}\\ \text{a liquid acts due to relative motion between liquid}\\ \text{and solid body.}\end{array}$