Question

A spherical ball of mass $4m$, density $\sigma$ and radius $r$ is attached to a pulley-mass system as shown in the figure. The ball is released in a beaker with a liquid of coefficient of viscosity $\eta$ and density $\rho (<\frac{\sigma }{2})$. If the length of the liquid column in the beaker is sufficiently long, the terminal velocity attained by the ball is given by: (Assume all pulleys to be massless and strings to be massless and inextensible):

• A. $\frac{2}{9}\frac{r^2(2\sigma -\rho )g}{\eta }$
• B. $\frac{r^2(\sigma -2\rho )g}{9\eta }$
• C. $\frac{2}{9}\frac{r^2(\sigma -4\rho )g}{\eta }$
• D. $\frac{2}{9}\frac{r^2(\sigma -3\rho )g}{\eta }$

Answer 1

The correct option is B $\frac{r^2(\sigma -2\rho )g}{9\eta }$

String is light and inextensible and tension in the string will be the same everywhere i.e $T_0$.
Pulley is also massless, hence tension in the string connecting $4m$ and the pulley will be $T=2T_0$.
When mass $4m$ attains $v_T$, thereafter it moves with a constant velocity and $F_{\text{net}}=0$ :
$\Rightarrow T+F_v+F_B-4mg=0...\left(i\right)$

Applying equilibrium for mass $m$,
$T_0=mg...\left(ii\right)$
The viscous force on $4m$,
$F_v=6\pi r\eta v_T$
Buoyant force, $F_B=V\rho g$
where $V=\frac{4}{3}\pi r^3$, is the volume of submerged portion of the body.
For mass $4m$, we can write
$4m=V\times \sigma =\frac{4}{3}\pi r^3\sigma$

Diagramatic representation of forces:


On substituting in Eq.$\left(i\right)$,
$\Rightarrow F_v+F_B=4mg-T=2mg=\frac{4mg}{2}$
$\Rightarrow F_v=\frac{4mg}{2}-F_B$
$\Rightarrow 6\pi r\eta v_T=\frac{4}{6}\pi r^3\sigma g-\frac{4}{3}\pi r^3\rho g$
$\Rightarrow 6\pi r\eta v_T=\frac{4}{3}\pi r^{3}g(\frac{\sigma }{2}-\rho )$
$\Rightarrow v_T=\frac{r^2(\sigma -2\rho )g}{9\eta }$