Question

Two spheres P and Q of equal radii have densities ${\rho }_1$ and ${\rho }_2$, respectively. The spheres are connected by a massless string and placed in liquids $L_1$ and $L_2$ of densities ${\sigma }_1$ and ${\sigma }_2$ and viscosities ${\eta }_1$ and ${\eta }_2$, respectively. They float in equilibrium with the sphere P in $L_1$ and sphere Q in $L_2$ and the string being taut. If sphere P along in $L_2$ has terminal velocity ${\overrightarrow{V}}_P$ and Q along in $L_1$ has terminal velocity ${\overrightarrow{V}}_Q$, then

• A. $\frac{|{\overrightarrow{V}}_P|}{|{\overrightarrow{V}}_Q|}=\frac{{\eta }_1}{{\eta }_2}$
• B. $\frac{|{\overrightarrow{V}}_P|}{|{\overrightarrow{V}}_Q|}=\frac{{\eta }_2}{{\eta }_1}$
• C. ${\overrightarrow{V}}_P.{\overrightarrow{V}}_Q>0$
• D. ${\overrightarrow{V}}_P.{\overrightarrow{V}}_Q<0$

Answer 1

Answer: (A), (D)

${\overrightarrow{V}}_P.{\overrightarrow{V}}_Q<0$

For the body to float
$({\rho }_1+{\rho }_2)V=({\sigma }_1+{\sigma }_2)V$
$\begin{array}{}{\rho }_1+{\rho }_2={\sigma }_1+{\sigma }_2& \text{(1)}\end{array}$
As the strings are taut
${\rho }_1<{\sigma }_1$ and ${\rho }_2>{\sigma }_2$
Now using Stokes law
$\begin{array}{}{V}_P=\frac{2}{9}\frac{({\sigma }_2-{\rho }_1)r^{2}g}{{\eta }_2}& \text{(2)}\end{array}$
or
$\begin{array}{}{V}_Q=\frac{2}{9}\frac{({\sigma }_1-{\rho }_2)r^{2}g}{{\eta }_1}& \text{(3)}\end{array}$
From $\left(1\right)$
${\sigma }_2-{\rho }_1=-({\sigma }_1-{\rho }_2)$
Dividing $\left(1\right)$ and $\left(2\right)$
$\left|\frac{V_P}{V_Q}\right|=\frac{{\eta }_1}{{\eta }_2}$
The dot product $V_P.V_Q<0$ as $V_P$ and $V_Q$ are opposite.