Two spheres P and Q of equal radii have densities $ρ_{1}$ and $ρ_{2}$ , respectively. The spheres are connected by a massless string and placed in liquids $L_{1}$ and $L_{2}$ of densities $σ_{1}$ and $σ_{2}$ and viscosities $η_{1}$ and $η_{2}$ , respectively. They float in equilibrium with the sphere P in $L_{1}$ and sphere Q in $L_{2}$ and the string being taut (see figure). If sphere P alone in $L_{2}$ has terminal velocity ${\vec{V}}_{P}$ and Q alone in $L_{1}$ has terminal velocity ${\vec{V}}_{Q}$ , then

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$\frac{|{\vec{V}}_{P}|}{|{\vec{V}}_{Q}|} = \frac{η_{1}}{η_{2}}$

$\frac{|{\vec{V}}_{P}|}{|{\vec{V}}_{Q}|} = \frac{η_{2}}{η_{1}}$

${\vec{V}}_{P} . {\vec{V}}_{Q} > 0$

${\vec{V}}_{P} . {\vec{V}}_{Q} < 0$

answer is A, D.

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Detailed Solution

$\begin{array}{l} F o r e q u i l i b r i u m o f P : σ_{1} V g = ρ_{1} V g + T \\ F o r e q u i l i b r i u m o f Q : σ_{2} V g + T = ρ_{2} V g \\ A d d i n g, \\ σ_{1} + σ_{2} = ρ_{1} + ρ_{2} \\ \Rightarrow σ_{2} - ρ_{1} = ρ_{2} - σ_{1} \\ T e r m i n a l v e l o c i t y o f P : V_{P} = \frac{2 r^{2} g}{9 η_{2}} (σ_{2} - ρ_{1}) \to U p ward terminal velocity \\ T e r m i n a l v e l o c i t y o f Q : V_{Q} = \frac{2 r^{2} g}{9 η_{1}} (ρ_{2} - σ_{1}) \to downword terminal velocity \\ r a t i o = \frac{V_{P}}{V_{Q}} = \frac{η_{1}}{η_{2}} \end{array}$