Gujarati
9-1.Fluid Mechanics
normal

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 (see figure). If sphere $P$ alone in $L _2$ has terminal velocity $\overrightarrow{ V }_{ P }$ and $Q$ alone in $L _1$ has terminal velocity $\overrightarrow{ V }_{ Q }$, then

$(A)$ $\frac{\left|\overrightarrow{ V }_{ P }\right|}{\left|\overrightarrow{ V }_{ Q }\right|}=\frac{\eta_1}{\eta_2}$ $(B)$ $\frac{\left|\overrightarrow{ V }_{ P }\right|}{\left|\overrightarrow{ V }_{ Q }\right|}=\frac{\eta_2}{\eta_1}$

$(C)$ $\overrightarrow{ V }_{ P } \cdot \overrightarrow{ V }_{ Q } > 0$ $(D)$ $\overrightarrow{ V }_{ P } \cdot \overrightarrow{ V }_{ Q } < 0$

A

$(B,D)$

B

$(B,C)$

C

$(A,C)$

D

$(A,D)$

(IIT-2015)

Solution

From the given conditions, $\rho_1<\sigma_1<\sigma_2<\rho_2$

From equilibrium, $\sigma_1+\sigma_2=\rho_1+\rho_2$

$V _{ p }=\frac{2}{9}\left(\frac{\rho_1-\sigma_2}{\eta_2}\right) g \text { and } V _{ Q }=\frac{2}{9}\left(\frac{\rho_2-\sigma_1}{\eta_1}\right) g$

So, $\frac{\left|\overrightarrow{ V }_{ P }\right|}{\left|\overrightarrow{ V }_{ Q }\right|}=\frac{\eta_1}{\eta_2}$ and $\overrightarrow{ V }_{ P } \cdot \overrightarrow{ V }_{ Q }<0$

Standard 11
Physics

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