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
$(A)$ $\frac{\left|\overrightarrow{ V }_{ P }\right|}{\left|\overrightarrow{ V }_{ Q }\right|}=\frac{\eta_1}{\eta_2}$ $(B)$ $\frac{\left|\overrightarrow{ V }_{ P }\right|}{\left|\overrightarrow{ V }_{ Q }\right|}=\frac{\eta_2}{\eta_1}$
$(C)$ $\overrightarrow{ V }_{ P } \cdot \overrightarrow{ V }_{ Q } > 0$ $(D)$ $\overrightarrow{ V }_{ P } \cdot \overrightarrow{ V }_{ Q } < 0$
$(B,D)$
$(B,C)$
$(A,C)$
$(A,D)$
Define coefficient of viscosity.
An air bubble of diameter $6\,mm$ rises steadily through a solution of density $1750\,kg / m ^3$ at the rate of $0.35\,cm / s$. The co-efficient of viscosity of the solution (neglect density of air) is $..........\,Pas$ (given, $g =10\,ms ^{-2}$)
The diameter of an air bubble which was initially $2\,mm$, rises steadily through a solution of density $1750\,kg\,m\,m ^{-3}$ at the rate of $0.35\,cms ^{-1}$. The coefficient of viscosity of the solution is poise (in nearest integer). (the density of air is negligible).
Two uniform solid balls of same density and of radii $r$ and $2r$ are dropped in air and fall vertically downwards. The terminal velocity of the ball with radius $r$ is $1\,cm\,s^{-1}$ , then find the terminal velocity of the ball of radius $2r$ (neglect bouyant force on the balls.) ........... $cm\,s^{-1}$
How coefficient of liquid and gas depend on temperature ?