$1$ poiseille $=$ ………. poise

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 ball of mass $m$ and radius $ r $ is gently released in a viscous liquid. The mass of the liquid displaced by it is $m' $ such that $m > m'$. The terminal velocity is proportional to

A spherical ball of radius $1 \times 10^{-4} \mathrm{~m}$ and density $10^5$ $\mathrm{kg} / \mathrm{m}^3$ falls freely under gravity through a distance $h$ before entering a tank of water, If after entering in water the velocity of the ball does not change, then the value of $h$ is approximately:

(The coefficient of viscosity of water is $9.8 \times 10^{-6}$ $\left.\mathrm{N} \mathrm{s} / \mathrm{m}^2\right)$

An object falling through a fluid is observed to have acceleration given by $a = g -bv$ where $g =$ gravitational acceleration and $b$ is constant. After a long time of release, it is observed to fall with constant speed. The value of constant speed is