$A$ ball rises to the surface of a liquid with constant velocity. The density of the liquid is four times the density of the material of the ball. The viscous force of the liquid on the rising ball is greater than the weight of the ball by a factor of

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
$(A)$ $\frac{|\overrightarrow{V}_{P}|}{|\overrightarrow{V}_{Q}|}=\frac{\eta_1}{\eta_2}$
$(B)$ $\frac{|\overrightarrow{V}_{P}|}{|\overrightarrow{V}_{Q}|}=\frac{\eta_2}{\eta_1}$
$(C)$ $\overrightarrow{V}_{P} \cdot \overrightarrow{V}_{Q} > 0$
$(D)$ $\overrightarrow{V}_{P} \cdot \overrightarrow{V}_{Q} < 0$

Small water droplets of radius $0.01 \,mm$ are formed in the upper atmosphere and falling with a terminal velocity of $10 \,cm/s$. Due to condensation, if $8$ such droplets are coalesced to form a larger drop, the new terminal velocity will be ........... $cm/s$.

$A$ solid sphere falls with a terminal velocity of $10 \, cm/s$ in the Earth's gravitational field. If it is allowed to fall freely in a region outside the gravitational field of the Earth,the terminal velocity will be:

If an air bubble of diameter $2 \text{ mm}$ rises steadily through a liquid of density $2000 \text{ kg/m}^3$ at a rate of $0.5 \text{ cm/s}$,then the coefficient of viscosity of the liquid is . . . . . . $\text{Poise}$. (Take $g = 10 \text{ m/s}^2$)

The terminal velocity of a copper ball of radius $2.0 \; mm$ falling through a tank of oil at $20 \; ^{\circ}C$ is $6.5 \; cm \; s^{-1}$. Compute the viscosity of the oil at $20 \; ^{\circ}C$. Density of oil is $1.5 \times 10^{3} \; kg \; m^{-3}$,density of copper is $8.9 \times 10^{3} \; kg \; m^{-3}$.