A raindrop with radius $R=0.2\, {mm}$ fells from a cloud at a height $h=2000\, {m}$ above the ground. Assume that the drop is spherical throughout its fall and the force of buoyance may be neglected, then the terminal speed attainde by the raindrop is : (In ${ms}^{-1}$)

[Density of water $f_{{w}}=1000\;{kg} {m}^{-3}$ and density of air $f_{{a}}=1.2\; {kg} {m}^{-3}, {g}=10 \;{m} / {s}^{2}$ Coefficient of viscosity of air $=18 \times 10^{-5} \;{Nsm}^{-2}$ ]

An air bubble of radius $r$ rises steadily through a liquid of density $\rho $ with velocity $v$ . The coefficient of viscosity of liquid is

Why not rain drops do not posses greater velocity than some velocity ? Explain.

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

Which of the following is not the property of an ideal fluid?