Two small spherical metal balls, having equal masses, are made from materials of densities $\rho_{1}$ and $\rho_{2}\left(\rho_{1}=8 \rho_{2}\right)$ and have radii of $1\; \mathrm{mm}$ and $2\; \mathrm{mm}$, respectively. They are made to fall vertically (from rest) in a viscous medum whose coefficient of viscosity equals $\eta$ and whose denstry is $0.1 \mathrm{\rho}_{2} .$ The ratio of their terminal velocitites would be 

What is the velocity $v$  of a metallic ball of radius $r$  falling in a tank of liquid at the instant when its acceleration is one-half that of a freely falling body ? (The densities of metal and of liquid are $\rho$ and $\sigma$ respectively, and the viscosity of the liquid is $\eta$).

A small metal sphere of radius $a$ is falling with a velocity $v$ through a vertical column of a viscous liquid. If the coefficient of viscosity of the liquid is $\eta $ , then the sphere encounters an opposing force of

A ball of radius $r $ and density $\rho$ falls freely under gravity through a distance $h$ before entering water. Velocity of ball does not change even on entering water. If viscosity of water is $\eta$, the value of $h$ is given by

A thin square plate of side $2\ m$ is moving at the interface of two very viscous liquids of viscosities ${\eta _1} = 1$ poise and ${\eta _2} = 4$ poise respectively as shown in the figure. Assume a linear velocity distribution in each fluid. The liquids are contained between two fixed plates. $h_1 + h_2 = 3\ m$ . A force $F$ is required to move the square plate with uniform velocity $10\ m/s$ horizontally then the value of minimum applied force will be …….. $N$