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 ........ $N$

The terminal velocity of a small sphere of radius $a$ in a viscous liquid is proportional to

A spherical body of radius $R$ consists of a fluid of constant density and is in equilibrium under its own gravity. If $P ( r )$ is the pressure at $r ( r < R )$, then the correct option$(s)$ is(are)

$(A)$ $P ( I =0)=0$ $(B)$ $\frac{ P ( r =3 R / 4)}{ P ( r =2 R / 3)}=\frac{63}{80}$

$(C)$ $\frac{ P ( r =3 R / 5)}{ P ( r =2 R / 5)}=\frac{16}{21}$ $(D)$ $\frac{ P ( r = R / 2)}{ P ( r = R / 3)}=\frac{20}{27}$

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

If a ball of steel (density $\rho=7.8 \;gcm ^{-3}$) attains a terminal velocity of $10 \;cms ^{-1}$ when falling in a tank of water (coefficient of viscosity $\eta_{\text {water }}=8.5 \times 10^{-4} \;Pa - s$ ) then its terminal velocity in glycerine $\left(\rho=12 gcm ^{-3}, \eta=13.2\right)$ would be nearly

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 $9.8 \times 10^{-6}$ $\left.\mathrm{N} \mathrm{s} / \mathrm{m}^2\right)$