$A$ thin plate is placed between two fixed plates,separating two liquids with coefficients of viscosity $\eta$ and $4\eta$,as shown in the figure. If the plate is to be pulled by applying the minimum force,then the ratio $\frac{d_2}{d_1}$ is:

$A$ cylindrical metal box whose flat surface has an area of $0.01 \ m^2$ rests on a liquid layer of $0.3 \ mm$ thickness. If upon applying a horizontal force of magnitude $\frac{1}{3} \ N$,the box slides with a constant speed of $0.09 \ m/s$,the coefficient of viscosity of the liquid is nearly:

Water in a river $20 \ m$ deep is flowing at a speed of $10 \ ms^{-1}$. The shearing stress between the horizontal layers of water in the river in $Nm^{-2}$ is (Coefficient of viscosity of water $= 10^{-3} \ SI \ units$)

As the temperature of water increases,its viscosity

The speed of the water in a river is $V$ near the surface. If the coefficient of viscosity of water is $\eta$ and the depth of the river is $H$,then the shearing stress between the horizontal layers of water is

$A$ thin square plate of side $2\ m$ is moving at the interface of two very viscous liquids of viscosities $\eta_1 = 1 \text{ poise}$ and $\eta_2 = 4 \text{ poise}$ 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 a uniform velocity $10\ m/s$ horizontally. The value of the minimum applied force will be ........ $N$.