$A$ river $10\,m$ deep is flowing at $5\,m/s$. The shearing stress between horizontal layers of the river is :- $(\eta = 10^{-3}\,SI\,unit)$

$A$ cubical block of side $a$ and density $\rho$ slides over a fixed inclined plane with constant velocity $v$. There is a thin film of viscous fluid of thickness $t$ between the plane and the block. If the angle of inclination is $37^{\circ}$,then the coefficient of viscosity of the thin film will be:

$A$ metal block of base area $0.2\; m^{2}$ is connected to a $0.02\; kg$ mass via a string that passes over an ideal pulley as shown in the figure. $A$ liquid film of thickness $0.6\; mm$ is placed between the block and the table. When released,the block moves to the right with a constant speed of $0.17\; m/s$. The coefficient of viscosity of the liquid is:

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

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 $= 10^{-3} \ SI \ units$)

$A$ metal cube of side $10 \,cm$ rests on a film of a liquid of thickness $0.2 \,mm$. If upon applying a horizontal force $\overrightarrow{F}$ of magnitude $0.1 \,N$ the cube slides with a constant speed of $0.08 \,m/s$, then the coefficient of viscosity is