A spherical ball of radius \(r\) is falling in a viscous fluid of viscosity \(\eta\) with a velocity \(v.\) The retarding viscous force acting on the spherical ball is:

The relative velocity of two adjacent layers of a liquid is 6 cm/s and the perpendicular distance between layers is 0.1 mm. The velocity gradient for liquid (in per second) is:

The velocity of a small ball of mass m and density $\mathrm{\rho }$ when dropped in a container filled with glycerin of density $\sigma$ becomes constant after sometime. The viscous force acting on the ball in the final stage is:-

1. $\mathrm{mg}\left(\frac{\mathrm{\sigma }}{\mathrm{\rho }}\right)$

2. $\mathrm{mg}\left(1+\frac{\mathrm{\sigma }}{\mathrm{\rho }}\right)$

3. $\mathrm{mg}\left(1-\frac{\mathrm{\sigma }}{\mathrm{\rho }}\right)$

4. mg

With an increase in temperature, the viscosity of liquid and gas, respectively will:

A metal block of area \(0.10~\text{m}^{2}\) is connected to a \(0.010\) kg mass via a string that passes over an ideal pulley (considered massless and frictionless), as in the figure below. A liquid film with a thickness of \(0.30\) mm is placed between the block and the table. When released the block moves to the right with a constant speed of \(0.085\) m/s. The coefficient of viscosity of the liquid is:

          
1. \(4.45 \times 10^{-2}~\text{Pa-s}\)
2. \(4.45 \times 10^{-3}~\text{Pa-s}\)
3. \(3.45 \times 10^{-2}~\text{Pa-s}\)
4. \(3.45 \times 10^{-3}~\text{Pa-s}\)

A square plate of 0.1 m side moves parallel to a second plate with a velocity of 0.1 m/s, both plates being immersed in water. If the viscous force is 0.002 N and the coefficient of viscosity is 0.01 poise, then the distance between the plates in m is: