If an air bubble of diameter $2 \text{ mm}$ rises steadily through a liquid of density $2000 \text{ kg/m}^3$ at a rate of $0.5 \text{ cm/s}$,then the coefficient of viscosity of the liquid is . . . . . . $\text{Poise}$. (Take $g = 10 \text{ m/s}^2$)

The terminal velocity of a copper ball of radius $2.0 \; mm$ falling through a tank of oil at $20 \; ^{\circ}C$ is $6.5 \; cm \; s^{-1}$. Compute the viscosity of the oil at $20 \; ^{\circ}C$. Density of oil is $1.5 \times 10^{3} \; kg \; m^{-3}$,density of copper is $8.9 \times 10^{3} \; kg \; m^{-3}$.

Two drops of equal radius are falling with a terminal velocity of $5 \, cm/s$. If they coalesce to form a single larger drop,what will be the terminal velocity of the new drop?

$A$ solid metal sphere released in a vertical liquid column has attained terminal velocity in the downward direction. The magnitudes of viscous force,buoyant force,and gravitational force acting on it are $F_{v}$,$F_{B}$,and $F_{W}$ respectively. Then,the correct relation between them is:

$A$ small spherical ball of radius $r$,falling through a viscous medium of negligible density has terminal velocity $v$. Another ball of the same mass but of radius $2r$,falling through the same viscous medium will have terminal velocity:

The graph between terminal velocity $(v_t)$ (along $y$-axis) and the square of the radius $(r^2)$ (along $x$-axis) of a spherical body of density $\rho$ falling through a fluid of density $\sigma$ is: