$A$ copper ball of radius $r$ travels with a uniform speed $v$ in a viscous fluid. If the ball is changed with another ball of radius $2r$,then the new uniform speed will be

$A$ solid steel ball of diameter $3.6 \ mm$ acquires a terminal velocity of $2.45 \times 10^{-2} \ m/s$ while falling under gravity through an oil of density $925 \ kg/m^3$. Take the density of steel as $7825 \ kg/m^3$ and $g$ as $9.8 \ m/s^2$. The viscosity of the oil in $SI$ units is:

$A$ spherical ball of density $\rho$ and radius $0.003 \ m$ is dropped into a tube containing a viscous fluid filled up to the $0 \ cm$ mark as shown in the figure. The viscosity of the fluid $\eta = 1.260 \ N \cdot s \cdot m^{-2}$ and its density $\rho_L = \rho/2 = 1260 \ kg \cdot m^{-3}$. Assume the ball reaches a terminal speed by the $10 \ cm$ mark. The time taken by the ball to traverse the distance between the $10 \ cm$ and $20 \ cm$ mark is $(g = 10 \ m \cdot s^{-2})$

$A$ metal sphere of radius $R$ and density $\rho_1$ is dropped in a liquid of density $\sigma$ and moves with terminal velocity $V$. Another metal sphere of the same radius and density $\rho_2$ is dropped in the same liquid. Its terminal velocity will be:

$A$ 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 sphere of radius $r$ falls from rest in a viscous liquid. As a result,heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity,is proportional to