$A$ 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

Two rain drops of same radius $r$ falling with terminal velocity $V$ merge and form a bigger drop with radius $R$. The terminal velocity of the bigger drop is:

$A$ metal sphere of radius $R$,density $\rho_1$ moves with terminal velocity $V_1$ through a liquid of density $\sigma$. Another sphere of same radius but density $\rho_2$ moves through the same liquid. Its terminal velocity is $V_2$. The ratio $V_1: V_2$ is

Eight spherical rain drops of the same mass and radius are falling down with a terminal speed of $6 \ cm \ s^{-1}$. If they coalesce to form one big drop,what will be the terminal speed of the bigger drop (in $cm \ s^{-1}$)? (Neglect the buoyancy of the air)

The velocity of a small ball of mass $M$ and density $d_1$ when dropped in a container filled with glycerin becomes constant after some time. If the density of glycerin is $d_2$,the viscous force acting on the ball is ($g$ = acceleration due to gravity).

$A$ small spherical ball of radius $0.1 \,mm$ and density $10^{4} \,kg \,m^{-3}$ falls freely under gravity through a distance $h$ before entering a tank of water. If after entering the water the velocity of the ball does not change and it continues to fall with the same constant velocity inside the water,then the value of $h$ will be $m$. (Given $g = 10 \,m \,s^{-2}$,viscosity of water $\eta = 1.0 \times 10^{-5} \,N \,s \,m^{-2}$,density of water $\rho_w = 10^3 \,kg \,m^{-3}$)