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 new uniform speed will be

What is the velocity $v$  of a metallic ball of radius $r$  falling in a tank of liquid at the instant when its acceleration is one-half that of a freely falling body ? (The densities of metal and of liquid are $\rho$ and $\sigma$ respectively, and the viscosity of the liquid is $\eta$).

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. Viscosity of the fluid $=$ $1.260$ $N.m^{-2}$ and its density $\rho_L=\rho/2$ $=$ $1260$ $kg.m^{-3}$. 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

( $g$ $ =$ acceleration due to gravity $= 10$ $ ms^{^{-2}} )$

Which falls faster, big rain drops or small rain drops ?

A spherical body of radius $R$ consists of a fluid of constant density and is in equilibrium under its own gravity. If $P ( r )$ is the pressure at $r ( r < R )$, then the correct option$(s)$ is(are)

$(A)$ $P ( I =0)=0$ $(B)$ $\frac{ P ( r =3 R / 4)}{ P ( r =2 R / 3)}=\frac{63}{80}$

$(C)$ $\frac{ P ( r =3 R / 5)}{ P ( r =2 R / 5)}=\frac{16}{21}$ $(D)$ $\frac{ P ( r = R / 2)}{ P ( r = R / 3)}=\frac{20}{27}$

The terminal velocity of a small sphere of radius $a$ in a viscous liquid is proportional to