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}} )$
$500$ $ \mu s$
$50$ $ ms$
$0.5$ $ s$
$5 $ $ s$
The terminal velocity of a small sphere of radius $a$ in a viscous liquid is proportional to
Write the equation of terminal velocity.
$Assertion :$ Falling raindrops acquire a terminal velocity.
$Reason :$ A constant force in the direction of motion and a velocity dependent force opposite to the direction of motion, always result in the acquisition of terminal velocity.
A solid sphere, of radius $R$ acquires a terminal velocity $\nu_1 $ when falling (due to gravity) through a viscous fluid having a coefficient of viscosity $\eta $. The sphere is broken into $27$ identical solid spheres. If each of these spheres acquires a terminal velocity, $\nu_2$, when falling through the same fluid, the ratio $(\nu_1/\nu_2)$ equals
An air bubble of diameter $6\,mm$ rises steadily through a solution of density $1750\,kg / m ^3$ at the rate of $0.35\,cm / s$. The co-efficient of viscosity of the solution (neglect density of air) is $..........\,Pas$ (given, $g =10\,ms ^{-2}$)