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
$27$
$1/27$
$9$
$1/9$
Mercury is filled in a tube of radius $2 \mathrm{~cm}$ up to a height of $30 \mathrm{~cm}$. The force exerted by mercury on the bottom of the tube is. . . . . . $\mathrm{N}$.
(Given, atmospheric pressure $=10^5 \mathrm{Nm}^{-2}$, density of mercury $=1.36 \times 10^4 \mathrm{~kg} \mathrm{~m}^3, \mathrm{~g}=10 \mathrm{~ms}^2$, $\left.\pi=\frac{22}{7}\right)$
The graph between terminal velocity (along $y-axis$ ) and square of radius (along $x-axis$ ) of spherical body of density $\rho $ allowed to fall through a fluid of density $\rho $ is $a$
Why not rain drops do not posses greater velocity than some velocity ? Explain.
Spherical balls of radius $ 'r'$ are falling in a viscous fluid of viscosity '$\eta$' with a velocity $ 'v'. $ The retarding viscous force acting on the spherical ball is
Velocity of water in a river is