A table tennis ball has radius $(3 / 2) \times 10^{-2} m$ and mass $(22 / 7) \times 10^{-3} kg$. It is slowly pushed down into a swimming pool to a depth of $d=0.7 m$ below the water surface and then released from rest. It emerges from the water surface at speed $v$, without getting wet, and rises up to a height $H$. Which of the following option(s) is (are) correct?

[Given: $\pi=22 / 7, g=10 ms ^{-2}$, density of water $=1 \times 10^3 kg m ^{-3}$, viscosity of water $=1 \times 10^{-3} Pa$-s.]

$(A)$ The work done in pushing the ball to the depth $d$ is $0.077 J$.

$(B)$ If we neglect the viscous force in water, then the speed $v=7 m / s$.

$(C)$ If we neglect the viscous force in water, then the height $H=1.4 m$.

$(D)$ The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is $500 / 9$.

Two drops of same radius are falling through air with steady velocity of $v $ $cm/s$. If the two drops coalesce, what would be the terminal velocity?

In Millikan's oll drop experiment, what is the terminal speed of an uncharged drop of radius $2.0 \times 10^{-5} \;m$ and density $1.2 \times 10^{3} \;kg m ^{-3} .$ Take the viscosity of air at the temperature of the experiment to be $1.8 \times 10^{-5}\; Pa\; s$. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to atr.

A spherical solid ball of volume $V$ is made of a material of density $\rho_1$ . It is falling through a liquid of density $\rho_2 (\rho_2 < \rho_1 )$. Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed $v$, i.e., $F_{viscous}= -kv^2 (k >0 )$,The terminal speed of the ball is 

Sixty four spherical rain drops of equal size are falling vertically through air with terminal velocity $1.5\, m/s$. All of the drops coalesce to form a big spherical drop, then terminal velocity of big drop is ……….. $m/s$