Why bubbles rise in soda water bottle ?

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 $500 / 9$.

If a ball of steel (density $\rho=7.8 \;gcm ^{-3}$) attains a terminal velocity of $10 \;cms ^{-1}$ when falling in a tank of water (coefficient of viscosity $\eta_{\text {water }}=8.5 \times 10^{-4} \;Pa - s$ ) then its terminal velocity in glycerine $\left(\rho=12 gcm ^{-3}, \eta=13.2\right)$ would be nearly

A steel ball is dropped in a viscous liquid. The distance of the steel ball from the top of the liquid is shown below. The terminal velocity of the ball is closest to .......... $m/s$

A small metal sphere of radius $a$ is falling with a velocity $v$ through a vertical column of a viscous liquid. If the coefficient of viscosity of the liquid is $\eta $ , then the sphere encounters an opposing force of

Small water droplets of radius $0.01 \mathrm{~mm}$ are formed in the upper atmosphere and falling with a terminal velocity of $10 \mathrm{~cm} / \mathrm{s}$. Due to condensation, if $8 \mathrm{such}$ droplets are coalesced and formed a larger drop, the new terminal velocity will be ........... $\mathrm{cm} / \mathrm{s}$.