An air bubble of radius $r$ in water is at depth $h$ below the water surface. If $P$ is the atmospheric pressure, and $d$ and $T$ are the density and surface tension of water respectively, then the pressure inside the bubble will be:

An ice cube of edge $1 \ cm$ melts in a gravity-free container. The approximate surface area of the water formed is (water is in the form of a spherical drop)

Consider an air bubble of radius $2 \,mm$ in a liquid at a depth of $5 \,cm$ below the free surface. The density of the liquid is $1000 \,kg/m^3$ and the surface tension is $0.1 \,N/m$. Find the pressure inside the air bubble relative to the pressure at the free surface of the liquid. (Take $g = 10 \,m/s^2$) (in $\,Pa$)

$64$ small drops of mercury,each of radius $r$ and charge $q$,coalesce to form a single large drop. What is the ratio of the surface charge density of a small drop to that of the large drop?

$A$ number of small water droplets of surface tension '$T$',each of radius '$r$',are combined to form a single drop of radius '$R$'. If the released energy is converted into kinetic energy,then the velocity acquired by the bigger drop is . . . . . . ($\rho$ - density of water)

The excess pressure inside the first soap bubble of radius $R_1$ is three times that inside the second soap bubble of radius $R_2$. The ratio of volumes of the first to second bubble is