$8$ mercury drops coalesce to form one mercury drop. The energy changes by a factor of:

The excess pressure inside the first soap bubble of radius $R_{1}$ is two times that inside the second soap bubble of radius $R_{2}$. The ratio of the volumes of the first bubble to that of the second bubble is:

If two soap bubbles each of radius $2 \ cm$ combine in vacuum under isothermal conditions,then the radius of the new bubble formed is

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)

$A$ capillary tube of radius $r$ is dipped in a liquid of density $\rho$ and surface tension $S$. If the angle of contact is $\theta$,what is the pressure difference between the two surfaces in the beaker and the capillary?

What is the excess pressure inside a bubble of soap solution of radius $5.00 \; mm$,given that the surface tension of soap solution at the temperature $(20 \; ^{\circ}C)$ is $2.50 \times 10^{-2} \; N m^{-1}$? If an air bubble of the same dimension were formed at a depth of $40.0 \; cm$ inside a container containing the soap solution (of relative density $1.20$),what would be the pressure inside the bubble? ($1$ atmospheric pressure is $1.01 \times 10^{5} \; Pa$).