There is a small hole in a hollow sphere. Water enters the sphere when it is taken to a depth of $40 \, cm$ in water. The diameter of the hole is ....... $mm$ (Surface tension of water $= 0.07 \, N/m$):

$A$ large number of liquid drops each of radius $r$ coalesce to form a single drop of radius $R$. The energy released in the process is converted into kinetic energy of the big drop so formed. The speed of the big drop is (given,surface tension of liquid $T$,density $\rho$):

$A$ soap bubble $(S.T = 30\, dyne/cm)$ has a radius of $1\, cm$. The work done in doubling its radius would be ........ $ergs$.

When a big drop of water is formed from $n$ small drops of water, the energy loss is $3E$, where $E$ is the energy of the bigger drop. If $R$ is the radius of the bigger drop and $r$ is the radius of the smaller drop, then the number of smaller drops $(n)$ is:

Consider a water droplet of diameter $0.2 \,mm$ where the outside pressure is $1.5 \,N / cm^2$ at $25^{\circ} C$. The pressure inside the droplet, when the surface tension at $25^{\circ} C$ is $0.08 \,N / m$ is

The work done in splitting a drop of water of $1\, mm$ radius into $10^6$ droplets is (surface tension of water $T = 72 \times 10^{-3}\, N/m$):