Question

There is a small hole in a hollow sphere. When it is taken to a depth of $20\, cm$ under water, water enters in it. If the surface tension of water is $0.07\, N / m$, the radius of the hole in $\mu m$ is______.
(Take $g=10\, m / s ^{2}$ )

Answer 1

When water enters the hollow sphere, force of surface tension is balanced by normal force exerted by the liquid column,
i.e., $T \times 2 \pi r =\pi r ^{2} h \rho g$
where, $r$ is radius of hole of the sphere.
$\therefore r=\frac{2 T }{ h \rho g}=\frac{2 \times 0.07}{20 \times 10^{-2} \times 10^{3} \times 10}$
$=70 \times 10^{-6} m$
$\therefore r=70\, \mu m$