# Question 86931

Nov 22, 2016

When the hollow sphere with a small hole is immersed in water a thin film of water is formed at the mouth of the hole.This film acts as a membrane and prevents water to enter in the hollow sphere.

With increase in depth of immersion the pressure on this film goes to increase and slowly it acquires spherical shape and when a spherical droplet of diameter equal to the diameter of the hole is formed, water starts to enter into the cavity of hollow sphere. At this moment the excess pressure within the the spherical droplet of water becomes equal to the hyodrostatic pressure of water column at the depth of water that hole reaches.

The extra pressure within the droplet due to surface tension ${P}_{\text{st}} = \frac{2 \gamma}{r}$
Where
$\gamma \to \text{surface tension of water} = 0.07 \frac{N}{m}$

=(0.07xx10^5 "dyne")/(100cm)=70"dyne/cm"

r->"radius of the hole"=?#

The hydrostatic pressure at h depth of water

${P}_{\text{hy}} = h \times d \times g$

where
$h \to \text{depth of water} = 40 c m$

$d \to \text{density of water} = 1 g c {m}^{-} 3$

$g \to \text{acceleration due to gravity} = 980 c m {s}^{-} 2$

By the condition

${P}_{\text{st"=P_"hy}}$

$\implies \frac{2 \gamma}{r} = h \times d \times g$

$\implies r = \frac{2 \gamma}{h \times d \times g}$

$\implies r = \frac{2 \times 70}{40 \times 1 \times 980} c m \approx 3.57 \times {10}^{-} 3 c m$

So diameter of the hole $= 2 \times r = 7.14 \times {10}^{-} 3 c m$