Question

There is a hole of radius $r$ in a cylindrical glass pot. To what depth in the sea can it be immersed so that water may not enter it? (Surface tension of water is $T$)

• A. $\frac{2T}{r}$
• B. $\frac{2T}{rgd}$
• C. $\frac{T}{rgd}$
• D. $\frac{rgd}{2}$

Answer 1

The correct option is B $\frac{2T}{rgd}$

Pressure due to sea water at depth $h$ is $dgh$,
where $d$ is density of sea water
$g$ is acceleration due to gravity
$h$ is depth of sea water

so, force at point is $dgh\times \pi r^2$
$F_w=dgh\pi r^2$
and force due to surface tension is $F_T=2\pi rT\cos \theta$ (taking $\theta =0^o$ we can write $\cos \theta =1$)
= $2\pi rT$
So, to stop water from entering to pot we should have $F_w=F_T$
$dgh\pi r^2=2RrT$
$h=\frac{2T}{rgd}$
Hence, pot can be immersed upto $\frac{2T}{rgd}$ depth