A long cylindrical glass vessel has a pinhole of diameter \(0.2 \mathrm{~mm}\) at its bottom. The depth to…
A long cylindrical glass vessel has a pinhole of diameter \(0.2 \mathrm{~mm}\) at its bottom. The depth to which the vessel can be lowered vertically in a deep water bath without the water entering into the vessel is (surface tension of water, \(T=0.07 \mathrm{Nm}^{-1}\), acceleration due to gravity, \(g=10 \mathrm{~ms}^{-2}\) )
\(14 \mathrm{~cm}\)
\(7 \mathrm{~cm}\)
\(21 \mathrm{~cm}\)
\(28 \mathrm{~cm}\)
Solution
Given, diameter of a pinhole, \(d=0.2 \mathrm{~mm}\)
\(\therefore\) radius of pinhole, \(r=\frac{0.2}{2}=0 \mathrm{l} \mathrm{mm}=0 . \mathrm{I} \times 10^{-3} \mathrm{~m}\)
When the hydrostatic pressure is equal to the excess pressure, then water cannot enter the pinhole
Hence, \(h \rho g=\frac{2 T}{r}\)
where, \(h=\) depth up to which cylinder is immersed.
\(\rho=\) density of water
\(T=\) surface tension of water
and \(r=\) radius of a pinhole
\(\begin{aligned}
\therefore \quad h & =\frac{2 T}{\rho g r}=\frac{2 \times 0.07}{10^3 \times 10 \times 0.1 \times 10^{-3}} \\
& =0.14 \mathrm{~m}=14 \mathrm{~cm}
\end{aligned}\)
Hence, the depth to which the vessel can be lowered vertically in a deep water bath without the water entering into the vessel is \(14 \mathrm{~cm}\).