A small hollow vessel which has a small hole in it is immersed in water to a depth of $40$ cm before water enters into the vessel. Calculate the radius of the hole. [Surface tension of water $= 70 \times 10^{- 3}$ N/m, density of water $= 10^{3}$ kg $/ m^{3}$ ]

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Solution

When water starts entering and forms a hemispherical surface, the pressure inside hemisphere $= p = 2 \frac{s u r f a c e t e n s i o n (T)}{r a d i u s (r)}$

But pressure at depth $= p = d e p t h (h) \times d e n s i t y (d) \times a c c e l e r a t i o n d u e t o g r a v i t y (g)$

$\frac{2 T}{r} = h d g$

$r = \frac{2 T}{h d g} = \frac{2 \times 70 \times 10^{- 3}}{10^{3} \times 0.4 \times 9.8} = 3.571 \times 10^{- 5} m$

the radius of the hole = $3.571 \times 10^{- 5} m$ or $0.03571 m m$

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A small hollow vessel which has a small hole in it is immersed in water to a depth of 40cm before water enters into the vessel. Calculate the radius of the hole. [Surface tension of water =70×10−3N/m, density of water =103 kg/m3]

A small hollow vessel which has a small hole in it is immersed in water to a depth of $40$ cm before water enters into the vessel. Calculate the radius of the hole. [Surface tension of water $= 70 \times 10^{- 3}$ N/m, density of water $= 10^{3}$ kg $/ m^{3}$ ]