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

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Solution

Since pressure = $h ϱ g = \frac{2 T}{R}$
$R = \frac{2 T}{h ϱ g} = \frac{2 \times 70 \times 10^{- 3}}{0.4 \times 10^{3} \times 10}$
Radius of hole R = $35 \times 10^{- 6}$
R = 0.035 mm

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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 has the vessel. Calculate the radius of the hole. [Surface tension of water =70×10−3N/m, density of water =103kg/m3].

Since pressure = hϱg=2TR R=2Thϱg=2×70×10−30.4×103×10Radius of hole R = 35×10−6 R = 0.035 mm

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

Since pressure = $h ϱ g = \frac{2 T}{R}$
$R = \frac{2 T}{h ϱ g} = \frac{2 \times 70 \times 10^{- 3}}{0.4 \times 10^{3} \times 10}$
Radius of hole R = $35 \times 10^{- 6}$
R = 0.035 mm