A small hole made at bottom of a hollow sphere. THe water enters into it when it is taken to a depht of 40cm under water. If the surface tension of water is. 07N/m..diameter of tge hole is?

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Solution

A small hollow sphere which has a small hole in it is immersed in water to a depth of 40cm before any water is penetrated into it. if the surface tension of water is 0.073N/m, find radius of the hole. in this question why have we taken only the pressure due to liquid?the atmospheric pressure is also acting on the liquid.so it would had been 2T/r= atmospheric pressure + hydrostatic pressure.

Water ruches into the hollow sphere through the sphere and start to fill it. Now, an air bubble will be formed surrounding the hole when it is fully immersed in water.

Under equilibrium condition the pressure exerted by water would be equal to the excess pressure inside the bubble.

so,

excess pressure = hydrostatic pressure

or

2T/r = hρg

so, the radius of hole will be

r = 2T/hρg

here

T = 0.075 N/m

h = 40cm = 0.4 m

ρ = 1000 kg/m3

g = 9.81 m/s2

so, we have

r = 2 x 0.075 / (0.4 x 1000 x 9.81)

= 0.15 / 3924

thus, we have

r = 3.8 x 10-5m

So diameter = 2r

= 0.078 mm

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