Question

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 = hrhog

so, the radius of hole will be

r = 2T/hrhog

here

T = 0.075 N/m

h = 40cm = 0.4 m

rho = 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 -5 m

Answer 1

The radius of the hole is  r = 3.8 x 10-5 m

Explanation:

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

Excess pressure = Hydro-static pressure

or

2T/r = hρg

The radius of hole will be

r = 2T/hρg

Here

T = 0.073 N/m

h = 40 cm = 0.4 m

ρ = 1000 kg/m^3

g = 9.81 m/s^2

We have

r = 2 x 0.073 / (0.4 x 1000 x 9.81)   = 0.15 / 3924  

r = 3.8 x 10-5 m

Thus the radius of the hole is  r = 3.8 x 10-5 m

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