There is a small hole in a hollow sphere. Wat | vedclass

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(B) The pressure due to the depth of water is given by $P = h \rho g$.For water to enter the hole,the pressure exerted by the water must overcome the

Vedclass · Accepted Answer

$0.07$

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(B) The pressure due to the depth of water is given by $P = h ho g$.For water to enter the hole,the pressure exerted by the water must overcome the capillary pressure (excess pressure) at the hole,given by $P = \frac{2T}{r}$.Equating the two,we have $h ho g = \frac{2T}{r}$.Solving for the radius $r$,we get $r = \frac{2T}{h ho g}$.The diameter $d$ of the hole is $d = 2r = \frac{4T}{h ho g}$.Given: $h = 40 \, cm = 0.4 \, m$,$T = 0.07 \, N/m$,$ho = 10^3 \, kg/m^3$,and $g = 9.8 \, m/s^2$.Substituting the values: $d = \frac{4 	imes 0.07}{0.4 	imes 10^3 	imes 9.8} = \frac{0.28}{3920} = 7.14 	imes 10^{-5} \, m \approx 0.07 \, mm$ (using $g \approx 10 \, m/s^2$ gives $d = \frac{0.28}{4000} = 0.07 	imes 10^{-3} \, m = 0.07 \, mm$).

There is a small hole in a hollow sphere. Water enters the sphere when it is taken to a depth of $40 \, cm$ in water. The diameter of the hole is ....... $mm$ (Surface tension of water $= 0.07 \, N/m$):

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