A hollow sphere which has a small hole in it is immersed in water to a depth of 40 cm before any water penetrates into it
Answers
radius of hole is 0.00375 cm
your full question is -> A canister has a small hole at its bottom. Water penetrates into the cylinder when its base is at a depth of 40 cm from the surface of water. If surface tension of water is 73.5 dyne/cm, find the radius of the hole: a) 0.00375 cm b) 0.375 cm c) 3.75 cm d) 37.5 cm
solution : at equilibrium,
excess pressure = hydraulic pressure
⇒2T/r = hρg
where, T is surface tension, r is radius of hole , h is depth , ρ is density and g is acceleration due to gravity.
given, T = 73.5 Dyne/cm , h = 40cm, g = 980 cm/s² and ρ = 1g/cm³
then, 2 × 73.5/r = 40 × 1 × 980
r = 147/(40 × 980) = 0.00375 cm
hence, radius of hole is 0.00375 cm.
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