At what depth below the surface of water will pressure be equal to twice the atmospheric pressure? The atmospheric pressure is 10 N/m^2, density of water = 10^3 kg/m^3, g = 9.8 m/s.

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Solution

$P r e s s u r e a t t h e d e p t h h b e l o w t h e s u r f a c e o f t h e w a t e r = A t m o s p h e r i c p r e s s u r e + P r e s s u r e d u e t o w a t e r c o l u m n o f h e i g h t h I f t h e P r e s s u r e a t t h e d e p t h h b e l o w t h e s u r f a c e o f t h e w a t e r = 2 \times A t m o s p h e r i c p r e s s u r e t h e n P = 2 P_{0} S o, P = P_{0} + h ρ g 2 P_{0} = P_{0} + h ρ g P_{0} = h ρ g 10 = h \times 10^{3} \times 9.8 h = \frac{10}{10^{3} \times 9.8} = 0.001 m$

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