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(A) The absolute pressure $P$ at a depth $h$ is given by the formula: $P = P_{atm} + \rho gh$.Given values are: $h = 1 \,km = 1000 \,m$,$\rho = 10^{3}

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$1.011 	imes 10^{7} \,N/m^{2}$

Vedclass · Answer

(A) The absolute pressure $P$ at a depth $h$ is given by the formula: $P = P_{atm} + ho gh$.Given values are: $h = 1 \,km = 1000 \,m$,$ho = 10^{3} \,kg/m^{3}$,$g = 10 \,m/s^{2}$,and $P_{atm} = 1.01 	imes 10^{5} \,N/m^{2}$.Calculating the gauge pressure (pressure due to water column): $P_{gauge} = ho gh = 10^{3} 	imes 10 	imes 1000 = 10^{7} \,N/m^{2}$.Now,adding the atmospheric pressure: $P = 1.01 	imes 10^{5} + 10^{7} = 0.0101 	imes 10^{7} + 10^{7} = 1.0101 	imes 10^{7} \,N/m^{2}$.Rounding to the appropriate significant figures,we get $P \approx 1.011 	imes 10^{7} \,N/m^{2}$.

What would be the absolute pressure at a depth of $1 \,km$ below the ocean surface? [Given: density of water $\rho = 10^{3} \,kg/m^{3}$,$g = 10 \,m/s^{2}$,$1 \,atm = 1.01 \times 10^{5} \,N/m^{2}$]

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