The density of sea water is \(1030 kg / m ^{3}\). What is the pressure at a depth of \(80 m\) below sea surface? [Atmospheric pressure is \(1.013 \times 10^{5} Pa\) and acceleration due to gravity is ____________ \(\left.10 m / s ^{2}\right]\).
A. \(9.25 \times 10^{5} Pa\) B. \(8.24 \times 10^{5} Pa\) C. \(7.23 \times 10^{5} Pa\) D. \(8.34 \times 10^{9} Pa\)
Correct Answer: A
Explanation
Given: \(\rho_{s}=1030 kg / m ^{3}, h=80 m\), \(\rho_{\text {atm }}=1.013 \times 105 Pa , g =10 m / s ^{2}\) required pressure \(P =\) ? correlating equation \(P=P_{\text {atm }}+\rho g h\) \(P=1.013 \times 10^{5}+1030 \times 10 \times 80\) \(=1.013 \times 10^{5}+824000\) \(P=1.013 \times 10^{5}+8.24 \times 10^{5}\) \(=9.25 \times 105 Pa\) or \(N / m ^{2}\)
