The pressure at a depth of 200 m below the ocean surface (sea water density ≈ 1030 kg/m³, $g = 10$ m/s², $p_0 = 10^5$ Pa) is approximately:
A$2.06 \times 10^6$ Pa
B$2.16 \times 10^6$ Pa
C$1.03 \times 10^5$ Pa
D$1.0 \times 10^7$ Pa
Answer & Solution
Correct answer: B. $2.16 \times 10^6$ Pa
$p = p_0 + h\rho g = 10^5 + 200 \times 1030 \times 10 = 10^5 + 2.06\times10^6 = 2.16\times10^6$ Pa.
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