At what depth below the sea-level would one experience a change of pressure equal to one atmosphere? Density of sea water = 1013kg-3, one atmosphere = 0.01 x 105Nm-2, g = 10ms-2
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Correct Answer: Option C
Explanation:
H = \(\frac{P}{e_g}\)
H = \(\frac{1.01 \times 10^5}{1013 \times 10}\)
H = 9.97
H = \(\frac{P}{e_g}\)
H = \(\frac{1.01 \times 10^5}{1013 \times 10}\)
H = 9.97