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JAMB Physics 2023 Paper

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Question : 6

Total: 65

Calculate the absolute pressure at the bottom of a lake at a depth of 32.8 m. Assume the density of the water is

1 \times 10^{3}\ \mathrm{kg\,m^{-3}}

and the air above is at a pressure of 101.3 kPa.

g = 9.8\ \mathrm{m\,s^{-2}}

]

422.7
220.14
464.53
321.74

Solution:

Concept:

Absolute pressure at the bottom is the sum of atmospheric pressure and the gauge pressure due to the water depth.

Explanation:

Gauge pressure is given by

P = h \rho g

.

Substitute

h = 32.8\ \mathrm{m}

,

\rho = 1 \times 10^{3}\ \mathrm{kg\,m^{-3}}

, and

g = 9.8\ \mathrm{m\,s^{-2}}

:

P = 32.8 \times 1 \times 10^{3} \times 9.8

P = 321440\ \mathrm{Pa} = 321.44\ \mathrm{kPa}

Atmospheric pressure is

P_a = 101.3\ \mathrm{kPa}

.

Absolute pressure:

P_T = P + P_a = 321.44 + 101.3

P_T = 422.74\ \mathrm{kPa} \approx 422.7\ \mathrm{kPa}

Answer:

422.7\ \mathrm{kPa}

(Option A).

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