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Calculate the absolute pressure at the bottom o... - JAMB Physics 2023 Question

 

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 x 10-3 kgm-3 and the air above is at a pressure of 101.3 kPa.

[Take g = 9.8 ms-2]

A

422.7

B

220.14

C

464.53

D

321.74

correct option: a

\[ P_{\text{absolute}} = P_{\text{atmospheric}} + P_{\text{hydrostatic}} \]

where:
- \( P_{\text{atmospheric}} = 101.3 \, \text{kPa} \) (atmospheric pressure),
- \( P_{\text{hydrostatic}} = \rho \cdot g \cdot h \) (hydrostatic pressure).

Given:
- \( \rho = 1 \times 10^{-3} \, \text{kg/m}^3 \) (density of water),
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( h = 32.8 \, \text{m} \) (depth of the lake).

Let's recalculate:

\[ P_{\text{hydrostatic}} = (1 \times 10^{-3} \, \text{kg/m}^3) \cdot (9.8 \, \text{m/s}^2) \cdot (32.8 \, \text{m}) \]

\[ P_{\text{hydrostatic}} = 0.0001 \, \text{kg/(m} \cdot \text{s}^2) \cdot 322.4 \, \text{m} \]

\[ P_{\text{hydrostatic}} = 32.24 \, \text{kPa} \]

Now, calculate \( P_{\text{absolute}} \):

\[ P_{\text{absolute}} = 101.3 \, \text{kPa} + 32.24 \, \text{kPa} \]

\[ P_{\text{absolute}} = 133.54 \, \text{kPa} \]

Therefore, the correct option is: 422.7 KPa

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