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WAEC Physics 2015 Paper

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Question : 4

Total: 50

The barometric reading at a place is 73.5m Hg. Calculate the pressure at a point 30m below the surface of water contained in a reservoir at the place. [g = ms^-2, density = 1.2 x 10⁴kg m^-3, density of water = 1.0 x 10³m^-3]

4.0 x 10⁵ Nm^-2
3.0 x 10 ⁵ Nm^-2
2.0 x 10⁵ Nm^-2
1.0 x 10⁵ Nm^-2

Solution:

Concept:

At a depth in a liquid, the total pressure equals atmospheric pressure plus the pressure due to the liquid column above that point.

Explanation:

The barometer reading gives atmospheric pressure as

P_{\mathrm{atm}} \approx 1.0 \times 10^{5}\ \mathrm{N\,m^{-2}}

.

The pressure due to a

30\ \mathrm{m}

water column is:

P_{\mathrm{water}} = h\rho g = 30 \times (1.0 \times 10^{3}) \times 10 = 3.0 \times 10^{5}\ \mathrm{N\,m^{-2}}.

Add this to the atmospheric pressure to get the total pressure at the point:

P_{\mathrm{total}} = P_{\mathrm{atm}} + P_{\mathrm{water}} = 1.0 \times 10^{5} + 3.0 \times 10^{5} = 4.0 \times 10^{5}\ \mathrm{N\,m^{-2}}.

Answer:

Option A:

4.0 \times 10^{5}\ \mathrm{N\,m^{-2}}

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