physics TBR

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Jun 2, 2014
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At a specific depth in a swimming pool, a barometer measures the total pressure to be twice that of atmospheric pressure. if the barometer is now submerged to a depth that is twice its initial depth, by how much does the total pressure increase?
A by 50%
B by 100%
C by 200%
D by 300%

Answer is A

Im not sure because k so P(Total)= P(atm) + P(gauge)

here we know that P(Total)= 2P(atm) and this is at the inital specfic depth

Then final pressure we have an increase in depth by 2 and so P(final)= 2P(atm) + 2P(gauge)

and now Im lost :(

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Ptotal=Patm + pgh. If twice deeper Ptotal*=Patm+2pgh. since the initial pressure contributed by water is P(atm). The final pressure contributed by water is just 2P(atm) . So the final total pressure would just be 3Patm. The pressure contributed by atmosphere in both cases does not change. The overall increase is just 1P(atm)/
so 1/2=50% increase.