What is the pressure on a swimmer $20 \,m$ below the surface of water ........ $atm$
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(d)
Let atmospheric pressure $= P _3=1.05 \times 10^5 \,Pa$
Pressure $20\,m$ below surface of lake
$P = P _0+\int gh$
$=1.05 \times 10^5+1000 \times 9.8 \times 20$
$=3.01 \times 10^5 Pa \left(\int \rightarrow \text { water density }=1000 kg / m ^3\right)$
$=3\,atm$
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