9-1.Fluid Mechanics
normal

What is the pressure on a swimmer $20 \,m$ below the surface of water is ..... $atm$

A

$1$

B

$2$

C

$3$

D

$4$

Solution

(c)

Let atmospheric pressure $= P _3=1.05 \times 10^5\,Pa$

Pressure $20\, m$ below surface of lake

$P = P _0+\rho gh$

$=1.05 \times 10^5+1000 \times 9.8 \times 20$

$=3.01 \times 10^5 Pa \left(\rho \rightarrow \text { water density }=1000 kg / m ^3\right)$

$=3\,atm$

Standard 11
Physics

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