The pressure on a swimmer 10 m below the surface lake is:(Atmospheric pressure= $1.01 \times 10^{5}$ Pa,Density of water $= 1000 k g / m^{3}$ )

10 a t m

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5 a t m

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15 a t m

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2 a t m

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Solution

The correct option is D $2 a t m$
Given : $h = 10 m$ , $ρ = 1000 k g / m^{3}$ , $g = 10 m / s^{2}$
Pressure on swimmer $=$ pressure of atmosphere + pressure of water
$= 1 a t m + ρ g h \times 10^{- 5} a t m$
$= 1 a t m + 1000 \times 10 \times 10 \times 10^{- 5} a t m$
$= 2 a t m$

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10 m

g = 10 m s^{- 2}; 1 a t m = 1.01 \times 10^{5} P a

= 1000 k g / m^{3}

___ $P a$ is the pressure at a depth of $9 m$ below the surface of the water tank. The density of water is $1000 k g / m^{3}$ .

Calculate the depth at which pressure on a swimmer is 100 Pa. Given, density of water = 1000 kg/m³

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