At a depth of 1000 m in the ocean ,what is the gauge pressure when density of sea water is given $1.03 \times 10^{3}$ $k g m^{- 3}$ ? Take $g = 10 m s^{- 2} .$

675 atm

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103 atm

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203 atm

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244 atm

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Solution

The correct option is B 103 atm
Given : $h = 1000$ m, $ρ = 1030$ $k g / m^{3}$
Gauge pressure at a depth $h$ , $P_{g} = \frac{ρ g h}{10^{5}}$ atm
$∴$ $P_{g} = \frac{1030 \times 10 \times 1000}{10^{5}} = 103$ atm

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