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

300 atm

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210 atm

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104 atm

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205 atm

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Solution

The correct option is B 104 atm
Given : $h = 1000$ m $ρ = 1030$ $k g / m^{3}$ $P_{o} = 1$ atm
Pressure at a depth $h$ , $P = P_{o} + \frac{ρ g h}{10^{5}}$ atm
$∴$ $P = 1 + \frac{1030 \times 10 \times 1000}{10^{5}} = 104$ atm

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