A submarine experience a pressure of 5-05 - 106 Pa at a depth of d1 in a sea- When it goes further to a depth of d2- it experiences a pressure of 8-08 - 106 Pa- Then d2 - d1 is approximately density of water - 103 kg-m3 and acceleration due to gravity - 10 ms-2

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Applying NLM along 3 Axis

A submarine e...

Question

A submarine experience a pressure of $5.05 \times 10^{6}$ Pa at a depth of $d_{1}$ in a sea. When it goes further to a depth of $d_{2}$ , it experiences a pressure of $8.08 \times 10^{6} P a$ ., Then $d_{2} - d_{1}$ is approximately (density of water = $10^{3} k g / m^{3}$ and acceleration due to gravity = $10 m s^{- 2}$ )

500 m

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400 m

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300 m

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600 m

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Solution

The correct option is C $300 m$
$P_{0} + ρ g d_{1} = P_{1}$
$P_{0} + ρ g d_{2} = P_{2}$
$ρ g (d_{2} - d_{1}) = P_{2} - P_{1}$
$10^{3} \times 10 (d_{2} - d_{1}) = 3.03 \times 10^{6}$
$d_{2} - d_{1} = 303 m$
$≃ 300 m$

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A submarine experience a pressure of 5.05×106 Pa at a depth of d1 in a sea. When it goes further to a depth of d2, it experiences a pressure of 8.08×106Pa., Then d2−d1 is approximately (density of water = 103kg/m3 and acceleration due to gravity = 10ms−2)

The correct option is C 300mP0+ρgd1=P1P0+ρgd2=P2ρg(d2−d1)=P2−P1103×10(d2−d1)=3.03×106d2−d1=303m≃300m

The correct option is C $300 m$
$P_{0} + ρ g d_{1} = P_{1}$
$P_{0} + ρ g d_{2} = P_{2}$
$ρ g (d_{2} - d_{1}) = P_{2} - P_{1}$
$10^{3} \times 10 (d_{2} - d_{1}) = 3.03 \times 10^{6}$
$d_{2} - d_{1} = 303 m$
$≃ 300 m$