A submarine experiences 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 m-s2

Name: CENGA - Grade 11 - Physics - Mechanical properties of Fluids - Q291
Uploaded: 2022-07-04T10:36:42+05:30
Description: We know that nbsp; P=P_0+ρ gh Given, pressure at nbsp;d_1 depth P_1=5.05×10^6 Pa P_0+ρ gd_1=P_1=5.05×10^6 Pa And pressure at nbsp;d_2 depth P_2=8.08 ...

We know that nbsp; P=P_0+ρ gh Given, pressure at nbsp;d_1 depth P_1=5.05×10^6 Pa P_0+ρ gd_1=P_1=5.05×10^6 Pa And pressure at nbsp;d_2 depth P_2=8.08 ...

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Standard IX

Physics

Pressure at a Depth in a Liquid

A submarine e...

Question

A submarine experiences a pressure of $5.05 \times 10^{6} P a$ 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 $= 10 m / s^{2}$ )

400 m

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300 m

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600 m

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500 m

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Solution

The correct option is B $300 m$
We know that

$P = P_{0} + ρ g h$

Given, pressure at $d_{1}$ depth

$P_{1} = 5.05 \times 10^{6} P a$

$P_{0} + ρ g d_{1} = P_{1} = 5.05 \times 10^{6} P a$

And pressure at $d_{2}$ depth

$P_{2} = 8.08 \times 10^{6} P a$

$P_{0} + ρ g d_{2} = P_{2} = 8.08 \times 10^{6} P a$

So, $ρ g (d_{2} - d_{1}) = P_{2} - P_{1}$
$10^{3} \times 10 (d_{2} - d_{1})$

$= (8.08 \times 10^{6} - 5.05 \times 10^{6})$

$= 3.03 \times 10^{6}$

$d_{2} - d_{1} = 303 m \approx 300 m$

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Pressure at a Depth in a Liquid

Standard IX Physics

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A submarine experiences 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 m/s2)

The correct option is B 300 mWe know that P=P0+ρgh Given, pressure at d1 depth P1=5.05×106 Pa P0+ρgd1=P1=5.05×106 Pa And pressure at d2 depth P2=8.08×106 Pa P0+ρgd2=P2=8.08×106 Pa So, ρg(d2−d1)=P2−P1 103×10(d2−d1) =(8.08×106−5.05×106) =3.03×106 d2−d1=303 m≈300 m