A submarine experiences a pressure of 5-05 - 106 Pa at 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 approximatelydensity of water -103 kg-m3 and acceleration due to gravity - 10 ms-2

Given: Pressure nbsp;at depth of (d_1), P_1 =5.05 × 10^6 Pa Pressure nbsp;at depth of (d_2), P_2=8.08 ×10^6 Pa. Density of water =10^3 kg/m^3 Now, absolute ...

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Standard XII

Physics

Accelerating Containers

A submarine e...

Question

A submarine experiences a pressure of $5.05 \times 10^{6} Pa$ at depth of $d_{1}$ in a sea. When it goes further to a depth of $d_{2}$ , it experiences a pressure of $8.08 \times 10^{6} Pa$ . Then $d_{2} - d_{1}$ is approximately

(density of water $= 10^{3} {kg/m}^{3}$ and acceleration due to gravity $= 10 {ms}^{- 2})$

400 m

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600 m

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500 m

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300 m

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Solution

The correct option is D $300 m$
Given:
Pressure at depth of $(d_{1}), P_{1} = 5.05 \times 10^{6} Pa$
Pressure at depth of $(d_{2}), P_{2} = 8.08 \times 10^{6} Pa$ .
Density of water $= 10^{3} {kg/m}^{3}$

Now, absolute pressure at some depth in sea,
$P_{1} = P_{0} + ρ g d_{1}$
$P_{2} = P_{0} + ρ g d_{2}$
Pressure difference,

$Δ P = P_{2} - P_{1} = ρ g (d_{2} - d_{1}) = ρ g Δ d$

$\Rightarrow 3.03 \times 10^{6} = 10^{3} \times 10 \times Δ d$

$\Rightarrow Δ d ≃ 300 m$

Hence option $(A)$ is correct.

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A submarine experiences a pressure of 5.05×106 Pa at 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)