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A submarine experiance 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 $8.08\times 10^{6}Pa$ .Then $d_{2}$ $-$ $d_{1}$ is approximately (density of water = $10^{3}Kg/m^{3}$ and acceleration due to gravity $=10ms^{-2}$ ):

Option 1)
300m

Option 2)
400m

Option 3)
600m

Option 4)
500m

Answers (1)

best_answer

Absolute Pressure -

$P= P_{0}+\rho\: gh$

- wherein

$P\rightarrow hydrostatics \: Pressure$

$P_{0}\rightarrow atmospheric \: Pressure$

$P_{0}+ \rho gd_{1}=P_{1}...........................1$

$P_{0}+ \rho gd_{2}=P_{2}...........................2$

$\Rightarrow 2................1$

$\rho g \left ( d_{2} -d_{1}\right )=P_{2}-P_{1}$

$\Rightarrow 1000\times 10\left ( d_{2}-d_{1} \right )=3.03\times 10^{6}$

$\Rightarrow d_{2}-d_{1} \right )=303m$

$\simeq 300m$

Option 1)

300m

Option 2)

400m

Option 3)

600m

Option 4)

500m

Posted by

Aadil

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