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)

 

 

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

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