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JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Afternoon)

  • question_answer
    A submarine experiences 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 \[=10m{{s}^{-2}}\]) [JEE Main 10-4-2019 Afternoon]

    A) 500 m             

    B) 400 m

    C) 300 m

    D) 600 m

    Correct Answer: C

    Solution :

                \[{{P}_{0}}+\rho g{{d}_{1}}={{P}_{1}}\]           \[{{P}_{0}}+\rho g{{d}_{2}}={{P}_{2}}\]           \[\rho g({{d}_{2}}-{{d}_{1}})={{P}_{2}}-{{P}_{1}}\]           \[{{10}^{3}}\times 10\left( {{d}_{2}}{{d}_{1}} \right)=3.03\times {{10}^{6}}\] \[{{d}_{2}}{{d}_{1}}=303\text{ }m\] \[\simeq 300m\]          


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