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  5. Fluid Mechanics, Surface Tension & Viscosity / द्रव यांत्रिकी, भूतल तनाव और चिपचिपापन

JEE Main & Advanced Physics Fluid Mechanics, Surface Tension & Viscosity / द्रव यांत्रिकी, भूतल तनाव और चिपचिपापन JEE PYQ-Fluid Mechanics, Surface Tension and Viscosity

  • 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 :

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


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