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 ........ $m$ (density of water $= 10^3\,kg/m^3$ and acceleration due to gravity $= 10\,ms^{-2}$ )
JEE MAIN 2019, Medium
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${P_1} = 5.05 \times {10^6}\,;\,{P_2} = 8.08 \times {10^6}$
${P_2} - {P_1} = \rho g\left( {{d_2} - {d_1}} \right)$
${d_2} - {d_1} = \frac{{3.03 \times {{10}^6}}}{{{{10}^3} \times 10}} = 3.03 \times {10^2} = 303$
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