JEE Mains · Physics · STD 11 - 9.1 fluid mechanics
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 \(= 10\,ms^{-2}\) )
- A \(400\)
- B \(500\)
- C \(600\)
- D \(303\)
Answer & Solution
Correct Answer
(D) \(303\)
Step-by-step Solution
Detailed explanation
\({P_1} = 5.05 \times {10^6}\,;\,{P_2} = 8.08 \times {10^6}\) \({P_2} - {P_1} = \rho g\left( {{d_2} - {d_1}} \right)\) \({d_2} - {d_1} = \frac{{3.03 \times {{10}^6}}}{{{{10}^3} \times 10}} = 3.03 \times {10^2} = 303\)
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