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JEE Main 2019 (10 Apr, Shift 2) · Pressure in a Fluid

Pressure in a Fluid question from JEE Main 2019 (10 Apr, Shift 2)

Q21JEE Main 2019 · 10 APR · SHIFT 2Medium

Pressure in a Fluid

A submarine experiences a pressure of 5.05 × 106 Pa at a depth of d1 in a sea. When it goes further to a depth of d2, it experiences a pressure of 8.08 × 106 Pa. Then d2 –d1 is approximately (density of water = 103 kg/m3 and acceleration due to gravity = 10 ms–2):

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Answer: (c)

  1. P2−P1=ρg(d2−d1)P_2 - P_1 = \rho g (d_2 - d_1)
  2. d2−d1=(8.08−5.05)×106103×10=303 md_2 - d_1 = \dfrac{(8.08 - 5.05)\times10^6}{10^3\times10} = 303\text{ m}
  3. So the answer is (c) about 300 m.
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