Verified Solution • 2025–26 Session moderate

QJEEM21Ph8b60: JEE Main 2021 Physics - Class 11 JEEM21Ph8b60 Solution: The pressure acting on a submarine is 3 10 5 Pa at a certain…

The pressure acting on a submarine is 3 ×\times 105 Pa at a certain depth. If the depth is doubled, the percentage increase in the pressure acting on the submarine would be :

(Assume that atmospheric pressure is 1 ×\times 105 Pa density of water is 103 kg m −- 3, g = 10 ms −- 2)

Multiple Choice Options

  • A 2005{{200} \over 5} %
  • B 2003{{200} \over 3} %
  • C 3200{{3} \over 200} %
  • D 5200{{5} \over 200} %

Step-by-Step Verified Solution

P = P0 + h ρ\rho g = 3 ×\times 105 Pa

⇒\Rightarrow h ρ\rho g = 3 ×\times 105 −- 1 ×\times 105

⇒\Rightarrow h ρ\rho g = 2 ×\times 105

∴\therefore 2h ρ\rho g = 4 ×\times 105

∴\therefore P' = P0 + 4 ×\times 105

∴\therefore P' = 5 ×\times 105 Pa

∴\therefore % increase in pressure = P′−PP×100{{P' - P} \over P} \times 100

=(5−3)×1053×105×100= {{(5 - 3) \times {{10}^5}} \over {3 \times {{10}^5}}} \times 100

=2003= {{200} \over 3} %

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