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JEE Main 202116 MarchMorning Shift
Exam
JEE Main
Year
2021
Date
16 March
Shift
Morning Shift
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Main 2021 · 16 March · Morning Shift · Physics · Mechanical Properties of Fluids

The pressure acting on a submarine is 3 × 10 ^ 5 Pa at a certain...

The pressure acting on a submarine is 3×105Pa3 \times {10}^{5}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 1×105Pa1 \times {10}^{5}Pa density of water is 103kgm−3,g=10ms−2{10}^{3}kg{m}^{- 3}, g = 10{ms}^{- 2})

Options

  1. A2003\dfrac{200}{3} %
  2. B2005\dfrac{200}{5} %
  3. C5200\dfrac{5}{200} %
  4. D3200\dfrac{3}{200} %

Correct answer

A. 2003\dfrac{200}{3} %

Step-by-step solution

P1=ρgd+P0=3×105Pa{P}_{1} = ρgd + {P}_{0} = 3 \times {10}^{5}Pa

∴ρgd=2×105Pa∴ ρgd = 2 \times {10}^{5}Pa

P2=2ρgd+P0{P}_{2} = 2ρgd + {P}_{0}

=4×105+105=5×105Pa= 4 \times {10}^{5} + {10}^{5} = 5 \times {10}^{5}Pa

% increase =P2−P1P1×100= \dfrac{{P}_{2} - {P}_{1}}{{P}_{1}} \times 100

=5×105−3×1053×105×100=2003= \dfrac{5 \times {10}^{5} - 3 \times {10}^{5}}{3 \times {10}^{5}} \times 100 = \dfrac{200}{3} %

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