Fluid Mechanics

JEE Main level, single correct, medium

13 of 192

Question 13 of 192

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})

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Answer: (A) 2003\dfrac{200}{3}%

P1=ρgd+P0=3×105Pa{P}_{1}=\rho gd+{P}_{0}=3\times {10}^{5}Pa
∴ρgd=2×105Pa∴\rho gd=2\times {10}^{5}Pa
P2=2ρgd+P0{P}_{2}=2\rho gd+{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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