Fluid Mechanics

JEE Main level, single correct, medium

17 of 192

Question 17 of 192

A submarine experiences a pressure of 5.05×106Pa5.05\times {10}^{6}Pa at a depth of d1{d}_{1} in a sea. When it goes further to a depth of d2{d}_{2} , it experiences a pressure of 8.08×106Pa8.08\times {10}^{6}Pa . Then d2−d1{d}_{2}-{d}_{1} is approximately (density of water =103kg/m3={10}^{3}kg/{m}^{3} and acceleration due to gravity =10ms−2=10m{s}^{-2} ):

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Answer: (C) 300m300m

The pressure at a depth dd is p=p0+ρgdp={p}_{0}+\rho gd, here p0{p}_{0} is the atmospheric pressure.
∴p1=p0+ρgd1∴{p}_{1}={p}_{0}+\rho g{d}_{1}
and p2=p0+ρgd2{p}_{2}={p}_{0}+\rho g{d}_{2}
The difference in pressure is p2−p1=ρg(d2−d1){p}_{2}-{p}_{1}=\rho g\left({d}_{2}-{d}_{1}\right)
Putting the values in the above equation
8.08×106−5.05×106=103×10(d2−d1)8.08\times {10}^{6}-5.05\times {10}^{6}={10}^{3}\times 10\left({d}_{2}-{d}_{1}\right)
d2−d1=3.03×106104{d}_{2}-{d}_{1}=\dfrac{3.03\times {10}^{6}}{{10}^{4}}
=303m=303 m

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