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Q.

The pressure acting on a submarine is  3×105 Pa at a certain depth. If the depth is doubled, the percentage increase in the pressure acting on the submarine would be:(Assume that atmosphere pressure is 1×105  Pa; density of water is 103kgm−3 , g = 10ms−2 )

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a

5200%

b

2005%

c

3200%

d

2003%

answer is D.

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Detailed Solution

Po+ρgh=3×105 PaIf depth is doubled,new pressure on submarine will be,⇒P'=P0+ρg2h=105+4×105=5×105Pa⇒ ΔPP×100=P'-PP×100=5-33×100=2003%⇒ρgh=2×105 Pa
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