P1=ρgd+P0=3×105Pa
∴ρgd=2×105Pa
P2=2ρgd+P0
=4×105+105=5×105Pa
increase =P1P2−P1×100
=3×1055×105−3×105×100=3200
The pressure acting on a submarine is 3×105Pa 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 1×105Pa density of water is 103kgm−3,g=10ms−2)
Held on 16 Mar 2021 · Verified 6 Jul 2026.
3200
5200
2005
2003
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