Physics · Mechanical Properties Of Fluids · NEET
It depends on what the question gives. The total (absolute) pressure pushing on the hull is P = P0 + rho g h. This includes the atmosphere pressing on the sea surface plus the water column above the submarine. If the question says the hull withstands an absolute pressure of 100 atm, you must subtract the 1 atm of atmosphere before finding depth: rho g h = P - P0 = 100 atm - 1 atm = 99 atm. Forgetting to subtract P0 is the most common mistake.
Fluid pressure depends only on depth h, density rho, and g, not on the total amount of water. From P = P0 + rho g h, a submarine 100 m deep in a small deep lake feels the same water pressure as one 100 m deep in the ocean. Pressure at a point is caused by the height of the fluid column directly above it, so a wide sea and a narrow pipe at the same depth give the same pressure.
Put the numbers in: rho g h = 1000 x 10 x 10 = 100000 Pa = 1 x 10^5 Pa, which is about 1 atm. So each 10 m of water adds roughly one atmosphere. At 990 m the water alone adds about 99 atm, and with the 1 atm atmosphere on top the total is 100 atm.
No. The bottom of the submarine is deeper than the top, so it feels slightly more pressure. But because a submarine is small compared with its depth, we usually treat the pressure as the value at its depth h. Fluid pressure at a given depth acts equally in all directions, so it pushes inward on the sides too, not just downward.
It means the hull can survive a total outside pressure of 100 atm before it is crushed. You set the maximum allowed absolute pressure equal to P0 + rho g h and solve for h. That h is the deepest the submarine can safely go.
A submarine is designed to withstand an absolute pressure of 100 atm. How deep can it go below the water surface? (density of water = 1000 kg/m3, 1 atm = 1 x 10^5 Pa, g = 10 m/s2)
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The absolute pressure is P = P0 + rho g h, where P0 is atmospheric pressure, rho is the water density, g is gravity, and h is the depth. This total pressure pushes inward on the hull.
Because the hull limit of 100 atm is absolute pressure. The atmosphere already supplies 1 atm at the surface, so only 99 atm is available for the water column. 99 atm of water equals 990 m, not 1000 m.
Yes, real sea water is denser (about 1030 kg/m3), so the same pressure is reached at a slightly smaller depth. NEET problems usually give you the density to use, so always plug in the value stated in the question.
Yes. rho g h = 1000 x 10 x 10 = 1 x 10^5 Pa, which is about 1 atm. This is a useful check: divide the extra pressure in atm by 1 and multiply by 10 to estimate depth in metres.
