Question

if pressure at the bottom of the vessel is n 10^5 Pa, then the value of n is (given atmospheric pressure = 1*10^5 Pa take g = 10ms-2)​

Answer 1

If the pressure at the bottom of the vessel is n 10 ^ 5 Pa, the value of n is:

• Mercury concentration, S = 4.65 × 10-1 N m-1.
• Radius of mercury drop, r = 3.00 mm = 3 × 10-3 m.
• Atmospheric pressure, P 0 = 1.01 × 10 5 Pa.
• We know: Total pressure inside mercury drop = Extra pressure inside mercury + Atmospheric pressure.
• Initially, the surface water is atmospheric pressure (1.01 × 10 5 P a) and the underwater pressure gauge is 2500 Pa.
• Then more air is added, which increases the air pressure above the water by 1500 Pa.
• Seawater density is 1.03 X 10 3 kg / m 3 and atmospheric pressure is 1.01 x 10 5 N / m 2.
• P liquid = r g h = (1.03 x10 3 kg / m 3) (9.8 m / s 2) (12 m) = 1.21 x 10 5 Newtons / m 2.
• P total = P atmosphere + P liquid = (1.01 x 10 5) + (1.21 x 10 5) Pa = 2.22 x 10 2 kPa (kilo Pascals).
• The atmosphere was originally a unit related to air pressure at sea level. It was later described as 1.01325 x 10 5 Pa. atm to Pa Problem.
• The pressure under the sea rises by about 0.1 atm meters.