Two drivers G and H are at depths 20m and 40m, respectively below the water surface in a lake. The pressure on G is O1 1 while the pressure on H is P 2 , If...

Two drivers G and H are at depths 20m and 40m, respectively below the water surface in a lake. The pressure on G is O1 ${}_1$ while the pressure on H is P ${}_2$, If the atmospheric pressure is equivalent to 10m² ${}^2$ of water, then the value of P1 ${}_{\mathrm{_{<span\; style="font-size:11.0pt;line-height:107\%;font-family:"Calibri","sans-serif";\; mso-ascii-theme-font:minor-latin;mso-fareast-font-family:Calibri;mso-fareast-theme-font:\; minor-latin;mso-hansi-theme-font:minor-latin;mso-bidi-font-family:"Iskoola\; Pota";\; mso-bidi-theme-font:minor-bidi;mso-ansi-language:EN-US;mso-fareast-language:\; EN-US;mso-bidi-language:SI-LK">2</span>}}}$/P ${}_{\mathrm{_{<span\; style="font-size:11.0pt;line-height:107\%;font-family:"Calibri","sans-serif";\; mso-ascii-theme-font:minor-latin;mso-fareast-font-family:Calibri;mso-fareast-theme-font:\; minor-latin;mso-hansi-theme-font:minor-latin;mso-bidi-font-family:"Iskoola\; Pota";\; mso-bidi-theme-font:minor-bidi;mso-ansi-language:EN-US;mso-fareast-language:\; EN-US;mso-bidi-language:SI-LK">1</span>}}}$ 2 is

Answer Details

The pressure at a depth h below the surface of a liquid can be calculated using the equation: P = P_atmosphere + ρgh where P_atmosphere is the atmospheric pressure, ρ is the density of the liquid, g is the acceleration due to gravity, and h is the depth below the surface. Given that the atmospheric pressure is 10m^2, the pressure at a depth of 20m and 40m below the surface of the lake would be: P1 = 10 + ρgh = 10 + 1000 * 9.8 * 20 = 10 + 1960 P2 = 10 + ρgh = 10 + 1000 * 9.8 * 40 = 10 + 3920 So, P2 / P1 = 3920 / 1960 = 2.00 Therefore, the value of P2 / P1 2 is 2.00.