Please answer part a)ii) and a)iii) in full detail

Q5 a) Figure Q5 shows a manometer consisting of two vertical arms of equal cross-sectional area, a, connected to form a U-tube. The cylindrical reservoirs at the top of each arm are identical and of cross-sectional area A. The fluid in the right arm is water with density pw, and the fluid in the left arm is oil with density po which is less than pw. The position of the oil/water interface is used to measure the difference between the pressures P, and P2 for a gas of negligible density in contact with the liquids in the reservoirs. (i) Show that if P, and P2 are equal the difference in height Hi between the reservoir levels is Ho (1-po/pw) where Ho is the height of the oil column measured above the oil/water interface when P, and P2 are equal. [3] (ii) If the interface between the water and oil moves up a vertical distance h in the small diameter tube, derive an equation for the corresponding height change o of the liquids in each reservoir [3] (iii) Show that if the interface moves up a vertical distance h from the position shown in Figure Q5, the corresponding pressure difference p2 - P1 is given by 12 - P1 - [ PWE ( 1 + A ) - POE ( 1 - 2 ) ] n [6] b ) (i) The oil in the manometer has a density of 870 kg/m and the manometer tube has an internal diameter of 4 mm. Calculate the reservoir diameter if the manometer reading h is to be 150 mm for a pressure difference of 2 mbar. [5] (ii) If the reservoirs were replaced by tubing of the same internal diameter as the manometer tube, calculate the manometer reading h for a pressure difference of 2 mbar. [3]P1 P 2 H1 WATER OIL Ho Figure Q5P P2 final water level . I am solving this problem using another D final oil level initial water level method so that you can understand easily . initial oil level . Here AB water level moves down to A B and G |CD oil level moves to c'D . iF G/ H Whereas EF interface moves to E'F . Now for equilibrium pressure at FF and GH level should be equal . Hences P2 + ( Pressure decrease due to AB to A'B water level change ) = P t ( Pressme increase due to CD to C'D' vid level change ) + (Pressure difference due to - Eff'E sil level changed with water ) Hency 12 - op wg = P1 + 8 Pog + ( h P wg - h pog ) . Hena P2 - P1 = 8 Pwg + 8 Pog + h ewg - h pog Henc P2 - PI = Pwg ( h + ah) - pog ( h ah ) where 8 = ah Hens P 2 - P 1 = Pug ( 1 + a ) - pog (1 - a )