Question $2(35$ points, $20$ minutes$)$: An oil $(SG=0\mathrm{.}7,=0\mathrm{.}38N\frac{s}{m^2})$ inclined manometer is used

to monitor the pressure between the tanks in the figure. A closed spherical tank of radius $9$ m contains

water. The pressure gauge at the top of the of the spherical tank reads $27$ kPa $.$ Take the atmospheric

pressure to be $101$ kPa $.$

a$)$ Write out an expression describing the pressure variation between the open tank and the

pressurized spherical tank $(10$ pts $).$

b$)$ Find the level of water in the spherical tank above the entrance in the manometer $(10$ pts $).$

c$)$ If$,$ in the opened tank, the water is replaced with a $3$ kN weight:

a$.$ Draw $($on next page$)$ the new pressure levels in the manometer

b$.$ Write the new pressure variation expression and calculate the new differential

reading in the manometer. Assume the levels of fluid in the tanks are about the

same. Assume that the cylindrical container and the spherical tank are large

enough so that the heights of the oil$/$water inside them stay the same and changes in

fluid levels occur only inside the thin manometer tube. $(15$ points$)$