Question

A cylindrical container with a cross-sectional area of 70.2cm2 holds a fluid of density 796kg/m3. At the bottom of the container, their pressure is 117kpa. Assume pat=101kpa. what is the depth of the fluid? Find the pressure at the bottom Of the container after an additional 2.35×10^-3m³.

Answer 1

$p=p_0+\rho gh\to h=(p-p_0)/(\rho g)=$

$=(117000-101000)/(796\cdot9.8)=2.051\ (m)$ . Answer

$h_1=V/A=0.00235/0.00702=0.335\ (m)$

$p_1=p_0+\rho gH=p_0+\rho g (h+h_1)=$

$=101000+796\cdot9.8\cdot(2.051+0.335)=119613\ (Pa)$ . Answer