Radius of neck (r1) = 2/2 = 1 cm
Area of neck (A1) = π (1)2 = π cm2
Radius of bottom (r2) = 10/2 = 5 cm
Area of bottom (A2) = π (5)2 = 25π cm2
{F_1\over A_1 } = {F_2\over A_2}⇒ {1.2\over π } = {F_2\over 25π}
⇒ F2 = {1.2\over π }× 25π = 1.2 × 25 = 30 kgf
Hence, force exerted at the bottom of the neck = 30 kgf.
(b) The Pascal’s law is applied to solve the part (a) which states that the pressure exerted anywhere in a confined liquid is transmitted equally and undiminished in all directions throughout the liquid.
Hence,
{F_1\over A_1 } = {F_2\over A_2}