Question
Question: A solid hemisphere is just pressed below the liquid, the value of $\frac{F_1}{F_2}$ is (where $F_1$ ...
A solid hemisphere is just pressed below the liquid, the value of F2F1 is (where F1 and F2 are the hydrostatic forces acting on the curved and flat surfaces of the hemisphere) (Neglect atmospheric pressure).

21
32
31
none of these
none of these
Solution
Let R be the radius of the hemisphere, ρ be the density of the liquid, and g be the acceleration due to gravity. Let the flat surface of the hemisphere be at a depth h from the free surface of the liquid.
The hydrostatic force on the flat surface (F2) is given by: F2=Pavg×A, where Pavg is the average pressure on the flat surface and A is the area of the flat surface. The pressure at depth h is P=ρgh. Since the flat surface is at a uniform depth h, the pressure is uniform. A=πR2 So, F2=(ρgh)×(πR2)=ρghπR2.
The hydrostatic force on the curved surface (F1) is the resultant upward force exerted by the liquid pressure on the curved surface. This resultant force is equal to the buoyant force acting on the hemisphere. The volume of the hemisphere is V=32πR3. The buoyant force (FB) is given by Archimedes' principle: FB=ρgV=ρg(32πR3). So, F1=FB=32ρgπR3.
The ratio F2F1 is: F2F1=ρghπR232ρgπR3=3h2R.
The problem states that the hemisphere is "just pressed below the liquid". This implies that the flat surface could be at the free surface of the liquid (h=0). In this case, F2=0, and the ratio F2F1 would be undefined.
If we interpret "just pressed below the liquid" to mean that the entire hemisphere is submerged and its flat surface is at some depth h>0, the ratio F2F1=3h2R depends on the depth h and the radius R. Since the options provided are constant values, and the ratio depends on h, none of the given options can be universally correct without a specific value for h relative to R.
Therefore, the correct answer is "none of these".
