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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 F1F2\frac{F_1}{F_2} is (where F1F_1 and F2F_2 are the hydrostatic forces acting on the curved and flat surfaces of the hemisphere) (Neglect atmospheric pressure).

A

12\frac{1}{2}

B

23\frac{2}{3}

C

13\frac{1}{3}

D

none of these

Answer

none of these

Explanation

Solution

Let RR be the radius of the hemisphere, ρ\rho be the density of the liquid, and gg be the acceleration due to gravity. Let the flat surface of the hemisphere be at a depth hh from the free surface of the liquid.

The hydrostatic force on the flat surface (F2F_2) is given by: F2=Pavg×AF_2 = P_{avg} \times A, where PavgP_{avg} is the average pressure on the flat surface and AA is the area of the flat surface. The pressure at depth hh is P=ρghP = \rho g h. Since the flat surface is at a uniform depth hh, the pressure is uniform. A=πR2A = \pi R^2 So, F2=(ρgh)×(πR2)=ρghπR2F_2 = (\rho g h) \times (\pi R^2) = \rho g h \pi R^2.

The hydrostatic force on the curved surface (F1F_1) 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=23πR3V = \frac{2}{3} \pi R^3. The buoyant force (FBF_B) is given by Archimedes' principle: FB=ρgV=ρg(23πR3)F_B = \rho g V = \rho g (\frac{2}{3} \pi R^3). So, F1=FB=23ρgπR3F_1 = F_B = \frac{2}{3} \rho g \pi R^3.

The ratio F1F2\frac{F_1}{F_2} is: F1F2=23ρgπR3ρghπR2=2R3h\frac{F_1}{F_2} = \frac{\frac{2}{3} \rho g \pi R^3}{\rho g h \pi R^2} = \frac{2R}{3h}.

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=0h=0). In this case, F2=0F_2 = 0, and the ratio F1F2\frac{F_1}{F_2} 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>0h > 0, the ratio F1F2=2R3h\frac{F_1}{F_2} = \frac{2R}{3h} depends on the depth hh and the radius RR. Since the options provided are constant values, and the ratio depends on hh, none of the given options can be universally correct without a specific value for hh relative to RR.

Therefore, the correct answer is "none of these".