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Capillary Action

A solid hemis...

Question

A solid hemisphere is just pressed below the liquid, the value of $\frac{F_{1}}{F_{2}}$ is (where $F_{1}$ and $F_{2})$ are the hydrostatic force acting on the curved and flat surfaces of the hemisphere)(Neglect atmospheric pressure)

\frac{1}{2}

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\frac{2}{3}

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\frac{1}{3}

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Solution

The correct option is C $\frac{1}{3}$
In the given figure.
$\begin{matrix} f_{2} = P \times A = (ρ g R) \times π R^{2} \end{matrix}$
$f_{1} =$ weight of liquid above curved surface
$\begin{matrix} f_{1} = ρ (π R^{2} - R - \frac{2 π R^{3}}{3}) . g = \frac{1}{3} ρ g π R^{3} \frac{f_{1}}{f_{2}} = \frac{1}{3} \end{matrix}$
So,
Option $C$ is correct answer.

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A solid hemisphere is just pressed below the liquid, the value of F1F2 is (where F1 and F2) are the hydrostatic force acting on the curved and flat surfaces of the hemisphere)(Neglect atmospheric pressure)

The correct option is C 13In the given figure.f2=P×A=(ρgR)×πR2f1= weight of liquid above curved surfacef1=ρ(πR2−R−2πR33).g=13ρgπR3f1f2=13So,Option C is correct answer.

A solid hemisphere is just pressed below the liquid, the value of $\frac{F_{1}}{F_{2}}$ is (where $F_{1}$ and $F_{2})$ are the hydrostatic force acting on the curved and flat surfaces of the hemisphere)(Neglect atmospheric pressure)