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# in the bottom of a vessel with mercury of density $\rho$ there is a round hole of the radius $r$. At what maximum height of the mercury layer will the liquid still not flow out through this hole. (Surface tension =$T$).A) $\dfrac{T}{{r\rho g}}$ B) $\dfrac{T}{{2r\rho g}}$C) $\dfrac{{2T}}{{r\rho g}}$D) $\dfrac{{4T}}{{r\rho g}}$

Last updated date: 13th Jun 2024
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Hint:First we have to define the hydrostatic pressure and then we construct an equation.
After that we apply internal excess pressure, and we use the condition for the mercury in the given data.
Finally we get the required answer.

Formula used:
$P = h \times \rho \times g$.
${p_{ex}} = \dfrac{{2T}}{r}$.

The pressure that is exerted by a fluid at equilibrium position at a given point within it due to the gravitational force is called the hydrostatic pressure.
This pressure is directly proportional to the depth measured from the surface because the weight of fluid applies a downward force from the above.
Let us consider the hydrostatic pressure be $P$, the depth measured from the surface is $h$, the density of the fluid is $\rho$ and the acceleration due to gravity $g$,
Therefore, $P = h \times \rho \times g...\left( 1 \right)$.
Let the surface tension of a liquid drop is $T$ and the radius is $r$.
Every particle of the surface of the liquid drop is realized as a net force (perpendicular) inside due to the surface tension.
The internal excess pressure of the liquid drop applies a force outside of the drop which is normal to the surface. These two forces diminish each other and create an equilibrium position.
If the excess pressure is ${p_{ex}}$,
So we can write it as,${p_{ex}} = \dfrac{{2T}}{r}...\left( 2 \right)$ .
In the question state as, the density of mercury= $\rho$,Radius= $r$
$T$ = Surface tension
h= maximum height,
g= acceleration due to gravity.
Now we have to find the maximum height of the mercury layer.
Also, we can get the limiting condition for the mercury does not flow out through this hole,$P = {p_{ex}}....\left( 3 \right)$
Putting the values $\left( 1 \right)$and $\left( 2 \right)$in $\left( 3 \right)$ we get
$\therefore h \times \rho \times g = \dfrac{{2T}}{r}$
Here we take h as LHS and remaining as RHS in take of divided, and we get
$\Rightarrow h = \dfrac{{2T}}{{r \times \rho \times g}}$
Therefore the maximum height, $h = \dfrac{{2T}}{{r\rho g}}$.

Hence the right answer is in option (C).

Notes:The excess pressure inside an air bubble in a liquid is also ${p_{ex}} = \dfrac{{2T}}{r}$.
But, The excess pressure inside a soap bubble in a liquid is ${p_{ex}} = \dfrac{{4T}}{r}$.