
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) $\dfrac{{4T}}{{r\rho g}}$
Answer
474.9k+ views
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:
$P = h \times \rho \times g$.
${p_{ex}} = \dfrac{{2T}}{r}$.
Complete step by step answer:
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 ${p_{ex}} = \dfrac{{4T}}{r}$.
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:
$P = h \times \rho \times g$.
${p_{ex}} = \dfrac{{2T}}{r}$.
Complete step by step answer:
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 ${p_{ex}} = \dfrac{{4T}}{r}$.
Recently Updated Pages
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE

The highest possible oxidation states of Uranium and class 11 chemistry CBSE

Find the value of x if the mode of the following data class 11 maths CBSE

Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE

A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE

Statement I Reactivity of aluminium decreases when class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

How many valence electrons does nitrogen have class 11 chemistry CBSE
