In the bottom of a vessel with mercury of density `rho` there is a round hole of radius r. At what maximum height of the mercury layer will the liquid still not flow out through this hole. (Surface tension = T)–

Options

A. `T/(rrhog)`
B. `T/(2rrhog)`
C. `(2T)/(rrhog)`
D. `(4T)/(rrhog)`

Answer

To solve the problem of determining the maximum height of the mercury layer that will not flow out through a hole at the bottom of a vessel, we can follow these steps:

### Step 1: Understand the forces acting on the liquid
The liquid (mercury) experiences two main forces:
1. The downward pressure due to the height of the mercury column.
2. The upward force due to surface tension at the hole.

### Step 2: Define the pressure at the hole
At a height \( h \) of the mercury, the pressure at the hole can be expressed as:
\[ P_A = P_{\text{atm}} + \rho g h \]
where:
- \( P_A \) is the pressure at the hole,
- \( P_{\text{atm}} \) is the atmospheric pressure,
- \( \rho \) is the density of mercury,
- \( g \) is the acceleration due to gravity,
- \( h \) is the height of the mercury column.

### Step 3: Calculate the upward force due to surface tension
The surface tension force acting on the liquid at the hole can be calculated using:
\[ F_s = T \cdot L \]
where:
- \( T \) is the surface tension,
- \( L \) is the length around the hole, which is the circumference of the hole, \( L = 2 \pi r \).

Thus, the surface tension force becomes:
\[ F_s = T \cdot 2 \pi r \]

### Step 4: Calculate the upward pressure force
The upward force due to the pressure difference can be expressed as:
\[ F_p = (P_{\text{atm}} - P_A) \cdot A \]
where \( A \) is the area of the hole, given by \( A = \pi r^2 \). Therefore:
\[ F_p = (P_{\text{atm}} - (P_{\text{atm}} + \rho g h)) \cdot \pi r^2 = -\rho g h \cdot \pi r^2 \]

### Step 5: Set the forces equal for equilibrium
For the liquid to not flow out, the upward surface tension force must balance the downward pressure force:
\[ F_s = |F_p| \]
This gives us:
\[ T \cdot 2 \pi r = \rho g h \cdot \pi r^2 \]

### Step 6: Simplify the equation
Cancel \( \pi r \) from both sides:
\[ 2T = \rho g h r \]

### Step 7: Solve for height \( h \)
Rearranging the equation to find \( h \):
\[ h = \frac{2T}{\rho g r} \]

### Final Answer
Thus, the maximum height of the mercury layer that will not flow out through the hole is:
\[ h = \frac{2T}{\rho g r} \]