An expansible balloon filled with air is floating on the top surface of a lake, with `2//3` of its volume is submerged in water. How deep should it be sunk in water without changing the temperature of air in it, in order that it is just in equilibrium, neither sinking nor rising further? Height of water barometer `= 10 m`.

Explanation

To solve the problem of how deep the balloon should be sunk in water to achieve equilibrium, we can follow these steps:

### Step 1: Understand the Initial Conditions
The balloon is floating on the surface of the lake, with \( \frac{2}{3} \) of its volume submerged. This means that the buoyant force acting on the balloon is equal to the weight of the balloon when it is in this state.

### Step 2: Apply Archimedes' Principle
According to Archimedes' Principle, the buoyant force \( F_b \) acting on the balloon is equal to the weight of the water displaced by the submerged part of the balloon.

For the initial state (State A):
\[ F_b = \text{Weight of the balloon} = mg \]
The volume of the submerged part of the balloon is \( \frac{2}{3} V \), where \( V \) is the total volume of the balloon. Thus, we can express the buoyant force as:
\[ F_b = \left(\frac{2}{3} V\right) \cdot \rho \cdot g \]
where \( \rho \) is the density of water and \( g \) is the acceleration due to gravity.

### Step 3: Set Up the Equilibrium Condition
At equilibrium, the weight of the balloon is balanced by the buoyant force:
\[ mg = \left(\frac{2}{3} V\right) \cdot \rho \cdot g \]
We can simplify this by canceling \( g \) from both sides:
\[ m = \frac{2}{3} V \cdot \rho \]

### Step 4: Consider the New State (State B)
Now, we want to find out how deep the balloon can be sunk without changing the temperature of the air inside it. When the balloon is fully submerged, let’s denote the new volume of the balloon as \( V_2 \) and the pressure inside the balloon as \( P_2 \).

Using Boyle's Law (since the temperature is constant):
\[ P_1 V_1 = P_2 V_2 \]
Where \( P_1 \) is the initial pressure and \( V_1 \) is the initial volume of the balloon.

### Step 5: Relate the Volumes
From the problem, we know that:
\[ V_2 = V \quad \text{(when fully submerged)} \]
And since \( V_1 = V \) (the volume of the balloon does not change), we can express the relationship:
\[ P_1 V = P_2 V \]
This simplifies to:
\[ P_1 = P_2 \]

### Step 6: Calculate the Pressure at Depth
The pressure at a depth \( h \) in a fluid is given by:
\[ P = P_0 + \rho g h \]
Where \( P_0 \) is the atmospheric pressure (which we can assume is equivalent to the height of the water barometer, \( 10 \, m \)).

### Step 7: Find the New Pressure
At the surface, the pressure is:
\[ P_1 = 10 \, m \, \text{of water} \]
At depth \( h \), the pressure becomes:
\[ P_2 = 10 + \rho g h \]

### Step 8: Set Up the Equilibrium Condition
From the equilibrium condition derived earlier, we have:
\[ P_1 = \frac{2}{3} P_2 \]
Substituting \( P_2 \):
\[ P_1 = \frac{2}{3} (10 + \rho g h) \]
Substituting \( P_1 = 10 \):
\[ 10 = \frac{2}{3} (10 + \rho g h) \]

### Step 9: Solve for Depth \( h \)
Now, we can solve for \( h \):
\[ 30 = 2(10 + \rho g h) \]
\[ 30 = 20 + 2 \rho g h \]
\[ 10 = 2 \rho g h \]
\[ h = \frac{10}{2 \rho g} \]

### Step 10: Substitute Values
Given that the height of the water barometer is \( 10 \, m \), we can find the additional depth:
\[ h = 5 \, m \]

### Final Answer
Thus, the balloon should be sunk an additional \( 5 \, m \) into the water to achieve equilibrium.

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