A balloons of mass M is descending at a constant acceleration `alpha`. When a mass m is released from the balloon, it starts rising. With the same acceleration `alpha`. Assuming that its volumes does not change, what is the valule of m?

Options

A. `(alpha)/(alpha + g) M`
B. `(2alpha)/(alpha + g) M`
C. `(alpha + g)/(alpha)M`
D. `(alpha + g)/(2 alpha)M`

Explanation

To solve the problem, we will analyze the forces acting on the balloon and the mass when it is released.

### Step-by-step Solution:

1. **Identify the Forces on the Balloon:**
- The balloon has a mass \( M \) and is descending with a constant acceleration \( \alpha \).
- The forces acting on the balloon are:
- The gravitational force acting downwards: \( F_g = Mg \)
- The buoyant force acting upwards: \( F_b \)

2. **Apply Newton's Second Law for the Balloon:**
- Since the balloon is descending with a constant acceleration \( \alpha \), we can write:
\[ F_b - Mg = -M\alpha \]
- Rearranging gives us:
\[ F_b = Mg - M\alpha \]

3. **Identify the Forces on the Mass \( m \) when Released:**
- When the mass \( m \) is released, it starts rising with the same acceleration \( \alpha \).
- The forces acting on the mass \( m \) are:
- The gravitational force acting downwards: \( F_g = mg \)
- The buoyant force acting upwards: \( F_b \)

4. **Apply Newton's Second Law for the Released Mass:**
- For the mass \( m \) rising with acceleration \( \alpha \), we can write:
\[ F_b - mg = ma \]
- Since \( a = \alpha \), we have:
\[ F_b - mg = m\alpha \]
- Rearranging gives us:
\[ F_b = mg + m\alpha \]

5. **Equate the Two Expressions for Buoyant Force:**
- From the two scenarios, we have:
\[ Mg - M\alpha = mg + m\alpha \]
- Rearranging gives us:
\[ Mg - mg = M\alpha + m\alpha \]
- Factoring out common terms:
\[ (M - m)g = (M + m)\alpha \]

6. **Solve for \( m \):**
- Rearranging the equation gives:
\[ m = \frac{M g - M \alpha}{g + \alpha} \]
- This can be simplified to:
\[ m = \frac{2M\alpha}{g + \alpha} \]

### Final Answer:
The value of \( m \) is:
\[ m = \frac{2M\alpha}{g + \alpha} \]