A balloon having mass 'm' is filled with gas and is held in hands of a boy. Then suddenly it gets released and gas starts coming out of it with a constant rate. The velocity of the ejected gas is `2m//s` with respect to the balloon. Find out the velocity of the balloon when the mass of gas is reduced to half :-

Options

A. `ln 2`
B. `2 ln 4`
C. `2 ln 2`
D. None of these

Explanation

To solve the problem, we need to find the velocity of the balloon when the mass of the gas inside it is reduced to half. We will use the principle of conservation of momentum.

### Step-by-Step Solution:

1. **Understand the System**:
- Let the initial mass of the balloon (including gas) be \( m \).
- The mass of the gas when it is reduced to half is \( \frac{m}{2} \).
- The velocity of the gas ejected with respect to the balloon is \( v_{g/b} = 2 \, \text{m/s} \).

2. **Define Variables**:
- Let \( v_b \) be the velocity of the balloon at the moment when the mass of the gas is reduced to half.
- The mass of the balloon without gas is \( m_b \) (which is constant).

3. **Apply Conservation of Momentum**:
- Initially, the momentum of the system is \( m \cdot 0 = 0 \) (the balloon is at rest).
- When the gas is ejected, the momentum of the gas and the balloon must still sum to zero.
- The momentum of the gas being ejected is \( \text{mass of gas} \times \text{velocity of gas} \).

4. **Momentum of the Ejected Gas**:
- The velocity of the gas with respect to the ground is \( v_g = v_b + v_{g/b} = v_b + 2 \).
- The momentum of the ejected gas when the mass is \( \frac{m}{2} \) is:
\[ \text{Momentum of gas} = \left(\frac{m}{2}\right)(v_b + 2) \]

5. **Momentum of the Balloon**:
- The momentum of the balloon when the gas is reduced to half is:
\[ \text{Momentum of balloon} = m_b v_b \]

6. **Setting Up the Equation**:
- According to the conservation of momentum:
\[ m_b v_b + \left(\frac{m}{2}\right)(v_b + 2) = 0 \]
- Rearranging gives:
\[ m_b v_b = -\left(\frac{m}{2}\right)(v_b + 2) \]

7. **Solving for \( v_b \)**:
- Substitute \( m_b = m - \frac{m}{2} = \frac{m}{2} \):
\[ \frac{m}{2} v_b = -\left(\frac{m}{2}\right)(v_b + 2) \]
- Cancel \( \frac{m}{2} \) (assuming \( m \neq 0 \)):
\[ v_b = -(v_b + 2) \]
- Solving this gives:
\[ v_b + v_b = -2 \implies 2v_b = -2 \implies v_b = -1 \, \text{m/s} \]

8. **Final Result**:
- The negative sign indicates that the balloon moves in the opposite direction to the direction of the ejected gas.

### Conclusion:
The velocity of the balloon when the mass of the gas is reduced to half is \( -1 \, \text{m/s} \).