Gravitation on moon is `(1)/(6)` th of that on earth. When a balloon filled with hydrogen is released on moon then this

Options

A. will rise with an acceleration less then `(g)/(6)`
B. will rise with acceleration `(g)/(6)`
C. will fail down with an acceleration less than `(5g)/(6)`
D. will fall down with acceleration `(g)/(6)`

Explanation

To solve the problem, we need to analyze the forces acting on the hydrogen-filled balloon when it is released on the Moon.

### Step-by-Step Solution:

1. **Understand the Gravitational Force on the Moon**:
The gravitational acceleration on the Moon is given as \( g_{moon} = \frac{1}{6} g_{earth} \). This means that any object on the Moon will experience a gravitational force that is one-sixth of that on Earth.

2. **Identify the Forces Acting on the Balloon**:
When the balloon is released, it experiences two main forces:
- **Weight (Downward Force)**: This is the force due to gravity acting on the balloon, which can be calculated as:
\[ F_{weight} = m \cdot g_{moon} = m \cdot \left(\frac{1}{6} g_{earth}\right) \]
- **Buoyant Force (Upward Force)**: The buoyant force is the upward force exerted by the surrounding fluid (in this case, air). However, on the Moon, there is no atmosphere (no air).

3. **Determine the Buoyant Force**:
Since there is no atmosphere on the Moon, the density of air (\( \rho_{air} \)) is effectively zero. The buoyant force can be calculated using the formula:
\[ F_{buoyant} = \rho_{air} \cdot V \cdot g \]
Since \( \rho_{air} = 0 \), it follows that:
\[ F_{buoyant} = 0 \]

4. **Calculate the Net Force on the Balloon**:
The net force acting on the balloon when it is released is simply the weight of the balloon, as the buoyant force is zero:
\[ F_{net} = F_{weight} - F_{buoyant} = m \cdot \left(\frac{1}{6} g_{earth}\right) - 0 = m \cdot \left(\frac{1}{6} g_{earth}\right) \]

5. **Determine the Acceleration of the Balloon**:
According to Newton's second law, the acceleration (\( a \)) of the balloon can be calculated as:
\[ F_{net} = m \cdot a \]
Substituting for \( F_{net} \):
\[ m \cdot \left(\frac{1}{6} g_{earth}\right) = m \cdot a \]
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ a = \frac{1}{6} g_{earth} \]

6. **Conclusion**:
The balloon will fall with an acceleration of \( \frac{1}{6} g_{earth} \) on the Moon due to the absence of buoyant force.

### Final Answer:
The balloon filled with hydrogen will fall with an acceleration of \( \frac{1}{6} g_{earth} \) on the Moon.