The value of g at moon is `(1/6)^(th)`of that on earth. A balloon filled with hydrogen will

Options

A. fall with acceleration g on the moon
B. rise with acceleration g on the moon
C. rise with acceleration g/6 on the moon
D. fall with acceleration g/6 on the moon

Explanation

To solve the problem of what happens to a balloon filled with hydrogen on the Moon, we can follow these steps:

### Step 1: Understand the gravitational acceleration on the Moon
The problem states that the value of gravitational acceleration (g) on the Moon is \( \frac{1}{6} \)th of that on Earth. Therefore, if we denote the gravitational acceleration on Earth as \( g \), then on the Moon it is:

\[ g_{\text{moon}} = \frac{g}{6} \]

### Step 2: Analyze the forces acting on the balloon
When the balloon is on the Moon, the only force acting on it is the gravitational force. Since there is no atmosphere on the Moon, there is no buoyant force acting on the balloon. The gravitational force acting on the balloon can be expressed as:

\[ F_{\text{gravity}} = m \cdot g_{\text{moon}} = m \cdot \frac{g}{6} \]

where \( m \) is the mass of the balloon.

### Step 3: Determine the acceleration of the balloon
According to Newton's second law, the acceleration \( a \) of the balloon can be calculated using the formula:

\[ F = m \cdot a \]

Since the only force acting on the balloon is the gravitational force, we can set the gravitational force equal to the mass times the acceleration:

\[ m \cdot a = m \cdot \frac{g}{6} \]

Dividing both sides by \( m \) (assuming \( m \neq 0 \)), we find:

\[ a = \frac{g}{6} \]

### Step 4: Conclusion
The balloon filled with hydrogen will fall towards the Moon with an acceleration of \( \frac{g}{6} \).

### Final Answer
The balloon will fall with an acceleration of \( \frac{g}{6} \) on the Moon.

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