The weight of an object on the moon is one-sixth of its weight on Earth. The ratio of the kinetic energy of a body on Earth moving with speed V to that of the same body moving with speed V on the moon is :

Options

A. `6 : 1`
B. `36 : 1`
C. `1 : 1`
D. `1 : 6`

Explanation

To solve the problem, we need to find the ratio of the kinetic energy of a body moving with speed \( V \) on Earth to that of the same body moving with speed \( V \) on the Moon.

### Step-by-Step Solution:

1. **Understanding Kinetic Energy**:
The formula for kinetic energy (KE) is given by:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the object and \( v \) is its velocity.

2. **Kinetic Energy on Earth**:
For an object moving with speed \( V \) on Earth, the kinetic energy \( KE_{E} \) is:
\[ KE_{E} = \frac{1}{2} m V^2 \]

3. **Kinetic Energy on the Moon**:
The mass of the object remains the same regardless of location (Earth or Moon). Therefore, the kinetic energy \( KE_{M} \) of the same object moving with speed \( V \) on the Moon is:
\[ KE_{M} = \frac{1}{2} m V^2 \]

4. **Finding the Ratio**:
Now, we find the ratio of the kinetic energy on Earth to the kinetic energy on the Moon:
\[ \text{Ratio} = \frac{KE_{E}}{KE_{M}} = \frac{\frac{1}{2} m V^2}{\frac{1}{2} m V^2} \]
Simplifying this gives:
\[ \text{Ratio} = \frac{1}{1} = 1 \]

### Final Answer:
The ratio of the kinetic energy of a body on Earth moving with speed \( V \) to that of the same body moving with speed \( V \) on the Moon is \( 1:1 \).