Moon is the natural satellite of the earth and weight of an object on the moon is one sixth times the weight of same body on the earth. If a body is raised through height h on the surface of earth and the energy spent is E then for the some amount of energy E the body on the surface of moon will rise through the height of

Options

A. 2h
B. 6h
C. 4h
D. 12h

Explanation

To solve the problem, we need to understand the relationship between the energy spent in raising a body on Earth and on the Moon. Here’s a step-by-step solution:

### Step 1: Understand the Energy Spent on Earth
When a body of mass \( m \) is raised to a height \( h \) on Earth, the energy \( E \) spent is given by the formula:
\[ E = mgh \]
where \( g \) is the acceleration due to gravity on Earth.

### Step 2: Determine the Weight on the Moon
The weight of the same body on the Moon is one-sixth of its weight on Earth. Therefore, the acceleration due to gravity on the Moon \( g' \) can be expressed as:
\[ g' = \frac{g}{6} \]

### Step 3: Energy Spent on the Moon
When the same body is raised to a height \( h' \) on the Moon, the energy spent \( E \) is given by:
\[ E = m g' h' \]
Substituting \( g' \) from the previous step, we have:
\[ E = m \left(\frac{g}{6}\right) h' \]

### Step 4: Equate the Energies
Since the energy spent on both Earth and Moon is the same, we can set the two expressions for \( E \) equal to each other:
\[ mgh = m \left(\frac{g}{6}\right) h' \]

### Step 5: Simplify the Equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gh = \frac{g}{6} h' \]

### Step 6: Solve for \( h' \)
To find \( h' \), we can rearrange the equation:
\[ h' = 6h \]

### Conclusion
Thus, for the same amount of energy \( E \) spent, the body on the surface of the Moon will rise through a height of:
\[ h' = 6h \]