The value of `g` at a certain height `h` above the free surface of the earth is `x//4` where `x` is the value of `g` at the surface of the earth. The height `h` is

Options

A. `R`
B. `2R`
C. `3R`
D. `4R`

Explanation

To solve the problem, we need to find the height \( h \) above the Earth's surface where the acceleration due to gravity \( g \) is \( \frac{x}{4} \), where \( x \) is the acceleration due to gravity at the Earth's surface.

### Step-by-Step Solution:

1. **Understanding the formula for \( g \)**:
The acceleration due to gravity at the surface of the Earth is given by the formula:
\[ g = \frac{GM}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

2. **Setting up the equation at height \( h \)**:
At a height \( h \) above the Earth's surface, the acceleration due to gravity \( g' \) can be expressed as:
\[ g' = \frac{GM}{(R + h)^2} \]
According to the problem, at height \( h \), \( g' = \frac{x}{4} \).

3. **Substituting \( x \)**:
Since \( x = g \), we can substitute \( g \) in the equation:
\[ \frac{GM}{(R + h)^2} = \frac{1}{4} \cdot \frac{GM}{R^2} \]

4. **Canceling \( GM \)**:
We can cancel \( GM \) from both sides (assuming \( GM \neq 0 \)):
\[ \frac{1}{(R + h)^2} = \frac{1}{4R^2} \]

5. **Cross-multiplying**:
Cross-multiplying gives us:
\[ 4R^2 = (R + h)^2 \]

6. **Expanding the right side**:
Expanding the right side:
\[ 4R^2 = R^2 + 2Rh + h^2 \]

7. **Rearranging the equation**:
Rearranging the equation gives:
\[ 4R^2 - R^2 = 2Rh + h^2 \]
\[ 3R^2 = 2Rh + h^2 \]

8. **Rearranging to form a quadratic equation**:
Rearranging this into standard quadratic form:
\[ h^2 + 2Rh - 3R^2 = 0 \]

9. **Using the quadratic formula**:
We can use the quadratic formula \( h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 2R, c = -3R^2 \):
\[ h = \frac{-2R \pm \sqrt{(2R)^2 - 4 \cdot 1 \cdot (-3R^2)}}{2 \cdot 1} \]
\[ h = \frac{-2R \pm \sqrt{4R^2 + 12R^2}}{2} \]
\[ h = \frac{-2R \pm \sqrt{16R^2}}{2} \]
\[ h = \frac{-2R \pm 4R}{2} \]

10. **Calculating the possible values of \( h \)**:
This gives us two possible solutions:
\[ h = \frac{2R}{2} = R \quad \text{(taking the positive root)} \]
or
\[ h = \frac{-6R}{2} = -3R \quad \text{(not physically meaningful)} \]

Thus, the height \( h \) is equal to the radius of the Earth \( R \).

### Final Answer:
\[ h = R \]