If radius of earth is `R`, then the height h at which the value of `g` becomes (1/49)th of its value at the surface is

Options

A. 2R
B. 3R
C. 6R
D. 4R

Explanation

To solve the problem, we need to find the height \( h \) at which the acceleration due to gravity \( g' \) becomes \( \frac{1}{49} \) of its value at the surface of the Earth \( g \).

### Step-by-Step Solution:

1. **Understanding the relationship between \( g \) and \( g' \)**:
The formula for the acceleration due to gravity at a height \( h \) above the Earth's surface is given by:
\[ g' = \frac{g}{(1 + \frac{h}{R})^2} \]
where \( R \) is the radius of the Earth.

2. **Setting up the equation**:
We know that \( g' = \frac{1}{49} g \). Therefore, we can write:
\[ \frac{g}{(1 + \frac{h}{R})^2} = \frac{1}{49} g \]

3. **Cancelling \( g \) from both sides**:
Since \( g \) is not zero, we can cancel it from both sides:
\[ \frac{1}{(1 + \frac{h}{R})^2} = \frac{1}{49} \]

4. **Cross-multiplying**:
This leads to:
\[ (1 + \frac{h}{R})^2 = 49 \]

5. **Taking the square root**:
Taking the square root of both sides gives:
\[ 1 + \frac{h}{R} = 7 \quad \text{(since we take the positive root)} \]

6. **Solving for \( h \)**:
Rearranging the equation to solve for \( h \):
\[ \frac{h}{R} = 7 - 1 \]
\[ \frac{h}{R} = 6 \]
\[ h = 6R \]

Thus, the height \( h \) at which the value of \( g \) becomes \( \frac{1}{49} \) of its value at the surface is:
\[ \boxed{6R} \]