The height a which the weight of a body becomes `1//16th` its weight on the surface of earth (radius `R`) is

Options

A. `4R`
B. `15R`
C. `3R`
D. `5R`

Answer

To find the height at which the weight of a body becomes \( \frac{1}{16} \) of its weight on the surface of the Earth, we can follow these steps:

### Step 1: Understand the relationship between weight and gravitational force
The weight \( W \) of a body is given by the formula:
\[ W = mg \]
where \( m \) is the mass of the body and \( g \) is the acceleration due to gravity.

### Step 2: Determine the gravitational force at height \( h \)
The gravitational force \( g' \) at a height \( h \) above the surface of the Earth is given by the formula:
\[ g' = \frac{g}{(1 + \frac{h}{R})^2} \]
where \( R \) is the radius of the Earth.

### Step 3: Set up the equation for the weight at height \( h \)
According to the problem, we need to find the height \( h \) where the weight becomes \( \frac{1}{16} \) of its weight on the surface:
\[ mg' = \frac{1}{16} mg \]
Since \( m \) is constant, we can simplify this to:
\[ g' = \frac{1}{16} g \]

### Step 4: Substitute \( g' \) into the equation
Now substituting \( g' \) into the equation gives:
\[ \frac{g}{(1 + \frac{h}{R})^2} = \frac{1}{16} g \]
We can cancel \( g \) from both sides (assuming \( g \neq 0 \)):
\[ \frac{1}{(1 + \frac{h}{R})^2} = \frac{1}{16} \]

### Step 5: Cross-multiply and simplify
Cross-multiplying gives:
\[ 16 = (1 + \frac{h}{R})^2 \]

### Step 6: Take the square root
Taking the square root of both sides results in:
\[ 4 = 1 + \frac{h}{R} \]

### Step 7: Solve for \( h \)
Now, isolate \( h \):
\[ \frac{h}{R} = 4 - 1 \]
\[ \frac{h}{R} = 3 \]
\[ h = 3R \]

### Conclusion
Thus, the height at which the weight of the body becomes \( \frac{1}{16} \) of its weight on the surface of the Earth is:
\[ h = 3R \]