If R is the radius of earth , the height ,at which the weight of a body becomes `1/4` of its weight on the surface of earth , is

Options

A. 2R
B. R
C. R/2
D. R/4

Answer

To find the height at which the weight of a body becomes one-fourth of its weight on the surface of the Earth, we can follow these steps:

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

### Step 2: Define the weight at height \( h \)
At a height \( h \) above the surface of the Earth, the acceleration due to gravity \( g' \) can be expressed as:
\[ g' = \frac{g}{(1 + \frac{h}{R})^2} \]
where \( R \) is the radius of the Earth.

### Step 3: Set up the equation for weight at height \( h \)
The weight of the body at height \( h \) is:
\[ W' = mg' = m \cdot \frac{g}{(1 + \frac{h}{R})^2} \]

### Step 4: Relate the weights
According to the problem, the weight at height \( h \) is one-fourth of the weight on the surface:
\[ W' = \frac{W}{4} \]
Substituting for \( W \):
\[ mg' = \frac{mg}{4} \]
This simplifies to:
\[ \frac{g}{(1 + \frac{h}{R})^2} = \frac{g}{4} \]

### Step 5: Cancel \( g \) from both sides
Since \( g \) is common on both sides, we can cancel it out:
\[ \frac{1}{(1 + \frac{h}{R})^2} = \frac{1}{4} \]

### Step 6: Solve for \( (1 + \frac{h}{R})^2 \)
Taking the reciprocal of both sides gives:
\[ (1 + \frac{h}{R})^2 = 4 \]

### Step 7: Take the square root
Taking the square root of both sides results in:
\[ 1 + \frac{h}{R} = 2 \]
or
\[ 1 + \frac{h}{R} = -2 \] (we discard this solution as height cannot be negative)

### Step 8: Isolate \( h \)
From the equation \( 1 + \frac{h}{R} = 2 \):
\[ \frac{h}{R} = 2 - 1 \]
\[ \frac{h}{R} = 1 \]
Thus, we find:
\[ h = R \]

### Final Answer
The height \( h \) at which the weight of a body becomes one-fourth of its weight on the surface of the Earth is:
\[ h = R \]

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