The depth at which the effective value of acceleration due to gravity is `g/4` is (R=radius of the earth)

Options

A. `R`
B. `(3R)/4`
C. `R/2`
D. `R/4`

Explanation

To find the depth at which the effective value of acceleration due to gravity is \( \frac{g}{4} \), we can follow these steps:

### Step 1: Understand the formula for gravity variation
The acceleration due to gravity at a depth \( d \) below the Earth's surface can be expressed as:
\[ g' = g \left(1 - \frac{d}{R}\right) \]
where:
- \( g' \) is the acceleration due to gravity at depth \( d \),
- \( g \) is the acceleration due to gravity at the surface,
- \( R \) is the radius of the Earth.

### Step 2: Set up the equation for the given condition
We want to find the depth \( d \) where \( g' = \frac{g}{4} \). Substituting this into the formula gives:
\[ \frac{g}{4} = g \left(1 - \frac{d}{R}\right) \]

### Step 3: Simplify the equation
Dividing both sides by \( g \) (assuming \( g \neq 0 \)):
\[ \frac{1}{4} = 1 - \frac{d}{R} \]

### Step 4: Rearranging the equation
Rearranging the equation to isolate \( \frac{d}{R} \):
\[ \frac{d}{R} = 1 - \frac{1}{4} \]
\[ \frac{d}{R} = \frac{3}{4} \]

### Step 5: Solve for \( d \)
Multiplying both sides by \( R \) to find \( d \):
\[ d = \frac{3R}{4} \]

### Final Answer
Thus, the depth at which the effective value of acceleration due to gravity is \( \frac{g}{4} \) is:
\[ d = \frac{3R}{4} \]