The depth `d`, at which the value of acceleration due to gravity becomes 1/n times the value at the surface is (R = radius of the earth)

Options

A. `(R)/(n)`
B. `R((n-1)/(n))`
C. `(R)/n^(2)`
D. `R((n)/(n+1))`

Explanation

To find the depth \( d \) at which the value of acceleration due to gravity becomes \( \frac{1}{n} \) times the value at the surface, we can follow these steps:

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

### Step 2: Set up the equation for the given condition
According to the problem, we need to find the depth \( d \) where the acceleration due to gravity \( g' \) is \( \frac{1}{n} g \):
\[ g' = \frac{1}{n} g \]

### Step 3: Substitute \( g' \) into the equation
Substituting the expression for \( g' \) from Step 1 into the equation from Step 2 gives:
\[ g \left(1 - \frac{d}{R}\right) = \frac{1}{n} g \]

### Step 4: Simplify the equation
We can cancel \( g \) from both sides (assuming \( g \neq 0 \)):
\[ 1 - \frac{d}{R} = \frac{1}{n} \]

### Step 5: Solve for \( d \)
Rearranging the equation to isolate \( d \):
\[ \frac{d}{R} = 1 - \frac{1}{n} \]
\[ d = R \left(1 - \frac{1}{n}\right) \]
\[ d = R \left(\frac{n - 1}{n}\right) \]

### Step 6: Final expression for depth \( d \)
Thus, the depth \( d \) at which the value of acceleration due to gravity becomes \( \frac{1}{n} \) times the value at the surface is:
\[ d = \frac{R(n - 1)}{n} \]

### Summary
The depth \( d \) is given by:
\[ d = \frac{R(n - 1)}{n} \]

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