What is depth at which value of acceleration due to gravity becomes 1/n times value that at surface of earth

Options

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

Answer

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

### Step 1: Understand the relation of gravity with depth
The acceleration due to gravity \( g' \) at a depth \( h \) below the surface of the Earth is given by the formula:
\[ g' = g \left(1 - \frac{h}{R}\right) \]
where \( g \) is the acceleration due to gravity at the surface, and \( R \) is the radius of the Earth.

### Step 2: Set up the equation
According to the problem, we want to find the depth \( h \) where:
\[ g' = \frac{g}{n} \]
Substituting this into the equation from Step 1 gives:
\[ \frac{g}{n} = g \left(1 - \frac{h}{R}\right) \]

### Step 3: Simplify the equation
We can cancel \( g \) from both sides (assuming \( g \neq 0 \)):
\[ \frac{1}{n} = 1 - \frac{h}{R} \]

### Step 4: Rearranging the equation
Now, rearranging this equation to isolate \( h \):
\[ \frac{h}{R} = 1 - \frac{1}{n} \]
\[ h = R \left(1 - \frac{1}{n}\right) \]

### Step 5: Final expression for depth
To express this in a more convenient form:
\[ h = R \left(\frac{n-1}{n}\right) \]

### Conclusion
Thus, the depth \( h \) at which the acceleration due to gravity becomes \( \frac{1}{n} \) times the value at the surface of the Earth is:
\[ h = R \left(\frac{n-1}{n}\right) \]