Due to certain mass distribution, the gravitational field along X-axis is given as `I = (K)/(r^(4))` (where K is a constant). Considering the value of gravitational potential to be zero at infinity, its value at a distance x will be

Options

A. `(K)/(2x^(2))`
B. `(K)/(3) x^(2)`
C. `(K)/(2) x^(3)`
D. `(K)/(3x^(3))`

Explanation

To find the gravitational potential at a distance \( x \) from a mass distribution where the gravitational field intensity \( I \) is given by \( I = \frac{K}{r^4} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between gravitational field and potential**:
The gravitational field intensity \( I \) is related to the gravitational potential \( V \) by the equation:
\[ I = -\frac{dV}{dr} \]
This indicates that the gravitational field is the negative gradient of the gravitational potential.

2. **Substitute the given gravitational field expression**:
From the problem, we know:
\[ I = \frac{K}{r^4} \]
Therefore, we can write:
\[ -\frac{dV}{dr} = \frac{K}{r^4} \]

3. **Rearranging the equation**:
Rearranging gives us:
\[ dV = -\frac{K}{r^4} dr \]

4. **Integrate to find the potential**:
We need to find the potential \( V \) at a distance \( x \), considering the potential at infinity to be zero. We set up the integral:
\[ V(x) - V(\infty) = \int_{\infty}^{x} -\frac{K}{r^4} dr \]
Since \( V(\infty) = 0 \), we have:
\[ V(x) = -\int_{\infty}^{x} \frac{K}{r^4} dr \]

5. **Calculate the integral**:
The integral can be computed as follows:
\[ V(x) = -\left[-\frac{K}{3r^3}\right]_{\infty}^{x} \]
Evaluating this gives:
\[ V(x) = \left[-\frac{K}{3x^3} + 0\right] = \frac{K}{3x^3} \]

6. **Final result**:
Thus, the value of gravitational potential at a distance \( x \) is:
\[ V(x) = -\frac{K}{3x^3} \]

### Summary:
The gravitational potential at a distance \( x \) from the mass distribution is given by:
\[ V(x) = -\frac{K}{3x^3} \]