The gravitational potential in a region is given by V = (3x + 4y + 12z) J/kg. The modulus of the gravitational field at (x = 1, y = 0,z = 3) is

The correct option is 13 N kg⁻¹.

Step-by-step Explanation:

The relationship between the gravitational potential $V$ and the gravitational field intensity vector $\mathbf{E}$ is given by the negative gradient of the potential:
$\mathbf{E}=-\nabla V$
In terms of components along the x, y, and z axes, this relation can be written as:
$E_x=-\frac{\partial V}{\partial x}$
$E_y=-\frac{\partial V}{\partial y}$
$E_z=-\frac{\partial V}{\partial z}$

Given the potential function:
$V=3x+4y+12z$

We calculate the partial derivatives of $V$ with respect to x, y, and z:
$\frac{\partial V}{\partial x}=\frac{d}{dx}\left(3x\right)=3$
$\frac{\partial V}{\partial y}=\frac{d}{dy}\left(4y\right)=4$
$\frac{\partial V}{\partial z}=\frac{d}{dz}\left(12z\right)=12$

Substituting these values back to find the components of the gravitational field:
$E_x=-3{\text{N kg}}^{-1}$
$E_y=-4{\text{N kg}}^{-1}$
$E_z=-12{\text{N kg}}^{-1}$

Since the components of the gravitational field are constant, the field is uniform throughout the region. This means the field strength at the specific point $(x=1,y=0,z=3)$ is the same as anywhere else in the region.

Now, we find the modulus (magnitude) of the gravitational field $\left|\mathbf{E}\right|$:
$\left|\mathbf{E}\right|=\sqrt{{E_x}^2+{E_y}^2+{E_z}^2}$
$\left|\mathbf{E}\right|=\sqrt{{(-3)}^2+{(-4)}^2+{(-12)}^2}$
$\left|\mathbf{E}\right|=\sqrt{9+16+144}$
$\left|\mathbf{E}\right|=\sqrt{169}=13{\text{N kg}}^{-1}$

Therefore, the modulus of the gravitational field is 13 N kg^-1.