A circular plot of land is divided into two regions by a chord of length 10√3 meters such that the chord subtends an angle of 120^◦ at the center. Then, the area, in square meters, of the smaller region is:

The correct option is 25 (4π/3 -√3).

To find the area of the smaller region, which is the minor segment of the circle, we first need to determine the radius of the circle. Let the radius of the circle be represented by $r$ meters.

We are given that a chord subtends an angle of ${120}^{\circ }$ at the center. The relation between the chord length ($L$), the radius ($r$), and the central angle ($\theta$) is given by the formula:
$$L=2r\sin \left(\frac{\theta }{2}\right)$$

Substitute the given values, where $L=10\sqrt{3}$ meters and $\theta ={120}^{\circ }$:
$$10\sqrt{3}=2r\sin \left({60}^{\circ }\right)$$

Since $\sin \left({60}^{\circ }\right)=\frac{\sqrt{3}}{2}$, we can write:
$$10\sqrt{3}=2r\left(\frac{\sqrt{3}}{2}\right)$$
Simplifying this equation gives:
$$10\sqrt{3}=r\sqrt{3}$$
Thus, the radius is:
$$r=10\text{ meters}$$

The area of the smaller region (minor segment) is calculated by subtracting the area of the triangle formed by the chord and the radii from the area of the corresponding sector of the circle:
$$\text{Area of smaller region}=\text{Area of sector}-\text{Area of triangle}$$

First, let's calculate the area of the sector. The formula for the area of a sector with angle $\theta$ in degrees is:
$${\text{Area}}_{\text{sector}}=\frac{\theta }{{360}^{\circ }}\times \pi r^2$$
Substituting $r=10$ and $\theta ={120}^{\circ }$:
$${\text{Area}}_{\text{sector}}=\frac{120}{360}\times \pi \times {10}^2=\frac{100\pi }{3}$$

Next, let's find the area of the triangle formed by the chord and the radii. The area of a triangle with two sides of length $r$ and an included angle $\theta$ is given by:
$${\text{Area}}_{\text{triangle}}=\frac{1}{2}{r}^2\sin \theta$$
Substituting the values:
$${\text{Area}}_{\text{triangle}}=\frac{1}{2}\times {10}^2\times \sin \left({120}^{\circ }\right)$$
Since $\sin \left({120}^{\circ }\right)=\sin \left({60}^{\circ }\right)=\frac{\sqrt{3}}{2}$:
$${\text{Area}}_{\text{triangle}}=50\times \frac{\sqrt{3}}{2}=25\sqrt{3}$$

Now, we subtract the area of the triangle from the area of the sector to find the area of the smaller region:
$$\text{Area of smaller region}=\frac{100\pi }{3}-25\sqrt{3}$$
Factoring out 25 from both terms, we get:
$$\text{Area of smaller region}=25\left(\frac{4\pi }{3}-\sqrt{3}\right)$$