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Given that, a chord length $5 \; \text{cm}$ subtends an angle of $60^{\circ}$ at the centre of a circle.

Let the radius of a circle be $`r\text{’} \; \text {cm}.$

So, $\boxed{OA = OB = OC = r}$



In $\triangle OAB,$ two sides are equal $(OA = OB = r).$

So, $\angle A = \angle B = 60^{\circ}$

Thus, $\triangle ABC$ is an equilateral triangle.

$OABC$ is a rhombus.

Area of rhombus $= \frac{d_{1} \times d_{2}}{2} \; \text{sq. units},$ where $d_{1}, d_{2}$ are the diagonal of the rhombus.

Diagonals are perpendicular, to each other. $ \boxed{d^{2}_{1} + d^{2}_{2} = 4a^{2}} $

So, $ (OB)^{2} + (AC)^{2} = 4r^{2} $

$ \Rightarrow (5)^{2} + (AC)^{2} = 4(5)^{2} $

$ \Rightarrow 25 + (AC)^{2} = 100 $

$ \Rightarrow (AC)^{2} = 75 $

$ \Rightarrow AC = \sqrt{75} = \sqrt{25 \times 3} $

$ \Rightarrow AC = 5 \sqrt{3} \; \text{cm} $

$\therefore$ The length of a chord that subtends an angle of $120^{\circ}$ at the center of the circle is $5 \sqrt{3} \; \text{cm}.$

Correct Answer $:\text{C}$

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