This might not be the first thing that comes to mind, but it might be to some people in the stem field especially!

Since we are working with a circle, we know that the arc length is \( s = r \times dθ \), where dθ is the change of the angle between AOB.

The triangle is also equilateral, so the angle ABC is 60 degrees. Thus, AOB is twice that, so 120 degrees (they target the same arc, but AOB is central, where ABC is on the circle)

If we write 120 degrees as \(\frac{2π}{3}\), we can now integrate from 0 to \(\frac{2π}{3}\), as this would account for the 120 degree angle.
This is an especially easy integral that just equals \(\frac{2π}{3}\). Now plug in \(r = 2\) (don't forget that!) and we conclude that \(s = \frac{4π}{3}\) which is of course larger than π.

As a one liner, we got: \(r\int_0^\frac{2π}{3}dθ = \frac{4\pi}{3} > π\)

Sir, why dont you subtract the area of triangle AOC from arc taken from centroid?

You want to find the Arc Length by subtracting the Area?? I would like to see that happening. Kindly explain Sir!

Given that ABC is an equilateral triangle, so if you join the points A and C to O to make the angle AOC
then Angle AOC will be 120 degree. (As all three angles at the center will be equal so \(\frac{360}{3}\) = 120 degree

Watch this video to know how

Length of Minor Arc AC = \(\frac{120}{360}\) * 2\(\pi\)r = \(\frac{1}{3}\) * 2\(\pi\)2 = \(\frac{4}{3}\) * \(\pi\)

Clearly Quantity A(\(\frac{4}{3}\) * \(\pi\)) is greater than Quantity B (\(\pi\))

So, Answer will be A
Hope it helps!

Watch the following video to Learn Properties of Equilateral Triangle inscribed in a Circle

Circumference of entire circle is equal to 2*pi*r, r = 2. Because the triangle is an equilateral triangle, you can say that each arc measures 1/3 of the circumference.

So take (2*pi*2) and multiply by 1/3 and you get 4pi/3