A sportsman runs around a circular track of radius $ r $ such that he traverses the path ABAB. The distance travelled and displacement, respectively, are:

Displacement is the straight-line distance from the initial point to the final point.
Since the sportsman runs around the circular track and ends up at the same position (A), the displacement is the straight-line distance through the circle’s center. Therefore: \[ \text{Displacement} = 2r \] The distance travelled is the total path length covered by the sportsman, which consists of two complete laps around the circular track. Thus, the total distance is: \[ \text{Distance} = 2\pi r + \pi r = 3\pi r \] Thus, the correct answer is: \[ 3\pi r, 2r \]
A sportsman runs on a circular track of radius \(r\) such that he goes along the path A → B → A → B.
In circular motion problems, distance is the total length of the path covered, while displacement is the straight-line distance between the initial and final positions.
For points A and B on opposite ends of a diameter of the circle, the arc length between A and B (half the circle) is \(\pi r\), and the straight-line distance (displacement between A and B) is the diameter \(2r\).
Step 1: The sportsman moves A → B → A → B.
So he goes from A to B (one semicircular arc), then back to A (another semicircular arc), and again to B (one more semicircular arc).
Step 2: Distance for each semicircular path = \(\pi r\).
Thus total distance = \(3 \times \pi r = 3\pi r.\)
Step 3: The initial point is A, and final point is B.
Hence displacement = straight line distance between A and B = diameter = \(2r.\)
Answer: Distance = \(3\pi r\), Displacement = \(2r\).
A body of mass $100 \;g$ is moving in a circular path of radius $2\; m$ on a vertical plane as shown in the figure. The velocity of the body at point A is $10 m/s.$ The ratio of its kinetic energies at point B and C is: (Take acceleration due to gravity as $10 m/s^2$)


Consider the following sequence of reactions : 
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).