RRB Technician Grade I 13 March 2026 Paper

Concept:When the ball is dropped, it displaces its own volume of water, so the total volume inside the hemisphere becomes the sum of the initial water volume and the ball volume.Explanation:Volume of water in the bowl $= \frac{992\pi}{3}\,\text{cm}^3$.Radius of the ball $r = 2\,\text{cm}$.Volume of the ball $= \frac{4}{3}\pi r^{3} = \frac{4}{3}\pi (2)^{3} = \frac{32\pi}{3}\,\text{cm}^3$.After dropping the ball, the water reaches the brim, so the total volume equals the volume of the hemispherical bowl.Total volume $= \frac{992\pi}{3} + \frac{32\pi}{3} = \frac{1024\pi}{3}\,\text{cm}^3$.Volume of a hemisphere of radius $R$ is $\frac{2}{3}\pi R^{3}$.So, $\frac{2}{3}\pi R^{3} = \frac{1024\pi}{3}$.Cancelling $\pi$ and simplifying: $2R^{3} = 1024$.Thus, $R^{3} = 512$, giving $R = 8\,\text{cm}$.Therefore, the radius of the bowl is $8\,\text{cm}$.Answer:The radius of the bowl is $\boxed{8\,\text{cm}}$.So, the correct option is A.