RRB Group D 1 Dec 2025 Shift 2 Solved Paper

Concept:Volume remains constant when metal is melted and recast.The number of smaller spheres is found from the ratio of volumes.The required ratio is total surface area of all smaller spheres to surface area of the original sphere.Explanation:Radius of original sphere $R = 10.5\ \text{cm}$.Radius of each smaller sphere $r = 3.5\ \text{cm}$.Volume of a sphere is $\frac{4}{3}\pi r^3$.Number of smaller spheres $n = \frac{\text{Volume of original}}{\text{Volume of one smaller}} = \frac{R^3}{r^3}$.$n = \left(\frac{10.5}{3.5}\right)^3 = (3)^3 = 27$.Surface area of a sphere is $4\pi r^2$.Total surface area of all smaller spheres $= n \times 4\pi r^2$.Surface area of original sphere $= 4\pi R^2$.Required ratio = $\frac{n \times 4\pi r^2}{4\pi R^2} = n \times \left(\frac{r}{R}\right)^2$.$= 27 \times \left(\frac{3.5}{10.5}\right)^2 = 27 \times \left(\frac{1}{3}\right)^2 = 27 \times \frac{1}{9} = 3$.Thus, the ratio is $3 : 1$.Answer:$3 : 1$ (Option D).