A solid spherical ball made of iron is melted and two new balls are made whose diameters are in the ratio of 1:2. The ratio of the volume of the smaller new ball to the original ball is
1:3
1:5
2:9
1:9
Answer is Wrong!
Answer is Right!
This question was previously asked in
UPSC CAPF – 2022
Let the radius of the original ball be $R$. Its volume is $V_{orig} = \frac{4}{3}\pi R^3$. This ball is melted to make two new balls whose diameters are in the ratio 1:2. Let their radii be $r_1$ and $r_2$. The ratio of radii is the same as the ratio of diameters, so $r_1 : r_2 = 1:2$, or $r_2 = 2r_1$. The volumes of the new balls are $V_1 = \frac{4}{3}\pi r_1^3$ and $V_2 = \frac{4}{3}\pi r_2^3 = \frac{4}{3}\pi (2r_1)^3 = \frac{4}{3}\pi (8r_1^3)$. Since the volume is conserved, $V_{orig} = V_1 + V_2 = \frac{4}{3}\pi r_1^3 + \frac{4}{3}\pi (8r_1^3) = \frac{4}{3}\pi (r_1^3 + 8r_1^3) = \frac{4}{3}\pi (9r_1^3)$. The question asks for the ratio of the volume of the smaller new ball ($V_1$) to the original ball ($V_{orig}$). This ratio is $\frac{V_1}{V_{orig}} = \frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi (9r_1^3)} = \frac{r_1^3}{9r_1^3} = \frac{1}{9}$. The ratio is 1:9.
When a solid is melted and recast into new shapes, the total volume remains constant. The volume of a sphere is proportional to the cube of its radius ($V \propto r^3$).