A right circular cylinder is partially filled with water. Two iron spherical balls are completely immersed in the water so that the height of the water in the cylinder rises by 4 cm. If the radius of one ball is half of the other and the diameter of the cylinder is 18 cm, then the radii of the spherical balls are

Solution (By Examveda Team)

Total volume of 2 balls = volume of cylinder of height = 4 cm
$$\eqalign{ & \frac{4}{3}\pi r_1^3 + \frac{4}{3}\pi r_2^3 = \pi {9^2} \times 4 \cr & {r_1} = 2{r_2}{\text{ }}\left( {{\text{given}}} \right) \cr & \frac{4}{3}\pi \left( {8r_2^3 + r_2^3} \right) = \pi \times 9 \times 9 \times 4 \cr & 9r_2^3 = 9 \times 9 \times 3 \cr & {r_2} = 3 \cr & {r_1} = 2{r_2} = 2 \times 3 = 6 \cr} $$