Let the radius of the ball be \( r \) cm. Therefore, the volume of water displaced by the ball = volume of the ball \[ = \frac{4}{3}\pi r^3 \text{ cubic cm} \] The radius of the cylindrical vessel = \( \frac{12}{2} \) cm = 6 cm So, the volume of the raised water column \[ = \pi \times 6^2 \times 1 = 36\pi \text{ cubic cm} \] According to the question: \[ \frac{4}{3}\pi r^3 = 36\pi \] \[ \Rightarrow r^3 = \frac{36 \times 3}{4} = 27 \] \[ \Rightarrow r = 3 \] Therefore, the diameter of the ball \[ = 3 \times 2 = 6 \text{ cm} \]