A spherical iron ball is dropped into a cylindrical vessel of base diameter 14 cm containing water. The water level is increased by 9 $\dfrac{1}{3}$ cm. The radius of the ball is :

1. 3.5 cm

2. 7 cm

3. 9 cm

4. 12 cm

Let the radius of the sphere be r cm.

Radius of cylinder, R = $\dfrac{\text{diameter}}{2} = \dfrac{14}{2}$ = 7 cm

Since, a spherical iron ball is dropped into the vessel.

Height of water raised by 9 $\dfrac{1}{3} \text{cm} = \dfrac{28}{3}$ cm

Volume of water rise in cylinder = Volume of sphere

$\Rightarrow π\text{R}^2\text{h} = \dfrac{4}{3}π\text{r}^3 \\[1em] \Rightarrow \text{R}^2\text{h} = \dfrac{4}{3}\text{r}^3 \\[1em] \Rightarrow \text{r}^3 = \dfrac{3}{4} \times \text{R}^2 \times \text{h} \\[1em] \Rightarrow \text{r}^3 = \dfrac{3}{4} \times 7^2 \times \dfrac{28}{3} \\[1em] \Rightarrow \text{r}^3 = \dfrac{3}{4} \times 49 \times \dfrac{28}{3} \\[1em] \Rightarrow \text{r}^3 = \dfrac{4116}{12} \\[1em] \Rightarrow \text{r}^3 = 343 \\[1em] \Rightarrow \text{r} = \sqrt[3]{343} \\[1em] \Rightarrow \text{r} = 7 \text{ cm.}$

Hence, option 2 is the correct option.