A conical iron piece having diameter 28 cm and height 30cm is totally immersed into the water of a cylindrical vessel, resulting in the rise of water level by 6.4 cm. The diameter, in cm, of the vessel is : (1) $3.5$

Radius of cylindrical vessel = r cm. (let)

Volume of conical piece of iron $=\frac{1}{3} \pi R^{2} h$

$=\left(\frac{1}{3} \pi \times 14 \times 14 \times 30\right)$ cu. $\mathrm{cm}$.

Volume of raised water $=\pi \mathrm{r}^{2} \times 6.4 \mathrm{cu} . \mathrm{cm} .$

$\therefore \pi r^{2} \times 6.4$

$=\frac{1}{3} \pi \times 14 \times 14 \times 30$

$\Rightarrow r^{2}=\frac{14 \times 14 \times 10}{6.4}$

$\Rightarrow r^{2}=\frac{14^{2} \times 10^{2}}{8^{2}}$

$\Rightarrow r=\frac{14 \times 10}{8}$

$\Rightarrow 2 r=\frac{2 \times 14 \times 10}{8}=35 \mathrm{~cm}=$ diameter