A cylindrical vessel with internal diameter 10 cm and height 10.5 cm

Solution:

We have,

Internal radius of the cylindrical vessel, $R=\frac{10}{2}=5 \mathrm{~cm}$,

Height of the cylindrical vessel, $H=10.5 \mathrm{~cm}$,

Radius of the solid cone, $r=\frac{7}{2}=3.5 \mathrm{~cm}$ and

Height of the solid cone, $h=6 \mathrm{~cm}$

(i)

Volume of water displaced out of the cylinder $=$ Volume of the solid cone

$=\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 6$

$=77 \mathrm{~cm}^{3}$

(ii) $\mathrm{As}$,

Volume of the cylindrical vessel $=\pi R^{2} H$

$=\frac{22}{7} \times 5 \times 5 \times 10.5$

$=825 \mathrm{~cm}^{3}$

So, the volume of water left in the cylindrical vessel = Volume of the cylindrical vessel - Volume of the solid cone

$=825-77$

$=748 \mathrm{~cm}^{3}$