Question

A cylindrical vessel with internal diameter 10 cm and height 10.5 cm is full of water. A solid cone of base diameter 7 cm and height 6 cm is completely immersed in water. Find the volume of water (i) displaced out of the cylinder (ii) left in the cylinder.

Answer 1

We have,

Internal radius of the cylindrical vessel, R = 10/2= 5 cm,

Height of the cylindrical vessel, H = 10.5 cm,

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

Height of the solid cone, h = 6 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 $c{m}^3$

(ii) As, Volume of the cylindrical vessel = $\pi R^{2}H$

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

= 825 $c{m}^3$

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

= 825 - 77 = 748 $c{m}^3$