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 in the vessel. Find the volume of water left in the cylinder.
(Take $\pi = \dfrac{22}{7}$)

Given,

For cylindrical vessel :

Diameter = 10 cm

Radius (R) = $\dfrac{10}{2}$ = 5 cm.

Height (H) = 10.5 cm

For cone :

Diameter = 7 cm

Radius (r) = $\dfrac{7}{2}$ = 3.5 cm.

Height (h) = 6 cm

Volume of water left = Volume of cylinder - Volume of cone

= πR²H - $\dfrac{1}{3}$πr²h

$= \dfrac{22}{7} \times 5^2 \times 10.5 - \dfrac{1}{3} \times \dfrac{22}{7} \times (3.5)^2 \times 6 \\[1em] = 22 \times 25 \times 1.5 - 22 \times 0.5 \times 3.5 \times 2 \\[1em] = 825 - 77 \\[1em] = 748 \text{ cm}^3.$

Hence, volume of water left in cylinder = 748 cm³.