An open cylindrical vessel of internal diameter 7 cm and height 8 cm stands on a horizontal table. Inside this is placed a solid metallic right circular cone, the diameter of whose base is $3\dfrac{1}{2}$ cm and height 8 cm. Find the volume of water required to fill the vessel.

If this cone is replaced by another cone, whose height is $1\dfrac{3}{4}$ cm and the radius of whose base is 2 cm, find the drop in the water level.

Given,

Diameter of cylindrical vessel = 7 cm

Radius of cylindrical vessel (R) = $\dfrac{7}{2}$ = 3.5 cm

Height of cylindrical vessel (H) = 8 cm

Diameter of base of cone = $3\dfrac{1}{2} = \dfrac{7}{2}$ cm

Radius of base of cone (r) = $\dfrac{7}{4}$ cm

Height of cone (h) = 8 cm

Volume of cylindrical vessel = $πR^2H = \dfrac{22}{7} \times (3.5)^2 \times 8$

= $22 \times 0.5 \times 3.5 \times 8$

= 308 cm³.

Volume of original cone = $\dfrac{1}{3}πr^2h$

= $\dfrac{1}{3} \times \dfrac{22}{7} \times \Big(\dfrac{7}{4}\Big)^2 \times 8$

= $\dfrac{1}{3} \times 22 \times \dfrac{1}{4} \times \dfrac{7}{4} \times 8$

= $\dfrac{308}{12}$ cm³

Volume of water required to fill the vessel = Volume of cylindrical vessel - Volume of original cone

$= 308 - \dfrac{308}{12} \\[1em] = \dfrac{12 \times 308 - 308}{12} \\[1em] = \dfrac{3388}{12} \\[1em] = 282.33 \text{ cm}^3.$

Given,

Radius of new cone (r₁) = 2 cm

Height of new cone (h₁) = $\dfrac{7}{4}$ cm

Volume of new cone = $\dfrac{1}{3}πr$1^2h1

$= \dfrac{1}{3} \times \dfrac{22}{7} \times 2^2 \times \dfrac{7}{4} \\[1em] = \dfrac{1}{3} \times 22 \times 4 \times \dfrac{1}{4} \\[1em] = \dfrac{22}{3} \text{ cm}^3$

Volume of water which comes down = Volume of original cone - Volume of new cone

= $\dfrac{308}{12} - \dfrac{22}{3}$

= $\dfrac{77}{3} - \dfrac{22}{3} = \dfrac{55}{3}$ cm³.

Let drop in height of water be h₂ cm.

Drop in volume of water = $\dfrac{55}{3}$ cm³

$\therefore πR^2h$2 = \dfrac{55}{3} \\[1em] \Rightarrow \dfrac{22}{7} \times (3.5)^2 \times h2 = \dfrac{55}{3} \\[1em] \Rightarrow 22 \times 0.5 \times 3.5 \times h2 = \dfrac{55}{3} \\[1em] \Rightarrow 38.5 \times h2 = \dfrac{55}{3} \\[1em] \Rightarrow h2 = \dfrac{55}{115.5} \\[1em] \Rightarrow h2 = \dfrac{10}{21} \text{ cm}. \\[1em]

Hence, volume of water required to fill the vessel = 282.33 cm³ and drop in level of water = $\dfrac{10}{21}$ cm.