Question

An open cylindrical vessel of internal diameter $7cm$ and $8cm$ stands on a horizontal table. Inside this is placed a solid metallic right circular cone, the diameter of whose base is $3\frac{1}{2}cm$ and height $8cm$. Find the volume of water required to fill the vessel. If the one is replaced by another cone, whose height is $1\frac{3}{4}cm$ and the radius of whose base is $2cm$, find the drop in the water level.

Answer 1

Let $r$ be the radius of the cylindrical vessel, $r=\frac{7}{2}$.

Let $r_1$ be the radius of the cone, $r_1=\frac{\frac{7}{2}}{2}=\frac{7}{4}$.

The volume of water is equal to the difference in the volume of cylindrical vessel and the volume of cone.

$V_1=\pi {\left(\frac{7}{2}\right)}^2\times 8-\frac{1}{3}\times \pi {\left(\frac{7}{4}\right)}^2\times 8$

$=\frac{22}{7}\times {\left(\frac{7}{2}\right)}^2\times 8-\frac{1}{3}\times \frac{22}{7}\times {\left(\frac{7}{4}\right)}^2\times 8$

$=\frac{847}{3}c{m}^3$

Therefore, the volume of water required to fill the vessel is $\frac{847}{3}c{m}^3$.

As the cone is replaced by new cone which has the radius as 2 cm and height as $\frac{7}{4}$ cm, so the volume of water with the new cone will be,

$V_2=\pi {\left(\frac{7}{2}\right)}^2\times 8-\frac{1}{3}\times \pi {\left(2\right)}^2\times \frac{7}{4}$

$=\frac{22}{7}\times {\left(\frac{7}{2}\right)}^2\times 8-\frac{1}{3}\times \frac{22}{7}\times {\left(2\right)}^2\times \frac{7}{4}$

$=\frac{902}{3}c{m}^3$

Let $h$ be the drop in the water level, then the drop in the volume of the water is,

$V=V_2-V_1$

$\pi {\left(\frac{7}{2}\right)}^2\times h=\frac{902}{3}-\frac{847}{3}$

$\frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}\times h=\frac{55}{3}$

$h=\frac{10}{21}cm$

Therefore, the drop in the water level is $\frac{10}{21}cm$.