Question

A solid is in the form of a right circular cone mounted on a hemisphere.The radius of the hemisphere is $3.5$cm and the height of the cone is $4$cm.The solid is placed in a cylindrical vessel ,full of water ,in such a way that the whole solid is submerged in water.If the radius of the cylindrical vessel is $5$cm and its height is $10.5$cm ,find the volume of water left in the cylindrical vessel ?

Answer 1

Given, radius of the hemisphere,$r=3.5cm$

Now, since the solid is in the form of a right circular cone mounted on a hemisphere, then radius of base of the cone

$=$radius of the hemisphere

$\Rightarrow$ radius of the base of the cone$=r=3.5cm$

Height of the cone$=h=4cm$

So,

volume of the solid=volume of the cone$+$ volume of the hemisphere

$\Rightarrow$ volume of the solid=$\frac{1}{3}\pi r^{2}h+\frac{2}{3}\pi r^3$

$\Rightarrow$ volume of the solid=$\frac{1}{3}\pi r^2(h+2r)$

$\Rightarrow$ volume of the solid=$\frac{1}{3}\times \frac{22}{7}\times (4+7)=141.16{cm}^3$

Now, radius of the base of the cylindrical vessel=$r_1=5cm$

Height of the cylindrical vessel,$h_1=10.5cm$

$\therefore$ Volume of the water in the cylindrical vessel $=\frac{22}{7}\times 25\times 10.5=825{cm}^3$

Now, when the solid is completely submerged in the cylindrical vessel full of water,

then

volume of the water displaced by the solid= volume of solid

Hence, volume of the water left in the vessel= volume of the water in the vessel- volume of solid

=$(825-141.16){cm}^3$

=$683.84{cm}^3$