Question

# A cylindrical vessel with internal diameter ${\text{10}} cm$ and height ${\text{10}}{\text{.5}} cm$ is full of water. A solid cone of base diameter ${\text{7}} cm$ and height ${\text{6}} cm$ is completely immersed in water. Find the value of water (1) displaced out of the cylinder (2) left in the cylinder. $\left( {{\text{Take }}\pi {\text{ = }}\dfrac{{22}}{7}} \right)$

As it is given radius of cone $r = \dfrac{7}{2} cm$ and height $h = 6 cm$ and
radius of cylinder $R = \dfrac{10}{2} cm$ and height $H = 10.5 cm$
Then, Volume of water displaced $=$ volume of cone.
$\Rightarrow \dfrac{1}{3}\pi {r^2}h = \dfrac{1}{3} \times \dfrac{{22}}{7} \times \dfrac{7}{2} \times \dfrac{7}{2} \times 6 \\ \Rightarrow 77c{m^3} = 0.077{\text{litre}} \\$
And volume of water left $=$ volume of cylinder$- 77 c{m^3}$
$\Rightarrow \pi {R^2}H - 77 \\ \Rightarrow \dfrac{{22}}{7} \times 5 \times 5 \times 10.5 - 77 \\ \Rightarrow 825 - 77 \\ \Rightarrow 748c{m^3}{\text{or 0}}{\text{.748 litre}} \\$
Note: - Whenever this type of question appears first note what is given in the question. Volume of Cone is $\dfrac{1}{3}\pi {r^2}h$ and volume of cylinder is $\pi {R^2}H$.The most important thing in this question is Volume of water displaced is nothing but volume of cone.