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# 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. $\left( {{\text{Take }}\pi {\text{ = }}\dfrac{{22}}{7}} \right)$  Hint: - As the solid cone is immersed inside the water, the volume of water displaced out of the cylinder is equal to the volume of the cone.

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.
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