A cylindrical vessel with internal diameter 10 cm

Solution:

We have a cylindrical vessel in which a cone is inserted. We have,

Radius of the cylinder $\left(r_{1}\right)=5 \mathrm{~cm}$

Radius of cone $\left(r_{2}\right)=3.5 \mathrm{~cm}$

Height of cylinder $(h)=10.5 \mathrm{~cm}$

 

Height of cone $(l)=6 \mathrm{~cm}$

(i) We have to find the volume of water displaced from the cylinder when cone is inserted.

So,

Volume of water displaced $=$ Volume of cone

So volume of water displaced,

$=\frac{1}{3} \pi r_{2}^{2} l$

$=\frac{1}{3}\left(\frac{22}{7}\right)(12.25)(6) \mathrm{cm}^{3}$

$=77 \mathrm{~cm}^{3}$

(ii) We have to find the volume of water remaining in the cylinder.

Volume of water left = Volume of cylinder - Volume of cone

So volume of the water left in the cylinder,

$=\left[\left(\frac{22}{7}(25)(10.5)\right)-(77)\right] \mathrm{cm}^{3}$

$=(825-77) \mathrm{cm}^{3}$

$=748 \mathrm{~cm}^{3}$