A solid toy is in the form of a hemisphere surmounted

Question:

A solid toy is in the form of a hemisphere surmounted by a right circular cone. height of the cone is 2 cm and the diameter of the base is 4 cm. If a right circular cylinder circumscribes the toy, findĀ  how much more space it will cover.

Solution:

We have to find the remaining volume of the cylinder when the toy is inserted into it. The toy is a hemisphere surmounted by a cone.

Radius of cone, cylinder and hemisphere $(r)=2 \mathrm{~cm}$

Height of $\operatorname{cone}(l)=2 \mathrm{~cm}$

Height of the cylinder $(h)=4 \mathrm{~cm}$

So the remaining volume of the cylinder when the toy is inserted into it,

$=\pi r^{2} h-\left(\frac{1}{3} \pi r^{2} l+\frac{2}{3} \pi r^{3}\right)$

Put the values to get,

$=16 \pi-\left(\frac{8 \pi}{3}+\frac{16 \pi}{3}\right)$

$=16 \pi-8 \pi$

$=8 \pi$

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now