If the radius and the height of a cone are increased by 10 %/then the volume of the cone is - Maths - Surface Areas and Volumes

Question

If the radius and the height of a cone are increased by 10
%/then the volume of the cone is increased by how much?

Aakash Sharma · Accepted Answer

Dear Student!

Here is the answer to your query.

Let the radius and height of the cone be r and h respectively.

Then the volume of the cone = $\frac{1}{3} π r^{2} h$

Given : The radius and height are increased by 10%.

∴ New radius and height of the cone are

$r + \frac{10}{100} r and h + \frac{10}{100} h = \frac{11}{10} r and \frac{11}{10} h$

$\Rightarrow New volume of the cone = \frac{1}{3} π (\frac{11}{10} r)^{2} \frac{11}{10} h = (\frac{11}{10})^{3} \frac{1}{3} π r^{2} h$

$Now increase in volume of cone = (\frac{11}{10})^{3} \frac{1}{3} π r^{2} h - \frac{1}{3} π r^{2} h = \frac{1}{3} π r^{2} h ((\frac{11}{10})^{3} - 1) = \frac{1}{3} π r^{2} h (\frac{{(11)}^{3} - (10)^{3}}{(10)^{3}}) = \frac{1}{3} π r^{2} h (\frac{1331 - 1000}{1000}) = \frac{1}{3} π r^{2} h \times \frac{331}{1000} Now increased % in volume = \frac{\frac{1}{3} π r^{2} h \times \frac{331}{1000}}{\frac{1}{3} π r^{2} h} \times 100 % = \frac{331}{1000} \times 100 % = 33.1 %$

Cheers!

If the radius and the height of a cone are increased by 10 %/then the volume of the cone is increased by how much?