Question

If the radius of a sphere is increased by 10% ,prove that it's volume will be in creased by 33.1%

Answer 1

Given

• Radius Increased by 10%
• Volume Increased by 33.1%

Explanation:

Let the original radius of the sphere be r

Then, Volume be ${\sf{ \left( \dfrac{4}{3} πr^3 \right) Unit^3}}$

Now, Radius of the New Sphere = (110% of r)

$\colon\implies{\sf{ \left( \dfrac{110}{100} \times r \right) unit }} \\ \\ \\ \colon\implies{\sf{ \left( \dfrac{11r}{10} \times r \right) unit }}$

Volume of the New Sphere :-

$\colon\implies{\sf{ \dfrac{4}{3} π \times \left( \dfrac{11r}{10} \right)^3 }} \\ \\ \\ \colon\implies{\sf{ \left( \dfrac{4}{3} πr^3 \times \dfrac{1331}{1000} \right) Unit^3}}$

Now, We can Find increase in Volume as :-

$\colon\implies{\sf{ \left( \dfrac{4}{3} πr^3 \times \dfrac{1331}{1000} \right) - \left( \dfrac{4}{3} πr^3 \right) }} \\ \\ \\ \colon\implies{\sf{ \dfrac{4}{3} πr^3 \times \left( \dfrac{1331}{1000} -1 \right) }} \\ \\ \\ \colon\implies{\sf{ \dfrac{4}{3} πr^3 \times \left( \dfrac{1331-1000}{1000} \right) }} \\ \\ \\ \colon\implies{\sf{ \dfrac{4}{3} πr^3 \times \left( \dfrac{331}{1000} \right) Unit^3 }}$

$\maltese \ \ {\large{\pmb{\underline {\sf\gray{Percentage \ Increase \ in \ Volume :}}}}}$

$\colon\implies{\sf{ \left[ \dfrac{ \cancel{ \dfrac{4}{3} πr^3} \times \dfrac{331}{1000} }{ \cancel{ \dfrac{4}{3} πr^3 } } \times 100 \right] \% }} \\ \\ \\ \colon\implies{\sf{ \left[ \dfrac{331}{10 \cancel{00} } \times \cancel{100} \right] \% }} \\ \\ \\ \colon\implies{\sf{ \left[ \dfrac{331}{10} \right] \% }} \\ \\ \\ \colon\implies{\sf\large\red{ 33.1 \% }}$

Hence, Proved !!