Question

The radius of a sphere is increased by $10\%$. Prove that the volume is increased by $33.1\%.$

Answer 1

Let the radius of first sphere is $r_1$ and second sphere with increased radius is $r_2$.

$r_1=r$

When radius is incresed by $10\%$

$r_2=r+\left(\frac{10r}{100}\right)=1.1r$

We know that the volume of sphere is given by $V=\frac{4}{3}\pi r^3$

$V_1=\left(\frac{4}{3}\right)\pi r^3$

$V_2=\left(\frac{4}{3}\right)\pi (1.1r{)}^3$

$\frac{V_1}{V_2}=\frac{1}{(1.1{)}^3}=\frac{1}{1.331}$
% increase in volume $=\left[\frac{(V_2-V_1)}{V_1}\right]\times 100$

$=\left[\right(\frac{V_2}{V_1})-1]\times 100=[1.331-1]\times 100=0.331\times 100=33.1$ %

Hence, the volume will increase by $33.1\%$ when radius is increased by $10\%$ in sphere.