Question

If the radius of a sphere is increased by
20 %, by what per cent does its volume
increase?​

Answer 1

$\large\underline{\bold{Solution-}}$

Let radius of sphere be 'r' units.

So,

Volume of sphere is given by

$\tt \: Volume_{(sphere)} \: = V_{1} = \: \dfrac{4}{3} \: \pi \: {r}^{3} - - - (1)$

Now,

it is given that Radius is increased by 20 %.

So,

$\tt \: New \: radius, R \: = \: r \: + \dfrac{20}{100} r$

$\tt \: New \: radius, R \: = \: r \: + \: \dfrac{1}{5} r$

$\tt \: New \: radius, R \: = \: \dfrac{5r + r}{5}$

$\tt \: New \: radius, R \: = \: \dfrac{6r}{5}$

Now,

New volume of sphere is given by

$\tt \: Volume_{(sphere)} = V_{2} = \dfrac{4}{3} \pi \: {\bigg(\dfrac{6r}{5} \bigg) }^{3}$

$\tt \: Volume_{(sphere)} = V_{2} =\dfrac{4}{3} \times \dfrac{216}{125} \pi \: {r}^{3} - - (2)$

Now,

%age increase in Volume of sphere is given by

$\tt \: \%age \: increase \: = \: \dfrac{V_{2} - V_{1}}{V_{1}} \times 100\%$

$\tt \: \%age \: increase \: = \: \dfrac{\dfrac{4}{3} \times \dfrac{216}{125}\pi \: {r}^{3} - \dfrac{4}{3}\pi \: {r}^{3} }{\dfrac{4}{3}\pi \: {r}^{3} } \times 100\%$

$\tt \: \%age \: increase \: = \: \dfrac{\cancel{\dfrac{4}{3}\pi \: {r}^{3}} \times \bigg( \dfrac{216}{125} - 1 \bigg) }{\cancel{\dfrac{4}{3}\pi \: {r}^{3}}} \times 100\%$

$\tt \: \%age \: increase \: = \: \dfrac{216 - 125}{125} \times 100\%$

$\tt \: \%age \: increase \: = \: \dfrac{91}{5} \times 4 \: \%$

$\tt \: \%age \: increase \: = \: 72.8 \: \%$

Additional Information :-

$1. \: \: \: \boxed{ \tt{Surface Area_{(sphere)} \: = \: 4 \: \pi \: {r}^{2} }}$

$2. \: \: \: \boxed{ \tt{Surface Area_{(hemi \: - \: sphere)} = 2 \: \pi \: {r}^{2} }}$

$3. \: \: \: \boxed{ \tt{Total \: Surface Area_{(sphere)} = 3\pi \: {r}^{2} }}$

$4. \: \: \: \boxed{ \tt{Volume_{(hemi - sphere)} = \dfrac{2}{3}\pi \: {r}^{3}}}$