Question

If the radius of a sphere is increased by p% then its surface area will be increased by

Answer 1

$let \: the \: radius \: be \: 100$
$then \: increased \: radius =( 100 + \frac{100p}{100} ) = \frac{10000 + 100p}{100}$

$surface \: area \: of \: original \: sphere \: = 4\pi \times {r}^{2} \\ \\ = 4 \times \frac{22}{7} \times 100^{2} \\ = \frac{880000}{7}$

And,
$surface \: area \: of \: new \: increased \: sphere = 4\pi \times {r}^{2} \\ \\ = 4 \times \frac{22}{7} \times ( \frac{10000 + 100p}{100} )^{2} \\ = \frac{88}{7} \times \frac{100000000 + 200000p + 10000p}{10000 {p}^{2} }$
Now,
$required \: increment = \frac{880000}{7} \div (\frac{88}{7} \times \frac{100000000 + 200000p + 10000 {p}^{2} }{10000} \\ \\ = \frac{880000 \times 7 \times 100000}{7 \times 88 \times (100000000 + 200000p + 10000 {p}^{2}) }$