Question
The surface area of a solid sphere is increased by $21\%$ without changing its shape. Find the percentage increase in its: volume
✓
Answer
Let the volume of the sphere be $V$
Let the new volume of the sphere be $V'.$
$V=\frac{4}{3} \pi r^3 $
$ V^I=\frac{4}{3} \pi r_1^3$
$ \Rightarrow V^I=\frac{4}{3} \pi\left(\frac{11 r}{10^3}\right) $
$ \Rightarrow V^I=\frac{4}{3} \pi \frac{1331}{1000} r^3$
$\Rightarrow v^I=\frac{4}{3} \pi r^3 \frac{1331}{1000}$
$ \Rightarrow V^I=\frac{1331}{1000} V $
$\Rightarrow V^I=v+\frac{1331}{1000} v$
$\Rightarrow V^I-v=\frac{331}{1000} v$
$\therefore$ Change in volume $=\frac{331}{10000} v$
Percentage change in volume $=\frac{\text { Change in volume }}{\text { original volume }} \times 100$
$ =\frac{\frac{331}{1000} V }{V} \times 100 $
$ =\frac{331}{10} \\ =33.1$
Percentage change in volume $=33.1 \%$
Let the new volume of the sphere be $V'.$
$V=\frac{4}{3} \pi r^3 $
$ V^I=\frac{4}{3} \pi r_1^3$
$ \Rightarrow V^I=\frac{4}{3} \pi\left(\frac{11 r}{10^3}\right) $
$ \Rightarrow V^I=\frac{4}{3} \pi \frac{1331}{1000} r^3$
$\Rightarrow v^I=\frac{4}{3} \pi r^3 \frac{1331}{1000}$
$ \Rightarrow V^I=\frac{1331}{1000} V $
$\Rightarrow V^I=v+\frac{1331}{1000} v$
$\Rightarrow V^I-v=\frac{331}{1000} v$
$\therefore$ Change in volume $=\frac{331}{10000} v$
Percentage change in volume $=\frac{\text { Change in volume }}{\text { original volume }} \times 100$
$ =\frac{\frac{331}{1000} V }{V} \times 100 $
$ =\frac{331}{10} \\ =33.1$
Percentage change in volume $=33.1 \%$
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