Question
The surface area of a solid sphere is increased by $12\%$ without changing its shape. Find the percentage increase in its: radius .
✓
Answer
Let the radius of the sphere be $'r\ '.$
Total surface area the sphere, $S=4 \pi r^2$
New surface area of the sphere, $S^{\prime}$
$=4 \pi r^2+\frac{21}{100} \times 4 \pi r^2$
$=\frac{121}{1004} \pi r^2$
$(1)$ let the new radius be $r^1$
$S^I=4 p r_1^2$
$S^I=\frac{121}{1004} \pi r^2$
$\Rightarrow 4 \pi r_1^2=\frac{121}{100} 4 \pi r^2$
$\Rightarrow r_1^2=\frac{121}{100} r^2$
$\Rightarrow r_1=\frac{11}{10} r$
$\Rightarrow r_1=r+\frac{r}{10}$
$\Rightarrow r_1-r=\frac{r}{10}$
$\Rightarrow \text { change in radius }=\frac{r}{10}$
Percentage change in radius $=\frac{\text { change in radius }}{\text { change in radius }} \times 100$
$\Rightarrow \frac{\frac{r}{10}}{r} \times 100$
Percentage change in radius $=10\%$
Total surface area the sphere, $S=4 \pi r^2$
New surface area of the sphere, $S^{\prime}$
$=4 \pi r^2+\frac{21}{100} \times 4 \pi r^2$
$=\frac{121}{1004} \pi r^2$
$(1)$ let the new radius be $r^1$
$S^I=4 p r_1^2$
$S^I=\frac{121}{1004} \pi r^2$
$\Rightarrow 4 \pi r_1^2=\frac{121}{100} 4 \pi r^2$
$\Rightarrow r_1^2=\frac{121}{100} r^2$
$\Rightarrow r_1=\frac{11}{10} r$
$\Rightarrow r_1=r+\frac{r}{10}$
$\Rightarrow r_1-r=\frac{r}{10}$
$\Rightarrow \text { change in radius }=\frac{r}{10}$
Percentage change in radius $=\frac{\text { change in radius }}{\text { change in radius }} \times 100$
$\Rightarrow \frac{\frac{r}{10}}{r} \times 100$
Percentage change in radius $=10\%$
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