The surface area of a solid sphere is increased by 21% without changing its shape.Find the percentage increase in its:

(i) radius (ii) volume

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Solution

Sol:
Let the original and increased surface areas be A₁ and A₂.
$\frac{A_{2}}{A_{1}} = {\frac{r_{2}}{r_{1}}}^{2}$
1.21 $\frac{A_{1}}{A_{1}} = {\frac{r_{2}}{r_{1}}}^{2}$

1.1 = $\frac{r_{2}}{r_{1}}$

∴ r₂ = 1.1r₁

∴ Increase percentage = $\frac{r_{2} - r_{1}}{r_{1}}$ x 100 % = 10 %

Hence, the percentage increase in diameter is 10 %

Let the original and increased volume be V₁ and V₂.
$\frac{V_{1}}{V_{2}} = {\frac{r_{1}}{r_{2}}}^{3}$

$\frac{V_{2}}{V_{1}} = {\frac{11}{10}}^{3}$

∴ V₂ = 1.331 V₁

Increase % = $\frac{V_{2} - V_{1}}{V_{1}}$ x 100 %

= 33.1%

Hence, the percentage increase in volume is 33.1 %

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Similar questions

21

The surface area of a solid sphere is increased by 21% without changing it's shape. Find the percentage increase in its:

I)radius

2) volume

r a d i u s

V o l u m e

20 %

10 %

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