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If the volume of a sphere is increased by 25%, what is the percentage increase in the surface area, correct to the nearest whole?

1 Answer

Shwetank Mauria

Jul 29, 2017

Percentage incease in surface area is $16.04 %$ $16.04%$

Explanation:

If radius of sphere is $r$ $r$ , volume $V$ $V$ is given by $\frac{4}{3} π r^{3}$ $4/3pir^3$ and surface area $S$ $S$ is $4 π r^{2}$ $4pir^2$

Hence $V \propto r^{3}$ $Vpropr^3$ and $S \propto r^{2}$ $Spropr^2$ i.e. $r \propto S^{\frac{1}{2}}$ $rpropS^(1/2)$

and hence $V \propto S^{\frac{3}{2}}$ $VpropS^(3/2)$ or $S \propto V^{\frac{2}{3}}$ $SpropV^(2/3)$

as volume rises by $25 %$ $25%$ , $V \to 1.25 V$ $V->1.25V$

hence $S \to S \times {(1.25)}^{\frac{2}{3}} = 1.1604 S$ $S->Sxx(1.25)^(2/3)=1.1604S$

i.e. percentage incease in surface area is $16.04 %$ $16.04%$

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